Organic Chemistry – Some Basic Principles and Techniques — NCERT Solutions | Class 11 Chemistry | Academia Aeternum

Ch 8 · Q–

Class 11 Chemistry Exercise NCERT Solutions Olympiad Board Exam

Chapter 8

Organic Chemistry – Some Basic Principles and Techniques

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

40 Questions

90–130 min Ideal time

Q1 Now at

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📘 Concept & Theory Theory ›

Hybridisation is the process of mixing atomic orbitals of nearly equal energy to form equivalent hybrid orbitals. The hybridisation of a carbon atom depends upon the number of electron domains (sigma bonds and lone pairs) around it.

For carbon, the common hybridisation states are:

sp : Two electron domains, linear geometry, bond angle approximately 180°.

sp² : Three electron domains, trigonal planar geometry, bond angle approximately 120°.

sp³ : Four electron domains, tetrahedral geometry, bond angle approximately 109.5°.

Remember that only sigma (σ) bonds are counted while determining hybridisation. A double bond contains one σ and one π bond, whereas a triple bond contains one σ and two π bonds.

The carbon atom participating in resonance (such as benzene) remains sp² hybridised.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the complete structural formula.
Identify every carbon atom separately.
Count the number of sigma bonds around each carbon atom.
Determine the steric number (number of sigma bonds plus lone pairs).
Assign the corresponding hybridisation.
Write the final answer for each carbon atom individually.

✏️ Solution Complete Solution ›

Step-by-step Solution · 43 steps

(i) CH₂=C=O (Ketene)
CH₂=C=O
There are two carbon atoms.
Carbon-1 (CH₂)
This carbon forms:
- One σ bond with the second carbon.
- Two σ bonds with two hydrogen atoms.
- Total σ bonds = 3
- Therefore, its steric number is: 3
Hence, Carbon-1 is sp² hybridised.
Carbon-2 (=C=O)
This carbon forms:
- One σ bond with Carbon-1.
- One σ bond with Oxygen.
- Total σ bonds = 2
- herefore, its steric number i: 2
Hence, Carbon-2 is sp hybridised.
Answer:

Carbon-1 : sp²

Carbon-2 : sp
(ii) CH₃CH=CH₂ (Propene)
CH₃—CH=CH₂
There are three carbon atoms.
Carbon-1 (CH₃) forms:
- Three σ bonds with hydrogen atoms.
- One σ bond with Carbon-2.
- Total σ bonds = 4
Hence, Carbon-1 is sp³ hybridised.
Carbon-2 (CH) forms
- One σ bond with Carbon-1.
- One σ bond with Carbon-3.
- One σ bond with Hydrogen.
Total σ bonds = 3
Hence, Carbon-2 is sp² hybridised.
Carbon-3 (CH₂) forms:
- One σ bond with Carbon-2.
- Two σ bonds with Hydrogen atoms.
- Total σ bonds = 3
Hence, Carbon-3 is sp² hybridised.
Answer:

Carbon-1 : sp³

Carbon-2 : sp²

Carbon-3 : sp²
(iii) (CH₃)₂CO (Acetone)
CH₃—C(=O)—CH₃
There are three carbon atoms.
Left CH₃ carbon
Total σ bonds = 4
Hybridisation = sp³
Central carbonyl carbon forms
- One σ bond with left CH₃.
- One σ bond with right CH₃.
- One σ bond with oxygen.
- Total σ bonds = 3
Hybridisation = sp²
Right CH₃ carbon
Total σ bonds = 4
Hybridisation = sp³
Answer:

CH₃ carbon : sp³

Carbonyl carbon : sp²

CH₃ carbon : sp³
(iv) CH₂=CHCN (Acrylonitrile)
CH₂=CH—C≡N
There are three carbon atoms.
Carbon-1 (CH₂)
Total σ bonds = 3
Hybridisation = sp²
Carbon-2 (CH) forms
- One σ bond with Carbon-1.
- One σ bond with Carbon-3.
- One σ bond with Hydrogen.
- Total σ bonds = 3
Hybridisation = sp²
strong>Carbon-3 (Cyano carbon) forms
- One σ bond with Carbon-2.
- One σ bond with Nitrogen.
- Total σ bonds = 2
Hybridisation = sp
Answer:

Carbon-1 : sp²

Carbon-2 : sp²

Carbon-3 : sp
(v) C₆H₆ (Benzene)
Each carbon atom in benzene forms:
- One σ bond with one hydrogen atom.
- Two σ bonds with adjacent carbon atoms.
- Total σ bonds = 3
- The unhybridised p orbital of each carbon overlaps sideways with neighbouring p orbitals to form a delocalised π-electron cloud over the entire ring.
Therefore, every carbon atom is sp² hybridised.
Answer:

All six carbon atoms are sp² hybridised.

🎯 Exam Significance Exam Significance ›

This question is frequently asked in CBSE board examinations to test conceptual understanding of hybridisation.

JEE Main and NEET regularly include questions requiring identification of hybridisation from molecular structures.

Knowledge of hybridisation helps predict molecular geometry, bond angle and bond strength.

Hybridisation is the foundation for understanding resonance, conjugation and aromaticity in Organic Chemistry.

Many reaction mechanisms in higher classes require correct identification of hybridisation.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Always determine hybridisation by counting only sigma bonds and lone pairs.
A carbon involved in a triple bond is generally sp hybridised.
A carbon involved in a double bond is generally sp² hybridised.
A carbon forming four sigma bonds is sp³ hybridised.
Every carbon atom in benzene is sp² hybridised because of resonance and delocalised π electrons.

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📘 Concept & Theory Theory / Concept ›

A covalent bond is formed by the overlap of atomic orbitals. Depending upon the mode of overlap, covalent bonds are classified into sigma (σ) bonds and pi (π) bonds.

A sigma (σ) bond is formed by the head-on (axial) overlap of orbitals. Every single bond is a sigma bond.
A pi (π) bond is formed by the sideways (lateral) overlap of two parallel p-orbitals.
A double bond consists of one sigma bond and one pi bond.
A triple bond consists of one sigma bond and two pi bonds.

Useful rules for counting bonds:

Single bond = 1σ
Double bond = 1σ + 1π
Triple bond = 1σ + 2π

🗺️ Solution Roadmap Step-by-step Plan ›

Draw the complete structural formula of the molecule.
Identify all single, double and triple bonds.
Replace every single bond by one sigma bond.
Replace every double bond by one sigma bond and one pi bond.
Replace every triple bond by one sigma bond and two pi bonds.
Add the total number of sigma and pi bonds separately.

✏️ Solution Complete Solution ›

Step-by-step Solution · 32 steps

(i) C₆H₆ (Benzene)
Step 1. Benzene contains six carbon atoms arranged in a hexagonal ring.
It has:
- Six C—C sigma bonds forming the ring.
- Three pi bonds due to the three alternating double bonds.
- Six C—H sigma bonds.
Total sigma bonds
6 (C—C) + 6 (C—H) = 12
Total pi bonds: 3
Answer:

σ bonds = 12

π bonds = 3
(ii) C₆H₁₂ (Cyclohexane)
Cyclohexane contains only single bonds.
It has:
- Six C—C single bonds.
- Twelve C—H single bonds.
Total sigma bonds: 6+12=18
There are no double or triple bonds.
Answer:

σ bonds = 18

π bonds = 0
(iii) CH₂Cl₂ (Dichloromethane)
The central carbon atom forms four single bonds.
- Two C—H bonds
- Two C—Cl bonds
All are single bonds.
Total sigma bonds: 2+2=4
There are no pi bonds.
Answer:

σ bonds = 4

π bonds = 0
(iv) CH₂=C=CH₂ (Allene)
The structure is CH₂=C=CH₂
There are:
- Two C=C double bonds.
- Four C—H single bonds.
Each double bond contributes
- One sigma bond.
- One pi bond.
Total sigma bonds: 2 (from C=C) + 4 (C—H) = 6
Total pi bonds: 2
Answer:

σ bonds = 6

π bonds = 2
(v) CH₃NO₂ (Nitromethane)
The structure may be represented as: CH₃—N(=O)—O
Considering the usual Lewis structure:
- Three C—H sigma bonds.
- One C—N sigma bond.
- One N=O double bond contributes one sigma and one pi bond.
- One N—O single bond contributes one sigma bond.
Total sigma bonds: 3 + 1 + 1 + 1 = 6
Total pi bonds: 1
Answer:

σ bonds = 6

π bonds = 1
Note: The π bond is delocalised due to resonance between the two oxygen atoms. However, the total number of π bonds remains one.
(vi) HCONHCH₃ (N-Methylformamide)
The structure is H—C(=O)—NH—CH₃

Count the sigma bonds one by one.

Bond	Number of σ Bonds
H—C	1
C=O	1
C—N	1
N—H	1
N—C	1
Three C—H bonds	3

Total sigma bonds: 1+1+1+1+1+3=8
The C=O double bond contributes one pi bond.
Answer:

σ bonds = 8

π bonds = 1

🎯 Exam Significance Exam Significance ›

This question tests the fundamental understanding of sigma and pi bonds, a core concept in Organic Chemistry.

It is frequently asked in CBSE board examinations as short-answer or objective-type questions.

JEE Main, NEET and CUET often include bond-counting problems involving sigma and pi bonds.

Correct bond counting helps in determining hybridisation, molecular geometry and predicting chemical reactivity.

Understanding sigma and pi bonds is essential before studying resonance, conjugation, aromaticity and reaction mechanisms.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Every single bond is a sigma bond.
Every double bond contains one sigma and one pi bond.
Every triple bond contains one sigma and two pi bonds.
Always count sigma bonds first and then count pi bonds separately.
Resonance changes the distribution of electrons but does not change the total number of sigma and pi bonds.

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📘 Concept & Theory Theory / Concept ›

Bond line formula (also called skeletal formula or line-angle formula) is a simplified representation of an organic compound. Instead of writing every carbon and hydrogen atom, only the carbon skeleton is represented using zig-zag lines.

Each end point and each vertex represents one carbon atom.
Hydrogen atoms attached to carbon are not shown explicitly.
Hydrogen atoms attached to heteroatoms (O, N, S, etc.) are shown.
Multiple bonds and functional groups are always shown explicitly.
Branches are represented by short lines emerging from the main carbon chain.

Bond line formula makes complex organic molecules easier to draw and interpret. It is extensively used in Organic Chemistry, reaction mechanisms and competitive examinations.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the parent carbon chain from the IUPAC or common name.
Locate the principal functional group.
Identify the position of substituents, if any.
Draw the carbon skeleton as a zig-zag chain.
Attach the functional group and substituents at their correct positions.

✏️ Solution Complete Solution ›

Step-by-step Solution · 14 steps

(i) Isopropyl Alcohol
Step 1. The IUPAC name of isopropyl alcohol is propan-2-ol.
The parent chain contains three carbon atoms.
The hydroxyl (-OH) group is attached to the second carbon atom.
Condensed structural formula
CH₃CH(OH)CH₃
Bond line formula
(ii) 2,3-Dimethylbutanal
The parent chain is butanal, which contains four carbon atoms including the aldehyde carbon.
Numbering starts from the aldehyde carbon.
A methyl group is attached to Carbon-2 and another methyl group is attached to Carbon-3.
Condensed structural formula
CHO–CH(CH₃)–CH(CH₃)–CH₃
Bond line formula
(iii) Heptan-4-one
The parent chain contains seven carbon atoms.
The ketone (>C=O) group is present on Carbon-4.
Condensed structural formula
CH₃CH₂CH₂COCH₂CH₂CH₃
Bond line formula

🎯 Exam Significance Exam Significance ›

Bond line formulas are frequently asked in CBSE Board examinations to test understanding of IUPAC nomenclature and structural representation.

JEE Main, NEET and CUET regularly include questions requiring conversion between condensed, expanded and bond line formulas.

Drawing correct bond line structures is essential for studying reaction mechanisms, stereochemistry and organic synthesis.

A clear understanding of skeletal formulae reduces errors while identifying parent chains, functional groups and substituents.

Mastery of bond line notation greatly improves speed in solving Organic Chemistry problems in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Every vertex and line end represents one carbon atom.
Hydrogen atoms attached to carbon are omitted in bond line formula.
Functional groups such as –OH, –CHO and >C=O must always be shown explicitly.
Branches are represented by short lines emerging from the main chain.
Bond line formulas are the standard representation used in higher Organic Chemistry.

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📘 Concept & Theory Concept/Theory ›

The IUPAC name of an organic compound is assigned according to systematic rules. The most important steps are:

Identify the principal functional group.
Select the longest carbon chain (or ring) containing the principal functional group.
Number the parent chain to give the functional group the lowest possible locant.
Assign the positions of substituents.
Arrange substituents alphabetically.
Write the complete IUPAC name.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the parent chain or parent ring.
Locate the principal functional group.
Number the carbon atoms according to IUPAC rules.
Identify all substituents and their positions.
Arrange substituents alphabetically and write the complete IUPAC name.

✏️ Solution Complete Solution ›

Step-by-step Solution · 40 steps

Part (a)
Parent Chain / Ring: A benzene ring ($\text{C}_6\text{H}_5-$).
Substituent: A 3-carbon straight chain $\text{CH}_3-\text{CH}_2-\text{CH}_2-$, which is a propyl group attached at position 1.
IUPAC Preferred Name: Propylbenzene
Systematic IUPAC Name: 1-Phenylpropane
Part (b)
The functional group is nitrile (-CN).
The nitrile carbon is always Carbon-1.
Count the longest chain including the nitrile carbon.
The parent chain contains five carbon atoms.
A methyl group is attached to Carbon-3.
IUPAC Name: 3-Methylpentanenitrile
Part (c)
The longest continuous carbon chain contains seven carbon atoms.
Hence, the parent hydrocarbon is heptane.
Two methyl groups are present.
Number the chain from the left to obtain the lowest set of locants.
The methyl groups are attached at Carbon-2 and Carbon-5.
IUPAC Name: 2,5-Dimethylheptane
Part (d)
The longest carbon chain contains six carbon atoms.
Therefore, the parent hydrocarbon is hexane.
Both chlorine and bromine are attached to Carbon-3.
Arrange substituents alphabetically.
'Bromo' is written before 'chloro'.
IUPAC Name: 3-Bromo-3-chlorohexane
Part (e)
The functional group is aldehyde (-CHO).
The aldehyde carbon is always Carbon-1.
The parent chain contains three carbon atoms.
Hence, the parent compound is propanal.
A chlorine atom is attached to Carbon-3.
IUPAC Name: 3-Chloropropanal
Part (f)
Cl₂CHCH₂OH<
The principal functional group is the alcohol (-OH).
The parent chain contains two carbon atoms.
Hence, the parent compound is ethan-1-ol.
Number the chain from the carbon carrying the hydroxyl group.
Both chlorine atoms are attached to Carbon-2.
IUPAC Name: 2,2-Dichloroethan-1-ol

🎯 Exam Significance Exam Significance ›

This question develops the ability to apply IUPAC nomenclature rules systematically.

It is frequently asked in CBSE Board examinations as short-answer and objective questions.

JEE Main, NEET and CUET often test the identification of the parent chain, functional groups and substituent positions.

Understanding nomenclature is essential for studying reaction mechanisms and isomerism in later chapters.

Regular practice of such questions improves speed and accuracy in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Always give priority to the principal functional group while numbering.
The carbon of the nitrile and aldehyde functional groups is always included in the parent chain.
Choose the longest possible parent chain containing the principal functional group.
When multiple substituents are present, use the lowest set of locants.
Write substituent names in alphabetical order irrespective of prefixes like di-, tri- and tetra-.

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📘 Concept & Theory Theory / Concept ›

IUPAC nomenclature follows internationally accepted rules to assign a unique name to every organic compound.

The parent chain must be the longest chain containing the principal functional group.
The parent chain is numbered so that the functional group gets the lowest possible locant.
If there is no functional group, numbering should provide the lowest possible set of locants to substituents.
When two numbering schemes give different locants, the one having the lower number at the first point of difference is preferred.
If two different substituents receive identical locants from opposite directions, alphabetical order decides which substituent gets the lower number.
The position of every substituent must always be mentioned. Omitting a locant makes the name incomplete.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the parent chain.
Locate the principal functional group, if present.
Apply the lowest set of locants rule.
Apply alphabetical order whenever required.
Select the name that satisfies all IUPAC rules.

✏️ Solution Complete Solution ›

Step-by-step Solution · 30 steps

(a) 2,2-Dimethylpentane or 2-Dimethylpentane
The parent chain is pentane.
Two methyl groups are attached to Carbon-2.
Whenever more than one identical substituent is present, the position of each substituent must be specified.
Therefore, both locants must be written.
Correct IUPAC Name: 2,2-Dimethylpentane
Reason: The name 2-Dimethylpentane is incorrect because it does not indicate the position of both methyl groups.
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
The parent chain contains eight carbon atoms.
Compare the two sets of locants.
First set: \[2,\;4,\;7\]
Second set: \[2,\;5,\;7\]
According to the lowest set of locants rule, compare the numbers one by one.
First locant: \[2=2\]
Second locant: \[4<5\]
Therefore, the first numbering is preferred.
Correct IUPAC Name: 2,4,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
The parent chain is pentane.
Numbering from either end gives the same locant set. \[2,\;4\]
Since both numbering schemes give identical locants, the substituent whose name comes first alphabetically receives the lower number.
Alphabetically, Chloro comes before Methyl.
Therefore, chlorine receives locant 2.
Correct IUPAC Name: 2-Chloro-4-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne
The principal functional group is the alcohol (-OH).
Number the parent chain so that the hydroxyl group gets the lowest possible locant.
The hydroxyl group is located at Carbon-1.
The triple bond is therefore located between Carbon-3 and Carbon-4.
In IUPAC nomenclature, the suffixes are written in the following order:
Parent chain + unsaturation + principal functional group
Hence, the correct format is: But-3-yn-1-ol
Correct IUPAC Name: But-3-yn-1-ol
The name But-4-ol-1-yne violates IUPAC numbering rules because the hydroxyl group must receive the lowest possible locant.

🎯 Exam Significance Exam Significance ›

Questions based on correct IUPAC names are frequently asked in CBSE Board examinations as objective and short-answer questions.

JEE Main, NEET and CUET regularly test the application of the lowest set of locants rule and alphabetical priority.

Understanding the priority of functional groups is essential for naming compounds containing multiple functional groups.

These concepts form the foundation for studying isomerism, reaction mechanisms and advanced organic chemistry.

Mastering nomenclature rules improves both accuracy and speed in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Every identical substituent must have its own locant.
Always choose the lowest possible set of locants.
When locant sets are identical, alphabetical order determines which substituent gets the lower number.
The principal functional group always receives the lowest possible locant.
In the suffix, unsaturation is written before the principal functional group.

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📘 Concept & Theory Theory / Concept ›

General characteristics of a homologous series are:

All members possess the same functional group.
Successive members differ by one –CH₂– unit.
The molecular mass increases by 14 u from one member to the next.
Physical properties show a gradual change, while chemical properties remain similar.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the homologous series of the given compound.
Identify the functional group.
Add one –CH₂– group successively to obtain the next members.
Write the structural formula and IUPAC name of each member.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(a) Beginning with H–COOH

Member	Structural Formula	IUPAC Name
1	HCOOH	Methanoic acid
2	CH₃COOH	Ethanoic acid
3	CH₃CH₂COOH	Propanoic acid
4	CH₃CH₂CH₂COOH	Butanoic acid
5	CH₃CH₂CH₂CH₂COOH	Pentanoic acid

(b) Beginning with CH₃COCH₃
The given compound is propanone. It belongs to the ketone homologous series.

Member	Structural Formula	IUPAC Name
1	CH₃COCH₃	Propanone
2	CH₃COCH₂CH₃	Butan-2-one
3	CH₃COCH₂CH₂CH₃	Pentan-2-one
4	CH₃COCH₂CH₂CH₂CH₃	Hexan-2-one
5	CH₃COCH₂CH₂CH₂CH₂CH₃	Heptan-2-one

(c) Beginning with H₂C=CH₂
The given compound is ethene. It belongs to the alkene homologous series.

Member	Structural Formula	IUPAC Name
1	CH₂=CH₂	Ethene
2	CH₂=CHCH₃	Propene
3	CH₂=CHCH₂CH₃	But-1-ene
4	CH₂=CHCH₂CH₂CH₃	Pent-1-ene
5	CH₂=CHCH₂CH₂CH₂CH₃	Hex-1-ene

🎯 Exam Significance Exam Significance ›

Questions based on homologous series are frequently asked in CBSE Board examinations to test conceptual understanding of organic compounds.

JEE Main, NEET and CUET often require identification of homologous series and prediction of successive members.

Knowledge of homologous series is essential for understanding IUPAC nomenclature, physical properties and organic reaction mechanisms.

This concept forms the basis for studying hydrocarbons and functional group chemistry in higher classes.

Regular practice improves speed in writing molecular and structural formulae during competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Successive members of a homologous series differ by one –CH₂– group.
Each successive member has a molecular mass 14 u greater than the previous member.
All members possess the same functional group.
Chemical properties remain similar throughout a homologous series.
Physical properties change gradually with increasing molecular mass.

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📘 Concept & Theory Theory / Concept ›

A condensed structural formula represents the arrangement of atoms in a compact form, whereas a bond line (skeletal) formula represents only the carbon skeleton. Carbon atoms and the hydrogen atoms attached to them are not shown explicitly in bond line formula.

The functional group is the atom or group of atoms responsible for the characteristic chemical properties of an organic compound.

Alkanes contain no functional group.
Alcohols contain the hydroxyl group (-OH).
Carboxylic acids contain the carboxyl group (-COOH).
Aldehydes contain the formyl group (-CHO).

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the parent chain from the IUPAC name.
Locate all substituents and functional groups.
Write the condensed structural formula.
Draw the corresponding bond line formula.
Identify the functional group(s) present.

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps

(a) 2,2,4-Trimethylpentane
The parent chain contains five carbon atoms.
Two methyl groups are attached to Carbon-2 and one methyl group is attached to Carbon-4.
Condensed Structural Formula:
CH₃C(CH₃)₂CH₂CH(CH₃)CH₃
Bond Line Formula
Functional Group: No functional group is present. It is a branched alkane.
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
The parent chain contains three carbon atoms.
Carboxyl groups are attached to Carbon-1, Carbon-2 and Carbon-3.
A hydroxyl group is attached to Carbon-2.
This compound is commonly known as citric acid.
Condensed Structural Formula
HOOCCH₂C(OH)(COOH)CH₂COOH
Bond Line Formula
Functional Groups
- Three carboxylic acid groups (-COOH)
- One hydroxyl group (-OH)
(c) Hexanedial
The parent chain contains six carbon atoms.
The suffix -dial indicates that aldehyde groups are present at both ends of the carbon chain.
Condensed Structural Formula:
OHC(CH₂)₄CHO
or
CHOCH₂CH₂CH₂CH₂CHO
Bond Line Formula
Functional Groups: Two aldehyde groups (-CHO).

🎯 Exam Significance Exam Significance ›

This question develops the ability to convert IUPAC names into structural representations.

CBSE Board examinations frequently test condensed structural formulae and identification of functional groups.

JEE Main, NEET and CUET often include questions requiring conversion between IUPAC names, condensed structures and bond line formulae.

Recognition of functional groups is essential for understanding nomenclature, chemical properties and reaction mechanisms.

These concepts serve as the foundation for studying alcohols, aldehydes, ketones and carboxylic acids in later chapters.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Condensed structural formula shows atoms in a compact form.
Bond line formula represents only the carbon skeleton.
Hydrogen atoms attached to carbon are omitted in bond line formula.
The suffix of the IUPAC name helps identify the functional group.
A compound may contain more than one functional group.

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📘 Concept & Theory Theory / Concept ›

A functional group is an atom or group of atoms that determines the characteristic chemical properties of an organic compound. A single organic molecule may contain one or more functional groups.

Some common functional groups are:

Functional Group	Representation	Suffix / Prefix
Aldehyde	–CHO	-al / formyl
Alcohol (Hydroxyl)	–OH	-ol / hydroxy
Ether	–O–	alkoxy
Amine	–NH₂, –NHR, –NR₂	-amine / amino
Ester	–COOR	-oate
Nitro	–NO₂	nitro
Alkene	C=C	-ene

🗺️ Solution Roadmap Step-by-step Plan ›

Observe the structural formula carefully.
Locate every characteristic atom or group of atoms.
Identify each functional group using standard IUPAC notation.
List all functional groups present in the molecule.

✏️ Solution Complete Solution ›

Step-by-step Solution · 25 steps

Part (a)
The compound contains a benzene ring substituted with three different groups.
The group –CHO is present at the top of the ring.
This is an aldehyde functional group.
The group –OH is directly attached to the aromatic ring.
This is a phenolic hydroxyl (phenol) functional group.
The group –OCH₃ (–OMe) is attached to the ring.
This is a methoxy (ether) functional group.
Functional Groups Present
- Aldehyde (-CHO)
- Phenolic hydroxyl (-OH)
- Ether / Methoxy (-OCH₃)
Part (b)
The compound contains an aromatic ring substituted with amino and ester groups.
The group –NH₂ is attached to the benzene ring.
This is a primary amino functional group.
The group –COOCH₂CH₂N(C₂H₅)₂ contains the linkage –COO–.
This is an ester functional group.
The terminal nitrogen N(C₂H₅)₂ is bonded to three carbon atoms.
Hence, it is a tertiary amine.
Functional Groups Present
- Primary amine (-NH₂)
- Ester (-COO-)
- Tertiary amine (-NR₃)
Part (c)
The compound contains a benzene ring attached to the side chain CH=CHNO₂
The side chain contains a carbon-carbon double bond.
This is an alkene functional group.
The group –NO₂ is attached to the terminal carbon.
This is a nitro functional group.
Functional Groups Present
- Alkene (C=C)
- Nitro (-NO₂)

🎯 Exam Significance Exam Significance ›

Identification of functional groups is one of the most fundamental skills in Organic Chemistry.

CBSE Board examinations frequently ask students to recognize functional groups from structural formulae.

JEE Main, NEET and CUET regularly include questions involving compounds containing multiple functional groups.

Correct identification of functional groups is essential for IUPAC nomenclature, prediction of chemical reactions and understanding reaction mechanisms.

This concept forms the basis for the study of alcohols, aldehydes, ketones, carboxylic acids, amines and biomolecules in later chapters.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

A single organic compound may contain more than one functional group.
Phenolic –OH differs from the alcohol functional group because it is directly attached to an aromatic ring.
The ester functional group is identified by the linkage –COO–.
The nitro group (-NO₂) is different from amino (-NH₂).
Always examine the complete structure before identifying functional groups.

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Q9 →

📘 Concept & Theory Theory / Concept ›

The stability of negatively charged species (anions) depends upon several factors such as resonance effect, inductive effect, electronegativity and hybridisation.

An electron-withdrawing group stabilises a negative charge by pulling electron density away through the inductive (-I) effect.
An electron-donating group destabilises a negative charge by increasing electron density.
The nitro group (–NO₂) is one of the strongest electron-withdrawing groups and exhibits a strong -I effect.
Greater stabilisation of the negative charge results in a more stable anion.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the negatively charged atom.
Identify the substituent attached to the carbon chain.
Determine whether the substituent is electron-withdrawing or electron-donating.
Compare the stability of the two alkoxide ions based on the inductive effect.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Consider the first ion. O₂NCH₂CH₂O⁻
This ion contains a nitro group (-NO₂), which is a strong electron-withdrawing group.
The nitro group withdraws electron density through the carbon chain by its strong -I (negative inductive) effect.
As a result, the negative charge present on the oxygen atom is dispersed and stabilised.
Consider the second ion. CH₃CH₂O⁻
The ethyl group (–CH₂CH₃) exhibits a weak +I (positive inductive) effect.
It pushes electron density towards the negatively charged oxygen atom.
This increases the electron density on oxygen, making the anion comparatively less stable.
Compare the stability.
The nitro group stabilises the negative charge, whereas the ethyl group slightly destabilises it
Therefore,
O₂NCH₂CH₂O⁻ is more stable than CH₃CH₂O⁻.

📊 Graph / Figure Graph / Figure ›

🎯 Exam Significance Exam Significance ›

This question tests the application of inductive effect in comparing the stability of organic ions.

CBSE Board examinations frequently ask conceptual questions based on electron-withdrawing and electron-donating groups.

JEE Main, NEET and CUET regularly include problems involving the relative stability of carbocations, carbanions and alkoxide ions.

A clear understanding of inductive effect is essential for studying acidity, basicity, reaction mechanisms and organic reaction intermediates.

This concept forms the foundation for many advanced topics in Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The nitro group (-NO₂) is a strong electron-withdrawing group.
The ethyl group exhibits a weak electron-releasing (+I) effect.
Electron-withdrawing groups stabilise negatively charged species.
Electron-donating groups generally destabilise negatively charged species.
Greater stabilisation of charge corresponds to greater stability of the ion.

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Q10 →

📘 Concept & Theory Theory / Concept ›

When an alkyl group is attached to a carbon atom containing a π bond or an aromatic ring, it donates electron density towards the π system. This electron-releasing behaviour arises mainly due to the positive inductive (+I) effect and hyperconjugation.

The +I (positive inductive) effect is the tendency of an alkyl group to push electron density through sigma (σ) bonds.
Hyperconjugation, also known as no-bond resonance, is the delocalisation of electrons from the σ bond of an α C–H bond into an adjacent π bond or vacant p-orbital.
Both effects increase the electron density of the π system.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the effect produced by an alkyl group.
Explain the positive inductive effect.
Describe hyperconjugation.
State how these effects increase electron density in the π system.

📊 Graph / Figure Graph / Figure ›

Fig. 1 — Free body diagram

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

Alkyl groups exhibit a positive inductive (+I) effect.
Carbon atoms in an alkyl group are slightly electron-releasing in nature. Therefore, they push electron density through sigma (σ) bonds towards the carbon atom attached to the π system.
Alkyl groups also exhibit hyperconjugation.
In hyperconjugation, the electrons of the σ bond between the α-carbon and hydrogen overlap with the adjacent π orbital.
This results in the delocalisation of σ electrons into the π system.
Due to these two effects, the electron density of the π system increases.
As a result, the π system becomes more electron-rich and is stabilised.
Example
Toluene is more electron-rich than benzene because the methyl group donates electron density to the benzene ring through the +I effect and hyperconjugation.
Similarly, alkyl-substituted alkenes are generally more stable than unsubstituted alkenes because hyperconjugation stabilises the double bond.

🎯 Exam Significance Exam Significance ›

This question tests the understanding of inductive effect and hyperconjugation, which are fundamental concepts in Organic Chemistry.

CBSE Board examinations frequently ask conceptual questions on electron-donating and electron-withdrawing groups.

JEE Main, NEET and CUET regularly include questions involving hyperconjugation, stability of alkenes, carbocations and aromatic compounds.

Understanding electron displacement effects is essential for predicting acidity, basicity, orientation in aromatic substitution and reaction mechanisms.

This concept is extensively used throughout higher Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Alkyl groups exhibit a positive inductive (+I) effect.
Hyperconjugation involves delocalisation of σ-electrons into an adjacent π system.
Both effects increase the electron density of the π system.
Electron donation by alkyl groups stabilises alkenes, carbocations and aromatic compounds.
Hyperconjugation is also known as "no-bond resonance".

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Q11 →

Q11

NUMERIC3 marks

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

$\ce{C6H5OH}$
$\ce{C6H5NO2}$
$\ce{CH3CH=CHCHO}$
$\ce{C6H5–CHO}$
$\mathrm{C_6H_5-\overset{+}{C}H_2}$
$\ce{CH3CH=CH \overset{+}{C}H2}$

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

(a) $\ce{C6H5OH}$
(b) $\ce{C6H5NO2}$
(c) $\ce{CH3CH=CHCHO}$
(d) $\ce{C6H5–CHO}$
(e) $\mathrm{C_6H_5-\overset{+}{C}H_2}$
(f) $\ce{CH3CH=CH \overset{+}{C}H2}$

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Q12 →

📘 Concept & Theory Theory / Concept ›

Most organic reactions occur because one species donates an electron pair while another species accepts it. Based on this behaviour, reacting species are classified as nucleophiles and electrophiles.

According to the Lewis theory, a nucleophile acts as a Lewis base, whereas an electrophile acts as a Lewis acid.

🗺️ Solution Roadmap Step-by-step Plan ›

Define nucleophiles.
List the important characteristics of nucleophiles.
Give suitable examples of nucleophiles.
Define electrophiles.
List the important characteristics of electrophiles.
Give suitable examples of electrophiles.
Compare the two species.

📊 Graph / Figure Graph / Figure ›

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Nucleophiles
A nucleophile is an electron-rich atom, ion or molecule that donates an electron pair to an electron-deficient species to form a covalent bond.

Since nucleophiles donate an electron pair, they are also known as Lewis bases.

Characteristics of Nucleophiles
- They possess one or more lone pairs of electrons or π-electrons.
- They may carry a negative charge or be electrically neutral.
- They attack positively charged or electron-deficient centres.
- They donate an electron pair during a chemical reaction.
Examples of Nucleophiles
- $\ce{OH^-}$
- $\ce{CN^-}$
- $\ce{Cl^-}$
- $\ce{Br^-}$
- $\ce{NH3}$
- $\ce{H2O}$
- $\ce{RO^-}$ (Alkoxide ion)
Example Reaction

\[ \ce{CH3Br + OH^- -> CH3OH + Br^-} \]

In this reaction, $\ce{OH^-}$ donates an electron pair to the carbon atom and behaves as a nucleophile.
Electrophiles
An electrophile is an electron-deficient atom, ion or molecule that accepts an electron pair from another species to form a covalent bond.

Since electrophiles accept an electron pair, they are known as Lewis acids.

Characteristics of Electrophiles
- They possess a positive charge or a vacant orbital.
- They are electron deficient.
- They accept an electron pair from nucleophiles.
- They attack electron-rich regions such as π-bonds and lone pairs.
Examples of Electrophiles
- $\ce{H+}$
- $\ce{NO2+}$
- $\ce{CH3CO+}$
- $\ce{BF3}$
- $\ce{AlCl3}$
- $\ce{SO3}$
- $\ce{R+}$ (Carbocation)
Example Reaction

\[ \ce{CH2=CH2 + H+ -> CH3-CH2+} \]

Here, $\ce{H+}$ accepts an electron pair from the π-bond of ethene and behaves as an electrophile.
Difference Between Nucleophiles and Electrophiles

Property	Nucleophile	Electrophile
Nature	Electron-rich species	Electron-deficient species
Behaviour	Donates an electron pair	Accepts an electron pair
Lewis Concept	Lewis Base	Lewis Acid
Charge	Usually negative or neutral	Usually positive or neutral with a vacant orbital
Examples	$\ce{OH^-}$, $\ce{CN^-}$, $\ce{NH3}$	$\ce{H+}$, $\ce{NO2+}$, $\ce{BF3}$

🎯 Exam Significance Exam Significance ›

Electrophiles and nucleophiles form the foundation of Organic Chemistry reaction mechanisms.

CBSE Board examinations frequently ask definitions, examples and differences between nucleophiles and electrophiles.

JEE Main, NEET and CUET regularly include questions involving identification of nucleophiles and electrophiles in reaction mechanisms.

Understanding these species is essential for studying substitution, addition, elimination and aromatic substitution reactions.

This concept is repeatedly applied throughout higher Organic Chemistry and is one of the most important topics for competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Nucleophiles donate an electron pair and are Lewis bases.
Electrophiles accept an electron pair and are Lewis acids.
Nucleophiles are usually negatively charged or possess lone pairs.
Electrophiles are positively charged or possess vacant orbitals.
Organic reactions generally involve the interaction of nucleophiles with electrophiles.

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📘 Concept & Theory Theory / Concept ›

A nucleophile is an electron-rich species that donates an electron pair to an electron-deficient atom to form a covalent bond.

An electrophile is an electron-deficient species that accepts an electron pair from a nucleophile.

General characteristics are:

Nucleophiles possess lone pairs, negative charge or π-electrons.
Electrophiles possess a positive charge or an incomplete octet.
Nucleophiles are Lewis bases.
Electrophiles are Lewis acids.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the reagent written in bold.
Determine whether it donates or accepts an electron pair.
Classify it as a nucleophile or an electrophile.
Give the reason based on its electronic nature.

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Part (a)
CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O
The reagent in bold is the hydroxide ion, HO⁻.
Hydroxide ion possesses a negative charge and a lone pair of electrons on oxygen.
It donates an electron pair to the acidic hydrogen atom of acetic acid.
Therefore, HO⁻ behaves as a nucleophile.
Answer: HO⁻ is a nucleophile.
Part (b)
$\ce{CH3COCH3+ \overset{-}{C}N → (CH3)2C(CN)(OH)}$
The reagent in bold is the cyanide ion, $\mathrm{\overset{-}{C}N}$.
Cyanide ion carries a negative charge and has a lone pair of electrons.
It attacks the electron-deficient carbonyl carbon of acetone by donating its electron pair.
Hence, CN⁻ acts as a nucleophile.
Answer: CN⁻ is a nucleophile.
Part (c)
$\mathrm{\ce{C6H6 + CH3\overset{+}{C}O → C6H5COCH3}}$
The reagent in bold is the acetylium ion, CH₃CO⁺.
It carries a positive charge and is electron deficient.
It accepts an electron pair from the π-electron cloud of benzene during Friedel-Crafts acylation.
Therefore, CH₃CO⁺ behaves as an electrophile.
Answer: CH₃CO⁺ is an electrophile.

🎯 Exam Significance Exam Significance ›

Identification of nucleophiles and electrophiles is one of the most important concepts in Organic Chemistry.

CBSE Board examinations frequently ask direct questions on classifying reagents as nucleophiles or electrophiles.

JEE Main, NEET and CUET regularly include reaction mechanism problems based on nucleophilic and electrophilic attack.

Understanding these species is essential for studying substitution, addition and elimination reactions.

This concept forms the basis of almost every organic reaction mechanism studied in higher classes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Nucleophiles donate an electron pair.
Electrophiles accept an electron pair.
Negatively charged ions are generally strong nucleophiles.
Positively charged species are generally strong electrophiles.
The carbonyl carbon is an electrophilic centre because of the polar C=O bond.

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Q14 →

📘 Concept & Theory Theory / Concept ›

Organic reactions are generally classified into four major types depending upon the change taking place in the molecule.

Substitution Reaction: One atom or group is replaced by another atom or group.
Addition Reaction: Atoms or groups are added across a multiple bond.
Elimination Reaction: Two atoms or groups are removed from adjacent carbon atoms to form a multiple bond.
Rearrangement Reaction: The carbon skeleton is reorganised to form a more stable product.

🗺️ Solution Roadmap Step-by-step Plan ›

Observe the reactants and products carefully.
Identify whether any atom or group is replaced, added, removed or rearranged.
Compare the reaction with the standard reaction types.
Classify the reaction accordingly.

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻
Bromine is attached to the ethyl group in the reactant.
In the product, bromine has been replaced by the HS group.
Since one group replaces another group, the reaction is a substitution reaction.
Classification: Nucleophilic Substitution Reaction

b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂CCl–CH₃
The reactant contains a carbon-carbon double bond.

HCl adds across the double bond.

The double bond disappears and a saturated product is formed.

Classification: Addition Reaction (Electrophilic Addition).
c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻
A hydrogen atom and a bromine atom are removed from adjacent carbon atoms.
A hydrogen atom and a bromine atom are removed from adjacent carbon atoms.
A carbon-carbon double bond is formed.
Such reactions are called elimination reactions.
d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBrCH₂CH₃ + H₂O
The alcohol first reacts with HBr.
During the reaction, the carbon skeleton changes before bromine is attached.
Formation of a rearranged carbon skeleton indicates molecular rearrangement.
Classification: Rearrangement Reaction.

🎯 Exam Significance Exam Significance ›

Classification of organic reactions is one of the most important conceptual topics in Organic Chemistry.

CBSE Board examinations frequently ask students to identify the type of organic reaction from a given equation.

JEE Main, NEET and CUET regularly test substitution, addition, elimination and rearrangement reactions through reaction mechanism questions.

Understanding reaction types is essential before studying reaction mechanisms, named reactions and synthetic pathways.

Correct classification helps predict products and reaction conditions in advanced Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Substitution reactions involve replacement of one atom or group by another.
Addition reactions occur across double or triple bonds.
Elimination reactions produce unsaturated compounds by removing atoms or groups.
Rearrangement reactions involve migration of atoms or groups resulting in a new carbon skeleton.
Recognising reaction types is the first step in understanding organic reaction mechanisms.

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Q15 →

📘 Concept & Theory Theory / Concept ›

Two organic compounds may have the same molecular formula but differ in the arrangement of atoms or electrons. Such compounds may be related as structural isomers, geometrical isomers or resonance contributors.

Structural Isomers: They have the same molecular formula but different connectivity (arrangement) of atoms.
Geometrical Isomers: They have the same connectivity but differ in the spatial arrangement of groups around a double bond or a ring.
Resonance Contributors: They differ only in the distribution of electrons. The positions of atoms remain unchanged.

🗺️ Solution Roadmap Step-by-step Plan ›

Compare the molecular formula of both structures.
Check whether the connectivity of atoms is the same.
If connectivity differs, identify them as structural isomers.
If connectivity is the same but spatial arrangement differs around a double bond, identify them as geometrical isomers.
If only electron distribution changes while atom positions remain unchanged, identify them as resonance contributors.

✏️ Solution Complete Solution ›

Step-by-step Solution · 22 steps

Both compounds contain the same molecular formula and the same functional group (ketone).
In the first structure, the carbonyl group is attached to two ethyl groups.
This compound is pentan-3-one.
In the second structure, the carbonyl carbon is attached to one methyl group and one propyl group.
This compound is pentan-2-one.
The position of the carbonyl group has changed, resulting in different connectivity.
Conclusion: The two compounds are structural (positional) isomers.
Both compounds have the same molecular formula and the same connectivity.
The double bond restricts free rotation.
In the first structure, the two deuterium (D) atoms lie on opposite sides of the double bond.
This is the trans (E) arrangement.
In the second structure, both deuterium atoms lie on the same side of the double bond.
This is the cis (Z) arrangement.
Conclusion: The two compounds are geometrical (cis-trans) isomers.
In both structures, the positions of all atoms remain exactly the same.
Only the positions of electrons and the positive charge are different.
One structure shows the positive charge on oxygen, while the other shows the positive charge on carbon.
Since only electron distribution changes without changing atomic positions, these structures are resonance forms.
Conclusion: The two structures are resonance contributors (canonical forms).

🎯 Exam Significance Exam Significance ›

This question tests the ability to distinguish between different types of isomerism and resonance.

CBSE Board examinations frequently ask students to identify structural, geometrical and resonance relationships.

JEE Main, NEET and CUET regularly include questions based on positional isomerism, cis-trans isomerism and resonance.

Understanding these concepts is essential for studying stereochemistry, reaction mechanisms and molecular stability.

Correct identification of these relationships helps in predicting physical properties and chemical behaviour of organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Structural isomers differ in the connectivity of atoms.
Geometrical isomers differ only in the spatial arrangement around a double bond or ring.
Resonance contributors differ only in the distribution of electrons.
In resonance, atoms do not change their positions.
A double bond prevents free rotation, giving rise to geometrical isomerism.

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Q16 →

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps

a)
Bond Cleavage Type: Homolysis (Homolytic cleavage)
The single covalent oxygen–oxygen ($\text{O-O}$) bond breaks symmetrically, where each oxygen atom retains one electron from the shared bonding pair.
Curved-Arrow Electron Flow:
- Use two single-barbed (fishhook) arrows:
  - One fishhook arrow starts at the center of the $\text{O-O}$ bond and points to the left oxygen atom.
  - The second fishhook arrow starts at the center of the $\text{O-O}$ bond and points to the right oxygen atom.
- Reactive Intermediate Produced: Free Radical (Specifically, two methoxy radicals, $\text{CH}_3\dot{\text{O}}$.
b)
Bond Cleavage Type: Heterolysis (Heterolytic cleavage)
The $\text{C-H}$ bond ($\alpha$-position to the carbonyl) breaks unsymmetrically. The bonding pair of electrons is retained completely by the carbon atom as the hydroxide ion abstracts the proton $\text{H}^+$.
Curved-Arrow Electron Flow:
- Use two double-barbed arrows:
- - One arrow starts from a lone pair on the hydroxide oxygen ($^{-}\text{OH}$) and points directly to the $alpha$-hydrogen atom.
  - The second arrow starts from the $text{C-H}$ single bond and points directly onto the $alpha$-carbon atom.
- Reactive Intermediate Produced: Carbanion (Specifically, an resonance-stabilized enolate/carbanion).
c)
Bond Cleavage Type: Heterolysis (Heterolytic cleavage)
- The carbon–bromine ($\text{C-Br}$) bond breaks unsymmetrically because bromine is significantly more electronegative than carbon, taking both bonding electrons with it.
Curved-Arrow Electron Flow:
- Use a single double-barbed arrow:
- Reactive Intermediate Produced: Carbocation (Specifically, the tert-butyl carbocation, a $3^\circ$ carbocation).
d)
Bond Cleavage Type: Heterolysis (Heterolytic cleavage)
- One of the carbon–carbon $\pi$-bonds ($\text{C=C}$) in the benzene ring breaks unsymmetrically. The $\pi$-electron pair shifts to form a new $\text{C-E}$ $\sigma$-bond with the electrophile, leaving the adjacent ring carbon electron-deficient.
Curved-Arrow Electron Flow:
- Use a single double-barbed arrow:
  - It starts from the $\pi$-bond of the benzene ring and points directly to the positively charged electrophile ($\text{E}^+$).
Reactive Intermediate Produced: Carbocation (Specifically, an arenium ion / Wheland intermediate, which is a resonance-stabilized cyclohexadienyl cation).

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📘 Concept & Theory Theory / Concept ›

Inductive Effect (I-effect)

The inductive effect is the permanent displacement of sigma (σ) electrons along a carbon chain due to the difference in electronegativity between bonded atoms.

Electron-withdrawing groups exhibit the negative inductive (-I) effect.
Electron-donating groups exhibit the positive inductive (+I) effect.
The inductive effect decreases rapidly with increasing distance from the substituent.
It is a permanent effect and operates through sigma bonds only.

Electromeric Effect (E-effect)

The electromeric effect is the complete transfer of the shared pair of π-electrons of a multiple bond to one of the bonded atoms under the influence of an attacking reagent.

It is a temporary effect.
It operates only in compounds containing multiple bonds.
The effect disappears as soon as the attacking reagent is removed.
It involves complete transfer of π-electrons.

🗺️ Solution Roadmap Step-by-step Plan ›

Define the inductive effect.
Define the electromeric effect.
Identify the electron-displacement effect responsible for acidity.
Compare the electron-withdrawing and electron-donating abilities of the substituents.
Explain the observed order of acidity.

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

Inductive Effect
The inductive effect is the permanent displacement of sigma (σ) electrons through a chain of atoms due to differences in electronegativity.

Electron-withdrawing substituents pull electron density away from the carbon chain and exhibit a negative inductive (-I) effect.

Electron-donating substituents push electron density towards the carbon chain and exhibit a positive inductive (+I) effect.

Electromeric Effect

The electromeric effect is the temporary and complete transfer of π-electrons of a multiple bond in the presence of an attacking reagent.

It operates only during the course of a reaction and disappears once the reagent is removed.
(a) Acidity Order
Chlorine atoms are highly electronegative.
Each chlorine atom exerts a strong negative inductive (-I) effect

More chlorine atoms withdraw more electron density from the carboxyl group.

This stabilises the conjugate base (carboxylate ion) formed after loss of H⁺.

Greater stability of the conjugate base results in stronger acidity.

Therefore,
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
is explained by the negative inductive (-I) effect.

(b) Acidity Order
CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C–COOH

Alkyl groups exhibit a positive inductive (+I) effect.

Increasing the number of alkyl groups increases electron donation towards the carboxyl group.

Electron donation destabilises the negatively charged carboxylate ion.

Lesser stability of the conjugate base results in lower acidity.

The +I effect increases in the order

CH₃CH₂ < (CH₃)₂CH < (CH₃)₃C

Therefore, acidity decreases in the order
CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C–COOH

This order is explained by the positive inductive (+I) effect.

🎯 Exam Significance Exam Significance ›

Inductive effect is one of the most important electron displacement effects in Organic Chemistry.

CBSE Board examinations frequently ask conceptual questions based on acidity and basicity using the inductive effect.

JEE Main, NEET and CUET regularly include comparison-based questions involving the relative acidity of substituted carboxylic acids.

Understanding inductive and electromeric effects is essential for studying resonance, reaction mechanisms and stability of reaction intermediates.

These concepts are repeatedly used throughout higher Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

The inductive effect is a permanent effect transmitted through sigma bonds.

The electromeric effect is a temporary effect involving complete transfer of π-electrons.

Electron-withdrawing groups (-I) increase the acidity of carboxylic acids.

Electron-donating alkyl groups (+I) decrease the acidity of carboxylic acids.

The stability of the conjugate base determines the strength of an acid.

Technique	Principle	Example
Crystallisation	Difference in solubility of a substance in a solvent at different temperatures.	Purification of benzoic acid from hot water.
Distillation	Difference in boiling points (volatility) of liquids.	Obtaining pure water from salt solution.
Chromatography	Difference in distribution between stationary and mobile phases.	Separation of pigments in black ink by paper chromatography.

Technique	Principle	Suitable For	Example
Distillation	Difference in boiling points.	Volatile liquids and liquids with a large boiling point difference.	Pure water from salt solution.
Distillation under Reduced Pressure	Boiling point decreases when pressure is reduced.	High-boiling and heat-sensitive liquids.	Purification of glycerol.
Steam Distillation	Immiscible liquids boil together below their individual boiling points.	Steam-volatile, water-insoluble organic compounds.	Extraction of essential oils.

Halogen	Reaction	Observation
Chlorine	\[ \mathrm{NaCl + AgNO_3 \rightarrow AgCl\downarrow + NaNO_3} \]	White precipitate
Bromine	\[ \mathrm{NaBr + AgNO_3 \rightarrow AgBr\downarrow + NaNO_3} \]	Cream precipitate
Iodine	\[ \mathrm{NaI + AgNO_3 \rightarrow AgI\downarrow + NaNO_3} \]	Yellow precipitate

Basis	Dumas Method	Kjeldahl's Method
Principle	Nitrogen is converted into nitrogen gas (N₂).	Nitrogen is converted into ammonium sulphate and finally into ammonia (NH₃).
Measurement	Volume of nitrogen gas is measured.	Amount of ammonia liberated is determined by acid-base titration.
Reagent Used	Copper(II) oxide and heated copper.	Concentrated H₂SO₄ followed by NaOH.
Suitable For	Almost all nitrogen-containing organic compounds.	Most organic compounds except nitro, azo and diazo compounds.
Not Suitable For	No major limitation for ordinary nitrogen compounds.	Nitro, azo and diazo compounds.

Element	Method	Converted Into	Final Precipitate
Halogens (Cl, Br, I)	Carius Method	Silver Halides	AgCl, AgBr or AgI
Sulphur	Carius Method	Sulphuric Acid	BaSO₄
Phosphorus	Carius Method	Phosphoric Acid	Ammonium Phosphomolybdate

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Q18 →

Q18

NUMERIC3 marks

Give a brief description of the principles of the following techniques taking an example in each case.

Crystallisation

Distillation

Chromatography

📘 Concept & Theory Theory / Concept ›

Pure compounds are essential for studying the physical and chemical properties of substances. Various purification techniques are employed depending upon the nature of the impurities and the physical properties of the compounds.

The commonly used purification techniques are crystallisation, distillation and chromatography. Each technique is based on a different physical property of the substances present in a mixture.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the purification technique.

State the basic principle of the technique.

Briefly describe how the technique works.

Give one suitable example of its application.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps
a) Crystallisation
Principle

Crystallisation is based on the difference in the solubility of a compound in a suitable solvent at different temperatures.

Generally, a compound is highly soluble in a hot solvent but only sparingly soluble in the same solvent when it is cold.

Working

The impure solid is dissolved in the minimum quantity of hot solvent.

Insoluble impurities are removed by filtration.

The hot saturated solution is allowed to cool slowly.

Pure crystals separate out, while most impurities remain dissolved in the mother liquor.

The crystals are filtered, washed and dried.

Example

Benzoic acid can be purified from its impure sample by crystallisation using hot water.

b) Distillation
Principle

Distillation is based on the difference in the boiling points or volatility of the components of a liquid mixture.

The component having the lower boiling point vaporises first and is then condensed to obtain the pure liquid.

Working

The liquid mixture is heated.

The more volatile component boils first.

The vapours pass through a condenser.

The vapours condense into the liquid state.

The condensed liquid is collected separately.

Example

Pure water is obtained from salt solution by simple distillation.

Fractional distillation is used to separate benzene and toluene because their boiling points are close to each other.

c) Chromatography
Principle

Chromatography is based on the different distribution of the components of a mixture between a stationary phase and a mobile phase.

Components having a greater affinity for the stationary phase move more slowly, whereas those having a greater affinity for the mobile phase move faster.

Working

The sample is placed on the stationary phase.

A suitable solvent (mobile phase) is allowed to move.

Different components travel at different speeds.

As a result, the components separate into distinct bands or spots.

Example

Paper chromatography is used to separate coloured pigments present in black ink.

Chromatography is also widely used for separating amino acids, plant pigments and drugs.

Technique Principle Example

Crystallisation Difference in solubility of a substance in a solvent at different temperatures. Purification of benzoic acid from hot water.

Distillation Difference in boiling points (volatility) of liquids. Obtaining pure water from salt solution.

Chromatography Difference in distribution between stationary and mobile phases. Separation of pigments in black ink by paper chromatography.

🎯 Exam Significance Exam Significance ›

Purification techniques are frequently asked in CBSE Board examinations as short-answer and long-answer questions.

JEE Main, NEET and CUET regularly test the principles and applications of crystallisation, distillation and chromatography.

Understanding these techniques is essential for laboratory work and analytical chemistry.

These methods are widely used in pharmaceutical industries, forensic science, environmental analysis and food quality testing.

Knowledge of the correct purification method helps in selecting the appropriate technique for separating different types of mixtures.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Crystallisation separates solids based on differences in solubility.

Distillation separates liquids based on differences in boiling points.

Chromatography separates components based on their different affinities for stationary and mobile phases.

Chromatography is one of the most sensitive techniques for separating complex mixtures.

The choice of purification technique depends on the physical properties of the substances involved.

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Q19 →

Q19

NUMERIC3 marks

Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

📘 Concept & Theory Theory / Concept ›

When two compounds have different solubilities in the same solvent, they can be separated by fractional crystallisation. This technique is based on the difference in the solubilities of the compounds at different temperatures.

The compound that is less soluble crystallises out first when the hot saturated solution is cooled, while the more soluble compound remains dissolved in the solvent.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the difference in solubility of the two compounds.

Prepare a hot saturated solution using solvent S.

Allow the solution to cool slowly.

Separate the crystals by filtration.

Recover the second compound from the filtrate by further concentration or cooling.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
The appropriate method is fractional crystallisation.

Dissolve the mixture of the two compounds in the minimum amount of hot solvent S to prepare a hot saturated solution.

If any insoluble impurities are present, remove them by hot filtration.

Allow the hot solution to cool slowly without disturbing it.

The compound having lower solubility crystallises first, while the more soluble compound remains dissolved in the solvent.

Separate the crystals by filtration.

Concentrate the filtrate by gentle evaporation and cool it again to obtain crystals of the second compound.

Wash and dry both sets of crystals separately to obtain the purified compounds.

Principle
Fractional crystallisation works because different compounds have different solubilities in the same solvent. The less soluble compound crystallises first, whereas the more soluble compound remains in the mother liquor.

Example
A mixture of potassium nitrate (KNO₃) and sodium chloride (NaCl) can be separated by fractional crystallisation because their solubilities in water differ significantly with temperature.

🎯 Exam Significance Exam Significance ›

Fractional crystallisation is a frequently asked purification technique in CBSE Board examinations.

JEE Main, NEET and CUET often test the principle and applications of crystallisation-based separation methods.

Understanding this technique helps students select the appropriate purification method based on differences in physical properties.

Fractional crystallisation is widely used in chemical industries, pharmaceutical laboratories and analytical chemistry.

This concept forms the basis for the purification of many inorganic salts and organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Fractional crystallisation separates compounds based on differences in solubility.

The less soluble compound crystallises first from a hot saturated solution.

Slow cooling produces larger and purer crystals.

The more soluble compound remains in the mother liquor.

The method is commonly used for the purification of both inorganic and organic compounds.

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Q20 →

Q20

NUMERIC2 marks

What is the difference between distillation, distillation under reduced pressure and steam distillation ?

📘 Concept & Theory Theory / Concept ›

Distillation is one of the most important techniques used for the purification and separation of liquids. Depending upon the nature of the liquid and its boiling point, different types of distillation are employed.

The choice of the distillation method depends upon the boiling point, thermal stability and miscibility of the liquid with water.

🗺️ Solution Roadmap Step-by-step Plan ›

Define each distillation technique.

State the principle on which it is based.

Mention when the technique is used.

Give one suitable example for each method.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps
(a) Distillation
Principle

Distillation is based on the difference in the boiling points of liquids.

Method

The liquid is heated until it boils.

The vapours formed are condensed using a condenser.

The condensed liquid (distillate) is collected separately.

When Used

To separate a volatile liquid from non-volatile impurities.

To separate two liquids having a sufficiently large difference in boiling points (generally greater than 25 K).

Example

Preparation of pure water from a salt solution.

(b) Distillation under Reduced Pressure (Vacuum Distillation)
Principle

Lowering the external pressure decreases the boiling point of a liquid.

Therefore, liquids that decompose at their normal boiling points can be distilled safely at lower temperatures under reduced pressure.

Method

The pressure inside the apparatus is reduced using a vacuum pump.

The liquid boils at a temperature lower than its normal boiling point.

The vapours are condensed and collected.

When Used

For high-boiling liquids.

For liquids that decompose on strong heating.

Example

Purification of glycerol and many organic oils that decompose at high temperatures.

(c) Steam Distillation
Principle

Steam distillation is based on the fact that two immiscible liquids boil together at a temperature lower than the boiling point of either liquid.

The compound is carried over with steam and can be separated after condensation.

Method

Steam is passed through the organic substance.

The organic compound vaporises along with steam.

The mixed vapours are condensed.

The organic layer is separated from water.

When Used

For water-insoluble volatile organic compounds.

For compounds that are steam volatile.

For substances that decompose at their normal boiling points.

Example

Extraction of essential oils such as clove oil, eucalyptus oil and turpentine oil.

Technique Principle Suitable For Example

Distillation Difference in boiling points. Volatile liquids and liquids with a large boiling point difference. Pure water from salt solution.

Distillation under Reduced Pressure Boiling point decreases when pressure is reduced. High-boiling and heat-sensitive liquids. Purification of glycerol.

Steam Distillation Immiscible liquids boil together below their individual boiling points. Steam-volatile, water-insoluble organic compounds. Extraction of essential oils.

🎯 Exam Significance Exam Significance ›

Distillation techniques are among the most frequently asked purification methods in CBSE Board examinations.

JEE Main, NEET and CUET regularly include conceptual questions based on the principles and applications of different types of distillation.

Students should be able to choose the correct distillation technique based on the boiling point and thermal stability of the compound.

These purification methods are widely used in pharmaceutical industries, petroleum refining, perfume manufacture and chemical laboratories.

A clear understanding of these techniques helps in solving both theoretical and application-based questions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Simple distillation separates liquids based on differences in boiling points.

Vacuum distillation is used for high-boiling or heat-sensitive liquids.

Steam distillation is suitable for steam-volatile, water-insoluble compounds.

Reducing pressure lowers the boiling point of a liquid.

Steam distillation prevents decomposition by allowing distillation at temperatures below the normal boiling point.

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Q21

NUMERIC3 marks

Discuss the chemistry of Lassaigne’s test.

📘 Concept & Theory Theory / Concept ›

Organic compounds contain carbon and hydrogen along with elements such as nitrogen, sulphur and halogens. Since these elements are covalently bonded to carbon, they cannot be detected directly by ordinary inorganic tests.

Lassaigne's test, also called the Sodium Fusion Test, converts these covalently bonded elements into water-soluble ionic compounds by fusing the organic compound with metallic sodium. These ionic compounds can then be identified using suitable chemical tests.

🗺️ Solution Roadmap Step-by-step Plan ›

State the principle of Lassaigne's test.

Explain the role of sodium fusion.

Write the reactions involved in the conversion of covalent compounds into ionic salts.

Describe the tests for nitrogen, sulphur and halogens.

Mention the special case when both nitrogen and sulphur are present together.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps
Principle of Lassaigne's Test

When an organic compound is fused with metallic sodium, the elements nitrogen, sulphur and halogens combine with sodium to form water-soluble ionic salts.

These salts dissolve in water to form the Lassaigne's extract (Sodium Fusion Extract), which is used for qualitative analysis.

Formation of Sodium Salts

Nitrogen

\[ \mathrm{Na + C + N \longrightarrow NaCN} \]

Nitrogen present in the organic compound is converted into sodium cyanide.

Sulphur

\[ \mathrm{2Na + S \longrightarrow Na_2S} \]

Sulphur is converted into sodium sulphide.

Halogens

\[ \mathrm{Na + X \longrightarrow NaX} \]

where

\[ \mathrm{X = Cl,\ Br,\ I} \]

Halogens are converted into sodium halides.

Nitrogen and Sulphur Present Together

If both nitrogen and sulphur are present in the same compound, sodium thiocyanate is formed.

\[ \mathrm{Na + C + N + S \longrightarrow NaSCN} \]

Test for Nitrogen
The sodium fusion extract containing sodium cyanide is treated with freshly prepared ferrous sulphate solution. \[\mathrm{FeSO_4 + 6NaCN \longrightarrow Na_4[Fe(CN)_6] + Na_2SO_4}\]

On boiling, cooling and acidifying with dilute sulphuric acid, some ferrous ions are oxidised to ferric ions.

Ferric ions react with sodium ferrocyanide to produce ferric ferrocyanide (Prussian blue). \[\mathrm{3Na_4[Fe(CN)_6] + 4Fe^{3+}\longrightarrow Fe_4[Fe(CN)_6]_3 + 12Na^+}\]
Observation

Appearance of a Prussian blue precipitate confirms the presence of nitrogen.

Test for Sulphur
The sodium fusion extract containing sodium sulphide is treated with sodium nitroprusside solution. \[\mathrm{Na_2S + Na_2[Fe(CN)_5NO]\longrightarrow Na_4[Fe(CN)_5NOS]}\]
Observation

A violet or purple colour confirms the presence of sulphur.

Alternatively, sulphide ions react with lead acetate solution to form black lead sulphide.
\[\mathrm{Na_2S + Pb(CH_3COO)_2\longrightarrow PbS \downarrow + 2CH_3COONa}\]

Test for Halogens
The sodium fusion extract is first acidified with dilute nitric acid.
This removes interfering ions such as CN⁻ and S²⁻.

Silver nitrate solution is added
The following precipitates are obtained. tbody>

Halogen Reaction Observation

Chlorine \[ \mathrm{NaCl + AgNO_3 \rightarrow AgCl\downarrow + NaNO_3} \] White precipitate

Bromine \[ \mathrm{NaBr + AgNO_3 \rightarrow AgBr\downarrow + NaNO_3} \] Cream precipitate

Iodine \[ \mathrm{NaI + AgNO_3 \rightarrow AgI\downarrow + NaNO_3} \] Yellow precipitate

🎯 Exam Significance Exam Significance ›

Lassaigne's test is one of the most important qualitative analysis topics in Organic Chemistry.

CBSE Board examinations frequently ask the principle, reactions and observations involved in sodium fusion tests.

JEE Main, NEET and CUET regularly include questions based on the chemistry of sodium fusion extract and the identification of nitrogen, sulphur and halogens.

Students should remember the characteristic colours and precipitates produced during each test, as these are commonly tested in objective examinations.

A thorough understanding of Lassaigne's test is essential for qualitative organic analysis and practical chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Lassaigne's test converts covalently bonded elements into ionic sodium salts.

Nitrogen forms sodium cyanide (NaCN).

Sulphur forms sodium sulphide (Na₂S).

Halogens form sodium halides (NaX).

Prussian blue confirms nitrogen, violet colour confirms sulphur, and AgCl, AgBr and AgI precipitates confirm chlorine, bromine and iodine respectively.

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Q22

NUMERIC3 marks

Differentiate between the principle of estimation of nitrogen in an organic compound by:

Dumas Method

Kjeldahl's Method

📘 Concept & Theory Theory / Concept ›

Nitrogen is one of the most important elements present in organic compounds such as proteins, amino acids, dyes and pharmaceuticals. The percentage of nitrogen present in an organic compound can be determined quantitatively by different analytical methods.

The two commonly used methods are Dumas Method and Kjeldahl's Method. Although both methods estimate nitrogen, their principles and applications are different.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain the principle of Dumas method.

Explain the principle of Kjeldahl's method.

Compare the two methods based on their working principles.

Mention the limitations of each method.

✏️ Solution Complete Solution ›

Step-by-step Solution · 3 steps
(i) Dumas Method
Principle

In Dumas method, the organic compound is heated strongly with excess copper(II) oxide in an atmosphere of carbon dioxide.

The carbon and hydrogen present in the compound are oxidised to carbon dioxide and water, whereas nitrogen is converted into molecular nitrogen.

The oxides of nitrogen, if formed, are reduced to nitrogen gas by passing the gases over heated copper.

The nitrogen gas evolved is collected and its volume is measured. From the measured volume of nitrogen, the percentage of nitrogen present in the compound is calculated.

Main Reactions

\[ \mathrm{Organic\ Compound + CuO \longrightarrow CO_2 + H_2O + N_2} \]

If nitrogen oxides are formed, they are reduced by hot copper.

\[ \mathrm{2Cu + 2NO \longrightarrow N_2 + 2CuO} \]

Suitable For

Almost all nitrogen-containing organic compounds.

Compounds containing nitro, azo and diazo groups.

(ii) Kjeldahl's Method
Principle

In Kjeldahl's method, the organic compound is heated with concentrated sulphuric acid.

The nitrogen present in the compound is converted into ammonium sulphate.

\[ \mathrm{Organic\ Nitrogen \xrightarrow{Conc.\ H_2SO_4} (NH_4)_2SO_4} \]

The reaction mixture is then treated with excess sodium hydroxide solution.

Ammonia gas is liberated.

\[ \mathrm{(NH_4)_2SO_4 + 2NaOH \longrightarrow 2NH_3 + Na_2SO_4 + 2H_2O} \]

The ammonia evolved is absorbed in a known excess of standard acid.

The unused acid is finally determined by titration with a standard alkali.

From the amount of ammonia evolved, the percentage of nitrogen is calculated.

Suitable For

Kjeldahl's method is suitable only for compounds in which nitrogen can be converted into ammonium sulphate.

class="text-warning"Limitation

This method cannot be used for compounds containing nitrogen in the form of:

Nitro group (-NO₂)

Azo group (-N=N-)

Diazo group (-N=N⁺-)

Nitrogen directly bonded to oxygen or another nitrogen atom.

Difference Between Dumas Method and Kjeldahl's Method

Basis Dumas Method Kjeldahl's Method

Principle Nitrogen is converted into nitrogen gas (N₂). Nitrogen is converted into ammonium sulphate and finally into ammonia (NH₃).

Measurement Volume of nitrogen gas is measured. Amount of ammonia liberated is determined by acid-base titration.
Reagent Used Copper(II) oxide and heated copper. Concentrated H₂SO₄ followed by NaOH.

Suitable For Almost all nitrogen-containing organic compounds. Most organic compounds except nitro, azo and diazo compounds.

Not Suitable For No major limitation for ordinary nitrogen compounds. Nitro, azo and diazo compounds.

🎯 Exam Significance Exam Significance ›

Dumas and Kjeldahl methods are among the most important quantitative analytical techniques in Organic Chemistry.

CBSE Board examinations frequently ask students to compare these two methods and explain their principles.

JEE Main, NEET and CUET regularly include conceptual questions based on the reagents, principles and limitations of nitrogen estimation methods.

Students should remember that Dumas method estimates nitrogen as nitrogen gas, whereas Kjeldahl's method estimates nitrogen through ammonia formation.

The limitations of Kjeldahl's method regarding nitro, azo and diazo compounds are frequently tested in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Dumas method converts nitrogen into nitrogen gas.

Kjeldahl's method converts nitrogen into ammonium sulphate and finally into ammonia.

Dumas method involves measurement of nitrogen gas volume.

Kjeldahl's method involves acid-base titration of ammonia.

Kjeldahl's method cannot be used for nitro, azo and diazo compounds.

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Q23

NUMERIC3 marks

Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

📘 Concept & Theory Theory / Concept ›

In quantitative organic analysis, elements such as halogens, sulphur and phosphorus present in an organic compound are estimated by converting them into stable inorganic compounds. These inorganic products are then quantitatively analysed using gravimetric methods.

The estimation is based on the law of conservation of mass. The amount of precipitate formed is proportional to the amount of the element present in the original organic compound.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain the principle of estimation of halogens.

Explain the principle of estimation of sulphur.

Explain the principle of estimation of phosphorus.

Mention the important reactions involved.

State the precipitate obtained and its significance.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps
(a) Estimation of Halogens (Carius Method)
Principle

The organic compound is heated with fuming nitric acid (HNO₃) in the presence of silver nitrate inside a sealed Carius tube.

During oxidation, carbon and hydrogen are converted into carbon dioxide and water, while the halogens are converted into silver halides.

The silver halide precipitate is filtered, washed, dried and weighed.

The mass of the silver halide formed is used to calculate the percentage of halogen present in the compound.

Reactions

For chlorine

\[\mathrm{R{-}Cl + AgNO_3 \longrightarrow AgCl\downarrow}\]

For bromine

\[\mathrm{R{-}Br + AgNO_3 \longrightarrow AgBr\downarrow}\]

For iodine

\[\mathrm{R{-}I + AgNO_3 \longrightarrow AgI\downarrow}\]

Observation

AgCl : White precipitate

AgBr : Pale yellow (cream) precipitate

AgI : Yellow precipitate

(b) Estimation of Sulphur (Carius Method)
Principle

The organic compound is heated with fuming nitric acid in a sealed Carius tube.

All sulphur present in the compound is oxidised to sulphuric acid.

The sulphuric acid formed is treated with barium chloride solution to precipitate barium sulphate.

The barium sulphate precipitate is filtered, dried and weighed.

The percentage of sulphur is calculated from the mass of barium sulphate obtained.

Reactions

\[\mathrm{S \xrightarrow{HNO_3} H_2SO_4}\]

\[\mathrm{H_2SO_4 + BaCl_2\longrightarrow BaSO_4\downarrow + 2HCl}\]

Observation

A heavy white precipitate of barium sulphate (BaSO₄) is obtained.

(c) Estimation of Phosphorus (Carius Method)
Principle

The organic compound is heated with fuming nitric acid in a sealed Carius tube.

Phosphorus present in the compound is oxidised to phosphoric acid.

The phosphoric acid is treated with ammonium molybdate solution.

A yellow precipitate of ammonium phosphomolybdate is formed.

The precipitate is filtered, washed, dried and weighed to determine the phosphorus content.

Reactions

\[\mathrm{P \xrightarrow{HNO_3} H_3PO_4}\]

\[\mathrm{H_3PO_4 + (NH_4)_2MoO_4\longrightarrow (NH_4)_3PO_4\cdot12MoO_3\downarrow}\]

Observation

A yellow precipitate of ammonium phosphomolybdate confirms the presence of phosphorus.

Summary Table

Element Method Converted Into Final Precipitate

Halogens (Cl, Br, I) Carius Method Silver Halides AgCl, AgBr or AgI

Sulphur Carius Method Sulphuric Acid BaSO₄

Phosphorus Carius Method Phosphoric Acid Ammonium Phosphomolybdate

🎯 Exam Significance Exam Significance ›

The Carius method is one of the most important quantitative estimation techniques in Organic Chemistry.

CBSE Board examinations frequently ask the principles, reactions and precipitates involved in the estimation of halogens, sulphur and phosphorus.

JEE Main, NEET and CUET often include conceptual questions based on quantitative elemental analysis and the identification of the final precipitates.

Students should remember the characteristic precipitates formed during each estimation, as these are commonly tested in objective and assertion-reason questions.

The principles of elemental estimation provide the foundation for analytical chemistry and laboratory analysis of organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Halogens, sulphur and phosphorus are estimated by the Carius method.

Halogens are estimated as silver halides (AgCl, AgBr and AgI).

Sulphur is estimated as barium sulphate (BaSO₄).

Phosphorus is estimated as ammonium phosphomolybdate.

The mass of the final precipitate is used to calculate the percentage of the corresponding element.

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Q24 →

Q24

NUMERIC3 marks

Explain the principle of paper chromatography.

📘 Concept & Theory Theory / Concept ›

Chromatography is one of the most important separation techniques used in chemistry. It is employed for the separation, identification and purification of the components of a mixture.

Paper chromatography is a type of partition chromatography in which a sheet of special filter paper acts as the stationary phase, while a suitable solvent acts as the mobile phase.

🗺️ Solution Roadmap Step-by-step Plan ›

Define paper chromatography.

State the principle on which it is based.

Explain the role of stationary and mobile phases.

Describe how the separation occurs.

Give suitable examples and applications.

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps
Principle of Paper Chromatography
Paper chromatography is based on the partition of the components of a mixture between two phases.

The stationary phase is the thin film of water adsorbed on the cellulose fibres of the filter paper.

The mobile phase is a suitable solvent or a mixture of solvents that moves upward through the paper by capillary action.

Each component of the mixture distributes itself differently between the stationary phase and the mobile phase.

Components having a greater affinity for the stationary phase move slowly, whereas components having a greater affinity for the mobile phase move faster.

As a result, the components travel different distances and become separated into distinct spots on the chromatographic paper.

Working of Paper Chromatography
Step 1. A small spot of the mixture is placed near one end of the chromatography paper using a capillary tube.

Step 2. The lower end of the paper is dipped into a suitable solvent, ensuring that the sample spot remains above the solvent level.

Step 3. The solvent rises through the paper by capillary action.

Step 4. The different components of the mixture move upward at different rates because of their different partition coefficients.

Step 5. After sufficient separation, the paper is removed, dried and the separated spots are observed.

Retention Factor (R_f Value)
The movement of a component on chromatographic paper is expressed by its retention factor (R_f value).

\[R_f=\frac{\text{Distance travelled by the substance}}{\text{Distance travelled by the solvent front}}\]

The value of

\[R_f<1\]

because the solvent front always travels farther than the solute.

The R_f value is characteristic for a given compound under fixed experimental conditions and is therefore useful for identification.

Applications

Separation of coloured pigments present in ink.

Separation of amino acids.

Identification of sugars.

Detection of drugs in pharmaceutical analysis.

Analysis of plant pigments such as chlorophyll and carotenoids.

Forensic analysis of dyes and inks.

Example
When black ink is subjected to paper chromatography, it separates into several coloured dyes because each dye has a different affinity for the stationary and mobile phases.

Final Answer

Paper chromatography is based on the principle of differential partition of the components of a mixture between a stationary phase (water adsorbed on filter paper) and a mobile phase (solvent). Components having greater affinity for the stationary phase move slowly, whereas those having greater affinity for the mobile phase move faster. Consequently, the components travel different distances and get separated into distinct spots on the paper.

🎯 Exam Significance Exam Significance ›

Paper chromatography is one of the most frequently asked purification and separation techniques in CBSE Board examinations.

JEE Main, NEET and CUET regularly ask conceptual questions on the principle, stationary phase, mobile phase and applications of paper chromatography.

Students should remember that paper chromatography is a type of partition chromatography and that separation occurs because of different partition coefficients.

The R_f value is an important numerical quantity used for identifying compounds and is frequently tested in competitive examinations.

Paper chromatography has extensive applications in analytical chemistry, pharmaceuticals, forensic science and biochemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Paper chromatography is based on the principle of partition.

The stationary phase is water adsorbed on cellulose fibres of filter paper.

The mobile phase is a suitable solvent that moves by capillary action.

Different components travel different distances because of their different affinities for the two phases.

The R_f value is characteristic of a compound under fixed experimental conditions.

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Q25

NUMERIC3 marks

Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

📘 Concept & Theory Theory / Concept ›

In Lassaigne's test for halogens, the sodium fusion extract may contain not only sodium halides but also sodium cyanide (NaCN) and sodium sulphide (Na₂S) if the organic compound contains nitrogen or sulphur.

Both cyanide and sulphide ions interfere with the silver nitrate test because they also react with silver nitrate to form insoluble precipitates. Therefore, these interfering ions must be removed before testing for halogens.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the impurities present in the sodium fusion extract.

Explain how these impurities interfere with the silver nitrate test.

Describe the role of nitric acid.

State why silver nitrate is added only after acidification.

✏️ Solution Complete Solution ›

Step-by-step Solution · 3 steps
During sodium fusion, if the organic compound contains nitrogen or sulphur, the sodium extract may contain:

Sodium cyanide (NaCN)

Sodium sulphide (Na₂S)

Sodium halides (NaCl, NaBr or NaI)

If silver nitrate is added directly, cyanide ions and sulphide ions also react with silver nitrate to produce insoluble precipitates.

Formation of silver cyanide:

\[ \mathrm{NaCN + AgNO_3 \longrightarrow AgCN\downarrow + NaNO_3} \]

Formation of silver sulphide:

\[ \mathrm{Na_2S + 2AgNO_3 \longrightarrow Ag_2S\downarrow + 2NaNO_3} \]

These precipitates may be mistaken for silver halides, leading to incorrect conclusions.

Before adding silver nitrate, the sodium extract is boiled with dilute nitric acid.

Nitric acid decomposes sodium cyanide and sodium sulphide, converting them into volatile products that escape from the solution.

For cyanide:

\[ \mathrm{NaCN + HNO_3 \longrightarrow HCN\uparrow + NaNO_3} \]

Hydrogen cyanide gas escapes from the solution.

For sulphide:

\[ \mathrm{Na_2S + 2HNO_3 \longrightarrow H_2S\uparrow + 2NaNO_3} \]

Hydrogen sulphide gas also escapes from the solution.

After removal of cyanide and sulphide ions, only halide ions remain in the sodium extract.

Now, addition of silver nitrate produces only the corresponding silver halide.

\[ \mathrm{NaCl + AgNO_3 \longrightarrow AgCl\downarrow + NaNO_3} \]

\[ \mathrm{NaBr + AgNO_3 \longrightarrow AgBr\downarrow + NaNO_3} \]

\[ \mathrm{NaI + AgNO_3 \longrightarrow AgI\downarrow + NaNO_3} \]

Thus, the test gives correct and reliable results.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions from Lassaigne's test in CBSE Board examinations.

JEE Main, NEET and CUET often test the purpose of adding nitric acid before silver nitrate and the role of interfering ions.

Students should remember that nitric acid is used to remove cyanide and sulphide ions, not the halide ions.

Knowledge of the interfering reactions is important for understanding qualitative organic analysis.

This concept is also useful in laboratory practical examinations involving elemental analysis.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Sodium fusion extract may contain CN⁻, S²⁻ and halide ions.

CN⁻ and S²⁻ interfere with the silver nitrate test.

Nitric acid converts CN⁻ into HCN gas and S²⁻ into H₂S gas.

After acidification, only halide ions react with silver nitrate.

Characteristic precipitates of AgCl, AgBr and AgI confirm the presence of chlorine, bromine and iodine respectively.

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Q26

NUMERIC3 marks

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

📘 Concept & Theory Theory / Concept ›

Organic compounds generally contain carbon and hydrogen along with elements such as nitrogen, sulphur and halogens. These elements are covalently bonded to carbon and therefore cannot be detected directly by ordinary inorganic qualitative tests.

To identify these elements, the covalent bonds must first be broken and the elements converted into water-soluble ionic compounds. This conversion is achieved by fusing the organic compound with metallic sodium. The process is known as Lassaigne's sodium fusion.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain why nitrogen, sulphur and halogens cannot be detected directly.

Describe the role of metallic sodium.

Write the reactions showing the formation of sodium salts.

Explain why the sodium fusion extract is used for qualitative analysis.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps
In an organic compound, nitrogen, sulphur and halogens are covalently bonded to carbon atoms.

Because of their covalent nature, these elements do not produce ions in aqueous solution.

Therefore, they cannot be detected by the ordinary qualitative tests used for inorganic salts.

The organic compound is fused with metallic sodium.

Metallic sodium is highly reactive and breaks the covalent bonds present in the organic compound.

As a result, the elements combine with sodium to form water-soluble ionic salts.

The following sodium salts are formed during fusion.

Nitrogen

\[\mathrm{Na + C + N \longrightarrow NaCN}\]

Nitrogen is converted into sodium cyanide (NaCN).

Sulphur

\[\mathrm{2Na + S \longrightarrow Na_2S}\]

Sulphur is converted into sodium sulphide (Na₂S).

Halogens

\[\mathrm{Na + X \longrightarrow NaX}\]

where

\[\mathrm{X = Cl,\ Br,\ I}\]

Halogens are converted into sodium halides (NaCl, NaBr or NaI).

If both nitrogen and sulphur are present in the same compound, sodium thiocyanate is formed.

\[\mathrm{Na + C + N + S \longrightarrow NaSCN}\]

The sodium salts formed are soluble in water.

They dissolve to produce the Lassaigne's extract (Sodium Fusion Extract), which is used for testing nitrogen, sulphur and halogens by suitable qualitative reactions.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions on Lassaigne's test in CBSE Board examinations.

JEE Main, NEET and CUET regularly test the principle of sodium fusion and the reason for converting covalent compounds into ionic salts.

Students should remember the sodium salts formed during fusion, as they are commonly asked in objective and assertion-reason questions.

The concept forms the basis of qualitative elemental analysis in Organic Chemistry practical examinations.

A clear understanding of sodium fusion helps in explaining all subsequent tests for nitrogen, sulphur and halogens.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Nitrogen, sulphur and halogens are covalently bonded in organic compounds.

Covalently bonded elements cannot be detected directly by ordinary inorganic tests.

Metallic sodium converts these elements into water-soluble ionic sodium salts.

The sodium fusion extract is used for qualitative analysis.

Lassaigne's test is based on the conversion of covalent compounds into ionic compounds.

← Q25

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Q27 →

Q27

NUMERIC3 marks

Name a suitable technique for the separation of the components from a mixture of calcium sulphate and camphor.

📘 Concept & Theory Theory / Concept ›

Different separation techniques are employed depending on the physical properties of the substances present in a mixture. One such property is sublimation.

Sublimation is the process in which a solid changes directly into vapour on heating without passing through the liquid state. On cooling, the vapour changes directly back into the solid state.

Only substances that sublime can be separated from non-sublimable substances using this technique.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the physical properties of both components.

Determine whether either component undergoes sublimation.

Select the appropriate separation technique based on this property.

Explain the separation process.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps
Camphor is a sublimable solid. On heating, it changes directly from the solid state into vapour.

Calcium sulphate (CaSO₄) is a non-sublimable solid. It does not vaporise on heating under ordinary conditions.

Therefore, the mixture is heated gently in a sublimation apparatus.

Camphor sublimes and its vapours come in contact with a cold surface, where they condense to form pure solid camphor.

Hence, the two components are separated successfully.

🎯 Exam Significance Exam Significance ›

Questions based on selecting an appropriate separation technique are frequently asked in CBSE Board examinations.

JEE Main, NEET and CUET often test the applications of sublimation for separating mixtures containing sublimable and non-sublimable solids.

Students should remember common sublimable substances such as camphor, ammonium chloride, iodine and naphthalene.

Understanding the physical properties of substances helps in selecting the correct purification or separation method.

This concept has practical applications in laboratory purification and industrial processing of volatile solids.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Sublimation is used to separate sublimable solids from non-sublimable solids.

Camphor undergoes sublimation on heating.

Calcium sulphate does not sublime under ordinary conditions.

During sublimation, the solid changes directly into vapour without becoming a liquid.

The vapours condense on a cold surface to give the purified solid.

← Q26

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Q28 →

Q28

NUMERIC3 marks

Explain why an organic liquid vaporises at a temperature below its boiling point in steam distillation.

📘 Concept & Theory Theory / Concept ›

Steam distillation is a special distillation technique used for separating steam-volatile, water-insoluble organic compounds that decompose at or near their normal boiling points.

The method is based on the behaviour of immiscible liquids. Water and the organic liquid do not mix with each other; therefore, each liquid exerts its own vapour pressure independently.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain the behaviour of immiscible liquids.

State Dalton's law of partial pressures.

Relate the total vapour pressure to boiling.

Explain why boiling occurs below the normal boiling point.

State the advantage of steam distillation.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps
In steam distillation, the organic liquid and water are immiscible. Therefore, each liquid exerts its own vapour pressure independently.

According to Dalton's law of partial pressures, the total vapour pressure of the mixture is equal to the sum of the vapour pressures of water and the organic liquid. \[\mathrm{P_{\text{total}} = P_{\text{water}} + P_{\text{organic}}}\]

A liquid boils when its vapour pressure becomes equal to the external atmospheric pressure.

In steam distillation, the combined vapour pressure reaches atmospheric pressure even though the individual vapour pressures of water and the organic liquid are each lower than atmospheric pressure. \[\mathrm{P_{\text{water}} + P_{\text{organic}} = P_{\text{atmospheric}}}\]

Since the combined vapour pressure becomes equal to atmospheric pressure at a lower temperature, the mixture boils below the normal boiling point of the organic liquid.

Thus, the organic liquid vaporises along with steam without being heated to its own boiling point.

This prevents decomposition of heat-sensitive organic compounds

Example
Essential oils such as clove oil, eucalyptus oil and turpentine oil are extracted by steam distillation because they decompose if heated to their normal boiling points.

🎯 Exam Significance Exam Significance ›

This question is frequently asked in CBSE Board examinations to test the principle of steam distillation.

JEE Main, NEET and CUET often include conceptual questions based on vapour pressure, Dalton's law and purification techniques.

Students should remember that steam distillation is applicable only to steam-volatile, water-insoluble compounds.

The concept of total vapour pressure is fundamental for understanding why heat-sensitive organic compounds can be purified safely.

Steam distillation is widely used in the extraction of perfumes, essential oils and medicinal compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Steam distillation is used for steam-volatile, water-insoluble organic compounds.

Water and the organic liquid are immiscible.

The total vapour pressure is the sum of the vapour pressures of both liquids.

The mixture boils when the total vapour pressure equals atmospheric pressure.

The organic liquid vaporises below its normal boiling point, preventing thermal decomposition.

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Q29 →

Q29

NUMERIC3 marks

Will $\mathrm{CCl_4}$ give a white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

📘 Concept & Theory Theory / Concept ›

Silver nitrate gives a white precipitate of silver chloride (AgCl) only when chloride ions (Cl⁻) are present in the solution.

In inorganic compounds such as sodium chloride, chlorine exists as chloride ions and readily reacts with silver nitrate.

However, in organic compounds, halogens are generally covalently bonded to carbon atoms. Such covalent halogens do not ionise in solution and therefore do not react directly with silver nitrate.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the nature of the carbon-chlorine bond in carbon tetrachloride.

Determine whether chloride ions are produced.

State the condition required for the formation of silver chloride.

Draw the final conclusion.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps
In carbon tetrachloride, chlorine atoms are covalently bonded to the carbon atom. \[\mathrm{CCl_4}\]

On heating with silver nitrate solution, carbon tetrachloride does not dissociate to produce chloride ions.\[\mathrm{CCl_4 \;\not\rightarrow\; C^{4+} + 4Cl^-}\]

Silver nitrate forms a white precipitate only when free chloride ions are present.\[\mathrm{Ag^+ + Cl^- \longrightarrow AgCl\downarrow}\]

Since no chloride ions are produced from carbon tetrachloride, silver chloride cannot be formed.

Therefore, carbon tetrachloride does not give a white precipitate with silver nitrate.

Note: If carbon tetrachloride is first subjected to Lassaigne's sodium fusion, the covalent carbon-chlorine bonds are broken and chlorine is converted into sodium chloride.\[\mathrm{Na + Cl \longrightarrow NaCl}\]
Now, chloride ions are available in the sodium fusion extract. \[\mathrm{NaCl + AgNO_3 \longrightarrow AgCl\downarrow + NaNO_3}\] Hence, the white precipitate of silver chloride is obtained only after sodium fusion and not by directly treating $\mathrm{CCl_4}$ with silver nitrate.

Final Answer

No. Carbon tetrachloride does not give a white precipitate of silver chloride on heating with silver nitrate because the chlorine atoms are covalently bonded to carbon and do not produce chloride ions in solution. Since silver nitrate reacts only with free chloride ions, no AgCl precipitate is formed. The chloride ions become available only after sodium fusion in Lassaigne's test.

🎯 Exam Significance Exam Significance ›

This is a frequently asked conceptual question in CBSE Board examinations based on Lassaigne's test and qualitative organic analysis.

JEE Main, NEET and CUET often test the difference between covalent halogens and ionic halides.

Students should remember that silver nitrate detects only free halide ions and not covalently bonded halogens.

The role of sodium fusion in converting covalent halogens into ionic sodium halides is an important concept in Organic Chemistry.

This principle is widely used in laboratory identification of halogens present in organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

In carbon tetrachloride, chlorine is covalently bonded to carbon.

Covalent chlorine does not ionise to produce chloride ions.

Silver nitrate reacts only with free chloride ions.

Therefore, carbon tetrachloride does not give a white precipitate of AgCl directly.

After Lassaigne's sodium fusion, chlorine is converted into sodium chloride, which gives the AgCl precipitate with silver nitrate.

← Q28

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Q30 →

Q30

NUMERIC2 marks

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

📘 Concept & Theory Theory / Concept ›

The estimation of carbon in an organic compound is generally carried out by the Liebig's combustion method. In this method, the organic compound is heated with excess copper(II) oxide, which oxidises carbon to carbon dioxide and hydrogen to water.

The carbon dioxide produced is passed through a weighed absorption bulb containing potassium hydroxide (KOH) solution. The increase in the mass of the absorption bulb corresponds to the mass of carbon dioxide absorbed, from which the percentage of carbon in the organic compound is calculated.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain how carbon is converted into carbon dioxide during combustion.

State the role of potassium hydroxide solution.

Write the chemical reaction involved.

Explain why the increase in mass is measured.

Relate the absorbed carbon dioxide to the estimation of carbon.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps
During Liebig's combustion, the organic compound is heated with excess copper(II) oxide.
All the carbon present in the compound is oxidised to carbon dioxide.\[ \mathrm{C + O_2 \longrightarrow CO_2}\]

The carbon dioxide produced is passed through a weighed absorption bulb containing aqueous potassium hydroxide solution.

Potassium hydroxide readily absorbs carbon dioxide to form potassium carbonate. \[\mathrm{2KOH + CO_2 \longrightarrow K_2CO_3 + H_2O}\] If excess carbon dioxide is present, potassium bicarbonate may also be formed. \[\mathrm{K_2CO_3 + CO_2 + H_2O \longrightarrow 2KHCO_3}\]

As carbon dioxide is absorbed, the mass of the potassium hydroxide absorption bulb increases.

The increase in mass is equal to the mass of carbon dioxide produced during combustion.

Since every mole of carbon dioxide contains one mole of carbon, the mass o carbon present in the original organic compound is calculated from the mass of carbon dioxide obtained. \[\mathrm{Mass\ of\ Carbon}=\frac{12}{44}\times\mathrm{Mass\ of\ CO_2}\] Hence, the percentage of carbon is calculated as \[\%\mathrm{C}=\frac{\mathrm{Mass\ of\ Carbon}}{\mathrm{Mass\ of\ Organic\ Compound}}\times100\]

🎯 Exam Significance Exam Significance ›

Liebig's method for the estimation of carbon and hydrogen is an important topic in quantitative organic analysis.

CBSE Board examinations frequently ask the role of potassium hydroxide and copper(II) oxide in carbon estimation.

JEE Main, NEET and CUET regularly include conceptual and numerical questions based on the estimation of carbon from the mass of carbon dioxide produced.

Students should remember that potassium hydroxide is used specifically to absorb carbon dioxide, whereas anhydrous calcium chloride is used to absorb water vapour.

This concept forms the basis of elemental analysis of organic compounds in analytical chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Carbon is oxidised to carbon dioxide during combustion.

Potassium hydroxide solution absorbs carbon dioxide quantitatively.

The increase in the mass of the KOH bulb equals the mass of carbon dioxide absorbed.

The mass of carbon is calculated using the relation $ \frac{12}{44} \times \text{Mass of CO}_2 $.

Liebig's combustion method is widely used for the quantitative estimation of carbon in organic compounds.

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Q31 →

Q31

NUMERIC3 marks

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

📘 Concept & Theory Theory / Concept ›

In Lassaigne's test, sulphur present in an organic compound is converted into sodium sulphide (Na₂S) by fusion with metallic sodium. The sulphide ions present in the sodium fusion extract are detected using lead acetate solution.

Before adding lead acetate, the sodium extract is acidified. The acid used must not itself produce any precipitate with lead ions; otherwise, the sulphur test will give incorrect results.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain the purpose of acidifying the sodium fusion extract.

Describe the reaction of sulphide ions with lead acetate.

Explain why sulphuric acid cannot be used.

State why acetic acid is the preferred acid.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps
During sodium fusion, sulphur present in the organic compound is converted into sodium sulphide. \[\mathrm{2Na + S \longrightarrow Na_2S}\]

The sodium fusion extract containing sulphide ions is acidified before adding lead acetate solution.

When lead acetate is added, sulphide ions react with lead(II) ions to form a black precipitate of lead sulphide. \[\mathrm{Na_2S + Pb(CH_3COO)_2\longrightarrow PbS\downarrow + 2CH_3COONa}\] The appearance of a black precipitate of PbS confirms the presence of sulphur.

If sulphuric acid is used for acidification, sulphate ions are introduced into the solution. Lead acetate reacts with sulphate ions to produce a white precipitate of lead sulphate. \[\mathrm{Pb(CH_3COO)_2 + H_2SO_4\longrightarrow PbSO_4\downarrow + 2CH_3COOH}\] This white precipitate interferes with the sulphur test and may mask or prevent the observation of the black precipitate of lead sulphide.

Acetic acid does not produce any insoluble precipitate with lead acetate.
Therefore, only sulphide ions react with lead(II) ions to form the characteristic black precipitate of lead sulphide.
>Hence, acetic acid is used instead of sulphuric acid.

🎯 Exam Significance Exam Significance ›

This is a frequently asked conceptual question from Lassaigne's test in CBSE Board examinations.

JEE Main, NEET and CUET often test the choice of reagents used in qualitative organic analysis and the reason for avoiding sulphuric acid.

Students should remember that sulphuric acid introduces sulphate ions, which interfere with the lead acetate test.

Understanding reagent selection helps explain the chemistry behind qualitative analysis rather than merely memorising the procedure.

This concept is also important in practical examinations involving the detection of sulphur in organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Sodium fusion converts sulphur into sodium sulphide.

Lead acetate detects sulphur by forming black lead sulphide (PbS).

Sulphuric acid cannot be used because it forms white lead sulphate (PbSO₄).

Acetic acid does not interfere with the lead acetate test.

The appearance of a black precipitate confirms the presence of sulphur.

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Q32 →

Q32

NUMERIC3 marks

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

📘 Concept & Theory Theory / Concept ›

During the complete combustion of an organic compound, all the carbon present is converted into carbon dioxide, while all the hydrogen is converted into water.

The masses of carbon dioxide and water produced are calculated from the percentages of carbon and hydrogen present in the compound using stoichiometric relationships.

The required conversion factors are:
\[\mathrm{12\ g\ of\ C \longrightarrow 44\ g\ of\ CO_2}\] \[\mathrm{2\ g\ of\ H \longrightarrow 18\ g\ of\ H_2O}\]

🗺️ Solution Roadmap Step-by-step Plan ›

Calculate the mass of carbon present in the given sample.

Calculate the mass of hydrogen present in the sample.

Use stoichiometric ratios to determine the mass of carbon dioxide produced.

Use stoichiometric ratios to determine the mass of water produced.

Write the final answers with proper units.

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps
Given
Mass of organic compound $=\mathrm{=0.20\ g}$

Percentage of carbon $=\mathrm{=69\%}$

Percentage of hydrogen $=\mathrm{=4.8\%}$

Caluculation
Calculate the mass of carbon present. \[\begin{aligned}\mathrm{Mass\ of\ carbon}&=\frac{69}{100}\times0.20 \\&=0.138\ \mathrm{g}\end{aligned}\]

Calculate the mass of carbon dioxide produced. \[12\ \mathrm{g\ C}\longrightarrow 44\ \mathrm{g\ CO_2}\] \[\begin{aligned}0.138\ \mathrm{g\ C}\longrightarrow& \frac{44}{12}\times0.138\\&=0.506\ \mathrm{g}\end{aligned}\]

Therefore, \[\boxed{\mathrm{Mass\ of\ CO_2=0.506\ g}}\]

Calculate the mass of hydrogen present. \[ \begin{aligned} \mathrm{Mass\ of\ hydrogen}=\frac{4.8}{100}\times0.20\\ &=0.0096\ \mathrm{g} \end{aligned} \]

Calculate the mass of water produced. \[ \begin{aligned} 2\ \mathrm{g\ H} \longrightarrow& 18\ \mathrm{g\ H_2O}\\ 0.0096\ \mathrm{g\ H} \longrightarrow &\frac{18}{2}\times0.0096\\ &=0.0864\ \mathrm{g} \end{aligned} \]

Therefore,\[\boxed{\bbox[2pt]{\mathrm{Mass\ of\ H_2O=0.0864\ g}}}\]

🎯 Exam Significance Exam Significance ›

This numerical is based on Liebig's combustion method, an important topic in quantitative organic analysis.

CBSE Board examinations frequently ask numerical problems involving the calculation of carbon dioxide and water produced during combustion.

JEE Main, NEET and CUET often include stoichiometric calculations based on elemental composition and combustion analysis.

Students should remember the conversion factors between carbon and carbon dioxide, and between hydrogen and water.

Such numericals strengthen the application of percentage composition and mole concepts in Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

All carbon present in an organic compound is converted into carbon dioxide during complete combustion.

All hydrogen present is converted into water.

Use the relation $12\ \mathrm{g\ C} \rightarrow 44\ \mathrm{g\ CO_2}$.

Use the relation $2\ \mathrm{g\ H} \rightarrow 18\ \mathrm{g\ H_2O}$.

Always calculate the mass of the element first before applying stoichiometric ratios.

← Q31

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Q33 →

Q33

NUMERIC3 marks

A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M $\ce{H2SO4}$. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

📘 Concept & Theory Theory / Concept ›

In Kjeldahl's method, nitrogen present in an organic compound is converted into ammonium sulphate by heating with concentrated sulphuric acid.

On adding sodium hydroxide, ammonia gas is liberated.

The ammonia is absorbed in a known excess of standard sulphuric acid. The unused sulphuric acid is then determined by back titration with standard sodium hydroxide.

The amount of sulphuric acid neutralised by ammonia is used to calculate the amount of nitrogen present in the compound.

🗺️ Solution Roadmap Step-by-step Plan ›

Calculate the initial moles of sulphuric acid.

Calculate the moles of sodium hydroxide used in back titration.

Determine the moles of sulphuric acid remaining after absorption of ammonia.

Calculate the moles of sulphuric acid neutralised by ammonia.

Calculate the moles and mass of nitrogen present.

Determine the percentage of nitrogen in the compound.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps
Given
Mass of organic compound \[ \mathrm{=0.50\ g} \] Volume of sulphuric acid \[ \mathrm{=50\ mL=0.050\ L} \] Molarity of sulphuric acid \[ \mathrm{=0.5\ M} \] Volume of sodium hydroxide \[ \mathrm{=60\ mL=0.060\ L} \] Molarity of sodium hydroxide \[ \mathrm{=0.5\ M} \]

Solution
Calculate the initial moles of sulphuric acid. \[\mathrm{Moles=Molarity\times Volume}\] \[\mathrm{Moles\ of\ H_2SO_4}=0.5\times0.050=0.025\ mol\]

Calculate the moles of sodium hydroxide used. \[\mathrm{Moles\ of\ NaOH}=0.5\times0.060=0.030\ mol\]

Calculate the moles of sulphuric acid remaining.

The neutralisation reaction is \[\mathrm{H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O}\]

Therefore, \[\mathrm{1\ mol\ H_2SO_4}=\mathrm{2\ mol\ NaOH}\] \[\mathrm{Moles\ of\ residual\ H_2SO_4}=\frac{0.030}{2}=0.015\ mol\]

Calculate the moles of sulphuric acid neutralised by ammonia. \[\mathrm{Moles\ of\ H_2SO_4\ consumed}=0.025-0.015=0.010\ mol\]

Calculate the moles of ammonia produced.
The reaction is \[\mathrm{H_2SO_4+2NH_3\rightarrow (NH_4)_2SO_4}\] \[\mathrm{1\ mol\ H_2SO_4}=\mathrm{2\ mol\ NH_3}\] \[\mathrm{Moles\ of\ NH_3}=2\times0.010=0.020\ mol\]

Calculate the mass of nitrogen.
Each mole of ammonia contains one mole of nitrogen. \[\mathrm{Moles\ of\ N}=0.020\ mol\] \[\mathrm{Mass\ of\ N}=0.020\times14=0.28\ g\]

Calculate the percentage of nitrogen. \[\begin{aligned}\%\mathrm{N}&=\frac{0.28}{0.50}\times100\\&=56\%\end{aligned}\]

🎯 Exam Significance Exam Significance ›

This is one of the most important numerical problems based on Kjeldahl's method in quantitative organic analysis.

CBSE Board examinations frequently ask back-titration problems involving nitrogen estimation.

JEE Main, NEET and CUET regularly include numerical questions requiring mole calculations and stoichiometric relationships in Kjeldahl's method.

Students should remember the stoichiometric relations between H₂SO₄, NaOH and NH₃ while solving such problems.

Mastering this method improves problem-solving skills in volumetric analysis and elemental estimation.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

In Kjeldahl's method, ammonia is absorbed in a known excess of sulphuric acid.

The excess sulphuric acid is determined by back titration with sodium hydroxide.

One mole of H₂SO₄ reacts with two moles of NH₃.

The number of moles of nitrogen is equal to the number of moles of ammonia produced.

Always determine the acid consumed by ammonia before calculating the percentage of nitrogen.

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Q34 →

Q34

NUMERIC3 marks

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

📘 Concept & Theory Theory / Concept ›

The percentage of halogens in an organic compound is estimated by the Carius method. During the estimation, the organic compound is heated with fuming nitric acid in the presence of silver nitrate in a sealed Carius tube.

The chlorine present in the compound is completely converted into insoluble silver chloride (AgCl).

The mass of silver chloride obtained is used to calculate the mass and percentage of chlorine present in the original organic compound.

The stoichiometric relationship used is

\[ \mathrm{143.5\ g\ AgCl\ contains\ 35.5\ g\ Cl} \]

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given data.

Calculate the mass of chlorine present in the silver chloride obtained.

Calculate the percentage of chlorine in the organic compound.

Write the final answer.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps
Given
Mass of organic compound

\[\mathrm{=0.3780\ g}\]

Mass of silver chloride obtained

\[\mathrm{=0.5740\ g}\]

Calculation
Calculate the mass of chlorine present in silver chloride.

From the molar masses, \[\mathrm{Molar\ mass\ of\ AgCl}=108+35.5=143.5\] \[143.5\ \mathrm{g\ AgCl}\longrightarrow 35.5\ \mathrm{g\ Cl}\] \[0.5740\ \mathrm{g\ AgCl}\longrightarrow \frac{35.5}{143.5}\times0.5740\] \[=0.1420\ \mathrm{g}\]

Therefore,\[\boxed{\mathrm{Mass\ of\ Cl}=0.1420\ \mathrm{g}}\]

Calculate the percentage of chlorine. \[\%\mathrm{Cl}=\frac{\mathrm{Mass\ of\ Cl}}{\mathrm{Mass\ of\ Organic\ Compound}}\times100\] \[=\frac{0.1420}{0.3780}\times100\] \[=37.57\%\]

Therefore,\[\boxed{\%\mathrm{Cl}=37.57\%\approx37.6\%}\]

🎯 Exam Significance Exam Significance ›

This numerical is based on the Carius method for the quantitative estimation of halogens, an important topic in elemental analysis.

CBSE Board examinations frequently ask calculations involving the percentage of chlorine, bromine or iodine from the mass of silver halide obtained.

JEE Main, NEET and CUET often include similar stoichiometric problems based on gravimetric analysis.

Students should remember that the mass of chlorine is calculated from the mass of AgCl using the molar mass ratio.

Understanding this method helps in solving numerical problems related to elemental analysis accurately and quickly.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Carius method estimates halogens gravimetrically.

Chlorine is precipitated as silver chloride (AgCl).

Use the relation $143.5\ \mathrm{g\ AgCl} \rightarrow 35.5\ \mathrm{g\ Cl}$.

The percentage of chlorine is calculated from the mass of chlorine and the original sample mass.

Always use stoichiometric ratios based on molar masses for gravimetric estimations.

← Q33

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Q35 →

Q35

NUMERIC3 marks

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

📘 Concept & Theory Theory / Concept ›

In the Carius method, sulphur present in an organic compound is oxidised into sulphuric acid by heating the compound with fuming nitric acid in a sealed Carius tube.

The sulphuric acid formed is then treated with barium chloride solution to produce insoluble barium sulphate (BaSO₄).

The mass of barium sulphate obtained is used to calculate the amount of sulphur present in the original organic compound.

The stoichiometric relationship is

\[\mathrm{233\ g\ BaSO_4\ contains\ 32\ g\ S}\]

🗺️ Solution Roadmap Step-by-step Plan ›

Write the given data.

Calculate the mass of sulphur present in the barium sulphate.

Calculate the percentage of sulphur in the organic compound.

Write the final answer.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps
Given
Mass of organic compound

\[\mathrm{=0.468\ g}\]

Mass of barium sulphate obtained

\[\mathrm{=0.668\ g}\]

Calculations
Calculate the molar mass of barium sulphate. \[\mathrm{BaSO_4=137+32+64=233}\] \[\mathrm{Molar\ mass\ of\ BaSO_4=233\ g\ mol^{-1}}\]

Calculate the mass of sulphur present. \[233\ \mathrm{g\ BaSO_4}\longrightarrow 32\ \mathrm{g\ S}\] \[0.668\ \mathrm{g\ BaSO_4}\longrightarrow \frac{32}{233}\times0.668\]\] \[=0.09173\ \mathrm{g}\]

Therefore, \[\boxed{\mathrm{Mass\ of\ S}=0.0917\ \mathrm{g}}\]

Calculate the percentage of sulphur. \[\%\mathrm{S}=\frac{\mathrm{Mass\ of\ S}}{\mathrm{Mass\ of\ Organic\ Compound}}\times100\] \[\frac{0.09173}{0.468}\times100\] \[=19.60\%\]

Therefore, \[\boxed{\%\mathrm{Sulphur}=19.6\%}\]

🎯 Exam Significance Exam Significance ›

This numerical is based on the Carius method for the quantitative estimation of sulphur and is frequently asked in CBSE Board examinations.

JEE Main, NEET and CUET regularly include gravimetric analysis problems involving the calculation of sulphur from the mass of barium sulphate.

Students should remember that sulphur is estimated as barium sulphate and the molar mass ratio is used to calculate the mass of sulphur.

Knowledge of stoichiometric conversions is essential for solving elemental analysis numericals accurately.

This concept forms an important part of quantitative organic analysis and laboratory chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

In the Carius method, sulphur is estimated as barium sulphate (BaSO₄).

Use the relation $233\ \mathrm{g\ BaSO_4} \rightarrow 32\ \mathrm{g\ S}$.

First calculate the mass of sulphur present in the precipitate.

Then calculate the percentage of sulphur using the sample mass.

Always use stoichiometric ratios based on molar masses in gravimetric analysis.

← Q34

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Q36 →

Q36

NUMERIC3 marks

In the organic compound

\[\mathrm{CH_2=CH-CH_2-CH_2-C\equiv CH}\]

the pair of hybridised orbitals involved in the formation of the C₂–C₃ bond is

(a) sp – sp²

(b) sp – sp³

(c) sp² – sp³

(d) sp³ – sp³

📘 Concept & Theory Theory / Concept ›

The type of hybrid orbitals involved in bond formation depends upon the hybridisation of the bonded carbon atoms.

A carbon atom involved in a single bond only is generally sp³ hybridised.

A carbon atom involved in one double bond is sp² hybridised.

A carbon atom involved in one triple bond is sp hybridised.

Therefore, the first step is to determine the hybridisation of C₂ and C₃.

🗺️ Solution Roadmap Step-by-step Plan ›

Number the carbon atoms.

Determine the hybridisation of C₂.

Determine the hybridisation of C₃.

Identify the orbitals overlapping to form the C₂–C₃ σ bond.

Select the correct option.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

Number the carbon atoms. \[\mathrm{CH_2^{(1)}=CH^{(2)}-CH_2^{(3)}-CH_2^{(4)}-C^{(5)\equiv}CH^{(6)}}\]

Determine the hybridisation of C₂.
Carbon-2 is part of a carbon-carbon double bond.

Hence, it is sp² hybridised.

\[\boxed{\mathrm{C_2:\ sp^2}}\]
Determine the hybridisation of C₃.
Carbon-3 is attached through only single bonds.

It forms four σ bonds.

Hence, it is sp³ hybridised.

\[\boxed{\mathrm{C_3:\ sp^3}}\]
Identify the orbitals forming the C₂–C₃ bond.
The σ bond between C₂ and C₃ is formed by the overlap of

\[\boxed{\mathrm{sp^2-sp^3}}\]

hybrid orbitals.

Final Answer

The correct answer is

\[\boxed{\textbf{(c)\ sp^2-sp^3}}\]
Explanation of Options
- (a) sp – sp² is incorrect because C₂ is not sp hybridised.
- (b) sp – sp³ is incorrect because C₂ is sp² hybridised.
- (c) sp² – sp³ is correct because C₂ is sp² hybridised and C₃ is sp³ hybridised.
- (d) sp³ – sp³ is incorrect because C₂ is not sp³ hybridised.

🎯 Exam Significance Exam Significance ›

Hybridisation-based objective questions are frequently asked in CBSE Board examinations.

JEE Main, NEET and CUET regularly test the ability to identify the hybridisation of carbon atoms in organic compounds.

Students should always determine the hybridisation of each carbon before identifying the orbitals involved in bond formation.

Questions involving σ and π bond formation and orbital overlap are common in competitive examinations.

A clear understanding of hybridisation is essential for studying molecular geometry, bond angles and reaction mechanisms.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Carbon in a double bond is sp² hybridised.
Carbon attached through only single bonds is sp³ hybridised.
The C₂–C₃ σ bond is formed by sp²–sp³ overlap.
Always identify the hybridisation of each carbon before selecting the orbitals.
The correct option is (c).

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Q37 →

📘 Concept & Theory Theory / Concept ›

In Lassaigne's test, nitrogen present in an organic compound is first converted into sodium cyanide (NaCN) by fusion with metallic sodium.

The sodium fusion extract is then treated with freshly prepared ferrous sulphate solution. Sodium cyanide reacts with ferrous ions to form sodium ferrocyanide.

On acidification and oxidation, ferric ions react with ferrocyanide ions to produce the deep blue coloured compound called Prussian blue (ferric ferrocyanide).

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the compound formed after sodium fusion.
Write the reaction with ferrous sulphate.
Explain the formation of Prussian blue.
Select the correct option.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

Formation of sodium cyanide.
During sodium fusion, nitrogen is converted into sodium cyanide.

\[\mathrm{Na + C + N \longrightarrow NaCN}\]
Formation of sodium ferrocyanide.
Sodium cyanide reacts with ferrous sulphate in alkaline medium.

\[\mathrm{FeSO_4 + 6NaCN\longrightarrow Na_4[Fe(CN)_6] + Na_2SO_4}\]

Thus, sodium ferrocyanide is formed.
Formation of Prussian blue.
On boiling and subsequent acidification, a part of the ferrous ions is oxidised to ferric ions.

The ferric ions react with ferrocyanide ions to produce ferric ferrocyanide (Prussian blue).

\[\mathrm{4Fe^{3+}+3[Fe(CN)_6]^{4-}\longrightarrow Fe_4[Fe(CN)_6]_3\downarrow}\]

This deep blue precipitate confirms the presence of nitrogen in the organic compound.
Final Answer
The Prussian blue colour is produced due to the formation of
\[\boxed{\mathrm{Fe_4[Fe(CN)_6]_3}}\] Hence, the correct option is \[\boxed{\textbf{(b)\ Fe_4[Fe(CN)_6]_3}}\]
Explanation of Incorrect Options
- (a) Na₄[Fe(CN)₆] is sodium ferrocyanide, an intermediate formed during the test. It is not responsible for the Prussian blue colour.
- (b) Fe₄[Fe(CN)₆]₃ is ferric ferrocyanide (Prussian blue). This is the correct answer.
- (c) Fe₂[Fe(CN)₆] is not the compound responsible for the characteristic blue colour.
- (d) Fe₃[Fe(CN)₆]₄ has an incorrect stoichiometric formula and is not formed in Lassaigne's test.

🎯 Exam Significance Exam Significance ›

Lassaigne's test is one of the most important topics in qualitative organic analysis for CBSE Board examinations.

JEE Main, NEET and CUET frequently ask objective questions on the sequence of reactions involved in the nitrogen test.

Students should remember that sodium ferrocyanide is only an intermediate, whereas ferric ferrocyanide produces the characteristic Prussian blue colour.

The formula of Prussian blue is a frequently tested fact in multiple-choice and assertion-reason questions.

A clear understanding of the reaction mechanism helps distinguish between intermediate and final products in qualitative analysis.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Sodium fusion converts nitrogen into sodium cyanide (NaCN).
NaCN reacts with ferrous sulphate to form sodium ferrocyanide.
Ferric ions react with ferrocyanide ions to form Prussian blue.
Prussian blue is ferric ferrocyanide, Fe₄[Fe(CN)₆]₃.
The correct option is (b).

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Q38 →

📘 Concept & Theory Theory / Concept ›

A carbocation is an electron-deficient species in which a carbon atom carries a positive charge.

The stability of carbocations depends mainly on:

Hyperconjugation
Positive inductive (+I) effect of alkyl groups
Resonance (if present)

For simple alkyl carbocations, the stability order is

\[\boxed{\mathrm{3^\circ > 2^\circ > 1^\circ > CH_3^+}}\]

This is because alkyl groups donate electron density through the +I effect and hyperconjugation, thereby reducing the positive charge on the carbon atom.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the degree of each carbocation.
Compare the number of alkyl groups attached to the positively charged carbon.
Apply the stability order of carbocations.
Select the most stable carbocation.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

Option (a)
\[(\mathrm{CH_3})_3\mathrm{C{-}\overset{+}{CH_2}}\]
The positively charged carbon is attached to only one carbon atom.

This is a primary (1°) carbocation.
Option (b)
\[(\mathrm{CH_3})_3\mathrm{\overset{+}{C}}\]
The positively charged carbon is attached to three methyl groups.

It is a tertiary (3°) carbocation.

Three alkyl groups donate electron density through the +I effect and provide maximum hyperconjugation.

Hence, it is the most stable.
Option (c)
\mathrm{CH_3CH_2\overset{+}{CH_2}}
The positively charged carbon is attached to only one carbon atom.

Therefore, it is a primary (1°) carbocation.
Option (d)
\[\mathrm{CH_3\overset{+}{CH}CH_2CH_3}\]
The positively charged carbon is attached to two alkyl groups.

Therefore, it is a secondary (2°) carbocation.
Comparison of Stability
\[\mathrm{(b)\ 3^\circ > (d)\ 2^\circ > (a)\ 1^\circ \approx (c)\ 1^\circ}\]
Hence, option (b) is the most stable.
Final Answer
The most stable carbocation is

\[\boxed{\textbf{(b)\ (\mathrm{CH_3})_3\mathrm{\overset{+}{C}}}}\]

It is a tertiary carbocation, stabilised by the maximum +I effect and the largest number of hyperconjugative structures.
Explanation of Incorrect Options
- (a) is a primary carbocation and is much less stable than a tertiary carbocation.
- (b) is a tertiary carbocation and is the most stable due to three alkyl groups providing maximum hyperconjugation and +I effect.
- (c) is also a primary carbocation and is less stable.
- (d) is a secondary carbocation. It is more stable than primary carbocations but less stable than a tertiary carbocation.

🎯 Exam Significance Exam Significance ›

Carbocation stability is one of the most fundamental concepts in Organic Chemistry and is frequently tested in CBSE Board examinations.

JEE Main, NEET and CUET regularly ask multiple-choice questions based on the relative stability of carbocations.

Students should remember that hyperconjugation and the +I effect of alkyl groups are the major factors responsible for stabilising alkyl carbocations.

Understanding carbocation stability is essential for predicting reaction mechanisms such as electrophilic addition, nucleophilic substitution and elimination reactions.

The stability order of carbocations is one of the most frequently repeated concepts throughout Organic Chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Tertiary carbocations are the most stable among simple alkyl carbocations.
Alkyl groups stabilise carbocations through the +I effect and hyperconjugation.
The stability order is $3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^+}$.
Option (b) is a tertiary carbocation.
The correct answer is (b).

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Q39 →

📘 Concept & Theory Theory / Concept ›

Organic compounds are often obtained as mixtures containing impurities or several closely related compounds. Various separation techniques such as crystallisation, distillation and sublimation are used depending on the physical properties of the substances.

Among all these methods, chromatography is the most advanced, versatile and efficient technique for the separation, purification and identification of organic compounds.

Modern chromatography is extensively used in chemical industries, pharmaceutical laboratories, forensic science, environmental analysis and biochemical research.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the purpose of each separation technique.
Identify the technique that provides the highest efficiency and resolution.
Compare chromatography with crystallisation, distillation and sublimation.
Select the correct option.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Option (a): Crystallisation
Crystallisation is used to purify solid compounds based on differences in their solubility.

Although it is an effective purification technique, it cannot separate complex mixtures containing many components.

Hence, it is not the best or latest technique.
Option (b): Distillation
Distillation is used to separate liquids having different boiling points.

It is suitable for volatile liquids but cannot separate compounds having very similar boiling points or non-volatile substances efficiently.

Hence, it is not the best or latest technique.
Option (c): Sublimation
Sublimation is applicable only to solids that sublime on heating, such as camphor, iodine and naphthalene.

Its applications are limited to a few substances.

Hence, it is not the best or latest technique.
Option (d): Chromatography
Chromatography is capable of separating, purifying and identifying even very small quantities of closely related compounds.

It provides high accuracy, high sensitivity and excellent resolution.

Several modern chromatographic techniques are widely used, such as:
- Paper Chromatography
- Thin Layer Chromatography (TLC)
- Column Chromatography
- Gas Chromatography (GC)
- High Performance Liquid Chromatography (HPLC)
Therefore, chromatography is regarded as the best and the latest technique among the given options.
Final Answer
The correct answer is \[\boxed{\textbf{(d)\ Chromatography}}\]
Explanation of Incorrect Options
- (a) Crystallisation is useful only for purification of solids based on solubility differences.
- (b) Distillation separates liquids based on differences in boiling points.
- (c) Sublimation is applicable only to sublimable solids.
- (d) Chromatography is the most versatile technique and is used for separation, purification and identification of complex organic mixtures with high precision.

🎯 Exam Significance Exam Significance ›

Questions based on purification and separation techniques are frequently asked in CBSE Board examinations.

JEE Main, NEET and CUET often test the applications and advantages of chromatography over conventional separation methods.

Students should remember that chromatography is considered the most efficient modern analytical technique for separation and purification.

Knowledge of different chromatographic techniques such as TLC, GC and HPLC is useful for higher studies in chemistry and pharmaceutical sciences.

Chromatography is extensively used in research laboratories, food analysis, forensic science, biotechnology and environmental monitoring.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Chromatography is the most advanced separation technique among the given options.
It is used for isolation, purification and identification of organic compounds.
It can separate even very small quantities of closely related compounds.
Common chromatographic techniques include Paper Chromatography, TLC, Column Chromatography, GC and HPLC.
The correct option is (d) Chromatography.

← Q38

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Q40 →

📘 Concept & Theory Theory / Concept ›

Organic reactions are broadly classified into substitution, addition, elimination and rearrangement reactions.

In a nucleophilic substitution reaction, a nucleophile attacks an electron-deficient carbon atom and replaces a leaving group such as a halide ion.

In this reaction, the hydroxide ion (OH⁻) acts as the nucleophile, while iodide ion (I⁻) acts as the leaving group.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the substrate and the reagent.
Determine the role of hydroxide ion.
Identify the leaving group.
Classify the reaction based on the change occurring.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

The given reaction is \[\mathrm{CH_3CH_2I + KOH_{(aq)}\longrightarrow CH_3CH_2OH + KI}\]
In aqueous potassium hydroxide, hydroxide ions are produced. \[\mathrm{KOH \longrightarrow K^+ + OH^-}\] The hydroxide ion has a lone pair of electrons and behaves as a nucleophile.
The nucleophile attacks the carbon atom bonded to iodine.
The iodide ion leaves the molecule. \[\mathrm{CH_3CH_2I\longrightarrow CH_3CH_2OH}\] Thus, the iodine atom is replaced by the hydroxyl group.
Since one group (I) is substituted by another group (OH) through the attack of a nucleophile, the reaction is a nucleophilic substitution reaction.
Final Answer
The correct answer is \[\boxed{\textbf{(b)\ Nucleophilic\ substitution}}\]
Explanation of Incorrect Options
- (a) Electrophilic substitution is incorrect because the attacking species is a nucleophile (OH⁻), not an electrophile.
- (b) Nucleophilic substitution is correct because OH⁻ replaces I⁻ in the molecule.
- (c) Elimination is incorrect because no small molecule such as HX or H₂O is eliminated and no multiple bond is formed.
- (d) Addition is incorrect because no atoms are added across a double or triple bond.

🎯 Exam Significance Exam Significance ›

Classification of organic reactions is one of the most frequently tested topics in CBSE Board examinations.

JEE Main, NEET and CUET regularly include objective questions asking students to identify substitution, addition and elimination reactions.

Students should remember that aqueous KOH favours nucleophilic substitution, whereas alcoholic KOH generally favours elimination.

Recognising the role of the reagent and the leaving group is the quickest way to classify organic reactions.

This concept forms the basis for understanding reaction mechanisms studied in higher classes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

OH⁻ acts as the nucleophile.
I⁻ acts as the leaving group.
Iodine is replaced by the hydroxyl group.
The reaction is a substitution reaction involving a nucleophile.
The correct option is (b) Nucleophilic substitution.

← Q39

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Organic Chemistry - Some Basic Principles and Techniques — Learning Resources

Frequently Asked Questions

Yes. These solutions are based on the latest NCERT textbook and include theory, solution roadmap, detailed explanations, key takeaways, and exam-oriented points for CBSE preparation.

Yes. The solutions strengthen conceptual understanding of nomenclature, hybridisation, electronic effects, reaction mechanisms, purification techniques and organic analysis, which are frequently tested in competitive exams.

Theory helps students understand the underlying concept before solving the question, improving conceptual clarity instead of rote learning.

The roadmap outlines the logical steps required to solve the problem, making lengthy numerical and conceptual questions easier to understand and revise.

Yes. Every numerical solution includes all intermediate calculations, relevant formulas, unit conversions and final answers without skipping any steps.

Yes. MCQ solutions include explanations for both the correct option and the incorrect alternatives to help eliminate common misconceptions.

They explain the reasoning behind every answer using concepts such as hybridisation, resonance, inductive effect, reaction mechanisms, chromatography, Lassaigne's test and elemental analysis.

Yes. Wherever required, balanced chemical equations, reaction mechanisms, electron movement and stepwise explanations are included using NCERT-level concepts.

They highlight frequently tested concepts, common mistakes and important facts that students should remember for Board and competitive examinations.

Yes. The structured format with theory, roadmap, detailed solution, exam significance and key takeaways makes them ideal for last-minute revision as well as regular study.

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Organic Chemistry – Some Basic Principles and Techniques

Nucleophiles

Electrophiles

Difference Between Nucleophiles and Electrophiles

Electromeric Effect

Principle of Lassaigne's Test

Formation of Sodium Salts

Final Answer

Final Answer

Final Answer

Chapter Complete!

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Organic Chemistry - Some Basic Principles and Techniques — Learning Resources

Frequently Asked Questions

Q 01 Are these NCERT Class 11 Chemistry Chapter 8 exercise solutions sufficient for CBSE Board exams?

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Q 05 Are numerical problems solved step by step?

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Q 07 How do these solutions improve conceptual understanding of Organic Chemistry?

Q 08 Are reaction mechanisms and chemical equations explained in detail?

Q 09 Why are exam significance and key takeaways included after every solution?

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