What do the NCERT Class 11 Chemistry Chapter 9 Hydrocarbons exercise solutions include?

The solutions include detailed, step-by-step answers to every NCERT textbook exercise along with explanations, reaction equations, and important concepts.

Are these Hydrocarbons exercise solutions based on the latest NCERT syllabus?

Yes, all solutions are prepared according to the latest NCERT Class 11 Chemistry syllabus and follow the textbook closely.

Do the solutions explain the concepts behind each answer?

Yes, every solution begins with the relevant theory, explains the concept involved, and then presents a systematic solution.

Are reaction mechanisms included wherever required?

Yes, important reaction mechanisms such as free radical substitution, electrophilic substitution, and addition reactions are explained wherever applicable.

How can these solutions help in CBSE board exam preparation?

These solutions follow the NCERT pattern, include exam-oriented explanations, key takeaways, and important points that help students score well in board examinations.

Are these solutions useful for JEE Main, JEE Advanced, and NEET?

Yes, the solutions strengthen conceptual understanding and cover topics that are frequently tested in JEE Main, JEE Advanced, NEET, and other competitive examinations.

Do the solutions include chemical equations and structural formulas?

Yes, balanced chemical equations, structural formulas, IUPAC names, and reaction conditions are included wherever necessary for better understanding.

Can beginners understand these NCERT Hydrocarbons solutions?

Yes, the explanations are written in simple language with step-by-step reasoning, making them suitable for beginners as well as advanced learners.

Why should I practice NCERT Chapter 9 Hydrocarbons exercise questions?

Practicing NCERT exercise questions improves conceptual clarity, reaction knowledge, problem-solving skills, and confidence for both school and competitive exams.

Do the solutions contain revision-friendly summaries?

Yes, every solution includes theory, a solution roadmap, final answer, key takeaways, exam significance, and diagrams or illustrations wherever appropriate.

Hydrocarbons NCERT Class 11 Exercise Solutions

📘 Concept & Theory Concept/Theory ›

Chlorination of methane is an example of a free radical substitution reaction. The reaction takes place in the presence of sunlight or ultraviolet (UV) light. Under these conditions, chlorine molecules undergo homolytic cleavage to generate highly reactive chlorine free radicals.

A free radical is an electrically neutral species containing one unpaired electron. Because of this unpaired electron, free radicals are highly reactive and readily participate in chain reactions.

The chlorination of methane proceeds through three stages:

Initiation: Formation of chlorine free radicals by homolytic cleavage.

Propagation: Chlorine radicals abstract hydrogen from methane to produce methyl radicals. The methyl radicals further react with chlorine molecules to continue the chain.

Termination: Two free radicals combine to form stable molecules, thereby terminating the chain reaction.

The formation of ethane occurs during the termination step, where two methyl radicals combine together.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the type of reaction occurring during chlorination of methane.
Recall that UV light causes homolytic cleavage of chlorine molecules.
Write the initiation reaction producing chlorine radicals.
Write the propagation reactions showing formation of methyl radicals.
Identify the termination step in which two methyl radicals combine.
Conclude that this coupling reaction produces ethane.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

During chlorination, methane reacts with chlorine in the presence of sunlight or ultraviolet light.
The reaction begins with the homolytic cleavage of chlorine molecules.
Initiation Step
\[\mathrm{Cl_2 \xrightarrow{h\nu} \overset{\bullet}{C}l + \overset{\bullet}{C}l}\]
The chlorine radicals formed are highly reactive.
Propagation Step 1
A chlorine radical removes one hydrogen atom from methane to produce a methyl radical.
\[\mathrm{CH_4 + \overset{\bullet}{C}l \rightarrow \overset{\bullet}{C}H_3 + HCl}\]
Propagation Step 2
The methyl radical reacts with another chlorine molecule to form chloromethane and regenerate a chlorine radical.
\[\mathrm{\overset{\bullet}{C}H_3 + Cl_2 \rightarrow CH_3Cl + \overset{\bullet}{C}l}\]
The regenerated chlorine radical continues the chain reaction.
Termination Step
Sometimes, instead of reacting with chlorine, two methyl radicals collide and combine.
\[\mathrm{\overset{\bullet}{C}H_3 + \overset{\bullet}{C}H_3 \rightarrow C_2H_6}\]
The product formed is ethane.
Thus, ethane is obtained because two methyl free radicals produced during chlorination combine together during the termination step of the free radical chain reaction.

🎯 Exam Significance Exam Significance ›

This question is frequently asked to test the free radical substitution mechanism.

Students should remember the three stages of the chain reaction: initiation, propagation, and termination.

The examiner expects the correct free radical notation and the termination reaction leading to ethane.

Writing balanced equations for each stage improves presentation and fetches full marks.

Significance for JEE / NEET and Other Competitive Examinations

Free radical mechanisms are fundamental concepts in organic chemistry and are repeatedly tested in entrance examinations.

Questions may ask the role of UV light, homolytic bond cleavage, or the products formed during termination reactions.

Students should be able to identify intermediates such as chlorine radicals and methyl radicals.

This concept also forms the basis for understanding halogenation of higher alkanes and radical polymerization reactions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 6 points

Chlorination of methane is a free radical substitution reaction.
Ultraviolet light causes homolytic cleavage of chlorine molecules.
Chlorine radicals initiate the chain reaction.
Methyl radicals are formed during the propagation stage.
Ethane is produced when two methyl radicals combine during the termination stage.
Termination reactions stop the free radical chain process.

📘 Concept & Theory Theory / Concept ›

The International Union of Pure and Applied Chemistry (IUPAC) has developed a systematic method for naming organic compounds so that every compound has a unique and universally accepted name. IUPAC nomenclature eliminates confusion caused by common or trivial names and clearly represents the structure of the compound.

While naming hydrocarbons, the first step is to identify the longest continuous carbon chain, which becomes the parent hydrocarbon. The suffix of the parent name depends on the type of carbon-carbon bond present, such as -ane for single bonds, -ene for double bonds, and -yne for triple bonds.

If a compound contains both double and triple bonds, the parent chain should include both multiple bonds. Numbering is done in such a way that the multiple bonds receive the lowest possible set of locants. If there is a tie, the double bond is assigned the lower number than the triple bond.

Substituent groups such as alkyl groups are identified, numbered, and written as prefixes before the parent name. When more than one identical substituent is present, prefixes such as di-, tri-, tetra- are used.

For aromatic compounds, the benzene ring is considered the parent structure if it has higher priority than the attached carbon chain. Otherwise, the benzene ring is treated as a phenyl substituent attached to the parent hydrocarbon.

The general procedure for writing IUPAC names can be summarized as follows.

Identify the longest continuous carbon chain.

Select the parent hydrocarbon based on the number of carbon atoms.

Identify all multiple bonds and functional groups present.

Number the parent chain to give the lowest possible locants to multiple bonds and substituents.

Identify and name all substituents.

Arrange substituents alphabetically, ignoring multiplicative prefixes such as di-, tri-, and tetra-.

Write the complete IUPAC name by combining prefixes, parent name, multiple bond locants, and suffix according to IUPAC rules.

🗺️ Solution Roadmap Step-by-step Plan ›

Observe the complete structural formula carefully and identify the longest continuous carbon chain.
Determine whether the compound is an alkane, alkene, alkyne, diene, aromatic hydrocarbon, or contains a combination of multiple bonds.
Select the parent hydrocarbon by choosing the chain that contains the maximum number of multiple bonds and, if possible, the maximum number of carbon atoms.
Number the parent chain from the end that gives the lowest possible locants to the double bond(s), triple bond(s), or both according to IUPAC priority rules.
Identify all alkyl groups and other substituents attached to the parent chain.
Assign the correct position number (locant) to each substituent.
If more than one identical substituent is present, use the appropriate prefixes such as di-, tri-, or tetra-.
Arrange different substituents in alphabetical order while ignoring multiplicative prefixes such as di-, tri-, and tetra-.
Write the parent hydrocarbon name using the correct root word and appropriate suffix (-ane, -ene, -yne, or combinations such as -diene and -en-yne).
For aromatic compounds, determine whether benzene is the parent compound or whether it should be treated as a phenyl substituent attached to the parent chain.
Combine the locants, prefixes, parent name, and suffix according to IUPAC nomenclature rules to obtain the final systematic name.
Finally, verify that the numbering, spelling, commas, and hyphens are correctly written according to IUPAC conventions.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

a)
\(\ce{CH3CH=C(CH3)2}\)
\[\begin{aligned}\overset{4}{\text{C}}\text{H}_3 -\overset{3}{\text{C}}\text{H} = &\;\overset{2}{\text{C}}-\overset{1}{\text{C}}\text{H}_3\\&\;|\\&\text{CH}_3\end{aligned}\]
Longest carbon chain: 4 carbons
Parent word root: but
Principal functional group: Double bond at position 2: but-2-ene
Numbering (right to left): Gives the methyl substituent the lowest locant at C2.
IUPAC Name: 2-Methylbut-2-ene
b)
\(\ce{CH2=CH-C≡C-CH3}\)
\[\overset{1}{\text{C}}\text{H}_2 = \overset{2}{\text{C}}\text{H} - \overset{3}{\text{C}} \equiv \overset{4}{\text{C}} - \overset{5}{\text{C}}\text{H}_3\]
Longest carbon chain: 5 carbons
Parent word root: pent
Multiple bonds: Double bond and triple bond present.
Numbering (left to right): Gives lowest locants (1 for double bond, 3 for triple bond).}
((Number the chain from the end that gives the lowest set of locants to the unsaturated bonds (considering double and triple bonds together, without preference to either)
IUPAC Name: Pent-1-en-3-yne
c)
Longest carbon chain: 4 carbons
Parent word root: but
Multiple bonds: Two double bonds at positions 1 and 3: -1,3-diene
IUPAC Name: Buta-1,3-diene
d)
Principal chain: Aliphatic chain with double bond (4 carbons): but-1-ene
Numbering (right to left): Start from double bond carbon (C₁).
Substituent: Phenyl group attached at position C₄.
IUPAC Name: 4-Phenylbut-1-ene
e)
Principal functional group: -OH- attached to benzene: -Phenol (C₁).
Substituent: Methyl group at position C₂.
IUPAC Name: 2-Methylphenol or 2-Methylbenzenol
f)
\[\begin{aligned}\text{CH}_3(\text{CH}_2)_4-&\text{CH}-(\text{CH}_2)_3\text{CH}_3\\&\;|\\&\text{CH}_2-\text{CH}(\text{CH}_3)_2\end{aligned}\]
\[\begin{aligned} \overset{10}{\text{C}}\text{H}_3-\overset{9}{\text{C}}\text{H}_2-\overset{8}{\text{C}}\text{H}_2-\overset{7}{\text{C}}\text{H}_2-\overset{6}{\text{C}}\text{H}_2-&\overset{5}{\text{C}}\text{H}-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H}_2-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{H}_3 \\&| \\ &\overset{1'}{\text{C}}\text{H}_2-\overset{2'}{\text{C}}\text{H}\left(\overset{3'}{\text{C}}\text{H}_3\right)_2 \end{aligned}\]
Longest continuous chain: 10 carbons
Parent alkane: Decane
Numbering: From right end so substituent gets lowest locant (C₅ instead of C₆)
Substituent at C₅:
\(-\text{CH}_2-\text{CH}(\text{CH}_3)_2\rightarrow \text{2-methylpropyl (or isobutyl)}\)

IUPAC Name: 5-(2-Methylpropyl)decane
g)
\[\begin{aligned}\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}=\text{CH}-&\text{CH}-\text{CH}_2-\text{CH}=\text{CH}_2\\&\;|\\&\text{C}_2\text{H}_5\end{aligned}\]
\[\begin{aligned}\overset{10}{\text{C}}\text{H}_3-\overset{9}{\text{C}}\text{H}=\overset{8}{\text{C}}\text{H}-\overset{7}{\text{C}}\text{H}_2-\overset{6}{\text{C}}\text{H}=\overset{5}{\text{C}}\text{H}-&\overset{4}{\text{C}}\text{H}-\overset{3}{\text{C}}\text{H}_2-\overset{2}{\text{C}}\text{H}=\overset{1}{\text{C}}\text{H}_2 \\&\;| \\ &\text{C}_2\text{H}_5 \quad (\text{or } -\overset{1'}{\text{C}}\text{H}_2-\overset{2'}{\text{C}}\text{H}_3)\end{aligned}\]
Longest chain with maximum double bonds: 10 carbons: deca-
Numbering (right to left): Gives lower locants for double bonds (1, 5, 8)
Substituent: Ethyl group:
\((-\text{C}_2\text{H}_5)\)
at position C₄
IUPAC Name: 4-Ethyldeca-1,5,8-triene.

🎯 Exam Significance Exam Significance ›

IUPAC nomenclature is one of the most frequently tested topics in Class 11 Chemistry examinations.

Students are expected to identify the correct parent chain, assign proper numbering, and write the complete IUPAC name without spelling errors.

Questions are commonly asked on compounds containing double bonds, triple bonds, branched chains, and aromatic rings.

Board examiners often award stepwise marks for correctly identifying the parent chain, numbering, substituents, and final IUPAC name.

Mastering nomenclature also helps in writing chemical equations accurately throughout Organic Chemistry.

Significance for JEE, NEET and Other Competitive Examinations

IUPAC nomenclature is a fundamental topic for JEE Main, JEE Advanced, NEET, CUET, Olympiads, and other competitive examinations.

Competitive examinations frequently test compounds having multiple substituents, branched carbon chains, alkenes, alkynes, cycloalkanes, and aromatic hydrocarbons.

Students should be able to determine the correct parent chain even when it is not the longest visible chain but the one containing the maximum number of multiple bonds.

Questions often require distinguishing between common names and IUPAC names or identifying the correct structure from a given IUPAC name.

A strong understanding of nomenclature forms the foundation for learning reaction mechanisms, stereochemistry, isomerism, and biomolecules in later chapters.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 8 points

IUPAC nomenclature provides a unique and internationally accepted name for every organic compound.
The longest carbon chain containing the maximum number of multiple bonds is selected as the parent chain.
Multiple bonds receive priority while assigning numbering.
Double bonds receive lower locants than triple bonds when both cannot be numbered equally.
Substituents are named as prefixes and arranged alphabetically.
Benzene may act as either the parent hydrocarbon or a phenyl substituent depending on the structure.
Correct numbering and proper placement of hyphens and commas are essential for writing accurate IUPAC names.
Strong knowledge of nomenclature is essential for understanding isomerism, reaction mechanisms, and advanced organic chemistry.

📘 Concept & Theory Theory / Concept ›

Isomers are compounds having the same molecular formula but different arrangements of atoms. Alkenes exhibit both chain isomerism and position isomerism, whereas alkynes mainly exhibit chain and position isomerism.

For this question, only those isomers containing exactly one carbon-carbon double bond or one carbon-carbon triple bond are to be considered. Geometrical (cis-trans/E-Z) isomers are not included because the question asks only for structural isomers.

🗺️ Solution Roadmap Step-by-step Plan ›

Calculate the degree of unsaturation from the molecular formula.
Identify whether the compound is an alkene or an alkyne.
Draw every possible carbon skeleton.
Move the double bond or triple bond to every distinct position.
Eliminate duplicate structures obtained by renumbering.
Assign the correct IUPAC name to each remaining isomer.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

a)
Molecular Formula: C₄H₈ (One Double Bond)
C₄H₈ corresponds to alkenes having one C=C double bond.
Step 1: Straight Chain Isomers
Isomer 1
Structural Formula
\[\ce{CH2=CH−CH2−CH3}\]
IUPAC Name: But-1-ene

Isomer 2
Structural Formula
\[\ce{CH3−CH=CH−CH3}\]
IUPAC Name: But-2-ene
Step 2: Branched Chain Isomer

Isomer 3
Structural Formula
\[\begin{aligned}&\mathrm{CH_3}\\&\;|\\\mathrm{\overset{1}CH_2=\;}&\mathrm{\overset{2}C-\overset{3}CH_3}\end{aligned}\]
IUPAC Name: 2-Methylprop-1-ene
b)
Molecular Formula: C₅H₈ (One Triple Bond)
Step 1: Straight Chain Isomers
C₅H₈ corresponds to alkynes containing one carbon-carbon triple bond.

Isomer 1
Structural Formula
\[\ce{H - \overset{1}{C} # \overset{2}{C} - \overset{3}{C}H2 - \overset{4}{C}H2 - \overset{5}{C}H3}\]
IUPAC Name: Pent-1-yne
Isomer 2
Structural Formula
\[\ce{\overset{1}{C}H3 - \overset{2}{C} # \overset{3}{C} - \overset{4}{C}H2 - \overset{5}{C}H3}\]
IUPAC Name: Pent-2-yne
Step 2: Branched Chain Isomer

strong>Isomer 3
Structural Formula
\[\begin{aligned}&\mathrm{CH_3}\\&\;|\\\mathrm{\overset{1}CH=\overset{2}C-\;}&\mathrm{\overset{3}CH-\overset{4}CH_3}\end{aligned}\]
IUPAC Name: 3-Methylbut-1-yne

🎯 Exam Significance Exam Significance ›

This question tests the ability to identify all structural isomers for a given molecular formula.

Students should draw every possible carbon skeleton before assigning the position of the multiple bond.

Correct IUPAC nomenclature and proper numbering are essential for securing full marks.

Avoid writing duplicate structures obtained by numbering the carbon chain from the opposite end.

Significance for JEE / NEET and Other Competitive Examinations

Structural isomerism is a frequently tested concept in Organic Chemistry.

Competitive examinations often ask students to determine the number of possible isomers rather than merely naming them.

A clear understanding of chain and position isomerism helps solve higher-level questions involving reaction mechanisms and stereochemistry.

This concept also forms the foundation for learning alkadienes, aromatic hydrocarbons, and functional group isomerism.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

C₄H₈ with one double bond has three structural isomers.
C₅H₈ with one triple bond has three structural isomers.
Number the parent chain so that the double bond or triple bond receives the lowest possible locant.
Always examine both straight-chain and branched-chain carbon skeletons.
Do not count duplicate structures formed by reversing the numbering of the carbon chain.

📘 Concept & Theory Theory / Concept ›

Ozonolysis is an important reaction of alkenes. In this reaction, ozone (O₃) attacks the carbon-carbon double bond to form an ozonide, which on reductive hydrolysis (using Zn/H₂O) breaks the double bond into two carbonyl compounds.

The nature of the carbonyl product depends upon the number of alkyl groups attached to each carbon of the double bond.

If a double-bond carbon has at least one hydrogen atom, it forms an aldehyde.

If a double-bond carbon is attached to two carbon atoms, it forms a ketone.

Thus, ozonolysis is frequently used to determine the position of the double bond in an alkene.

🗺️ Solution Roadmap Step-by-step Plan ›

Locate the carbon-carbon double bond.
Imagine breaking the double bond into two separate fragments.
Add an oxygen atom to each broken carbon to form carbonyl compounds.
Determine whether each product is an aldehyde or a ketone depending upon the presence or absence of hydrogen.
Write the structural formula and IUPAC name of each product.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

(i) Pent-2-ene
Structure
\[\mathrm{CH_3-CH=CH-CH_2-CH_3}\]
Step 1: Break the double bond.
\[\mathrm{CH_3-CH \quad || \quad CH-CH_2-CH_3}\]

Step 2: Convert each carbon into a carbonyl carbon.

\[\mathrm{CH_3CHO + CH_3CH_2CHO}\]
IUPAC NamesEthanal & Propanal
(ii) 3,4-Dimethylhept-3-ene
Structure:
\[\mathrm{CH_3CH_2C(CH_3)=C(CH_3)CH_2CH_2CH_3}\]
Step 1: Break the double bond.
Left fragment
\[\mathrm{CH_3CH_2COCH_3}\]
Right fragment

\[\mathrm{CH_3COCH_2CH_2CH_3}\]
Since neither carbon of the double bond contains hydrogen, both products are ketones.

IUPAC Names: Butan-2-one and Pentan-2-one
(iii) 2-Ethylbut-1-ene
Structure
\[\mathrm{CH_2=C(CH_2CH_3)CH_2CH_3}\]
Step 1: Break the double bond.

The terminal carbon is CH₂; therefore it produces methanal.
\[\mathrm{HCHO}\]
The second carbon forms a ketone.
\[\mathrm{CH_3CH_2COCH_2CH_3}\]
IUPAC Names: Methanal and Pentan-3-one
(iv) 1-Phenylbut-1-ene
Structure
\[\mathrm{C_6H_5CH=CHCH_2CH_3}\]
Step 1: Break the double bond.<

The phenyl-substituted carbon contains one hydrogen and therefore forms an aldehyde.
\[\mathrm{C_6H_5CHO}\]
The second fragment also contains one hydrogen.
\[\mathrm{CH_3CH_2CHO}\]
IUPAC Names: Benzenecarbaldehyde (Common name: Benzaldehyde) and Propanal

🎯 Exam Significance Exam Significance ›

Ozonolysis is one of the most important reactions of alkenes and is frequently asked in CBSE board examinations.

Students should learn to identify the position of the double bond before predicting the products.

Remember that carbons containing hydrogen produce aldehydes, whereas carbons attached to two carbon atoms produce ketones.

Writing both the structural formulas and IUPAC names of the products helps in obtaining full marks.

Significance for JEE / NEET and Other Competitive Examinations

Questions based on ozonolysis are common in JEE Main, JEE Advanced and NEET because they test understanding of alkene reactions.

Reverse ozonolysis, where the reactants are identified from the products, is a frequently asked application.

Students should practice predicting carbonyl products rapidly by mentally cleaving the double bond.

This reaction is also useful for determining the structure of unknown alkenes.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ozonolysis cleaves the carbon-carbon double bond.
Each double-bond carbon is converted into a carbonyl group.
Carbons containing hydrogen produce aldehydes.
Carbons without hydrogen produce ketones.
Ozonolysis is an important method for locating the position of a double bond.

📘 Concept & Theory Theory / Concept ›

Ozonolysis is the oxidative cleavage of a carbon-carbon double bond. Under reductive conditions (O₃/Zn, H₂O), the double bond breaks and each carbon atom of the double bond is converted into a carbonyl group.

The carbonyl compounds obtained help us determine the structure of the original alkene. This process is known as reverse ozonolysis.

While reconstructing the alkene:

Identify the carbon atoms of the carbonyl groups.

Remove the oxygen atom from each carbonyl group.

Join the two carbon atoms by a carbon-carbon double bond.

The resulting compound is the required alkene.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the structures of the ozonolysis products.
Identify the carbonyl carbon in each product.
Remove the oxygen atom from both carbonyl groups.
Join the two carbonyl carbons with a double bond.
Write the structural formula of the alkene.
Assign the correct IUPAC name.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

Step 1: Write the products obtained after ozonolysis.

Product 1: Ethanal
\[\mathrm{CH_3CHO}\]
Product 2: Pentan-3-one
\[\mathrm{CH_3CH_2COCH_2CH_3}\]
Step 2: Locate the carbonyl carbons.

Ethanal contains the carbonyl carbon shown below.
\[\mathrm{CH_3-\boxed{CHO}}\]
Pentan-3-one contains the carbonyl carbon shown below.
\[\mathrm{CH_3CH_2-\boxed{CO}-CH_2CH_3}\]
Step 3: Remove the oxygen atom from each carbonyl carbon.

The carbonyl carbon of ethanal becomes
\[\mathrm{CH_3-CH}\]
The carbonyl carbon of pentan-3-one becomes
\[\mathrm{CH_3CH_2-C-CH_2CH_3}\]
Step 4: Join the two carbon atoms by a double bond.
\[\mathrm{CH_3CH=C(CH_2CH_3)CH_2CH_3}\]
Step 5: Determine the parent chain.
The longest chain containing the double bond has six carbon atoms.

Numbering is done from the end nearer to the double bond.

The double bond is between carbon-2 and carbon-3.

An ethyl substituent is present on carbon-3.

🎯 Exam Significance Exam Significance ›

Reverse ozonolysis is a commonly asked reasoning question in CBSE examinations.

Students should practice reconstructing the alkene from the carbonyl products step by step.

Writing the intermediate reconstruction steps makes the solution easier to understand and earns better marks.

Always verify the answer by performing ozonolysis mentally and checking whether the products match the question.

Significance for JEE / NEET and Other Competitive Examinations

Reverse ozonolysis is one of the most important applications of alkene reactions in competitive examinations.

JEE and NEET frequently ask students to identify unknown alkenes from the products of ozonolysis.

Questions may involve multiple possible structures, making proper reconstruction and IUPAC nomenclature essential.

This concept is also useful in determining the structures of unknown organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ozonolysis cleaves the carbon-carbon double bond.
Reverse ozonolysis reconstructs the original alkene from carbonyl products.
The oxygen atoms of the carbonyl compounds are removed and the carbonyl carbons are joined by a double bond.
The parent chain must contain the double bond and receive the lowest possible locant.
The required alkene is 3-Ethylhex-2-ene.

📘 Concept & Theory Theory/Concept ›

An alkene contains one carbon-carbon double bond consisting of one σ bond and one π bond.

During ozonolysis, the carbon-carbon double bond is cleaved and each carbon of the double bond is converted into a carbonyl group.

If a carbon of the double bond possesses at least one hydrogen atom, an aldehyde is formed after ozonolysis.

If both carbon atoms of the double bond are identical, ozonolysis produces two identical molecules of the same aldehyde or ketone.

The number of σ bonds present in the molecule helps determine the number of carbon atoms in the alkene.

🗺️ Solution Roadmap Step-by-step Plan ›

Determine the molecular formula from the given σ and π bonds.
Identify the aldehyde from its molar mass.
Use reverse ozonolysis to reconstruct the original alkene.
Verify that the reconstructed alkene has the required number of σ and π bonds.
Write the correct IUPAC name.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Determine the molecular formula.

The compound contains

Three C–C σ bonds.

Eight C–H σ bonds.

One C–C π bond.

Three C–C σ bonds indicate that the molecule contains four carbon atoms.

Eight C–H σ bonds indicate eight hydrogen atoms.

Therefore, the molecular formula is

\[\mathrm{C_4H_8}\]

This corresponds to an alkene.
Step 2: Identify the aldehyde.

The aldehyde has molar mass 44 u.

The aldehyde having molar mass 44 u is

\[\mathrm{CH_3CHO}\]

This compound is ethanal.
Step 3: Interpret the ozonolysis products.

The alkene gives two moles of ethanal.

This means both halves of the alkene are identical.

Hence, the original alkene must be obtained by joining two molecules of ethanal through their carbonyl carbons.
Step 4: Reconstruct the alkene.
Removing the oxygen atoms from the two molecules of ethanal and joining the carbonyl carbons by a double bond gives

\[\mathrm{CH_3CH=CHCH_3}\]
Step 5: Verification.
Ozonolysis of this alkene is

\[\mathrm{CH_3CH=CHCH_3\xrightarrow[Zn/H_2O]{O_3}2CH_3CHO}\]

Thus, two molecules of ethanal are obtained exactly as stated in the question.

The molecule also satisfies the given bond count.

C–C σ bonds = 3

C–H σ bonds = 8

C–C π bond = 1
Final Answer
Structure
\[\mathrm{CH_3CH=CHCH_3}\]
IUPAC Name: But-2-ene

🎯 Exam Significance Exam Significance ›

This question combines bond counting with ozonolysis and reverse ozonolysis, making it an important application-based problem.

Students should first identify the molecular formula from the number of σ and π bonds before reconstructing the alkene.

Always verify the answer by checking both the ozonolysis products and the bond count.

Writing the reconstruction process step by step improves presentation and helps secure full marks.

Significance for JEE / NEET and Other Competitive Examinations

Questions involving σ-bond counting, molecular formula determination and reverse ozonolysis are frequently asked in competitive examinations.

Students should develop the ability to deduce an unknown alkene from the products obtained after ozonolysis.

Such problems also strengthen concepts of structural determination and reaction mechanisms.

This type of reasoning is useful in multi-concept organic chemistry questions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Three C–C σ bonds indicate a four-carbon chain.
Eight C–H σ bonds correspond to the molecular formula C₄H₈.
An aldehyde of molar mass 44 u is ethanal.
Formation of two identical molecules of ethanal indicates a symmetrical alkene.
The required alkene is But-2-ene.

📘 Concept & Theory Theory / Concept ›

Ozonolysis is one of the most important reactions of alkenes. During ozonolysis, the carbon-carbon double bond is cleaved, and each carbon atom of the double bond is converted into a carbonyl group.

When the products of ozonolysis are known, the structure of the original alkene can be determined by reverse ozonolysis.

The procedure is simple:

Identify the carbonyl carbon in each product.

Remove the oxygen atom from each carbonyl group.

Join the two carbonyl carbons by a carbon-carbon double bond.

The resulting compound is the required alkene.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the structures of propanal and pentan-3-one.
Locate the carbonyl carbon in each molecule.
Remove the oxygen atom from each carbonyl group.
Join the two carbonyl carbons with a carbon-carbon double bond.
Verify the answer by writing the ozonolysis reaction.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Write the structures of the products.

Propanal

\[\mathrm{CH_3CH_2CHO}\]

Pentan-3-one

\[\mathrm{CH_3CH_2COCH_2CH_3}\]
Step 2: Identify the carbonyl carbons.
In propanal, the carbonyl carbon is

\[\mathrm{CH_3CH_2-\boxed{CHO}}\]

In pentan-3-one, the carbonyl carbon is

\[\mathrm{CH_3CH_2-\boxed{CO}-CH_2CH_3}\]
Step 3: Remove the oxygen atoms.
After removing the oxygen atoms, the fragments become

\[\mathrm{CH_3CH_2CH}\]

and

\[\mathrm{CH_3CH_2CCH_2CH_3}\]
Step 4: Join the two carbon atoms by a double bond.
\[\mathrm{CH_3CH_2CH=C(CH_2CH_3)CH_2CH_3}\]
Step 5: Verification.
On ozonolysis, the alkene undergoes cleavage at the double bond.

\[\mathrm{CH_3CH_2CH=C(CH_2CH_3)CH_2CH_3\xrightarrow[Zn/H_2O]{O_3}CH_3CH_2CHO + CH_3CH_2COCH_2CH_3}\]

The products obtained are propanal and pentan-3-one, exactly as given in the question.
Final Answer

Structural Formula

\[\mathrm{CH_3CH_2CH=C(CH_2CH_3)CH_2CH_3}\]

IUPAC Name

3-Ethylhept-3-ene

🎯 Exam Significance Exam Significance ›

Reverse ozonolysis is an important application of alkene reactions and is frequently tested in board examinations.

Students should always reconstruct the alkene by joining the carbonyl carbons after removing the oxygen atoms.

Writing the verification reaction strengthens the answer and helps in securing full marks.

Correct IUPAC nomenclature should always accompany the structural formula.

Significance for JEE / NEET and Other Competitive Examinations

Questions based on reverse ozonolysis are common in JEE Main, JEE Advanced and NEET.

Competitive examinations often provide carbonyl products and ask students to identify the original alkene.

Understanding carbon skeleton reconstruction is essential for solving higher-level organic chemistry problems.

This concept is also useful in structural elucidation of unknown organic compounds.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ozonolysis cleaves the carbon-carbon double bond.
Reverse ozonolysis reconstructs the original alkene from carbonyl compounds.
The carbonyl carbons are joined by a carbon-carbon double bond after removing oxygen.
Verification of the reconstructed alkene should always be performed.
The required alkene is 3-Ethylhept-3-ene.

📘 Concept & Theory Theory / Concept ›

Combustion is an oxidation reaction in which a hydrocarbon reacts with oxygen to produce carbon dioxide and water along with the evolution of a large amount of heat and light.

The general combustion reaction of a hydrocarbon is

\[\mathrm{Hydrocarbon + O_2 \rightarrow CO_2 + H_2O + Heat}\]

For complete combustion, sufficient oxygen must be available. If oxygen is insufficient, incomplete combustion occurs, producing carbon monoxide or carbon (soot).

The balancing of combustion reactions is done in the following order:

Balance carbon atoms.

Balance hydrogen atoms.

Finally, balance oxygen atoms.

If fractional coefficients appear, multiply the entire equation by the appropriate number to obtain whole-number coefficients.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the molecular formula of the hydrocarbon.
Write the products as carbon dioxide and water.
Balance carbon atoms first.
Balance hydrogen atoms next.
Balance oxygen atoms at the end.
Convert fractional coefficients into whole numbers if necessary.

✏️ Solution Complete Solution ›

Step-by-step Solution · 24 steps

(i) Butane
Step 1: Molecular Formula
\[\mathrm{C_4H_{10}}\]
Step 2: Write the unbalanced equation.
\[\mathrm{C_4H_{10}+O_2\rightarrow CO_2+H_2O}\]
Step 3: Balance carbon atoms.
\[\mathrm{C_4H_{10}+O_2\rightarrow4CO_2+H_2O}\]
Step 4: Balance hydrogen atoms.
\[\mathrm{C_4H_{10}+O_2\rightarrow4CO_2+5H_2O}\]
Step 5: Balance oxygen atoms.
\[\mathrm{2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O}\]
Balanced Combustion Equation
\[\boxed{\mathrm{2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O}}\]
(ii) Pentene
Step 1: Molecular Formula
\[\mathrm{C_5H_{10}}\]
Step 2: Write the unbalanced equation.
\[\mathrm{C_5H_{10}+O_2\rightarrow CO_2+H_2O}\]
Step 3: Balance carbon atoms.
\[\mathrm{C_5H_{10}+O_2\rightarrow5CO_2+H_2O}\]
Step 4: Balance hydrogen atoms.
\[\mathrm{C_5H_{10}+O_2\rightarrow5CO_2+5H_2O}\]
Step 5: Balance oxygen atoms.
\[\mathrm{2C_5H_{10}+15O_2\rightarrow10CO_2+10H_2O}\]
Balanced Combustion Equation
\[\boxed{\mathrm{2C_5H_{10}+15O_2\rightarrow10CO_2+10H_2O}}\]
(iii) Hexyne
Step 1: Molecular Formula
\[\mathrm{C_6H_{10}}\]
Step 2: Write the unbalanced equation.
\[\mathrm{C_6H_{10}+O_2\rightarrow CO_2+H_2O}\]
Step 3: Balance carbon atoms.
\[\mathrm{C_6H_{10}+O_2\rightarrow6CO_2+H_2O}\]
Step 4: Balance hydrogen atoms.
\[\mathrm{C_6H_{10}+O_2\rightarrow6CO_2+5H_2O}\]
Step 5: Balance oxygen atoms.
\[\mathrm{2C_6H_{10}+17O_2\rightarrow12CO_2+10H_2O}\]
Balanced Combustion Equation
\[\boxed{\mathrm{2C_6H_{10}+17O_2\rightarrow12CO_2+10H_2O}}\]
(iv) Toluene
Step 1: Molecular Formula
\[\mathrm{C_7H_8}\]
Step 2: Write the unbalanced equation.
\[\mathrm{C_7H_8+O_2\rightarrow CO_2+H_2O}\]
Step 3: Balance carbon atoms.
\[\mathrm{C_7H_8+O_2\rightarrow7CO_2+H_2O}\]
Step 4: Balance hydrogen atoms.
\[\mathrm{C_7H_8+O_2\rightarrow7CO_2+4H_2O}\]
Step 5: Balance oxygen atoms.
\[\mathrm{C_7H_8+9O_2\rightarrow7CO_2+4H_2O}\]
Balanced Combustion Equation
\[\boxed{\mathrm{C_7H_8+9O_2\rightarrow7CO_2+4H_2O}}\]

🎯 Exam Significance Exam Significance ›

Combustion reactions are among the most frequently asked balancing questions in CBSE examinations.

Students should always balance carbon atoms first, followed by hydrogen and finally oxygen.

Whenever fractional coefficients appear, multiply the entire equation to obtain whole-number coefficients.

Writing balanced equations with correct molecular formulae helps secure full marks.

Significance for JEE / NEET and Other Competitive Examinations

Combustion reactions are frequently used in stoichiometry and thermochemistry problems.

Students should be able to write balanced combustion equations quickly without trial and error.

Knowledge of the molecular formulae of alkanes, alkenes, alkynes and aromatic hydrocarbons is essential.

These reactions are also used in questions involving oxygen requirement, gaseous products and calorific value.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Complete combustion produces carbon dioxide and water.
Combustion is an exothermic oxidation reaction.
Carbon atoms should always be balanced before hydrogen atoms.
Oxygen atoms are balanced in the final step.
Whole-number coefficients should be used in the final balanced equation.

📘 Concept & Theory Theory / Concept ›

Hex-2-ene exhibits geometrical (cis-trans) isomerism because each carbon atom of the carbon-carbon double bond is attached to two different groups.

The carbon-carbon double bond contains one σ bond and one π bond. Rotation about the π bond is restricted, so the positions of the substituents remain fixed, giving rise to geometrical isomers.

In the cis isomer, similar or higher-priority groups are present on the same side of the double bond.

In the trans isomer, similar or higher-priority groups are present on opposite sides of the double bond.

The cis isomer possesses a small permanent dipole moment because the bond dipoles reinforce each other. The trans isomer is nearly non-polar because the bond dipoles largely cancel each other.

As a result, intermolecular dipole-dipole attractions are stronger in the cis isomer, giving it a slightly higher boiling point.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the structure of hex-2-ene.
Locate the carbon-carbon double bond between carbon-2 and carbon-3.
Draw the cis arrangement by placing the larger alkyl groups on the same side.
Draw the trans arrangement by placing the larger alkyl groups on opposite sides.
Compare their polarity and intermolecular forces.
Identify the isomer having the higher boiling point.

✏️ Solution Complete Solution ›

Step-by-step Solution · 3 steps

Structure of Hex-2-ene
\[\mathrm{CH_3CH=CHCH_2CH_2CH_3}\]

The double bond lies between carbon-2 and carbon-3.
Cis-Hex-2-ene
In the trans isomer, the two hydrogen atoms (or the alkyl groups) are on opposite sides of the double bond.

Comparison of Boiling Points

The cis isomer has the higher boiling point.

Reason:

In cis-hex-2-ene, the molecular dipoles do not cancel completely.

Therefore, the molecule possesses a net dipole moment.

As a result, molecules attract one another through dipole-dipole interactions in addition to London dispersion forces.

In trans-hex-2-ene, the dipoles almost cancel because of the symmetrical arrangement of the substituents.

Hence, intermolecular attraction is weaker, leading to a comparatively lower boiling point.
Trans Structure
Boiling Point:

Cis-hex-2-ene has a higher boiling point than trans-hex-2-ene because it is more polar and experiences stronger intermolecular dipole-dipole attractions.

🎯 Exam Significance Exam Significance ›

Geometrical isomerism is one of the most important topics in the Hydrocarbons chapter.

Students should know the conditions required for cis-trans isomerism, namely restricted rotation about the double bond and two different groups attached to each double-bond carbon.

The difference between the physical properties of cis and trans isomers, especially boiling point and melting point, is frequently tested.

Well-labelled diagrams are essential for obtaining full marks.

Significance for JEE / NEET and Other Competitive Examinations

Competitive examinations frequently ask students to identify cis and trans isomers or determine whether geometrical isomerism is possible.

Questions also test the relationship between polarity and physical properties.

Students should remember that cis isomers generally possess higher boiling points due to greater polarity, whereas trans isomers generally possess higher melting points because of better crystal packing.

This concept serves as the foundation for E-Z nomenclature introduced in advanced organic chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Hex-2-ene exhibits geometrical isomerism because rotation about the double bond is restricted.
Cis isomers have similar groups on the same side of the double bond.
Trans isomers have similar groups on opposite sides.
Cis-hex-2-ene is more polar than trans-hex-2-ene.
Cis-hex-2-ene has a higher boiling point due to stronger dipole-dipole interactions.

📘 Concept & Theory Theory / Concept ›

At first sight, benzene appears to contain three carbon-carbon double bonds alternating with three single bonds. If this were true, benzene should behave like an alkene and readily undergo addition reactions.

However, experiments show that benzene is much more stable than expected and prefers substitution reactions rather than addition reactions.

The extraordinary stability of benzene is explained by the concept of aromaticity and resonance.

Each carbon atom in benzene is sp²-hybridised.

Every carbon contributes one unhybridised p-orbital perpendicular to the plane of the ring.

These six p-orbitals overlap sideways to form a continuous cyclic cloud of delocalised π-electrons above and below the plane of the benzene ring.

The six π-electrons are not confined between two carbon atoms. Instead, they are equally distributed over all six carbon atoms.

This delocalisation lowers the total energy of the molecule, making benzene much more stable than a hypothetical cyclohexatriene containing three isolated double bonds.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the bonding in benzene.
Explain the hybridisation of carbon atoms.
Describe the formation of the delocalised π-electron cloud.
Discuss resonance and aromaticity.
Conclude why benzene is unusually stable.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

Step 1: Structure of Benzene
Benzene consists of six carbon atoms arranged in a planar hexagonal ring.

Each carbon atom is bonded to two neighbouring carbon atoms and one hydrogen atom.

Each carbon atom is sp²-hybridised.
Step 2: Formation of the π-System
Each carbon possesses one unhybridised p-orbital.

These six p-orbitals overlap continuously around the ring.

As a result, a continuous π-electron cloud is formed above and below the plane of the ring.
Step 3: Delocalisation of π-Electrons
The six π-electrons are shared equally by all six carbon atoms.

They are not localised between individual pairs of carbon atoms.

This phenomenon is called electron delocalisation.
Step 4: Resonance
Benzene is represented by two equivalent Kekulé structures.

The actual structure is not either one of them but a resonance hybrid of both.

Therefore, all six carbon-carbon bonds are identical.

The bond length of every carbon-carbon bond is approximately

\[\mathrm{1.39\ AA}\]

This value lies between a typical single bond and a typical double bond.
Step 5: Aromatic Stability
Benzene contains six delocalised π-electrons.

It satisfies Hückel's rule for aromaticity.

\[\mathrm{4n+2}\]

where

\[\mathrm{n=1}\]

Therefore,

\[\mathrm{4(1)+2=6}\]

Since benzene possesses six π-electrons, it is an aromatic compound.

The aromatic π-electron cloud provides exceptional thermodynamic stability.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions from the Hydrocarbons chapter.

Students should mention resonance, delocalisation of π-electrons and aromaticity while answering.

Writing that all carbon-carbon bonds in benzene are identical and have equal bond lengths helps in securing full marks.

Mentioning Hückel's rule further strengthens the answer.

Significance for JEE / NEET and Other Competitive Examinations

Aromaticity is one of the most important concepts in Organic Chemistry and is frequently tested in JEE Main, JEE Advanced and NEET.

Questions often ask students to identify aromatic, anti-aromatic and non-aromatic compounds using Hückel's rule.

Students should understand the relationship between resonance energy, bond length and chemical stability.

This concept forms the basis for electrophilic aromatic substitution reactions studied in later chapters.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Benzene contains six sp²-hybridised carbon atoms.
The six p-orbitals overlap to form a continuous delocalised π-cloud.
All carbon-carbon bonds in benzene are identical.
Benzene satisfies Hückel's
\(4n+2\)
rule with six π-electrons.
Resonance and aromaticity are responsible for the extraordinary stability of benzene.

📘 Concept & Theory Theory / Concept ›

Aromatic compounds are exceptionally stable cyclic compounds due to the delocalisation of π-electrons over the entire ring.

Not every cyclic compound is aromatic. A compound must satisfy certain structural and electronic requirements to exhibit aromaticity.

These requirements are collectively known as the conditions for aromaticity.

🗺️ Solution Roadmap Step-by-step Plan ›

Define aromaticity.
State the structural conditions.
Explain the requirement of conjugation.
Apply Hückel's rule.
Conclude with examples.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

A compound is aromatic only if it satisfies all of the following conditions.
1. The Molecule Must Be Cyclic
The atoms should form a closed ring.

An open-chain compound cannot be aromatic because continuous cyclic delocalisation of electrons is not possible.
2. The Molecule Must Be Planar
All the atoms in the ring should lie in the same plane.

Planarity allows effective sideways overlap of adjacent p-orbitals.

If the ring is non-planar, continuous overlap of p-orbitals is interrupted and aromaticity is lost.
3. The Ring Must Be Completely Conjugated
Every atom in the ring must possess an unhybridised p-orbital.

These p-orbitals overlap continuously around the ring to form a delocalised π-electron cloud.

Any interruption in conjugation prevents aromatic character.
4. The Molecule Must Obey Hückel's Rule
The number of delocalised π-electrons must satisfy
\[\mathrm{4n+2}\]

where
\[\mathrm{n=0,1,2,3,\ldots}\]

Examples of aromatic systems are:

\[\mathrm{2,\ 6,\ 10,\ 14,\ 18,\ldots}\]

π-electrons.

For benzene,

\[\mathrm{n=1}\]

Therefore,

\[\mathrm{4(1)+2=6}\]

Hence, benzene is aromatic.
Summary of Necessary Conditions

Condition	Requirement
Shape	Cyclic (closed ring)
Geometry	Planar structure
Bonding	Continuous conjugation with overlapping p-orbitals
Electron Count	\(\mathrm{4n+2}\) π-electrons (Hückel's rule)

Examples
Aromatic Compounds

Benzene (6 π-electrons)

Naphthalene (10 π-electrons)

Pyridine (6 π-electrons)

Non-Aromatic Compounds

Cyclooctatetraene (non-planar)

Cyclohexane (no conjugated π-system)
Answer
A system is aromatic if it satisfies all of the following conditions:

The molecule must be cyclic.

The molecule must be planar.

There must be continuous conjugation with overlapping p-orbitals around the ring.

The number of delocalised π-electrons must satisfy Hückel's rule, i.e.,

\[\boxed{\mathrm{4n+2\ \pi\ electrons}}\]

where
\(n = 0,1,2,\ldots\)
.

🎯 Exam Significance Exam Significance ›

This is a very important theory question from aromatic hydrocarbons.

Students should remember all four conditions in the correct order.

Mentioning Hückel's rule with examples helps in obtaining full marks.

A neat table summarising the conditions makes the answer more effective.

Significance for JEE / NEET and Other Competitive Examinations

Questions on aromaticity are frequently asked in JEE Main, JEE Advanced and NEET.

Students are often required to determine whether a compound is aromatic, anti-aromatic or non-aromatic.

Application of Hückel's rule and recognition of conjugation are essential problem-solving skills.

This concept forms the basis for electrophilic aromatic substitution and heterocyclic chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Aromatic compounds are cyclic.
The ring must be planar.
Continuous conjugation is essential.
The molecule must contain
\(\mathrm{4n+2}\)
delocalised π-electrons.
Benzene is the simplest aromatic compound because it satisfies all four conditions.

📘 Concept & Theory Theory / Concept ›

A compound is aromatic only if it satisfies all of the following conditions.

The molecule must be cyclic.

The molecule must be planar.

There must be continuous conjugation, i.e., every atom in the ring should possess an unhybridised p-orbital.

The molecule must contain

\(\mathrm{(4n+2)}\)

π-electrons (Hückel's rule).

If even one of these conditions is not fulfilled, the compound is not aromatic.

🗺️ Solution Roadmap Step-by-step Plan ›

Examine each structure individually.
Check whether the ring is planar.
Check whether conjugation is continuous.
Count the number of π-electrons.
Identify which aromaticity condition is violated.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

(i)
The compound is a cyclohexadiene ring containing an exocyclic methylene group.

The π-electrons are cross-conjugated rather than continuously conjugated around the six-membered ring.

The exocyclic double bond interrupts cyclic delocalisation of π-electrons.

Since a continuous cyclic π-electron cloud is not formed, the molecule does not satisfy the requirement of complete conjugation.

Therefore, the compound is not aromatic.
(ii)
The compound is cyclopenta-1,3-diene.

One carbon atom of the ring is sp³-hybridised.

This carbon does not possess an unhybridised p-orbital.

Hence, continuous overlap of p-orbitals around the ring is not possible.

Because conjugation is interrupted, the compound is non-aromatic.
(iii)
The compound is cyclooctatetraene.

Although it contains eight π-electrons, it does not remain planar.

Instead, it adopts a tub-shaped (non-planar) conformation.

This prevents continuous overlap of all p-orbitals.

If it were planar, it would contain

\[\mathrm{8\ \pi\ electrons = 4n\ (n=2)}\]

which would make it anti-aromatic.

To avoid this instability, the molecule becomes non-planar and consequently loses aromatic character.

Therefore, cyclooctatetraene is non-aromatic.
Summary Table

System	Reason for Non-aromatic Nature
(i)	Continuous cyclic conjugation is absent because of cross-conjugation caused by the exocyclic double bond.
(ii)	Contains one sp³-hybridised carbon; conjugation is interrupted.
(iii)	Adopts a non-planar tub-shaped structure; continuous overlap of p-orbitals is not possible.

Answer
(i) It is not aromatic because it lacks continuous cyclic conjugation due to the exocyclic double bond.

(ii) It is not aromatic because one carbon atom is sp³-hybridised, interrupting conjugation.

(iii) It is not aromatic because it is non-planar and therefore does not have continuous overlap of p-orbitals.

🎯 Exam Significance Exam Significance ›

Students should identify which condition for aromaticity is violated instead of merely stating that the compound is non-aromatic.

Keywords such as planarity, continuous conjugation and Hückel's rule should always be included in the explanation.

Drawing attention to the sp³-hybridised carbon in cyclopentadiene and the non-planar structure of cyclooctatetraene earns better marks.

Significance for JEE / NEET and Other Competitive Examinations

Questions based on aromatic, anti-aromatic and non-aromatic compounds are frequently asked in JEE Main, JEE Advanced and NEET.

Students should first check cyclic structure, planarity, conjugation and finally apply Hückel's rule before deciding whether a compound is aromatic.

Cyclopentadiene and cyclooctatetraene are classic examples used in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Aromatic compounds must be cyclic, planar and completely conjugated.
Interrupted conjugation destroys aromaticity.
Non-planarity prevents effective overlap of p-orbitals.
Cross-conjugation does not produce aromatic stabilisation.
Failure to satisfy even one condition makes a compound non-aromatic.

📘 Concept & Theory Theory / Concept ›

Benzene undergoes electrophilic aromatic substitution (EAS) reactions. During these reactions, one hydrogen atom of benzene is replaced by an electrophile while the aromaticity of the ring is retained.

The position of substitution depends upon the directing effect of the substituent already present on the benzene ring.

Ortho-para directing groups: –CH₃, –OH, –NH₂, –Cl, –Br.

Meta directing groups: –NO₂, –CHO, –COOH, –SO₃H.

Therefore, the order of introducing substituents is very important in organic synthesis.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the directing effect of the substituent.
Select the reaction that should be carried out first.
Perform nitration, halogenation or Friedel-Crafts reaction in the correct sequence.
Verify that the desired product is obtained.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

(i) Conversion of Benzene into p-Nitrobromobenzene
Step 1: Bromination of benzene

\[\mathrm{C_6H_6 \xrightarrow{Br_2/FeBr_3} C_6H_5Br}\]

Bromine is an ortho-para directing group.

Step 2: Nitration

\[\mathrm{C_6H_5Br \xrightarrow{Conc.\ HNO_3/Conc.\ H_2SO_4}p\text{-}BrC_6H_4NO_2}\]

The para isomer is obtained as the major product because of less steric hindrance.

Conversion:

\[ \boxed{\mathrm{C_6H_6 \rightarrow C_6H_5Br \rightarrow p\text{-}BrC_6H_4NO_2}} \]
(ii) Conversion of Benzene into m-Nitrochlorobenzene
Step 1: Nitration of benzene

\[\mathrm{C_6H_6\xrightarrow{Conc.\ HNO_3/Conc.\ H_2SO_4}C_6H_5NO_2}\]

Nitro group is a strong meta-directing group.
Step 2: Chlorination

\[\mathrm{C_6H_5NO_2\xrightarrow{Cl_2/FeCl_3}m\text{-}ClC_6H_4NO_2}\]
Conversion:

\[\boxed{\mathrm{C_6H_6\rightarrow C_6H_5NO_2\rightarrow m\text{-}ClC_6H_4NO_2}}\]
(iii) Conversion of Benzene into p-Nitrotoluene
Step 1: Friedel-Crafts methylation

\[\mathrm{C_6H_6\xrightarrow{CH_3Cl/AlCl_3}C_6H_5CH_3}\]

Toluene contains a methyl group, which is an ortho-para directing group.
Step 2: Nitration

\[\mathrm{C_6H_5CH_3\xrightarrow{Conc.\ HNO_3/Conc.\ H_2SO_4}p\text{-}CH_3C_6H_4NO_2}\]

The para isomer is obtained as the major product.
Conversion:

\[\boxed{\mathrm{C_6H_6\rightarrow C_6H_5CH_3\rightarrow p\text{-}CH_3C_6H_4NO_2}}\]
(iv) Conversion of Benzene into Acetophenone
Acetophenone is prepared by Friedel-Crafts acylation.

\[\mathrm{C_6H_6\xrightarrow{CH_3COCl/AlCl_3}C_6H_5COCH_3}\]

The product obtained is acetophenone.
Conversion:

\[\boxed{\mathrm{C_6H_6\xrightarrow{CH_3COCl/AlCl_3}C_6H_5COCH_3}}\]
Summary Table

Target Compound	Sequence of Reactions
p-Nitrobromobenzene	Bromination → Nitration
m-Nitrochlorobenzene	Nitration → Chlorination
p-Nitrotoluene	Friedel-Crafts Methylation → Nitration
Acetophenone	Friedel-Crafts Acylation

Answer
(i)

\[\mathrm{C_6H_6\xrightarrow{Br_2/FeBr_3}C_6H_5Br\xrightarrow{HNO_3/H_2SO_4}p\text{-}BrC_6H_4NO_2}\]

(ii)

\[\mathrm{C_6H_6\xrightarrow{HNO_3/H_2SO_4}C_6H_5NO_2\xrightarrow{Cl_2/FeCl_3}m\text{-}ClC_6H_4NO_2}\]

(iii)

\[\mathrm{C_6H_6\xrightarrow{CH_3Cl/AlCl_3}C_6H_5CH_3\xrightarrow{HNO_3/H_2SO_4}p\text{-}CH_3C_6H_4NO_2}\]

(iv)

\[\mathrm{C_6H_6\xrightarrow{CH_3COCl/AlCl_3}C_6H_5COCH_3}\]

🎯 Exam Significance Exam Significance ›

Conversions involving benzene are among the most frequently asked questions in CBSE examinations.

Students should remember the directing effects of substituents before deciding the order of reactions.

The sequence of introducing substituents is crucial for obtaining the desired isomer.

Friedel-Crafts alkylation and acylation are important named reactions that should be memorised.

Significance for JEE / NEET and Other Competitive Examinations

Electrophilic aromatic substitution reactions are fundamental topics in JEE Main, JEE Advanced and NEET.

Competitive examinations frequently test directing effects, reaction sequences and synthetic pathways.

Students should understand why nitration is performed before chlorination in meta substitution and after bromination or methylation in para substitution.

Mastery of these conversions is essential for solving multi-step organic synthesis problems.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Reaction sequence determines the final product.
Bromine and methyl groups are ortho-para directors.
Nitro group is a strong meta director.
Friedel-Crafts alkylation introduces an alkyl group.
Friedel-Crafts acylation introduces an acyl group without rearrangement.

📘 Concept & Theory Theory / Concept ›

Carbon atoms are classified according to the number of other carbon atoms directly attached to them.

Primary (1°) Carbon: A carbon atom attached to only one other carbon atom.

Secondary (2°) Carbon: A carbon atom attached to two other carbon atoms.

Tertiary (3°) Carbon: A carbon atom attached to three other carbon atoms.

Quaternary (4°) Carbon: A carbon atom attached to four other carbon atoms and therefore has no hydrogen atom attached to it.

🗺️ Solution Roadmap Step-by-step Plan ›

Number all the carbon atoms.
Determine how many carbon atoms are attached to each carbon.
Classify each carbon as 1°, 2°, 3° or 4°.
Count the hydrogen atoms attached to each type of carbon.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps

Step 1: Number the carbon atoms.

\[\mathrm{CH_3^{(1)}-CH_2^{(2)}-C^{(3)}(CH_3^{(4)})(CH_3^{(5)})-CH_2^{(6)}-CH^{(7)}(CH_3^{(8)})(CH_3^{(9)})}\]

The molecule contains 9 carbon atoms.

Step 2: Classify each carbon atom

Carbon Atom	Type of Carbon	Hydrogen Atoms Attached
C-1	1°	3
C-2	2°	2
C-3	4°	0
C-4	1°	3
C-5	1°	3
C-6	2°	2
C-7	3°	1
C-8	1°	3
C-9	1°	3

Step 3: Count the number of each type
Primary (1°) carbon atoms

C-1, C-4, C-5, C-8 and C-9

Number of 1° carbon atoms = 5

Each primary carbon is a CH₃ group.

Hydrogen atoms attached to each 1° carbon = 3

Total hydrogen atoms attached to all primary carbons

\[ \mathrm{5\times3=15} \]

Secondary (2°) carbon atoms

C-2 and C-6

Number of 2° carbon atoms = 2

Each secondary carbon is a CH₂ group.

Hydrogen atoms attached to each 2° carbon = 2

Total hydrogen atoms attached to all secondary carbons

\[ \mathrm{2\times2=4} \]

Tertiary (3°) carbon atom

C-7

Number of 3° carbon atoms = 1

It is a CH group.

Hydrogen atoms attached = 1

Quaternary (4°) carbon atom

C-3

It is attached to four carbon atoms.

Hydrogen atoms attached = 0
Summary Table

Type of Carbon	Carbon Atoms	Number of Carbon Atoms	Hydrogen Atoms on Each
1°	C-1, C-4, C-5, C-8, C-9	5	3
2°	C-2, C-6	2	2
3°	C-7	1	1
4°	C-3	1	0

🎯 Exam Significance Exam Significance ›

Classification of carbon atoms is a frequently asked conceptual question in CBSE examinations.

Students should distinguish between the classification of carbon atoms and the classification of hydrogen atoms.

Counting the number of carbon atoms attached to a given carbon is the simplest method for identifying its type.

Remember that a quaternary carbon atom never carries a hydrogen atom.

Significance for JEE / NEET and Other Competitive Examinations

Identification of primary, secondary, tertiary and quaternary carbon atoms is essential in reaction mechanisms involving free radicals, carbocations and carbanions.

Competitive examinations often combine carbon classification with IUPAC nomenclature and reaction mechanisms.

Students should practise identifying carbon types directly from condensed structural formulae.

This concept is fundamental to understanding stability orders in organic chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Primary carbon is attached to one carbon atom.
Secondary carbon is attached to two carbon atoms.
Tertiary carbon is attached to three carbon atoms.
Quaternary carbon is attached to four carbon atoms and has no hydrogen.
The given molecule contains 5 primary, 2 secondary, 1 tertiary and 1 quaternary carbon atom.

📘 Concept & Theory Theory / Concept ›

The boiling point of an alkane depends mainly on the strength of the intermolecular forces (London dispersion or van der Waals forces) between its molecules.

These intermolecular forces increase with the surface area available for contact between molecules.

Straight-chain (unbranched) alkanes have a larger surface area than their branched-chain isomers. Therefore, the intermolecular attractions between straight-chain molecules are stronger.

As branching increases, the molecules become more compact and nearly spherical. This decreases the surface area available for intermolecular contact.

Consequently, the van der Waals forces become weaker, requiring less energy to separate the molecules during boiling.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the factor affecting the boiling point of alkanes.
Compare the shapes of straight-chain and branched-chain alkanes.
Relate molecular shape to surface area.
Explain the effect on intermolecular forces.
State the trend in boiling points.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Nature of intermolecular forces
Alkanes are non-polar molecules.

Therefore, the only intermolecular forces present are London dispersion (van der Waals) forces.
Step 2: Effect of branching

Branching makes the molecule more compact.

The effective surface area of contact between neighbouring molecules decreases.
Step 3: Effect on intermolecular attraction

Because the surface area decreases, the van der Waals forces between molecules become weaker.

Hence, less heat energy is required to separate the molecules.
Step 4: Effect on boiling point

As the degree of branching increases, the boiling point decreases.
Illustration

Compound	Nature of Chain	Relative Boiling Point
n-Pentane	Straight chain	Highest
2-Methylbutane (Isopentane)	Branched chain	Intermediate
2,2-Dimethylpropane (Neopentane)	Highly branched	Lowest

The trend is

\[\mathrm{n\text{-}Pentane > Isopentane > Neopentane}\]
in terms of boiling point.

🎯 Exam Significance Exam Significance ›

This is a frequently asked conceptual question in CBSE examinations.

Students should mention that branching decreases the surface area, which weakens intermolecular forces and lowers the boiling point.

Quoting an example such as n-pentane, isopentane and neopentane strengthens the answer.

Significance for JEE / NEET and Other Competitive Examinations

Competitive examinations often ask students to arrange isomeric alkanes in increasing or decreasing order of boiling points.

Understanding the relationship between branching, molecular shape and intermolecular forces is essential for solving physical property questions.

This concept is also useful while comparing melting points and molecular packing.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Alkanes are non-polar molecules.
Their boiling points depend on London dispersion forces.
Branching reduces molecular surface area.
Reduced surface area weakens intermolecular attractions.
Therefore, increased branching lowers the boiling point of alkanes.

📘 Concept & Theory Theory / Concept ›

Addition of hydrogen bromide (HBr) to an unsymmetrical alkene can proceed by two different mechanisms.

In the absence of peroxide, the reaction follows an electrophilic addition mechanism and obeys Markovnikov's rule.

In the presence of benzoyl peroxide, the reaction proceeds through a free radical chain mechanism and follows the anti-Markovnikov (Peroxide or Kharasch) effect.

The peroxide effect is observed only with HBr because the radical chain mechanism is energetically favourable only in this case.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the reaction without peroxide.
Explain Markovnikov's rule.
Write the electrophilic addition mechanism.
Write the reaction in the presence of benzoyl peroxide.
Explain the free radical chain mechanism.
Compare the products obtained.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

(A) Addition of HBr in the Absence of Peroxide (Markovnikov Addition)
Reaction
\[\mathrm{CH_3CH=CH_2 + HBr \longrightarrow CH_3CHBrCH_3}\]
The major product formed is 2-bromopropane
Reason (Markovnikov's Rule)

According to Markovnikov's rule, during the addition of HX to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom already carrying more hydrogen atoms, while the halide ion attaches to the carbon atom carrying fewer hydrogen atoms.

This pathway proceeds through the formation of the more stable secondary carbocation.
Mechanism
Step 1: Protonation of the double bond
\[\mathrm{CH_3CH=CH_2 + H^+\longrightarrow CH_3\overset{+}{CH}CH_3}\]
A secondary carbocation is formed.
Step 2: Attack by bromide ion
\[\mathrm{CH_3\overset{+}{CH}CH_3 + Br^-\longrightarrow CH_3CHBrCH_3}\]

The final product is 2-bromopropane.
(B) Addition of HBr in the Presence of Benzoyl Peroxide (Anti-Markovnikov Addition)
Reaction

\[\mathrm{CH_3CH=CH_2 + HBr\xrightarrow{\text{Benzoyl Peroxide}}CH_3CH_2CH_2Br}\]

The major product formed is 1-bromopropane.

This reaction is called the Peroxide Effect or Kharasch Effect.
Free Radical Mechanism
Initiation Step

Benzoyl peroxide decomposes on heating to produce free radicals.

\[ \mathrm{(C_6H_5COO)_2 \longrightarrow 2C_6H_5COO^\bullet} \]

The benzoyloxy radicals generate bromine radicals.

\[ \mathrm{C_6H_5COO^\bullet + HBr \longrightarrow C_6H_5COOH + Br^\bullet} \]
Propagation Step 1

The bromine radical attacks the terminal carbon atom of propene.

\[ \mathrm{CH_3CH=CH_2 + Br^\bullet \longrightarrow CH_3\overset{\bullet}{CH}CH_2Br} \]

A secondary free radical is formed, which is more stable than a primary radical.
Propagation Step 2

The secondary radical abstracts a hydrogen atom from another molecule of HBr.

\[ \mathrm{CH_3\overset{\bullet}{CH}CH_2Br + HBr \longrightarrow CH_3CH_2CH_2Br + Br^\bullet} \]

A new bromine radical is regenerated, and the chain reaction continues.
Termination Step

The reaction stops when two free radicals combine.

Examples:

\[ \mathrm{Br^\bullet + Br^\bullet \longrightarrow Br_2} \]

\[ \mathrm{R^\bullet + Br^\bullet \longrightarrow RBr} \]

Why Are Different Products Formed?

Condition	Mechanism	Major Product	Reason
Without Peroxide	Electrophilic Addition	2-Bromopropane	Formation of the more stable secondary carbocation (Markovnikov addition)
With Benzoyl Peroxide	Free Radical Addition	1-Bromopropane	Formation of the more stable secondary free radical (Anti-Markovnikov addition)

Answer
In the absence of benzoyl peroxide, HBr adds to propene according to Markovnikov's rule through a carbocation mechanism, producing 2-bromopropane.

In the presence of benzoyl peroxide, the reaction proceeds by a free radical chain mechanism (Peroxide Effect or Kharasch Effect). The bromine radical adds first to form the more stable secondary radical, ultimately producing 1-bromopropane.

🎯 Exam Significance Exam Significance ›

The peroxide effect is one of the most frequently asked named reactions in the Hydrocarbons chapter.

Students should clearly distinguish between Markovnikov addition and anti-Markovnikov addition.

Writing the initiation, propagation and termination steps fetches full marks.

Remember that the peroxide effect is observed only with HBr and not with HCl or HI.

Significance for JEE / NEET and Other Competitive Examinations

Questions on the peroxide effect, carbocation stability and free radical stability are very common in JEE Main, JEE Advanced and NEET.

Students should understand why bromine attaches to the terminal carbon in the radical mechanism.

The reaction mechanism is frequently tested in assertion-reason and product prediction questions.

This concept is fundamental for studying free radical reactions in advanced organic chemistry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Without peroxide, HBr follows Markovnikov's rule.
With benzoyl peroxide, HBr follows anti-Markovnikov addition.
The peroxide effect proceeds through a free radical chain mechanism.
The major products are 2-bromopropane (without peroxide) and 1-bromopropane (with peroxide).
The peroxide effect is unique to HBr.

📘 Concept & Theory Theory/Concept ›

Ozonolysis cleaves carbon-carbon double bonds to form carbonyl compounds.

According to Kekulé's structure, benzene contains three alternate carbon-carbon double bonds.

If the benzene ring is subjected to ozonolysis, each double bond is cleaved, producing smaller carbonyl compounds.

The products obtained from ozonolysis of substituted benzene provide experimental evidence in support of Kekulé's proposed structure.

🗺️ Solution Roadmap Step-by-step Plan ›

Draw the structure of o-xylene.
Locate the three alternate double bonds according to Kekulé's structure.
Break each double bond by ozonolysis.
Write the products formed.
Explain how these products support the Kekulé structure of benzene.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Structure of o-Xylene
o-Xylene is 1,2-dimethylbenzene.

\[\mathrm{C_6H_4(CH_3)_2}\]

It contains two adjacent methyl groups attached to the benzene ring.
Step 2: Ozonolysis

Ozone cleaves all three carbon-carbon double bonds present in the Kekulé structure.

After reductive work-up, the ring opens to form carbonyl compounds.
Step 3: Products Formed

The ozonolysis of 1,2-dimethylbenzene gives:

\[\boxed{\mathrm{2\ CH_3COCHO+OHCCHO}}\]

That is,

Two molecules of methylglyoxal (2-oxopropanal)

and

One molecule of glyoxal (ethanedial).

The overall reaction may be represented as

\[\mathrm{C_6H_4(CH_3)_2\xrightarrow[\mathrm{Zn/H_2O}]{O_3}2CH_3COCHO+OHCCHO}\]
How Does This Support Kekulé's Structure?
Kekulé proposed that benzene contains three alternate carbon-carbon double bonds.

During ozonolysis, each of these double bonds undergoes cleavage.

The observed products can be explained only if three alternating double bonds are present in the benzene ring.

Thus, the formation of two molecules of methylglyoxal and one molecule of glyoxal is consistent with the cleavage pattern predicted by Kekulé's structure.

Therefore, the experimental results obtained from ozonolysis provide strong evidence in favour of Kekulé's structural formula of benzene.
Summary Table

,tbody>

Reactant	Products of Ozonolysis
1,2-Dimethylbenzene (o-Xylene)	2 Methylglyoxal (2-Oxopropanal) + 1 Glyoxal (Ethanedial)

Answer
Products of ozonolysis:

\[\boxed{\mathrm{C_6H_4(CH_3)_2\xrightarrow[\mathrm{Zn/H_2O}]{O_3}2CH_3COCHO+OHCCHO}}\]

Thus, the products are:

Two molecules of methylglyoxal (2-oxopropanal).

One molecule of glyoxal (ethanedial).

Support for Kekulé's structure:

The products are formed due to cleavage of the three alternate double bonds proposed by Kekulé. Hence, the ozonolysis products support the Kekulé structure of benzene.

🎯 Exam Significance Exam Significance ›

This is an important theory question related to aromatic hydrocarbons.

Students should remember both the products of ozonolysis and their significance in verifying Kekulé's structure.

Writing both the reaction and the explanation helps in obtaining full marks.

Use the terms alternate double bonds, ozonolysis and Kekulé structure in the answer.

Significance for JEE / NEET and Other Competitive Examinations

Questions on aromaticity, Kekulé's structure and ozonolysis are common in JEE Main, JEE Advanced and NEET.

Students should understand how experimental observations support structural theories.

This concept links aromatic chemistry with oxidation reactions and structural elucidation.

Knowledge of the ozonolysis products of substituted benzenes is useful for advanced organic chemistry problems.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ozonolysis cleaves carbon-carbon double bonds.
o-Xylene produces two molecules of methylglyoxal and one molecule of glyoxal.
The products arise from cleavage of three alternate double bonds.
The experimental results support Kekulé's structure of benzene.
Ozonolysis provides evidence for the presence of alternating double bonds in benzene.

📘 Concept & Theory Theory / Concept ›

The acidity of a hydrocarbon depends upon how easily it can lose a proton (H⁺).

When a proton is removed, a negatively charged species (carbanion) is formed.

The greater the stability of the carbanion, the greater is the acidity of the hydrocarbon.

The stability of a carbanion depends on the hybridisation of the carbon atom bearing the negative charge.

The percentage of s-character in different hybrid orbitals is:

Hybridisation	s-Character
sp	50%
sp²	33%
sp³	25%

A carbon atom with greater s-character is more electronegative and holds the negative charge more strongly, thereby stabilising the carbanion.

Hence, acidity increases with increasing s-character.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the hybridisation of the carbon atom attached to the acidic hydrogen.
Compare the percentage of s-character.
Determine the stability of the corresponding carbanions.
Arrange the compounds in decreasing order of acidity.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

1. Ethyne
Formula:

\[\mathrm{HC\equiv CH}\]

The acidic hydrogen is attached to an sp-hybridised carbon atom.

s-character = 50%.

The resulting acetylide ion is relatively stable.

Therefore, ethyne shows the highest acidity among the given compounds.
2. Benzene
Formula:

\[\mathrm{C_6H_6}\]

Each carbon atom is sp²-hybridised.

s-character = 33%.

The phenyl carbanion formed after removal of a proton is less stable than the acetylide ion.

Hence, benzene is less acidic than ethyne.
3. n-Hexane
Formula:

\[\mathrm{CH_3(CH_2)_4CH_3}\]

All carbon atoms are sp³-hybridised.

s-character = 25%.

The corresponding alkyl carbanion is highly unstable.

Therefore, n-hexane is the least acidic.

Decreasing Order of Acidic Behaviour

\[\boxed{\mathrm{Ethyne > Benzene > n\text{-}Hexane}}\]
Reason
The acidity depends upon the stability of the conjugate base (carbanion).

The stability of the carbanion increases with increasing s-character.

\[\mathrm{sp > sp^2 > sp^3}\]

Therefore,

\[\boxed{\mathrm{Ethyne > Benzene > n\text{-}Hexane}}\]
Summary Table

Compound	Hybridisation	s-Character	Relative Acidity
Ethyne	sp	50%	Highest
Benzene	sp²	33%	Intermediate
n-Hexane	sp³	25%	Lowest

💡 Answer Final Answer ›

The decreasing order of acidic behaviour is

\[\boxed{\mathrm{Ethyne > Benzene > n\text{-}Hexane}}\]

This order is due to the increasing stability of the conjugate base (carbanion) with increasing s-character of the hybrid orbital.

Since the s-character follows

\[\mathrm{sp > sp^2 > sp^3}\]

ethyne is the most acidic, while n-hexane is the least acidic.

🎯 Exam Significance Exam Significance ›

This is an important conceptual question based on acidity and hybridisation.

Students should relate acidity to the stability of the conjugate base rather than simply memorising the order.

Mentioning the percentage of s-character strengthens the explanation and helps secure full marks.

Remember that greater s-character leads to greater electronegativity and higher acidity.

Significance for JEE / NEET and Other Competitive Examinations

Questions on the acidity of hydrocarbons are frequently asked in JEE Main, JEE Advanced and NEET.

Students should understand the relationship between hybridisation, electronegativity, carbanion stability and acidity.

Similar concepts are also applied while comparing alcohols, phenols, terminal alkynes and substituted hydrocarbons.

This concept forms the basis for many reaction mechanisms involving carbanions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Acidity depends on the stability of the conjugate base.
Greater s-character stabilises the negative charge.
Hybridisation order is sp > sp² > sp³.
Terminal alkynes are more acidic than benzene and alkanes.
The decreasing order of acidity is: Ethyne > Benzene > n-Hexane.

📘 Concept & Theory Theory / Concept ›

Benzene is an aromatic compound containing six delocalised π-electrons spread uniformly over the six carbon atoms of the ring.

This delocalisation gives benzene exceptional stability, known as aromatic stability or resonance stabilisation.

Any reaction involving benzene tends to preserve this aromatic character.

Therefore, reactions that temporarily disturb aromaticity but restore it in the final product are favoured, whereas reactions that permanently destroy aromaticity are generally unfavourable.

🗺️ Solution Roadmap Step-by-step Plan ›

Explain the electron-rich nature of the benzene ring.
Discuss why electrophiles are attracted to benzene.
Describe the mechanism of electrophilic substitution.
Explain why nucleophiles are repelled by benzene.
State why nucleophilic substitution is difficult.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

Why Benzene Undergoes Electrophilic Substitution Easily
Step 1: Electron-rich aromatic ring

The benzene ring contains a cloud of delocalised π-electrons above and below the plane of the ring.

These π-electrons create a region of high electron density.

Hence, benzene readily attracts electrophiles (electron-deficient species).
Step 2: Formation of the σ-complex

The electrophile attacks the π-electron cloud to form a non-aromatic intermediate called the σ-complex (arenium ion).

\[\mathrm{C_6H_6 + E^+ \longrightarrow \sigma\text{-complex}}\]
Step 3: Restoration of aromaticity

Loss of a proton from the σ-complex restores the aromatic ring.

\[\mathrm{\sigma\text{-complex} \longrightarrow C_6H_5E + H^+}\]

Since aromaticity is regained, the overall reaction is energetically favourable.
Why Benzene Undergoes Nucleophilic Substitution with Difficulty
Step 1: Electron-rich ring repels nucleophiles

Nucleophiles are electron-rich species.

Because the benzene ring is also electron-rich, strong electrostatic repulsion exists between the nucleophile and the π-electron cloud.
Step 2: Loss of aromatic stability

If a nucleophile attacks benzene directly, the aromatic system would be disrupted.

The resulting intermediate is highly unstable because aromaticity is lost.
Step 3: No favourable driving force

Unlike electrophilic substitution, nucleophilic attack does not readily regenerate aromaticity under ordinary conditions.

Therefore, nucleophilic substitution occurs only under special conditions, such as when the benzene ring contains strong electron-withdrawing groups or a suitable leaving group.
Comparison

Electrophilic Substitution	Nucleophilic Substitution
Benzene is rich in π-electrons and attracts electrophiles.	Benzene repels electron-rich nucleophiles.
Aromaticity is restored after substitution.	Aromaticity is difficult to restore.
Reaction occurs readily.	Reaction occurs with difficulty.
Common reactions: Nitration, Halogenation, Sulphonation, Friedel-Crafts reactions.	Occurs only under special conditions with activated aromatic compounds.

💡 Answer Final Answer ›

Benzene undergoes electrophilic substitution reactions easily because its delocalised π-electron cloud attracts electrophiles. Although aromaticity is temporarily disturbed during the reaction, it is restored after substitution, making the reaction favourable.

Benzene undergoes nucleophilic substitution with difficulty because the electron-rich benzene ring repels nucleophiles. Moreover, nucleophilic attack destroys the aromatic character, producing an unstable intermediate. Therefore, such reactions occur only under special conditions.

🎯 Exam Significance Exam Significance ›

This is one of the most important conceptual questions on aromatic hydrocarbons.

Students should explain both the electron-rich nature of benzene and the importance of aromatic stability.

Mentioning restoration of aromaticity during electrophilic substitution helps in securing full marks.

Using terms such as π-electron cloud, σ-complex, aromaticity and resonance stabilisation makes the answer more effective.

Significance for JEE / NEET and Other Competitive Examinations

Electrophilic aromatic substitution is one of the most frequently tested topics in JEE Main, JEE Advanced and NEET.

Students should understand the role of aromatic stabilisation in determining the reactivity of benzene.

Questions often compare electrophilic substitution with nucleophilic substitution and addition reactions.

This concept also forms the basis for understanding directing effects and substitution mechanisms in substituted benzene derivatives.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Benzene contains a delocalised π-electron cloud.
Electrophiles are attracted to the electron-rich aromatic ring.
Electrophilic substitution restores aromaticity after the reaction.
Nucleophiles are repelled by the electron-rich benzene ring.
Aromatic stability is the main reason why benzene readily undergoes electrophilic substitution but resists nucleophilic substitution.

📘 Concept & Theory Theory / Concept ›

Benzene can be prepared from various hydrocarbons through suitable chemical reactions. The conversion depends upon the functional group and the carbon skeleton of the starting compound.

The important reactions involved are:

Polymerisation (Trimerisation) of ethyne.

Dehydrogenation of cyclohexane.

Cyclisation (Aromatisation) of n-hexane.

When the starting compound does not possess six carbon atoms (such as ethene), additional reactions are required to first produce a six-carbon intermediate.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the number of carbon atoms in the starting compound.
Increase the carbon chain wherever necessary.
Carry out cyclisation or dehydrogenation to obtain benzene.
Write the balanced reaction and conditions.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

(i) Conversion of Ethyne into Benzene
Step 1: Trimerisation of Ethyne

Three molecules of ethyne polymerise at high temperature in the presence of a red-hot iron tube or suitable catalyst to form benzene.

\[\boxed{\mathrm{3HC\equiv CH\xrightarrow[\;873\,K\;]{Red\;hot\;Fe}C_6H_6}}\]

This reaction is called the cyclic polymerisation (trimerisation) of ethyne.
(ii) Conversion of Ethene into Benzene
Ethene contains only two carbon atoms. Therefore, it is first converted into ethyne, which is then polymerised to benzene.
Step 1: Convert ethene into 1,2-dibromoethane

\[\mathrm{CH_2=CH_2\xrightarrow{Br_2}BrCH_2CH_2Br}\]
Step 2: Double dehydrohalogenation

\[\mathrm{BrCH_2CH_2Br\xrightarrow{Alcoholic\;KOH}HC\equiv CH}\]
Step 3: Trimerisation of ethyne

\[\mathrm{3HC\equiv CH\xrightarrow[\;873\,K\;]{Red\;hot\;Fe}C_6H_6}\]
Overall Conversion

\[\boxed{\mathrm{CH_2=CH_2\rightarrow BrCH_2CH_2Br\rightarrow HC\equiv CH\rightarrow C_6H_6}}\]
(iii) Conversion of Hexane into Benzene
n-Hexane undergoes cyclisation followed by dehydrogenation in the presence of chromium oxide or platinum catalyst.

\[\boxed{\mathrm{CH_3(CH_2)_4CH_3\xrightarrow[\;773\,K\;]{Cr_2O_3/Al_2O_3}C_6H_6+4H_2}}\]

This process is known as aromatisation or dehydrocyclisation.
Summary Table

Starting Compound	Reaction	Product
Ethyne	Trimerisation	Benzene
Ethene	Bromination → Dehydrohalogenation → Trimerisation	Benzene
n-Hexane	Dehydrocyclisation (Aromatisation)	Benzene

💡 Answer Final Answer ›

(i)

\[\mathrm{3HC\equiv CH\xrightarrow[\;873\,K\;]{Red\;hot\;Fe}C_6H_6}\]

(ii)

\[\mathrm{CH_2=CH_2\xrightarrow{Br_2}BrCH_2CH_2Br\xrightarrow{Alcoholic\;KOH}HC\equiv CH\xrightarrow[\;873\,K\;]{Red\;hot\;Fe}C_6H_6}\]

(iii)

\[\mathrm{CH_3(CH_2)_4CH_3\xrightarrow[\;773\,K\;]{Cr_2O_3/Al_2O_3}C_6H_6+4H_2}\]

🎯 Exam Significance Exam Significance ›

Conversions between aliphatic and aromatic hydrocarbons are frequently asked in CBSE examinations.

Students should remember the names of the reactions such as trimerisation, dehydrohalogenation and aromatisation.

Writing both the reaction conditions and catalysts helps in scoring full marks.

Flow-chart style reaction sequences make conversion questions easier to present.

Significance for JEE / NEET and Other Competitive Examinations

Multi-step organic conversions are common in JEE Main, JEE Advanced and NEET.

Students should understand why ethene cannot be converted directly into benzene and why an intermediate such as ethyne is required.

Knowledge of dehydrocyclisation of alkanes is useful in petroleum chemistry and catalytic reforming.

Catalysts and reaction conditions are frequently tested in competitive examinations.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ethyne forms benzene by cyclic trimerisation.
Ethene is first converted into ethyne before forming benzene.
n-Hexane gives benzene by catalytic dehydrocyclisation.
Red-hot iron catalyses the trimerisation of ethyne.
Chromium oxide/alumina or platinum catalysts are used for aromatisation of n-hexane.

📘 Concept & Theory Theory / Concept ›

Hydrogenation is the addition of hydrogen across a carbon-carbon double bond in the presence of catalysts such as Ni, Pt or Pd.

\[\mathrm{RCH=CHR + H_2 \xrightarrow{Ni/Pt/Pd} RCH_2CH_2R}\]

During hydrogenation, the carbon skeleton remains unchanged. Only the double bond is converted into a single bond.

Therefore, to determine the possible alkenes, identify all possible positions where a C=C double bond can be introduced into the carbon skeleton of 2-methylbutane without changing its carbon framework.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the structure of 2-methylbutane.
Introduce one double bond at every possible position.
Remove duplicate structures arising from symmetry.
Verify that hydrogenation of each alkene gives 2-methylbutane.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Structure of 2-Methylbutane
\[\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}\]
Step 2: Introduce the double bond at all possible positions
(i) 2-Methylbut-1-ene

\[\mathrm{CH_2=C(CH_3)-CH_2-CH_3}\]

Hydrogenation:

\[\mathrm{CH_2=C(CH_3)-CH_2-CH_3\xrightarrow{H_2/Ni}CH_3-CH(CH_3)-CH_2-CH_3}\]
(ii) 3-Methylbut-1-ene

\[\mathrm{CH_2=CH-CH(CH_3)-CH_3}\]

Hydrogenation:

\[\mathrm{CH_2=CH-CH(CH_3)-CH_3\xrightarrow{H_2/Ni}CH_3-CH_2-CH(CH_3)-CH_3}\]

This product is identical to 2-methylbutane after proper numbering.
(iii) 2-Methylbut-2-ene

\[\mathrm{CH_3-C(CH_3)=CH-CH_3}\]

Hydrogenation:

\[\mathrm{CH_3-C(CH_3)=CH-CH_3\xrightarrow{H_2/Ni}CH_3-CH(CH_3)-CH_2-CH_3}\]
Summary Table

Alkene	Structure	Hydrogenation Product
2-Methylbut-1-ene	\[\mathrm{CH_2=C(CH_3)-CH_2-CH_3}\]	\[\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}\]
3-Methylbut-1-ene	\[\mathrm{CH_2=CH-CH(CH_3)-CH_3}\]	\[\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}\]
2-Methylbut-2-ene	\[\mathrm{CH_3-C(CH_3)=CH-CH_3}\]	\[\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}\]

🎯 Exam Significance Exam Significance ›

Questions involving hydrogenation and identification of all possible alkene isomers are frequently asked in CBSE examinations.

Students should remember that hydrogenation changes only the double bond into a single bond without altering the carbon skeleton.

Systematically introducing the double bond at all possible positions helps avoid missing any isomer.

Correct IUPAC names should always accompany the structures.

Significance for JEE / NEET and Other Competitive Examinations

Competitive examinations often ask students to identify all possible alkenes that yield a particular alkane on hydrogenation.

Such questions test knowledge of constitutional isomerism, IUPAC nomenclature and catalytic hydrogenation.

Students should be able to recognise identical structures obtained by renumbering the parent chain.

This concept is also useful in reaction mechanism and organic synthesis problems.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Hydrogenation converts a C=C bond into a C–C bond.
The carbon skeleton remains unchanged during hydrogenation.
All possible positions of the double bond must be considered.
Three different alkenes produce 2-methylbutane on hydrogenation.
Always eliminate duplicate structures generated by symmetry.

📘 Concept & Theory Theory / Concept ›

Electrophilic aromatic substitution (EAS) occurs more readily when the benzene ring has a high electron density.

Substituents attached to the benzene ring influence its reactivity by either donating or withdrawing electrons.

Electron-donating groups (EDGs) activate the benzene ring towards electrophilic substitution.

Electron-withdrawing groups (EWGs) deactivate the benzene ring by decreasing the electron density.

The common effects are:

Substituent	Effect	Influence on EAS
–CH₃	Electron-donating (+I effect)	Activates the ring
–Cl	Strong –I effect, weak +R effect	Deactivates the ring
–NO₂	Strong –I and –R effects	Strongly deactivates the ring

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the substituents present on each benzene ring.
Determine whether each substituent activates or deactivates the ring.
Compare the combined electronic effects.
Arrange the compounds in decreasing order of reactivity towards electrophilic substitution.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

(a) Chlorobenzene, p-Nitrochlorobenzene and 2,4-Dinitrochlorobenzene
Step 1: Chlorobenzene

Chlorine withdraws electrons by the –I effect, making the ring less reactive than benzene.

However, chlorine also donates electron density through resonance (+R effect), so its deactivating effect is only moderate.
Step 2: p-Nitrochlorobenzene

Besides chlorine, the molecule contains one nitro group.

The nitro group is a very strong electron-withdrawing group due to both –I and –R effects.

Therefore, the ring becomes much less reactive than chlorobenzene.
Step 3: 2,4-Dinitrochlorobenzene

This compound contains two nitro groups, both of which strongly withdraw electron density.

The combined effect greatly reduces the electron density of the aromatic ring.

Hence, it is the least reactive towards electrophilic substitution.
Step 4: Decreasing Order of Reactivity
\[\boxed{\mathrm{Chlorobenzene > p\text{-}Nitrochlorobenzene > 2,4\text{-}Dinitrochlorobenzene}}\]
Toluene, p-Methylnitrobenzene and p-Dinitrobenzene
Step 1: Toluene

The methyl group donates electrons through the +I effect and hyperconjugation.

It increases the electron density of the benzene ring and activates it towards electrophilic substitution.
Step 2: p-Methylnitrobenzene

This compound contains both a methyl group (activating) and a nitro group (strongly deactivating).

The deactivating effect of the nitro group is stronger than the activating effect of the methyl group.

Hence, its reactivity is lower than that of toluene.
Step 3: p-Dinitrobenzene

This compound contains two nitro groups, both of which strongly withdraw electron density.

The combined effect greatly reduces the electron density of the aromatic ring.

Hence, it is the least reactive towards electrophilic substitution.
Step 4: Decreasing Order of Reactivity
\[\boxed{\mathrm{Toluene > p\text{-}Methylnitrobenzene > p\text{-}Dinitrobenzene}}\]
Summary Table

Compound	Substituents	Effect on Reactivity
Chlorobenzene	–Cl	Moderately deactivating
p-Nitrochlorobenzene	–Cl, –NO₂	Strongly deactivating
2,4-Dinitrochlorobenzene	–Cl, 2×–NO₂	Very strongly deactivating
Toluene	–CH₃	Activating
p-Methylnitrobenzene	–CH₃, –NO₂	Slightly deactivating (due to –NO₂)
p-Dinitrobenzene	2×–NO₂	Very strongly deactivating

Final Answer
(a)
\[\boxed{\mathrm{Chlorobenzene > p\text{-}Nitrochlorobenzene > 2,4\text{-}Dinitrochlorobenzene}}\]

(b)
\[\boxed{\mathrm{Toluene > p\text{-}Methylnitrobenzene > p\text{-}Dinitrobenzene}}\]
Reason
Electron-donating groups increase the electron density of the aromatic ring and enhance electrophilic substitution.

Electron-withdrawing groups decrease the electron density of the ring and reduce its reactivity.

The nitro group is one of the strongest deactivating substituents because it withdraws electrons through both the inductive (–I) and resonance (–R) effects.

🎯 Exam Significance Exam Significance ›

Questions on the effect of substituents on electrophilic aromatic substitution are frequently asked in CBSE examinations.

Students should remember that the nitro group is a strong deactivating group, whereas the methyl group is an activating group.

Although chlorine is ortho-para directing, it is overall a deactivating group because its –I effect dominates over its +R effect.

Always justify the order using electronic effects rather than memorising it.

Significance for JEE / NEET and Other Competitive Examinations

Electronic effects and aromatic substitution are among the most important topics in JEE Main, JEE Advanced and NEET.

Questions often require arranging substituted benzenes according to their reactivity towards electrophiles or nucleophiles.

Students should understand the combined influence of multiple substituents on the electron density of the aromatic ring.

Mastering inductive and resonance effects is essential for solving advanced organic chemistry problems.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Electrophiles attack electron-rich aromatic rings more readily.
The methyl group activates the benzene ring.
Chlorine is an ortho-para director but overall deactivates the ring.
The nitro group is a powerful deactivating substituent.
Increasing the number of nitro groups progressively decreases the reactivity of benzene towards electrophilic substitution.

📘 Concept & Theory Theory / Concept ›

Nitration of benzene is an electrophilic aromatic substitution (EAS) reaction in which the electrophile is the nitronium ion,

\[\mathrm{NO_2^+}\]

The ease of nitration depends upon the electron density of the benzene ring.

Substituents already present on the ring may either:

Activate the ring by donating electrons.

Deactivate the ring by withdrawing electrons.

An electron-rich ring reacts more rapidly with the electrophile, whereas an electron-deficient ring reacts more slowly.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the substituent present on each aromatic compound.
Determine whether the substituent activates or deactivates the benzene ring.
Compare the electron density of the three compounds.
Arrange them according to their ease of nitration.

✏️ Solution Complete Solution ›

Step-by-step Solution · 5 steps

1. Toluene
Toluene contains a methyl group (–CH₃).

The methyl group donates electron density to the benzene ring through the +I (inductive) effect and hyperconjugation.

This increases the electron density of the aromatic ring.

Therefore, toluene is more reactive than benzene towards electrophilic substitution.
2. Benzene
Benzene has no activating or deactivating substituent.

Its electron density is intermediate.

Hence, its rate of nitration is lower than toluene but higher than m-dinitrobenzene.
3. m-Dinitrobenzene
m-Dinitrobenzene contains two nitro groups (–NO₂).

Each nitro group withdraws electrons strongly through both the –I (inductive) and –R (resonance) effects.

The presence of two nitro groups greatly decreases the electron density of the aromatic ring.

Consequently, the ring becomes highly deactivated towards electrophilic substitution.

Therefore, m-dinitrobenzene undergoes nitration with the greatest difficulty.
Order of Ease of Nitratio
\[\boxed{\mathrm{Toluene > Benzene > m\text{-}Dinitrobenzene }}\]
Reason

Compound	Electronic Effect	Reactivity Towards Nitration
Toluene	+I effect and hyperconjugation increase electron density	Highest
Benzene	No substituent	Intermediate
m-Dinitrobenzene	Two –NO₂ groups strongly withdraw electrons (–I and –R effects)	Lowest

💡 Answer Final Answer ›

Toluene undergoes nitration most easily.

The decreasing order of ease of nitration is

\[\boxed{\mathrm{Toluene>Benzene>m\text{-}Dinitrobenzene}}\]

This is because the methyl group activates the benzene ring by increasing its electron density through the +I effect and hyperconjugation, whereas the nitro groups strongly deactivate the ring by withdrawing electrons through both the –I and –R effects.

🎯 Exam Significance Exam Significance ›

Students should remember that electron-donating groups activate the benzene ring towards electrophilic substitution, while electron-withdrawing groups deactivate it.

The methyl group is an activating substituent, whereas the nitro group is one of the strongest deactivating substituents.

Justifying the order using electronic effects helps in securing full marks.

Always mention both the +I effect of the methyl group and the –I and –R effects of the nitro group.

Significance for JEE / NEET and Other Competitive Examinations

Questions involving the relative reactivity of substituted benzenes are common in JEE Main, JEE Advanced and NEET.

Students should understand how inductive effects, resonance effects and hyperconjugation influence electrophilic aromatic substitution.

Such concepts are also useful for predicting directing effects and the products of aromatic substitution reactions.

Electronic effects form the foundation of advanced organic reaction mechanisms.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Nitration is an electrophilic aromatic substitution reaction.
Toluene is more reactive than benzene because the methyl group donates electrons.
Nitro groups strongly deactivate the aromatic ring.
Electron-rich aromatic rings undergo nitration more readily.
The order of ease of nitration is: Toluene > Benzene > m-Dinitrobenzene.

📘 Concept & Theory Theory / Concept ›

Ethylation of benzene is carried out by the Friedel-Crafts alkylation reaction.

In this reaction, a Lewis acid catalyst is required to generate the electrophile from an alkyl halide.

A Lewis acid is an electron-pair acceptor.

It accepts a lone pair of electrons from the halogen atom of the alkyl halide, thereby producing a positively charged electrophile (or electrophile-like species) that attacks the benzene ring.

Although anhydrous aluminium chloride (AlCl₃) is the most commonly used catalyst, several other Lewis acids can also catalyse the reaction.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the catalyst used in Friedel-Crafts alkylation.
Identify another compound that behaves as a Lewis acid.
Write the reaction using the alternative catalyst.

✏️ Solution Complete Solution ›

Step-by-step Solution · 2 steps

One commonly used Lewis acid other than anhydrous aluminium chloride is anhydrous ferric chloride (FeCl₃).

It acts as an electron-pair acceptor and helps in generating the electrophile required for electrophilic aromatic substitution.

The ethylation reaction may be represented as:

\[\mathrm{C_6H_6 + C_2H_5Cl\xrightarrow{FeCl_3}C_6H_5C_2H_5 + HCl}\]
Other Lewis acids that can also be used in Friedel-Crafts reactions include:

• Anhydrous ferric chloride (FeCl₃)

• Boron trifluoride (BF₃)

• Ferric bromide (FeBr₃)

• Zinc chloride (ZnCl₂)

However, FeCl₃ is the standard alternative generally expected in NCERT examinations.

💡 Answer Final Answer ›

Anhydrous ferric chloride (FeCl₃) can be used as a Lewis acid catalyst in place of anhydrous aluminium chloride during the ethylation of benzene.

\[\boxed{\mathrm{C_6H_6 + C_2H_5Cl\xrightarrow{FeCl_3}C_6H_5C_2H_5 + HCl}}\]

🎯 Exam Significance Exam Significance ›

Students should remember that Friedel-Crafts alkylation requires a Lewis acid catalyst.

Anhydrous ferric chloride (FeCl₃) is the most commonly accepted alternative to anhydrous aluminium chloride in NCERT examinations.

Knowing the role of the catalyst in generating the electrophile helps in answering conceptual questions.

This is a short, one-mark question that is frequently asked in board examinations.

Significance for JEE / NEET and Other Competitive Examinations

Competitive examinations often test the role of Lewis acids in electrophilic aromatic substitution.

Students should know the commonly used Lewis acid catalysts such as AlCl₃, FeCl₃, BF₃ and FeBr₃.

Understanding the mechanism of electrophile generation is essential for solving reaction mechanism questions.

This concept is widely used in Friedel-Crafts alkylation and acylation reactions.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Ethylation of benzene is a Friedel-Crafts alkylation reaction.
A Lewis acid catalyst is required to generate the electrophile.
Anhydrous FeCl₃ is a suitable alternative to anhydrous AlCl₃.
Lewis acids are electron-pair acceptors.
Lewis acid catalysts facilitate electrophilic aromatic substitution.

📘 Concept & Theory Theory / Concept ›

Wurtz reaction is a coupling reaction in which two molecules of an alkyl halide react with sodium metal in the presence of dry ether to form a higher alkane.

The general reaction is:

\[\boxed{\mathrm{2R-X + 2Na\overset{Dry\ Ether}{\longrightarrow}R-R + 2NaX}}\]

If the same alkyl halide is used, a symmetrical alkane containing an even number of carbon atoms is obtained.

To prepare an alkane containing an odd number of carbon atoms, two different alkyl halides must be used. However, this leads to the formation of multiple coupling products, making the reaction unsuitable.

🗺️ Solution Roadmap Step-by-step Plan ›

Recall the Wurtz reaction.
Understand why odd-carbon alkanes require two different alkyl halides.
Write all the possible coupling reactions.
Explain why a mixture of products is obtained.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Step 1: Preparation of an Odd-Carbon Alkane
Suppose we want to prepare propane (C₃H₈).

This requires the reaction between methyl bromide (CH₃Br) and ethyl bromide (C₂H₅Br).

Ideally, the desired reaction is

\[\mathrm{CH_3Br + C_2H_5Br + 2Na\overset{Dry\ Ether}{\longrightarrow}CH_3CH_2CH_3 + 2NaBr}\]
Step 2: Formation of Side Products
Since two different alkyl halides are present, sodium cannot distinguish between them.

As a result, three different coupling reactions occur simultaneously.
(i) Desired Product

\[\mathrm{CH_3Br + C_2H_5Br\longrightarrow CH_3CH_2CH_3}\]
(ii) Coupling of Two Methyl Groups

\[\mathrm{2CH_3Br\longrightarrow CH_3CH_3}\]

Product: Ethane
(iii) Coupling of Two Ethyl Groups

\[\mathrm{2C_2H_5Br\longrightarrow CH_3CH_2CH_2CH_3}\]

Product: n-Butane

Step 3: Result

Instead of obtaining only propane, a mixture of three alkanes is formed.

Reactants	Product Obtained
\(\mathrm{CH_3Br + CH_3Br}\)	Ethane (C₂H₆)
\(\mathrm{CH_3Br + C_2H_5Br}\)	Propane (C₃H₈)
\(\mathrm{C_2H_5Br + C_2H_5Br}\)	n-Butane (C₄H₁₀)

Because a mixture of products is formed, separation becomes difficult and the yield of the desired odd-carbon alkane is very low.

💡 Answer Final Answer ›

Wurtz reaction is not preferred for the preparation of alkanes containing an odd number of carbon atoms because two different alkyl halides must be used, which undergo both self-coupling and cross-coupling reactions.

This results in a mixture of alkanes rather than a single product.

For example, when methyl bromide and ethyl bromide are treated with sodium in dry ether, the products obtained are:

\[\boxed{\begin{aligned}\mathrm{2CH_3Br} &\mathrm{\rightarrow C_2H_6}\\[4pt]\mathrm{CH_3Br + C_2H_5Br} &\mathrm{\rightarrow C_3H_8}\\[4pt]\mathrm{2C_2H_5Br} &\mathrm{\rightarrow C_4H_{10}}\end{aligned}}\]

Hence, Wurtz reaction is suitable mainly for the preparation of symmetrical alkanes containing an even number of carbon atoms.

🎯 Exam Significance Exam Significance ›

This is one of the most frequently asked conceptual questions from the chapter on hydrocarbons.

Students should remember that Wurtz reaction is best suited for preparing symmetrical alkanes.

Always explain that the use of two different alkyl halides leads to both self-coupling and cross-coupling.

Illustrating the answer with the preparation of propane helps in securing full marks.

Significance for JEE / NEET and Other Competitive Examinations

Questions on the limitations of Wurtz reaction are common in JEE Main, JEE Advanced and NEET.

Students should be able to predict all possible products formed when two different alkyl halides are used.

This concept is also useful in solving questions on organic reaction mechanisms and synthetic strategies.

Understanding coupling reactions helps distinguish between Wurtz reaction and other alkane preparation methods.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Wurtz reaction couples two alkyl halides using sodium in dry ether.
Using identical alkyl halides gives symmetrical alkanes with an even number of carbon atoms.
Using different alkyl halides produces multiple coupling products.
Odd-carbon alkanes are obtained in poor yield due to product mixtures.
Therefore, Wurtz reaction is not preferred for preparing odd-carbon alkanes.

Carbon Atom	Type of Carbon	Hydrogen Atoms Attached
C-1	1°	3
C-2	2°	2
C-3	4°	0
C-4	1°	3
C-5	1°	3
C-6	2°	2
C-7	3°	1
C-8	1°	3
C-9	1°	3

Carbon Atom	Type of Carbon	Hydrogen Atoms Attached
C-1	1°	3
C-2	2°	2
C-3	4°	0
C-4	1°	3
C-5	1°	3
C-6	2°	2
C-7	3°	1
C-8	1°	3
C-9	1°	3

Hydrocarbons

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE, NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Comparison of Boiling Points

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Reason (Markovnikov's Rule)

Initiation Step

Propagation Step 1

Propagation Step 2

Termination Step

Why Are Different Products Formed?

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Decreasing Order of Acidic Behaviour

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Step 1: Chlorobenzene

Step 2: p-Nitrochlorobenzene

Step 3: 2,4-Dinitrochlorobenzene

Step 1: Toluene

Step 2: p-Methylnitrobenzene

Step 3: p-Dinitrobenzene

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Significance for JEE / NEET and Other Competitive Examinations

Chapter Complete!

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Hydrocarbons — Learning Resources

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Carbon Atom	Type of Carbon	Hydrogen Atoms Attached
C-1	1°	3
C-2	2°	2
C-3	4°	0
C-4	1°	3
C-5	1°	3
C-6	2°	2
C-7	3°	1
C-8	1°	3
C-9	1°	3