The above diagram represents electron sharing between atoms forming a covalent bond.
Chapter 4 · CBSE Class X
📘 Definition
📌 Note
Carbon’s Electronic Configuration
🎨 SVG Diagram
Octet Completion through Sharing
🗂️ Types / Category
Types of Covalent Bonds
Single Covalent Bond
Formed by sharing one pair of electrons.
Example: Methane \(\ce{CH4}\)
Double Covalent Bond
Formed by sharing two pairs of electrons.
Example: Ethene \(\ce{C2H4}\)
Triple Covalent Bond
Formed by sharing three pairs of electrons.
Example: Ethyne \(\ce{C2H2}\)
Coordinate (Dative) Bond
Both electrons in the shared pair are donated by the same atom.
Example: \(\ce{NH4+}\)
🎨 SVG Diagram
Visualization of Bond Types
🏷️ Properties
Properties of Covalent Compounds
Properties of Covalent Compounds
Melting & Boiling Points
Generally low for covalent molecules; only weak intermolecular forces need to be overcome, not the strong covalent bonds inside the molecules.
Electrical Conductivity
Usually non-conducting in both solid and liquid states because there are no mobile ions or delocalized electrons.
Solubility Profile
Tend to dissolve in non-polar or weakly polar organic solvents (like benzene or ether), but are mostly insoluble in polar solvents such as water, following the rule "like dissolves like".
Critical Exceptions
Graphite conducts electricity due to delocalized pi-electrons between layers, and diamond and silica have very high melting and boiling points because they form giant covalent network structures.
Nature of Force
Atoms within each molecule are held by strong covalent bonds, while the molecules themselves are held together by weak intermolecular forces.
🔢 Formula
Key Concept (Octet Rule)
✏️ Example
Why does carbon form covalent compounds?
Roadmap:
Electronic configuration → energy consideration → bonding type.
Solution:
Carbon has 4 valence electrons and needs 4 more to complete its octet. Gaining 4 electrons to form C⁴⁻ or losing 4 electrons to form C⁴⁺ both require very high energy and would be unstable. Instead, carbon shares its 4 valence electrons with other atoms, forming strong covalent bonds and achieving a stable octet at low energy cost.
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound does not conduct electricity and has low melting point.
Question: Identify bonding type and justify.
Answer: Covalent compound, due to absence of free ions and weak intermolecular forces.
🗒️ Definitiion
The versatile nature of carbon refers to its unique ability to form a vast number of compounds due to tetravalency, catenation, multiple bonding, and isomerism. This makes carbon the backbone of organic chemistry.
📘 Definition
Electronic Basis of Versatility
🏷️ Properties
Core Reasons for Carbon’s Versatility
Core Reasons for Carbon’s Versatility
Tetravalency
Carbon has 4 valence electrons and can form four covalent bonds with the same or different atoms, allowing highly diverse and complex molecules.
Catenation
Carbon atoms can link to one another forming long straight chains, branched chains, and rings, due to strong C–C bonds and small atomic size.
Multiple Bond Formation
Carbon forms single \(\mathrm{C{-}C}\), double \(\mathrm{C{=}C}\), and triple \(\mathrm{C{≡}C}\) bonds, increasing structural variety and reactivity patterns.
Isomerism
Isomerism is the phenomenon where different compounds have the same molecular formula but different structural arrangements or spatial orientations.
Strong Covalent Bonds
Due to its small atomic size and suitable electronegativity, carbon forms strong C–C bonds that give organic compounds high thermal and chemical stability.
🎨 SVG Diagram
Visualization of Catenation
Carbon chains can be straight, branched, or cyclic, explaining enormous compound diversity.
✏️ Example
1
Example
Ethane \(\ce{C2H6}\) → single bond (alkane)
2
Example
Ethene \(\ce{C2H4}\) → double bond (alkene)
3
Example
Ethyne \(\ce{C2H2}\) → triple bond (alkyne)
4
Example
Butane and Isobutane \(\ce{C4H10}\)
🔢 Formula
General Formula (Hydrocarbons)
Alkanes:
\[
C_nH_{2n+2}
\]
Alkenes:
\[
C_nH_{2n}
\]
Alkynes:
\[
C_nH_{2n-2}
\]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound has long carbon chains and shows different structures with same formula.
Question: Identify two properties responsible.
Answer: Catenation and isomerism.
📐 Derivation
Derivation Insight (Alkanes)
Each carbon forms 4 bonds. In a chain:
Internal carbons bond with 2 carbons → remaining 2 bonds with H
Terminal carbons bond with 1 carbon → remaining 3 bonds with H
Result: \[ C_nH_{2n+2} \]
📘 Definition
🎨 SVG Diagram
Classification Overview
Single bond → Saturated | Double/Triple bond → Unsaturated
🏷️ Properties
Saturated Carbon Compounds (Alkanes)
Saturated Carbon Compounds (Alkanes)
Bond Type
Consist only of single \(\ce{C-C}\) and \(\ce{C-H}\) bonds; all bonds are strong sigma (\(\sigma\)) bonds with no double or triple bonds.
General Formula
For straight‑chain alkanes: \(\ce{C_nH_{2n+2}}\), where \(n\) is the number of carbon atoms.
Reactivity
Relatively less reactive compared to other hydrocarbons because all bonds are strong sigma bonds and the molecules are non‑polar or only weakly polar.
Type of Reaction
Undergo mainly substitution reactions, especially with halogens in the presence of sunlight, where hydrogen atoms are replaced by halogen atoms.
Stability
Highly stable due to strong \(\sigma\) bonds and the absence of reactive multiple bonds; this makes alkanes relatively inert under normal conditions.
✏️ Example
1
Example
Methane \(\ce{CH4}\): Simplest alkane, tetrahedral structure.
2
Example
Ethane \(\ce{C2H6}\): Two carbons linked by single bond.
3
Example
Butane \(\ce{C4H10}\): Exists in isomeric forms.
📘 Definition
Unsaturated Carbon Compounds
Alkenes
Unsaturated hydrocarbons containing at least one carbon–carbon double bond (\(\ce{C=C}\)); general formula \(\ce{C_nH_{2n}}\) for straight‑chain alkenes with one double bond.
Alkynes
Unsaturated hydrocarbons containing at least one carbon–carbon triple bond (\(\ce{C#C}\)); general formula \(\ce{C_nH_{2n-2}}\) for straight‑chain alkynes with one triple bond.
Properties
Bond Type
Unsaturated hydrocarbons containing at least one carbon–carbon double or triple bond; double bonds are \(\ce{C=C}\) and triple bonds are \(\ce{C#C}\).
Reactivity
More reactive than alkanes because the pi bonds in double or triple bonds are electron‑rich sites that readily undergo attack by electrophiles.
Reaction Type
Undergo mainly addition reactions (e.g., with hydrogen, halogens, or hydrogen halides), where atoms add across the multiple bond without removing any atom.
Stability
Less stable than alkanes under typical conditions because pi bonds are weaker and more easily broken than sigma bonds, even though the total bond energy of multiple bonds is higher.
✏️ Example
4
Example
Ethene \(\ce{C2H4}\): Double bond compound
5
Example
Ethyne \(\ce{C2H2}\): Triple bond compound
🔁 Reaction
Important Reaction (Hydrogenation)
Unsaturated hydrocarbons can be converted into saturated hydrocarbons by adding hydrogen across the multiple bond in a reaction called catalytic hydrogenation.
\[ \ce{C2H4 + H2 ->[{\text{Ni} \atop \Delta}] C2H6} \]
This addition of hydrogen converts ethene (unsaturated) into ethane (saturated); the catalyst used is usually finely divided nickel (Ni), and the reaction is carried out at about 300 °C.
📊 Comparison Table
Satuarted Vs Unsaturated
| Property | Saturated | Unsaturated |
|---|---|---|
| Bond Type | Single | Double/Triple |
| Formula | \(C_nH_{2n+2}\) | \(C_nH_{2n}\), \(C_nH_{2n-2}\) |
| Reactivity | Low | High |
| Test | No change with bromine water | Decolorizes bromine water |
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A hydrocarbon decolorizes bromine water instantly.
Question: Identify type and justify.
Answer: Unsaturated hydrocarbon due to presence of multiple bond.
📘 Definition
💡 Concept
Conceptual Foundation
🗂️ Types / Category
Straight Chain Compounds
In straight‑chain compounds, carbon atoms are linked in a continuous, unbranched sequence where each carbon is bonded to at most two other carbon atoms.
Example: Butane \(\ce{C4H10}\), which has a linear carbon backbone.
Branched Chain Compounds
In branched‑chain compounds, one or more carbon atoms form side branches (alkyl groups) attached to the main carbon chain.
Example: Isobutane \(\ce{C4H10}\) (2‑methylpropane), a branched isomer of butane.
Link with Isomerism
Straight‑chain and branched‑chain compounds with the same molecular formula but different arrangements of the carbon chain are called chain isomers.
This is a key example of structural isomerism in organic chemistry.
Cyclic (Ring) Compounds
In cyclic or ring‑compounds, carbon atoms join end‑to‑end to form closed rings such as cycloalkanes, aromatics, and heterocycles.
Example: Cyclohexane \(\ce{C6H12}\); these cyclic structures are highly important in both aliphatic and aromatic chemistry.
🎨 SVG Diagram
🎨 SVG Diagram
🗂️ Types / Category
Types of Carbon Rings
Aromatic Rings
Aromatic rings contain a closed ring (often six‑membered) with alternating single and double bonds, giving resonance and extra stability.
Example: Benzene \(\ce{C6H6}\), the classic aromatic hydrocarbon.
Alicyclic (Aliphatic Rings)
Alicyclic rings are closed carbon rings that behave like open‑chain aliphatic compounds; they contain only single bonds and no aromatic resonance.
Example: Cyclohexane \(\ce{C6H12}\), a saturated alicyclic hydrocarbon.
Heterocyclic Rings
Heterocyclic rings contain one or more atoms other than carbon (commonly N, O, or S) in the ring, while still forming a closed cyclic structure.
Example: Pyridine \(\ce{C5H5N}\), a nitrogen‑containing heterocyclic aromatic compound.
🔢 Formula
General Formula Insight
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
Two compounds have same molecular formula \(\ce{C4H10}\) but different structures.
Question: Identify concept and types.
Answer: Chain isomerism → straight chain (butane) and branched chain (isobutane)
📘 Definition
💡 Concept
Conceptual Understanding
🗒️ Importance of Functional Groups
Importance of Functional Groups
- Classifies organic compounds into families
- Determines chemical reactivity
- Predicts physical properties (boiling point, solubility)
- Forms basis of nomenclature in organic chemistry
🎨 SVG Diagram
Basic Structural Visualization
Hydrocarbon chain (R) + Functional Group (FG) → Organic Compound
📊 Comparison Table
Common Functional Groups
| Group | Functional Group | General Formula | Example | Structure |
|---|---|---|---|---|
| Alcohol | \(-OH\) | \(R-OH\) | Ethanol \(\ce{C2H5OH}\) | |
| Aldehyde | \(-CHO\) | \(R-CHO\) | Ethanal \(\ce{CH3CHO}\) | |
| Ketone | \(>C=O\) | \(R-CO-R'\) | Propanone \(\ce{CH3COCH3}\) | |
| Carboxylic Acid | \(-COOH\) | \(R-COOH\) | Ethanoic acid \(\ce{CH3COOH}\) | |
| Haloalkane | \(-X\) | \(R-X\) | Chloroethane \(\ce{C2H5Cl}\) |
🏷️ Properties
Reactivity & Behaviour
Reactivity & Behaviour
Alcohols
React with sodium to liberate hydrogen gas; contain the hydroxyl \(-OH\) group; are polar and may show hydrogen bonding.
Carboxylic acids
Show acidic nature; turn blue litmus red; react with bases and carbonates; contain the carboxyl \(-COOH\) group.
Aldehydes and ketones
Contain the carbonyl \(>C=O\) group; undergo important oxidation and addition reactions; aldehydes are generally more easily oxidised than ketones.
Haloalkanes
Contain a halogen atom \(-X\); commonly undergo substitution reactions; are useful intermediates in organic synthesis.
✏️ Example
Identify the functional group in \(\ce{CH3COOH}\) and name the class of compound.
Find the characteristic group attached to the carbon chain → note the presence of \(\ce{-COOH}\) → classify the compound accordingly.
The functional group is \(\ce{-COOH}\), so \(\ce{CH3COOH}\) belongs to the carboxylic acids.
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound reacts with sodium and releases hydrogen gas.
Question: Identify functional group.
Answer: Alcohol group (–OH)
📐 Derivation
Concept Insight (Why Properties Change)
Functional groups alter:
- Electron distribution
- Polarity of molecule
- Intermolecular forces
Hence, physical and chemical properties change significantly.
📘 Definition
A homologous series is a family of organic compounds having the same functional group,
similar chemical properties, and where each successive member differs by a \(-CH_2-\) unit.
General Representation
CH4 → C2H6 → C3H8 → C4H10
Note
Difference between consecutive members:
\[
+CH_2 \quad (\text{mass difference } = 14u)
\]
Each step adds one \(-CH_2-\) unit to the carbon chain.
🏷️ Properties
Key Characteristics
Key Characteristics
Same Functional Group
All members have identical functional group, hence similar chemical properties (e.g., all alcohols react similarly)
Example: Alkanes (- single bond)
Example: Alkanes (- single bond)
Common General Formula
Follow same general formula
Example: Alkanes \[ \ce{C_nH_{2n+2}} \] (n ≥ 1)
Example: Alkanes \[ \ce{C_nH_{2n+2}} \] (n ≥ 1)
Successive Difference
Consecutive members differ by \[ -\ce{CH2}- \] unit (14 u)
Example: CH₄ (methane) to C₂H₆ (ethane)
Example: CH₄ (methane) to C₂H₆ (ethane)
Gradation in Physical Properties
Regular increase in boiling point, melting point, density with molecular mass
(Due to increased van der Waals forces)
(Due to increased van der Waals forces)
Regular Increase in Molecular Mass
+14 u per successive member (due to \( \ce{CH2} \))
Similar Chemical Reactions
Undergo same type of reactions due to same functional group
(Exam Q: Why do homologues show similar chemistry?)
(Exam Q: Why do homologues show similar chemistry?)
✏️ Example
Alkane Series
1
Example
\(\mathrm{\underset{\text{Methane}}{\ce{CH4}}}\) → \(\mathrm{\underset{Ethane}{\ce{C2H6}}}\) → \(\mathrm{\underset{Propane}{\ce{C3H8}}}\) → \(\mathrm{\underset{Butane}{\ce{C4H10}}}\)
Alcohol Series
2
Example
\(\mathrm{\underset{\text{Methanol}}{\ce{CH3OH}}}\) → \(\mathrm{\underset{\text{Ethanol}}{\ce{C2H5OH}}}\) → \(\mathrm{\underset{\text{Prpanol}}{\ce{C3H7OH}}}\)
🏷️ Properties
Trend in Properties
Trend in Properties
Boiling Point
Increases gradually with molecular mass
\[ \text{n-pentane} < \text{n-hexane} < \text{n-heptane} \]
(↑ van der Waals forces)
\[ \text{n-pentane} < \text{n-hexane} < \text{n-heptane} \]
(↑ van der Waals forces)
Melting Point
Increases with chain length
\[ \ce{CH3(CH2)4CH3} < \ce{CH3(CH2)6CH3} \]
(Exam note: Even carbon # affects regularity)
\[ \ce{CH3(CH2)4CH3} < \ce{CH3(CH2)6CH3} \]
(Exam note: Even carbon # affects regularity)
Solubility in Water
Decreases down the series
\[ \ce{CH3OH > C2H5OH > C3H7OH} \]
(Hydrocarbon chain ↑, polarity ↓)
\[ \ce{CH3OH > C2H5OH > C3H7OH} \]
(Hydrocarbon chain ↑, polarity ↓)
Density
Increases slightly with chain length
\[ 0.79 \, \text{g/mL (pentane)} \to 0.83 \, \text{g/mL (octane)} \]
\[ 0.79 \, \text{g/mL (pentane)} \to 0.83 \, \text{g/mL (octane)} \]
🔢 Formula
General formula of alkanes
✏️ Example
Why do members of a homologous series show similar chemical properties?
All members contain the same functional group which determines their chemical reactivity.
Example: All alcohols (-OH group) undergo similar reactions: \[ \ce{R-OH + HBr -> R-Br + H2O} \] \[ \ce{CH3OH, C2H5OH, C3H7OH} \text{ all react similarly}\]
Example: All alcohols (-OH group) undergo similar reactions: \[ \ce{R-OH + HBr -> R-Br + H2O} \] \[ \ce{CH3OH, C2H5OH, C3H7OH} \text{ all react similarly}\]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
Two compounds differ by 14u in molecular mass but have same functional group.
Question: Identify the concept.
Answer: Homologous series.
📘 Definition
🌟 Importance
Why Nomenclature is Important
🔢 Formula
Basic Structure of IUPAC Name
Prefix + Root Word + Primary Suffix + Secondary Suffix
Step-by-Step Rules
1. Identify Longest Carbon Chain
This determines the root name.
| No. of Carbon | Root |
|---|---|
| 1 | Meth |
| 2 | Eth |
| 3 | Prop |
| 4 | But |
| 5 | Pent |
| 6 | Hex |
| 7 | Hept |
| 8 | Oct |
| 9 | Non |
| 10 | Dec |
2. Identify Bond Type
| Bond | Suffix |
|---|---|
| Single | -ane |
| Double | -ene |
| Triple | -yne |
3. Identify Functional Group
| Group | Suffix |
|---|---|
| -OH | -ol |
| -CHO | -al |
| >C=O | -one |
| -COOH | -oic acid |
4. Number the Chain
Number from the end that gives: lowest number to functional group → multiple bond → substituent
5. Identify Substituents
Common substituents: \(\ce{-CH3}\) = Methyl, \(\ce{-C2H5}\) = Ethyl
Always choose numbering that gives lowest possible locants.
Priority Order
- Functional group (highest priority)
-
Multiple bonds:
- Numbering → give lowest possible number (double/triple whichever gets lowest locant)
- Naming → double bond (-ene) is written before triple bond (-yne)
-
Substituents (Prefixes):
- Written in alphabetical order
- Ignore prefixes like di-, tri-, tetra- while arranging
- Example:
\[ \text{Ethyl + Methyl → Ethyl comes first} \] Correct name: 3-ethyl-2-methylpentane
✏️ Example
1
Example
Structure: \(\ce{CH3-CH2-OH}\)
Step 1: Longest chain = 2 carbons → Eth
Step 2: Single bond → ane
Step 3: Functional group = –OH → ol
Final Name: Ethanol
2
Example
Structure:
\(\ce{CH3-CH2-CH(CH3)-CH2-OH}\)
Step 1: Identify Longest Chain
Main chain = 4 carbons (butane)
Count from OH carbon: C1(OH)-C2-C3-C4 → butan
Step 2: Functional Group Priority
–OH (alcohol) = highest priority
Suffix: -ol, numbered from OH end → butan-1-ol
Step 3: Identify & Number Substituents
Methyl (–CH₃) group at carbon 2
Name: 2-methylbutan-1-ol
Final IUPAC Name
2-methylbutan-1-ol
Exam Note: Always number chain from functional group end giving lowest numbers to substituents (Rule #1 in CBSE marking scheme)
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound contains both –OH and double bond.
Question: Which gets priority?
Answer: –OH group (functional group priority)
📌 Note
Overview
📌 Note
Combustion
Combustion is the reaction of carbon compounds with oxygen producing carbon dioxide, water, and energy.
- \(\ce{C + O2 -> CO2}\) + Heat + Light
- \(\ce{CH4 + 2O2 -> CO2 + 2H2O}\) + Heat
- \(\ce{C2H5OH + 3O2 -> 2CO2 + 3H2O}\) + Heat
Observation
Key Observations
- Saturated compounds → clean blue flame
- Unsaturated compounds → yellow sooty flame
- Incomplete combustion produces CO (poisonous)
📌 Note
Oxidation
Oxidation involves addition of oxygen or removal of hydrogen from a compound.
Conversion of alcohol to acid:
⚗️ Oxidation✔ Balanced
Oxidation of Ethanol to Ethanoic Acid
\[ \ce{CH3CH2OH \xrightarrow[\text{Heat}]{KMnO_4 / K_2Cr_2O_7} CH3COOH} \]
Conditions: Acidified KMnO₄ or K₂Cr₂O₇, Heat
ΔH = −1370 kJ/mol
Oxidising Agents
- Alkaline \(\ce{KMnO4}\)
- Acidified \(\ce{K2Cr2O7}\)
📌 Note
Addition Reaction
Unsaturated hydrocarbons undergo addition reactions where atoms are added across double/triple bonds.
⚗️ Addition✔ Balanced
Hydrogenation of Ethene to Ethane
\[ \ce{C2H4 + H2 ->[Ni][\text{300°C, 30 atm}] C2H6} \]
Conditions: Ni catalyst, 300°C, 30 atm pressure
ΔH = -137 kJ/mol
- Called hydrogenation
- Used in conversion of vegetable oils to saturated fats
- Catalyst: Nickel/Palladium/Platinum
📌 Note
Substitution Reaction
In saturated hydrocarbons, one atom (usually hydrogen) is replaced by another atom.
⚗️ Substitution✔ Balanced
Free Radical Chlorination of Methane
\[ \ce{CH4 + Cl2 ->[hv][\text{Sunlight}] CH3Cl + HCl} \]
Conditions: UV light / Sunlight, Room temperature
ΔH = -104 kJ/mol
Key Features
- Occurs in presence of sunlight (UV light)
- Typical for alkanes
- Chain reaction mechanism
📊 Comparison Table
Reaction Types of Carbon Compounds (NCERT Class 10)
| Reaction Type | Reactants | Products | Conditions/Catalyst | Example Equation |
|---|---|---|---|---|
| Combustion | All C-compounds (CxHy) | CO₂ + H₂O + Heat/Energy | O₂, Heat | CH₄ + 2O₂ → CO₂ + 2H₂O |
| Oxidation | Alcohols (R-OH) | Carboxylic acids (R-COOH) | K₂Cr₂O₇/KMnO₄, Heat | C₂H₅OH → CH₃COOH |
| Addition | Unsaturated (C=C, C≡C) | Saturated compounds | Ni/Pt/Pd catalyst | C₂H₄ + H₂ → C₂H₆ |
| Substitution | Saturated (Alkanes) | Halogenated compounds | UV light/Sunlight | CH₄ + Cl₂ → CH₃Cl + HCl |
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound reacts with hydrogen in presence of nickel catalyst.
Question: Identify reaction type.
Answer: Addition reaction (hydrogenation)
📘 Definition
💡 Concept
🎨 SVG Diagram
🔎 Key Fact
Characteristics of Catalysts
🗂️ Types / Category
Types of Catalysts
Positive Catalyst
Speeds up reaction rate without being consumed
Example: Fe in Haber's process (N₂ + 3H₂ ⇌ 2NH₃)
Example: Fe in Haber's process (N₂ + 3H₂ ⇌ 2NH₃)
Negative Catalyst
Slows down reaction rate
Example: Glycerine in H₂O₂ decomposition (2H₂O₂ → 2H₂O + O₂)
Example: Glycerine in H₂O₂ decomposition (2H₂O₂ → 2H₂O + O₂)
Enzyme (Biocatalyst)
Protein catalysts in living cells, highly specific
Example: Ptyalin in saliva (starch → maltose)
Example: Ptyalin in saliva (starch → maltose)
Auto Catalyst
One product acts as catalyst
Example: NO in lead chamber process
Example: NO in lead chamber process
Promoter
Increases activity of catalyst
Example: Molybdenum in Haber's process
Example: Molybdenum in Haber's process
✏️ Example
Hydrogenation (Addition Reaction)
\[
\ce{C2H4 + H2 ->[Ni][300°C, 30 atm] C2H6}
\]
Ni catalyst converts unsaturated alkene to saturated alkaneH₂O₂ Decomposition (Catalase reaction)
\[
\ce{2H2O2 ->[MnO2][Room temp] 2H2O + O2 ^}
\]
MnO₂ lowers activation energy, black solid catalystBiological Catalysts (Enzymes)
\[
\ce{(C6H10O5)_n + (n)H2O ->[Ptyalin][Saliva] n C6H12O6}
\]
Starch → Maltose (amylase enzyme, highly specific)Haber's Process (Industrial)
\[
\ce{N2 + 3H2 <=> [Fe][450°C, 200 atm] 2NH3}
\]
Iron catalyst with Mo promoter, equilibrium reaction
🌟 Importance
Importance in Daily Life
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A reaction occurs faster in presence of nickel without nickel being consumed.
Question: Identify the role of nickel.
Answer: Catalyst.
📌 Note
Ethanol (Ethyl Alcohol)
🎨 SVG Diagram
Structure of Ethanol
Functional group –OH determines its chemical properties.
🏷️ Properties
Physical Properties
Properties
Physical State
Colorless, volatile liquid at room temp (b.p. 78°C)
Odour
Characteristic wine-like smell, burning taste
Flammability
Highly flammable (flash point 13°C)
Solubility
∞ miscible with water & organic solvents (H-bonding)
Density
0.789 g/mL (less dense than water)
🏷️ Properties
Chemical Propertiwes
Properties
Combustion
\(\scriptsize\ce{C2H5OH + 3O2 ->[heat] 2CO2 + 3H2O ^[+ Heat]}\)
Blue flame, complete oxidation
Blue flame, complete oxidation
Reaction with Sodium
\(\scriptsize\ce{2C2H5OH + 2Na -> 2C2H5ONa + H2 ^}\}\)
Hydrogen gas evolves (characteristic test)
Hydrogen gas evolves (characteristic test)
Oxidation (Primary alcohol)
\(\scriptsize\ce{C2H5OH ->[K2Cr2O7/H2SO4][heat] CH3COOH}\}\)
Orange → Green (chromic acid test)
Orange → Green (chromic acid test)
Dehydration (Intramolecular)
\(\scriptsize\ce{C2H5OH ->[conc.H2SO4][443K] C2H4 + H2O}\)
Ethene formation (elephant trunk smell)
Ethene formation (elephant trunk smell)
Dehydration (Intermolecular)
\(\scriptsize\ce{2C2H5OH ->[conc.H2SO4][443K] C2H5OC2H5 + H2O}\)
Diethyl ether (sweet smell)
Diethyl ether (sweet smell)
📌 Note
Uses of Ethanol
📘 Definition
Ethanoic Acid (Acetic Acid)
Ethanoic acid is a carboxylic acid with functional group \(-COOH\).
Molecular formula: \[ \ce{CH3COOH} \]
Structure of Ethanoic Acid
🏷️ Properties
Chemical Properties
Properties
Nature
Weak organic acid (pKa = 4.76, 1% ionized)
Taste & Smell
Sour taste, pungent vinegar odour
Physical State
Freezes at 290 K (17°C) → glacial acetic acid (ice-like solid)
Solubility
∞ miscible in water (forms H-bonds)
Density
1.05 g/mL (heavier than water)
Household Use
5–8% solution in water = vinegar (preservative)
🗂️ Types / Category
Chemical Reactions
Esterification
\[\scriptsize\begin{aligned}\ce{CH3COOH + C2H5OH \quad\\<=>[H^+] CH3COOC2H5 + H2O}\end{aligned}\]
Saponification
\[\scriptsize\begin{aligned}\ce{CH3COOC2H5 + NaOH \quad\\-> C2H5OH + CH3COONa} \end{aligned}\]
Reaction with Base
\[\scriptsize\begin{aligned}\ce{CH3COOH + NaOH \quad\\-> CH3COONa + H2O} \end{aligned}\]
Reaction with Carbonates
\[\scriptsize\begin{aligned}\ce{2CH3COOH + Na2CO3 \quad\\-> 2CH3COONa + CO2 + H2O} \end{aligned}\]
Reaction with Hydrogencarbonate
\[\scriptsize\begin{aligned}\ce{CH3COOH + NaHCO3 \quad\\-> CH3COONa + CO2 + H2O} \end{aligned}\]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A compound reacts with sodium carbonate releasing CO₂.
Question: Identify compound type.
Answer: Carboxylic acid (Ethanoic acid)
📘 Definition
Soaps
🔁 Reaction
Saponification Reaction
Formation of soap by reaction of fats/oils with a strong base:
\[
\ce{Fat/Oil + NaOH -> Soap + Glycerol}
\]
- NaOH: Hard soaps
- KOH: Soft soaps
🗒️ Svg Iamge
Structure of Soap Molecule
- Hydrophilic head: Water-attracting (–COO⁻Na⁺)
- Hydrophobic tail: Oil-attracting hydrocarbon chain
🎨 SVG Diagram
Cleansing Action of Soap (Micelle Formation)
Soap molecules arrange themselves into micelles where:
- Hydrophobic tails trap grease inside
- Hydrophilic heads face water
- Dirt is lifted and washed away
📌 Note
Limitations of Soaps
📘 Definition
Detergents
🎨 SVG Diagram
Structure of Detergents
- Hydrophobic tail: Long hydrocarbon chain
- Hydrophilic head: Sulfate/Sulfonate group
📊 Comparison Table
Difference: Soap vs Detergent
| Property | Soap | Detergent |
|---|---|---|
| Raw Materials | Natural (vegetable oils/animal fats) | Synthetic (petroleum derivatives) |
| Structure | Sodium salt of fatty acids | Sodium salt of alkyl benzene sulphonic acid |
| Hard Water | Ineffective (forms insoluble scum) | Effective (no scum formation) |
| pH | Alkaline (~9-10) | Neutral to slightly alkaline |
| Biodegradability | Readily biodegradable | Some non-biodegradable (branched chain) |
| Cost | Higher | Lower |
| Mechanism | Reduces surface tension + precipitates Ca²⁺/Mg²⁺ | Reduces surface tension (no ppt.) |
⚡ Exam Tip
⚠️ Warning
Common Mistake
📋 Case Study
A cleansing agent works effectively in hard water without forming scum.
Question: Identify type.
Answer: Detergent.
🗒️ Important Points
Important Points
Important Points (Quick Revision – Carbon and Its Compounds)
Core Concepts
- Carbon is the basis of life due to its ability to form a vast number of stable compounds.
- Tetravalency (valency = 4) allows carbon to form four covalent bonds.
- Catenation enables carbon to form long chains, branches, and rings.
Bonding & Structure
- Covalent bonds involve sharing of electrons.
- Carbon forms bonds with H, O, N, S, Cl and itself.
- Carbon forms single, double, and triple bonds.
- Structures include straight chains, branched chains, and cyclic compounds.
Homologous Series & Functional Groups
- Homologous series differ by \(-CH_2\) unit and have similar chemical properties.
- Functional groups determine reactivity and properties.
- Important groups: –OH, –CHO, –COOH, >C=O.
Chemical Properties
- Carbon compounds undergo combustion, oxidation, addition, and substitution.
- Saturated compounds → substitution reactions.
- Unsaturated compounds → addition reactions.
Important Compounds
- Ethanol: Alcohol, used as fuel, solvent, antiseptic.
- Ethanoic Acid: Weak acid, present in vinegar.
- Esterification forms esters (fruity smell).
Soaps and Detergents
- Soap molecules contain hydrophilic head + hydrophobic tail.
- Cleaning occurs via micelle formation.
- Soaps are ineffective in hard water; detergents are effective.
Exam Focus (High Yield)
-
>Reason-based questions on tetravalency & catenation
>Functional group identification
>Reaction types (addition vs substitution)
>Esterification & saponification reactions
>Soap cleansing mechanism (diagram-based)
Ultra-Quick Revision Table
| Concept | Key Idea |
|---|---|
| Tetravalency | Carbon forms 4 bonds |
| Catenation | Carbon forms chains |
| Functional Group | Determines properties |
| Homologous Series | Difference of CH₂ |
| Soap Action | Micelle formation |
NCERT Class X · Chapter 4
Carbon & Its Compounds
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Core Concepts
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Carbon (C) has atomic number 6 and electronic configuration 2, 4 — meaning it has 4 valence electrons. This gives carbon two extraordinary properties:
- Tetravalency: Carbon can form four covalent bonds simultaneously with other atoms (C, H, O, N, S, halogens etc.), creating enormously diverse molecules.
- Catenation: Carbon atoms can bond with other carbon atoms to form long chains, branched chains, and closed rings — a property almost unique to carbon.
Why Carbon?
Carbon–Carbon bond is very stable due to its intermediate electronegativity (2.5). Silicon also catenates but Si–Si bonds are weaker, explaining why life is carbon-based, not silicon-based.
The combination of catenation + tetravalency results in over 10 million known organic compounds — more than all other elements combined.
Hydrocarbons are compounds of carbon and hydrogen only.
Alkanes: CnH2n+2 | Alkenes: CnH2n | Alkynes: CnH2n−2
General formulas for saturated & unsaturated hydrocarbons
- Alkanes (saturated): Only single C–C bonds. CH₄ (methane), C₂H₆ (ethane), C₃H₈ (propane).
- Alkenes (unsaturated): One C=C double bond. C₂H₄ (ethene/ethylene), C₃H₆ (propene).
- Alkynes (unsaturated): One C≡C triple bond. C₂H₂ (ethyne/acetylene), C₃H₄ (propyne).
Combustion Clue
Saturated hydrocarbons burn with a clean (blue) flame; unsaturated ones burn with a sooty (yellow) flame due to incomplete combustion from high C:H ratio.
A homologous series is a group of organic compounds sharing the same general formula and functional group, with each successive member differing by –CH₂– (14u).
Characteristics:
- Same general molecular formula
- Same functional group → same chemical properties
- Physical properties (MP, BP) change gradually with increasing carbon chain length
- Each member is called a homologue
Example Series
Alkane series: CH₄ → C₂H₆ → C₃H₈ → C₄H₁₀ — each step adds exactly CH₂ (= 14u) to molecular mass.
A functional group is an atom or group of atoms that determines the characteristic chemical properties of an organic compound.
| Functional Group | Formula | Class of Compound |
|---|---|---|
| Hydroxyl | –OH | Alcohol |
| Aldehyde | –CHO | Aldehyde |
| Ketone | –CO– | Ketone |
| Carboxyl | –COOH | Carboxylic Acid |
| Halogen | –X (F,Cl,Br,I) | Haloalkane |
Carbon compounds undergo three main types of reactions:
1. Combustion: Organic compounds burn in oxygen to produce CO₂ and H₂O with release of heat and light.
CH₄ + 2O₂ → CO₂ + 2H₂O + Heat
2. Addition Reaction: Unsaturated compounds (alkenes/alkynes) add atoms across the double or triple bond to become saturated.
CH₂=CH₂ + H₂ → CH₃–CH₃ (Ni catalyst, heat)
3. Substitution Reaction: In saturated compounds, one H atom is replaced by another atom (e.g., halogen) in presence of sunlight or UV light.
CH₄ + Cl₂ → CH₃Cl + HCl (sunlight)
Memory Key
ADD to unsaturated, SUBSTITUTE in saturated. Unsaturated → becomes saturated via addition. Saturated → can only substitute.
Ethanol (C₂H₅OH) — the common alcohol. It is a colourless liquid, soluble in water in all proportions, with BP 78°C.
- Reaction with Na: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑ (sodium ethoxide + hydrogen gas)
- Reaction with conc. H₂SO₄ (170°C): Dehydration → C₂H₄ (ethene)
- Oxidation: CH₃CH₂OH → CH₃COOH (acetic acid) using alkaline KMnO₄ or acidified K₂Cr₂O₇
- Esterification: Reacts with acetic acid + conc. H₂SO₄ to form ethyl acetate (ester)
Denatured Alcohol
Industrial ethanol is denatured (methanol added) to make it poisonous and unfit for drinking. Methanol causes blindness and death even in small doses — do NOT confuse with ethanol.
Uses: Beverages, solvent, antiseptic, fuel (petrol blends), manufacture of perfumes, medicines.
Ethanoic acid (CH₃COOH) — also called acetic acid. Vinegar contains 5–8% acetic acid in water. Pure acetic acid is called glacial acetic acid (BP 118°C; freezes at 16.6°C).
- Reaction with NaHCO₃: CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
- Reaction with Na: 2CH₃COOH + 2Na → 2CH₃COONa + H₂↑
- Esterification: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄ catalyst)
- Saponification: Esters are hydrolysed by NaOH → sodium salt + alcohol
Weak Acid Test
Acetic acid is a weak acid — it reacts with NaHCO₃ to release CO₂, proving it is acidic. However, it does NOT affect litmus as strongly as strong acids like HCl.
Soaps are sodium or potassium salts of long-chain fatty acids (e.g., sodium stearate: C₁₇H₃₅COONa). Made by saponification of oils/fats with NaOH.
Cleansing Action: Soap molecules have a hydrophilic head (–COO⁻Na⁺) that attracts water, and a hydrophobic tail (long carbon chain) that repels water but attracts grease. When soap is added to dirty water, the hydrophobic tails cluster around oily dirt while hydrophilic heads face outward into water, forming micelles. These micelles are washed away.
Soap vs Hard Water
Soaps do NOT work in hard water because Ca²⁺ and Mg²⁺ ions react with soap to form an insoluble scum (calcium/magnesium stearate). Detergents work even in hard water because their calcium salts are soluble.
Detergents are sulphonate or sulphate salts of long-chain hydrocarbons. They are non-biodegradable (branched chain) or biodegradable (straight chain) and cause water pollution.
IUPAC names are assigned systematically using prefixes based on carbon chain length:
1C = Meth | 2C = Eth | 3C = Prop | 4C = But | 5C = Pent | 6C = Hex
Suffixes: –ane (single bond), –ene (double bond), –yne (triple bond), –ol (alcohol), –al (aldehyde), –one (ketone), –oic acid (carboxylic acid)
Naming Rule
Choose the longest carbon chain containing the functional group. Number from the end closest to the functional group. Functional group gets lowest possible locant.
Carbon exists in several allotropic forms:
- Diamond: Each C bonded to 4 others in a tetrahedral network → hardest natural substance, non-conductor of electricity (no free electrons), high melting point.
- Graphite: Each C bonded to 3 others in hexagonal layers. One free electron per C → good conductor. Layers can slide → lubricant. Used in pencil lead, electrodes.
- Fullerenes (C₆₀): Spherical cage of carbon atoms (Buckyballs). 60 C in soccer-ball shape. Used in drug delivery, superconductors.
Diamond vs Graphite
Both are pure carbon — difference is in bonding arrangement. Diamond: sp³ hybridized (4 bonds, 3D). Graphite: sp² hybridized (3 bonds, 2D layers).
Oxidising agents (alkaline KMnO₄, acidified K₂Cr₂O₇) convert alcohols to aldehydes/ketones and then to carboxylic acids. Reducing agents (H₂ + Ni catalyst) convert alkenes to alkanes and aldehydes to alcohols.
Ethanol →[Oxidation]→ Ethanal →[Oxidation]→ Ethanoic Acid
CH₃CH₂OH → CH₃CHO → CH₃COOH
CH₃CH₂OH → CH₃CHO → CH₃COOH
Oxidation State Trick
As carbon gets more O atoms bonded to it (and loses H), it is oxidised. Going from alcohol → acid = oxidation. Going acid → alcohol = reduction.
Isomers are compounds with the same molecular formula but different structural arrangements. They have different physical properties but may show similar chemical properties if the functional group is the same.
Example: C₄H₁₀ has two isomers:
n-Butane: CH₃–CH₂–CH₂–CH₃ (straight chain)
iso-Butane: CH₃–CH(CH₃)–CH₃ (branched chain)
iso-Butane: CH₃–CH(CH₃)–CH₃ (branched chain)
Exam Alert
Isomers have the same molecular formula but DIFFERENT structural formulas. Don't confuse with allotropes (same element, different forms) or isotopes (same element, different mass numbers).
⚗️
Complete Formula Reference
All key formulas, reactions, and general expressions — organised by category.
General Molecular Formulas
Alkanes (Saturated)
CnH2n+2
CH₄ (n=1), C₂H₆ (n=2), C₃H₈ (n=3)…
Alkenes (1 double bond)
CnH2n
C₂H₄ ethene, C₃H₆ propene…
Alkynes (1 triple bond)
CnH2n−2
C₂H₂ ethyne, C₃H₄ propyne…
Alcohols
CnH2n+1OH
CH₃OH methanol, C₂H₅OH ethanol…
Carboxylic Acids
CnH2n+1COOH
HCOOH formic, CH₃COOH acetic…
Homologue Mass Difference
Δm = 14 u (CH₂)
Each successive member differs by 14 u
Key Chemical Reactions
Complete Combustion
CH₄ + 2O₂ → CO₂ + 2H₂O
Methane combustion — releases heat & light
Hydrogenation (Addition)
CH₂=CH₂ + H₂ →Ni,Δ CH₃CH₃
Ethene → ethane (unsaturated → saturated)
Halogenation of Methane
CH₄ + Cl₂ →hν CH₃Cl + HCl
Substitution in sunlight
Oxidation of Ethanol
C₂H₅OH →[O] CH₃COOH
Alk. KMnO₄ or acid K₂Cr₂O₇
Esterification
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Conc. H₂SO₄ catalyst, reversible
Saponification
Ester + NaOH → Soap + Alcohol
Hydrolysis of ester; soap making
Ethanol + Sodium
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
Vigorous reaction; H₂ gas evolved
Acetic acid + NaHCO₃
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
Confirms acidic nature; CO₂ brisk effervescence
Dehydration of Ethanol
C₂H₅OH →conc.H₂SO₄, 170°C C₂H₄ + H₂O
Produces ethene (alkene)
Halogenation of Ethene
CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br
Bromine water decolourises → test for unsaturation
Soap Micelle Formation
Hydrophobic tail → grease
Hydrophilic head → water
→ Micelle formed
Hydrophilic head → water
→ Micelle formed
Surrounds oily dirt; washed away
Soap + Hard Water
2RCOO⁻Na⁺ + CaCl₂ → (RCOO)₂Ca↓ + 2NaCl
Insoluble scum; soap ineffective
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Concept-Building Q&A Bank
Rich, original questions with full step-by-step solutions — no repetition of textbook questions. Click to reveal solutions.
Q1
Carbon is said to show tetravalency. Explain what this means and why carbon does not form ionic bonds despite this.
Conceptual
Full Solution
1
Tetravalency defined: Carbon has 4 valence electrons (configuration 2,4). To achieve stability (noble gas config of 8), it needs 4 more electrons — so it forms exactly 4 covalent bonds. This is called tetravalency.
2
Why not ionic? To form C⁴⁺, carbon must lose 4 electrons — requiring enormous energy (very high ionisation energy). To form C⁴⁻, it must gain 4 electrons — the nuclear charge cannot hold 4 extra electrons stably in the outer shell.
3
Solution: The energy required for ionic bond formation is too large to be compensated by lattice energy. Therefore, carbon always forms covalent bonds by sharing electrons equally — which is energetically favourable.
✅ Carbon forms 4 covalent bonds (tetravalency). Forming C⁴⁺ or C⁴⁻ ions is energetically impossible, so carbon always uses electron sharing, not transfer.
Q2
Bromine water is used to test for unsaturation. Explain with a chemical equation why a saturated compound does NOT decolourise bromine water.
Application
Full Solution
1
How unsaturated compounds decolourise Br₂ water: Alkenes have a C=C double bond. Bromine adds across this double bond in an addition reaction: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br. The reddish-brown Br₂ is consumed → solution decolourises.
2
Why saturated compounds fail: Alkanes (e.g., ethane: CH₃CH₃) have only single C–C and C–H bonds. There is no π bond available for addition reaction. Br₂ cannot simply substitute in the absence of sunlight/UV. Without UV, no reaction occurs.
3
Test logic: Saturated → Br₂ water remains orange-red (no decolourisation). Unsaturated → Br₂ water turns colourless immediately.
✅ Saturated compounds have no π bond for addition, so they cannot react with Br₂ water at room temperature → no decolourisation.
Q3
Two liquids A and B both have molecular formula C₂H₆O. A reacts with Na to produce H₂ gas; B does not react with Na. Identify A and B and explain the structural difference.
Analysis
Full Solution
1
Molecular formula C₂H₆O has two isomers: Ethanol (CH₃CH₂OH) and Dimethyl ether (CH₃–O–CH₃).
2
Identifying A: A reacts with Na → releases H₂. This shows presence of active –OH (hydroxyl) group. So A = Ethanol (CH₃CH₂OH). Reaction: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
3
Identifying B: B does not react with Na → no active H on O. The oxygen is flanked by two carbon groups (ether linkage –O–). So B = Dimethyl ether (CH₃OCH₃).
4
Structural difference: In ethanol, –OH is the terminal group (O bonded to H and C). In dimethyl ether, O is bonded to two carbons only — no O–H bond present.
✅ A = Ethanol (has –OH, reacts with Na). B = Dimethyl ether (no –OH, inert to Na). Both have same formula C₂H₆O — they are isomers.
Q4
Why does a candle flame appear yellow and sooty, while an LPG flame appears blue? Relate your answer to carbon content.
HOTS
Full Solution
1
Candle wax: Is a long-chain hydrocarbon (high C:H ratio, e.g. C₂₅H₅₂). When burned, the carbon content is very high relative to hydrogen.
2
Incomplete combustion: High carbon content means insufficient oxygen reaches the inner flame to burn all carbon atoms. Unburned carbon particles get heated → emit yellow light (incandescence). These particles form soot (carbon black).
3
LPG (mainly propane/butane): These are gaseous fuels with lower C:H ratio and can mix more efficiently with oxygen. Complete combustion occurs → CO₂ + H₂O produced, no free carbon particles → clean blue flame.
4
General rule: Saturated hydrocarbons with lower C:H ratio → blue (complete combustion). Unsaturated / high C:H ratio → yellow sooty flame (incomplete combustion).
✅ Candle wax has high C:H ratio → incomplete combustion → hot carbon particles → yellow sooty flame. LPG has lower C:H ratio → complete combustion → blue flame.
Q5
Why is acetic acid a weaker acid than hydrochloric acid even though both donate H⁺? Use the concept of ionisation.
Conceptual
Full Solution
1
HCl ionisation: HCl is a strong acid — it ionises completely in water: HCl → H⁺ + Cl⁻ (100%). All molecules donate H⁺.
2
Acetic acid ionisation: CH₃COOH is a weak acid — it ionises only partially in water (less than 1% at typical concentrations): CH₃COOH ⇌ CH₃COO⁻ + H⁺. Most molecules remain undissociated.
3
Why partial ionisation? The O–H bond in the carboxyl group of acetic acid is stabilised by the adjacent C=O (resonance), making it harder to break. The bond between O and H in HCl is weaker and more polar → complete dissociation.
✅ HCl fully ionises → more H⁺ → stronger acid. Acetic acid partially ionises → fewer H⁺ → weaker acid. Strength ∝ degree of ionisation.
Q6
Explain the cleansing action of soap using the concept of hydrophilic and hydrophobic ends. Why does soap form a scum in hard water?
Application
Full Solution
1
Soap structure: Soap molecule has two parts — a long non-polar hydrocarbon tail (hydrophobic = water-fearing, oil-loving) and an ionic –COO⁻Na⁺ head (hydrophilic = water-loving).
2
Micelle formation: When soap is dissolved in water with oily dirt, the hydrophobic tails cluster inward around the oil droplet. The hydrophilic heads face outward into water. This spherical cluster is called a micelle.
3
Washing away: Micelles are suspended in water (emulsification). During washing/rinsing, micelles carrying dirt are removed with water.
4
Hard water problem: Hard water contains Ca²⁺ and Mg²⁺ ions. These react with soap: 2RCOO⁻Na⁺ + Ca²⁺ → (RCOO)₂Ca↓ + 2Na⁺. The calcium/magnesium stearate formed is insoluble → precipitates as scum (grey sticky deposit).
✅ Soap works via micelles: hydrophobic tails trap oil, hydrophilic heads face water. In hard water, Ca²⁺/Mg²⁺ form insoluble scum with soap → soap is wasted before cleaning.
Q7
Diamond is the hardest substance yet graphite is a lubricant — both are pure carbon. Reconcile this contradiction using bonding.
HOTS
Full Solution
1
Diamond structure: Each carbon atom is bonded to 4 other carbon atoms in a 3D tetrahedral network. This forms an extremely rigid, interlocked lattice with no weak points → hardest substance. No free electrons → non-conductor.
2
Graphite structure: Each carbon atom bonds to 3 others in a 2D hexagonal layer (sp² hybridised). The fourth electron is delocalised — this makes graphite a good electrical conductor.
3
Why graphite is a lubricant: Graphite layers are held together only by weak van der Waals forces. These layers can slide over each other very easily → slippery texture → acts as dry lubricant.
✅ Same element, different bonding → completely different properties. Diamond: 3D covalent network (hard). Graphite: 2D layers with weak interlayer forces (lubricant + conductor). This is allotropy.
Q8
A student performed esterification and observed a fruity smell. On adding excess NaOH and heating, the smell disappeared. Explain what happened in each step.
Analysis
Full Solution
1
Step 1 — Esterification: Ethanol + acetic acid (with conc. H₂SO₄ catalyst) form ethyl acetate ester: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Ethyl acetate has a sweet/fruity smell.
2
Why fruity smell? Esters are responsible for the characteristic fruity fragrances of fruits and flowers. They are used in artificial flavourings and perfumes.
3
Step 2 — Saponification: NaOH hydrolyses the ester: CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH. The ester is broken back into sodium acetate (odourless salt) and ethanol. The fruity smell disappears.
✅ Esterification creates fruity-smelling ester. Saponification (with NaOH) reverses this, destroying the ester → smell disappears. This is the reverse of esterification, used in soap-making.
Q9
What is the IUPAC name of CH₃–CH(OH)–CH₂–CH₃? Identify its functional group and homologous series.
Conceptual
Full Solution
1
Count carbons: CH₃–CH(OH)–CH₂–CH₃ has 4 carbon atoms in the longest chain. Parent chain = but (4C).
2
Identify functional group: –OH group is present → class = alcohol → suffix = –ol.
3
Number the chain: Number from the end closest to –OH. Starting from CH₃ end: C1–C2(OH)–C3–C4. The –OH is on C2. IUPAC name = Butan-2-ol.
4
Homologous series: Alcohol series. General formula CnH2n+1OH. This compound: n=4, C₄H₉OH ✓.
✅ IUPAC name: Butan-2-ol. Functional group: –OH (Hydroxyl). Belongs to: Alcohol homologous series.
Q10
Silicon (Si) is also tetravalent and can catenate. Why then does Si not form as many compounds as C? Why is life carbon-based?
HOTS
Full Solution
1
Si–Si bond vs C–C bond stability: C–C bond is very strong (347 kJ/mol) due to effective orbital overlap between small carbon atoms. Si–Si bond is weaker (226 kJ/mol) because Si atoms are larger → less effective overlap → Si–Si chains are less stable.
2
Si–O bond vs C–O: Si forms very strong Si–O bonds (silicates). In the presence of oxygen (atmosphere), Si tends to react with O to form SiO₂ rather than maintaining Si–Si chains. C–C bonds persist in air under normal conditions.
3
Multiple bonds: Carbon forms stable C=C and C≡C (π bonds). Silicon rarely forms stable Si=Si bonds due to its larger atomic size → less π-bond diversity.
4
Biological conclusion: Carbon's unique combination of stable catenation + multiple bond formation + tetravalency + compatibility with O, N, H → builds complex biomolecules (DNA, proteins, carbohydrates). Silicon cannot match this versatility → life is carbon-based.
✅ Si has weaker catenation (Si–Si < C–C), rarely forms π bonds, and preferentially forms Si–O bonds. Carbon's superior bonding versatility makes it the basis of all known life.
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Common Mistakes to Avoid
These are the mistakes most students make in exams — understand them deeply to avoid losing marks.
✗
Confusing addition & substitution reactions
Students often write addition reactions for saturated compounds (alkanes) or substitution for unsaturated compounds (alkenes). Remember: addition occurs in unsaturated (double/triple bond available), substitution in saturated (no π bond).
✓ Fix: Add to unsaturated (π bond used up). Substitute in saturated (H replaced by halogen with UV light).
✗
Writing wrong general formula for homologous series
Many students write CnH2n for alkanes (it's actually for alkenes) or forget the +2 in alkane formula.
✓ Fix: Alkane = CnH2n+2. Alkene = CnH2n. Alkyne = CnH2n-2. Each loses 2H per degree of unsaturation.
✗
Saying ethanol reacts with NaHCO₃
Ethanol does NOT react with NaHCO₃ (it is not acidic enough). Only carboxylic acids react with NaHCO₃ to produce CO₂ gas.
✓ Fix: Ethanol reacts with Na (active metal) but not NaHCO₃. Acetic acid reacts with BOTH Na and NaHCO₃. Use this to distinguish the two in lab/exam.
✗
Confusing esterification with saponification
Esterification = acid + alcohol → ester + water. Saponification = ester + NaOH → soap (salt) + alcohol. Students often mix up the reactants and products.
✓ Fix: Ester-ification MAKES esters. Saponi-fication BREAKS esters (hydrolysis). Saponification is the reverse of esterification with a base.
✗
Calling diamond a conductor (or graphite an insulator)
Diamond = non-conductor (all 4 electrons in bonds, no free electrons). Graphite = conductor (1 delocalised electron per carbon atom is free to move).
✓ Fix: Diamond: 4 bonds, 0 free electrons → insulator. Graphite: 3 bonds + 1 free electron per C → conductor. Graphite used as electrode for this reason.
✗
Forgetting catalyst and conditions in reactions
Writing CH₂=CH₂ + H₂ → CH₃CH₃ without Ni catalyst and heat — this is an incomplete equation and loses marks. Similarly forgetting conc. H₂SO₄ in esterification, or sunlight in halogenation.
✓ Fix: Always write conditions above the arrow. Hydrogenation: Ni/Pt, Δ. Halogenation (substitution): hν (UV/sunlight). Esterification: conc. H₂SO₄, Δ.
✗
Saying detergents are biodegradable
Branched-chain detergents are NOT biodegradable (bacteria cannot break the branched chain). Only straight-chain detergents are biodegradable. Confusion often arises here.
✓ Fix: Branched detergents = non-biodegradable (cause river foam). Straight-chain detergents = biodegradable (better for environment). Soaps are always biodegradable.
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Exam Tips & Tricks
Proven strategies to score maximum marks in carbon compounds questions.
01
The Bromine Water Test
To test for unsaturation: Add bromine water. If it decolourises → unsaturated compound (alkene/alkyne). If it stays orange → saturated compound. Always mention "bromine water decolourises" as the observation.
02
Homologous Series — Spotting Pattern
If a series member has formula C₄H₁₀, to find the 6th member: count from C1. If starting alkane is CH₄ (n=1), 6th member has n=6 → C₆H₁₄. General formula saves time.
03
IUPAC Naming — 5-Step Method
1. Find longest chain containing functional group. 2. Number from closest end to FG. 3. Identify FG and assign suffix. 4. Add substituent names as prefixes. 5. Write complete name. Never skip the functional group position number for chains of 3+ C.
04
Distinguishing Ethanol & Acetic Acid
Use NaHCO₃ — only acetic acid gives CO₂ effervescence. Use smell — ethanol has a pleasant smell; acetic acid has a pungent vinegar smell. For lab: litmus test (acid turns blue litmus red; ethanol does not).
05
Micelle — Draw & Explain
In 3-mark "cleansing action" questions, always draw a rough labelled micelle (circle with tails pointing in, heads pointing out). This gets you full credit. Mention: "dirt trapped in hydrophobic core, removed with water."
06
Memory Trick — Catenation
Catenation = "chain formation." Remember: C of Catenation = C of Carbon. Only carbon does it so extensively because C–C bond is very strong and stable under normal conditions.
07
Esterification is Reversible
The ⇌ sign is critical in esterification. In exams write: acid + alcohol ⇌ ester + water (conc. H₂SO₄ as catalyst AND dehydrating agent). H₂SO₄ shifts equilibrium forward by removing water.
08
Flame Colour & Saturation
Blue flame = complete combustion = saturated/lower carbon content. Sooty/yellow flame = incomplete combustion = unsaturated OR high carbon:hydrogen ratio. Candle (sooty) vs LPG (blue) = classic exam example.
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Carbon & Its Compounds Class 10 Notes (NCERT)
Carbon & Its Compounds Class 10 Notes (NCERT) — Complete Notes & Solutions · academia-aeternum.com
Carbon is one of the most remarkable elements on Earth, forming the basis of all living and many non-living substances around us. NCERT Class X Science Chapter 4, Carbon and Its Compounds, helps students explore why carbon can create such a vast variety of molecules—from simple gases to complex structures like proteins, plastics, fuels, and medicines. The chapter introduces the concepts of catenation, bonding, functional groups, homologous series, and versatile carbon compounds such as ethanol,…
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