Hydrocarbons — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Hydrocarbons" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Hydrocarbons" — 6 important questions with detailed answers for CBSE board exam prep…
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NCERT Solutions for Class 11 Chemistry — Hydrocarbons. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Hydrocarbons — All 6 Questions Solved
Q1. What are hydrocarbons? Classify them with examples and properties.
Hydrocarbons: Organic compounds containing only carbon and hydrogen atoms.
Formula: CnHm
Classification 1: Based on Saturation
1. Saturated Hydrocarbons (Alkanes)
- General formula: CnH(2n+2)
- Only single bonds between carbons
- Non-polar, hydrophobic
- Low reactivity, combustible
- Example: CH3CH2CH3 (Propane)
2. Unsaturated Hydrocarbons
a) Alkenes (C=C double bond)
Formula: CnH(2n)
Reactive, undergo addition reactions
Example: CH2=CHCH3 (Propene)
b) Alkynes (C≡C triple bond)
Formula: CnH(2n-2)
Highly reactive, acidic H
Example: HC≡CCH3 (Propyne)
c) Aromatic hydrocarbons (benzene ring)
Formula: CnH(2n-6) for monosubstituted benzene
Resonant structure, aromatic stability
Example: C6H6 (Benzene), C6H5CH3 (Toluene)
Classification 2: Based on Chain Structure
1. Aliphatic Hydrocarbons:
- Straight or branched chains
- Acyclic (open-chain)
- Examples: Alkanes, alkenes, alkynes
2. Aromatic Hydrocarbons:
- Contain benzene ring
- Cyclic with delocalized π electrons
- Examples: Benzene, naphthalene, toluene
Properties Comparison:
Alkanes Alkenes Alkynes Aromatic
Bond type Single C=C double C≡C triple Aromatic
Reactivity Low High High Moderate
Color Colorless Colorless Colorless Colorless
Odor Weak Pleasant Pleasant Pleasant
Chemical Properties:
- Alkanes: Combustion, halogenation, isomerization
- Alkenes: Addition (H2, HX, X2, H2O), oxidation, polymerization
- Alkynes: Addition reactions, acidic H in terminal alkynes
- Aromatic: Substitution (electrophilic), stability, resonance
Q2. Explain cracking of hydrocarbons. What is the difference between thermal and catalytic cracking?
Cracking: Process of breaking C-C bonds in long-chain hydrocarbons to produce shorter-chain hydrocarbons.
Conditions:
- Temperature: 400-900°C
- Purpose: Convert heavy fractions into lighter, more useful products
- Products: Alkanes (LPG, petrol) + Alkenes (ethene, propene)
Thermal Cracking:
- Temperature: 800-900°C
- Pressure: 10-20 atm
- No catalyst
- Mechanism: Free radical chain reaction
1. Initiation: C-C bond breaks into radicals
2. Propagation: Radical attack creates new radicals
3. Termination: Radicals combine
- Produces: Long-chain alkenes, alkanes, carbon black
- Products: Less specific (mixture)
- Disadvantage: Produces unwanted carbon black
Catalylic Cracking:
- Temperature: 400-500°C
- Pressure: 1-2 atm (atmospheric)
- Catalyst: Zeolites or silica-alumina
- Mechanism: Carbocation intermediate on catalyst surface
1. Alkane attaches to catalyst
2. C-C bond breaks to form carbocation
3. Carbocation rearranges (isomerization)
4. Product desorbs
- Products: More specific (branched alkanes, aromatic compounds)
- Advantage: More controlled, produces useful branched hydrocarbons
Comparison:
Thermal Catalytic
Temperature 800-900°C 400-500°C
Pressure High (10-20) Low (1-2)
Catalyst None Zeolites
Mechanism Free radical Carbocation
Products Unsaturated Branched alkanes
Yield of alkenes High Lower
Carbon black High Minimal
Cost Lower Higher (catalyst)
Example:
C16H34 (hexadecane) thermal cracking:
C16H34 -> C10H22 + C6H12 (nonene + hexane mix)
C16H34 catalytic cracking:
C16H34 -> C8H18 (2,2,4-trimethylpentane) + C8H16 (branched octene)
Q3. What are aromatic hydrocarbons? Explain structure and aromaticity of benzene (C6H6).
Aromatic Hydrocarbons: Cyclic, unsaturated hydrocarbons with special stability and properties.
Benzene (C6H6): Most common aromatic hydrocarbon
Structure:
- 6 carbon atoms in hexagonal ring
- Each C bonded to 1 H atom
- Each C undergoes sp2 hybridization
- Three unhybridized p-orbitals on each C
- These 6 p-orbitals overlap to form π electron cloud
Resonance Structures (Kekulé structures):
H C-C H
| || |
H-C C C-H
|| ||
H-C C-C-H
H
Actual Structure: Resonance hybrid
- All six C-C bonds equivalent (bond order = 1.5)
- Bond length: 1.40 Å (between single 1.54 Å and double 1.34 Å)
- Delocalized π electron cloud above and below ring
Aromaticity Criteria (Hückel's Rule):
1. Cyclic conjugated system
2. Planar structure
3. (4n + 2) π electrons, where n = 0, 1, 2, 3...
4. Aromatic compounds have special stability
For Benzene:
- 6 π electrons = 4(1) + 2 = follows Hückel's rule
- Aromatic stability = 150 kJ/mol
Resonance Energy (Stabilization Energy):
- Experimental ΔHc(benzene) = -3169 kJ/mol
- Calculated ΔHc(Kekulé) = -3220 kJ/mol
- Resonance energy = -3220 - (-3169) = -51 kJ/mol
Properties of Benzene:
1. Highly stable (does not undergo addition easily)
2. Planar hexagonal structure
3. Undergoes substitution (electrophilic aromatic substitution)
4. Does not decolorize Br2/CCl4 (unlike alkenes)
5. Delocalized π electrons give unusual stability
Comparison with Cyclohexatriene:
If benzene were Kekulé structure (C=C and C-C alternating):
- Would have 3 double bonds
- Should show alkene-like reactivity
- Should easily add Br2, HBr
- Would be less stable
But benzene is 150 kJ/mol more stable than Kekulé structure predicts.
This extra stability = resonance/aromatic stability.
Q4. Write reactions of benzene showing substitution patterns. Explain ortho-, meta-, and para-directors.
Benzene Reactions: Substitution (electrophilic aromatic substitution), not addition
1. Halogenation (Bromination)
C6H6 + Br2 ---FeBr3--> C6H5Br + HBr
catalyst
Bromobenzene (monobromination)
With excess Br2:
C6H5Br + Br2 ---FeBr3--> C6H4Br2 + HBr (o-, p-, m-isomers possible)
2. Nitration
C6H6 + HNO3 ---H2SO4--> C6H5NO2 + H2O
conc., 60°C
Nitrobenzene
3. Friedel-Crafts Alkylation
C6H6 + RCl ---AlCl3--> C6H5R + HCl
Alkylbenzene (toluene if R = CH3)
4. Friedel-Crafts Acylation
C6H6 + RCOCl ---AlCl3--> C6H5COR + HCl
Acylbenzene (acetophenone if R = CH3)
5. Sulfonation
C6H6 + H2SO4 ---conc., heat--> C6H5SO3H + H2O
100°C
Benzenesulfonic acid
6. Oxidation (harsh conditions only)
C6H6 + 3O2 ---Δ, catalyt--> 6CO2 + 3H2O
Complete combustion
Director Effects in Disubstituted Benzene:
Electron-Donating Groups (Activating, Ortho/Para-Directors):
-OH, -OR, -NH2, -NHR, -NR2, -CH3, -R (alkyl)
Mechanism: Donate electron density to ring, increase reactivity
Site: Positions 2,6 (ortho) and 4 (para) most reactive
Example: Toluene (C6H5-CH3)
CH3 donates electrons -> Ortho/Para-directing
Electron-Withdrawing Groups (Deactivating, Meta-Directors):
-NO2, -CN, -CHO, -COR, -COOH, -COOR, -SO3H, -N(CH3)3+, -F, -Cl, -Br, -I (halogens)
Mechanism: Withdraw electron density from ring, decrease reactivity
Site: Position 3 (meta) most reactive (ortho and para positions deactivated)
Example: Nitrobenzene (C6H5-NO2)
NO2 withdraws electrons -> Meta-directing
Ortho and para products are very minor
Reactivity Comparison:
1. Activated benzenes (with electron-donating groups)
Faster than benzene, undergo substitution easily
Example: Aniline (C6H5NH2) very reactive
2. Deactivated benzenes (with electron-withdrawing groups)
Slower than benzene, requires stronger conditions
Example: Nitrobenzene less reactive, requires harsher catalyst
3. Halogens: Deactivating but ORTHO/PARA-directing
Exception to typical trends
Reason: F, Cl, Br, I withdraw electrons (deactivate)
But π-donor through lone pairs (ortho/para directing)
Practical Synthesis:
To make 4-Bromonitrobenzene:
1. Start with benzene
2. Nitrate first: C6H6 + HNO3 -> C6H5NO2
(NO2 is meta-director, blocks ortho/para)
3. Brominate: C6H5NO2 + Br2 -> m-Bromination (unwanted)
Better approach:
1. Start with benzene
2. Brominate first: C6H6 + Br2 -> C6H5Br
(Br is ortho/para-director)
3. Nitrate: C6H5Br + HNO3 -> 4-Bromnitrobenzene + 2-Bromnitrobenzene
(Mixture of ortho/para, para is major due to steric hindrance at ortho)
Q5. Compare alkanes, alkenes, and alkynes. Give combustion reactions and bond energies.
Comparison of Alkanes, Alkenes, and Alkynes
1. Structure and Formula:
Alkanes (CnH(2n+2)):
- Only single C-C bonds
- Fully saturated
- Example: C2H6 (Ethane) CH3-CH3
Alkenes (CnH(2n)):
- One or more C=C double bonds
- Unsaturated (one degree of unsaturation per double bond)
- Example: C2H4 (Ethene) CH2=CH2
Alkynes (CnH(2n-2)):
- One or more C≡C triple bonds
- Highly unsaturated (two degrees of unsaturation per triple bond)
- Example: C2H2 (Ethyne) HC≡CH
2. Combustion Reactions:
Ethane Combustion:
2C2H6 + 7O2 -> 4CO2 + 6H2O
ΔHc° = -1560 kJ/mol
Complete combustion gives CO2 + H2O
Ethene Combustion:
C2H4 + 3O2 -> 2CO2 + 2H2O
ΔHc° = -1411 kJ/mol
Higher combustion enthalpy per mole C
Ethyne Combustion:
2C2H2 + 5O2 -> 4CO2 + 2H2O
ΔHc° = -2600 kJ/mol
Highest combustion enthalpy per mole C (most energy released)
3. Bond Energies:
C-C Bond Energy:
- Alkane C-C (σ bond): 348 kJ/mol
- Alkene C=C (σ + π): 610 kJ/mol (total, σ ≈ 350, π ≈ 260)
- Alkyne C≡C (σ + 2π): 835 kJ/mol (total, σ ≈ 350, 2π ≈ 485)
C-H Bond Energy:
- All similar: 410-420 kJ/mol
4. Chemical Reactivity:
Alkanes:
- Very unreactive toward addition
- Undergo substitution (especially with halogens)
- Combustion easily
- Free radical chemistry
- Example: CH4 + Cl2 ---light--> CH3Cl + HCl
Alkenes:
- Highly reactive toward addition
- React with Br2 (decolorize), H2SO4, HX, H2O, KMnO4
- Example: CH2=CH2 + Br2 -> CH2Br-CH2Br
Alkynes:
- Most reactive toward addition (two π bonds)
- React with 2 moles of Br2 or HX
- Terminal alkynes have acidic H
- Example: HC≡CH + 2Br2 -> CHBr2-CHBr2
5. Properties Comparison:
Alkanes Alkenes Alkynes
Formula CnH(2n+2) CnH(2n) CnH(2n-2)
Saturation Saturated Unsaturated Highly unsaturated
Reactivity Low High Very high
Color Colorless Colorless Colorless
Br2 test No change Decolorizes Decolorizes
KMnO4 test No change Decolorizes Decolorizes
Addition No Yes Yes (2 moles)
Aromatic No No Terminal H is acidic
Combs.E Lower Intermediate Highest (per C)
6. Addition Reaction Examples:
Alkene Addition:
CH2=CH2 + H2 --Ni,heat--> CH3-CH3 (Hydrogenation)
CH2=CH2 + HBr -> CH3-CH2Br (Hydrohalogenation)
Alkyne Addition:
HC≡CH + H2 --Ni,heat--> CH2=CH2
CH3C≡CH + H2 -> CH3-CH=CH2 (Markovnikov product)
HC≡CH + 2HBr -> CHBr2-CH3 (Markovnikov, two additions)
7. Stability Trend:
Alkynes generally less stable than corresponding alkenes due to sp hybridization, but triple bonds provide extra stability through σ + 2π bonding.
Q6. Explain addition reactions of alkenes. Give mechanism of HBr addition and Markovnikov's rule.
Addition Reactions of Alkenes: Reactions where two atoms or groups add across the C=C double bond.
General Pattern:
R2C=CR2 + A-B -> R2C(A)-C(B)R2
Types of Addition Reactions:
1. Hydrogenation (Addition of H2)
R2C=CR2 + H2 --catalyst(Ni,Pd,Pt)--> R2CH-CHR2
Example: CH2=CH2 + H2 --Ni, 150°C--> CH3-CH3
Requires catalyst (Ni, Pd, Pt, Rh)
2. Hydrohalogenation (Addition of HX where X = F, Cl, Br, I)
R2C=CR2 + HX -> R2C(X)-CHR2
Example: CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (Markovnikov product)
Follows Markovnikov's rule
3. Hydration (Addition of H2O)
R2C=CR2 + H2O --H+ catalyst--> R2C(OH)-CHR2
Example: CH2=CH2 + H2O --H2SO4--> CH3-CH2OH (Ethanol)
Follows Markovnikov's rule
4. Halogenation (Addition of X2)
R2C=CR2 + X2 -> R2C(X)-CX(R)R
Example: CH2=CH2 + Br2 -> CHBr2-CHBr2 (1,2-Dibromethane)
No catalyst required, free radical or ionic mechanism
5. Addition of KMnO4 (Oxidation)
R2C=CR2 + KMnO4 --cold, dilute--> R2C(OH)-C(OH)R2
Produces diols, double bond cleaved
Mechanism of HBr Addition (Electrophilic Addition):
Example: CH3-CH=CH2 + HBr -> CH3-CHBr-CH3
Step 1: Formation of Carbocation (Rate-determining)
- π electrons of C=C attack H+ from HBr
- Forms H-C bond
- Creates carbocation on other carbon
H+
| H
CH3-C=CH2 -> CH3-C+-CH3 + Br-
(π attacks H)
Secondary carbocation (2°)
Step 2: Nucleophilic Attack by Br- (Fast)
- Br- attacks carbocation
- Forms C-Br bond
- Restores neutral product
Br-
|
CH3-C+-CH3 -> CH3-CHBr-CH3
Final product: 2-Bromopropane
Markovnikov's Rule:
In addition of HX (or H2O) to unsymmetrical alkenes, H adds to the carbon with more hydrogen atoms, and X (or OH) adds to the carbon with fewer hydrogen atoms.
Reason: Forms more stable carbocation intermediate.
Example 1: CH3-CH=CH2 + HBr
Carbocations possible:
- Path 1: CH3-CH2+ (primary, 1°, unstable) - NOT formed
- Path 2: CH3-CHBr+-CH3 (secondary, 2°, stable) - FORMED
Product: CH3-CHBr-CH3 (2-Bromopropane, major, 100%)
Not: CH3-CH2-CH2Br (1-Bromopropane, minor)
Example 2: (CH3)2C=CH2 + HBr
Carbocations possible:
- Path 1: (CH3)2CH+ (secondary, 2°)
- Path 2: (CH3)3C+ (tertiary, 3°, most stable) - FORMED
Product: (CH3)3CBr (tert-Butyl bromide, exclusive)
Carbocation Stability Order:
3° (tertiary) > 2° (secondary) > 1° (primary) > CH3+ (methyl)
(Most stable) ...................... (Least stable)
Regioselectivity:
Markovnikov addition is highly regioselective (major product forms).
Minor product from rearrangement or competing mechanisms is usually <5%.
Stereochemistry:
Addition is NOT stereospecific. Carbocation intermediate is sp2 planar, so Br- can attack from either side (above or below), producing racemic mixture of enantiomers.
Hydrocarbons — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following represents an isomer of n-pentane?
- A. 2,2-dimethylpropane
- B. 2-methylbutane
- C. Neopentane
- D. All of the above
Answer: D. All of the above — All the given options (2,2-dimethylpropane, 2-methylbutane, and neopentane, which is another name for 2,2-dimethylpropane) have the molecular formula C5H12, same as n-pentane. They are structural isomers.
MCQ 2. Methane reacts with chlorine in the presence of sunlight to form methyl chloride. This reaction is an example of:
- A. Electrophilic addition
- B. Nucleophilic substitution
- C. Free radical substitution
- D. Elimination reaction
Answer: C. Free radical substitution — Halogenation of alkanes, such as the reaction of methane with chlorine in sunlight, proceeds via a free radical mechanism involving initiation, propagation, and termination steps. It is a substitution reaction where hydrogen is replaced by halogen.
MCQ 3. Dehydration of ethanol with concentrated H2SO4 at 443 K yields:
- A. Ethane
- B. Ethene
- C. Ethyne
- D. Diethyl ether
Answer: B. Ethene — When ethanol is heated with concentrated H2SO4 at 443 K (170°C), it undergoes intermolecular dehydration to form ethene. This is a common method for the preparation of alkenes from alcohols.
MCQ 4. What is the major product formed when propene reacts with HBr in the presence of peroxide?
- A. 1-bromopropane
- B. 2-bromopropane
- C. Propane
- D. 1,2-dibromopropane
Answer: A. 1-bromopropane — The reaction of propene with HBr in the presence of peroxides follows the anti-Markovnikov's rule, leading to the formation of 1-bromopropane as the major product. Peroxides promote a free radical addition mechanism.
MCQ 5. Which of the following alkynes is most acidic?
- A. Ethyne
- B. Propyne
- C. But-1-yne
- D. But-2-yne
Answer: A. Ethyne — Terminal alkynes (like ethyne, propyne, but-1-yne) are acidic due to the sp-hybridized carbon atoms, which makes the C-H bond more polarized. Ethyne, having two terminal acidic hydrogens, is generally considered the most acidic among the simple alkynes.
MCQ 6. According to Hückel's rule, an aromatic compound must have:
- A. (4n) π electrons
- B. (4n + 2) π electrons
- C. (2n + 2) π electrons
- D. (4n - 2) π electrons
Answer: B. (4n + 2) π electrons — Hückel's rule states that a cyclic, planar, fully conjugated molecule with (4n + 2) π electrons (where n is a non-negative integer) is aromatic. This rule helps in predicting the aromatic character of compounds.
MCQ 7. The nitration of benzene is an example of which type of reaction?
- A. Nucleophilic addition
- B. Free radical substitution
- C. Electrophilic substitution
- D. Elimination reaction
Answer: C. Electrophilic substitution — Nitration of benzene, along with halogenation, sulfonation, and Friedel-Crafts reactions, are characteristic electrophilic aromatic substitution reactions. The electrophile (NO2+) attacks the electron-rich benzene ring.
MCQ 8. The most stable conformation of n-butane is:
- A. Eclipsed
- B. Skew
- C. Anti
- D. Gauche
Answer: C. Anti — In n-butane, the anti conformation is the most stable because the two bulky methyl groups are oriented 180° apart, minimizing steric strain and electron-electron repulsion.
MCQ 9. What is the IUPAC name for the compound CH3-CH(CH3)-CH2-CH3?
- A. n-Pentane
- B. Isopentane
- C. 2-Methylbutane
- D. 3-Methylbutane
Answer: C. 2-Methylbutane — The longest carbon chain has four carbons (butane), and there is a methyl group at the second position. Hence, the IUPAC name is 2-Methylbutane.
MCQ 10. Ozonolysis of an alkene yields propanal and ethanal. What is the structure of the alkene?
- A. Pent-1-ene
- B. Pent-2-ene
- C. 2-Methylbut-1-ene
- D. 2-Methylbut-2-ene
Answer: B. Pent-2-ene — Ozonolysis cleaves the C=C double bond, and the carbons involved in the double bond are oxidized to carbonyl groups. If propanal (CH3CH2CHO) and ethanal (CH3CHO) are formed, the alkene must be CH3CH2CH=CHCH3, which is Pent-2-ene.
MCQ 11. Which of the following reagents is used for Friedel-Crafts acylation of benzene?
- A. CH3Cl / AlCl3
- B. CH3COCl / AlCl3
- C. H2SO4 / HNO3
- D. Cl2 / FeCl3
Answer: B. CH3COCl / AlCl3 — Friedel-Crafts acylation involves the reaction of an acyl halide (like acetyl chloride, CH3COCl) or an acid anhydride with an aromatic compound in the presence of a Lewis acid catalyst like anhydrous AlCl3 to form a ketone.
MCQ 12. Wurtz reaction is used for the preparation of:
- A. Alkenes
- B. Alkynes
- C. Symmetrical alkanes
- D. Aromatic hydrocarbons
Answer: C. Symmetrical alkanes — Wurtz reaction involves the coupling of two alkyl halides with sodium metal in dry ether to form a higher alkane, preferably symmetrical alkanes (R-R). For unsymmetrical alkanes, it leads to a mixture of products.
MCQ 13. What is the hybridization of carbon atoms in a benzene ring?
- A. sp3
- B. sp2
- C. sp
- D. dsp2
Answer: B. sp2 — Each carbon atom in benzene is bonded to two other carbon atoms and one hydrogen atom, with a double bond character due to resonance. This arrangement requires sp2 hybridization for each carbon.
MCQ 14. Which of the following compounds can exhibit cis-trans isomerism?
- A. But-1-ene
- B. 2-Methylpropene
- C. But-2-ene
- D. 2,3-Dimethylbut-2-ene
Answer: C. But-2-ene — Cis-trans isomerism (geometric isomerism) occurs in alkenes when each carbon of the double bond is attached to two different groups. But-2-ene (CH3CH=CHCH3) has a methyl group and a hydrogen on each double-bonded carbon, allowing for cis and trans forms.
MCQ 15. Dehydrohalogenation of vicinal dihalides in the presence of strong base like alcoholic KOH followed by sodamide (NaNH2) produces:
- A. Alkanes
- B. Alkenes
- C. Alkynes
- D. Aromatic compounds
Answer: C. Alkynes — Vicinal dihalides undergo double dehydrohalogenation to form alkynes. The first step uses alcoholic KOH to form an alkene, and the second step with a stronger base like NaNH2 removes the second molecule of HX to form an alkyne.
MCQ 16. Which of the following groups is an ortho-para director and an activating group in electrophilic aromatic substitution?
- A. -NO2
- B. -COOH
- C. -OH
- D. -CN
Answer: C. -OH — Groups like -OH, -NH2, -OCH3, and alkyl groups are activating groups and ortho-para directors because they donate electron density to the ring, stabilizing the carbocation intermediate at ortho and para positions.
MCQ 17. Complete combustion of methane produces:
- A. CO2 and H2
- B. CO and H2O
- C. CO2 and H2O
- D. C and H2O
Answer: C. CO2 and H2O — Complete combustion of any hydrocarbon in the presence of sufficient oxygen yields carbon dioxide and water. For methane: CH4 + 2O2 → CO2 + 2H2O.
MCQ 18. Addition of bromine water to an alkene results in the disappearance of the reddish-brown color. This reaction is used as a test for:
- A. Alkanes
- B. Alkynes
- C. Saturated hydrocarbons
- D. Unsaturation
Answer: D. Unsaturation — Alkenes (and alkynes) decolorize bromine water due to the addition of bromine across the double (or triple) bond. This is a common chemical test to detect the presence of unsaturation.
MCQ 19. Ethyne reacts with water in the presence of HgSO4 and H2SO4 to form:
- A. Ethanol
- B. Ethanal
- C. Ethanoic acid
- D. Ethene
Answer: B. Ethanal — Hydration of ethyne (acetylene) in the presence of mercuric sulfate and sulfuric acid follows Markovnikov's rule, initially forming an enol (vinyl alcohol), which immediately tautomerizes to ethanal (acetaldehyde).
MCQ 20. Benzene can be prepared from phenol by heating it with:
- A. Zn dust
- B. NaOH
- C. CH3Cl
- D. Concentrated H2SO4
Answer: A. Zn dust — Phenol can be converted to benzene by distillation with zinc dust. Zinc reduces phenol by removing the oxygen atom, forming benzene.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.