Practice MCQs for JEE Chemistry Hydrocarbons
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Multiple Choice Questions
(a) 3,3-Dimethylbutan-1-ol
(b) 3,3-Dimethylbutan-2-ol
(c) 2,3-Dimethylbutan-2-ol
(d) 2-Methylpentan-2-ol
Show Answer & Explanation
Answer: (c) 2,3-Dimethylbutan-2-ol
Explanation:
1. Protonation gives the secondary carbocation (CH3)3C–CH+–CH3.
2. A 1,2-methyl shift converts it into the more stable tertiary cation (CH3)2C+–CH(CH3)2.
3. Water attacks this cation, giving 2,3-dimethylbutan-2-ol.
4. Answer: (c)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) Buta-1,3-diene
(b) Buta-1,2-diene (CH2=C=CH–CH3)
(c) But-2-ene
(d) But-1-ene
Show Answer & Explanation
Answer: (b) Buta-1,2-diene (CH2=C=CH–CH3)
Explanation:
1. In an allene the central carbon is joined to two carbons by double bonds, and both bonds are cleaved.
2. The central carbon is left with two C=O bonds, which is CO2.
3. The terminal carbons give HCHO and CH3CHO.
4. Answer: (b)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) CH3CH2CH=C(CH2CH3)2
(b) CH3CH=C(CH2CH3)2
(c) (CH3CH2)2C=C(CH3)2
(d) CH3CH=CH–CH(CH3)CH2CH2CH3
Show Answer & Explanation
Answer: (a) CH3CH2CH=C(CH2CH3)2
Explanation:
1. Cleaving C=C in option (a) gives CH3CH2CHO from one end and (CH3CH2)2C=O from the other.
2. The other options give ethanal, propanone or other carbonyl compounds.
3. Answer: (a)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) Decarboxylation of sodium benzoate
(b) Deoxygenation of phenol
(c) Reduction of benzenediazonium chloride
(d) Catalytic hydrogenation of acetylene
Show Answer & Explanation
Answer: (d) Catalytic hydrogenation of acetylene
Explanation:
1. Soda lime on sodium benzoate, zinc dust on phenol and H3PO2 on a diazonium salt all give benzene.
2. Benzene arises from acetylene only by cyclic trimerisation through a red-hot tube.
3. Hydrogenation of acetylene gives ethene or ethane.
4. Answer: (d)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) Ethylbenzene
(b) Cumene
(c) Mesitylene
(d) sec-Butylbenzene
Show Answer & Explanation
Answer: (c) Mesitylene
Explanation:
1. Any alkylbenzene with a benzylic hydrogen is oxidised to benzoic acid.
2. Mesitylene has three methyl groups, so all three are oxidised to give benzene-1,3,5-tricarboxylic acid.
3. Answer: (c)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) Benzoic acid
(b) Acetophenone
(c) Ethyl phenyl ketone
(d) Benzophenone
Show Answer & Explanation
Answer: (b) Acetophenone
Explanation:
1. This is Friedel–Crafts acylation, with the acylium ion CH3CO+ as the electrophile.
2. It gives C6H5COCH3, acetophenone.
3. Answer: (b)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) Phenylacetic acid
(b) Salicylic acid
(c) o-Toluic acid
(d) Benzoic acid
Show Answer & Explanation
Answer: (b) Salicylic acid
Explanation:
1. Phenylacetic acid gives toluene, o-toluic acid gives toluene and benzoic acid gives benzene.
2. Salicylic acid gives phenol, which is an aromatic alcohol and not a hydrocarbon.
3. Answer: (b)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) A racemic mixture
(b) A mixture of diastereomers (cis- and trans-1,2-dimethylcyclohexane)
(c) A single meso compound only
(d) Structural isomers of 1,2-dimethylcyclohexane
Show Answer & Explanation
Answer: (b) A mixture of diastereomers (cis- and trans-1,2-dimethylcyclohexane)
Explanation:
1. The stereocentre at C-2 is untouched.
2. Hydrogen adds to C-1 from either face, which creates a new stereocentre and gives cis and trans-1,2-dimethylcyclohexane.
3. These differ in configuration at only one centre, so they are diastereomers and form in unequal amounts.
4. Answer: (b)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
(a) NH2NH2/KOH only
(b) Zn–Hg/HCl only
(c) Red P + HI only
(d) Any of these
Show Answer & Explanation
Answer: (d) Any of these
Explanation:
1. Wolff–Kishner (NH2NH2/KOH), Clemmensen (Zn–Hg/HCl) and red P + HI all reduce C=O to CH2.
2. None of them touches a saturated hydrocarbon skeleton, so any one can be used.
3. Answer: (d)
Teacher's Note:
1. For alkenes and arenes, decide first which bond or position reacts, then write the product.
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Practice MCQs for JEE Chemistry Hydrocarbons
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FAQs
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Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.
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