JEE Chemistry Hydrocarbons MCQs Set 42

Practice MCQs for JEE Chemistry Hydrocarbons

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Multiple Choice Questions

Question 1: 3,3-Dimethylbut-1-ene is treated with dilute acid (H+/H2O). The major product is:
(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.

Question 2: Which of the following hydrocarbons gives CO2 on reductive ozonolysis?
(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.

Question 3: Propanal and pentan-3-one are the two ozonolysis products (reductive work-up) of:
(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.

Question 4: Benzene is obtained by all of the following reactions except:
(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.

Question 5: Oxidation of which of the following arenes with hot acidic KMnO4 does not give benzoic acid?
(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.

Question 6: When benzene is heated with acetic anhydride in the presence of anhydrous AlCl3, the product formed is:
(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.

Question 7: Which of the following acids does not give an aromatic hydrocarbon on decarboxylation (heating with soda lime)?
(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.

Question 8: A single enantiomer of 2-methyl-1-methylenecyclohexane is hydrogenated over Pd. The products are:
(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.

Question 9: A saturated diketone has a ring C=O group and an acetyl group, –COCH3, on a polycyclic skeleton. Both are converted to –CH2– and –CH2CH3. Reagent X can be:
(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.

Practice MCQs for JEE Chemistry Hydrocarbons

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FAQs

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