JEE Chemistry Hydrocarbons MCQs Set 34

Practice MCQs for JEE Chemistry Hydrocarbons

Access targeted multiple-choice questions for Hydrocarbons designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Hydrocarbons Questions and Solutions

Navigate directly to the 50 objective questions for Hydrocarbons using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: Benzene does not readily undergo
(a) halogenation
(b) nitration
(c) sulphonation
(d) oxidation
Show Answer & Explanation

Answer: (d) oxidation

Explanation:
1. Benzene is aromatic and resists oxidation; the ring is not oxidised by common oxidants such as KMnO4.
2. It readily undergoes electrophilic substitutions such as halogenation, nitration and sulphonation, which keep the aromatic sextet.

Teacher's Note:
1. Aromatic stability makes the ring resist addition and oxidation.

Question 2: Which of the following is a free radical addition reaction?
(a) CH3CH=CH2 + HBr in the presence of R2O2 giving CH3CH2CH2Br
(b) CH2=CH2 + HBr giving CH3CH2Br in the dark
(c) CH3CH=CH2 + HCl giving CH3CHClCH3
(d) CH3CH=CH2 + HOBr giving CH3CH(OH)CH2Br
Show Answer & Explanation

Answer: (a) CH3CH=CH2 + HBr in the presence of R2O2 giving CH3CH2CH2Br

Explanation:
1. Peroxides produce Br· from HBr, so the addition is a radical chain (the peroxide effect) and gives the anti-Markovnikov product.
2. The other additions are polar electrophilic additions.

Teacher's Note:
1. Peroxide plus HBr: radical addition.
2. HCl and HI do not show the peroxide effect.

Question 3: Addition of bromine to cis-hex-3-ene gives
(a) a meso dibromide
(b) an optically active single enantiomer
(c) a mixture of diastereomeric dibromides
(d) a racemic dibromide
Show Answer & Explanation

Answer: (d) a racemic dibromide

Explanation:
1. Br2 adds anti through a bromonium ion. Anti addition to a cis alkene gives the (3R,4R) and (3S,4S) 3,4-dibromohexanes in equal amounts.
2. Equal amounts of enantiomers make a racemic mixture (the trans alkene would give the meso compound).

Teacher's Note:
1. cis + anti addition: racemic.
2. trans + anti addition: meso.

Question 4: Butane reacts with Cl2 in light. The major monochloro product is
(a) 2-chlorobutane
(b) 1-chlorobutane
(c) 1,2-dichlorobutane
(d) 2,2-dichlorobutane
Show Answer & Explanation

Answer: (a) 2-chlorobutane

Explanation:
1. Chlorine is only mildly selective: the secondary C-H is about 3.8 times as reactive as the primary C-H, and there are 4 secondary and 6 primary hydrogens.
2. Relative yield: secondary 4 x 3.8 = 15.2, primary 6 x 1 = 6, so about 72 percent 2-chlorobutane.

Teacher's Note:
1. Rate = number of hydrogens x reactivity per hydrogen.

Question 5: The peroxide effect (addition of HBr to an alkene in presence of peroxide) occurs by
(a) a free radical mechanism
(b) an ionic mechanism
(c) heterolytic fission of the double bond
(d) a carbocation mechanism
Show Answer & Explanation

Answer: (a) a free radical mechanism

Explanation:
1. Peroxides decompose into radicals, which generate Br· from HBr.
2. A chain of radical steps follows: Br· adds to the alkene and the resulting radical abstracts H from HBr.

Teacher's Note:
1. Peroxide means radicals.

Question 6: 3-Methylpentane undergoes monochlorination with Cl2/light; the relative reactivities of 3°:2°:1° hydrogens are 6:4:1. The relative amount of 3-chloro-3-methylpentane in the product is
(a) 16/31
(b) 6/31
(c) 3/31
(d) 9/31
Show Answer & Explanation

Answer: (b) 6/31

Explanation:
1. 3-Methylpentane has 1 tertiary H (reactivity 6), 4 secondary H (4 x 4 = 16), 6 primary H on the two ethyl CH3 groups (6 x 1 = 6) and 3 primary H on the methyl group (3 x 1 = 3).
2. Total = 6 + 16 + 6 + 3 = 31. 3-Chloro-3-methylpentane comes from the tertiary H: 6/31.

Teacher's Note:
1. Rate share = (number of H x reactivity) / total.

Question 7: The relative rates of mononitration of C6H5-R, where R = CH3, NO2, OH and Cl, follow the order
(a) CH3 > OH > NO2 > Cl
(b) OH > CH3 > Cl > NO2
(c) OH > Cl > CH3 > NO2
(d) OH > CH3 > NO2 > Cl
Show Answer & Explanation

Answer: (b) OH > CH3 > Cl > NO2

Explanation:
1. -OH is a strong activator (+M), CH3 a weak activator (+I and hyperconjugation).
2. Chlorine deactivates mildly (-I > +M), and NO2 deactivates strongly (-I and -M), so nitrobenzene reacts slowest.

Teacher's Note:
1. Strong activator > weak activator > halogen > strong deactivator.

Question 8: Allylic bromination of an alkene with NBS is a
(a) nucleophilic substitution
(b) electrophilic substitution
(c) electrophilic addition
(d) free radical substitution
Show Answer & Explanation

Answer: (d) free radical substitution

Explanation:
1. NBS provides a very low concentration of Br2; light or peroxide gives Br·, which abstracts an allylic hydrogen to form a resonance-stabilised allyl radical.
2. The allyl radical takes Br from Br2; one H is replaced by Br in a radical chain: free radical substitution.

Teacher's Note:
1. Low Br2 concentration prevents addition to the double bond.

Question 9: In the nitrating mixture of concentrated HNO3 and concentrated H2SO4, HNO3 acts as a
(a) base
(b) acid
(c) reducing agent
(d) catalyst
Show Answer & Explanation

Answer: (a) base

Explanation:
1. The stronger acid H2SO4 protonates HNO3 to H2NO3+, which loses water to give the nitronium ion NO2+.
2. In this step HNO3 accepts a proton, so it behaves as a base, and H2SO4 is the acid.

Teacher's Note:
1. Who gives the proton and who takes it?
2. H2SO4 is the stronger acid.

Question 10: Which alkene adds Br2 most readily?
(a) CH2=CH2
(b) CH3CH=CH2
(c) (CH3)2C=CH-CH3
(d) CH3CH=CHCH3
Show Answer & Explanation

Answer: (c) (CH3)2C=CH-CH3

Explanation:
1. Electrophilic addition is faster when the double bond carries more electron-releasing alkyl groups, which stabilise the bromonium ion or carbocation.
2. 2-Methylbut-2-ene is trisubstituted, so it is the most reactive.

Teacher's Note:
1. More substituted C=C means more reactive towards electrophiles.

Question 11: Which compound is nitrated with the most difficulty?
(a) nitrobenzene
(b) phenol
(c) quinol (hydroquinone)
(d) m-dinitrobenzene
Show Answer & Explanation

Answer: (d) m-dinitrobenzene

Explanation:
1. Two nitro groups strongly deactivate the ring (-I, -M), more than one.
2. Phenol and quinol have activating OH groups, and nitrobenzene has only one deactivating group.

Teacher's Note:
1. Count the deactivating groups.

Question 12: The chlorination of toluene in the presence of ferric chloride gives predominantly
(a) benzyl chloride
(b) m-chlorotoluene
(c) o- and p-chlorotoluene
(d) benzal chloride
Show Answer & Explanation

Answer: (c) o- and p-chlorotoluene

Explanation:
1. FeCl3 is a Lewis acid that generates the electrophile Cl+, so the ring is substituted (electrophilic substitution).
2. The methyl group activates and directs to the ortho and para positions. Side-chain chlorination (benzyl chloride, benzal chloride) needs light or heat.

Teacher's Note:
1. Lewis acid: ring substitution.
2. Light: side-chain substitution.

Question 13: The major product of the addition of HBr to 1-methylcyclohexene is
(a) 1-bromo-2-methylcyclohexane
(b) 1-bromo-1-methylcyclohexane
(c) 3-bromo-1-methylcyclohexene
(d) 6-bromo-1-methylcyclohexene
Show Answer & Explanation

Answer: (b) 1-bromo-1-methylcyclohexane

Explanation:
1. H+ adds to C-2 so that the tertiary carbocation forms at C-1.
2. Bromide adds to C-1: Markovnikov product, 1-bromo-1-methylcyclohexane.

Teacher's Note:
1. Tertiary carbocation is the most stable.

Question 14: In a Friedel-Crafts reaction using AlCl3, which of benzene and nitrobenzene can be used as the solvent?
(a) benzene but not nitrobenzene
(b) both benzene and nitrobenzene
(c) neither of them
(d) nitrobenzene but not benzene
Show Answer & Explanation

Answer: (d) nitrobenzene but not benzene

Explanation:
1. Benzene would itself be alkylated or acylated.
2. Nitrobenzene is strongly deactivated and does not undergo Friedel-Crafts reactions, so it is an inert solvent.

Teacher's Note:
1. A solvent must not react under the conditions.

Question 15: Which statement about ortho-para directing groups in electrophilic aromatic substitution is correct?
(a) They are mostly deactivating groups
(b) They are mostly activating groups
(c) They are mostly neutral groups
(d) They are never activating
Show Answer & Explanation

Answer: (b) They are mostly activating groups

Explanation:
1. Groups with lone pairs or alkyl groups donate electrons and direct ortho-para while activating the ring.
2. The halogens are the main exception: deactivating but ortho-para directing.

Teacher's Note:
1. +M and +I groups activate.

Question 16: The compound which does not react with bromine easily at room temperature is
(a) chlorobenzene
(b) phenol
(c) but-2-yne
(d) pent-1-ene
Show Answer & Explanation

Answer: (a) chlorobenzene

Explanation:
1. Alkenes and alkynes add Br2 readily, and phenol is brominated rapidly because OH activates the ring.
2. Chlorobenzene is deactivated and needs a Lewis acid catalyst.

Teacher's Note:
1. Benzene derivatives need FeBr3 unless strongly activated.

Question 17: The reaction of 1-phenylpropane with a limited amount of chlorine in the presence of light mainly gives
(a) 1-chloro-1-phenylpropane
(b) 3-chloro-1-phenylpropane
(c) 2-chloro-1-phenylpropane
(d) 4-chloropropylbenzene
Show Answer & Explanation

Answer: (a) 1-chloro-1-phenylpropane

Explanation:
1. Under light the reaction is a radical substitution on the side chain.
2. The benzylic hydrogen gives the benzylic radical, stabilised by resonance with the ring, so substitution occurs at the carbon next to the ring.

Teacher's Note:
1. Benzylic radical is the most stable.

Question 18: The major product when HBr adds to but-1-ene in the presence of benzoyl peroxide is
(a) 2-bromobutane
(b) 1-bromobut-1-ene
(c) 1-bromobutane
(d) 2-bromobut-2-ene
Show Answer & Explanation

Answer: (c) 1-bromobutane

Explanation:
1. Peroxide gives Br· which adds to the terminal carbon to produce the more stable secondary radical.
2. The radical abstracts H from HBr: anti-Markovnikov product CH3CH2CH2CH2Br.

Teacher's Note:
1. Peroxide effect applies only to HBr.

Question 19: The rates of ortho-nitration of toluene (I), 2-deuteriotoluene (II) and 2,6-dideuteriotoluene (III) are in the order
(a) I > II > III
(b) III > I > II
(c) the rate is the same for all three
(d) II > I > III
Show Answer & Explanation

Answer: (c) the rate is the same for all three

Explanation:
1. In electrophilic substitution the rate-determining step is attack of the electrophile on the ring (formation of the sigma complex).
2. C-H or C-D bond breaking occurs in the fast second step, so there is no kinetic isotope effect.

Teacher's Note:
1. No isotope effect when the C-H cleavage is not rate determining.

Question 20: The major product of the reaction of HOBr with propene is
(a) 2-bromopropan-1-ol
(b) 3-bromopropan-1-ol
(c) 2-bromopropan-2-ol
(d) 1-bromopropan-2-ol
Show Answer & Explanation

Answer: (d) 1-bromopropan-2-ol

Explanation:
1. Br+ adds to the terminal carbon, giving the bromonium ion or the secondary carbocation.
2. Water attacks the more substituted carbon, so OH ends on C-2: CH3CH(OH)CH2Br.

Teacher's Note:
1. Halohydrin formation: halogen on the less substituted carbon, OH on the more substituted carbon.

Question 21: In Friedel-Crafts acylation, the amount of AlCl3 that must be taken is
(a) a catalytic amount
(b) exactly 0.1 equivalent
(c) more than one equivalent
(d) any amount; it does not matter
Show Answer & Explanation

Answer: (c) more than one equivalent

Explanation:
1. The product ketone is a Lewis base and forms a stable complex with AlCl3, which therefore is not free to act as a catalyst.
2. So at least one equivalent (usually a slight excess) is needed.

Teacher's Note:
1. The ketone product complexes the Lewis acid.

Free study material for Chemistry

Multiple Choice Questions (MCQs) for JEE Chemistry Hydrocarbons

JEE Chemistry Hydrocarbons Objective Test Questions

Test your conceptual understanding of Hydrocarbons with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Chemistry, these problem sets build accuracy and prepare students for objective exams.

NCERT-Aligned Objective Questions and Solutions

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Chemistry to ensure absolute clarity across all sub-topics in this chapter.

Next Steps in Your Exam Preparation

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Chemistry Hydrocarbons MCQs Set 34?

You can get most exhaustive JEE Chemistry Hydrocarbons MCQs Set 34 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Hydrocarbons MCQs Set 34 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Hydrocarbons MCQs Set 34, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Hydrocarbons MCQs Set 34?

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.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Hydrocarbons MCQs Set 34 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.