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Organic Chemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

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Difficulty: All Easy Medium Hard 21–30 of 49
Topics in JEE Chemistry
Q.21 Medium Organic Chemistry
Identify the compound with molecular formula C6H12 that shows geometrical isomerism:
A Cyclohexane
B hex-3-ene
C 2-methylpent-2-ene
D Hexane
Correct Answer:  B. hex-3-ene
EXPLANATION

hex-3-ene (CH3CH2CH=CHCH2CH3) has different groups on each carbon of the double bond, allowing cis-trans (E-Z) isomerism. Cyclohexane and alkanes have no double bonds, and 2-methylpent-2-ene is not a correct formula for C6H12.

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Q.22 Medium Organic Chemistry
The product formed by the hydroboration-oxidation of 1-butene is:
A 1-butanol
B 2-butanol
C Butanal
D Butanone
Correct Answer:  A. 1-butanol
EXPLANATION

Hydroboration-oxidation follows anti-Markovnikov's rule with syn addition. The OH adds to the less substituted carbon (primary), giving 1-butanol (butan-1-ol).

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Q.23 Medium Organic Chemistry
Which statement about the aldol condensation is correct?
A It requires an aldehyde and ketone in equal amounts
B It produces a β-hydroxy carbonyl compound as the intermediate product
C It only works with aromatic aldehydes
D It requires acidic conditions exclusively
Correct Answer:  B. It produces a β-hydroxy carbonyl compound as the intermediate product
EXPLANATION

Aldol condensation produces a β-hydroxy carbonyl compound (aldol) initially, which can further dehydrate under heating to form an α,β-unsaturated carbonyl compound.

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Q.24 Medium Organic Chemistry
In the bromination of toluene with Br2/FeBr3, the major product is:
A 2-bromotoluene only
B 4-bromotoluene only
C ortho and para-bromotoluene (major)
D meta-bromotoluene only
Correct Answer:  C. ortho and para-bromotoluene (major)
EXPLANATION

The methyl group (-CH3) is an alkyl group, which is an electron-donating group that activates the benzene ring and is ortho/para-directing in electrophilic aromatic substitution.

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Q.25 Medium Organic Chemistry
The rate-determining step in an E1 elimination reaction is:
A Removal of hydrogen by base
B Formation of the carbocation
C Protonation of the alkene
D Formation of the alkyl halide
Correct Answer:  B. Formation of the carbocation
EXPLANATION

E1 elimination occurs in two steps: slow carbocation formation followed by fast deprotonation. The carbocation formation is the rate-determining step.

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Q.26 Medium Organic Chemistry
Which of the following will give a positive Tollens test?
A Acetone
B Benzaldehyde
C Diethyl ketone
D tert-Butyl alcohol
Correct Answer:  B. Benzaldehyde
EXPLANATION

Tollens test detects aldehydes. Benzaldehyde contains an aldehyde group (-CHO) and will be oxidized to benzoate ion, giving a positive test (silver mirror).

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Q.27 Medium Organic Chemistry
In the addition of HBr to propene, the major product is 2-bromopropane because of:
A Le Chatelier's principle
B Markovnikov's rule
C Anti-Markovnikov addition
D Hydrogen preference
Correct Answer:  B. Markovnikov's rule
EXPLANATION

Markovnikov's rule states that in addition to unsymmetrical alkenes, the hydrogen adds to the carbon with more hydrogen atoms, and the addendum to the carbon with fewer hydrogens, forming the more stable carbocation intermediate.

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Q.28 Medium Organic Chemistry
The reactivity order of alkyl halides in SN1 reactions is:
A Primary > Secondary > Tertiary
B Tertiary > Secondary > Primary
C Secondary > Primary > Tertiary
D All are equally reactive
Correct Answer:  B. Tertiary > Secondary > Primary
EXPLANATION

SN1 proceeds through carbocation formation. Tertiary carbocations are most stable (hyperconjugation and inductive effects), followed by secondary, then primary.

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Q.29 Medium Organic Chemistry
A chiral compound with the structure CH3-CHBr-CH(OH)-CH3 will have how many stereoisomers?
A 2
B 4
C 8
D 16
Correct Answer:  B. 4
EXPLANATION

The molecule has two chiral centers (carbons bearing Br and OH). Each chiral center can have R or S configuration, giving 2² = 4 possible stereoisomers (diastereomers and enantiomers).

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Q.30 Medium Organic Chemistry
In the conversion of benzene to benzoic acid through the Kolbe reaction, the carboxylic acid is ultimately derived from:
A The phenolic oxygen
B CO2 fixation on the aromatic ring
C Oxidation of a side chain
D Displacement of a leaving group on the ring
Correct Answer:  B. CO2 fixation on the aromatic ring
EXPLANATION

In the Kolbe carboxylation of phenols, CO2 is directly incorporated into the aromatic ring ortho to the hydroxyl group under high temperature and pressure with alkali, forming salicylic acid derivatives.

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