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JEE Chemistry
Organic Chemistry

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

100 Q 5 Topics Take Test
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Difficulty: All Easy Medium Hard 31–40 of 100
Topics in JEE Chemistry
The enolate ion formed from a ketone in basic conditions attacks an alkyl halide in the:
A Carbon atom (α-position)
B Oxygen atom
C Through radical mechanism
D Through concerted mechanism
Correct Answer:  A. Carbon atom (α-position)
EXPLANATION

Although the enolate has negative charge on both oxygen and carbon (resonance), the carbon is more nucleophilic toward alkyl halides. The SN2 attack occurs at the alkyl halide carbon, not the enolate O.

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In the Robinson annulation reaction, the product formed is:
A An aromatic compound
B A bicyclic ketone
C A linear diene
D A carboxylic acid
Correct Answer:  B. A bicyclic ketone
EXPLANATION

The Robinson annulation combines a Michael addition followed by an aldol condensation, forming a bicyclic ketone product with a new six-membered ring fused to an existing ring system.

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The directing effect of the amino group (-NH2) in electrophilic aromatic substitution is:
A Deactivating, meta-directing
B Deactivating, ortho/para-directing
C Activating, ortho/para-directing
D Activating, meta-directing
Correct Answer:  C. Activating, ortho/para-directing
EXPLANATION

The amino group is a strong electron-donating group (by resonance) that activates the benzene ring and directs incoming electrophiles to ortho/para positions due to resonance stabilization of the intermediate.

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In the reduction of ketones and aldehydes using DIBAL-H, the product is:
A An alkane
B An alcohol
C A carboxylic acid
D An alkene
Correct Answer:  B. An alcohol
EXPLANATION

DIBAL-H (Diisobutylaluminum hydride) reduces carbonyl compounds to primary alcohols (from aldehydes) or secondary alcohols (from ketones). It's milder than LiAlH4.

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When propyne undergoes hydration in the presence of Hg2+/H2SO4, the major product is:
A Propanol
B Propanal
C Acetone
D Propanoic acid
Correct Answer:  C. Acetone
EXPLANATION

Propyne (CH3C≡CH) hydrates to form an unstable enol intermediate. The enol tautomerizes to acetone (CH3COCH3), following Markovnikov's rule with H adding to the carbon with more H atoms.

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The mechanism of the esterification reaction between a carboxylic acid and alcohol is:
A SN1 mechanism
B SN2 mechanism
C Nucleophilic acyl substitution via tetrahedral intermediate
D Elimination-addition
Correct Answer:  C. Nucleophilic acyl substitution via tetrahedral intermediate
EXPLANATION

Esterification follows the nucleophilic acyl substitution mechanism: the OH of the alcohol attacks the carbonyl carbon after protonation, forming a tetrahedral intermediate that collapses to form the ester.

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The rearrangement that occurs during the dehydration of 3-methylbutan-2-ol under acidic conditions is:
A Pinacol rearrangement
B Methyl shift (Wagner-Meerwein rearrangement)
C Hofmann rearrangement
D Beckmann rearrangement
Correct Answer:  B. Methyl shift (Wagner-Meerwein rearrangement)
EXPLANATION

The initially formed secondary carbocation rearranges via methyl shift to form a more stable tertiary carbocation, producing 2-methylbut-2-ene as major product.

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In the crossed Cannizzaro reaction, the aldehyde that acts as both reducing and oxidizing agent is:
A Benzaldehyde
B Formaldehyde
C p-Methoxybenzaldehyde
D Trimethylacetaldehyde
Correct Answer:  B. Formaldehyde
EXPLANATION

Formaldehyde is the only aldehyde without an α-hydrogen, so it undergoes the Cannizzaro reaction where it acts as both oxidizing and reducing agent, forming formate and methanol.

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The decarboxylation of malonic acid derivatives proceeds through:
A Formation of a free carbocation
B Formation of a cyclic intermediate with loss of CO2
C Nucleophilic attack by base
D Radical mechanism
Correct Answer:  B. Formation of a cyclic intermediate with loss of CO2
EXPLANATION

Malonic acid derivatives (with two electron-withdrawing groups) undergo decarboxylation through a cyclic transition state mechanism, facilitated by the stabilization of the resulting carbanion by the remaining electron-withdrawing group.

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Which of the following compounds will show optical isomerism?
A 1-bromo-2-methylpropane
B 2-bromo-2-methylpropane
C 2-bromobutane
D 1-bromobutane
Correct Answer:  C. 2-bromobutane
EXPLANATION

2-bromobutane (CH3CHBrCH2CH3) has a carbon with four different groups attached, making it chiral. The bromine-bearing carbon is a stereocenter.

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