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JEE Chemistry

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

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Topics in JEE Chemistry
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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A compound C5H10O2 with no C=O in IR but shows broad O-H stretch. 1H-NMR exhibits signals at δ 3.8 and δ 4.7 ppm. The compound is most likely:
A 2-methylbutane-1,4-diol
B 2,2-dimethyl-1,3-dioxolane
C Cyclopentanone
D Tetrahydrofurfuryl alcohol
Correct Answer:  B. 2,2-dimethyl-1,3-dioxolane
EXPLANATION

No C=O in IR rules out ketone/aldehyde. The signals at δ 3.8 (OCH2) and δ 4.7 suggest acetal carbons (characteristic of dioxolane ring). 2,2-dimethyl-1,3-dioxolane fits C5H10O2 with protected diol functionality.

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In the hydroboration-oxidation of alkenes, the reaction is stereospecific because:
A It involves a free radical mechanism
B It proceeds through a four-membered ring transition state with syn-addition
C The oxidation step is stereospecific
D The alkene must be planar
Correct Answer:  B. It proceeds through a four-membered ring transition state with syn-addition
EXPLANATION

Hydroboration occurs through a concerted mechanism involving a four-membered ring transition state, resulting in syn-addition of BH across the double bond.

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A compound with molecular formula C6H12O shows a broad O-H stretch in IR at 3300 cm⁻¹ and no C=O peak. The 1H-NMR shows a singlet at δ 3.3 ppm. The compound is likely:
A 1-hexanol
B 2,3-dimethyl-2-butanol
C Tetrahydropyran
D 1,2-hexanediol
Correct Answer:  B. 2,3-dimethyl-2-butanol
EXPLANATION

The singlet at δ 3.3 ppm suggests the OH is on a quaternary carbon (no neighboring H, hence no coupling). 2,3-dimethyl-2-butanol fits: the OH is on a quaternary carbon, explaining the singlet and broad O-H in IR.

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The compound that will undergo decarboxylation easily is:
A Benzoic acid
B Acetic acid
C β-Ketoacid (like acetoacetic acid)
D Formic acid
Correct Answer:  C. β-Ketoacid (like acetoacetic acid)
EXPLANATION

β-Ketoacids (RCOCH2COOH) undergo easy decarboxylation through a six-membered transition state, producing ketones. This is more facile than decarboxylation of regular carboxylic acids.

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