JEE Main 2026 — Chemistry Organic Chemistry

2026mcqhard

' $x$ ' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at $333$ K. ' $y$ ' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from ' $x$ ' and ' $y$ ' in the presence of barium hydroxide followed by heating is :

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
2-Methylpent-4-en-3-one
B
4-Methylpent-3-en-2-one
C
4-Methylpent-1-ene
D
2-Methylpent-3-one

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Solution

Hydrolysis of prop-1-yne in the presence of $HgSO_4$ and dilute $H_2SO_4$ yields acetone.

$CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CO-CH_3$

Thus, $x$ is acetone.

Reaction of ethane nitrile ( $CH_3CN$ ) with methyl magnesium bromide ( $CH_3MgBr$ ) followed by hydrolysis also yields acetone.

$CH_3-C \equiv N + CH_3MgBr \rightarrow CH_3-C(CH_3)=NMgBr \xrightarrow{H_3O^+} CH_3-CO-CH_3$

Thus, $y$ is acetone.

Reaction of $x$ and $y$ (both acetone) in the presence of $Ba(OH)_2$ followed by heating results in an aldol condensation.

$2 CH_3-CO-CH_3 \xrightarrow{Ba(OH)_2} CH_3-C(OH)(CH_3)-CH_2-CO-CH_3$

Heating causes dehydration to form an unsaturated ketone.

$CH_3-C(OH)(CH_3)-CH_2-CO-CH_3 \xrightarrow{\Delta} CH_3-C(CH_3)=CH-CO-CH_3$

Numbering the principal carbon chain to give the lowest locant to the ketone group, the IUPAC name of the product is 4-methylpent-3-en-2-one.

Answer: 4-Methylpent-3-en-2-one

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Solution

Hydrolysis of prop-1-yne in the presence of $HgSO_4$ and dilute $H_2SO_4$ yields acetone.

$CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CO-CH_3$

Thus, $x$ is acetone.

Reaction of ethane nitrile ( $CH_3CN$ ) with methyl magnesium bromide ( $CH_3MgBr$ ) followed by hydrolysis also yields acetone.

$CH_3-C \equiv N + CH_3MgBr \rightarrow CH_3-C(CH_3)=NMgBr \xrightarrow{H_3O^+} CH_3-CO-CH_3$

Thus, $y$ is acetone.

Reaction of $x$ and $y$ (both acetone) in the presence of $Ba(OH)_2$ followed by heating results in an aldol condensation.

$2 CH_3-CO-CH_3 \xrightarrow{Ba(OH)_2} CH_3-C(OH)(CH_3)-CH_2-CO-CH_3$

Heating causes dehydration to form an unsaturated ketone.

$CH_3-C(OH)(CH_3)-CH_2-CO-CH_3 \xrightarrow{\Delta} CH_3-C(CH_3)=CH-CO-CH_3$

Numbering the principal carbon chain to give the lowest locant to the ketone group, the IUPAC name of the product is 4-methylpent-3-en-2-one.

Answer: 4-Methylpent-3-en-2-one

JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Organic Chemistry in other years

More JEE Main Organic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Inorganic Chemistry

Keep practicing Organic Chemistry