Which one of the following is the product ({Z}) formed at the end of the given reaction: {aligned} & { Propan-1-ol }+{SOCl}₂ arrow[{X}] \\ & {[{X}]+{HC} {C}-{Na} arrow[{Y}]} \\ & {[{Y}]+{H}₂ {SO}₄ / {HgSO}₄, 330 {~K} arrow[{Z}]} {aligned}

Correct Answer is : Pentan-2-one.

Correct Answer is : Pentan-2-one.

Test Index

COMEDK 2025 Evening Solved Paper

Question : 115 of 180

Marks: +1, -0

(\mathrm{Z})

\begin{aligned} & \text { Propan-1-ol }+\mathrm{SOCl}_2 \rightarrow[\mathrm{X}] \\ & {[\mathrm{X}]+\mathrm{HC} \equiv \mathrm{C}-\mathrm{Na} \rightarrow[\mathrm{Y}]} \\ & {[\mathrm{Y}]+\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{HgSO}_4, 330 \mathrm{~K} \rightarrow[\mathrm{Z}]} \end{aligned}

Solution:

Concept:

Hydration of a terminal alkyne in acidic medium yields a methyl ketone.

Explanation:

Step 1: Propan-1‑ol reacts with

\mathrm{SOCl}_2

to form 1‑chloropropane (X).

\mathrm{CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl}

Step 2: 1‑Chloropropane undergoes

\mathrm{S_N2}

substitution with sodium acetylide (

\mathrm{HC \equiv C - Na}

) to produce 1‑pentyne (Y).

\mathrm{CH_3CH_2CH_2Cl + HC \equiv C - Na \rightarrow HC \equiv C - (CH_2)_2CH_3}

Step 3: Hydration of 1‑pentyne using

\mathrm{H_2SO_4 / HgSO_4}

at 330 K follows Markovnikov’s rule.

An enol is formed first, which then tautomerizes to the ketone.

\mathrm{HC \equiv C - (CH_2)_2CH_3 + H_2O \rightarrow CH_2 = C(OH)(CH_2)_2CH_3}

\mathrm{CH_2 = C(OH)(CH_2)_2CH_3 \rightarrow CH_3COCH_2CH_2CH_3}

The final product (Z) is pentan‑2‑one.

Answer:

Option D — Pentan‑2‑one.

Go to Question:

Prev Question

Next Question