Question

Which one of the following is the product (Z) formed at the end of the given reaction:
\[ \text{Propan-1-ol} + SOCl_2 \rightarrow [X] \]
\[ [X] + HC \equiv C^-Na^+ \rightarrow [Y] \]
\[ [Y] + H_2SO_4/HgSO_4,\; 330K \rightarrow [Z] \]

• A. Pentan-2-ol.
• B. Pent-2-ene.
• C. Pentanal.
• D. Pentan-2-one.

Hint:

Terminal alkynes on hydration with \( H_2SO_4/HgSO_4 \) give methyl ketones after enol-keto tautomerism.

Answer 1

The Correct Option is D

Step 1: Reaction of propan-1-ol with \( SOCl_2 \).
Propan-1-ol reacts with thionyl chloride \( SOCl_2 \) to form 1-chloropropane.
\[ CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl + SO_2 + HCl \]
Thus, compound \( [X] \) is 1-chloropropane.

Step 2: Reaction of \( [X] \) with sodium acetylide.
Sodium acetylide \( HC \equiv C^-Na^+ \) acts as a nucleophile.
It reacts with primary alkyl halide by substitution reaction and forms a higher alkyne.
\[ CH_3CH_2CH_2Cl + HC \equiv C^-Na^+ \rightarrow CH_3CH_2CH_2C \equiv CH + NaCl \]
Thus, compound \( [Y] \) is pent-1-yne.

Step 3: Hydration of pent-1-yne.
Pent-1-yne reacts with \( H_2SO_4/HgSO_4 \) at 330 K.
This is acid-catalysed hydration of terminal alkyne.
Water adds according to Markovnikov's rule to form an enol intermediate.

Step 4: Enol-keto tautomerism.
The enol formed during hydration is unstable and immediately converts into a ketone by keto-enol tautomerism.
For pent-1-yne, the final ketone formed is pentan-2-one.
\[ CH_3CH_2CH_2C \equiv CH \xrightarrow{H_2SO_4/HgSO_4} CH_3CH_2CH_2COCH_3 \]

Step 5: Identify product \( [Z] \).
The product \( CH_3CH_2CH_2COCH_3 \) is pentan-2-one.

Step 6: Conclusion.
Thus, the final product \( [Z] \) is:
\[ \boxed{\text{Pentan-2-one}} \]