The major product obtained in the following reaction, is H-C≡ C-CH₂OH (i) LiNH₂ (2 equiv.)/liq. NH₃ (ii)… · ABC26OR0654

Subject: Organic Chemistry · Chapter: Reagents, Reaction Mechanisms & Named Reactions · Topic: Reagents, Reaction Mechanisms & Named Reactions – General · Exam: GATE 2000-2022 · Marks: 2 · Difficulty: Medium

The major product obtained in the following reaction, is
\[ \mathrm{H-C\equiv C-CH_2OH}\ \xrightarrow{\ \substack{\text{(i) }\mathrm{LiNH_2}\text{ (2 equiv.)/liq. }\mathrm{NH_3}\\ \text{(ii) }\mathrm{C_2H_5Br}\text{ (1 equiv.)}\\ \text{(iii) }\mathrm{H_3O^{+}}}\ } \]
(a)$\mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}$
(b)$\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$
(c)$\mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}_{2}$
(d)$\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{NH}-\mathrm{CH}_{2} \mathrm{CH}_{3}$
Answer
Answer: A ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
Two equivalents of LiNH₂ deprotonate both the OH and the terminal C–H; the acetylide (the more nucleophilic anion) is ethylated by one equivalent of EtBr, giving 2-pentyn-1-ol, CH₃CH₂C≡C–CH₂OH (a).

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