Question

What are X and Z in the following reaction sequence? (X forms sodium alkynide) \[ X (C_4H_6) \xrightarrow{{partial reduction}} Y \xrightarrow{H_2O/H^+} Z ({major product}) \]

• A. \( \mathrm{CH_3C \equiv CCH_3}, \mathrm{CH_3CH_2CH(OH)CH_3} \)
• B. \( \mathrm{CH_3C \equiv CCH_3}, \mathrm{CH_3CH_2CH_2CH_2OH} \)
• C. \( \mathrm{CH_3CH_2C \equiv CH}, \mathrm{CH_3CH_2CH(OH)CH_3} \)
• D. \( \mathrm{CH_3CH_2C \equiv CH}, \mathrm{CH_3CH_2CH_2CH_2OH} \)

Hint:

Partial reduction of alkynes gives alkenes; acid hydration adds OH to the more substituted carbon (Markovnikov's rule).

Answer 1

The Correct Option is C

- Starting with 1-butyne \(X = \mathrm{CH_3CH_2C \equiv CH}\), partial reduction forms 1-butene \(Y\).
- Acid-catalyzed hydration of 1-butene produces the major product \(Z = \mathrm{CH_3CH_2CH(OH)CH_3}\), a secondary alcohol.