Question

An alkyne \(X\) \((C_4H_6)\) does not form sodium alkynide. Reaction of \(X\) with HBr gave \(Y\). Another reaction of \(X\) with \(Na/\text{liq. }NH_3\) gave \(Z\). Identify \(Y\) and \(Z\).

• A. \(Y=\) geminal dibromide ; \(Z=\) non-polar compound
• B. \(Y=\) geminal dibromide ; \(Z=\) polar compound
• C. \(Y=\) vicinal dibromide ; \(Z=\) polar compound
• D. \(Y=\) vicinal dibromide ; \(Z=\) non-polar compound

Hint:

Terminal alkynes: \[ RC\equiv CH \] form sodium alkynides because they contain acidic hydrogen. \[ Na/\text{liq. }NH_3 \] reduces alkynes to trans-alkenes. Addition of excess HX to alkynes gives geminal dihalides.

Answer 1

The Correct Option is A

Step 1: Identify the alkyne \(X\). Given: \[ X(C_4H_6) \] does not form sodium alkynide. Only terminal alkynes possess acidic hydrogen and form sodium alkynides. Therefore \(X\) must be an internal alkyne. Among \(C_4H_6\) alkynes, \[ CH_3CH_2C\equiv CH \] (1-butyne) is terminal, while \[ CH_3C\equiv CCH_3 \] (2-butyne) is internal. Hence, \[ \boxed{X=2\text{-butyne}} \]

Step 2: Reaction of \(X\) with HBr. Addition of two moles of HBr to an alkyne gives a geminal dibromide. \[ CH_3C\equiv CCH_3 \xrightarrow[2\,HBr]{} CH_3CBr_2CH_2CH_3 \] Thus, \[ \boxed{Y=\text{geminal dibromide}} \]

Step 3: Reaction of \(X\) with \(Na/\text{liq. }NH_3\). Dissolving metal reduction converts an alkyne into a trans-alkene. \[ CH_3C\equiv CCH_3 \xrightarrow{Na/NH_3} trans\text{-}CH_3CH=CHCH_3 \] which is trans-2-butene. Since the dipole moments cancel, \[ \mu=0. \] Therefore, \[ \boxed{Z=\text{non-polar compound}} \]

Final Answer: \[ \boxed{Y=\text{geminal dibromide}} \] \[ \boxed{Z=\text{non-polar compound}} \] \[ \boxed{\text{Answer = (A)}} \]