Sequential Transformation of Terminal Alkyne to Alkene Derivative

To determine the major product $\mathbf{X}$ formed in Scheme 1, let us analyze the reaction sequence step by step:

1. Step 1: Reaction of $\mathbf{M}$ with excess $\mathrm{NaNH}_2$ The starting material $\mathbf{M}$ is 3-butyn-1-ol ($\mathrm{HO-CH_2-CH_2-C\equiv CH}$), which possesses two acidic protons:

   • The alcoholic proton ($\mathrm{-OH}$, $\mathrm{p}K_a \approx 16$)
   • The terminal alkyne proton ($\mathrm{-C\equiv CH}$, $\mathrm{p}K_a \approx 25$)

   Treatment with excess strong base ($\mathrm{NaNH}_2$) deprotonates both acidic sites to form a dianion: $\mathrm{HO-CH_2-CH_2-C\equiv CH} \xrightarrow{\mathrm{NaNH}_2\text{ (excess)}} \mathrm{^{-}O-CH_2-CH_2-C\equiv C^{-}}$

2. Step 2: Reaction with 1 equivalent of $\mathrm{CH_3CH_2I}$ The dianion contains two nucleophilic centers: an alkoxide ion ($\mathrm{-O^-}$) and an acetylide carbanion ($\mathrm{-C\equiv C^-}$). Since carbon is less electronegative than oxygen, the acetylide carbanion is more nucleophilic than the alkoxide ion. Therefore, it preferentially undergoes nucleophilic substitution ($\mathrm{S_N2}$) with ethyl iodide ($\mathrm{CH_3CH_2I}$): $\mathrm{^{-}O-CH_2-CH_2-C\equiv C^{-}} + \mathrm{CH_3CH_2I} \longrightarrow \mathrm{^{-}O-CH_2-CH_2-C\equiv C-CH_2CH_3} + \mathrm{I^{-}}$

3. Step 3: Reaction with 1 equivalent of $\mathrm{CH_3I}$ The remaining nucleophilic alkoxide center then undergoes an $\mathrm{S_N2}$ reaction with methyl iodide ($\mathrm{CH_3I}$) to form a methyl ether: $\mathrm{^{-}O-CH_2-CH_2-C\equiv C-CH_2CH_3} + \mathrm{CH_3I} \longrightarrow \mathrm{CH_3O-CH_2-CH_2-C\equiv C-CH_2CH_3} + \mathrm{I^{-}}$

4. Step 4: Stereoselective Hydrogenation with $\mathrm{H_2}$, Lindlar's catalyst Hydrogenation of an internal alkyne in the presence of Lindlar's catalyst ($\mathrm{Pd/CaCO_3/Pb(OAc)_2}$ or quinoline) results in stereoselective syn-addition of hydrogen, producing the corresponding cis-alkene ((Z)-alkene): $\mathrm{CH_3O-CH_2-CH_2-C\equiv C-CH_2CH_3} \xrightarrow{\mathrm{H_2\text{, Lindlar's catalyst}}} \text{\textit{cis}-}\mathrm{CH_3O-CH_2-CH_2-CH=CH-CH_2CH_3}$

This structure corresponds to the molecule shown in Option (A).

Correct Answer: (A)