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Question: Identify the reactants (X) and (Y) for the following reaction, respectively. (X) + (Y) $\xrightarro...

Identify the reactants (X) and (Y) for the following reaction, respectively.

(X) + (Y) →NaNH2\xrightarrow{NaNH_2} 4- Decyne Alkyl halide Alkyne

A

CH3(CH2)4CH2-CI + CH3-C ≡\equiv CH

B

CH3(CH2)2CH2-CI + CH3-(CH2)2-C ≡\equiv CH

C

CH3(CH2)2CH2-CI + CH3(CH2)3-C≡\equivCH

D

CH3-CH2-CH2-CI + CH3(CH2)4-C≡\equivCH

Answer

D

Explanation

Solution

The reaction shown is the synthesis of an internal alkyne by the alkylation of a terminal alkyne using a strong base like NaNH2NaNH_2. The general reaction is:

R−C≡CH+R′−X→NaNH2R−C≡C−R′+NaX+NH3R-C\equiv CH + R'-X \xrightarrow{NaNH_2} R-C\equiv C-R' + NaX + NH_3

Here, (Y) is the terminal alkyne (R−C≡CHR-C\equiv CH) and (X) is the alkyl halide (R′−XR'-X). NaNH2NaNH_2 deprotonates the terminal alkyne to form the acetylide anion (R−C≡C−R-C\equiv C^-), which then acts as a nucleophile in an SN2 reaction with the alkyl halide R′−XR'-X to form the new carbon-carbon bond and produce the internal alkyne R−C≡C−R′R-C\equiv C-R'.

The product is given as 4-Decyne, which has the structure CH3−CH2−CH2−C≡C−CH2−CH2−CH2−CH2−CH3CH_3-CH_2-CH_2-C\equiv C-CH_2-CH_2-CH_2-CH_2-CH_3.

Option (D) yields 4-Decyne: (X) = CH3−CH2−CH2−ClCH_3-CH_2-CH_2-Cl (1-Propyl chloride) (Y) = CH3(CH2)4−C≡CHCH_3(CH_2)_4-C\equiv CH (1-Heptyne)