Question

Consider that 1,2-Dibromopropane on treatment with X moles of $NaN{{H}_{2}}$ followed by treatment with ${{C}_{2}}{{H}_{5}}Br$ gives a pentyne. The value of X is:
A.1
B.2
C.3
D.4

Answer 1

Here in this reaction, $NaN{{H}_{2}}$ participates as a strong nucleophile and will show elimination reaction at first. Then when there is no more possibility of elimination it acts as a strong base to abstract acidic protons and generate carbanions.

Complete step by step answer:
$NaN{{H}_{2}}$​ is a strong base and it readily abstracts the acidic protons. It mediates the formation of alkyne via elimination reaction. Treatment of 1,2-dibromopropane for the formation of pentyne requires 3 moles of $NaN{{H}_{2}}$​. In which 2 moles form a propyne by two consecutive eliminations of $HBr$ which further combines which sodium ions to give sodium bromide $NaBr$ These reactions are anti-elimination reactions. Whenever there is a formation of terminal alkyne, $NaN{{H}_{2}}$ takes up the acidic Hydrogen. The third mole of $NaN{{H}_{2}}$​ produces propyne carbanion by abstracting the acidic proton attaches to the sp-hybridised carbon in the terminal alkyne which further reacts with ${{C}_{2}}{{H}_{5}}Br$ and to form 2-Pentyne as the final product.

$C{{H}_{3}}CH(Br)C{{H}_{2}}(Br)\xrightarrow{2NaNH2}CH3-C\equiv CH\xrightarrow{NaNH2}C{{H}_{3}}-C\equiv {{C}^{-}}N{{a}^{+}}\xrightarrow{{{C}_{2}}{{H}_{5}}Br}C{{H}_{3}}C\equiv CC{{H}_{2}}C{{H}_{3}}$
Hence, correct answer is that 1,2-Dibromopropane uses 3 moles of $NaN{{H}_{2}}$ followed by treatment with ${{C}_{2}}{{H}_{5}}Br$ to give a pentyne, which is option C.

Note:
$NaN{{H}_{2}}$ is used for deprotonation of weak acids and also for elimination reactions. As a strong base, $NaN{{H}_{2}}$ readily deprotonated all the acidic functional groups like alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones. It can also be regarded as a very strong nucleophile.