Question

In the reaction
$H - C \equiv CH\xrightarrow[{\left( 2 \right)C{H_3}C{H_2}Br}]{{\left( 1 \right)NaN{H_2}/liq.N{H_3}}}X\xrightarrow[{\left( 2 \right)C{H_3}C{H_2}Br}]{{\left( 1 \right)NaN{H_2}/liq.N{H_3}}}Y$. X and Y are.
A) $X = 1 - Butyne$ , $Y = 3 - Hexyne$
B) $X = 2 - Butyne$ , $Y = 3 - Hexyne$
C) $X = 2 - Butyne$ , $Y = 2 - Hexyne$
D) $X = 1 - Butyne$ , $Y = 2 - Hexyne$

Answer 1

Hint: We have to remember that the terminal acetylenic compounds having acidic hydrogen react with sodium amide to form sodium acetylenes which may be converted to acids and higher acetylenes. We also remember that the sodamide is the most commonly used base for the formation of triple bonds from suitable alkyl halides.

Complete step by step answer:
We must remember that the sodium metal itself, Na, may be a one-electron reducer – it's oxidized to \[N{a^ + }\] within the process. When it reduces neutral hydrocarbons, charged species are formed, which devour a proton from the ammonia solvent. Overall, \[Na/N{H_3}\] may be a way of reducing some organic compounds. Sodium amide is a useful reagent and it is prepared by reacting metallic sodium with liquid ammonia in the presence of nitrate. It finds use in various reactions in which reacting species is the amide anion.
Sodium amide may be a strong base which reacts with terminal alkynes and removes the terminal acid hydrogen leaving alkynes as a carbanion, which in reaction with haloalkane produces longer alkynes by nucleophilic addition. Thus, $X = 1 - Butyne$ , $Y = 3 - Hexyne$.
We can write the given reaction as below,

Therefore, the option A is correct.

Note: Now we discuss about the dehydrohalogenation of alkyl halides as,
We need to know that the sodium amide induces the loss of two bromide ions from a vicinal dibromoalkane to give a carbon-carbon triple bond during a preparation of phenylacetylene. Generally two equivalents of sodium amide yield the specified alkynes. Three equivalents are necessary within the preparation of a terminal alkyne because the terminal \[CH\] of the resulting alkynes protonated the same amount of base.