In the reaction `CH_3-C-=C-Hoverset((i) NaNH_2//NH_3(l))rarr(A)overset(CH_3-CH_2-CH_2-Br)rarr(B)` The product B is

Options

A. `CH_3-C-=C-CH_2-CH_2-CH_3`
B. `CH_3-CH=CH_2`
C. `CH_3-CH_2-C-=C-CH_2-CH_3`
D. `CH_3-CH=C=CH-CH_2-CH_3`

Answer

To solve the given reaction step by step, let's break it down:

### Step 1: Identify the Starting Material
The starting material is 1-butyne, represented as:
\[ CH_3-C \equiv C-H \]

### Step 2: Reaction with Sodium Amide (NaNH₂) in Liquid Ammonia (NH₃)
When 1-butyne reacts with sodium amide in liquid ammonia, it acts as a strong base. Sodium amide will deprotonate the terminal hydrogen of the alkyne, generating a nucleophilic anion:
\[ CH_3-C \equiv C^- \]

### Step 3: Formation of the Nucleophile
The deprotonation leads to the formation of the anion:
\[ CH_3-C \equiv C^- \]
This anion can now act as a nucleophile due to the negative charge on the carbon atom.

### Step 4: Reaction with 1-Bromopropane (CH₃-CH₂-CH₂-Br)
The next step involves the nucleophile attacking the electrophilic carbon in 1-bromopropane:
\[ CH_3-CH_2-CH_2-Br \]
The nucleophile will attack the carbon bonded to bromine, leading to the displacement of bromine (a good leaving group).

### Step 5: Product Formation
The nucleophilic attack results in the formation of a new carbon-carbon bond. The product after the substitution reaction will be:
\[ CH_3-C \equiv C-CH_2-CH_2-CH_3 \]
This compound is 1-hexyne.

### Final Product
Thus, the product B is:
\[ \text{1-Hexyne} \]

### Summary of the Reaction
The overall transformation can be summarized as:
\[ CH_3-C \equiv C-H \xrightarrow{NaNH_2/NH_3} CH_3-C \equiv C^- \xrightarrow{CH_3-CH_2-CH_2-Br} CH_3-C \equiv C-CH_2-CH_2-CH_3 \]