What is the final product, C, of the following reaction sequence?
`CH_(3)-CH_(2)-C-=C-Hunderset((ii).CH_(3)CH_(2)Br)overset((i)NaNH_(2))toA`
`Aoverset(Na,NH_(3)(liq.))toB`
`Bunderset(CH_(2)Cl_(2))overset(Br_(2))toC`

Options

A.
B.
C.
D.

Answer

To determine the final product C of the given reaction sequence, we will follow the steps outlined in the question.

### Step 1: Reaction with NaNH₂
The starting compound is 1-butyne (CH₃-CH₂-C≡C-H). When it reacts with sodium amide (NaNH₂), the NaNH₂ acts as a strong base and abstracts a proton from the terminal alkyne, forming a sodium acetylide.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{C}\equiv\text{CH} + \text{NaNH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}^- \text{Na}^+ \]

This gives us the intermediate A:
\[ \text{A} = \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}^- \]

### Step 2: Reaction with CH₃CH₂Br
The next step involves the reaction of the sodium acetylide (A) with ethyl bromide (CH₃CH₂Br). The nucleophilic acetylide ion attacks the electrophilic carbon in the ethyl bromide, leading to the formation of a new carbon-carbon bond.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}\text{CH}_2\text{CH}_3 \]

This gives us the intermediate B:
\[ \text{B} = \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}\text{CH}_2\text{CH}_3 \]

### Step 3: Reaction with Na in liquid NH₃
The next step involves the reduction of the alkyne (B) using sodium in liquid ammonia (Na/NH₃). This reaction converts the alkyne into a trans-alkene.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{C}\equiv\text{C}\text{CH}_2\text{CH}_3 \xrightarrow{\text{Na/NH}_3} \text{trans-alkene} \]

The product will be:
\[ \text{trans-} \text{CH}_3\text{CH}_2\text{C}=\text{C}\text{CH}_2\text{CH}_3 \]

### Step 4: Reaction with Br₂ in CH₂Cl₂
Finally, the trans-alkene reacts with bromine (Br₂) in dichloromethane (CH₂Cl₂). This reaction leads to the formation of a vicinal dibromide.

**Reaction:**
\[ \text{trans-CH}_3\text{CH}_2\text{C}=\text{C}\text{CH}_2\text{CH}_3 + \text{Br}_2 \rightarrow \text{Br-CH}_3\text{CH}_2\text{C}(\text{Br})\text{C}\text{H}_2\text{CH}_3 \]

The final product C will have bromine atoms added to the double bond, resulting in:
\[ \text{C} = \text{Br-CH}_3\text{CH}_2\text{C}(\text{Br})\text{H}\text{CH}_2\text{CH}_3 \]

### Final Product
Thus, the final product C is:
\[ \text{C} = \text{Br-CH}_3\text{CH}_2\text{C}(\text{Br})\text{H}\text{CH}_2\text{CH}_3 \]