CH 3 ​ −CH 2 ​ −Br Alco.KOH Δ ​ B HBr ​ C Na/ether ​ D the compound D is

Answer

To solve the question step by step, let's break down the reactions involving the compound CH₃-CH₂-Br.

### Step 1: Reaction with Alcoholic KOH
The first step involves the reaction of CH₃-CH₂-Br with alcoholic KOH under heat (Δ). Alcoholic KOH promotes elimination reactions. Here, the bromine (Br) will be eliminated along with a hydrogen atom (H) from the adjacent carbon, leading to the formation of an alkene.

**Reaction:**
\[ \text{CH}_3\text{-CH}_2\text{-Br} + \text{Alco. KOH} \xrightarrow{\Delta} \text{CH}_3\text{-CH}=\text{CH}_2 + \text{HBr} \]

**Product:**
The product formed is propene (CH₃-CH=CH₂).

### Step 2: Reaction with HBr
Next, the alkene (propene) reacts with HBr. This is an electrophilic addition reaction where HBr adds across the double bond. The bromine will add to one of the carbons of the double bond, and a hydrogen will add to the other.

**Reaction:**
\[ \text{CH}_3\text{-CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{-CH}_2\text{-Br} \]

**Product:**
The product formed is again bromoethane (CH₃-CH₂-Br).

### Step 3: Reaction with Sodium in Ether (Wurtz Reaction)
Finally, the bromoethane undergoes a Wurtz reaction when treated with sodium in ether. In this reaction, two alkyl halides couple to form a new alkane.

**Reaction:**
\[ 2 \text{CH}_3\text{-CH}_2\text{-Br} + 2 \text{Na} \xrightarrow{\text{Ether}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_3 + 2 \text{NaBr} \]

**Product:**
The product formed is butane (CH₃-CH₂-CH₂-CH₃).

### Final Answer
The compound D is butane (C₄H₁₀).

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