Identify B in the following scheme :
`CH_3CH_2CH_2OH overset(PCl_5)(rarr)A overset(alc. KOH)(rarr)B`

Options

A. Propyne
B. Propene
C. Propanol
D. Propanone

Answer

To identify compound B in the reaction scheme provided, we will follow the steps of the reactions systematically.

### Step 1: Identify the starting compound
The starting compound is **CH₃CH₂CH₂OH**, which is **propan-1-ol** (or simply propanol).

### Step 2: Reaction with PCl₅
When propan-1-ol reacts with phosphorus pentachloride (PCl₅), it undergoes chlorination. The hydroxyl group (-OH) is replaced by a chlorine atom (-Cl).

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl} \]

Thus, the product A is **1-chloropropane (CH₃CH₂CH₂Cl)**.

### Step 3: Reaction with alcoholic KOH
Next, the compound A (1-chloropropane) reacts with alcoholic potassium hydroxide (alc. KOH). This reaction typically leads to an elimination reaction (dehydrohalogenation), where the chlorine atom is removed along with a hydrogen atom from the adjacent carbon, resulting in the formation of an alkene.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{alc. KOH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O} \]

The product B formed from this reaction is **propene (CH₃CH=CH₂)**.

### Final Answer
Thus, the compound B is **propene (CH₃CH=CH₂)**.

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