Consider the reaction `CH_(3)CH_(2)CH_(2)OH overset(PCl_(5))toAunderset(KOH)overset(alc)toB`. The compound 'B' is

Options

A. propane
B. propene
C. propyne
D. propanal

Explanation

To solve the question step by step, we will analyze the given reaction and identify the final compound 'B'.

### Step 1: Identify the starting compound
The starting compound is `CH₃CH₂CH₂OH`, which is propanol (specifically, 1-propanol).

### Step 2: Reaction with PCl₅
When 1-propanol reacts with phosphorus pentachloride (PCl₅), it undergoes a nucleophilic substitution reaction. The hydroxyl (-OH) group of propanol is replaced by a chlorine atom (Cl⁻) from PCl₅.

**Reaction:**
\[ CH₃CH₂CH₂OH + PCl₅ \rightarrow CH₃CH₂CH₂Cl + POCl₃ + HCl \]

The product of this reaction is **1-chloropropane** (or simply **chloropropane**), which we will denote as compound 'A'.

### Step 3: Reaction with alcoholic KOH
Next, 1-chloropropane (compound 'A') reacts with alcoholic potassium hydroxide (KOH). This reaction typically leads to an elimination reaction (dehydrohalogenation), where a hydrogen atom and a chlorine atom are removed, resulting in the formation of a double bond.

**Reaction:**
\[ CH₃CH₂CH₂Cl + KOH (alc) \rightarrow CH₃CH=CH₂ + KCl + H₂O \]

During this elimination reaction, the chlorine atom leaves along with a hydrogen atom from the adjacent carbon, resulting in the formation of propene.

### Step 4: Identify the final compound 'B'
The final product of this reaction is **propene** (C₃H₆), which can be represented as \( CH₃CH=CH₂ \).

### Conclusion
Thus, the compound 'B' is **propene**.

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