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

Options

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

Answer

To solve the problem, we need to analyze the given reaction step by step.

### Step 1: Identify the starting compound
The starting compound is 1-propanol (CH₃CH₂CH₂OH), which is a three-carbon alcohol.

### Step 2: Reaction with PCl₅
When 1-propanol is treated with phosphorus pentachloride (PCl₅), it undergoes a nucleophilic substitution reaction. PCl₅ converts the alcohol (-OH) group into a chloride (-Cl) group. The reaction can be summarized as follows:

\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl} \]

The product of this reaction is 1-chloropropane (CH₃CH₂CH₂Cl), which we will denote as compound A.

### Step 3: Reaction with alcoholic KOH
Next, we treat compound A (1-chloropropane) with alcoholic KOH. Alcoholic KOH promotes elimination reactions, specifically dehydrohalogenation, where HCl is eliminated. The reaction can be summarized as follows:

\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{alc. KOH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{HCl} \]

In this reaction, the chlorine atom (Cl) is removed along with a hydrogen atom (H) from the adjacent carbon atom, resulting in the formation of an alkene.

### Step 4: Identify the final product
The product formed from the elimination reaction is propene (CH₃CH=CH₂). This compound has a double bond between the first and second carbon atoms.

### Conclusion
Thus, the compound B formed in the reaction is propene.

### Final Answer
The compound 'B' is propene (CH₃CH=CH₂).

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