The 'Y' in the following sequence of reactions `CH_3-underset(OH)underset(|)(CH)-CH_3overset(PBr_3)rarr(X)overset(aq. NaOH)rarr(Y)` is

Options

A. propene
B. propan - 1 - ol
C. propan -2 - ol
D. propyne

Answer

To solve the question regarding the sequence of reactions leading to compound 'Y', we will break down the process step by step.

### Step 1: Identify the starting compound
The starting compound is given as CH₃-CH(OH)-CH₃, which is 2-propanol (an alcohol).

### Step 2: Reaction with PBr₃
When 2-propanol (CH₃-CH(OH)-CH₃) reacts with phosphorus tribromide (PBr₃), a nucleophilic substitution reaction occurs. In this reaction, the hydroxyl group (-OH) is replaced by a bromine atom (Br).

The reaction can be represented as:
\[ \text{CH}_3-\text{CH(OH)}-\text{CH}_3 + \text{PBr}_3 \rightarrow \text{CH}_3-\text{CH(Br)}-\text{CH}_3 + \text{HBr} \]

Here, the product 'X' is 2-bromopropane (CH₃-CH(Br)-CH₃).

### Step 3: Reaction with aqueous NaOH
Now, the compound 'X' (2-bromopropane) reacts with aqueous sodium hydroxide (NaOH). In this step, NaOH acts as a base and abstracts a proton from the adjacent carbon, leading to the elimination of the bromine atom as a leaving group.

The reaction can be represented as:
\[ \text{CH}_3-\text{CH(Br)}-\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3-\text{CH}=\text{CH}_2 + \text{NaBr} + \text{H}_2\text{O} \]

The product 'Y' formed in this reaction is propene (CH₃-CH=CH₂).

### Conclusion
Thus, the final product 'Y' is propene.

### Final Answer
Y = Propene (C₃H₆)

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