An alkyl halide (X) on reaction with ethanolic sodium hydroxide forms an alkene (Y) which on further reaction with HBr gives the same alkyl halide.The alkene (Y) on reaction with HBr/ peroxide followed by reaction with Mg metal followed by reaction with HCN produces an aldehyde (Z).Z is :

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we need to analyze the reactions described in the question.

### Step 1: Identify the Alkyl Halide (X)
Let’s assume the alkyl halide (X) is a simple alkyl halide, such as bromoethane (C2H5Br).

### Step 2: Reaction with Ethanolic Sodium Hydroxide
When alkyl halide (X) reacts with ethanolic sodium hydroxide (NaOH), it undergoes an elimination reaction (dehydrohalogenation) to form an alkene (Y).
- For example, bromoethane (C2H5Br) reacts with NaOH to form ethene (C2H4).

**Reaction:**
C2H5Br + NaOH (ethanol) → C2H4 (Y) + NaBr + H2O

### Step 3: Reaction of Alkene (Y) with HBr
The alkene (Y) then reacts with HBr to give back the same alkyl halide (X).
- In our case, ethene (C2H4) reacts with HBr to form bromoethane (C2H5Br).

**Reaction:**
C2H4 (Y) + HBr → C2H5Br (X)

### Step 4: Reaction of Alkene (Y) with HBr in Presence of Peroxide
Now, if the alkene (Y) reacts with HBr in the presence of peroxide, it follows the anti-Markovnikov addition rule, leading to the formation of a bromoalkane where bromine adds to the less substituted carbon.
- For ethene, this step is not applicable as it is symmetrical, but if we consider a more complex alkene, the product would be different.

**Example Reaction:**
C2H4 + HBr (in presence of peroxide) → C2H5Br (still the same for symmetrical alkenes)

### Step 5: Reaction with Magnesium Metal
The product from the previous step (bromoalkane) reacts with magnesium metal to form a Grignard reagent.
- For example, C2H5Br + Mg → C2H5MgBr

### Step 6: Reaction with HCN
The Grignard reagent (C2H5MgBr) then reacts with HCN to form an intermediate that eventually leads to the formation of an aldehyde (Z).
- The reaction with HCN introduces a cyanide group, and upon hydrolysis, it will convert to an aldehyde.

**Final Reaction:**
C2H5MgBr + HCN → C2H5-CN (intermediate) → C2H5CHO (aldehyde Z)

### Conclusion
The final product (aldehyde Z) is **ethyl aldehyde (acetaldehyde)**, which is represented as CH3CHO.

### Final Answer:
**Z is: Acetaldehyde (CH3CHO)**

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