The end product (Z) in the given sequence of reaction is
`CH_(3)CH=CHCHO overset(NaBH_(4))toXoverset(HCl)underset(ZnCl_(2))toY overset((i)KCN)underset((ii)H+)toZ`

Options

A. `CH_(3)CH=CHCH_(2)COOH`
B. `CH_(3)CH_(2)CH_(2)COOH`
C. `CH_(3)CH=CHCOOH`
D. `CH_(3)CH(Cl)CH_(2)COOH`

Explanation

To solve the question step by step, we will analyze each reaction in the sequence provided and identify the end product (Z).

### Step 1: Identify the starting compound
The starting compound is given as CH₃CH=CHCHO, which is an alpha, beta-unsaturated carbonyl compound.

### Step 2: Reduction with Sodium Borohydride (NaBH₄)
When CH₃CH=CHCHO is treated with sodium borohydride (NaBH₄), it acts as a mild reducing agent. It will reduce the carbonyl group (the aldehyde) to an alcohol while leaving the double bond intact.

The reaction can be represented as follows:
\[ \text{CH}_3\text{CH}=\text{CHCHO} + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2\text{OH} \]

This product is denoted as X:
\[ X = \text{CH}_3\text{CH}=\text{CH-CH}_2\text{OH} \]

### Step 3: Reaction with HCl and ZnCl₂ (Lucas Reagent)
Next, the compound X reacts with hydrochloric acid (HCl) and zinc chloride (ZnCl₂), which is known as Lucas reagent. The alcohol group (–OH) in X will be protonated to form a better leaving group (water), leading to the formation of a carbocation.

The reaction can be represented as:
1. Protonation of the alcohol:
\[ \text{CH}_3\text{CH}=\text{CH-CH}_2\text{OH} + \text{HCl} \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2\text{OH}_2^+ \]

2. Loss of water to form a carbocation:
\[ \text{CH}_3\text{CH}=\text{CH-CH}_2\text{OH}_2^+ \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2^+ + \text{H}_2\text{O} \]

3. Nucleophilic attack by chloride ion:
\[ \text{CH}_3\text{CH}=\text{CH-CH}_2^+ + \text{Cl}^- \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2\text{Cl} \]

This product is denoted as Y:
\[ Y = \text{CH}_3\text{CH}=\text{CH-CH}_2\text{Cl} \]

### Step 4: Reaction with KCN
The compound Y then reacts with potassium cyanide (KCN). In this step, the cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbon atom bonded to the chlorine atom.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH-CH}_2\text{Cl} + \text{KCN} \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2\text{CN} + \text{Cl}^- \]

This product is denoted as Z:
\[ Z = \text{CH}_3\text{CH}=\text{CH-CH}_2\text{CN} \]

### Step 5: Hydrolysis of the Nitrile
The final step involves the hydrolysis of the nitrile (–CN) group to convert it into a carboxylic acid (–COOH). This can be done by treating the compound Z with an acid (H⁺).

The hydrolysis reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH-CH}_2\text{CN} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}=\text{CH-CH}_2\text{COOH} \]

Thus, the final product Z is:
\[ Z = \text{CH}_3\text{CH}=\text{CH-CH}_2\text{COOH} \]

### Conclusion
The end product Z in the given sequence of reactions is:
\[ \text{Z} = \text{CH}_3\text{CH}=\text{CH-CH}_2\text{COOH} \]