What is compound Z :
`CH_(3)CH_(2)CH_(2)Broverset(NaCN)rarrXoverset(H_(2)O^(+))rarrYunderset(H^(+))overset(CH_(3)CH_(2)OH)rarrZ`

Options

A. `CH_(3)CH=Choverset(O)overset(||)COH`
B. `CH_(3)CH_(2)CH_(2)CH=NOCH_(2)CH_(3)`
C. `CH_(3)CH_(2)CH_(2)CH(OCH_(2)CH_(3))_(2)`
D. `CH_(3)CH_(2)CH_(2)overset(O)overset(||)COCH_(2)CH_(3)`

Answer

To determine the compound Z from the given reaction sequence, we will break down the steps as follows:

### Step 1: Identify the starting compound
The starting compound is **CH₃CH₂CH₂Br** (1-bromopropane).

### Step 2: Reaction with NaCN
When 1-bromopropane reacts with sodium cyanide (NaCN), a nucleophilic substitution occurs. The bromine atom is replaced by a cyanide group (CN).

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{NaBr} \]

Here, we form **X** which is **CH₃CH₂CH₂CN** (propionitrile).

### Step 3: Hydrolysis of X
Next, the compound X (propionitrile) undergoes hydrolysis in the presence of water (H₂O) and an acid (H⁺). This hydrolysis converts the nitrile group (CN) into a carboxylic acid (COOH).

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 2 \text{H}_2\text{O} \xrightarrow{H^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_3 \]

Here, we form **Y** which is **CH₃CH₂CH₂COOH** (propanoic acid).

### Step 4: Esterification of Y with Ethanol
The next step involves the reaction of the carboxylic acid (Y) with ethanol (CH₃CH₂OH). This is an esterification reaction where the carboxylic acid reacts with an alcohol to form an ester.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{H^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O} \]

Here, we form **Z** which is **CH₃CH₂CH₂COOCH₂CH₃** (ethyl propanoate).

### Conclusion
The final compound Z is **ethyl propanoate**.

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