In the reaction sequence `CH_3-CH_2-CH_2-Broverset(AgCN) rarr (X) overset (H_2O/ H^(+))rarr(Y)` (Y) will be

Options

A. `CH_3-CH_2-CH_2-NH_2`
B. `CH_3-CH_2-NH_2`
C. `CH_3(CH_2)_2-CH_2-NH_2`
D. `CH_3-(CH_2)_3-CH_2-NH_2`

Answer

To solve the problem step by step, we will analyze the reaction sequence provided:

1. **Identify the starting compound**: The starting compound is CH3-CH2-CH2-Br (1-bromopropane).

2. **Nucleophilic substitution with AgCN**:
- In the presence of AgCN, the bromine atom (Br) is replaced by a cyanide group (CN).
- The nucleophile CN⁻ attacks the carbon atom bonded to Br, leading to the formation of a new compound.
- The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{AgCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{AgBr} \]
- The product formed here is CH3-CH2-CH2-CN (1-cyanopropane), which we will denote as (X).

3. **Hydrolysis of (X)**:
- The next step involves the hydrolysis of (X) in the presence of water (H2O) and an acid (H⁺).
- The cyanide group (CN) undergoes hydrolysis to form a carboxylic acid (COOH) and an amine (NH2).
- The hydrolysis reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 2\text{H}_2\text{O} \xrightarrow{H^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_3 \]
- The product formed in this step is CH3-CH2-CH2-COOH (propanoic acid) and NH3 (ammonia).

4. **Final product (Y)**:
- The final product (Y) after the hydrolysis of (X) is propanoic acid (CH3-CH2-COOH).

Thus, the final answer is:
**(Y) will be propanoic acid (CH3-CH2-COOH).**