`(CH_(3))_(2)COoverset(NaCN)underset(HCI)rarr(A)underset(Delta)overset(H_(3)^(+)O)rarr(B)`
In the above sequence of reactions, (A) and (B) are :

Options

A. `(CH_(3))_(2)C(OH)CN. (CH_(3))_(2)C(OH)COOH`
B. `(CH_(3))_(2)C(OH)CNH_(2). (CH_(3))_(2)C(OH)_(2)`
C. `(CH_(3))_(2)C(OH)CN.(CH_(3))_(2)CHCOOH`
D. `(CH_(3))_(2)C(OH)CNH_(2).(CH_(3))_(2)C=0`

Answer

To solve the given question, we will analyze the reaction step by step.

### Step 1: Identify the starting compound
The starting compound is (CH₃)₂CO, which is acetone (a ketone). The structure of acetone is:
```
O
||
CH₃-C-CH₃
```

### Step 2: Nucleophilic attack by CN⁻
In the presence of NaCN, the cyanide ion (CN⁻) acts as a nucleophile. The carbonyl carbon of acetone is electrophilic due to the partial positive charge on it. The CN⁻ ion attacks the carbonyl carbon, leading to the formation of a cyanohydrin.

The reaction can be represented as follows:
```
(CH₃)₂CO + CN⁻ → (CH₃)₂C(OH)(CN)
```
This product can be denoted as A:
```
A: (CH₃)₂C(OH)(CN)
```

### Step 3: Protonation of the alkoxide
When the product A reacts with HCl, the alkoxide (O⁻) gets protonated:
```
(CH₃)₂C(OH)(CN) + H⁺ → (CH₃)₂C(OH)(CN)
```
This step does not change the structure of A but confirms that we have a hydroxyl group and a cyanide group attached to the same carbon.

### Step 4: Hydrolysis of the cyanide group
Now, when product A undergoes hydrolysis in the presence of heat (Δ), the CN group is converted into a carboxylic acid (COOH). This reaction can be represented as:
```
(CH₃)₂C(OH)(CN) + H₂O → (CH₃)₂C(OH)(COOH)
```
This product can be denoted as B:
```
B: (CH₃)₂C(COOH)(OH)
```

### Final Products
Thus, the final products are:
- (A) = (CH₃)₂C(OH)(CN)
- (B) = (CH₃)₂C(COOH)(OH)

### Summary of Products
- A: (CH₃)₂C(OH)(CN)
- B: (CH₃)₂C(COOH)(OH)