`(CH_(3))_(2) CO underset((HCl))overset("NaCN")(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)CN, `(CH_(3))_(2) C(OH)_(2)`
C. `(CH_(3))_(2)`C(OH)CN, `(CH_(3))_(2)` CHCOOH
D. `(CH_(3))_(2)` C(OH)CN, `(CH_(3))_(2)` C = O

Answer

To solve the problem, we need to identify the compounds A and B in the given reaction sequence. Let's break it down step by step.

### Step 1: Identify the starting compound
The starting compound is (CH₃)₂CO, which is acetone.

### Step 2: Reaction with HCl and NaCN
When acetone reacts with HCl, it forms a protonated intermediate. The presence of NaCN indicates that a nucleophilic addition reaction will occur. The cyanide ion (CN⁻) acts as a nucleophile and attacks the carbonyl carbon of acetone.

### Step 3: Formation of compound A
The nucleophilic attack leads to the formation of a cyanohydrin. The structure of compound A can be represented as follows:
- The carbonyl group (C=O) is converted into a hydroxyl group (C-OH) and a cyanide group (C≡N) is added to the carbon.
- Therefore, compound A is (CH₃)₂C(OH)CN, which can be written as CH₃C(OH)(CN)CH₃.

### Step 4: Acidic hydrolysis of compound A
Next, compound A undergoes acidic hydrolysis (indicated by Δ and H₃O⁺). During this process, the cyanide group (CN) is converted into a carboxylic acid (COOH).
- The reaction replaces the CN group with a COOH group, leading to the formation of compound B.

### Step 5: Formation of compound B
The structure of compound B is:
- The cyanide group (CN) in compound A is replaced by a carboxylic acid group (COOH).
- Thus, compound B is (CH₃)₂C(OH)COOH, which can be written as CH₃C(OH)COOHCH₃.

### Conclusion
- Compound A is (CH₃)₂C(OH)CN (a cyanohydrin).
- Compound B is (CH₃)₂C(OH)COOH (a carboxylic acid).

### Final Answer
- A: (CH₃)₂C(OH)CN
- B: (CH₃)₂C(OH)COOH