A compound(A) reacts with thionyl chloride to give compound (B). (B) reacts with magnesium to form Grignard reagent which is treated with acetone and the product is hydrolysed to give 2-methyl-2-butanol.
What is(A) compound?

Options

A. Butanol
B. Propanol
C. Methanol
D. Ethanol

Answer

To determine the compound (A) that reacts with thionyl chloride to yield compound (B), which subsequently forms a Grignard reagent and leads to the final product of 2-methyl-2-butanol, we can follow these steps:

### Step 1: Identify the Final Product
The final product given is 2-methyl-2-butanol. This compound has the following structure:

- **Structure of 2-methyl-2-butanol**:
- It has a total of 5 carbon atoms (C5).
- The structure can be represented as:
```
CH3
|
CH3- C -OH
|
CH2-CH3
```

### Step 2: Understand the Reaction Pathway
We know that compound (B) reacts with magnesium to form a Grignard reagent. Grignard reagents are typically formed from alkyl halides. Therefore, compound (B) must be an alkyl halide.

### Step 3: Determine Compound (B)
Since the final product is 2-methyl-2-butanol, we can deduce that compound (B) must be a halide that can lead to this alcohol upon reaction with acetone and hydrolysis.

- **Possible structure of compound (B)**:
- The alkyl halide that would lead to 2-methyl-2-butanol when reacted with acetone is 2-methyl-2-bromobutane (or a similar halide).

### Step 4: Identify Compound (A)
Compound (A) reacts with thionyl chloride (SOCl2) to form compound (B). Thionyl chloride typically converts alcohols into alkyl chlorides.

- Since we are looking for an alcohol that can yield 2-methyl-2-bromobutane (compound B) upon reaction with thionyl chloride, we can deduce that compound (A) must be an alcohol with a structure that can lead to the formation of a secondary or tertiary halide.

- **Possible structure of compound (A)**:
- The simplest alcohol that can lead to 2-methyl-2-bromobutane when treated with thionyl chloride is **2-methyl-2-butanol** itself, which upon losing water and reacting with SOCl2, can yield the corresponding alkyl halide.

### Conclusion
Thus, the compound (A) is **2-methyl-2-butanol**.