An acid chloride on reaction with ` H_2//Pd -BaSO_4` gives `(CH_3 )_2 CHCHO` this acid chloride on reaction with dimethyl cadmium in the presence of dry ether , gives :

Options

A. `(CH_3 )_3 COH `
B. `(CH_3)_2 CHOH `
C. `(CH_3 )_2 CHCOCH_3`
D. `(CH_3)_2 CHCHO`

Explanation

To solve the problem step by step, we will analyze the reactions mentioned in the question.

### Step 1: Identify the Acid Chloride
The question states that an acid chloride reacts with `H2/Pd-BaSO4` to give `(CH3)2CHCHO`. This product is isobutyraldehyde (or 2-methylpropanal).

- The structure of isobutyraldehyde is:
```
CH3
|
CH3-C-CHO
```

### Step 2: Determine the Structure of the Acid Chloride
Since the product of the reduction is isobutyraldehyde, we can deduce that the acid chloride must be derived from it. The acid chloride that would yield isobutyraldehyde upon reduction is isobutyryl chloride.

- The structure of isobutyryl chloride is:
```
CH3
|
CH3-C-COCl
```

### Step 3: Reaction with Dimethyl Cadmium
Next, we need to consider the reaction of isobutyryl chloride with dimethyl cadmium in the presence of dry ether. Dimethyl cadmium (C2H6Cd) is a nucleophile and will react with the acid chloride to form a tertiary alcohol.

- The reaction can be represented as:
```
CH3-C(=O)Cl + (CH3)2Cd → CH3-C(OH)(CH3)2 + CdCl2
```

### Step 4: Determine the Final Product
The product formed from this reaction is a tertiary alcohol, specifically 3-methyl-2-butanol.

- The structure of 3-methyl-2-butanol is:
```
CH3
|
CH3-C-CHOH
|
CH3
```

### Step 5: Nomenclature of the Final Product
The final product can be named as follows:
- The longest carbon chain contains 4 carbons (butane).
- The hydroxyl group (OH) is on the second carbon.
- There is a methyl group on the third carbon.

Thus, the IUPAC name of the product is **3-methyl-2-butanol**.

### Final Answer
The compound formed when isobutyryl chloride reacts with dimethyl cadmium in the presence of dry ether is **3-methyl-2-butanol**.

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