`(A)overset((i)BD_(3)//THF)underset((ii)CH_(3)COOH)larrPenten eoverset((i)BD_(3)//THF)underset((ii)CH_(3)COOD)rarr(B)`
`(A)` and `(B)` are `:`

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the reaction of pentene with BD3 in THF followed by treatment with CH3COOH and CH3COOD. Let's break down the steps:

### Step 1: Draw the structure of pentene
Pentene is an alkene with five carbon atoms. The simplest form is 1-pentene, which can be represented as follows:

```
H H H H
| | | |
H--C==C--C--C--C--H
| | | |
H H H H
```

### Step 2: React pentene with BD3 in THF
When pentene reacts with BD3 in THF, the double bond in pentene undergoes hydroboration. This results in the addition of boron (B) and deuterium (D) across the double bond. The product will have D attached to the carbon that was less substituted (following Markovnikov's rule).

The structure after this reaction can be represented as:
```
H H H H
| | | |
H--C--C--C--C--C--D
| | | |
D H H H
```

### Step 3: Treat the product with CH3COOH
When the product from Step 2 is treated with CH3COOH, the boron is replaced by a hydrogen atom (H) from the acid, resulting in the formation of compound A. The structure will look like this:

```
H H H H
| | | |
H--C--C--C--C--C--H
| | | |
D H H H
```

### Step 4: Treat the product with CH3COOD
Now, if we treat the product from Step 2 with CH3COOD instead, the boron is replaced by deuterium (D) from the deuterated acetic acid. The structure will look like this:

```
H H H H
| | | |
H--C--C--C--C--C--D
| | | |
D H H D
```

### Step 5: Identify compounds A and B
- Compound A (formed with CH3COOH) has one D and one H.
- Compound B (formed with CH3COOD) has one D and one D.

### Conclusion
Thus, the final structures of compounds A and B can be summarized as follows:
- Compound A: Contains H from CH3COOH and D from BD3.
- Compound B: Contains D from CH3COOD and D from BD3.