`CH_2=CH underset(OH)underset(|)CHCH_2CH_2OH overset(MnO_2)to A,` A is

Options

A. `CH_2=CH_2underset(O)underset(||)C""CH_2CH_2OH`
B. `CH_2=CHunderset(OH)underset(|)C""CH_2CHO`
C. `CH_2=CHunderset(O)underset(||)C""CH_2CHO`
D. `CH_2=CHunderset(O)underset(||)C""CH_2underset(O)underset(||)(COH)`

Answer

To solve the given question, we need to analyze the reaction of the compound with manganese dioxide (MnO2), which is a mild oxidizing agent. The compound in question is an allylic alcohol. Let's break down the steps to find the product A.

### Step-by-Step Solution:

1. **Identify the Structure of the Compound:**
The compound given is:
\[ CH_2=CH-CH(OH)-CH_2-CH_2OH \]
This compound has two hydroxyl (OH) groups, making it a diol.

2. **Recognize the Type of Alcohol:**
The alcohols present are:
- One is an allylic alcohol (the one adjacent to the double bond).
- The other is a primary alcohol (the one at the end of the chain).

3. **Understand the Reactivity with MnO2:**
Manganese dioxide (MnO2) is known to selectively oxidize allylic and benzylic alcohols. In this case, the allylic alcohol will be oxidized to a carbonyl compound (ketone).

4. **Determine the Reaction Mechanism:**
The allylic alcohol will undergo oxidation:
- The hydroxyl group (OH) on the allylic carbon will be converted to a carbonyl group (C=O).
- The structure after oxidation will be:
\[ CH_2=CH-CO-CH_2-CH_2OH \]

5. **Identify the Product A:**
The product A formed from the oxidation of the allylic alcohol is:
\[ CH_2=CH-CO-CH_2-CH_2OH \]
This compound contains a ketone functional group due to the oxidation of the allylic alcohol.

6. **Final Structure of Product A:**
The final structure can be represented as:
\[ CH_2=CH-CO-CH_2-CH_2OH \]
This is the product A.

### Conclusion:
The product A formed from the reaction of the given compound with MnO2 is:
\[ A = CH_2=CH-CO-CH_2-CH_2OH \]