When `CH_(2)=CH -CO OH` is reduced with `LiAlH_(4)` the compound obtained will be

Options

A. `CH_(3)-CH_(2)-COOH`
B. `CH_(2)=CH-CH_(2)OH`
C. `CH_(3)-CH_(2)-CH_(2)OH`
D. `CH_(3)-CH_(2)-CHO`

Answer

To solve the problem of what compound is obtained when `CH2=CH-COOH` is reduced with `LiAlH4`, we can follow these steps:

### Step 1: Identify the Reactant
The reactant given is `CH2=CH-COOH`, which is an unsaturated carboxylic acid (specifically, it is acrylic acid).

### Step 2: Understand the Reducing Agent
`LiAlH4` (Lithium Aluminum Hydride) is a very strong reducing agent. It can reduce various functional groups, including carboxylic acids, to alcohols.

### Step 3: Determine the Reduction Process
When `LiAlH4` is used to reduce a carboxylic acid, the carboxylic acid group (`-COOH`) is reduced to a primary alcohol. Therefore, the `-COOH` group in `CH2=CH-COOH` will be converted to `-CH2OH` (the alcohol form).

### Step 4: Analyze the Alkene Part
The alkene part of the molecule (`CH2=CH`) remains unchanged during this reduction process. This is because the double bond has a high electron density, which makes it less susceptible to reduction by `LiAlH4`.

### Step 5: Write the Final Product
After the reduction, the structure of the compound will be:
- The double bond remains as `CH2=CH`.
- The carboxylic acid group is converted to an alcohol, resulting in `-CH2OH`.

Thus, the final product is `CH2=CH-CH2OH`.

### Final Answer
The compound obtained when `CH2=CH-COOH` is reduced with `LiAlH4` is `CH2=CH-CH2OH`.

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