In the following reaction order , B is
`CH_(3)CH_(2)COOH overset (P//Br_(2)) to A overset (alc KOH) underset (H^(+)) to B`

Options

A. `CH_(2)=CH-COOH`
B. `CH_(2)-underset(Br)underset(|)CH-COOH`
C. `CH_(3)CH_(2)COBr`
D. `CH_(3)CH_(2)OH`

Explanation

To determine the product B in the given reaction sequence, we will analyze each step carefully.

### Step 1: Identify the starting compound
The starting compound is propionic acid, which has the formula:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with Bromine and Red Phosphorus
When propionic acid is treated with bromine in the presence of red phosphorus, a bromination reaction occurs at the alpha carbon (the carbon adjacent to the carboxylic acid group). This reaction is known as the Hell-Volhard-Zelinsky reaction.

In this case, one of the alpha hydrogens (attached to the alpha carbon) is replaced by a bromine atom. The product formed is:
\[ \text{CH}_3\text{CHBr}\text{COOH} \]
This compound is called 2-bromopropionic acid.

### Step 3: Reaction with Alcoholic KOH and H+
Next, the 2-bromopropionic acid undergoes elimination when treated with alcoholic KOH and an acid (H+). This elimination reaction involves the removal of HBr from the molecule.

The elimination of HBr leads to the formation of a double bond between the alpha and beta carbons. The resulting product is:
\[ \text{CH}_2=\text{CH}\text{COOH} \]
This compound is known as acrylic acid or prop-2-enoic acid.

### Conclusion
Thus, the final product B is:
\[ \text{CH}_2=\text{CH}\text{COOH} \]

### Answer
The correct answer is:
\[ \text{B} = \text{CH}_2=\text{CH}\text{COOH} \]