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)CH_(2)OH`

Explanation

To solve the problem step by step, we will analyze the reactions taking place in the given sequence.

### Step 1: Identify the starting compound
The starting compound is **CH₃CH₂COOH**, which is propanoic acid (a carboxylic acid).

**Hint:** Recognize the functional groups and structure of the starting compound.

### Step 2: Understand the first reaction
The first reaction involves **CH₃CH₂COOH** reacting with **P/Br₂**. This is known as the **Hale–Volhard–Zelinsky (HVZ) reaction**. In this reaction, the alpha-hydrogen of the carboxylic acid is replaced by a bromine atom.

**Hint:** Recall that in HVZ reaction, bromination occurs at the alpha position of carboxylic acids.

### Step 3: Perform the HVZ reaction
In the HVZ reaction, the alpha-hydrogen (the hydrogen attached to the carbon adjacent to the carboxylic acid) is replaced by bromine. Thus, the product A formed from this reaction will be **CH₃CHBrCOOH**.

**Hint:** Identify the position of the alpha-hydrogen and understand how it is replaced by bromine.

### Step 4: Identify the second reaction
The next step involves treating product A (**CH₃CHBrCOOH**) with **alc KOH** (alcoholic potassium hydroxide) followed by hydrolysis. This step typically leads to an elimination reaction.

**Hint:** Remember that alcoholic KOH promotes elimination reactions, leading to the formation of alkenes.

### Step 5: Perform the elimination reaction
In the elimination reaction, the bromine atom and a hydrogen atom from the adjacent carbon are removed, resulting in the formation of a double bond. The product formed will be **CH₂=CHCOOH**, which is **acrylic acid**.

**Hint:** Visualize the elimination process to see how the double bond is formed.

### Conclusion
The final product B after the reactions is **CH₂=CHCOOH** (acrylic acid).

**Final Answer:** Product B is **CH₂=CHCOOH** (acrylic acid).

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