Propanoic acid on warming with `Br_(2)` in presence of red P gives

Options

A. `CH_(3)CH_(2)COBr`
B. `CH_(3)CH_(2)Br`
C. `CH_(3)CHBrCOOH`
D. `CH_(2)BrCH_(2)COOH`

Answer

To solve the problem of what product is formed when propanoic acid is warmed with bromine in the presence of red phosphorus, we can follow these steps:

### Step 1: Understand the Reactants
We start with propanoic acid, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with Red Phosphorus and Bromine
When propanoic acid is treated with bromine (\( \text{Br}_2 \)) in the presence of red phosphorus, the red phosphorus reacts with bromine to form phosphorus tribromide (\( \text{PBr}_3 \)):
\[ 2 \text{P} + 3 \text{Br}_2 \rightarrow 2 \text{PBr}_3 \]

### Step 3: Formation of Bromopropanoic Acid
The phosphorus tribromide then reacts with propanoic acid, replacing the hydroxyl group (\( \text{-OH} \)) with a bromine atom. This results in the formation of bromopropanoic acid:
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{PBr}_3 \rightarrow \text{CH}_3\text{CH}_2\text{C(Br)O} \]

### Step 4: Tautomerization
The bromopropanoic acid can undergo tautomerization. In this process, the hydrogen atom adjacent to the carbonyl group can shift, leading to the formation of an enol form:
\[ \text{CH}_3\text{CH}_2\text{C(Br)=O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{C(OH)=C(Br)} \]

### Step 5: Final Product Formation
The enol form can then react further, where the bromine can abstract a hydrogen atom, leading to the formation of a double bond and the final product, which is 3-bromopropenoic acid:
\[ \text{CH}_3\text{CH}=\text{C(Br)COOH} \]

### Conclusion
The final product of the reaction of propanoic acid with bromine in the presence of red phosphorus is 3-bromopropenoic acid.

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