Bromination of PhCOMe in acetic acid medium produces mainly

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Explanation

To determine the main product of the bromination of PhCOMe (benzaldehyde) in acetic acid medium, we can follow these steps:

### Step 1: Understand the Reactants
PhCOMe is benzaldehyde, which has a carbonyl group (C=O) attached to a phenyl group (Ph) and a methyl group (Me). The reaction takes place in an acetic acid medium, which is a polar protic solvent.

### Step 2: Identify the Reaction Type
Bromination is an electrophilic substitution reaction. In the presence of acetic acid, the carbonyl group of benzaldehyde can be activated towards electrophilic attack.

### Step 3: Electrophilic Attack
In acetic acid, the bromine (Br2) can act as an electrophile. The carbonyl carbon in benzaldehyde is partially positive due to the electronegativity of the oxygen atom, making it susceptible to attack by the bromine.

### Step 4: Formation of the Product
The bromine will add to the carbon adjacent to the carbonyl group. In this case, the methyl group (Me) is the site of bromination because the carbonyl group stabilizes the formation of the intermediate. The product formed will be bromobenzaldehyde.

### Step 5: Identify the Main Product
The main product of the bromination of PhCOMe in acetic acid medium is 2-bromobenzaldehyde. This is because the bromine preferentially adds to the methyl group due to steric and electronic factors.

### Final Answer:
The main product of the bromination of PhCOMe in acetic acid medium is **2-bromobenzaldehyde**.

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