The compound `C_7H_8` undergoes the following reactions :
`C_7H_8 overset(3Cl_2//Delta)(to) A overset(Br_2//Fe)(to) B overset(Zn//HCl)(to)C`
The product C is :
1) m- bromotoluene
2) o-bromotoluene
3)3-bromo-2, 4, 6-trichlorotoluene
4) p-bromotoluene

Options

A. m-bromotoluene
B. o-bromotoluene
C. 3-bromo-2, 4, 6-trichlorotoluene
D. p-bromotoluene

Answer

To determine the product C from the given reactions involving the compound C7H8, we will analyze each step of the reaction process systematically.

### Step 1: Identify the starting compound
The compound C7H8 is toluene, which can be represented as:
\[ \text{C}_6\text{H}_5\text{CH}_3 \]
This structure consists of a benzene ring (C6H5) with a methyl group (CH3) attached.

**Hint:** Recognize that C7H8 corresponds to toluene, which is a methyl-substituted benzene.

### Step 2: Reaction with Cl2
The first reaction involves toluene reacting with 3 moles of Cl2 under heat:
\[ \text{C}_7\text{H}_8 + 3 \text{Cl}_2 \xrightarrow{\Delta} \text{CCl}_3\text{C}_6\text{H}_2 \]
In this reaction, three chlorine atoms replace three hydrogen atoms on the benzene ring, resulting in the formation of trichloromethylbenzene (CCl3C6H2).

**Hint:** Chlorination replaces hydrogen atoms on the aromatic ring, leading to the formation of a chlorinated product.

### Step 3: Reaction with Br2 and Fe
Next, the product A (trichloromethylbenzene) reacts with bromine (Br2) in the presence of iron (Fe):
\[ \text{CCl}_3\text{C}_6\text{H}_2 + \text{Br}_2 \xrightarrow{\text{Fe}} \text{CCl}_3\text{C}_6\text{HBr} \]
This reaction introduces a bromine atom at the meta position relative to the trichloromethyl group, yielding a compound known as meta-bromotrichloromethylbenzene.

**Hint:** The presence of the electron-withdrawing CCl3 group directs the bromination to the meta position.

### Step 4: Reduction with Zn and HCl
Finally, the product B (meta-bromotrichloromethylbenzene) undergoes reduction with zinc and hydrochloric acid (HCl):
\[ \text{CCl}_3\text{C}_6\text{HBr} \xrightarrow{\text{Zn/HCl}} \text{C}_6\text{H}_5\text{Br} + \text{CCl}_2 \]
In this reduction process, the CCl3 group is reduced, and the bromine remains at the meta position, resulting in the formation of meta-bromotoluene (C6H4Br).

**Hint:** Reduction typically removes halogen substituents, and the position of the remaining substituent is determined by the previous reactions.

### Conclusion
The final product C is:
**C: meta-bromotoluene**

### Answer Choices
1) m-bromotoluene (Correct)
2) o-bromotoluene
3) 3-bromo-2, 4, 6-trichlorotoluene
4) p-bromotoluene

**Final Answer:** The product C is **m-bromotoluene**.