Among the following, one which reacts most readily with ethanol is

Options

A. p-nitrobenzyl bromide
B. p-chlorobenzyl bromide
C. p-methoxybenzyl bromide.
D. p-methylbenzyl bromide.

Explanation

To solve the question of which compound reacts most readily with ethanol, we need to analyze the given options based on their structures and the effects of the substituents on the benzene ring.

### Step-by-Step Solution:

1. **Identify the Compounds**:
We have four compounds to analyze:
- A: Para nitrobenzyl bromide (C6H4(NO2)CH2Br)
- B: Para chlorobenzyl bromide (C6H4(Cl)CH2Br)
- C: Para methoxybenzyl bromide (C6H4(OCH3)CH2Br)
- D: Para methylbenzyl bromide (C6H4(CH3)CH2Br)

2. **Understand the Reactivity with Ethanol**:
Ethanol (C2H5OH) can react with these compounds through nucleophilic substitution. The reactivity depends on the stability of the carbocation formed after the leaving group (Br) departs.

3. **Analyze the Effects of Substituents**:
- **Para nitrobenzyl bromide (A)**: The nitro group (-NO2) is a strong electron-withdrawing group (-M effect), which deactivates the ring and makes it less reactive towards nucleophiles.
- **Para chlorobenzyl bromide (B)**: The chlorine atom (-Cl) has a -I effect (inductive) and a +M effect (resonance). It slightly activates the ring but not as effectively as other groups.
- **Para methoxybenzyl bromide (C)**: The methoxy group (-OCH3) has a +M effect (resonance) and a -I effect (inductive). It significantly activates the ring and stabilizes the carbocation formed after the bromine leaves.
- **Para methylbenzyl bromide (D)**: The methyl group (-CH3) has a +I effect (inductive) but no resonance effect. It provides some activation but is less effective than the methoxy group.

4. **Compare the Reactivity**:
- The nitro group in option A deactivates the compound, so it is eliminated.
- Option B (chlorobenzyl) has some activation but is weaker than option C.
- Option D (methylbenzyl) has a weak activating effect.
- Option C (methoxybenzyl) has the strongest activating effect due to resonance stabilization of the carbocation.

5. **Conclusion**:
Based on the analysis, **para methoxybenzyl bromide (C)** will react most readily with ethanol due to the strong activating effect of the methoxy group, which stabilizes the carbocation formed during the reaction.

### Final Answer:
**C: Para methoxybenzyl bromide**