The compound (A) when treated with methanol and few drops of conc.H \(_{2} \mathrm{SO}_{4}\) gives the smell of oil of winter green. The compound (A) is (a) phenol (b) benzene (c) \(\mathrm{p}\) -cresol (d) salicylic acid

Short Answer

Expert verified
Answer: Salicylic acid.

Step by step solution

01

Analyze the reaction conditions and products

To find the compound (A), let's first analyze the reaction conditions. Methanol and concentrated H\(_2\)SO\(_4\) are involved in esterification reactions, where an alcohol reacts with a carboxylic acid, producing an ester and water. The smell of oil of wintergreen (methyl salicylate) is an indication that an ester is being formed.
02

Identify the compounds that can potentially undergo esterification

Now, let's look at the given compounds: (a) Phenol - It is an alcohol, but does not have a carboxylic acid group. Hence, it cannot undergo esterification. (b) Benzene - It does not have an alcohol group or a carboxylic acid group. So, it cannot undergo esterification. (c) \(\mathrm{p}\)-Cresol - It has an alcohol group, but does not have a carboxylic acid group. Hence it cannot undergo esterification. (d) Salicylic acid - It has both an alcohol group and a carboxylic acid group. So, it can undergo esterification.
03

Determine the product of the esterification reaction

When salicylic acid reacts with methanol in the presence of concentrated H\(_2\)SO\(_4\), an esterification reaction takes place. The product is methyl salicylate, which has the smell of oil of wintergreen. The reaction can be represented as: Salicylic acid + Methanol -> Methyl salicylate + Water
04

Conclusion

Based on the above analysis, it is clear that the compound (A) that gives the smell of oil of wintergreen when treated with methanol and few drops of concentrated H\(_2\)SO\(_4\) is salicylic acid (d).

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