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Organic Compounds Containing Oxygen: Chemistry | JEE Main

In a reaction, reagents '

\mathrm{X}

' and '

\mathrm{Y}

' respectively are :

Salicylic acid reacts with Y to form methyl salicylate (oil of wintergreen) and with X to form acetylsalicylic acid (aspirin).

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A

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

and

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

B

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

and

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

C

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

and

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

D

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

and

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

Step-by-step solutionView

Correct answer

Reagent X is acetic anhydride in acid which acetylates the phenolic group to form aspirin, while reagent Y is methanol with acid and heat which esterifies the carboxylic acid group to form methyl salicylate.

Option analysis

Why each option works or fails

A ·

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

and

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

Believing that acetic anhydride can both acetylate the phenolic hydroxyl group and convert the carboxylic acid to a methyl ester. Acetic anhydride acts as an acylating agent for nucleophiles like $-\mathrm{OH}$ ; it cannot introduce a methyl ester ( $-\mathrm{COOCH}_3$ ) group onto a carboxylic acid.

B ·

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

and

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

Reversing the functional group transformations, assuming Fischer esterification produces aspirin while acylation produces methyl salicylate. Identify the changed functional groups in each product: aspirin retains $-\mathrm{COOH}$ and converts $-\mathrm{OH}$ to $-\mathrm{OCOCH}_3$ (requires $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}$ ), whereas methyl salicylate retains $-\mathrm{OH}$ and converts $-\mathrm{COOH}$ to $-\mathrm{COOCH}_3$ (requires $\mathrm{CH}_3\mathrm{OH}/\mathrm{H}^+$ ).

C ·

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

and

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

Assuming that acid-catalyzed reaction with methanol can simultaneously acetylate a phenol and esterify a carboxylic acid. Methanol can esterify the carboxylic acid group ( $-\mathrm{COOH} \rightarrow -\mathrm{COOCH}_3$ ), but it cannot add an acetyl group ( $-\mathrm{COCH}_3$ ) to the phenolic oxygen.

D ·

(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}

and

\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta

None. The choice correctly identifies the reagents for O-acetylation and Fischer esterification. Correct. Acetic anhydride under acidic conditions acetylates the phenolic $-\mathrm{OH}$ group to form aspirin (acetylsalicylic acid), while methanol with catalytic acid and heat esterifies the carboxylic acid group to yield methyl salicylate.

Reviewed route

Solution

StepWorking

01identify

The starting material is salicylic acid (2-hydroxybenzoic acid), containing both a phenolic $-\mathrm{OH}$ group and a carboxylic acid $-\mathrm{COOH}$ group. Reaction with reagent 'X' acetylates the phenolic $-\mathrm{OH}$ group to form aspirin (acetylsalicylic acid), leaving $-\mathrm{COOH}$ intact. Reaction with reagent 'Y' esterifies the $-\mathrm{COOH}$ group with methanol to form methyl salicylate (oil of wintergreen), leaving phenolic $-\mathrm{OH}$ intact.

02mechanism

To convert salicylic acid into aspirin (2-acetoxybenzoic acid), the phenolic $-\mathrm{OH}$ acts as a nucleophile attacking acetic anhydride in acid: $\mathrm{Ar-OH} + (\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} \xrightarrow{\mathrm{H}^+} \mathrm{Ar-OCOCH}_3 + \mathrm{CH}_3\mathrm{COOH}$ . Thus, reagent 'X' is $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} / \mathrm{H}^+$ .

03mechanism

To convert salicylic acid into methyl salicylate (methyl 2-hydroxybenzoate), Fischer esterification of the carboxylic acid group $-\mathrm{COOH}$ with methanol is performed: $\mathrm{Ar-COOH} + \mathrm{CH}_3\mathrm{OH} \xrightarrow{\mathrm{H}^+, \Delta} \mathrm{Ar-COOCH}_3 + \mathrm{H}_2\mathrm{O}$ . Thus, reagent 'Y' is $\mathrm{CH}_3\mathrm{OH} / \mathrm{H}^+, \Delta$ .

04product

Reagents 'X' and 'Y' are $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} / \mathrm{H}^+$ and $\mathrm{CH}_3\mathrm{OH} / \mathrm{H}^+, \Delta$ , respectively, corresponding to Option (3).

✓verify

Checking product structures: Aspirin contains $-\mathrm{OCOCH}_3$ and $-\mathrm{COOH}$ , formed by acetylation of $-\mathrm{OH}$ via $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}$ . Oil of wintergreen contains $-\mathrm{OH}$ and $-\mathrm{COOCH}_3$ , formed by methanol esterification of $-\mathrm{COOH}$ via $\mathrm{CH}_3\mathrm{OH}/\mathrm{H}^+$ . Order X, Y matches Option (3).

Hints that build this answer step by step

What functional group transformation converts salicylic acid into aspirin (acetylsalicylic acid)?
Acetylation of the phenolic $-\mathrm{OH}$ group to an acetate ester ( $-\mathrm{OCOCH}_3$ )
Which reagent pair accomplishes the transformation of salicylic acid to aspirin (reagent X) and to methyl salicylate (reagent Y)?
$\mathrm{X} = (\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} / \mathrm{H}^+$ and $\mathrm{Y} = \mathrm{CH}_3\mathrm{OH} / \mathrm{H}^+, \Delta$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does acetic anhydride acetylate the phenolic -OH rather than the carboxylic acid group?

Phenolic -OH is a better nucleophile than the carbonyl oxygen of -COOH; reaction with acetic anhydride selectively yields an ester at the phenolic position (aspirin).

Answer

Reagent X is acetic anhydride in acid which acetylates the phenolic group to form aspirin, while reagent Y is methanol with acid and heat which esterifies the carboxylic acid group to form methyl salicylate.

Why each option works or fails

A: $(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}$ and $(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}$ - Believing that acetic anhydride can both acetylate the phenolic hydroxyl group and convert the carboxylic acid to a methyl ester. Acetic anhydride acts as an acylating agent for nucleophiles like $-\mathrm{OH}$ ; it cannot introduce a methyl ester ( $-\mathrm{COOCH}_3$ ) group onto a carboxylic acid.
B: $\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta$ and $(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}$ - Reversing the functional group transformations, assuming Fischer esterification produces aspirin while acylation produces methyl salicylate. Identify the changed functional groups in each product: aspirin retains $-\mathrm{COOH}$ and converts $-\mathrm{OH}$ to $-\mathrm{OCOCH}_3$ (requires $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}$ ), whereas methyl salicylate retains $-\mathrm{OH}$ and converts $-\mathrm{COOH}$ to $-\mathrm{COOCH}_3$ (requires $\mathrm{CH}_3\mathrm{OH}/\mathrm{H}^+$ ).
C: $\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta$ and $\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta$ - Assuming that acid-catalyzed reaction with methanol can simultaneously acetylate a phenol and esterify a carboxylic acid. Methanol can esterify the carboxylic acid group ( $-\mathrm{COOH} \rightarrow -\mathrm{COOCH}_3$ ), but it cannot add an acetyl group ( $-\mathrm{COCH}_3$ ) to the phenolic oxygen.
D · correct: $(\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O} / \mathrm{H}^{+}$ and $\mathrm{CH}_{3}\mathrm{OH} / \mathrm{H}^{+}, \Delta$ - None. The choice correctly identifies the reagents for O-acetylation and Fischer esterification. Correct. Acetic anhydride under acidic conditions acetylates the phenolic $-\mathrm{OH}$ group to form aspirin (acetylsalicylic acid), while methanol with catalytic acid and heat esterifies the carboxylic acid group to yield methyl salicylate.

Step-by-step solution

identify: The starting material is salicylic acid (2-hydroxybenzoic acid), containing both a phenolic $-\mathrm{OH}$ group and a carboxylic acid $-\mathrm{COOH}$ group. Reaction with reagent 'X' acetylates the phenolic $-\mathrm{OH}$ group to form aspirin (acetylsalicylic acid), leaving $-\mathrm{COOH}$ intact. Reaction with reagent 'Y' esterifies the $-\mathrm{COOH}$ group with methanol to form methyl salicylate (oil of wintergreen), leaving phenolic $-\mathrm{OH}$ intact.
mechanism: To convert salicylic acid into aspirin (2-acetoxybenzoic acid), the phenolic $-\mathrm{OH}$ acts as a nucleophile attacking acetic anhydride in acid: $\mathrm{Ar-OH} + (\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} \xrightarrow{\mathrm{H}^+} \mathrm{Ar-OCOCH}_3 + \mathrm{CH}_3\mathrm{COOH}$ . Thus, reagent 'X' is $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} / \mathrm{H}^+$ .
mechanism: To convert salicylic acid into methyl salicylate (methyl 2-hydroxybenzoate), Fischer esterification of the carboxylic acid group $-\mathrm{COOH}$ with methanol is performed: $\mathrm{Ar-COOH} + \mathrm{CH}_3\mathrm{OH} \xrightarrow{\mathrm{H}^+, \Delta} \mathrm{Ar-COOCH}_3 + \mathrm{H}_2\mathrm{O}$ . Thus, reagent 'Y' is $\mathrm{CH}_3\mathrm{OH} / \mathrm{H}^+, \Delta$ .
product: Reagents 'X' and 'Y' are $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O} / \mathrm{H}^+$ and $\mathrm{CH}_3\mathrm{OH} / \mathrm{H}^+, \Delta$ , respectively, corresponding to Option (3).
verify: Checking product structures: Aspirin contains $-\mathrm{OCOCH}_3$ and $-\mathrm{COOH}$ , formed by acetylation of $-\mathrm{OH}$ via $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}$ . Oil of wintergreen contains $-\mathrm{OH}$ and $-\mathrm{COOCH}_3$ , formed by methanol esterification of $-\mathrm{COOH}$ via $\mathrm{CH}_3\mathrm{OH}/\mathrm{H}^+$ . Order X, Y matches Option (3).