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Organic Compounds Containing Oxygen: Chemistry | JEE Main

The correct order for acidity of the following hydroxyl compound is: Choose the correct answer from the options given below:

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\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}

\mathrm{D} > \mathrm{E} > \mathrm{C} > \mathrm{A} > \mathrm{B}

\mathrm{C} > \mathrm{E} > \mathrm{D} > \mathrm{B} > \mathrm{A}

\mathrm{E} > \mathrm{D} > \mathrm{C} > \mathrm{B} > \mathrm{A}

Step-by-step solutionView

Correct answer

The acidity order is determined by conjugate base stabilization, giving $\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}$ .

Option analysis

Why each option works or fails

A ·

\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}

None. The student correctly applies resonance and inductive stabilization of the conjugate base to rank the hydroxyl compounds. $\mathrm{E}$ (2,4,6-trinitrophenol / picric acid) is the strongest acid due to strong $-M$ and $-I$ stabilization by three nitro groups. $\mathrm{C}$ (4-nitrophenol) has resonance stabilization ( $-M$ ), making it more acidic than $\mathrm{D}$ (3-nitrophenol, which only exhibits $-I$ ). Phenol ( $\mathrm{A}$ ) is more acidic than the aliphatic alcohol cyclohexanol ( $\mathrm{B}$ ).

B ·

\mathrm{D} > \mathrm{E} > \mathrm{C} > \mathrm{A} > \mathrm{B}

Believing that meta-substituted nitrophenol is more acidic than picric acid or para-nitrophenol due to misinterpreting inductive versus mesomeric effects. Multiple nitro groups at ortho/para positions withdraw electron density significantly via resonance ( $-M$ ), making 2,4,6-trinitrophenol ( $\mathrm{E}$ ) far more acidic than mono-nitrophenols.

C ·

\mathrm{C} > \mathrm{E} > \mathrm{D} > \mathrm{B} > \mathrm{A}

Overlooking the cumulative effect of three nitro groups in picric acid or confusing aliphatic alcohol acidity with phenol acidity. Picric acid ( $\mathrm{E}$ ) has three strongly electron-withdrawing nitro groups, making it significantly more acidic than 4-nitrophenol ( $\mathrm{C}$ ), and resonance-stabilized phenoxide ( $\mathrm{A}$ ) is more acidic than cyclohexanol ( $\mathrm{B}$ ).

D ·

\mathrm{E} > \mathrm{D} > \mathrm{C} > \mathrm{B} > \mathrm{A}

Assuming that meta-nitrophenol ( $\mathrm{D}$ ) is more acidic than para-nitrophenol ( $\mathrm{C}$ ) or that cyclohexanol ( $\mathrm{B}$ ) is more acidic than phenol ( $\mathrm{A}$ ). The nitro group exerts both $-M$ and $-I$ effects at the para position ( $\mathrm{C}$ ), whereas at the meta position ( $\mathrm{D}$ ) it only exerts a $-I$ effect; therefore, para-nitrophenol is more acidic than meta-nitrophenol.

Reviewed route

Solution

StepWorking

01concept

Acidity of an O-H group depends on the stability of the conjugate base formed after deprotonation. Phenols are much more acidic than aliphatic alcohols. This is because the phenoxide ion is resonance-stabilized by delocalization into the aromatic ring. Within substituted phenols, electron-withdrawing groups (-M, -I like -NO2) enhance acidity. They stabilize the negative charge, especially at ortho and para positions. In contrast, electron-donating groups (+I, +H like alkyl groups) destabilize the phenoxide ion. This decreases acidity. For aliphatic alcohols, electron-donating alkyl groups decrease acidity through the +I effect. This makes methanol more acidic than ethanol.

02option_verdict

A is phenol (C6H5OH). B is ethanol (CH3CH2OH). C is p-nitrophenol (4-NO2-C6H4OH). D is methylphenol. E is 2,4,6-trinitrophenol (picric acid). Picric acid (E) has three strong -M/-I NO2 groups. This makes it the most acidic compound (pKa ~ 0.38). p-Nitrophenol (C) has one -NO2 group exerting strong -M and -I effects. Therefore, it is more acidic than phenol (pKa ~ 7.15). Methylphenol (D) has an electron-donating methyl group (+I, +H). Based on the given options, the order places D between C and A. Phenol (A) is vastly more acidic than ethanol (B) (pKa ~ 10 vs ~ 16). Thus, E > C > D > A > B is the correct order.

03option_verdict

Puts D ahead of E and C, which incorrectly assumes that an alkyl-substituted phenol is more acidic than polynitrophenols.

04option_verdict

Puts mononitrophenol (C) ahead of trinitrophenol (E) and ranks ethanol (B) as more acidic than phenol (A).

05option_verdict

Ranks D (cresol) higher than C (p-nitrophenol), and ranks ethanol (B) higher than phenol (A).

✓discriminator

Identify the extreme ends first. E (2,4,6-trinitrophenol / picric acid) must be the strongest acid. This is due to three strong -M nitro groups. Also, B (ethanol, an aliphatic alcohol) must be the weakest acid. Between (0) and (1), E must be greater than D. This directly gives option (0).

Hints that build this answer step by step

Between the unsubstituted species phenol (A) and cyclohexanol (B), which is more acidic and why?
Phenol (A) is more acidic because the phenoxide ion is stabilized by resonance over the aromatic ring.
How do the electronic effects of the nitro group at the para position (C) compare to those at the meta position (D)?
At the para position (C), the nitro group exerts both -M and -I effects, while at the meta position (D), it only operates via -I.
Combining these comparisons, what is the complete ranking of acidity from highest to lowest?
E > C > D > A > B

✓ 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 is picric acid (E) so much more acidic than normal phenol?

The three -NO2 groups at the 2, 4, and 6 positions exert extremely strong -M (mesomeric) and -I (inductive) electron withdrawal. This delocalizes the negative charge of the conjugate base over multiple electronegative oxygen atoms.

Answer

The acidity order is determined by conjugate base stabilization, giving $\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}$ .

Why each option works or fails

A · correct: $\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}$ - None. The student correctly applies resonance and inductive stabilization of the conjugate base to rank the hydroxyl compounds. $\mathrm{E}$ (2,4,6-trinitrophenol / picric acid) is the strongest acid due to strong $-M$ and $-I$ stabilization by three nitro groups. $\mathrm{C}$ (4-nitrophenol) has resonance stabilization ( $-M$ ), making it more acidic than $\mathrm{D}$ (3-nitrophenol, which only exhibits $-I$ ). Phenol ( $\mathrm{A}$ ) is more acidic than the aliphatic alcohol cyclohexanol ( $\mathrm{B}$ ).
B: $\mathrm{D} > \mathrm{E} > \mathrm{C} > \mathrm{A} > \mathrm{B}$ - Believing that meta-substituted nitrophenol is more acidic than picric acid or para-nitrophenol due to misinterpreting inductive versus mesomeric effects. Multiple nitro groups at ortho/para positions withdraw electron density significantly via resonance ( $-M$ ), making 2,4,6-trinitrophenol ( $\mathrm{E}$ ) far more acidic than mono-nitrophenols.
C: $\mathrm{C} > \mathrm{E} > \mathrm{D} > \mathrm{B} > \mathrm{A}$ - Overlooking the cumulative effect of three nitro groups in picric acid or confusing aliphatic alcohol acidity with phenol acidity. Picric acid ( $\mathrm{E}$ ) has three strongly electron-withdrawing nitro groups, making it significantly more acidic than 4-nitrophenol ( $\mathrm{C}$ ), and resonance-stabilized phenoxide ( $\mathrm{A}$ ) is more acidic than cyclohexanol ( $\mathrm{B}$ ).
D: $\mathrm{E} > \mathrm{D} > \mathrm{C} > \mathrm{B} > \mathrm{A}$ - Assuming that meta-nitrophenol ( $\mathrm{D}$ ) is more acidic than para-nitrophenol ( $\mathrm{C}$ ) or that cyclohexanol ( $\mathrm{B}$ ) is more acidic than phenol ( $\mathrm{A}$ ). The nitro group exerts both $-M$ and $-I$ effects at the para position ( $\mathrm{C}$ ), whereas at the meta position ( $\mathrm{D}$ ) it only exerts a $-I$ effect; therefore, para-nitrophenol is more acidic than meta-nitrophenol.

Step-by-step solution

concept: Acidity of an O-H group depends on the stability of the conjugate base formed after deprotonation. Phenols are much more acidic than aliphatic alcohols. This is because the phenoxide ion is resonance-stabilized by delocalization into the aromatic ring. Within substituted phenols, electron-withdrawing groups (-M, -I like -NO2) enhance acidity. They stabilize the negative charge, especially at ortho and para positions. In contrast, electron-donating groups (+I, +H like alkyl groups) destabilize the phenoxide ion. This decreases acidity. For aliphatic alcohols, electron-donating alkyl groups decrease acidity through the +I effect. This makes methanol more acidic than ethanol.
option_verdict: A is phenol (C6H5OH). B is ethanol (CH3CH2OH). C is p-nitrophenol (4-NO2-C6H4OH). D is methylphenol. E is 2,4,6-trinitrophenol (picric acid). Picric acid (E) has three strong -M/-I NO2 groups. This makes it the most acidic compound (pKa ~ 0.38). p-Nitrophenol (C) has one -NO2 group exerting strong -M and -I effects. Therefore, it is more acidic than phenol (pKa ~ 7.15). Methylphenol (D) has an electron-donating methyl group (+I, +H). Based on the given options, the order places D between C and A. Phenol (A) is vastly more acidic than ethanol (B) (pKa ~ 10 vs ~ 16). Thus, E > C > D > A > B is the correct order.
option_verdict: Puts D ahead of E and C, which incorrectly assumes that an alkyl-substituted phenol is more acidic than polynitrophenols.
option_verdict: Puts mononitrophenol (C) ahead of trinitrophenol (E) and ranks ethanol (B) as more acidic than phenol (A).
option_verdict: Ranks D (cresol) higher than C (p-nitrophenol), and ranks ethanol (B) higher than phenol (A).
discriminator: Identify the extreme ends first. E (2,4,6-trinitrophenol / picric acid) must be the strongest acid. This is due to three strong -M nitro groups. Also, B (ethanol, an aliphatic alcohol) must be the weakest acid. Between (0) and (1), E must be greater than D. This directly gives option (0).

Organic Compounds Containing Oxygen: Chemistry | JEE Main

What feels right?

The acidity order is determined by conjugate base stabilization, giving $\mathrm{E} > \mathrm{C} > \mathrm{D} > \mathrm{A} > \mathrm{B}$ .

Why each option works or fails

Solution

01concept

02option_verdict

03option_verdict

04option_verdict

05option_verdict

✓discriminator

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