JEE MainChemistry

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Organic Compounds Containing Oxygen: Chemistry | JEE Main

The descending order of acidity for the following carboxylic acid is : A.

\mathrm{CH}_{3}\mathrm{COOH}

\mathrm{F}_{3}\mathrm{C-COOH}

\mathrm{ClCH}_{2}-\mathrm{COOH}

\mathrm{FCH}_{2}-\mathrm{COOH}

\mathrm{BrCH}_{2}-\mathrm{COOH}

Choose the correct answer from the options given below :

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

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

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

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

Step-by-step solutionView

Correct answer

Electron-withdrawing groups ( $-I$ effect) stabilize the carboxylate anion, giving the acidity order: $\mathrm{CF_3COOH} > \mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{BrCH_2COOH} > \mathrm{CH_3COOH}$ (i.e., $\mathrm{B} > \mathrm{D} > \mathrm{C} > \mathrm{E} > \mathrm{A}$ ).

Option analysis

Why each option works or fails

A ·

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

Believing that monosubstituted haloacids are stronger than tri-substituted ones and confusing the order of inductive stabilization. Three electron-withdrawing fluorines exert a vastly stronger cumulative $-I$ effect than a single fluorine, making $\mathrm{CF_3COOH}$ (B) the most acidic.

B ·

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

Assuming that acidity increases down the halogen group rather than tracking electronegativity, placing bromine higher than fluorine. The $-I$ effect increases with increasing electronegativity of the halogen ( $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ ), so fluorinated acids are more acidic than brominated acids.

C ·

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

Incorrectly ordering chlorine as more electronegative than fluorine, ranking $\mathrm{ClCH_2COOH}$ ahead of $\mathrm{FCH_2COOH}$ . Fluorine is more electronegative than chlorine, so $\mathrm{FCH_2COOH}$ (D) stabilizes the conjugate base more strongly via inductive effect than $\mathrm{ClCH_2COOH}$ (C).

D ·

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

None. This is the correct descending order of carboxylic acid acidity. Rank by number of electron-withdrawing groups first ( $\mathrm{CF_3}$ > $\mathrm{CH_2X}$ ), then by electronegativity of halogen ( $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ ), with unsubstituted $\mathrm{CH_3COOH}$ being the least acidic.

Reviewed route

Solution

StepWorking

01concept

Acidity of carboxylic acids increases with the stability of the conjugate base (carboxylate anion). Electron-withdrawing groups ( $-I$ effect) stabilize the carboxylate anion by dispersing negative charge. Acidity increases with the number and electronegativity of $-I$ groups, while $+I$ alkyl groups decrease acidity relative to substituted ones.

02option_verdict

$\mathrm{CF_3COOH}$ (B) has three strongly electron-withdrawing $-\mathrm{F}$ groups, making it the strongest acid. For monosubstituted haloacetic acids, the $-I$ effect follows halogen electronegativity: $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ , so $\mathrm{FCH_2COOH}$ (D) > $\mathrm{ClCH_2COOH}$ (C) > $\mathrm{BrCH_2COOH}$ (E). Unsubstituted $\mathrm{CH_3COOH}$ (A) with a $+I$ methyl group is the weakest. The descending order is $\mathrm{B > D > C > E > A}$ .

03option_verdict

Incorrectly places monosubstituted D ahead of tri-substituted B, ignoring the cumulative additive nature of $-I$ groups.

04option_verdict

Inverts the electronegativity trend of halogens, placing $\mathrm{Br}$ (E) as the strongest $-I$ group instead of $\mathrm{F}$ .

05option_verdict

Places $\mathrm{Cl}$ (C) ahead of $\mathrm{F}$ (D), which contradicts the electronegativity order $\mathrm{F > Cl > Br}$ .

✓discriminator

Identify the extreme ends first: B ( $\mathrm{CF_3COOH}$ ) is clearly strongest due to three F atoms and A ( $\mathrm{CH_3COOH}$ ) is weakest (no halogen). Then order D, C, E by halogen electronegativity ( $\mathrm{F > Cl > Br}$ ).

✓ 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: 7 September 2026

Quick checks

Students also ask

Why does trifluoroacetic acid have a much lower pKa than monofluoroacetic acid?

Inductive effects are additive. Three fluorine atoms pull electron density away from the carboxylate group much more strongly than a single fluorine atom. This significantly enhances the stabilization of the conjugate base.

Answer

Electron-withdrawing groups ( $-I$ effect) stabilize the carboxylate anion, giving the acidity order: $\mathrm{CF_3COOH} > \mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{BrCH_2COOH} > \mathrm{CH_3COOH}$ (i.e., $\mathrm{B} > \mathrm{D} > \mathrm{C} > \mathrm{E} > \mathrm{A}$ ).

Why each option works or fails

A: $\mathrm{D} > \mathrm{B} > \mathrm{A} > \mathrm{E} > \mathrm{C}$ - Believing that monosubstituted haloacids are stronger than tri-substituted ones and confusing the order of inductive stabilization. Three electron-withdrawing fluorines exert a vastly stronger cumulative $-I$ effect than a single fluorine, making $\mathrm{CF_3COOH}$ (B) the most acidic.
B: $\mathrm{E} > \mathrm{D} > \mathrm{B} > \mathrm{A} > \mathrm{C}$ - Assuming that acidity increases down the halogen group rather than tracking electronegativity, placing bromine higher than fluorine. The $-I$ effect increases with increasing electronegativity of the halogen ( $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ ), so fluorinated acids are more acidic than brominated acids.
C: $\mathrm{B} > \mathrm{C} > \mathrm{D} > \mathrm{E} > \mathrm{A}$ - Incorrectly ordering chlorine as more electronegative than fluorine, ranking $\mathrm{ClCH_2COOH}$ ahead of $\mathrm{FCH_2COOH}$ . Fluorine is more electronegative than chlorine, so $\mathrm{FCH_2COOH}$ (D) stabilizes the conjugate base more strongly via inductive effect than $\mathrm{ClCH_2COOH}$ (C).
D · correct: $\mathrm{B} > \mathrm{D} > \mathrm{C} > \mathrm{E} > \mathrm{A}$ - None. This is the correct descending order of carboxylic acid acidity. Rank by number of electron-withdrawing groups first ( $\mathrm{CF_3}$ > $\mathrm{CH_2X}$ ), then by electronegativity of halogen ( $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ ), with unsubstituted $\mathrm{CH_3COOH}$ being the least acidic.

Step-by-step solution

concept: Acidity of carboxylic acids increases with the stability of the conjugate base (carboxylate anion). Electron-withdrawing groups ( $-I$ effect) stabilize the carboxylate anion by dispersing negative charge. Acidity increases with the number and electronegativity of $-I$ groups, while $+I$ alkyl groups decrease acidity relative to substituted ones.
option_verdict: $\mathrm{CF_3COOH}$ (B) has three strongly electron-withdrawing $-\mathrm{F}$ groups, making it the strongest acid. For monosubstituted haloacetic acids, the $-I$ effect follows halogen electronegativity: $\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$ , so $\mathrm{FCH_2COOH}$ (D) > $\mathrm{ClCH_2COOH}$ (C) > $\mathrm{BrCH_2COOH}$ (E). Unsubstituted $\mathrm{CH_3COOH}$ (A) with a $+I$ methyl group is the weakest. The descending order is $\mathrm{B > D > C > E > A}$ .
option_verdict: Incorrectly places monosubstituted D ahead of tri-substituted B, ignoring the cumulative additive nature of $-I$ groups.
option_verdict: Inverts the electronegativity trend of halogens, placing $\mathrm{Br}$ (E) as the strongest $-I$ group instead of $\mathrm{F}$ .
option_verdict: Places $\mathrm{Cl}$ (C) ahead of $\mathrm{B} > \mathrm{D} > \mathrm{C} > \mathrm{E} > \mathrm{A}$ (D), which contradicts the electronegativity order $\mathrm{F > Cl > Br}$ .
discriminator: Identify the extreme ends first: B ( $\mathrm{CF_3COOH}$ ) is clearly strongest due to three F atoms and A ( $\mathrm{CH_3COOH}$ ) is weakest (no halogen). Then order D, C, E by halogen electronegativity ( $\mathrm{F > Cl > Br}$ ).