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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}

B.

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

C.

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

D.

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

E.

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

Choose the correct answer from the options given below :

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A

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

B

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

C

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

D

\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 depends on the stability of the carboxylate conjugate base. Electron-withdrawing groups (-I effect) disperse the negative charge. This stabilizes the carboxylate anion and increases acidity. The -I strength of halogens follows electronegativity: $\mathrm{-F} > \mathrm{-Cl} > \mathrm{-Br}$ . The number of electron-withdrawing groups also magnifies the effect. Three fluorines ( $\-\mathrm{CF}_3$ ) exert a much stronger -I effect than a single fluorine ( $\-\mathrm{CH}_2\mathrm{F}$ ). Unsubstituted acetic acid ( $\-\mathrm{CH}_3$ ) has an electron-donating (+I) alkyl group. This makes it the least acidic.

02option_verdict

Places D ( $\-\mathrm{CH}_2\mathrm{F}$ ) before B ( $\-\mathrm{CF}_3$ ), ignoring the cumulative magnitude of three -I groups.

03option_verdict

Places E (bromoacetic acid) as the strongest acid, completely inverting the halogen electronegativity trend.

04option_verdict

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

05option_verdict

Correct descending order: $\mathrm{CF}_3\mathrm{COOH}$ (B) > $\mathrm{FCH}_2\mathrm{COOH}$ (D) > $\mathrm{ClCH}_2\mathrm{COOH}$ (C) > $\mathrm{BrCH}_2\mathrm{COOH}$ (E) > $\mathrm{CH}_3\mathrm{COOH}$ (A).

✓discriminator

Identify the extreme ends first: $\mathrm{CF}_3\mathrm{COOH}$ (B) must be the strongest and $\mathrm{CH}_3\mathrm{COOH}$ (A) must be the weakest, eliminating options 0 and 1 immediately. Then between B > D vs B > C, fluorine is more electronegative than chlorine, locking B > D > C > E > A.

Hints that build this answer step by step

What primary electronic effect determines the relative acidity of halo-substituted carboxylic acids?
The inductive electron-withdrawing effect ( $-I$ ), which disperses negative charge in the conjugate base.
How does the number and electronegativity of the halogen substituents rank the acids from most acidic to least acidic?
$\mathrm{CF_3COOH} > \mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{BrCH_2COOH} > \mathrm{CH_3COOH}$

✓ 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 higher acidity than fluoroacetic acid?

Inductive effects are additive. Three electronegative fluorine atoms pull electron density much more strongly than a single fluorine atom, greatly stabilizing 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 depends on the stability of the carboxylate conjugate base. Electron-withdrawing groups (-I effect) disperse the negative charge. This stabilizes the carboxylate anion and increases acidity. The -I strength of halogens follows electronegativity: $\mathrm{-F} > \mathrm{-Cl} > \mathrm{-Br}$ . The number of electron-withdrawing groups also magnifies the effect. Three fluorines ( $\-\mathrm{CF}_3$ ) exert a much stronger -I effect than a single fluorine ( $-I$ ). Unsubstituted acetic acid ( $\-\mathrm{CH}_3$ ) has an electron-donating (+I) alkyl group. This makes it the least acidic.
option_verdict: Places D ( $\-\mathrm{CH}_2\mathrm{F}$ ) before B ( $\-\mathrm{CF}_3$ ), ignoring the cumulative magnitude of three -I groups.
option_verdict: Places E (bromoacetic acid) as the strongest acid, completely inverting the halogen electronegativity trend.
option_verdict: Places C ( $\-\mathrm{Cl}$ ) ahead of D ( $\-\mathrm{F}$ ), contradicting the electronegativity order $\mathrm{F} > \mathrm{Cl}$ .
option_verdict: Correct descending order: $\mathrm{CF}_3\mathrm{COOH}$ (B) > $\mathrm{FCH}_2\mathrm{COOH}$ (D) > $\mathrm{ClCH}_2\mathrm{COOH}$ (C) > $\mathrm{BrCH}_2\mathrm{COOH}$ (E) > $\mathrm{CH}_3\mathrm{COOH}$ (A).
discriminator: Identify the extreme ends first: $\mathrm{CF}_3\mathrm{COOH}$ (B) must be the strongest and $\mathrm{CH}_3\mathrm{COOH}$ (A) must be the weakest, eliminating options 0 and 1 immediately. Then between B > D vs B > C, fluorine is more electronegative than chlorine, locking B > D > C > E > A.