What is the correct order of increasing acidic strength of the given compounds?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (III) < (IV) < (II) < (I)
An electron-withdrawing group stabilises the carboxylate ion and makes the acid stronger; an electron-donating group destabilises it and makes the acid weaker. A withdrawing group acts more strongly the closer it is to the –COOH group.
(III) \( \ce{(CH3)2CHCOOH} \) (2-methylpropanoic acid) — 1st (weakest). The (CH3)2CH– group is a stronger electron donor than CH3CH2CH2–, so it destabilises the carboxylate ion most.
(IV) \( \ce{CH3CH2CH2COOH} \) (butanoic acid) — 2nd. Its propyl group also donates electrons, but less than the isopropyl group, and there is no withdrawing group.
(II) \( \ce{CH3CH(Br)CH2COOH} \) (3-bromobutanoic acid) — 3rd. Br is electron withdrawing and raises the acidity, but it sits on the β-carbon, one carbon away from the –COOH group.
(I) \( \ce{CH3CH2CH(Br)COOH} \) (2-bromobutanoic acid) — 4th (strongest). Br is on the α-carbon, next to the –COOH group, so its pull on the carboxylate ion is strongest.
NCERT (2026-27 edition):
8.9.1 Reactions Involving Cleavage of O–H Bond – “… Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base … Conversely, electron donating groups decrease the acidity by destabilising the conjugate base.”
“FCH2COOH > ClCH2COOH > BrCH2COOH > HCOOH > ClCH2CH2COOH > …”
6.7.1 Reactions of Haloalkanes – “… because of greater electron donating inductive effect of (CH3)2CH- group.”