Question

Arrange the following compounds in increasing order of their acidic strengths : \[ CH_3CH(Br)CH_2COOH,\qquad CH_3CH(CH_3)COOH,\qquad CH_3CH_2CH(Br)COOH \]

Hint:

For carboxylic acids: \[ -I \text{ effect } \Rightarrow \text{ Acidity increases} \] \[ +I \text{ effect } \Rightarrow \text{ Acidity decreases} \] Also remember: \[ \alpha\text{-substituent} \gt \beta\text{-substituent} \gt \gamma\text{-substituent} \] in influencing acidity because the inductive effect decreases with distance.

Answer 1

Step 1: Key principle.
Acidity of carboxylic acids depends on stability of $RCOO^-$. Electron-withdrawing groups ($-I$ effect) stabilise the carboxylate and increase acidity; electron-donating groups ($+I$ effect) destabilise it and decrease acidity.
Step 2: Rank the methyl compound as least acidic.
$CH_3CH(CH_3)COOH$ has a methyl group ($+I$ effect) on the alpha carbon, donating electron density toward $-COOH$ and destabilising the carboxylate. It is the least acidic.
Step 3: Compare the two bromo-acids by distance of Br from COOH.
Bromine ($-I$ effect) increases acidity; the effect is stronger when Br is closer to $-COOH$. In $CH_3CH_2CH(Br)COOH$ the Br is on the alpha carbon (strongest $-I$); in $CH_3CH(Br)CH_2COOH$ it is on the beta carbon (weaker $-I$). \[ \boxed{CH_3CH(CH_3)COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH} \]