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

Concept: The acidic strength of carboxylic acids depends upon the stability of the carboxylate ion formed after the loss of a proton. \[ RCOOH \rightleftharpoons RCOO^- + H^+ \] Greater the stability of the carboxylate ion, greater is the acidic strength of the corresponding carboxylic acid. The stability of the carboxylate ion is affected by:

• Electron-withdrawing groups (\(-I\) effect)

• Electron-donating groups (\(+I\) effect)

• Distance of substituent from the carboxyl group
Electron-withdrawing groups increase acidity, whereas electron-donating groups decrease acidity.

Step 1: Identifying the substituents present in each compound. The given compounds are: \[ CH_3CH(Br)CH_2COOH \] \[ CH_3CH(CH_3)COOH \] \[ CH_3CH_2CH(Br)COOH \] The substituents attached are:

• Bromine atom \((Br)\) in the first and third compounds.

• Methyl group \((CH_3)\) in the second compound.

Step 2: Effect of the methyl group. In \[ CH_3CH(CH_3)COOH \] the methyl group exhibits a \(+I\) (electron-releasing) effect. This pushes electron density towards the carboxyl group and destabilizes the carboxylate ion. As a result, acidity decreases. Therefore, this compound is expected to be the least acidic among the given compounds. \[ \boxed{ CH_3CH(CH_3)COOH \text{ is least acidic} } \]

Step 3: Effect of bromine atom. Bromine exhibits a strong \(-I\) (electron-withdrawing) effect. It withdraws electron density through sigma bonds and stabilizes the carboxylate ion. Consequently, compounds containing bromine are more acidic than the compound containing the methyl group. Therefore, \[ CH_3CH(Br)CH_2COOH \] and \[ CH_3CH_2CH(Br)COOH \] are more acidic than \[ CH_3CH(CH_3)COOH \]

Step 4: Comparing the two bromo-substituted acids. The strength of the \(-I\) effect decreases with increasing distance from the carboxyl group. Consider: \[ CH_3CH_2CH(Br)COOH \] Here bromine is present on the carbon adjacent to the carboxyl group (\(\alpha\)-carbon). Therefore, the \(-I\) effect is very strong. Now consider: \[ CH_3CH(Br)CH_2COOH \] Here bromine is one carbon farther away from the carboxyl group (\(\beta\)-carbon). Therefore, the \(-I\) effect is weaker. Hence, \[ CH_3CH_2CH(Br)COOH \] is more acidic than \[ CH_3CH(Br)CH_2COOH \]

Step 5: Writing the increasing order of acidity. Combining all observations:

• Methyl group decreases acidity.

• Bromine increases acidity.

• Bromine nearer to the carboxyl group increases acidity more strongly.
Therefore, \[ \boxed{ CH_3CH(CH_3)COOH \lt CH_3CH(Br)CH_2COOH \lt CH_3CH_2CH(Br)COOH } \]

Final Answer: \[ \boxed{ CH_3CH(CH_3)COOH \lt CH_3CH(Br)CH_2COOH \lt CH_3CH_2CH(Br)COOH } \]