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Q8.12(ii):
Arrange the following compounds in increasing order of their property as indicated: (ii) \(CH_{3}CH_{2}CH(Br)COOH\), \(CH_{3}CH(Br)CH_{2}COOH\), \((CH_{3})_{2}CHCOOH\), \(CH_{3}CH_{2}CH_{2}COOH\) (acid strength)

Solution :

Step 1: Identifying the Structural Factors Influencing Acid Strength

The acid strength of carboxylic acids ($R-COOH$) is determined by the stability of the conjugate base, the carboxylate ion ($R-COO^-$). Stability is enhanced by the presence of electron-withdrawing groups (EWGs) and diminished by electron-donating groups (EDGs) via the Inductive Effect ($I$).

  • Electron-Withdrawing Effect ($-I$ Effect): Electronegative substituents like Bromine ($-Br$) withdraw electron density from the carboxylate group, dispersing the negative charge and increasing acidity.
  • Distance Dependence: The inductive effect decreases rapidly with distance. An EWG closer to the $-COOH$ group exerts a stronger acidic influence.
  • Electron-Donating Effect ($+I$ Effect): Alkyl groups (like $-CH_3$) are electron-releasing. They increase electron density on the carboxylate ion, destabilizing it and decreasing acid strength.

Step 2: Analysis of the Given Compounds

Let us analyze the four compounds based on their inductive effects:

Compound Analysis
$CH_3CH_2CH_2COOH$ Standard butyric acid; possesses only weak $+I$ effect from the propyl chain.
$(CH_3)_2CHCOOH$ Isobutyric acid; contains two methyl groups attached to the $\alpha$-carbon, increasing electron density via $+I$ effect compared to butyric acid.
$CH_3CH(Br)CH_2COOH$ $\beta$-Bromobutyric acid; contains an EWG ($-Br$) at the $\beta$-position.
$CH_3CH_2CH(Br)COOH$ $\alpha$-Bromobutyric acid; contains an EWG ($-Br$) at the $\alpha$-position.

Step 3: Comparative Ordering Logic

  1. Comparing Alkyl Acids: $(CH_3)_2CHCOOH$ is a weaker acid than $CH_3CH_2CH_2COOH$ because the branching increases the electron-donating $+I$ effect, destabilizing the conjugate base [Per Inductive Effect theory].
  2. Comparing Bromo-Substituted Acids: The $-Br$ group exerts a strong $-I$ effect. In $CH_3CH_2CH(Br)COOH$, the bromine is at the $\alpha$-carbon (distance = 2 bonds from $-COO^-$). In $CH_3CH(Br)CH_2COOH$, the bromine is at the $\beta$-carbon (distance = 3 bonds from $-COO^-$). Since the inductive effect is distance-dependent, $\alpha$-bromobutyric acid is significantly more acidic [By distance-dependence of $-I$ effect].
  3. Synthesis: The overall order is: $(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$.
Increasing Order of Acid Strength: (CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Final Solution: $(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$


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