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Q8.17 (b):
Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (b) $CH_3CH_2COOH > (CH_3)_2CHCOOH > (CH_3)_3C.COOH$

Solution :

1. Definitions of Electron Displacement Effects

Inductive Effect ($I$-Effect): The permanent displacement of shared electron pairs along a chain of carbon atoms due to the difference in electronegativity of substituent atoms or groups. It operates through $\sigma$-bonds and decreases rapidly with distance from the source [Per the principle of polarization in sigma-bonds].

  • Negative Inductive Effect ($-I$): Electron-withdrawing groups (e.g., $-NO_2, -F, -Cl$).
  • Positive Inductive Effect ($+I$): Electron-donating groups (e.g., alkyl groups like $-CH_3, -CH_2CH_3$).

Electromeric Effect ($E$-Effect): A temporary effect involving the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond at the requirement of an attacking reagent. It is represented by a curved arrow showing electron flow [Per the mechanism of nucleophilic/electrophilic addition].

2. Analysis of the Acidity Order

The given acidity order is: $CH_3CH_2COOH > (CH_3)_2CHCOOH > (CH_3)_3C.COOH$.

To determine the acidity of carboxylic acids ($R-COOH$), we examine the stability of the conjugate base, the carboxylate ion ($R-COO^-$). An electron-donating group (EDG) destabilizes the carboxylate ion by increasing the negative charge density on the carboxylate group, thereby decreasing the acidity of the parent acid.

3. Step-by-Step Inductive Effect Analysis

Acid Alkyl Group Inductive Effect
$CH_3CH_2COOH$ Ethyl ($-C_2H_5$) Weakest $+I$ effect
$(CH_3)_2CHCOOH$ Isopropyl ($-CH(CH_3)_2$) Moderate $+I$ effect
$(CH_3)_3C.COOH$ tert-Butyl ($-C(CH_3)_3$) Strongest $+I$ effect

Step-by-step logic:

  1. The acidity of $RCOOH$ depends on the ease of proton release to form $RCOO^-$.
  2. Alkyl groups exhibit a $+I$ effect. As the number of alkyl groups attached to the $\alpha$-carbon increases, the total electron-donating inductive effect increases.
  3. The tert-butyl group in $(CH_3)_3C.COOH$ provides the maximum electron density to the carboxylate carbon via the $+I$ effect. This destabilizes the $RCOO^-$ anion significantly compared to the ethyl group.
  4. Consequently, the conjugate base of propanoic acid ($CH_3CH_2COO^-$) is the most stable, and the conjugate base of pivalic acid ($(CH_3)_3CCOO^-$) is the least stable.

4. Visual Representation of Electron Displacement

R -COO⁻ +I Effect Direction As alkyl bulk increases, electron density push increases, destabilizing RCOO⁻.

Final Solution: The inductive effect ($+I$ effect) explains this order. As the number of electron-donating alkyl groups increases, the electron density on the carboxylate group increases, destabilizing the conjugate base and thus decreasing the acidity of the carboxylic acid.


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