The following table is from Solomons, Fryhle and Snyder Organic Chemistry, chapter 3, An Introduction to Organic Reactions and Their Mechanisms: Acids and Bases:

Table 3.8 Isotopic substituents and inductive effect $$ \small \begin{array}{ccc} \hline \text{Acid} & \ce{pK_a} & \text{Inductive effect of the substituent} \\ \hline \ce{CH3-COOH} & 4.75 & \text{+I} \\ \ce{CD3-COOH} & 4.60 & \text{+I} \\ \hline \end{array} $$ +I effect of deuterium is stronger than that of hydrogen. +I ($\ce{-CD3>-CH3}$)

Why is $\ce{CD3-COOH}$ more acidic than $\ce{CH3-COOH}$ despite being less electronegative? The book also said that $\ce{CD3} > \ce{CH3}$ in terms of + Inductive effect.

  • $\begingroup$ Which edition you referred to? In the global 12E, there is no table 3.8 and in the Indian 2017 edition, the value for $\ce{CD3COOH}$ is 4.85. $\endgroup$ – Apurvium Sep 23 '20 at 4:55

To compare acidity, you must have made the conjugate bases of these acids, that is, $\ce{CH3COO-}$ and $\ce{CD3COO-}$ . Now, the hyperconjugation effect is usually a more dominant electronic effect than the inductive effect, and as the $\ce{C-D}$ bond is stronger than $\ce{C-H}$ bond, so the reduced $\ce{+H}$ effect in the conjugate base of $\ce{CD3COOH}$ makes it more acidic.

  • $\begingroup$ But there are zero alpha hydrogens on the carbon atom next to the oxygen anion so how is hyperconjugation possible ? $\endgroup$ – Akshay Mendon Sep 24 '18 at 22:00
  • $\begingroup$ There are α-hydrogens and α-deuteriums present with respect to the $\ce{COO-}$ group, so to see the hyperconjugative structures, break the $\ce{C-H}$ bond and shift the electron pair so that you get two negatively charged oxygens on one carbon, which can protonate and revert to the original carboxylate ion. Anyway, hyperconjugation is a theoretical effect, and it is not defined clearly for carbanions due to absence of vacant orbitals, so I would advise you not to set much store by the actual mechanism of hyperconjugation here, as the di-negative carbanaion formed was extremely unstable. $\endgroup$ – Yusuf Hasan Sep 25 '18 at 0:23
  • $\begingroup$ Again, I would like to add here that since hyperconjugation is completely theoretical, so the only concrete thing we can say here is that by whichever path the $\ce{+H}$ maybe happening, it is clear that it will increase electron density on the $\ce{COO-}$ group, so please don't take these structures I just suggested as absolute $\endgroup$ – Yusuf Hasan Sep 25 '18 at 0:31

The following articles suggest that the data reported from Solomons, Fryhle and Snyder Organic Chemistry is likely incorrect.

These findings indicate that acetic acid-d3 is less acidic than acetic acid-h3. This would be consistent with the operation of a simple inductive effect where protium is more electronegative than deuterium (as expected). Hence, the more electron withdrawing $\ce{CH3}$ group is better able to stabilize the electron rich carboxylate anion in the the ionized acid as compared to the less electron withdrawing $\ce{CD3}$ group.

  • $\begingroup$ What I have read is that one 'only' needs to see the inductive effect of the chain attached to compare the acidic strength of carboxylic acids. However, at multiple places in Chemistry SE, people have written that hyperconjugation effect and inductive effect of alkyl groups are related, so one should also consider hyperconjugation along with inductive effect. What's your take on it? $\endgroup$ – Aumkaar Pranav Mar 1 '20 at 1:55
  • $\begingroup$ @ron Based on this, can we say that migratory aptitude in a carbocation species follows the order: $\ce{CH3 > CD3 > CT3}$? It is due to the fact that if methanide migrates then the other two isotopic methyl group (due to more $\pu{+I}$ effect) will be able to stabilize the positive charge of carbocation better in comparison to other two scenarios? $\endgroup$ – Apurvium Sep 23 '20 at 4:40
  • $\begingroup$ @Apurvium The $\ce{CH3}$ group is more electron withdrawing, hence it would be expected to migrate more slowly (stabilize the carbocation less in the TS). $\endgroup$ – ron Sep 24 '20 at 15:01
  • $\begingroup$ @ron $\ce{CH3}$ is more electron withdrawing, thus, formation of $\ce{CH3-}$ will be easier and in turn, it can migrate fast. $\endgroup$ – Apurvium Sep 24 '20 at 15:24
  • $\begingroup$ @Apurvium But it's all about the migrating groups ability to stabilize the cationic center. $\endgroup$ – ron Sep 24 '20 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.