Which is more stable?

  1. $\ce{^+C(CH3)3}$
  2. $\ce{^+C(CD3)3}$

I have read in Solomon's and Fryhle that the +I effect of D is more than that of H but here in this problem it is given that $\ce{(CH3)3C+}$ is more stable.

How is this possible? Since, deuterium enriches the electron density over the central carbon resulting in the diminishing of a positive charge eventually making the second structure more stable.

Second explanation can be on the basis of hyperconjugation which says that $\ce{-CH3}$ is better in showing hyperconjugation than $\ce{-CD3}$, but why is this so?


1 Answer 1


$\ce{C-D}$ bonds are stronger. Hyperconjugation is thus more easily seen with hydrogen rather than deuterium.
Hyperconjugation> Inductive effects at stabilization.I believe you can reason the rest out by yourself.
To see why $\ce{C-D}$ bonds are stronger.
See this

  • 5
    $\begingroup$ While true, this answer is extremely unsatisfying. The link from bond strength to hyperconjugation is not obvious to me at all. (And I'm not saying I have any better explanation. This is something which I have never seen a good explanation for.) $\endgroup$ Jan 28, 2018 at 18:33
  • 1
    $\begingroup$ @orthocresol Maybe this contains the required answer. $\endgroup$ Jan 29, 2018 at 5:51

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