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According to the lecture notes I made, a 1* carbanion is more stable than a 3* carbanion. I can't grasp exactly why. (I had a look at the similar questions, the answers didn't help)

With my very limited knowledge, there can be possibly two effects at play here:

  • The Inductive Effect
  • Hyper-conjugation

The mesomeric effect can't take place here since there are no pi-bonds, as far as I know.

Now, for the inductive effect, I know that CH3 is considered a +I group (electron-donating), but I don't know why, since the inductive effect operates through electronegativity, and two carbon atoms have no relative electronegativity difference. Also, my textbook tells that the electronegativity difference between carbon and hydrogen can be considered to be zero, so I can't see why the inductive effect would do anything here. (If anything, it seems to me that since there's a negative charge on the central carbon, it should spread out to the methyls, since there's no electronegativity difference between carbons, stabilizing the tertiary carbanion the most)

For hyperconjugation, I've read that it is the interaction between sigma-electrons and pi-electrons, or sigma electrons and empty p-orbitals(as in the case with carbocations). Since there's no pi bond in these compounds, and the central carbon's octet is full (3 bonds, 1 negative charge), I can't see why there would be any hyperconjugation in these compounds.

What's happening?

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    $\begingroup$ Please typeset the material in the photo - it is not searchable. $\endgroup$
    – Todd Minehardt
    Jul 19, 2021 at 20:34
  • $\begingroup$ Added a different picture @ToddMinehardt $\endgroup$
    – VVidyan
    Jul 19, 2021 at 20:38
  • $\begingroup$ Which textbook mentions electronegativity difference between carbon and hydrogen as zero? That's a fantastically wrong statement. $\endgroup$
    – TRC
    Jul 20, 2021 at 13:39
  • $\begingroup$ @TRC My coaching modules do, they're not standard textbooks, maybe that;s why the error. So is that the reason? $\endgroup$
    – VVidyan
    Jul 21, 2021 at 15:49
  • $\begingroup$ Electronegativity of H is 2.2, C is 2.5. So in a $\ce{CH3}$ group the electron density is pulled towards the carbon side - even if by a (non-negligible) small amount. This increased electron density on the methyl carbon in turns makes this more electron rich, and it tries to push its electrons towards the next carbon. Hence causing +I effect. In carbanion case, if the methyl carbon is already electron rich due to the three attached hydrogen, it wouldn't like to accept more negative charge from the neighbour, would it? $\endgroup$
    – TRC
    Jul 21, 2021 at 15:54

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