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(I know there are many related questions, so please tell me which posts can solve my doubts before you close this question, thank you.)

$1$. $\ce{^+CH2F}$ is very unstable. This is because $\ce{F}$ is very electronegative, pulling the electron density from the already electrophilic carbon (empty $\text{p}$ orbital), thus increasing the positive charge.

For $\ce{CH3CH2+}$: E.N. of $\ce{H}$ = $2.2$, $\ce{C}$ = $2.5$, so the carbon in the methyl group is partially negative - like the $\ce{F}$ in $\ce{^+CH2F}$. To my understanding, the methyl group should be electron-withdrawing instead of donating, thus decreasing its stability, but this is false, why?

Some say that the partially negative carbon can donate some electron to the carbocation to increase stability, if it is true, why can't $\ce{^+CH2F}$ do the same thing?

$2.$ I know hyper-conjugation can form partial orbital overlap ($sp^3$ - $s$($\sigma$) for $\ce{H}$ & empty $p$ orbital in carbocation), thus delocalizing the electron density giving stability. However, the $\ce{C-C}$ $\sigma$ bond is always rotating, so from my view, hyper-conjugation is a bond that is continuously forming and breaking. Therefore, how come this is the predominant reason for carbocation stability?

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  • $\begingroup$ Yes in hyperconjugation, the sigma bond of carbon with the carbon having positive charge rotate continuously and there is continuous bond forming and bond breaking. But as a result, the positive charge is not just on the one carbon, but on alpha hydrogens also. So there is more dispersion of positive charge and this stabilize the carbocation. $\endgroup$ – Manu Jul 20 '20 at 7:29
  • $\begingroup$ @Manu You mean in as there are more alpha hydrogens present to overlap the empty p orbital (positive), thus better stabilise the charge? But can this 'override' the seemingly de-stabilization effect from my first doubt? Please correct me if I am incorrect, thanks! $\endgroup$ – user96067 Jul 20 '20 at 9:01
  • $\begingroup$ In CH2F+, F is more EN than C, hence it attracts e-density, while in ethyl C+, both C have same EN, so they share excess charge to each other. $\endgroup$ – Rahul Verma Jul 20 '20 at 10:00
  • $\begingroup$ Inductive effect is weak as compared to hyperconjugation, so the instability caused by inductive effect is less than the stability imparted by the hyperconjugation, due to more dispersion of positive charge. $\endgroup$ – Manu Jul 20 '20 at 10:44
  • $\begingroup$ Can I clarify like this, the stability is achieved by 1. hyperconjugation 2. partial negative carbon on methyl group, donating electron density to positive charge? Thank you all :) $\endgroup$ – user96067 Jul 20 '20 at 13:40