4
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Which of the highlighted C–H bonds in the following compounds is weakest? (R = generic alkyl group)

Methyl/primary/secondary/tertiary C–H bonds

I personally thought that the answer was 1 but it is given as 4 in the answer key. I didn't understand how it is so.

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2
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I believe hyperconjugation is a possible cause for this phenomonen. In general, the greater the amount of electron donation to the antibonding MO of a bond, the weaker the bond. Note also that it is widely-acclaimed that $\ce {C-C}$ $\sigma$ bonding MOs are better hyperconjugative donors than $\ce {C-H}$ $\sigma$ bonding MOs.

In the case of methane, there is no hyperconjugation occuring. Thus, naturally, we would predict that it has the strongest of the $\ce {C-H}$ bonds. In the case of the $\ce {C-H}$ bond made to the tertiary carbon, there is the most numerous number of hyperconjugative donations of electron density to the $\ce {C-H}$ $\sigma$ antibonding MO due to there being the most number of adjacent $\ce {\alpha C - \beta C}$ and $\ce {\alpha C - \beta H}$ bonds acting as hyperconjugative donors.

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  • $\begingroup$ It is probably Cα–Hβ (or Cα–Cβ) σ MOs acting as the donor orbital, analogous to ethane as in the diagram here, not the immediately adjacent C–Cα which have very poor overlap with the C–H σ* $\endgroup$ – orthocresol Sep 12 '18 at 15:09
  • $\begingroup$ @orthocresol Yes I agree. I should clarify that in my answer. $\endgroup$ – Tan Yong Boon Sep 12 '18 at 22:58
4
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$\ce{X–H}$ bond dissociation energies usually correspond inversely to $\ce{X^.}$ radical stability, so another way of looking at it is to recall that the stability of radicals increases in the order methyl < primary < secondary < tertiary.

The root cause of this is hyperconjugation (from $\sigma_\ce{C–H}$ to singly filled $\mathrm{p}$ orbital on carbon), analogous to the stabilisation of carbocations. So, it is just the second side of the same coin, really.

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  • $\begingroup$ My approach of analysing the hyperconjugative interactions to the C-H sigma * MO in the molecule is equally valid right? $\endgroup$ – Tan Yong Boon Sep 12 '18 at 23:04
  • $\begingroup$ Indeed, that’s the same thing in the end. $\endgroup$ – orthocresol Sep 13 '18 at 0:37
1
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Consider ethane. The bond lengths are as follows: $\ce{C-H}$: $\pu{110 pm}$. Bond angle is: $\ce{H-C-H}$: $109.6^\circ$.

And, consider those for methane: lengths are $\ce{C-H}$: $\pu{109 pm}$ and bond angle is: $\ce{H-C-H}$: $109.5^\circ$.

Though small, the angle has risen from methane to ethane (0° to 1°). Also, the bond length has risen. These influences are (supposed to have been) caused by steric effects of a methyl group in ethane. And thereby causing a decrease in bond energy. Also, due to the presence of three alkyl groups, the electrophilic character of carbon (though small already) is reduced reflecting in its share of electrons in the $\ce{C-H}$ bond of a 3° alkane causing the bond to weaken (I emphasize, this effect has very very little share).

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