Inductive effect is the partial shifting of the bonded electrons towards more EN (take as electronegativity or electronegative in this text) atom so why we will say that alkyl groups will always show +I effect. As the case I stuck in is with boron (EN 2.05) while carbon EN is 2.55 so it should not show +I effect there but the -I effect.


1 Answer 1


You are right, if we attach a low electronegativity element (such as boron) to carbon, the alkyl group does show -I effect, withdrawing electrons from the less electronegative element.

However, we usually say that less electronegative element ($\ce{B}$, in this case) is showing +I effect because the alkyl moiety is treated as the primary and the other groups are treated as secondary or substituents. Thus, these effects are calculated with respect to the alkyl moiety.

  • $\begingroup$ Then BH3 should be more stable than BR3 where R is an alkyl group. But in Hydroboration of an alkene, B is breaking bond from H and forms with R, that should mean that B which is electron deficient is getting relatively more electron density. And if we consider that alkyl groups show +I effect then this fits the logic but we have discussed about their EN values suggesting -I effect of Alkyl groups here. $\endgroup$
    – D13G
    May 11 at 12:36
  • $\begingroup$ We also need to consider hyperconjugation for $\ce{BR3}$, for example, $\ce{B(CH3)3}$ has 9 hyperconjugative structures involving the empty $\mathrm{p}$ orbital of $\ce{B}$. $\endgroup$
    – ananta
    May 11 at 12:41

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