Almost everywhere I checked, the Inductive effect is described as a permanent effect with almost no mention of what that actually means.

  • What exactly is permanent about the effect?

  • If the effect is permanent, does that mean that even if the attached electron withdrawing/donating group is removed, the induced effect still continues to exist?


1 Answer 1


The effect is permanent because the group that does the 'induction' is always there in the molecule. For example, if the induction is caused by a t-butyl group, that group is held by a strong carbon to carbon ($\ce{C-C}$) bond that will not easily be broken. One could also consider the inductive effect of the Fluorine held by a very strong $\ce{C-F}$ bond.

A temporary effect might be caused by protontation of an oxygen atom where the protonation can catalyse a reaction. The protonation is reversible. The bond between the proton and oxygen would be much weaker, as $\ce{C=O^+-H}$ will be in equilibrium with $\ce{C=O + H^+}$.

Removal of the group that does the induction will remove the inductive effect.

  • $\begingroup$ Doesn't the same happen in Electromeric Effect? I mean, for the time that the reagent is present, the pi-bond is polarized. On its removal, the molecule returns to its original configuration. But then we say that Electromeric effect is temporary. What am I missing? $\endgroup$
    – Tabish Mir
    Commented May 30, 2015 at 9:36
  • $\begingroup$ @Technotoes - might be worth asking another question about that... wikipedia I think summarizes it neatly by saying "This effect is shown by those compounds containing multiple bonds. When a double bond or triple bond is exposed to an attack by a reagent, a pair of bonding electrons involved in the pi bond is transferred completely from one atom to another. This effect will remain as long as the attacking reagent is present. As soon as the reagent is removed, the polarized molecule will come back to the original state." - so because the attacking reagent causes it the effect is 'temporary' $\endgroup$
    – tom
    Commented May 30, 2015 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.