With an analogous mechanism: C bonded to X has a δ+ charge and attacks the pi electrons to form a carbocation intermediate, with X- as a leaving group. X- then adds to the carbocation to neutralize it.

Why doesn't this occur?

If steric hinderance is the answer, what about MeX adding to a terminal alkyne?
RCH2X + OH- → RCH2OH + X- proceeds (with comparable/less hinderance?) so it seems unlikely that this the reason.

H-X bonds are more polar than C-X, but the electronegativity difference between C and H is small enough that we consider C-H bonds nonpolar. Electron-deficient C can pull some electron density from adjacent carbons if there are any - but again, what about MeX? Is the presence of 3 hydrogens enough to stabilize the C-X bond to the extent that the addition doesn't take place?

  • $\begingroup$ Well, something like that happens in Friedel-Crafts alkylation, but you need a catalyst. To protonate alkene (or alkyne) you need a very strong acid. Similarly you need very strong alkylating agent and MeX isn't strong. $\endgroup$
    – Mithoron
    Sep 8 at 21:58
  • $\begingroup$ @Mithoron What factor necessitates use of a catalyst? Autoionization of water seems less favorable and occurs at RTP. Adding R-X has the potential to generate a more substituted carbocation than H-X, with the same leaving group. What makes the transition state for this reaction so high-energy? I am just trying to understand this conceptually. If I saw this on paper, how would I know no reaction occurs? $\endgroup$ Sep 8 at 22:46
  • 2
    $\begingroup$ "Autoionization of water seems less favorable"? Only seems. Oxonium cations are waaay more stable than carbenium cations (typical carbocations like CH3+). For starters they have full octet. $\endgroup$
    – Mithoron
    Sep 8 at 23:51
  • $\begingroup$ @Mithoron That's what I was missing. Thanks. $\endgroup$ Sep 9 at 2:37


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