I don't see any reason why the transition state of an SN2 reaction that happens after an internal nucleophilic group attacks(which is known as NGP) would be more stable. In fact, the nucleophilic group that attacks is often a strong nucleophile that is quite reactive (more than the leaving group which is usually a weak base) which would only destabilise the transition state when the external nucleophile does attack.

My question is: Why is the internal attack and the external attack combined faster than simply the external attack since the rate of such reactions where neighboring groups are involved is greater than normal SN2 reactions?

  • $\begingroup$ I suppose these neighboring groups switch on abilities like dipole in the bond and make way for easier substitution. $\endgroup$
    – Suraj S
    Apr 30, 2017 at 14:16
  • 2
    $\begingroup$ Think entropy, the neighboring group is in the same molecule and held in close proximity. $\endgroup$
    – ron
    Apr 30, 2017 at 14:29
  • $\begingroup$ Possible somewhat answered here: chemistry.stackexchange.com/a/65872/21296 $\endgroup$
    – NotEvans.
    May 1, 2017 at 17:29
  • $\begingroup$ @NotBaran No, that does not answer my question at all. I understand the reason NGP occurs, I just don't understand why it increases the overall rate of the reaction $\endgroup$
    – xasthor
    May 3, 2017 at 4:30
  • 1
    $\begingroup$ This is a bit of a tautology, as you would only hear about NGP if it increased the rate of the reaction. If there was a NGP pathway available that didn't increase the rate of the reaction, it would simply not be mentioned. $\endgroup$ Jul 6, 2017 at 12:55

1 Answer 1


In neighbouring group participation (NGP), an internal nucleophile first displaces a leaving group in an intramolecular sense, forming a ring (usually 3 or 5 membered) which is then opened by the incoming nucleophile intermolecularly.

General NGP Mechanism

General mechanism of neighbouring group participation (taken from March, Advanced Organic Chemistry, 2007

In the scheme above, Z first displaces X to form a 3 membered ring, which is then opened by Y. Your question then becomes why is Z faster than Y at substituting X?


In order for the substitution to take place, the nucleophile must come into alignment with the leaving group. In a general sense, intramolecular reactions are usually faster than intermolecular since the two reacting groups are tethered together – there is no requirement that they collide in order to react. This is especially evident for intramolecular formation of 3 and 5 reasons, since those ring sizes are especially favourable. (Note that intramolecular reactions isn't always favourable, for example in macrocyclisations of large rings, where high dilution is needed to close the ring rather than getting intermolecular reactivity).


The reaction between the substrate and the external nucleophile Y in an intermolecular fashion is accompanied by a decrease in entropy (two molecules are becoming one in the transition state). Whilst this also happens in the intramolecular sense (the molecule is becoming more rigid), it is significantly less than in the bimolecular case.

Its also worth pointing out that its not always 100% clear whether a molecule is reacting via an NGP type pathway, and indeed there are many cases where its been shown that both are operating at the same time, with the intra being faster than the inter, but not sufficiently rapid to prevent any direct intermolecular reaction.

  • $\begingroup$ My question was: why is the internal attack+the external attack combined faster than simply the external attack since the rate of such reactions where neighboring groups are involved is greater than normal Sn2 reactions $\endgroup$
    – xasthor
    May 7, 2017 at 3:53
  • $\begingroup$ You may wish to edit your post - that comment is a lot clearer than your original question $\endgroup$
    – NotEvans.
    May 7, 2017 at 8:27

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