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What reaction does the following enolate undergo?

enter image description here

Both are nucleophilic substitution reactions. Ordinarily, the intramolecular product should be favoured, but I know that the transition state has a geometrical constraint in that the incoming nucleophile must approach the C-Br bond from the opposite side. Would that favour the intermolecular reaction instead?

I went through March's Advanced Organic Chemistry and found this:

enter image description here

And this from Clayden's Organic Chemistry:

enter image description here

That's why I am not able to decide if the reaction is intermolecular or intramolecular.

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    $\begingroup$ The intramolecular reaction is strongly favored entropically, but you can drive the reaction in either direction by adjusting reaction conditions such as concentration. See this earlier answer to a similar question for more background. $\endgroup$ – ron Dec 6 '15 at 17:19
  • $\begingroup$ @ron , I have seen some reactions in which intermolecular reactions are favoured. I have added one in the edited question. I will take one more from claydon. $\endgroup$ – Aditya Dev Dec 7 '15 at 2:22
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It will be the first product. Generally, Intramolecular reactions happen much faster than Intermolecular reactions. Imagine you are in a room with a few people and Godzilla picks up the room and shakes it. As you go tumbling around the room are you more likely to smack your head into your own knee or someone else's knee first?

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  • $\begingroup$ In general, intramolecular reactions are faster. Additionally, I don't know which type the metaphor is suggesting is faster. $\endgroup$ – jerepierre Dec 6 '15 at 17:17
  • $\begingroup$ I apologize, you are right. Thats what I meant, but due to a brain fart I swapped the terms. the answer has been corrected. $\endgroup$ – A.K. Dec 6 '15 at 17:23
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    $\begingroup$ There are bulk molecules that cannot react between certain reaction sites or may do so with great energy input lke when making cyclobutane and cyclopropane. Though unlike a fat man molecules can twist. $\endgroup$ – A.K. Dec 6 '15 at 18:31
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    $\begingroup$ Examples are given in Marxch's advanced organic chemistry which show that it's sometimes easier to smack your head onto someone else's knees :) See the edited question for an example. $\endgroup$ – Aditya Dev Dec 7 '15 at 2:17
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    $\begingroup$ This "answer" pretty much offers no explanation, it is much more suited as a comment. If you can explain why an intramolecular reaction is faster in many cases, then this might be salvageable, but in its current state it only provides an opinion. $\endgroup$ – Martin - マーチン Oct 31 '16 at 3:48
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I'd also like to add that the intramolecular cyclisation that you proposed is a 5-exo-tet reaction according to Baldwin's classification.

  • The ring being formed is 5-membered.
  • The bond being broken (the $\ce{C-Br}$ bond) is outside the ring, or exocyclic.
  • The carbon being attacked is a $\mathrm{sp^3}$-hybridised carbon with tetrahedal geometry.

If you look up Baldwin's rules, you'll see that 5-exo-tet reactions are perfectly fine and allowed. Evidently the compound you drew can twist itself a little bit to accommodate the linear transition state. (I am using "linear" loosely to refer to the arrangement of the incoming nucleophile, the carbon, and the leaving group.) Therefore, in this case, the intramolecular reaction will predominate, as is usual.

The disfavoured reactions that you cited from those textbooks are 6-endo-tet reactions. In such compounds, having a linear transition state is an issue as it would lead to too much ring strain.

As an ending note: Baldwin's rules are nothing more than an empirical observation. You might ask why 5-exo-tet reactions are fine, but not 6-endo-tet: where is the threshold or boundary which makes something favoured or disfavoured? The answer is, they did the experiments and found out that that's just how it is. No more, no less.

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  • $\begingroup$ Why is this not the accepted and most highly voted answer? It is the correct answer, i.e., stereoelectronic effects of $\mathrm{S}_{N}2$ reactions. $\endgroup$ – Zhe Oct 31 '16 at 0:27
  • $\begingroup$ @Zhe I just came (very) late to the question. It takes time for votes to come in and also the accepted answer may not even change if OP doesn't frequent the site. C'est la vie - not that I need rep anyway. $\endgroup$ – orthocresol Oct 31 '16 at 1:20
  • $\begingroup$ @orthocrescol But someone still had to down-vote it! $\endgroup$ – Zhe Oct 31 '16 at 2:42
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    $\begingroup$ I like the warning at the end! The currently accepted answer is in my opinion a comment not an answer. Apart from the guidelines you have posted, I think that we also have to look at the intermolecular reaction. There we see that in order to have an SN2 kind of attack the bromide must be close to the negative charge. Unlikely. $\endgroup$ – Martin - マーチン Oct 31 '16 at 4:24

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