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After formation of the bromocarbocation the alkoxide anion attacks the bromocarbocation which leads to ring formation and can proceed through two pathways enter image description here

5 endo tet cyclization is disfavored but 4 exo tet cyclization is favored according to Baldwin's rules; however, the answer given in my text book is the 5 membered ring.

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Your reasoning is totally correct according to Baldwin's Rule - the 4-exo-product would be more stable than the 5-endo-product ( see : This for more details).

But your method may be wrong in the given question the product A was first formed and then a ring closure step was performed using $\ce{KOH}$ as shown below : enter image description here enter image description here

here the 5-exo-trig ring closure is favoured according to Baldwin's rule .

Let me also address something that may be a doubt ( It was mine ) i.e. why doesn't the oxygen atom attack the bromonium Ion in the first step ? this is because even if it will attack you will also form a acid $\ce{HBr}$ which will cause the cleavage in the cyclic ester we just formed returning us to the original reagent hence it doesn't attack.

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  • $\begingroup$ Upon closer reading, I do want to point out some inaccuracies. (1) Baldwin's rules do not say anything about the stability of products, but only whether the route for their formation is kinetically possible. The 4-exo-tet pathway is "favoured", and the 5-endo-tet is "disfavoured". Of course, it does so happen that the five-membered ring is more stable than the four-membered ring. That, however, is not a consequence of Baldwin's rules. (2) Also, there are no trig ring closures; SN2 takes place at a tetrahedral carbon, hence tet. $\endgroup$ – orthocresol Mar 17 at 2:04
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    $\begingroup$ (3) Finally, and maybe more critically: in the SN2 step, both possibilities - 4-exo-tet, as well as 5-exo-tet - are "favoured", according to Baldwin's rules. Therefore, invoking Baldwin's rules is not sufficient to explain which pathway is preferred, as the rules do not distinguish between which is "more favoured". The preference for 5-membered ring formation over 4-membered ring formation is explained by the balance between enthalpy and entropy of activation, see chemistry.stackexchange.com/a/107070/16683 as well as pp 805–807 in Clayden 2nd ed $\endgroup$ – orthocresol Mar 17 at 2:10

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