4
$\begingroup$

Could someone help me out with the most likely major products for each reaction?

1 eq. of reagent is used for both reactions, so my initial thought was that both processes would only occur at the C=C double bond on the 6-membered ring as that one is more substituted and therefore more electron rich compared to the C=C bond at the bottom. However, the C=C bond at the bottom is less hindered so I was wondering if maybe sterics was a more important factor when considering regioselectivity for either one or both reactions.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
4
$\begingroup$

Sterics is the deciding factor in hydroboration which occurs at the least hindered carbon, thus you get the primary alcohol from limonene. The hydroboration of limonene is discussed here

MCPBA selectively oxidises the more substituted alkene source here so with limonene the selectivity is the reverse of the hydroboration and the cyclohexene double bond is epoxidised.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. Is there a reason why sterics is the deciding factor for hydroboration but not for epoxidation? Especially as BH3 (less bulky) is used instead of BHR2 as the reagent. $\endgroup$
    – Sasha J
    Aug 30, 2022 at 10:43
  • $\begingroup$ @SashaJ Does this answer your question? chemistry.stackexchange.com/questions/74005/… $\endgroup$
    – Waylander
    Aug 30, 2022 at 11:10
  • $\begingroup$ That's very helpful, thank you for sharing. I think I understand why epoxidation occurs at the more substituted electron rich bond, however I am still unsure why hydroboration does not follow the same selectivity. Isn't BH3 also acting as an electrophile, so shouldn't it also be attracted to the more electron rich bond? I get why sterics would be the deciding factor if the hydroboration reagent used was bulkier, but in this case it isn't. $\endgroup$
    – Sasha J
    Aug 30, 2022 at 12:10
  • $\begingroup$ @SashaJ I cannot immediately answer this. I think it deserves to be a new question. $\endgroup$
    – Waylander
    Aug 30, 2022 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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