I have read from my textbook, that the opening of an epoxide ring with a strong nucleophile like a Grignard reagent, takes place from a less substituted side, with a higher SN2 character.

The acidic workup of the product obtained above gives alcohol.

Also, if ring-opening takes place in an acidic medium, it takes place with a higher SN1 character (even though carbocation formation doesn't really take place) and the epoxide ring opens from the more substituted side

My question is if one can protonate the epoxide in an acidic medium, then facilitate the reaction with Grignard Reagent (RMgX), would the regioselectivity be according to SN2 type or SN1 type, i.e. from the less substituted side, or more substituted side

I propose that since the epoxide ring is now protonated, it can easily leave than it's unprotonated form, and the Grignard Reagent can now attack from the more substituted side as well.

Edit: The question I was talking of in the comments.

According to my teacher, the H+ is the reminscent from the epoxidation reaction as benzoic acid is formed as a product. I can see the the epoxide gets protonated, but the subsequent attack of Grignard Reagent is what confuses me

enter image description here

Answer given is A part, which corresponds to ring-opening from more substituted side.

  • 4
    $\begingroup$ Yeah, no, you can’t do that. Grignard plus acid do not go well together. $\endgroup$
    – orthocresol
    Jul 3 at 20:57
  • 2
    $\begingroup$ There are some nice examples of using Lewis acids + Grignards to control/change selectivity of ring opening. But as above, protic acids are a no go $\endgroup$
    – NotEvans.
    Jul 4 at 5:29
  • $\begingroup$ I agree with you guy 1000%, even I understand acids turn Grignard reagents back to alkanes, the thing is, I have a question in my textbook which I have added in the edit above. $\endgroup$
    – Anmoldeep
    Jul 4 at 6:00
  • 1
    $\begingroup$ If you use excess Grignard to neutralise all the H+ then eventually you will get Grignard attack on the epoxide, but why do it this way? You will get opening to give B and are using MeMgBr as a base, using up a costly reagent for no gain. If you want the epoxide opening to give A use BF3 as the complexing agent on the epoxide $\endgroup$
    – Waylander
    Jul 4 at 9:50
  • 1
    $\begingroup$ @Waylander, thanks a tonne, I get it now $\endgroup$
    – Anmoldeep
    Jul 4 at 10:00

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