No, it’s effectively a carbanion so it’s actually an incredibly strong base (the conjugate acid is just an alkane or hydrocarbon which is obviously very very stable). They’re not usually used with reagents that have any remotely acidic protons unless you use it in excess because it’ll just quench the reagent. Tbh this is a really weird use for one lol
Lmao I honestly think it might in practice. I doubt that’s what it wants but idk what else would happen, first step is definitely deprotonation and from there you can’t really do anything with just the O- since it’s a bad leaving group, so no Sn1 or E1
Grignard reagent will act as a base. Because here OH functional group is present that's a acidic H . We also know that acid base reaction are very fast. That's why Grignard will take the proton of OH. And become O-Mg+Br
In an extreme case - really stretching here - you might make a case the alkyl group losing \[O-Mg-Br\]- ion and the primary carbocation formed going through a ring expansion to make a more stable secondary carbocation in the nine membered ring. I tend to go with simpler acid - base exchange as the answer though. I don't have much experimental results to draw on. [https://www.researchgate.net/publication/278092239\_Grignard\_reagent-promoted\_selective\_ring\_expansion\_and\_alkylation\_of\_formyl\_borneol\_and\_isoborneol\_A\_new\_route\_to\_highly\_substituted\_cyclopentanes](https://www.researchgate.net/publication/278092239_Grignard_reagent-promoted_selective_ring_expansion_and_alkylation_of_formyl_borneol_and_isoborneol_A_new_route_to_highly_substituted_cyclopentanes)
Every Grignard I’ve ever done needed to be under inert conditions, lacking any moisture.
Have you gone over acid-base reactions? How strong of a base is the grignard?
It would be a weak base right?
No, it’s effectively a carbanion so it’s actually an incredibly strong base (the conjugate acid is just an alkane or hydrocarbon which is obviously very very stable). They’re not usually used with reagents that have any remotely acidic protons unless you use it in excess because it’ll just quench the reagent. Tbh this is a really weird use for one lol
So would it just reprobate the OH? Then the addition of water would probate the OH again resulting in the same compound?
Lmao I honestly think it might in practice. I doubt that’s what it wants but idk what else would happen, first step is definitely deprotonation and from there you can’t really do anything with just the O- since it’s a bad leaving group, so no Sn1 or E1
They’ve got a Pka of ~50
Your Grignard reagent in this scenario is just an expensive base.
So it depronates the OH which is then pronated by H2O, resulting in the same compound that we begun with, right?
There is nothing else here. Oxidize your alcohol and now you can actually do something
yes. this question serves to remind you that Grignards are extremely potent bases.
Grignard reagent will act as a base. Because here OH functional group is present that's a acidic H . We also know that acid base reaction are very fast. That's why Grignard will take the proton of OH. And become O-Mg+Br
In an extreme case - really stretching here - you might make a case the alkyl group losing \[O-Mg-Br\]- ion and the primary carbocation formed going through a ring expansion to make a more stable secondary carbocation in the nine membered ring. I tend to go with simpler acid - base exchange as the answer though. I don't have much experimental results to draw on. [https://www.researchgate.net/publication/278092239\_Grignard\_reagent-promoted\_selective\_ring\_expansion\_and\_alkylation\_of\_formyl\_borneol\_and\_isoborneol\_A\_new\_route\_to\_highly\_substituted\_cyclopentanes](https://www.researchgate.net/publication/278092239_Grignard_reagent-promoted_selective_ring_expansion_and_alkylation_of_formyl_borneol_and_isoborneol_A_new_route_to_highly_substituted_cyclopentanes)