grignard reaction question

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yoyohomieg5432

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this is from AAMC 10.

The answer is C. I understand why the grignard attacks the carbonyl and forms 2 Ph bonds there. I'm confused on why the OH in the back carbon is replace with OMgBr? I can understand it for the front carbon because after the carbonyl is attacked a carbanion forms and a MgBr can bond to neutralize that.


Also, after treatment with H30+, what causes MgBr to leave and OMgBr ----> OH?

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this is from AAMC 10.

The answer is C. I understand why the grignard attacks the carbonyl and forms 2 Ph bonds there. I'm confused on why the OH in the back carbon is replace with OMgBr? I can understand it for the front carbon because after the carbonyl is attacked a carbanion forms and a MgBr can bond to neutralize that.


Also, after treatment with H30+, what causes MgBr to leave and OMgBr ----> OH?

Grignard reagents are super strong bases. The Grignard reagent will strip away the H+ in OH, leaving the MgBr cation ionically bonded to the O-. That's why you never use water as a solvent in Grignard reaction. I think diethyl ether is the standard solvent.

The final treatment with H3O+ will remove the MgBr cation and reform the OH bond.
 
Grignard reagents are super strong bases. The Grignard reagent will strip away the H+ in OH, leaving the MgBr cation ionically bonded to the O-. That's why you never use water as a solvent in Grignard reaction. I think diethyl ether is the standard solvent.

The final treatment with H3O+ will remove the MgBr cation and reform the OH bond.

sorry forgot to post the question. OP is updated.

IF the MgBr is a strong base, why doesn't it take the H from the OH and make HMgBr and O-?
 
sorry forgot to post the question. OP is updated.

IF the MgBr is a strong base, why doesn't it take the H from the OH and make HMgBr and O-?

You're mistaken, MgBr is not the Grignard reagent (strong base) -- PhMgBr is the Grignard reagent. You can effectively look at PhMgBr as a Ph- anion. MgBr is the cation that is left over when Ph- strips away the H+ in OH to form the phenyl hydrocarbon .
 
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