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How can Br- deprotonate the intermediate in SN1 since it's such a weak base?
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Emily_6585
#1 Posted : Tuesday, August 17, 2021 7:41:16 PM
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I'm just reviewing OChem 4 where we talked about Sn1 reactions occurring in 3 steps when we have a neutral nucleophile.

In the example Br was the leaving group, producing Br- which subsequently deprotonated the R-OH2+ (water was the electrophile) to produce a neutral product.

I don't understand how this could happen since Br- (as the conjugate base of a strong acid) is such a weak base.

Would it be more plausible to say that the water solvent would deprotonate the intermediate?

What if the solvent/neutral electrophile was an alcohol? Could the alcohol act as a base and deprotonate the intermediate?

Thanks!
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INSTR_Katrina_128
#2 Posted : Tuesday, August 17, 2021 9:53:44 PM
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Hi Emily,

I agree. Br- is a very weak base. But that protonated species REALLY wants to get rid of the excess proton, so it needs to go somewhere. It will either go to the Br- or to the solvent molecule. If the solvent was an alcohol, it could still deprotonate the OH2+.

It seems weird to produce a strong acid as a product, but it does happen.

KEEP THIS IN MIND: If the solvent isn't water, then HBr is just another molecule. HBr is only an acid when it DISSOLVES AND DISSOCIATES IN WATER. Sneaky sneaky!
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