[moondaughter]

Hello, I have a two questions actually, I have tried to write everything I know, but I do need some help also please and thank you!

1.) Report your percent yield and discuss how your percent yield could have been
improved.

Reaction: 4 isobutryophenone + 4NaBH4 ---> 4 2-methyl-1-phenyl-1-propanol

.249 g of sodium borohydride ((1 mol)/(37.83 g)) X ((4 mole)/(1 moles)) X ((150 g)/(1 moles))  =  3.94 g of product
.500g of isobutyrophenone ((1 mol)/148.20g) X ((4 mole)/(4 moles)) X ((150 g)/(1 moles)) =  .506 of product

so the limiting reagent is the isobutyrophenone
actual yield:

Weight of product: .328 g

Which means that the % yield (which is actual yield/ theoretical yield)
= .328 g/ .506  X 100 = 64.8%

Here I would just like to know if this is correct? But what I don't understand is that this is the weight of the final "product" in the flask, but was not necessarily pure....

5.) Using MgSO4 as a drying agent can cause degradation of the product.  This is
because MgSO4 is slightly acidic.  What degradation product might you expect
from the addition of an acidic compound to your product?

this is the question that I need the most help on.

I was thinking that the -OH group would be protonated and form a -OH2 group and since water is a good leaving group, it would leave?

But at the same time if all I have is the product (2-methyl-1-phenyl-1-propanol) and the acid in solution, I don't have a better nucleophile than water...

I was thinking that it could form a double bond but that would mean another hydrogen would have to be removed as well....

[Dan]

Yield calculation looks fine to me.

For the second part, you are on the right track:

I was thinking that it could form a double bond but that would mean another hydrogen would have to be removed as well....

What is the problem with removing another hydrogen?

[moondaughter]

I just looked up the mechanism of alcohols to alkenes (which admittedly I should have done earlier) so this would mean that the sulfate species that initially donated the proton would come back to pick up the second proton.

My only other question then would be: is there rearrangement?

After the water leaves, would it be favorable for the carbocation to be rearranged, via migration of a methyl group to the benzylic carbon? I know that this would actually form a secondary carbocation, but I was wondering if the methyl groups would hinder the picking up of the second proton. As a secondary carbocation, there would be less steric hindrance.... and either way the alkene formed would be tri-substituted