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Topic: Percent yield concerning a friedel-crafts alkylation  (Read 5248 times)

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Offline Alexis7

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Percent yield concerning a friedel-crafts alkylation
« on: February 04, 2013, 01:46:59 PM »
Hello. The question is:

Assuming the yield of Friedel-crafts reaction performed was 65%, how much biphenyl (mass) and t-butyl chloride (volume) are required to give 3.0 grams of the product?

So far what I have found out is that the balanced equation should be:

C12H10+ 2(CH3)3CCl → C20H26 + 2HCl

The product is known as 4'4' di-tert-butylbiphenyl (C20H26)

molar mass of Biphenyl is 154 g/mol

molar mass of t-butyl chloride is 92 g/mol 

molar mass of product is 266 g/mol

Now I'm assuming the 3.0 g is the actual yield. Theoretical yield would have to be around 4.62 g.

How should I approach the problem? Should I move backwards and use the 3.0 g to find the moles?

Any tips would be appreciated.

Offline Dan

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Re: Percent yield concerning a friedel-crafts alkylation
« Reply #1 on: February 04, 2013, 03:14:34 PM »
You have figured out that based on a 65% actual yield, you will have to aim for a theoretical (100%) mass yield of 4.62 g.

So what number of moles do you think you should be aiming for?
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