A chemist finds that the addition of (+)-epinephrine to the catal...

Question

A chemist finds that the addition of (+)-epinephrine to the catalytic reduction of butan-2-one (Figure 5-17 ) gives a product that is slightly optically active, with a specific rotation of +0.45°.

Calculate the percentages of (+)-butan-2-ol and (−)-butan-2-ol formed in this reaction.

Accepted Answer

Hey everyone. Let's do this problem. It says addition of a Cairo molecule to the catalytic reduction of Penton to own produces a slightly optically active product, determine the percentage of plus Penton to all and minus Penton to all formed in the reaction. If the specific rotation of the product is plus 0. degrees, it also gives us the rotation of pure minus Penton to all. At the same condition is 13 degrees. Okay, So we know we have a mixture that has a specific rotation and we know the rotation of pure minus Penton to all. So if pure minus Penton tool has a rotation of 13 degrees positive and we know the specific rotation of the mixture is positive, right? It's plus .54. Then we know that the minus in in tumor is in excess. Right? If the degrees were negative or less than zero, then the other and then tumor would be in excess. Okay, so let's see what we can do with this. We can calculate the percentage that this an anti tumor is in excess. And then by knowing how much more of this in and tomorrow we have in excess. We know there has to be 100% total of these to an anti MERS. Right? So using that difference of the an anti american excess, we can find the exact percentages of each. An anti in the mixture. Okay, so we can use the equation the observed rotation over the rotation of the pure and then tumor times 100. So our observed rotation is .54. Right? And the pure the rotation of the pure an anti is 13.0. But I highlighted in green. Okay, so let's plug in these numbers. We get .54°. It is the absolute value 13 degrees times equals 4.15 degrees. So that means that the negative or sorry, the minus Penton to all. An anti tumor is in 4.15% excess. Okay. four point excess of. Okay, so now we know the difference how much access we have in that one in and tumor. So we can actually set up a system of equations because we know That both of these percentages have to add up to 100. And we know that one of the Ananta MERS is 4.15% greater amount than the other. So in this case X would be the minus Penton actual. And why would be the posit the plus Penton to all. So let's solve for our variables here. Okay, so we have X plus x minus 4.15. We're using substitution substituting this value of Y into our first equation giving us two, X equals one oh 4.15, divide by two. We get X is 52.75%. And again that is r minus an anti tumor. So subtracting that from 100. We will get our other and and tumor Which is 47.9 of the plus an anti. So let's write this out. So we have 52.075% of our excess and anti more the minus Penton to all In 47.9 of the plus oops Pen tan tool, which makes be the correct answer. That's it for this problem. I hope this is helpful. And thanks for watching.

A chemist finds that the addition of (+)-epinephrine to the catalytic reduction of butan-2-one (Figure 5-17 ) gives a product that is slightly optically active, with a specific rotation of +0.45°. Calculate the percentages of (+)-butan-2-ol and (−)-butan-2-ol formed in this reaction.

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