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EK In-class Exam 2
Divya_4978
#1 Posted : Wednesday, June 10, 2020 2:11:41 AM
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Hello,
In the In-class exam 2, I do not understand passage 4 questions 37 and 40. I looked at the solutions however I do not understand the explanations.

Thank you
INSTR_Katerina_102
#2 Posted : Wednesday, June 10, 2020 3:20:48 AM
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Hi,

For 37, we see in the passage that we are adding a methyl group to the aromatic compound in reaction 1. Because CH3I is the only other reagent, it is safe to assume the methyl group came from CH3I. As CH3I is an electrophile, the aromatic compound must be made into a nucleophile, because all MCAT reactions occur between a nucleophile and an electrophile.

From this, we can eliminate A and B as both refer to electrophiles.

Next, we realize that a carbanion is nucleophilic, because it has an excess of electrons. We can eliminate D because carbocations are electrophilic, not nucleophilic.

The role NaH plays is to generate the carbocation. It deprotonates the carbon next to the ester forming a carbanion stabilized by resonance.

For 44, we recall that if R naproxen has a optical rotation of -63 degrees, S naproxen must have an optical rotation of +63 degrees. We want a mixture of R and S which has a specific rotation of 31.5 degrees. The question specifically asks us what percent of this mixture is the S enantiomer.

We recall that a racemic mixture (50/50 R/S) has a optical rotation of 0, therefore we can eliminate A. We also know that we need more S than R as the optical rotation is postiive (more closely resembling a pure S mixture). From this we can eliminate C, which implies there is more R than S).

This leaves us with B and D.

To distinguish between the two, we know that the optical rotation of a solution is equal to the speicific rotation * the concentration of the molecule in solution, or

Optical roatation = (rotation of pure enantiomer 1)(% enantiomer 1) + (rotation of pure enantiomer 2)(% enantiomer 2)

With the formula OR = 63S -63R, where S is % S enantiomer and R is % R enantiomer.

Since we only have 2 compounds R = 1-S, and OR = 31.5 = 63/2

And the formula simplifies to 62/2 = 63S - 63(1-S)

63/2 = 63S + 63S -63

1/2 = 2S -1

1.5 = 2S

0.75 = S

Therefore D is the correct answer.


Hayley_5145
#3 Posted : Friday, June 12, 2020 11:10:14 PM
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I also looked at the answer key for ICE #2 question 37. As well as, read your explanation above and tried to figure it out using my notes however I still don't understand the answer.

Do you perhaps have any other way of explaining why 37 is C? I am sorry I have read over your explanation many times and can't figure it out.

Thank you
INSTR_Katerina_102
#4 Posted : Friday, June 12, 2020 11:50:32 PM
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Hi,

All, good, some things are very visual. I'm sorry my explanation is still unclear and I will try my best to find a way to post a picture.
INSTR_Katerina_102
#5 Posted : Saturday, June 13, 2020 12:10:58 AM
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https://ibb.co/3pBhKxs

If you look at the image in the link I have provided, it shows that NaH is a base that deprotonates the acidic proton by the ester, forming a carbanion.

A carbanion has a lone pair (which I've omitted accidentallly but it is represented by the negative formal charge), and recall species with an excess of electrons are Negative and therefore Nucleophiles.Therefore the correct answer is C because you've formed a carbanion nucleophile (pink text in photo).

If it is unclear what the distinction between a carbanion is and a carbocation is, recall that cations are PAWsitive (positive).

I have omitted the full structure in the example in order to not post the full question, but the other parts of the structure can be seen in Reaction 1, they're just not significant in answering the question.

Please let me know if this is still unclear, I hope this helps!
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