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End of chapter question Chemistry EK lecture 2
Michelle_6624
#1 Posted : Wednesday, June 16, 2021 12:41:53 AM
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Hello!

I don't understand the solution to question 41 and 44 on page 69 of the chemistry EK textbook in lecture/chapter 2? Also are there EK video solutions to the practice questions at the end of each chapter in each book or just video solutions for the EK FL MCAT exams?
INSTR_Kailey_109
#2 Posted : Thursday, June 17, 2021 2:36:55 PM
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Hi Michelle,

We'll need to wait for a specialist to respond to the first part of your session, but I can jump in here to clarify that the EK video solutions are provided for the full-length exams (excluding CARS), not for the Study Guide content.
INSTR_Katrina_128
#3 Posted : Friday, June 18, 2021 10:55:45 PM
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Hi Michelle! I'll take a stab at the EK questions for you. 😀

Q41:

When you're looking for a compound that can exist in cis or trans form, FOCUS ONLY ON THE DOUBLE BOND.

To exist as cis/trans isomers, the carbons that are double-bonded must EACH be attached to two different constituents.

Take a look at C:

The carbon on the right side of the double bond is attached to a methyl group (CH3) and a hydrogen (not drawn, but implied). So far so good.

The carbon on the left side of the double bond is attached to the top of a six membered ring and the bottom of a six membered ring. Since this ring is symmetrical, this carbon is TECHNICALLY attached to "two of the same thing". So, we can eliminate C because it doesn't fit the criterion of "both carbons being attached to TWO different constituents".

Take a look at D:

The carbon on the right side of the double bond is attached to two hydrogens (not drawn, but implied). We can therefore eliminate D.

Draw out B:

https://upload.wikimedia...methyl-2-butene.svg.png

The carbon on the left side of the double bond is attached to two methyl groups (CH3). We can eliminate B.

Therefore, A is the best answer.


Q44:

When you are looking for a compound that IS optically active, you want it to have a chiral center. When a carbon is attached to four different constituents, it's a chiral center.

When you have TWO chiral centers in a molecule, there is the possibility that you have a meso compound. A meso compound is the chemistry equivalent of the spiderman meme:

https://i.redd.it/bekphnqftcb41.jpg

One half of the molecule is a mirror image of the other one.

If that's true, then EVEN THOUGH there are two chiral centers, the molecule IS NOT OPTICALLY ACTIVE (meaning that it doesn't "rotate light").

OK... so how do we spot the meso compound?

With Fischer projections, it's relatively simple. We need to look for the Fischer projection that has a "mirror plane". Something that looks like this:

https://chem.libretexts.org/@api/deki/files/1574/Meso2_(5).bmp?revision=1

But wait... none of them look like that. 😫

Recall that SINGLE bonds can SPIN. So, we need to try spinning the top half of each of the available answers, and then see if we get a mirror plane.

https://chem.libretexts.org/@api/deki/files/1577/rotated_(1).bmp?revision=1

Trying this out, we can then see that B is the only answer.

Oh boy... what a question. 😣
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