Orgo question :( dat bootcamp

[JDHK]

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I'm so bad at orgo, since it's been a while since I took the course.. this is a question from dat bootcamp, and it seems like one of the easier ones b/c i knew it was sn1 reaction and I knew that the water could attack from both sides of the plane, but when I draw it out, I don't see in any way how they are enantiomers...I drew the 3d way it would look when both sides are attacked, but they are not even mirror images..b/c of the other carbon..well, take a look pls. Thanks in advance!

Question is attached.

Had to fit the size of the pic so the question is: The following yields two isomeric substitution products. What is their relationship? The answer is enantiomer.

 

[MAdental]

Question is attached.

No it Aint 🙂

 

[Ari Rezaei]

I think the best way to attack this problem would be the process of elimination.

You can eliminate A because the back carbon isn't chiral, there are two methyl groups attached to it. Since there is only one chiral center, C can also be eliminated. Rotamers refer to different types conformations, like anti and gauche. E can be eliminated because there is only one place for water to attack, so we won't have multiple products connected in different ways. The only answer left is B.

 

[JDHK]

GatorD, thank you! it's really awesome to see that the maker of DAT bootcamp is actually proactively helping students like myself. I really appreciate it that the person behind the business is genuinely helpful. That said, I just have a quick question on this problem as I see what you mean but I just want clarification for future reference. I attached my thought process, but as you can see in the carbon I marked, there is no way that is a mirror image of the other product. Sorry for the extra quetsion, and thank you!

 

[Ari Rezaei]

I think you may be mistaking the chiral carbon. The chiral carbon is the one on the right, not the left. The carbon on the left is not chiral, it is attached to two identical methyl groups (one of them is a wedge as a distractor, but they're both still methyls).

To number this system, the OH gets highest priority, followed by the phenyl, then by the carbon you marked on the left, then finally the methyl attached to the right carbon.

The top image you drew is in S configuration, and the bottom in R configuration. Since they differ at one chiral center and that's all we know, these must be enantiomers.

 

[JDHK]

ohhh thank you. I was looking at the entire molecule as a whole by accident instead..looks like i am really really rusty..Thank you again.