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Question before the big day!

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_a_rose2238

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Okay Guys,

DAT tomorrow. I feel like I am as ready as I am going to be. I am obviously super nervous but you all suggested so much QR material that I have dedicated a ton of time to. I feel so much better about it now.

I was wondering if anyone had any insight on a question I wanted to get cleared up before going into the exam.

For Ochem:
Enantiomers have individual optical activity but they cancel each other out. I get this. But as I was going back through the bootcamp tests yesterday I found two questions that seem to contradict each other. One says that enantiomers are optically active and the other implies they are not optically active.
If a question simply asks: Are enantiomers optically active? The answer would be "no", right? I feel like this is something that is super likely to show up so I don't want to get it wrong for stupid reasons lol


I know there are a couple of people taking it tomorrow as well so GOOD LUCK :)
 
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Steeezy

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Good question! I'd say that unless specified that the mixture of enantiomers is racemic (50/50) then the answer would be that it is optically active. Enantiomers on their own are optically active, and when mixed together in unequal amounts, the mixture is also optically active.
 
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DDSDCMUU

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So I actually think it would optically inactive. The two enantiomers would cancel each other out. That is what Dr. Mike said on DAT BC videos if I remember correctly. Maybe @orgoman22 can clarify this?
 
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KatawareDoki

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I've come across those exact two scenarios. How I understand it is like this:

Individually, they are optically active and hold some "+" or "-" value in a particular rotation.
When you have them mixed together at equal amounts, their NET value would be "0" since one would rotate "+44.3" and one would be "-44.3" when we're talking about a pair of enantiomers.

Again, I could be wrong, but with this thought process, I've been able to come up with correct answer choices. Cheers!
 
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_a_rose2238

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I've come across those exact two scenarios. How I understand it is like this:

Individually, they are optically active and hold some "+" or "-" value in a particular rotation.
When you have them mixed together at equal amounts, their NET value would be "0" since one would rotate "+44.3" and one would be "-44.3" when we're talking about a pair of enantiomers.

Again, I could be wrong, but with this thought process, I've been able to come up with correct answer choices. Cheers!

This is what I was thinking. I'm hoping they won't ask it in a confusing way. Because technically they are but together they aren't. I guess with this logic it would simply be: No. enantiomers are not optically active.
 

DDSDCMUU

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This is what I was thinking. I'm hoping they won't ask it in a confusing way. Because technically they are but together they aren't. I guess with this logic it would simply be: No. enantiomers are not optically active.
Yeah I think it would have to say "pure enantiomer" or only s. Good luck on your DAT! You will rock it!
 
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_a_rose2238

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So I actually think it would optically inactive. The two enantiomers would cancel each other out. That is what Dr. Mike said on DAT BC videos if I remember correctly. Maybe @orgoman22 can clarify this?
I watched that as well! And that's is the logic I've been sticking with until I found question 12 on exam 9. In the description it says that they are active.
 
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deleted678432

Chiral compounds (besides meso compounds) are optically active. This means that an enantiomer is optically active. When there is a racemic mixture (50% R/50% S), then the enantiomers are STILL optically active, but the overall mixture would have NO rotation in a polarimeter. To reiterate.. say that molecule I has light rotation of +20. This means that its enantiomer has a light rotation of -20. When the two are mixed together in an exact 50/50 racemic mix, each molecule STILL exhibits optical activity INDIVIDUALLY but the overall solution would have an overall rotation of 0.00.
 
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DAT Destroyer

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Okay Guys,

DAT tomorrow. I feel like I am as ready as I am going to be. I am obviously super nervous but you all suggested so much QR material that I have dedicated a ton of time to. I feel so much better about it now.

I was wondering if anyone had any insight on a question I wanted to get cleared up before going into the exam.

For Ochem:
Enantiomers have individual optical activity but they cancel each other out. I get this. But as I was going back through the bootcamp tests yesterday I found two questions that seem to contradict each other. One says that enantiomers are optically active and the other implies they are not optically active.
If a question simply asks: Are enantiomers optically active? The answer would be "no", right? I feel like this is something that is super likely to show up so I don't want to get it wrong for stupid reasons lol


I know there are a couple of people taking it tomorrow as well so GOOD LUCK :)
Enantiomers are optically active substances that have the ability to rotate the plane of polarized light. They are indeed optically active. If a reaction gives you a reacemic mixture of these enantiomers, however it would be inactive. Truth be known, rarely in real life do we get a 50-50 mixture....often the numbers are a bit different. In advanced organic chemistry we like to call it a scalemic mixture. To answer your question......YES...they are indeed optically active. Hope this helps.
 
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