AAMC Section Bank C/P Q14

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Karolina
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Joined: Mon Dec 03, 2018 9:42 pm

AAMC Section Bank C/P Q14

Post by Karolina » Thu Jan 03, 2019 4:31 pm

I'm a little confused about how this separation would even work. After adding NaOH, we have only amides and carboxylic acids in the mixture. However, aren't amides equally as polar (if not more) as carboxylic acids? If so, why would amides separate into the phase with diethyl ether, which is only somewhat more polar than an alkane? Wouldn't both polar compounds end up in the polar aqueous phase?
NS_Tutor_Will
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Joined: Fri May 25, 2018 9:15 am

Re: AAMC Section Bank C/P Q14

Post by NS_Tutor_Will » Sun Jan 06, 2019 11:25 am

Thanks for the question!

First I'm actually going to copy a useful explanation written by another tutor, Alex:
First thing's first, how do we tackle the question stem? The question tells us we are reacting an amine with excess carboxylic anhydride. We should not only know what each of these molecules look like, we should also know what the product will be in this reaction (i.e. amides and carboxylic acids). Once we're that far, it's time to consider the resulting solution comprised of amides, carboxylic acids and carboxylic anhydride. We know that we must quench the unreacted anhydride because it shows up in every answer choice, but what will do? NaOH is the appropriate choice because it will turn the remaining anhydride into carboxylic acids. The next step is to consider the differences between amides and carboxylic acids in this environment. If we draw on our knowledge of acid/base chemistry as well as solubility, it's clear that the amide will be neutral and non-polar, where as the carboxylic acid will be soluble due to a negative charge and it's overall polar structure. We now use the difference between the two molecules to our advantage by adding the organic solvent as the amide will migrate to that layer due to it's insolubility. This allows us to separate the layers to obtain the desired product.
Ultimately, the key here is that the carboxylic acid and diethyl ether will move because they gain a charge and therefore want to be in the aqueous phase. The amide won't move because there is no change in its charge (so it's perfectly happy staying in the organic layer).

Good luck!
samanthasheffels
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Joined: Sun Jun 02, 2019 5:15 pm

Re: AAMC Section Bank C/P Q14

Post by samanthasheffels » Mon Aug 05, 2019 6:19 pm

Could you please elaborate on why the evaporation is the right choice? I know that the neutralization options are incorrect, but I don't recall why/how we would know that the amide wouldn't evaporate also. Is it because of the high MP that amides won't evap??
NS_Tutor_Mathias
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Re: AAMC Section Bank C/P Q14

Post by NS_Tutor_Mathias » Wed Aug 07, 2019 7:24 pm

Diethyl ether has a boiling point of about 35 degrees Celsius, or in other words: It is highly volatile. Any amide is guaranteed to have a much, much higher boiling point. You don't need to memorize boiling points, but probably having a general sense of where they lie is useful. For comparison, acetamide, a very small amide, has a boiling point of about ~222 degrees Celsius.

It also helps of course that the other option, C, is nonsense: The amide is functionally non-polar and would not be found in the aqueous layer.
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