C/P Section Bank #14 (Extraction)

[theonlytycrane]

Wouldn't (D) also work? After H+ the amide and carboxylic acid would go into the aqueous layer, then NaOH would deprotonate both molecules resulting in a carboxylate (aqueous) and neutral amide (ppt out).

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[salemstein]

I think if you add NaOH later in D, all that would do is deprotonate your "+" charged anhydride from the HCl treatment, turning it back to a neutral compound. It would then re-mix with your neutral amide.

 

[theonlytycrane]

The two anhydrides to consider are the fully unreacted molecule: R-CO-O-CO-R and the half unreacted molecule: R-CO-OH.

I believe R-CO-O-CO-R will stay in the aqueous layer and R-CO-OH is the molecule referred to as the "unreacted" anhydride in the answer choices.

 

[salemstein]

The two anhydrides to consider are the fully unreacted molecule: R-CO-O-CO-R and the half unreacted molecule: R-CO-OH.

I believe R-CO-O-CO-R will stay in the aqueous layer and R-CO-OH is the molecule referred to as the "unreacted" anhydride in the answer choices.

I don't think you can consider RCOOH an "anhydride" tho... wouldn't it be a carboxylic acid? I interpreted "quencing the anhydride" as reacting with the leftover RCO-O-COR that didnt react with R-NH2

 

[theonlytycrane]

The solution references both the carboxylic acid and completely unreacted anhydride, so it's a bit ambiguous. I wish they provided explanations to the incorrect answers as well.

 

[wizzed101]

Amide can only be obtained in the ether layer so C and D are out.
B is wrong because a strong acid like HCL will protonate the carboxylic group and make all reagents soluble in the ether layer.

Edit: oh you were wondering why can't the amide go into the aqueous layer? Coz it can't be protonated, just like the peptide bonds can be either hydrophobic or hydrolyzed.

 

[The8]

Amide can only be obtained in the ether layer so C and D are out.
B is wrong because a strong acid like HCL will protonate the carboxylic group and make all reagents soluble in the ether layer.

Edit: oh you were wondering why can't the amide go into the aqueous layer? Coz it can't be protonated, just like the peptide bonds can be either hydrophobic or hydrolyzed.

how the hell do you boil out the solvent though, the boiling point is so much higher for carb acids and water than the amide ether mix, so you would actually remove the amide and ether first....

 

[theonlytycrane]

@The8

Are you referring to choice (A)?

Carboxylic acid and the anhydride go into the aqueous layer and amide goes into the ether layer. At this point, you just let the ether sit out and it will easily evaporate.

 

[goodtime]

@The8

Are you referring to choice (A)?

Carboxylic acid and the anhydride go into the aqueous layer and amide goes into the ether layer. At this point, you just let the ether sit out and it will easily evaporate.

can you explain why amide would be in the ether layer? is it because after using a base the molecule is still neutral. This would mean it doesn't matter if amide is polar when determining if it it ends up in aqueous or organic layer. Also, lets say if we used HCL(acid) instead of the base in the question, in this case would the amide would be protonated and then end up in the aqeous layer right?

 

[theonlytycrane]

can you explain why amide would be in the ether layer? is it because after using a base the molecule is still neutral. This would mean it doesn't matter if amide is polar when determining if it it ends up in aqueous or organic layer. Also, lets say if we used HCL(acid) instead of the base in the question, in this case would the amide would be protonated and then end up in the aqeous layer right?

The amide is neither acidic nor basic enough to react- it will remain soluble in the organic solvent throughout the extraction process. The lone pair on the nitrogen atom is resonating and too busy to pick up a proton.

 

[aldol16]

can you explain why amide would be in the ether layer? is it because after using a base the molecule is still neutral. This would mean it doesn't matter if amide is polar when determining if it it ends up in aqueous or organic layer. Also, lets say if we used HCL(acid) instead of the base in the question, in this case would the amide would be protonated and then end up in the aqeous layer right?

Stop talking about the protonation of an amide. An amide, for most MCAT organic purposes, will remain protonated and uncharged.

 

[laczlacylaci]

@theonlytycrane

I asked this question some time ago, and wanted to just confirm my thoughts.

1. Amine+anyhydride (excess)=amide(NH3+) & carboxylic acid (COO-) & unreacted anhydride
2. Using NaOH or HCl to quench the unreacted anhydride both works, correct?

3a. Addition of HCl would protonate the original amide to (NH3+), if not already, and the carboxylic acid to (COOH), getting original amide (NH3+) and new amide (NH3+) into the aq layer and the COOH into the ether layer. (D is incorrect because of the NaOH neutralization?)
3b. Addition of NaOH would deprotonate the new amide into (NH2) and carboxylic acid would stay like (COO-), if the original amide was protonated (NH3+), it will become (NH2). Therefore, the original amide & new amide (both NH2) will go into the ether layer, and the carboxylic acid would go into the aq layer. (therefore, A is correct)

4. Should we assume the original amide was protonated (NH3+) or deprotonated (NH2)? Or does that not matter, since NaOH and HCl can easily change that. (sorry aldol16, I read your comment, but just wanted to hear more.)

 

[theonlytycrane]

@laczlacylaci

amine (NH3) + anhydride (excess) -> amide + COO- + unreacted anhydride. Note: the amide doesn't get protonated because the lone-pair on the nitrogen is busy in resonance. Adding base (Na+ -OH) keeps the COO- deprotonated and converts the unreacted anhydride into COO-. These all go to the aqueous layer and the amide goes in the organic layer.

 

[663697]

Does a carboxylic anhydride have R groups -OH or just random alkyl groups? When I search it up on google it doesn't specify if the ends are both COOH

 

[The8]

Does a carboxylic anhydride have R groups -OH or just random alkyl groups? When I search it up on google it doesn't specify if the ends are both COOH

A carboxylic anhydride has two random R groups on either side and an oxygen inbetween. When it is broken up through a hydrolysis reaction with water, it becomes two separate carboxylic acids with those R groups.

 

[aldol16]

Does a carboxylic anhydride have R groups -OH or just random alkyl groups? When I search it up on google it doesn't specify if the ends are both COOH

I think you have the definition of carboxylic anhydride confused. The anhydride is formed from the free carboxyl groups joining together so that there are no more free carboxyl groups. So from R1-COOH and R2-COOH, you get R1-(C=O)-O-(C=O)-R2. Those R groups can be alkyl groups or contain heteroatoms as well. It's called an anhydride because in the process of forming it from the free carboxylic acids, you lose a water.