Extraction Question

[pm1]

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I just ran into a passage that has the extraction procedure as shown in the figure I have attached.
A question asks "what is more likely to be found in layer D?"

This was my reasoning:
NaHCO3 as a weak base would extract strong acid. Since there is no strong acids, nothing would be extracted. Then, NaOH would extract weak acids. This step would extract Boric acid and ethanol (since they are weak acids). This would leave only ester in the layer D.

However, the answer is both ester and alcohol. Why is that? Wouldn't NaOH extract alcohols because they are weak acids?

Also, why is boric acid a weak acid?

Thank you!!

 

[SaintJude]

Seriously, MCAT looooves this extraction problems but not a pro yet... Yeah, I also thought it would only be ester. Where is this from? Barron's? Looking forward to someone else pitching their thoughts in.
Ethanol is actually a weak base not an acid. I believe the reason ethanol would end up in the aqueous layer is b/c it can hydrogen bond with water. Like dissolves like.

 

[circulus vitios]

Preface: I haven't began to study organic chemistry so this could be completely wrong.

A: Nothing.
B: Ester, alcohol, boric acid.
C: Boric acid.
D: Ester, alcohol.

I don't think a strong base would extract an alcohol. An alcohol would not willingly give up a proton because the electron donating nature of the R group would make the hydrogen even more tightly bound to the oxygen of the hydroxyl group.

 

[pfaction]

Alcohols, depending on the # of carbons, are soluble in water. So A could have alcohol in it. But let's pretend it's like dodecanol. Also, boric acid is soluble in water.

This passage is either dangerously weird, or I'm completely incorrect.

vitios is right on his extraction part then.

And NaOH cannot deprotonate an alcohol.

 

[pm1]

Preface: I haven't began to study organic chemistry so this could be completely wrong.

A: Nothing.
B: Ester, alcohol, boric acid.
C: Boric acid.
D: Ester, alcohol.

I don't think a strong base would extract an alcohol. An alcohol would not willingly give up a proton because the electron donating nature of the R group would make the hydrogen even more tightly bound to the oxygen of the hydroxyl group.

the only reason why I thought that ethanol would be extracted by a strong base is because EK talks about extraction saying that NaOH extracts weak acids, and then it gives an example, phenol. Thus I thought NaOH would extract alcohols. Is phenol different from other types of alcohol? Wouldn't phenol be even more basic because of the double bonds?

 

[pfaction]

Yeah, it's more acidic because of the conjugation.

Benzene-OH -> Benzene-O [phenoxide] is stabilized via resonance.

 

[circulus vitios]

the only reason why I thought that ethanol would be extracted by a strong base is because EK talks about extraction saying that NaOH extracts weak acids, and then it gives an example, phenol. Thus I thought NaOH would extract alcohols. Is phenol different from other types of alcohol? Wouldn't phenol be even more basic because of the double bonds?

Phenol is slightly acidic because of resonance:

The prompt didn't explicitly mention that the alcohol wasn't phenol, I was just assuming that it wasn't. Like I said, I could be completely wrong with my line of thinking.

 

[pm1]

Yeah, it's more acidic because of the conjugation.

Benzene-OH -> Benzene-O [phenoxide] is stabilized via resonance.

oh, I see. So if the alcohol was a phenol then it would be found on layer c?

 

[pfaction]

TPR exam 4:

a) Phenol
b) some 4 carbon ketone
c) methanol
d) cyclohexanone

EDIT: I don't think the explanation is right for one of them. NaOH IS a strong enough base to remove the acidic proton ortho to the ketone group in cyclohexanone; this is classic for an aldol condensation reaction. That product would have an alcohol.

 

[pm1]

Thank you guys!

 

[SaintJude]

phenol ≠ alcohol but whatever...

Anyway, thanks for answering . Back in February, there was thread where I learned that alcohols (& carbonyls) with fewer than 5 carbons are water soluble. Should have remembered that.

Edit: I think Princeton is right about cylcohexanone. I am not sure it undergoes aldol condensation and even if it did...aldol condensation of even a non-cyclic ketone occurs with a strong base AND high temperatures.

 

[pfaction]

Actually, can someone verify I'm right on me edit? Princeton review themselves uses cyclohexanone and NaOH in aldol condensation, but maybe I'm overthinking, AND to boot it follows their schematics....

 

[SaintJude]

Ok, if Princeton uses cyclohexanone in an aldol condensation, then it must be possible. And I guess that makes sense since cyclohexanone does indeed have an alpha hydrogen present.

But I still think Princeton is right, b/c the conditions are not right. Aldol condensation occurs under high temperatures and I'm assuming the extraction is at room temperature and not being heated.

 

[pfaction]

Yeah, I'm thinking that for this passage and the boxes that the logic they used is right. Their explanation should included "for this passage" or something like that.