Aromaticity Experiments

[SaintJude]

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Sigh....

What simple [ha!] chemical test could be used to distinguish between the following 2 compounds?

View attachment Picture 31.png

A. Compound 2's solubility in NaHCO3
B. Compound I solubility in NaOH
C. Compound's 1 ability to decolorize bromine solution
D. Compound 1's solubility in NaHCO3

Edit: Answer is B.....It's saturday night, and I have no clue what this answer is saying.

 

[MedPR]

Compound 2 isn't showing.

I have no idea, but I'm going to guess C. Seems to be like they would be equally soluble/insoluble because of the hydroxyl. I assume decolorization of bromine solvent has something to do with adding bromine to the oxygen, and the first one is stabilized by resonance so it is more willing to pickup a negative charge by losing a protonated OH. :/

 

[milski]

Compound 2 isn't showing.

Same as compound 1 but OH is attached to the methy group instead of the ring directly.

 

[MedPR]

What's the answer? You're killing me!

 

[SaintJude]

Sorry, MedPr, I edited in the post. It's B!...but even better I understand why now.

When the MCAT gives me 2 compounds, I need to learn to ID them super-fast!
Comp 1: a phenol , Comp 2: an alcohol

What do we know about phenols in relation to alcohols?
Phenols are much more acidic than alcohols.

Glancing at the answer choices. A gives a strong base, while B& D give a weak base--> It probably then has to do something with reactivity of strong vs. weak base with these compounds....
In order to protonate a fairly acidic phenol, you need a strong base. B/c of this phenols are quite soluble in a strong base, like NaOH. Alcohols on the other hand are not acidic and won't be soluble in aqueous NaOH. Thus you can use solubility differences to distinguish b/w these compounds--choice B.

About choice C: I don't think I would have ever noted this, had it not been in a question like this. But you said in a previous post, that aromatic compunds don't undergo electrophilic addition. Well, the addition of liquid to Bromine to an aromatic compounds would be exactly that. Bromine in solution is too mild a reagent to react with the stable aromatic ring.

 

[MedPR]

Interesting. Thanks for that explanation.

What is compound 2 called, btw?

 

[SaintJude]

It's Benzyl alcohol--See the CH2?!

They draw it weird like this on purpose-to trip us novices up! I didn't see it myself...

I think I saw "CH2OH" before in a problem, and I still didn't remember how it looked like. Not good.

 

[chiddler]

I don't understand. Hydroxide is a strong enough base to deprotonate either of those, isn't it?

Is it meant that when you mix them together then add NaOH, the phenol should be deprotonated and not the alcohol?

 

[MedPR]

It's Benzyl alcohol--See the CH2?!

They draw it weird like this on purpose-to trip us novices up! I didn't see it myself...

I think I saw "CH2OH" before in a problem, and I still didn't remember how it looked like. Not good.

I saw it, I just didn't know how to name it. Nomenclature is a weak spot for me in organic.

@chiddler, normally alcohols other than short chain aliphatic alcohols are only partially soluble in NaOH. I don't know why, I just know it for some reason. For some reason (also unknown to me), Phenol is extremely more acidic than aliphatic alcohols.

I still not understanding why benzyl alcohol isn't acidic enough though. You would think benzyl acohol and phenol would be pretty similar. I mean, the extra CH2 is going to be electron donating, so I guess reduces acidity a little. I'm sure resonance has something to do with it as well.

 

[chiddler]

The hydroxide can resonance stabilize its extra electrons in the aromatic ring. If you place a methyl in between the hydroxide and the ring, then it's no longer possible thus the difference in acidity.

Not satisfied with the explanation for this question though! Both have an O(-); shouldn't that make them both very similar in solubility?

 

[SaintJude]

I think you kind of answered your own question better than I could. The differences in acidity is what accounts for their solubility.

Kaplan says this:
Alcohols, including benzyl alcohol, are NOT acidic. [ I take it they mean, it won't want to give up it's hydrogen to pair with the hydroxyl group of NaOH] So they won't be soluble in aqueous NaOH. Now there is 1 exceptions: alcohols with fewer than 5 carbons are water soluble-so they will dissolve in any aqueous solution.

 

[SaintJude]

Review this superb problem b/c SN2ed's schedule recommends reviewing!

a.) Finally understand what electrophilic addition is:! Breaking of pi bond & formation of 2 sigma bonds and no. of hydrogens in the reactant are conserved
b.) Learned that aromatics very rarely undergo electrophilic addition (except "super conditions" e.g. high pressure ). The major mechanism for aromatic compounds is electrophilic aromatic substitution.
c.) What the "Br2/CCl4" Test is all about: The Decolorization of bromine solvent occurs via an electrophilic addition. So if you're given two compounds where one de-colorizes bromine and the other doesn't, it has do something with the absence or stability of a double bond that the bromine can react with. In this case, the double bonds of a stable aromatic ring won't react with bromine.
d.) Phenols are much more acidic than alcohols. But that's not saying much, because phenol is a very weak acid, and alcohols for purposes of MCAT, are NOT acidic (on why: http://www.chemguide.co.uk/organicprops/phenol/acidity.html)
e.) phenols will react with NaOH

The strong basic hydroxide ion removes the hydrogen ion from the weak phenol.

BUT nor the phenol or the benzyl alcohol are acidic enough to react with a weak base such as sodium bicarbonate
f. ?! ) Small alcohols (5 or less carbons) are water soluble and will dissolve in any aqueous solution.
Thanks for those who were brave enough to figure out this problem! 😉
Side note: I pity the fool (as in me...)who doesn't ID their compounds on the MCAT & know what "CH2OH" looks like

 

[chiddler]

thanks stjude