Ochem Destroyer Question

[DoctaWalnuts]

I have the 2008 Destroyer this is number 151 in Ochem..

If NaHCO3 was added, which compound would be in the aqueous layer?

a. Sodium benzoate
b. toulene
c. Sodium bicarbonate
d. Sodium phenoxide
e. anilline

Answer is A. But why wouldn't the phenoxide be in the aqueous layer as well?

Thank you!

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[nze82]

I have the 2008 Destroyer this is number 151 in Ochem..

If NaHCO3 was added, which compound would be in the aqueous layer?

a. Sodium benzoate
b. toulene
c. Sodium bicarbonate
d. Sodium phenoxide
e. anilline

Answer is A. But why wouldn't the phenoxide be in the aqueous layer as well?

Thank you!

I'm guessing since the Benzoate component of compound A is more polar than the Phenoxide component of compound B, it can compensate for the large carbon ring (The major hydrophobic portion of the compound) much better and therefore, facilitate the compound to dissolve in the aqueous layer to a greater extent.

 

[DoctaWalnuts]

Thank you for your response. I see what you are saying..but I don't think that would be intuitive for me at least. I guess I'm mainly focusing on the ability to hydrogen bond and thus would be in the aqueous layer.

Aniline would also be in the aqeous layer though because aniline the Nitrogen can H-bond..err....

 

[pbure9]

This is a lab type question and a good one I think. Benzoic acid is a much stronger ACID than phenol therefore only requires a weaker base to react with it. Phenol would need to react with a strong base to be deprotonated, something like KOH/NaOH.

And aniline is basic, so if we add HCl or acid, we'd find its ion in the aqueous solution.

 

[DoctaWalnuts]

This is a lab type question and a good one I think. Benzoic acid is a much stronger base than phenol therefore only requires a weaker base to react with it. Phenol would need to react with a strong base to be deprotonated, something like KOH/NaOH.

And aniline is basic, so if we add HCl or acid, we'd find its ion in the aqueous solution.

Ohhh good points!! Both aniline and phenol are basic. The OH group donates electrons to the ring just as the NH2 group does in phenol. Right?

But the Benzoic acid (before addition of the base)
was acidic because of the electron withdrawing group (COOH) right?

In that case, the aniline and phenol would require an acid! While the Benzoic acid would require a BASE!! Yay!!

 

[GiTsticker]

... Benzoic acid is a much stronger base than phenol...

I think he meant stronger acid. Indeed it is the strongest acid of the bunch and thus the most likely to be deprotonated by HCO3- thus leaving it with a negative charge which makes it much more polar.

In that case, the aniline and phenol would require an acid

I think aniline would need an acid to protonate the N and leave it with a positive charge and phenol would need a strong base to deprotonate the O and leave it with a negative charge.

 

[DoctaWalnuts]

yeah phenol isn't a base, it's an acid. But aniline isn't a base either. It's more basic than carboxylic acid, but it's not a base. Right..I get what he's saying though. It's a stronger acid so it can work with a weaker base.

The phenol and aniline are both weaker acids and would require stronger bases like NaOH or NaNH2. Thank you guys!

 

[pbure9]

I think he meant stronger acid. Indeed it is the strongest acid of the bunch and thus the most likely to be deprotonated by HCO3- thus leaving it with a negative charge which makes it much more polar.



I think aniline would need an acid to protonate the N and leave it with a positive charge and phenol would need a strong base to deprotonate the O and leave it with a negative charge.

woops i did mean acid lol. changed that.

and id say aniline is pretty basic as GiT said. It would need a proton thats why it would be reactive in acid not base.

 

[Adamsswag]

We can extract a phenol by using a base such as NaOH. However, NaHCO3 is a weaker base and can also be amphoteric. Therefore, think of NaHCO3 as more-so exclusive to the extraction of an acid; hence a reaction with the carboxy acid yielding sodium benzoate.