Glycolic acid and Sodium Hydroxide reaction

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Ramjams
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#1
Report Thread starter 10 years ago
#1
Hey guys, just a quick question

In a past paper I've just done, it gave the structural formula of glycolic acid as HOCH2COOH and it asked for the reaction between that and NaOH.

I put:

HOCH2COOH + 2NaOH --> NaOCH2COONa + 2H20

As I'm aware that NaOH can react with both alcohol groups and carboxylic acid groups, but the markscheme (which my school produced) has:

HOCH2COOH + NaOH --> HOCH2COONa + H20

Is mine wrong? Or are they both correct?

Cheers
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CM23925
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#2
Report 10 years ago
#2
I wouldn't know for sure, but I would guess that because NaOH is a base, it would neutralise with the carboxilic acid group more readily than it would react with the alcohol group. I guess if there was excess NaOH then you would get a mixture of yours and the mark schemes product, but I can't be sure.
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MaturePharm
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#3
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#3
My guess is that sodium hydroxide (certainly aqueous) would not be strong enough to form the disodium salt. It would easily deprotonate the acid but to form the alkoxide usually requires a stronger base (like NaH). Certainly if you leave most sodium alkoxides lying around they hydrolyse fairly readily back to NaOH + ROH, even with the moisture in air.
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Kyri
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#4
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#4
Alcohols have a fairly high pka value (compared to carboxyl groups) and are usually only partially deprotonated by hydroxide. But once you've deprotonated the carboxyl group in glycolic acid (most acidic group is always deprotonated first) the resulting negative charge makes the alcohol proton even less acidic so now it won't be deprotonated by hydroxide. You might find it easier to think that the negative hydroxide will find it more difficult to approach the already negative glycoate ion. So the mark scheme is in fact correct.
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Ramjams
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#5
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#5
(Original post by Kyri)
Alcohols have a fairly high pka value (compared to carboxyl groups) and are usually only partially deprotonated by hydroxide. But once you've deprotonated the carboxyl group in glycolic acid (most acidic group is always deprotonated first) the resulting negative charge makes the alcohol proton even less acidic so now it won't be deprotonated by hydroxide. You might find it easier to think that the negative hydroxide will find it more difficult to approach the already negative glycoate ion. So the mark scheme is in fact correct.
Ah, thanks for clearing that up; but I'm pretty sure that this isnt on our specification; how would I know that alcohols have a higher pka than carboxyl groups? Or anything like that? Is it basically saying that a base will react more "readily" with the carboxylic acid group than the alcohol, and hence the negative charge repels any further reaction?

Thanks again
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Kyri
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#6
Report 10 years ago
#6
(Original post by Ramjams)
Ah, thanks for clearing that up; but I'm pretty sure that this isnt on our specification; how would I know that alcohols have a higher pka than carboxyl groups? Or anything like that? Is it basically saying that a base will react more "readily" with the carboxylic acid group than the alcohol, and hence the negative charge repels any further reaction?

Thanks again
That is correct. Sorry, I should have just said alcohols are much less acidic than carboxyl groups. Higher pka means less acidic.
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hpp1
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#7
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#7
R-COOH + OH- ------> R-COO- + H2O(pka: 4)
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Pigster
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#8
Report 3 years ago
#8
(Original post by hpp1)
R-COOH + OH- ------> R-COO- + H2O(pka: 4)
And only seven years too late. Welcome to the site, by the way, but it is best to stick to replying to posts people have made recently, especially if you're after a "thank you".
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tofskisum
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#9
Report 2 years ago
#9
i'm pretty glad this is here though, even now, because that paper is still being used for past questions. it helps other people if there are further clarifications to posts
(Original post by Pigster)
And only seven years too late. Welcome to the site, by the way, but it is best to stick to replying to posts people have made recently, especially if you're after a "thank you".
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