AAMC sample C/P Q39

[deleted647690]

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I tried to figure out the mechanism for this, but I can't seem to translate what I found here on this website:
Conversion of carboxylic acids to esters using acid and alcohols (Fischer Esterification) — Master Organic Chemistry

I can't seem to translate this to draw out the mechanism for this question. Is this the correct mechanism? (Fischer esterification)

I think when I did this, I tried thinking through the potential mechanism at first, but then just crossed off every answer except C because only in C did the number of Oxygens remain constant. I failed to realize that oxygen would be lost in the mechanism.

 

[Green_Goose]

The important thing here is that you form a six-membered ring. You pretty much never have an oxygen attack another oxygen to form a ring.

What I think happens here is that 1) the O- gets protonated; it's in aqueous solution so there's protons floating around 2) the -OH at the end of the ring attacks the carbonyl (forming the six-membered ring), and that carbonyl oxygen gets protonated at some point as well 3) Now you have the same molecule as choice A), except instead of a carbonyl oxygen you have two single-bonded -OH 4) Finally one of the -OH groups gets protonated and the other -OH group kicks down, forming a double bond and kicking off water.

Sorry if this is a little convoluted, this is how I was taught to think about it back when I took Orgo.

 

[aldol16]

This is a cyclization of an ester. It's acid-catalyzed if there's an acid around that can bond in a Lewis acid fashion to the carbonyl to increase its electrophilicity. Basically (no pun intended) what happens is the alcohol attacks the carbonyl on the carboxylic acid and you form a tetrahedral intermediate. This tetrahedral intermediate has an oxyanion, which can be stabilized by putting a positive charge somewhere close to it in the active site. The oxyanion then collapses and kicks out -OH, which gains a proton somewhere in the process to form water. This forms the six-membered ring. Six-membered rings are quite stable and so that's why it exists in equilibrium with the extended form. You can think of this as an esterification but the generalized mechanism can be thought of as an addition-elimination - you will see many addition-eliminations in natural chemistry.

 

[deleted647690]

The important thing here is that you form a six-membered ring. You pretty much never have an oxygen attack another oxygen to form a ring.

What I think happens here is that 1) the O- gets protonated; it's in aqueous solution so there's protons floating around 2) the -OH at the end of the ring attacks the carbonyl (forming the six-membered ring), and that carbonyl oxygen gets protonated at some point as well 3) Now you have the same molecule as choice A), except instead of a carbonyl oxygen you have two single-bonded -OH 4) Finally one of the -OH groups gets protonated and the other -OH group kicks down, forming a double bond and kicking off water.

Sorry if this is a little convoluted, this is how I was taught to think about it back when I took Orgo.

This is a cyclization of an ester. It's acid-catalyzed if there's an acid around that can bond in a Lewis acid fashion to the carbonyl to increase its electrophilicity. Basically (no pun intended) what happens is the alcohol attacks the carbonyl on the carboxylic acid and you form a tetrahedral intermediate. This tetrahedral intermediate has an oxyanion, which can be stabilized by putting a positive charge somewhere close to it in the active site. The oxyanion then collapses and kicks out -OH, which gains a proton somewhere in the process to form water. This forms the six-membered ring. Six-membered rings are quite stable and so that's why it exists in equilibrium with the extended form. You can think of this as an esterification but the generalized mechanism can be thought of as an addition-elimination - you will see many addition-eliminations in natural chemistry.

Ah, I drew it out and that makes sense, thanks!