Is there some required stereochemistry for anhydride formation? Do both of them form the same product, or does one not even undergo dehydration?

cis- and trans-1,2-cyclohexanedicarboxylic acid

I looked up cyclohexane-1,2-dicarboxylic anhydride's structure and found two images:

Hexahydrophthalic anhydride

Does that mean both of them are formed?

  • $\begingroup$ Consider how close the two carboxyl groups are in the different strutures. Maybe try drawing them in 3D. $\endgroup$ – Waylander May 18 '20 at 8:43
  • $\begingroup$ Think about the stereochemistry of the starting material (cyclohexane-1,2-dicarboxylic anhydride) $\endgroup$ – Eli Jones May 18 '20 at 8:44
  • $\begingroup$ um so i think that the reaction wont happen in case of trans and in case of cis ill just eliminate a H2o.does that sound right? $\endgroup$ – shreya May 18 '20 at 9:45

The mechanism of formation of anhydride from a dicarboxylic acid is as follows: [1]

enter image description here

So as you can see, it involves a nucleophilic attack at the $\ce{C=O}$ of one of the carboxylic acids by the other one, following which internal rearrangement occurs and a water molecule is kicked out.

Stereo chemically, we are interested in finding out in which configuration can the lone pairs of oxygen from one of the $\ce{-COOH}$ attack the $\ce{π^* C=O}$ type antibonding orbital of the other $\ce{-COOH}$ group.

To achieve this, the nucleophile( oxygen in this case) has to align properly with the $\ce{π^*}$ orbital at the Bürgi–Dunitz angle. This configuration is necessary whenever a nucleophile has to successfully add to a carbonyl group. The interacting diagram will look like this for formaldehyde: [2]

(Green ball= nucleophile| Blue and Red balls= carbonyl)

We can also roughly estimate that the situation will be similar for the $\ce{C=O}$ part of the carboxylic acid group.

To see which of the two reactants in question will form the required anhydride, one can simply look at the 3-D conformers of both in the appropriate orientation, and compare to see which of the cases resembles the standard Burgi-Dunitz interaction more closely as illustrated above

Here is the cis configuration: [3]

enter image description here

And this is the trans configuration: [4] enter image description here

1= oxygen| 2= carbonyl of acid group

As you can see, the oxygen seems much more aligned with the carbonyl in the trans configuration as compared to the cis. So, 2 might form the anhydride more easily than 1 (that is, with a lower activation energy barrier due to reduced torsional strain)

But again, it's not a strictly inaccesible barrier, as a few bond rotations in the cis configuration can also lead to a somewhat favorable configuration for the reaction. So it may ultimately lead to two different kinds of anydrides by each of the isomers like you said, 2 can lead to formation of 4, while 1 can lead to formation of 3

  • $\begingroup$ In your mechanism the carbonyl oxygen is the more nucleophilic of the two. If this were a Fischer esterification, I'm sure you would protonate the carbonyl oxygen. $\endgroup$ – user55119 May 18 '20 at 12:46
  • $\begingroup$ @user55119 Well, in hybridization terms, I would expect the carbonyl oxygen to have more s character than the hydroxyl oxygen, thereby the former being more electronegative and less nucleophilic. From what I understand, we protonate the carbonyl oxygen in Fischer esterifcation as a last resort. The initally preferred protonation of the hydroxyl oxygen would lead to no useful result,and so the carbonyl oxygen gets protonated after that to make progress in the reaction $\endgroup$ – Yusuf Hasan May 18 '20 at 12:55
  • $\begingroup$ To each his own. $\endgroup$ – user55119 May 18 '20 at 15:02
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    $\begingroup$ Also I guess the more interesting point here is that both the cis and trans anhydride forms can be formed and can even be bought online for almost 95% purity here and here respectively. So your professor will have to specifically clarify what exactly does he means when he claims that "it happens in cis but not in trans" $\endgroup$ – Yusuf Hasan May 18 '20 at 17:47
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    $\begingroup$ Your professor has evoked a form of reasoning where the reactants are closer to each other and so it's perhaps easier for them to react, while I tried to evoke a stereo chemical perspective and analyse through better approach for the nucleophile. Both the reasonings are okay on a qualitative basis, but what can really settle matters is a published finding which shows the yield, and also the probable reason behind it found experimentally. Also, your prof's statement is pretty vague as of now because in my prev. comment, you can see that both the products are formed and even commercially sold $\endgroup$ – Yusuf Hasan May 18 '20 at 17:58

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