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This synthesis example is shown in the book Cycloaddition Reactions in Organic Synthesis by S. Kobayashi and K. A. Jorgensen (link), on p. 15:

enter image description here

8 is a chiral catalyst, which can be ignored, in the context of my question. The product shown is the exo product. However, I am unable to understand or visualise why it is the exo product. It does not seem apparent to me that the transition state leading to this product would have the aldehyde group being further away from the $\pi$ system of electrons. I would greatly appreciate assistance in helping me understand this.

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  • $\begingroup$ This post on the DA reaction may be of interest to you. $\endgroup$
    – user55119
    Jan 7, 2022 at 20:44

2 Answers 2

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The product you have drawn looks like an exo-product. Conformation 1 leads to an exo-transition state that produces exo 2 as the enantiomer shown. With an achiral catalyst ent-exo 2 would also be formed as a racemic pair. To obtain the racemic endo product, flip the diene 180o in conformation 1 to obtain endo 2. For the enantiomer, flip the unsaturated aldehyde 180o. If you have access to the literature, research papers by Herbert House or William Roush[1] on this type of Diels-Alder reaction. I believe they saw trans-fusions (endo products) for the ring system.


Addendum: (12/7/2022)

It is likely that the formation of the exo-product you cite is a typographical error.
In this example,[2] a chiral ruthenium catalyst forms the endo-product, which has the trans-fusion.

Reference:

  1. Roush, W. R.; Peseckis, S. M. Intramolecular Diels-Alder reactions: the angularly methylated trans-perhydroindan ring system. J. Am. Chem. Soc. 1981, 103, 6696-6704; doi 10.1021/ja00412a027.
  2. Thamapipol, S.; Bernardinelli, G.; Besnard, C.; Kundig, E. P. Chiral Ruthenium Acid Catalyzed Intramolecular Diels-Alder Reactions. Org. Lett. 2010, 12, 24, 5604-5607; doi 10.1021/ol1019103
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  • $\begingroup$ @Buttonwood: Thanks for the fix. The hour was getting late. ;) $\endgroup$
    – user55119
    Jan 7, 2022 at 14:38
  • $\begingroup$ Thanks for pointing out that the product shown is in fact the exo product. I have made the necessary edits to my question. $\endgroup$
    – Tan Yong Boon
    Jan 7, 2022 at 15:01
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    $\begingroup$ Though not part of every citation style, I perceive the provision of a more complete set of bibliographic data -- doi to get to the landing page of the publication plus indication of the title of the article -- as helpful for learning. In this regard (my point of view) chemistry.se benefits from the pattern equally seen e.g., on Master Organic Chemistry (example), or papers in J.Chem. Educ. (example). $\endgroup$
    – Buttonwood
    Jan 7, 2022 at 15:04
  • $\begingroup$ With the Chemistry Reference Resolver living in the browser, it often only is a single click to eventually get to a place to collect the missing data, too. $\endgroup$
    – Buttonwood
    Jan 7, 2022 at 15:05
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For me, the most foolproof way to identify the endo and exo products is to look at the stereochemistry in the product. Consider first a standard intermolecular Diels–Alder reaction:

Endo Diels–Alder

I labelled the substituents on the diene $\mathrm{R^t}$ and $\mathrm{R^c}$, for trans and cis respectively, to describe their position with respect to the single bond in the middle of the diene.

What we can see from this is that the endo product has the electron-withdrawing group (EWG) placed cis to $\mathrm{R^t}$ (and trans to $\mathrm{R^c}$). (OK, sorry, the naming is a bit confusing; but I'm sure you can make sense of it.) The opposite would be true of the exo product: in that case, the EWG would be cis to $\mathrm{R^c}$ (and trans to $\mathrm{R^t}$).

The intramolecular case is harder to draw a transition state for (user55119 has done it nicely!), but this doesn't change the conclusion drawn above. In your case, the diene only has a single substituent which is of the $\mathrm{R^t}$ type (the substituted double bond is trans configured). This substituent ends up becoming the five-membered ring:

Diels–Alder stereochemistry identification

It is quite clear from the final product that the EWG is trans to the $\mathrm{R^t}$ substituent. So, in agreement with user55119's analysis, the product shown is actually exo.

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    $\begingroup$ @TanYongBoon First comment: In cyclopentadiene, yes, the CH2 group would take the place of R_c. When drawn in this way, then yes, the bridge will always end up "above" the 6-membered ring. However, that's not necessarily true, because if the dienophile approaches from above the diene rather than below as I've drawn, then the bridge will actually end up below the six-membered ring. That, therefore, isn't a reliable way to remember this. What is reliable is the relative stereochemistry between the EWG and the groups. Regardless of where the dienophile approaches from, [...] $\endgroup$
    – orthocresol
    Jan 8, 2022 at 11:43
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    $\begingroup$ [...] the endo product will always have the EWG trans to R_c, i.e., the EWG is opposite the methylene bridge. Now, to be honest, I don't suggest memorising this at all. Why bother? You can draw it out whenever you need to. If you aren't comfortable with drawing out the TS of a D–A reaction and obtaining the stereochemistry from that, I'd strongly recommend getting better at that, before trying to look at more complicated examples. Understanding it is better than memorising odd mnemonics like mine. Lastly, if R_c = R_t, then there are no endo or exo diastereomers, there is only one product. $\endgroup$
    – orthocresol
    Jan 8, 2022 at 11:45
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    $\begingroup$ @TanYongBoon Yes, they have to rotate, but it's physically impossible to rotate "the other way". The moment the single bond is formed, the stereochemistry is fixed. You just need to "unfold" the six-membered ring, and there aren't two ways to go about it, there isn't any conrotatory/disrotatory thing going on here. $\endgroup$
    – orthocresol
    Jan 8, 2022 at 13:16
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    $\begingroup$ @TanYongBoon Whether the dienophile approaches from top or bottom affects which enantiomer you get. It doesn't affect which diastereomer you get, i.e. the relative stereochemistry, and that's what exo/endo are about, they are labels for diastereomers. The endo product will always be the same diastereomer, although in general you will get both enantiomers of it. I'm afraid, I cannot give you a tutorial on Diels–Alder stereochemistry. Please look it up in a textbook. It's an important enough topic that any decent textbook will cover it. I vaguely recall Clayden has a good section on this. $\endgroup$
    – orthocresol
    Jan 8, 2022 at 15:04
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    $\begingroup$ @TanYongBoon Maybe this might help: i.stack.imgur.com/NLxaf.jpg (1) Reactants. (2) The DA product. Note that I haven't dont any "rotation" of the groups yet; this is just what happens when you form the necessary bonds. (3) Same DA product but viewed from a different angle. (4) Rotated product / redrawn stereochemistry of the bottom carbon to match conventional chemical structures. At no point in this, is it possible for C and T to change places. In particular, before the reaction happens they can't interchange (double bond) and after the reaction happens the stereochem is fixed. $\endgroup$
    – orthocresol
    Jan 8, 2022 at 15:44

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