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Bicyclo[2.2.1]heptane (red box) has two planes of symmetry. It is achiral. Placing a phenyl ring at $\ce{C2}$ in A and B removes the symmetry and renders both structures chiral. A has the phenyl syn to the methano bridge while B has the phenyl syn to an ethano bridge. They have the same atom connectivity but different arrangements in space. They are ...


7

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 ...


6

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: 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 ...


3

From the conceptual perspective applying (CIP)-rules, you are right; beside the carbon atom both $\alpha$ to nitrogen and the carboxylate, the nitrogen may be either up or down the plane of the cycle -- and given the pattern of substitution (including the free electron pair on $\ce{N}$), there could be diastereomers of proline. However in praxi there are not....


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