Which of the two products will be formed (or will be formed in greater yield) de-
pends on the charge distribution in each of the reactants. To determine the charge dis-
tribution, we need to draw contributing resonance structures. In the preceding example,
the methoxy group of the diene is capable of donating electrons by resonance. As a
result, the terminal carbon atom bears a partial negative charge. The aldehyde group of
the dienophile, on the other hand, withdraws electrons by resonance, so its terminal
carbon has a partial positive charge.The partially positively charged carbon atom of the dienophile will bond preferential-
ly to the partially negatively charged carbon of the diene. Therefore, 2-methoxy-3-
cyclohexenecarbaldehyde will be the major product.PROBLEM 15What would be the major product if the methoxy substituent in the preceding reaction were
bonded to C-2 of the diene rather than to C-1?PROBLEM 16Give the products of each of the following reactions (ignore stereoisomers):a.b.Conformations of the Diene
We saw in Section 7.11 that a conjugated diene such as 1,3-butadiene is most stable
in a planar confomation. A conjugated diene can exist in two different planar
conformations: an s-cis conformationand an s-trans conformation. By “s-cis,”
we mean that the double bonds are cis about the single bond The
s-trans conformation is little more stable (2.3 kcal or 9.6 kJ) than the s-cis confor-
mation because the close proximity of the hydrogens causes some steric strain
(Section 2.10). The rotational barrier between the s-cis and s-trans conformations
or is low enough to allow them to interconvert rapidly
at room temperature.(4.9 kcal>mol 20.5 kJ>mole)(s=single).CH 2 CH C CH 2 + HC C C N∆CH 3CH 2 CH CH CH CH 3 + HC C C N∆CH 3 OCH 2CH CHOδ−δ+ δ−δ+CHO
CH 3 Oresonance contributors of the dieneCH 2 CH CH CH OCH 3 CH 2 CH CH CH OCH 3−O+resonance contributors of the dienophileCH 2 CH CH CH 2 CH CHO−
+318 CHAPTER 8 Reactions of Dienes • Ultraviolet and Visible Spectroscopy
Tutorial:
Diels–Alder reaction