Why couldn't 3 be a player? The answer was "only 1&5" but I don't understand why 3&5 wouldn't work too.
06/08 Q of the day
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The answer was 3 and 6. Where did you get 1 and 5? 5 would not work because that's a triple bond. After it attaches to the diene, the dienophile part will stay with a double bond. We don't want that. We want that part to be a single bond.
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My bad gn4 your right. But my question is why not also 1&6 work?
Those methyl groups on 1 should be free to rotate and wind up ok for the reaction.
Granted its not an option, but from the list of choices it sounds like only 3&6 would work and nothing more. Why not 1?
Mega shout-out to you GatorD...Awesome work!!
Those methyl groups on 1 should be free to rotate and wind up ok for the reaction.
Granted its not an option, but from the list of choices it sounds like only 3&6 would work and nothing more. Why not 1?
Mega shout-out to you GatorD...Awesome work!!
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I actually have a question about this problem too.
In the explanation you mention that if the diene is funky looking like it is in the problem, then if both of the bonds have same chemistry, ie are Z, Z; or E, E; then the substituents in the product will end up cis in the product. So, why wouldn't 2 work? Number two is a diene that is E, E, and number 3 is Z, Z. Why can't number two adopt a cisoid formation?
In the explanation you mention that if the diene is funky looking like it is in the problem, then if both of the bonds have same chemistry, ie are Z, Z; or E, E; then the substituents in the product will end up cis in the product. So, why wouldn't 2 work? Number two is a diene that is E, E, and number 3 is Z, Z. Why can't number two adopt a cisoid formation?
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