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zaitsevs rule and e2 elimination

would an alkyl halide that gives a non zaitsev product on e2 elimination be something like ch3ch2ch(br)ch2ch3, as it has an equal number of carbons either side of the bromine atom that is going to be eliminated?
when you say a non-Zaitsev product, what do you mean? One formed against the Zaitsev rule?

In the example you have given, then Zaitsev rule doesn't actually apply, because the molecule is symmetrical, so there can only be one elimination product.

Zaitsev is only used in cases where there could be more than one possible product.
Me-CH(Me)-CH(NMe3^+)-Me

The NMe^+ group(the leaving group) is highly electron withdrawing and will increase proton acidity at both of the possible beta positions. However at the beta position on the left the two Me group exert a +ve inductive effect which reduces the increase in acidity. No +ve inductive effect is present at the beta position on the right so protons are more acidic at this position. In addition, the NMe^+ group will stabilise the transition state more at this position for the same reasons.
These two factors lower the activation energy to the extent that elimination happens much more from this position than from the left-hand beta position. This gives the Hofmann product rather than the zaitsev product, despite the zaitsev product being more thermodynamically stable.
Original post by LearningMath
Me-CH(Me)-CH(NMe3^+)-Me

The NMe^+ group(the leaving group) is highly electron withdrawing and will increase proton acidity at both of the possible beta positions. However at the beta position on the left the two Me group exert a +ve inductive effect which reduces the increase in acidity. No +ve inductive effect is present at the beta position on the right so protons are more acidic at this position. In addition, the NMe^+ group will stabilise the transition state more at this position for the same reasons.
These two factors lower the activation energy to the extent that elimination happens much more from this position than from the left-hand beta position. This gives the Hofmann product rather than the zaitsev product, despite the zaitsev product being more thermodynamically stable.


Um, I think you're answering a different question. The OP was starting from 3-bromopentane.
Original post by Plato's Trousers
Um, I think you're answering a different question. The OP was starting from 3-bromopentane.


Ooops sorry didnt see the alkyl halide part! Just replace NMe3^+ with Fluorine... I got the impression she was trying to understand why an anti-zaitsev product might arise and wanted an example. :/
Original post by LearningMath
Ooops sorry didnt see the alkyl halide part! Just replace NMe3^+ with Fluorine... I got the impression she was trying to understand why an anti-zaitsev product might arise and wanted an example. :/


Good. It's not me going crazy then. That's a relief.

Merry Christmas!

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