Hoffman vs. Zaitzef?

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

DMD-2-B

Senior Member
10+ Year Member
7+ Year Member
15+ Year Member
Joined
Mar 13, 2005
Messages
223
Reaction score
1
I ran into an elimination problem with DAT Achiever, and I gave the Zitzef product since it was high heat and there was no bulky base. Apparently, the answer is Hoffman product, but it doesn't state why.

When is Hoffman elimination product favored over Zaitzef?
 
DMD-2-B said:
I ran into an elimination problem with DAT Achiever, and I gave the Zitzef product since it was high heat and there was no bulky base. Apparently, the answer is Hoffman product, but it doesn't state why.

When is Hoffman elimination product favored over Zaitzef?

Zaitsev's rule: based induced elimination reactions generally give the more highly substituted (more stable) alkene product. (e.g. 2-Bromobutane + sodium ethoxide + ethanol --> 2-Butene (Major product) + 1-Butene (Minor product)

[NOTE: Elimination reactions almost always give mixtures of alkene products and the best we can usually do is to predict which will be the major product.]

Hofmann elimination is a process where an amine is reacted to create a tertiary amine and an alkene by treatment with excess methyl iodide followed by treatment with silver oxide, water, and heat.

[NOTE: because an amide ion such as NH2- is such a poor leaving group it's first necessary to convert it into a better leaving group via methylation.]

RNH2 + CH3I (excess) --> RN+(CH3)3I-
RN+(CH3)3I- + Ag2O + H2O + Heat --> (R - 1C) = CH2
 
dat_student said:
Zaitsev's rule: based induced elimination reactions generally give the more highly substituted (more stable) alkene product. (e.g. 2-Bromobutane + sodium ethoxide + ethanol --> 2-Butene (Major product) + 1-Butene (Minor product)

[NOTE: Elimination reactions almost always give mixtures of alkene products and the best we can usually do is to predict which will be the major product.]

Hofmann elimination is a process where an amine is reacted to create a tertiary amine and an alkene by treatment with excess methyl iodide followed by treatment with silver oxide, water, and heat.

[NOTE: because an amide ion such as NH2- is such a poor leaving group it's first necessary to convert it into a better leaving group via methylation.]

RNH2 + CH3I (excess) --> RN+(CH3)3I-
RN+(CH3)3I- + Ag2O + H2O + Heat --> (R - 1C) = CH2

Wonderful explanation! Thank you! 👍
 
DMD-2-B said:
Wonderful explanation! Thank you! 👍

Glad it helped. 🙂 If you have any other questions please don't hesitate to send me a PM
 
When I saw high heat, I thought about thermodynamic product vs. kinetic product, but that's usually for 1,4 or 1,2 additions. Does anyone know if it would apply here as well with the higher heat giving the thermodynamic product in greater yield than kinetic?
 
Top