In the above reaction,the maximum Saytzeff pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In an $E_2$ elimination reaction,the nature of the leaving group $X$ significantly influences the product distribution.For alkyl halides,the trans

Vedclass · Accepted Answer

$-I$

Vedclass · Answer

(A) In an $E_2$ elimination reaction,the nature of the leaving group $X$ significantly influences the product distribution.For alkyl halides,the transition state for the $E_2$ mechanism has significant carbanion character.Fluorine $(-F)$ is the most electronegative halogen,which makes the $\beta$-hydrogen more acidic and stabilizes the transition state leading to the less substituted alkene (Hoffmann product).Conversely,better leaving groups like $-I$,$-Br$,and $-Cl$ facilitate the formation of the more stable,more substituted alkene (Saytzeff product) because the transition state resembles the product more closely.Among the given options,$-F$ is the poorest leaving group,which promotes the formation of the Hoffmann product,while $-I$ is the best leaving group,which promotes the formation of the Saytzeff product.Therefore,the maximum Saytzeff product is obtained when $X = -I$.

In the above reaction,the maximum Saytzeff product will be obtained when $X$ is:

Similar Questions

$S_N1$ reactions are generally carried out in which type of solvents?

Why is chloroform stored in dark-colored bottles?

Identify $(B)$ and $(C)$ and how are $(A)$ and $(C)$ related?

Which is most reactive for $S_{N}2$ reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module