The major product obtained from E_{2}-elimina | vedclass

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(D) In $E_{2}$-elimination,a base removes a proton from the $\beta$-carbon while the leaving group departs from the $\alpha$-carbon.For $3$-bromo-$2$-

Vedclass · Accepted Answer

$CH_3CH_2CH=C(F)CH_3$

Vedclass · Answer

(D) In $E_{2}$-elimination,a base removes a proton from the $\beta$-carbon while the leaving group departs from the $\alpha$-carbon.For $3$-bromo-$2$-fluoropentane $(CH_3-CH_2-CH(Br)-CH(F)-CH_3)$,there are two potential $\beta$-carbons: $C_{2}$ (bearing $F$) and $C_{4}$ (bearing $H$).$Br^-$ is a much better leaving group than $F^-$. Therefore,the elimination occurs primarily at the $C_{3}$ position (where $Br$ is attached) and the $C_{2}$ position (where $H$ is attached).The base abstracts the more acidic proton from $C_{2}$ (which is adjacent to the electron-withdrawing $F$ atom).This leads to the formation of $CH_3-CH_2-CH=C(F)-CH_3$ as the major product,which is a stable alkene with $5$ $\alpha$-hydrogens.

The major product obtained from $E_{2}$-elimination of $3$-bromo-$2$-fluoropentane is

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