The major product obtained on treatment of CH3 CH2 CH(F) CH3 with CH3 O^- / CH3 OH is (a) CH3 CH

step 1 In this problem, set -jaf rule will be used. So the reaction can be given by CH3, CH2, CHF, CHF,

The major product obtained on treatment of CH3 CH2 CH(F) CH3...

The major product obtained on treatment of $\mathrm{CH}_3$ $\mathrm{CH}_2 \mathrm{CH}(\mathrm{F}) \mathrm{CH}_3$ with $\mathrm{CH}_3 \mathrm{O}^{-} / \mathrm{CH}_3 \mathrm{OH}$ is (a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{OCH}_3\right) \mathrm{CH}_3$ (b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3$ (c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2$ (d) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OCH}_3$

The major product obtained on treatment of $\mathrm{CH}_3$ $\mathrm{CH}_2 \mathrm{CH}(\mathrm{F}) \mathrm{CH}_3$ with $\mathrm{CH}_3 \mathrm{O}^{-} / \mathrm{CH}_3 \mathrm{OH}$ is
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{OCH}_3\right) \mathrm{CH}_3$
(b) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3$
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2$
(d) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OCH}_3$

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Key Concepts

E2 Elimination Reaction

This reaction mechanism involves the simultaneous removal of a proton by a strong base and the departure of a leaving group, resulting in the formation of a double bond. In the context of the reaction presented, a strong base (CH3O–) abstracts a proton from the substrate while the leaving group (fluoride) departs in a concerted process, favoring the formation of an alkene rather than substitution.

Zaitsev's Rule

Zaitsev's rule states that, in an elimination reaction, the most substituted alkene is the favored product due to its increased stability. In the given problem, the base abstracts a proton in such a way as to form the more substituted internal alkene, which is energetically more favorable compared to less substituted double bonds.

Leaving Group Ability

The leaving group ability is critical in determining the course of a reaction. In elimination reactions, a good leaving group is one that can stabilize the negative charge when it departs. Although fluoride is normally a poor leaving group, under strongly basic conditions and with the right substrate, elimination can still occur, leading to the formation of an alkene.

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