Sodium tertiary butoxide on reaction with met | vedclass

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(B) Reaction $1$: Sodium $t-butoxide$ $((CH_3)_3CONa)$ is a strong base and a bulky nucleophile. It reacts with methyl bromide $(CH_3Br)$ via an $S_N2

Vedclass · Accepted Answer

$P$ is $t-butyl$ methyl ether and $Q$ is $2-methylpropene$

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(B) Reaction $1$: Sodium $t-butoxide$ $((CH_3)_3CONa)$ is a strong base and a bulky nucleophile. It reacts with methyl bromide $(CH_3Br)$ via an $S_N2$ mechanism to produce $t-butyl$ methyl ether $(CH_3-O-C(CH_3)_3)$ as the major product $(P)$.Reaction $2$: Sodium methoxide $(CH_3ONa)$ is a strong base. When it reacts with a tertiary substrate like $t-butyl$ bromide $((CH_3)_3CBr)$,the steric hindrance prevents $S_N2$ substitution. Instead,an $E2$ elimination reaction occurs,producing $2-methylpropene$ $((CH_3)_2C=CH_2)$ as the major product $(Q)$.

Sodium tertiary butoxide on reaction with methyl bromide produces the product $P$.
Sodium methoxide upon reaction with tertiary butyl bromide generates the product $Q$.
The products $P$ and $Q$ are

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Sodium tertiary butoxide on reaction with methyl bromide produces the product $P$.Sodium methoxide upon reaction with tertiary butyl bromide generates the product $Q$.The products $P$ and $Q$ are