Major product formed in the following reaction {CH}_3 {\CH}_3- \stackrel{|}{\C}^{\CH_3}_{|}_{\CH_3}- {\Br}+ {\NaOCH}_3 \to

Correct Answer is : CH3−C|CH3=CH2

{\CH}_3- \stackrel{|}{\C}^{\CH_3}_{|}_{\CH_3}- {\ONa}

{\CH}_3- \stackrel{|}{\C}^{\CH_3}_{|}_{\CH_3}- {\OCH_3}

{\CH}_3- \stackrel{|}{\C}^{\CH_3}_{|}_{\CH_3}- {\O }- \stackrel{|}{\C}^{\CH_3}_{|}_{\CH_3}-\CH_3

{\CH_3}- \stackrel{|}{\C}^{\CH_3}={\CH_2}

CBSE Class 12 Chemistry 2022 Term 1 Paper - Question 23

Question : 23 of 55

Marks: +1, -0

Major product formed in the following reaction

{CH}_3

{\CH}_3- {\C}↖{|}↖{\CH_3}↙{|}↙{\CH_3}- {\Br}+ {\NaOCH}_3 →

${\CH}_3- {\C}↖{|}↖{\CH_3}↙{|}↙{\CH_3}- {\ONa}$
${\CH}_3- {\C}↖{|}↖{\CH_3}↙{|}↙{\CH_3}- {\OCH_3}$
${\CH}_3- {\C}↖{|}↖{\CH_3}↙{|}↙{\CH_3}- {\O }- {\C}↖{|}↖{\CH_3}↙{|}↙{\CH_3}-\CH_3$
${\CH_3}- {\C}↖{|}↖{\CH_3}={\CH_2}$

Solution:

( {\CH}_3) _3 {\C}- {\Br}+ {\Na}- {\O}- {\Me} →

{\CH}_3 {{-\C}↙{|}↙{\CH_3}}↙{\iso-\butene}= {\CH}_2+ {\NaBr}+ {\CH}_3 {\OH}

Tert-halide with strong base favours elimination reaction not the substitution reaction.

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