2-chloro-2-methylpentane on reaction with sod | vedclass

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(C) $2-$chloro$-2-$methylpentane is a tertiary $(3^{\circ})$ alkyl halide.In the presence of a strong base/nucleophile like sodium methoxide $(CH_3ONa

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(C) $2-$chloro$-2-$methylpentane is a tertiary $(3^{\circ})$ alkyl halide.In the presence of a strong base/nucleophile like sodium methoxide $(CH_3ONa)$ in methanol $(CH_3OH)$,it can undergo both substitution and elimination reactions.$1.$ Substitution $(S_N1)$: The $Cl^-$ is replaced by $-OCH_3$ to form $2-$methoxy$-2-$methylpentane (Product $(1)$).$2.$ Elimination $(E2)$: Removal of $HCl$ from $\beta-$carbons.$-$ From the terminal methyl group,it forms $2-$methylpent$-1-$ene (Product $(2)$).$-$ From the internal methylene group,it forms $2-$methylpent$-2-$ene (Product $(3)$).Thus,all three products are obtained.

$2-$chloro$-2-$methylpentane on reaction with sodium methoxide in methanol yields:
$(1)\ C_2H_5-CH_2-C(CH_3)_2-OCH_3$
$(2)\ C_2H_5-CH_2-C(CH_3)=CH_2$
$(3)\ C_2H_5-CH=C(CH_3)_2$

Similar Questions

Arrange the following halides in decreasing order of $S_N1$ reactivity:
$(I)$ $CH_3CH_2CH_2Cl$
$(II)$ $CH_2=CHCHClCH_3$
$(III)$ $CH_3CH_2CHClCH_3$

What is the correct order of leaving group ability for the following groups?
$I. -OAc$
$II. -OMe$
$III. -OSO_2Me$
$IV. -OSO_2CF_3$

The rate of $\beta$-dehydrohalogenation in the following halides under treatment with a strong base follows the order:

Vinyl chloride reacts with $HCl$ to form

The major product of the following reaction is:

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$2-$chloro$-2-$methylpentane on reaction with sodium methoxide in methanol yields:$(1)\ C_2H_5-CH_2-C(CH_3)_2-OCH_3$$(2)\ C_2H_5-CH_2-C(CH_3)=CH_2$$(3)\ C_2H_5-CH=C(CH_3)_2$