Which among the following halide(s) will not | vedclass

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(D) $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.$(A)$ $H_2C

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$(B)$ only

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(D) $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity.$(A)$ $H_2C=CH-CH_2Cl$ forms an allyl carbocation $(H_2C=CH-CH_2^+)$,which is resonance stabilized,so it shows $S_{N}1$ reaction.$(B)$ $CH_3-CH=CHCl$ is a vinylic halide. The carbocation formed $(CH_3-CH=CH^+)$ is highly unstable due to the positive charge on an $sp$-hybridized carbon atom,so it does not show $S_{N}1$ reaction.$(C)$ $C_6H_5CH_2Cl$ (benzyl chloride) forms a benzyl carbocation $(C_6H_5CH_2^+)$,which is highly resonance stabilized,so it shows $S_{N}1$ reaction.$(D)$ $(CH_3)_2CHCl$ (isopropyl chloride) forms a secondary carbocation $((CH_3)_2CH^+)$,which is relatively stable due to inductive effect and hyperconjugation,so it shows $S_{N}1$ reaction.Therefore,only $(B)$ will not show $S_{N}1$ reaction.

Which among the following halide$(s)$ will not show $S_{N}1$ reaction:
$(A)$ $H_2C=CH-CH_2Cl$
$(B)$ $CH_3-CH=CHCl$
$(C)$ $C_6H_5CH_2Cl$
$(D)$ $(CH_3)_2CHCl$
Choose the $most\ appropriate$ answer from the options given below:

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Which among the following halide$(s)$ will not show $S_{N}1$ reaction:$(A)$ $H_2C=CH-CH_2Cl$$(B)$ $CH_3-CH=CHCl$$(C)$ $C_6H_5CH_2Cl$$(D)$ $(CH_3)_2CHCl$Choose the $most\ appropriate$ answer from the options given below: