The reaction of 4-methylcyclohexyl halide wit | vedclass

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(D) The $S_N2$ mechanism involves a backside attack by the nucleophile $(OH^-)$ on the carbon atom bonded to the leaving group (halide).This results i

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(D) The $S_N2$ mechanism involves a backside attack by the nucleophile $(OH^-)$ on the carbon atom bonded to the leaving group (halide).This results in the inversion of configuration at the chiral center.If the starting material is a $cis-4-methylcyclohexyl$ halide,the $OH^-$ will attack from the opposite side of the halide,leading to the $trans-4-methylcyclohexanol$ product.Conversely,if the starting material is a $trans-4-methylcyclohexyl$ halide,the product will be the $cis-4-methylcyclohexanol$.Since the specific stereochemistry of the starting halide is not provided in the prompt,and the options refer to general stereochemical outcomes of $S_N2$ reactions on substituted cyclohexanes,the inversion of configuration is the key principle.Given the options provided,the reaction proceeds with inversion of configuration.

The reaction of $4-methylcyclohexyl$ halide with $OH^-$ via the $S_N2$ mechanism produces $A$. Identify $A$.

Similar Questions

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

Which of the following reagents,when heated with ethyl chloride,forms ethylene?

Consider the reaction: $CH_3-CH_2-CH(F)-CH_3 \xrightarrow{CH_3O^{-}} X$. The alkene formed in major amount is:

$CH_3Br + Nu^{-} \rightarrow CH_3-Nu + Br^{-}$
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