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 $A$.

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By passing excess $Cl_{2(g)}$ in boiling toluene, which one of the following compounds is exclusively formed?

Considering the above reactions,the compound $A$ and compound $B$ respectively are:

The correct reactivity order for $S_N2$ reaction is:
$(i)$ $CH_3COCH_2Cl$
(ii) $CH_2=CH-CH_2Cl$
(iii) $CH_3CH_2CH_2Cl$

The decreasing order of the rate of $S_{N^1}$ reaction for the following compounds is:
$I$. $Ph-C(Br)(CH_3)-Ph$
$II$. $Ph-CH(Br)-Ph$
$III$. $Ph-CH(Br)-CH_3$
$IV$. $Ph-CH_2Br$

The decreasing order of reactivity towards $S_N 1$ reaction for the following compounds is:

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