Question

The reaction of Sodium methoxide with the ${R_2}CHX$ is:
A.$S{N^1}$
B.$S{N^2}$
C.Elimination
D.Substitution

Answer 1

The chemical formula for sodium methoxide is $C{H_3}ONa$. Sodium methoxide is a strong base. Sodium methoxide is a nucleophile. Nucleophiles are nucleus loving. They are electron-rich species so they donate their electron pair to electron-deficient species.

Complete step by step answer:
-The formula for sodium methoxide is $C{H_3}{O^{( - )}}N{a^{( + )}}$
-We know nucleophiles are nucleus loving. They are electron-rich and thus they donate their electron pairs to electron-deficient species.
${R_2}CHX$ is an alkyl halide where R represents alkane and X represents halogen.
-Here the alkyl halide ${R_2}CHX$ is a secondary alkyl halide.
-The reaction of sodium methoxide ${R_2}CHX$ follows $S{N^1}$ the reaction mechanism.
$S{N^1}$ the reaction is a substitution nucleophilic unimolecular reaction.
-These reactions follow first-order kinetics.
-Because the rate-determining step depends only on the concentration of alkyl halide and it does not depend on the concentration of nucleophiles involved.
-Thus in $S{N^1}$ reaction, rate $\alpha$ [alkyl halide]. Hence unimolecular.
$S{N^1}$ reactions involve two steps.
-In the first step, it is the formation of a carbocation.
-In the second step, the carbocation reacts with the nucleophile to give the product.
-The reaction of sodium methoxide with the ${R_2}CHX$ is as follows:
${R_2}CHX + C{H_3}ONa \to {R_2}CHOC{H_3}$
-We can write the $S{N^1}$ mechanism of the reaction of sodium methoxide with ${R_2}CHX$.
Step (I) Alkyl halide breaks the $C - X$ bond and gives the intermediate carbocation. This is the slow step since it is the rate-determining step.
$C{H_3}ONa \to C{H_3}{O^ - } + N{a^ + }$
${R_2}CHX \rightleftharpoons {R_2}C{H^ + } + {X^ - }$
$C{H_3}{O^ - }$ is the nucleophile.
${R_2}C{H^ + }$ is the carbocation formed.
We know that the stability of carbocation follows the order.
Tertiary carbocation $>$ secondary carbocation $>$ Primary carbocation
Step (II) The carbocation formed reacts with the nucleophile and gives the product in the second step as shown below:
${R_2}C{H^ + } + C{H_3}{O^ - } \to {R_2}CHOC{H_3}$
-This is the fastest step in the reaction.
-Thus the product formed when sodium methoxide reacts with ${R_2}CHX$ is ${R_2}CHOC{H_3}$.
-The reaction of sodium methoxide ${R_2}CHX$ is $S{N^1}$.
The answer to the question is an option (A).

Note: The intermediate carbocation is also called carbonium ion. Here the carbonium ion formed is stabilized by inductive effect and hyperconjugation. Therefore the $S{N^1}$ reaction takes place here. The concentration of the nucleophile involved in the reaction does not affect the rate of $S{N^1}$ reaction. $S{N^2}$ reactions take place for primary alkyl halides and undergo an inversion in the configuration.