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How does the ratio of substitution product to elimination product formed from the reaction of propyl Bromide with CH3O- in methanol change if the nucleophile is changed to CH3S-?

Short Answer

Expert verified

change if the nucleophile is changed to CH3S- to CH3O- , the ratio of substitution to elimination reaction increases due to the increase in nucleophilicity of CH3S-

Step by step solution

01

Definition of substitution reaction

Substitution reaction is a nucleophilic reaction where the Nucleophilic attack and bond breaking between carbon and electronegative group happens either in a step manner or simultaneously.

Example: SN1 and SN2 reaction.

Mechanism of SN2 reaction is given above.

02

Definition of elimination reaction

Elimination reaction is basically a reaction between alkyl bromide and a base in a step-by-step manner or in a concerted mechanism of bond breaking and making.

Example: E1, E2 E1cB mechanism.

Mechanism of E1 reaction is given above.

03

 Step 3: Change in reaction path with reactant

In substitution reaction, nucleophile attack the alkyl bromide but in case of elimination reaction, a base is used to undergo the reaction as it is better to abstract proton than nucleophile. So, the nucleophilic and basic character determines the reaction path.

Between CH3O- and CH3S-, CH3O-is more basic than CH3S- as O is more electronegative than S and as a matter of fact, in presence of CH3O-, elimination reaction is more favorable than substitution. Whereas in case of CH3S-, as Sulphur is more polarizable, it is more nucleophilic and favors the substitution reaction more. Thus, ratio of substitution to elimination reaction increases in presence of CH3S-.

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