Question

write the mechanism of convertion of
methyl chloride to convertion of methyl
chloride to methyl alcohol by treating
with KOH,​

Answer 1

See the attachment for the mechanism of the reaction

Explanation :-

• The reaction that takes place is $SN_2$.

• $CH_3Cl$ {methyl chloride) is a primary alkyl halide and is suitable for $SN_2$ reaction.

• Aqueous KOH is the source of nucleophile $OH^-$.

• Here, backside attack occurs and a transition state is formed in which C-OH forms and C-Cl bond breaks.

• Finally , $Cl^-$ leaves and it is known as leaving group.

Additional details for IIT-JEE

• Leaving tendency is inversely proportional to basic strength.This is because stronger base forms stronger bond and that is why it will be tough for them to break the bond.

• Let us understand this fact using halides as an example ,their leaving tendency follows the order :-

$I^-\:>\:Br^-\:>\:Cl^-\:>\:F^-$

• This is beacuse,Fluoride ion is the strongest base due to its smallest size.

Answer 2

Answer:

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• See in the attachment