Question

Methoxy Ethane can be prepared by the reactions :
(I) Ethyl Chloride + Sodium Methoxide or
(II) Methyl Chloride + Sodium Ethoxide
(A) Reaction I is faster and gives better yield
(B) Reaction I is faster and gives poorer yield
(C) Reaction Il is faster and gives poorer yield
(D) Reaction II is faster and gives better yield​

Answer 1

Answer:

It reacts with Lewis acids and forms salts (acid-base reaction). This reaction is a SN2 substitution reaction where the base sodium ethoxide is attacking the methyl iodide such that the leaning group (iodide ion) leaves. Hence ethyl methyl ether can be prepared from both diazomethane and methyl iodide.base, it can react with Lewis acids to form salts and reacts violently with oxidizing agents.

a) Preparation of Dimethyl ether (Methoxymethane) from methyl iodide :

When methyl iodide is heated with alcoholic sodium methoxide, it gives dimethyl ether.

CH

3

−O−Na + I−CH

3

Δ

CH

3

−O−CH

3

+ NaI

b) Methoxyethane from diazomethane :

When ethyl alcohol is treated with diazomethane in presence of fluoroboric acid, methoxyethane is fromed.

C

2

H

5

−OH + CH

2

N

2

Δ

HBF

4

C

2

H

5

−O−CH

3

+ N