Question

how will you obtain methoxy ethane from bromomethane​

Answer 1

Answer:

How would you prepare methoxyethane from ethyl bromide?

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EtBr + MeONa → EtOMe + NaBr

You should do this in a solvent of MeOH and heat the reaction.

MeONa (sodium methoxide) is basic and exists as MeO- Na+ (when in solution)

MeO- attacks the EtBr at the carbon atom bonded to the bromide and ‘kicks-out’ the bromide ion. This bromide ion goes on to form a salt with the sodium ion which will be floating in solution nearby.

If water was used as the solvent we would get ethanol (EtOH) as the product:

MeONa + H2O → MeOH + NaOH. This NaOH (sodium hydroxide) would react instead with the ethylbromide:

EtBr + NaOH → EtOH + NaBr,

This happens because H2O’s H (proton) is more ‘loosely bound’ to the oxygen atom than the MeOH’s H. (This means that H2O is a stronger bronsted-lowry acid than MeOH).

Explanation:

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Answer 2

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$\boxed{ \sf{EtBr + MeONa → EtOMe + NaBr}}$

You should do this in a solvent of MeOH and heat the reaction.

MeONa (sodium methoxide) is basic and exists as MeO- Na+ (when in solution)

MeO- attacks the EtBr at the carbon atom bonded to the bromide and ‘kicks-out’ the bromide ion. This bromide ion goes on to form a salt with the sodium ion which will be floating in solution nearby.

If water was used as the solvent we would get ethanol (EtOH) as the product:

$\boxed{ \sf{MeONa + H2O → MeOH + NaOH. }}$

This NaOH (sodium hydroxide) would react instead with the ethylbromide:

$\boxed{ \sf{ EtBr + NaOH → EtOH + NaBr}}$

This happens because H2O’s H (proton) is more ‘loosely bound’ to the oxygen atom than the MeOH’s H. (This means that H2O is a stronger bronsted-lowry acid than MeOH).