reaction of ethanoyl chloride with
1)ethylamine
2)n- methylaniline
Answers
Answer:
CH3NHCOCH3
Methyl amine reacts with acetyl chloride to form N-methylacetamide.
Explanation:
Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride.
unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ethylamine to give ethylammonium chloride.
Answer:
Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride
Explanation:
The reaction between ethanoyl chloride and ethylamine
The facts
Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride.
Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ethylamine to give ethylammonium chloride.
The mechanism
The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ethylamine.
The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.
That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways:
It might be removed by a chloride ion, producing HCl (which would immediately react with excess ethylamine to give ethylammonium chloride as above) . . .
and
. . . or it might be removed directly by an ethylamine molecule.
The ethylammonium ion, together with the chloride ion already there, makes up the ethylammonium chloride formed in the reaction.
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