So in the first instance you get ethanoic acid and an organic compound called an amide. \text{XNH}_2 + \text{HCl} \longrightarrow {\text{XNH}_3}^+\text{Cl}^-, \text{CH}_3\text{COOH} + \text{XNH}_2 \longrightarrow \text{CH}_3\text{COO}^-{}^+\text{NH}_3\text{X}, (\text{CH}_3\text{CO})_2\text{O} + \text{NH}_3 \longrightarrow \underset{\color{#467abf}{\text{ethanamide}}}{\text{CH}_3\text{CONH}_2} +~\text{CH}_3\text{COOH}, \text{CH}_3\text{COOH} + \text{NH}_3 \longrightarrow \underset{\color{#467abf}{\text{ammonium ethanoate}}}{\text{CH}_3\text{COO}^-{}^+\text{NH}_4}, (\text{CH}_3\text{CO})_2\text{O} + 2\text{NH}_3 \longrightarrow \underset{\color{#467abf}{\text{ethanamide}}}{\text{CH}_3\text{CONH}_2} + \underset{\color{#467abf}{\text{ammonium ethanoate}}}{\text{CH}_3\text{COO}^-{}^+\text{NH}_4}, \text{CH}_3\text{COCl} + 2\text{NH}_3 \longrightarrow \text{CH}_3\text{CONH}_2 + {\text{NH}_4}^+\text{Cl}^-, (\text{CH}_3\text{CO})_2\text{O} + \text{CH}_3\text{NH}_2 \longrightarrow \underset{\color{#467abf}{\text{N-methylethanamide}}}{\text{CH}_3\text{CONHCH}_3} +~\text{CH}_3\text{COOH}, \text{CH}_3\text{COOH} + \text{CH}_3\text{NH}_2 \longrightarrow \underset{\color{#467abf}{\text{methylammonium ethanoate}}}{\text{CH}_3\text{COO}^-{}^+\text{NH}_3\text{CH}_3}, (\text{CH}_3\text{CO})_2\text{O} + 2\text{CH}_3\text{NH}_2 \longrightarrow \text{CH}_3\text{CONHCH}_3 + \text{CH}_3\text{COO}^-{}^+\text{NH}_3\text{CH}_3, \text{CH}_3\text{COCl} + 2\text{CH}_3\text{NH}_2 \longrightarrow \text{CH}_3\text{CONHCH}_3 + {\text{CH}_3\text{NH}_3}^+\text{Cl}^-, (\text{CH}_3\text{CO})_2\text{O} + 2\text{C}_6\text{H}_5\text{NH}_2 \longrightarrow \text{CH}_3\text{CONHC}_6\text{H}_5 + \text{CH}_3\text{COO}^-{}^+\text{NH}_3\text{C}_6\text{H}_5. . written as: The ethanoic acid produced reacts with excess ammonia to give ammonium ethanoate. . The corresponding reaction with an acyl chloride is: Note: There isn't anything difficult involved in the fact that I have moved the position of the positive charge in the ammonium ion in this equation compared with the previous one.In each case I have placed it as close as possible to the negative charge in the ethanoate ion or chloride ion. This looks more difficult than the acyl chloride case because of the way the salt is written. Comparing the structures of ammonia and primary amines. These reactions are considered together because their chemistry is so similar. Note: The colour coding in these equations is to try to help you to see where everything ends up and where it came from, and to enable you to compare the two reactions more easily. Missed the LibreFest? Asked on December 30, 2019 by Mishra Kamra. The second stage of the reaction involves the formation of an ethanoate rather than a chloride. In this case, the "X" in the equations above is a hydrogen atom. The overall equation for the reaction is: \[ (CH_3CO)_2O + 2C_6H_5NH_2 \longrightarrow CH_3CONHC_6H_5 + CH_3COO^- \; ^+NH_3C_6H_5 \]. The "N" simply shows that the substitution is on the nitrogen atom, and not elsewhere in the molecule. We'll take methylamine as typical of primary amines where the -NH2 is attached to an alkyl group. written as: Then the ethanoic acid reacts with excess ammonia or amine to give a salt - this time an ethanoate. Acid anhydrides aren't so violently reactive as acyl chlorides, and the reactions normally need heating. All you need to know is that at each corner of the hexagon there is a carbon atom, together with a hydrogen atom apart from where the -NH2 group is attached. In a primary amine, it is attached to an alkyl group (shown by "R" in the diagram below) or a benzene ring. If you compare the structure with the amide produced in the reaction with ammonia, the only difference is that one of the hydrogens on the nitrogen has been substituted for a methyl group. written as: The ethanoic acid produced reacts with excess ammonia to give ammonium ethanoate. The second stage of the reaction involves the formation of an ethanoate rather than a chloride. In phenylamine, there isn't anything else attached to the ring as well. Again, the reaction happens in two stages. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. It isn't particularly important in the context of the current page. We'll take methylamine as typical of primary amines where the -NH2 is attached to an alkyl group. In phenylamine, there isn't anything else attached to the ring as well. Everything left over just gets joined together. There is also a great similarity between acid anhydrides and acyl chlorides (acid chlorides) as far as these reactions are concerned. The products are N-phenylethanamide and phenylammonium ethanoate. This is more usually (and more easily!) We'll take ethanoyl chloride as typical of the acyl chlorides. and you can combine all this together to give one overall equation: You need to follow this through really carefully, because the two products of the reaction overall can look confusingly similar. Again, the reaction happens in two stages. If this is the first set of questions you have done, please read the introductory page before you start. Note: For UK A-level purposes, this is probably the most likely context that you will meet this particular reaction in. The first product this time is called an N-substituted amide. To the menu of other organic compounds . This page looks at the reactions of acid anhydrides with ammonia and with primary amines. confusing if the phenylamine is drawn showing the benzene ring, and especially if the reaction is looked at from the point of view of the phenylamine. Watch the recordings here on Youtube! So the second stage of the reaction is: Ethanoic anhydride is the only one you are likely to come across for UK A level purposes. So in the first instance you get ethanoic acid and an organic compound called an amide. This looks more difficult than the acyl chloride case because of the way the salt is written. This reaction can sometimes look (even more!) written as: This is more usually (and more easily!) For example, the product molecule might be drawn looking like this: If you stop and think about it, this is obviously the same molecule as in the equation above, but it stresses the phenylamine part of it much more. So in the first instance you get ethanoic acid and an organic compound called an amide. Each substance contains an -NH2 group. Acid anhydrides aren't so violently reactive as acyl chlorides, and the reactions normally need heating. So in the first instance you get ethanoic acid and an organic compound called an amide. Initially, ethanoic acid is formed as the second product rather than hydrogen chloride gas. So in the first instance you get ethanoic acid and an organic compound called an amide. There is no essential difference between this reaction and the reaction with methylamine, but I just want to look at the structure of the N-substituted amide formed. Amides contain the group -CONH 2. In a primary amine, it is attached to an alkyl group (shown by "R" in the diagram below) or a benzene ring. You can think of the entire bit of the ethanoic anhydride shown in red as being exactly the equivalent of the chlorine atom in the acyl chloride as far as these reactions are concerned. We'll take ethanoyl chloride as typical of the acyl chlorides.

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