Ariana123
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full details of the preparation and purification
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Ariana123
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Please anyone any help at all?
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seanw
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Basically, the N-phenyl is produced by reacting phenyl amine with carboxylic acid. phenyl ammonium chloride and ethanoic anhydride are "stores" of these two. Add ethanoate salt to the phenyl ammonium chloride to produce your phenyl amine, and water to the anhydride to get carboxylic acid.

phenyl ammonium chloride + sodium ethanoate -> phenyl amine + HCl + ethanoic acid

ethanoic anyhride + water -> 2 ethanoic acid

ethanoic acid + phenyl amine -> N-phenylethanamide + water

You can prove, by calculating the moles of each reagent, and the moles produced, which substance is limiting, and therefore the moles produced... The phenyl salt is actually the limiting factor here (by a large amount).


So moles of salt = moles produced. Mr of salt = 129.5 g/mol. Mr of N-phenyl = 135 g/mol.

For 1 gram, 1.0425 grams of N-sub phenylethanamide are produced. (1/129.5)*135 = 1.0425

For 5 grams, 5.21 grams are produced (100% yield) and 3.649 grams (70% yield).

Hope that's of some help. It could be completely wrong, so I'd take it with a pinch of salt.

**edit- turns out this is an old AQA practical exam question. My answers are correct... paper1 on this page. http://www.a-levelchemistry.co.uk/AQ...t%20Papers.htm
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Ariana123
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Thankyou! it makes a lot more sense now!! you're a star!
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Lets get close...
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what aparatus?
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TomLeigh
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I know this thread is a little old but I was wondering, I thought it was simply ethanoic anhydride + phenyl amine -> N-phenylethanamide + ethanoic acid

This in between step concerning water isn't mentioned in my CGP guide
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Maham22
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Why is the Mr of N-phenylethanamide 135.0 and not 270 when looking at the balanced equation
(Original post by seanw)
Basically, the N-phenyl is produced by reacting phenyl amine with carboxylic acid. phenyl ammonium chloride and ethanoic anhydride are "stores" of these two. Add ethanoate salt to the phenyl ammonium chloride to produce your phenyl amine, and water to the anhydride to get carboxylic acid.

phenyl ammonium chloride + sodium ethanoate -> phenyl amine + HCl + ethanoic acid

ethanoic anyhride + water -> 2 ethanoic acid

ethanoic acid + phenyl amine -> N-phenylethanamide + water

You can prove, by calculating the moles of each reagent, and the moles produced, which substance is limiting, and therefore the moles produced... The phenyl salt is actually the limiting factor here (by a large amount).


So moles of salt = moles produced. Mr of salt = 129.5 g/mol. Mr of N-phenyl = 135 g/mol.

For 1 gram, 1.0425 grams of N-sub phenylethanamide are produced. (1/129.5)*135 = 1.0425

For 5 grams, 5.21 grams are produced (100% yield) and 3.649 grams (70% yield).

Hope that's of some help. It could be completely wrong, so I'd take it with a pinch of salt.

**edit- turns out this is an old AQA practical exam question. My answers are correct... paper1 on this page. http://www.a-levelchemistry.co.uk/AQ...t%20Papers.htm
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