Reactions with Derivatives of Ammonia

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You have a bottle of an unknown liquid, and you want to know whether it holds a carbonyl group at all. You add one reagent, and a yellow, orange or red crystalline solid comes out. Then its melting point tells you which compound it was. How can one reagent do all that? Today we find out, with aldehydes and ketones meeting derivatives of ammonia. This comes up often in JEE, so let's begin.

0:30Addition, then loss of water

So, here is the whole lesson in one line. On the left is the carbonyl group: the equals sign is a double bond, and the little angle in front of the C stands for the two other bonds on that carbon. Plus H two N, dash, Z, where the small two written low counts the hydrogens, the dash is a single bond, and Z is a group that changes from reagent to reagent. The two half arrows mean the reaction is reversible. Okay? It happens in two stages. First, addition: the nitrogen acts as the nucleophile, and H two N, dash, Z adds across the C double bond O, giving an adduct. Then elimination: the adduct loses water, H two O, and the carbonyl group becomes C double bond N, dash, Z, as on the right. Addition, then elimination: remember these two words. Right?

1:25Six reagents, six products

Now, each Z gives a product with its own name. First, a primary amine, RNH two, where R is an alkyl or aryl group, so Z is R itself, gives a substituted imine, also called a Schiff's base. Got it? Hydroxylamine, where Z is the O H group, gives an oxime. Oximes come up often. Okay? Hydrazine, where Z is N H two, gives a hydrazone. And phenylhydrazine, where Z is NHC six H five, with C six H five the phenyl group, gives a phenylhydrazone. See how the names follow a pattern? Just change the ending, simple! Next, two four dinitrophenylhydrazine, or two four DNP for short. Its Z is NHC six H three, then N O two in brackets with a small two outside, so two nitro groups on the ring. It gives a two four dinitrophenylhydrazone. With me? And semicarbazide, where Z is NHCONH two, gives a semicarbazone. Clear?

2:31Acid helps, but not too much

Now, the conditions. The reaction is catalysed by acid, and it is fastest at a p H of about three point five; the wavy equals sign means approximately. Got it? If acid speeds it up, then more acid should make it even faster, right? No, no, careful, it does not work that way! In strongly acidic solution, H two N, dash, Z is protonated, so it is no longer a nucleophile, and it cannot attack the carbonyl carbon. Okay? And in basic solution, the carbonyl oxygen is not protonated, so the carbonyl group is not activated. Too much acid spoils the nucleophile, and too little acid leaves the carbonyl group unactivated, so a p H of about three point five is best. With me?

3:17The 2,4-DNP test for a carbonyl group

Now, back to our puzzle. Two four DNP gives a yellow, orange or red crystalline precipitate, the two four dinitrophenylhydrazone. So that was our mystery reagent! Right? Now let me ask you this. Can two four DNP tell an aldehyde from a ketone? It cannot, because aldehydes and ketones both give the precipitate, so it is simply a test for the carbonyl group. And the melting point of the precipitate identifies the compound, the second half of our puzzle. Okay?

3:49Semicarbazide: which NH₂ reacts?

Now, a favourite reasoning question. Semicarbazide is H two N, then N H, then C O, the carbonyl group, then N H two, joined by single bonds, so it has an N H two group at each end. Right? So which one attacks the aldehyde or ketone: the left one, or the right one, attached straight to the C O group? The left one, the far one, yes, exactly! The lone pairs of the other two nitrogens are delocalised onto the C double bond O group, so they are busy. Only the far N H two has a lone pair free to attack. Got it?

4:28Isomers, enamines and urotropine

Now, three more facts from the course. Oximes, and other compounds with a C double bond N, show geometrical isomerism, but only when the carbon carries two different groups. The O H of an oxime can point towards one group or the other, and if the groups differ, those are two different molecules. Clear? Next, a primary amine, RNH two, gave an imine, but secondary amines give enamines instead. Okay? And methanal plus ammonia gives urotropine, also called hexamethylenetetramine: C H two in brackets with a small six outside, then N four. With me?

5:06Example 1: oxime, hydrazone, semicarbazone of propanone

Time for Example one, an easy one to warm up. Name the product of propanone with, part a, hydroxylamine, part b, hydrazine, and part c, semicarbazide. Pause the video and try it first. Done? Okay, let's check part a. Propanone is C H three in brackets, with a small two outside, then C double bond O, so two methyl groups on the carbonyl carbon. Hydroxylamine adds, water leaves, and the arrow gives C H three in brackets, two, C double bond NOH, called propanone oxime. Right? Part b is hydrazine, where Z is N H two. The oxygen is replaced by N, dash, N H two, so we get propanone hydrazone. Okay? Now part c, semicarbazide. Which end of it joins the carbon of propanone? The far N H two end, correct! So the carbon is joined to N, then NHCONH two, and the product is propanone semicarbazone. Three reagents, one pattern. Clear?

6:17Example 2: oxime C4H9NO, iodoform test, E and Z isomers

Now Example two, a harder one. An oxime, C four H nine N O, is formed from a carbonyl compound that gives the iodoform test and does not reduce Fehling's solution. Identify the carbonyl compound, and state the number of stereoisomeric oximes. Here is a hint: count the carbons first. Pause the video and try it. Done? Okay, let's check. Making an oxime adds hydroxylamine and removes water. So, going back, add water to the oxime and take away N H two O H; the minus sign means take away. We get C four H eight O, a carbonyl compound with four carbons. Okay? Next, it does not reduce Fehling's solution. Fehling's solution is reduced by aliphatic aldehydes, and with four carbons, any aldehyde would be aliphatic. So our compound is a ketone. Right? And it gives the iodoform test, which needs a C H three C O group; the dash means it is joined to the rest of the molecule. Four carbons and a methyl ketone: that is butanone, C H three COCH two C H three. Got it? Now, the stereoisomers. One ketone, so just one oxime, right? No, no, careful, look at the carbon of the C double bond N group! In the oxime, C H three and C two H five in brackets, it carries a methyl group and an ethyl group, and the equals sign with a slash means not equal. Two different groups, so geometrical isomerism is possible. Clear? So there are two stereoisomeric oximes, E and Z, the letters in italics; this Z is a label, not our group from before. And propanone oxime has none, because its carbon carries two methyl groups. That's it!

8:14Quick revision: C=N formation, pH 3.5, 2,4-DNP, isomers

So, quick revision. Carbonyl compound plus H two N, dash, Z: addition, then loss of water, gives the C double bond N compound. The reaction is fastest at a p H of about three point five, and the two four DNP precipitate shows a carbonyl group. Semicarbazide reacts through its far N H two, and C double bond N isomers need two different groups on the carbon. Now go and try the four practice problems on this page. All the best!