From Alkyl Halides

Transcript

0:00Introduction

Hello everyone! Okay, let me start with a small story. You want to make ethanamine. So you heat chloroethane with ammonia in a sealed tube. Simple, right? But when you open the tube, there is not one amine in it, but a whole mixture of products. So where did the extra ones come from? And if you really need a pure primary amine, what do you do then? By the end of this video you will answer both questions. Making amines from alkyl halides comes up often in JEE, so let's begin.

0:34Ammonolysis: alkyl halide plus ammonia

So, the first method is called ammonolysis. We take R X, an alkyl halide, where R is an alkyl group and X is a halogen atom, okay? We heat it with alcoholic ammonia, N H three, in a sealed tube at three hundred and seventy three kelvin. The small three written low is a subscript, and the capital K stands for kelvin. Clear? Now, ammonia meets the alkyl halide. What kind of reaction do you expect here? It is nucleophilic substitution, yes, exactly! Ammonia replaces the halogen, and the arrow means gives. But look at the product: RNH three plus, X minus. The plus and the minus written high are charges, so this is a salt, not the free amine. Right? So we add an alkali, sodium hydroxide. It turns the salt into RNH two, the free primary amine, along with the sodium halide and water. So the alkali sets the amine free from its salt. Got it?

1:37Does the reaction stop at the primary amine?

Now here is the catch. The RNH two we just made is itself a nucleophile, just like ammonia. So it attacks more R X, and the reaction does not stop at the primary amine. With me? So in the end we get a mixture. One degree, two degree and three degree amines, where the small circle written high means degree: primary, secondary and tertiary. And last, R four N plus X minus, the quaternary salt, with four alkyl groups on the nitrogen. So ammonolysis does not give a pure primary amine. Okay? So let me ask you this: how can we make this reaction give mainly the primary amine? Use a large excess of ammonia, very good! Then the alkyl halide meets ammonia far more often than the amine already formed, so the primary amine is favoured. It is favoured, but the product is still not pure, so do not fall into that trap. Clear?

2:36Which halides react?

Next, the order of reactivity. Alkyl iodides react fastest, then bromides, then chlorides. The pointed sign means greater than. So the alkyl iodide is greater than the bromide, and the bromide is greater than the chloride. Okay? Now here is a tricky one for you. If we heat chlorobenzene with ammonia, do we get aniline? We do not, under these conditions, and many students miss this. Aryl halides, where the halogen sits on a benzene ring, like chlorobenzene, C six H five C l, do not react here. Right? The reason is resonance with the ring. The carbon to chlorine bond, shown with a dash, gets partial double bond character, and that carbon is s p two, with the two written high. So nucleophilic substitution there is very difficult. Got it?

3:26Gabriel phthalimide synthesis

Now, the way to a pure primary amine: the Gabriel phthalimide synthesis. We treat phthalimide with ethanolic KOH, potassium hydroxide, written above the arrow. This gives potassium phthalimide: N minus, K plus, a negatively charged nitrogen with a potassium ion. Okay? Now let me ask you this. Why can potassium hydroxide remove the hydrogen on this nitrogen at all? It is because of its neighbours, exactly. This N H, with the dash for the bond, is flanked by two C double bond O groups, the carbonyl groups. The anion is stabilised by resonance with both of them, so the hydrogen is acidic. Clear? Next, this nitrogen anion attacks an alkyl halide, R X, and displaces the halide. We get N alkyl phthalimide, where the italic capital N in front says the alkyl group sits on the nitrogen. With me? Finally, hydrolysis with sodium hydroxide in water, where the letters a q in brackets mean aqueous, or with hydrazine. This releases RNH two, a pure primary amine, free of secondary and tertiary amines. So that answers the second question from our story. Right?

4:42Where the Gabriel synthesis fails

But this method has limits too. Aryl halides do not undergo nucleophilic substitution by the phthalimide anion. So aromatic primary amines, like aniline, cannot be made this way. Okay? And tertiary halides give elimination instead of substitution, so they are no good either. Keep both limits in mind. Right?

5:03Example 1: chloroethane with NH3 gives four products

Now, remember our story from the start? Let's settle it. Read it with me: chloroethane is heated with ammonia. Name all the organic products that can form. Pause the video and try it first. Done? Okay, let's check. First, ammonia, written above the arrow, replaces the chlorine of chloroethane, and we get C two H five N H two, ethanamine. The one degree in brackets says it is primary. Okay? So that is the answer, just ethanamine, right? Oh no, careful, that is exactly the trap here, because the story does not stop there. Ethanamine is itself a nucleophile, so it attacks another chloroethane. Now two ethyl groups sit on the nitrogen, and the bracket with a small two shows them. This is N ethylethanamine, a secondary amine. Clear? And it goes on. The secondary amine attacks one more chloroethane, so now there are three ethyl groups on the nitrogen. This is N, N diethylethanamine, a tertiary amine. Right? And one last time. The tertiary amine takes a fourth ethyl group, giving tetraethylammonium chloride, the quaternary salt. So there are four organic products in all, and that is exactly the mixture from our story. With me?

6:25Example 2: which amines Gabriel synthesis cannot make

Now a harder one. Which of these cannot be prepared by the Gabriel synthesis? Ethanamine, aniline, N methylethanamine, two methylpropan two amine, and butan one amine. Here is a hint: check the class of each amine, and the halide it would need. Pause the video and try it first. Done? Okay, let's check. First, the rule. Gabriel gives only primary amines, and only from halides that undergo S N two substitution, with the N written small and low. Okay? Ethanamine and butan one amine are both primary, and both come from primary halides. So both can be made. Right? Aniline would need an aryl halide, and aryl halides do not undergo this substitution. And N methylethanamine has a methyl and an ethyl group on the nitrogen, so it is secondary, and Gabriel never gives secondary amines. So neither of these can be made. Clear? Now the last one, two methylpropan two amine. Its N H two group is primary, so Gabriel should work, right? Oh no, careful, you have to look at the halide it needs. The N H two sits on a carbon that carries three methyl groups, shown by the bracket with a small three. So we would need a tertiary halide, and tertiary halides give elimination instead. So this one cannot be made either. Got it? So the answer is three compounds: aniline, N methylethanamine, and two methylpropan two amine. One aromatic amine, one secondary amine, and one that needs a tertiary halide. Check these three things every time. With me?

8:02Quick revision: ammonolysis, halide order, Gabriel limits

So, quick revision. Ammonolysis: an alkyl halide with ammonia at three hundred and seventy three kelvin gives a mixture of amines, and a large excess of ammonia favours the primary amine. Reactivity goes iodide, then bromide, then chloride, and aryl halides do not react. And Gabriel gives pure primary amines, but no aromatic amines, and nothing from tertiary halides. Now go and try the practice problems on this page. All the best!