The Hoffmann Elimination

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take two bromopentane, with the bromine on carbon two, and heat it with alcoholic potassium hydroxide. The main product is pent two ene, the more substituted alkene, just as Saytzeff's rule says. Now put a trimethylammonium group on that same carbon in place of the bromine, and heat its hydroxide. It is the same carbon chain, with the group in the same position. But now the main product is pent one ene, the less substituted alkene! So why does the leaving group flip the answer? Questions on this reaction do appear in JEE, so let's find out.

0:41Exhaustive methylation: three steps

So, the Hoffmann elimination runs in three steps. Step one: we treat the amine with excess iodomethane, C H three I, where the small three written low counts the hydrogens, and the arrow means gives. Methyl groups keep adding to the nitrogen until we get the quaternary ammonium iodide. Since we methylate all the way, this is called exhaustive methylation. Okay? Step two: moist silver oxide, A g two O, with the small two written low. It converts the iodide into the quaternary ammonium hydroxide. Right? And step three is simply heat. The hydroxide breaks down to give an alkene, a tertiary amine, and H two O, water. This last step is the elimination itself. Clear?

1:29How much iodomethane? Count the H on N

Now, how much iodomethane does one mole of amine use up? Take a primary amine, RNH two, where R is an alkyl group. It has two hydrogens on nitrogen. So how many moles of iodomethane will it take? Many students say two, but careful, it takes three moles! Two methyl groups replace the two hydrogens, and then the tertiary amine takes one more methyl to become the quaternary salt. The small raised circle is read as degree, so one degree means primary, and mol is short for moles. Okay? Now, the secondary amine, R two N H, two degree, has one hydrogen to replace, and then one more methyl makes it quaternary. So it takes two moles. Right? And the tertiary amine, R three N, has no hydrogen on nitrogen at all. It just takes one methyl group to become quaternary, so one mole. Clear? So here is a quick trick: the moles of iodomethane equal the hydrogens on nitrogen, plus one. Two plus one gives three, one plus one gives two, and zero plus one gives one. Got it?

2:40Which alkene? The Hoffmann rule

Now the big question: which alkene forms? To make the double bond, the base removes a hydrogen from a carbon next to the carbon carrying nitrogen. The Greek letter on screen is beta, so we call it a beta hydrogen. Okay? And look at what leaves: N, then C H three in brackets with a small three, which means three methyl groups on nitrogen, trimethylamine. It is a bulky leaving group, really big. With me? Now suppose there are beta hydrogens on two sides, one on a crowded carbon and one on an open methyl end. With this bulky group in the way, which one can the base reach most easily? Yes, exactly, it is the one on the open methyl end, the least hindered position. The base takes the most accessible beta hydrogen, and that is also the most acidic one. Right? So the double bond forms toward that end, and the major product is the less substituted alkene, with fewer alkyl groups on its double bond carbons. This is the Hoffmann rule. Clear? And it is the opposite of Saytzeff's rule, where the more substituted alkene wins, as it does when an alkyl bromide is heated with alcoholic potassium hydroxide. So our puzzle is solved: the bromide follows Saytzeff and gives pent two ene, but the ammonium hydroxide follows Hoffmann and gives pent one ene. Got it?

4:03No beta hydrogen? Then no alkene

Now one special case. Tetramethylammonium hydroxide has four methyl groups on nitrogen, shown by the small four after the bracket. The raised plus and minus are the charges, and O H minus is the hydroxide ion. Heat it, and tell me, which alkene will you get? You get no alkene at all, and that is the trap here! Each methyl carbon is joined only to nitrogen, so there is no beta carbon, and so no beta hydrogen. With me? So substitution takes the place of elimination. We get methanol, C H three O H, and trimethylamine. Okay?

4:42Example 1: major alkene from butan-2-amine

Now Example one, an easier one. Butan two amine is exhaustively methylated, treated with moist silver oxide, and heated. Name the major alkene and the amine formed. Pause the video and try it first. Done? Okay, let's check. Butan two amine is a chain of four carbons, with the N H two group, shown in brackets, on the second carbon. So nitrogen sits on carbon two, written C two on screen. Okay? Now find the beta hydrogens. They sit on the carbons next to carbon two: carbon one, the C H three, and carbon three, the C H two. With me? Losing a hydrogen from carbon three would give but two ene, the more substituted alkene. So the answer is but two ene, right? That would be Saytzeff thinking, and it is exactly the trap here! The leaving group is the bulky trimethylamine, so the base takes the most accessible hydrogen, on carbon one, the methyl end. Clear? So the double bond forms between carbon one and carbon two, and the equals sign on screen is that double bond. That is but one ene, the less substituted alkene. Got it? And the amine formed is the leaving group itself, trimethylamine.

6:00Example 2: tertiary amine C5H13N that gives only ethene

Right? And now Example two, a harder one. Read it with me: a tertiary amine, C five H thirteen N, is treated with iodomethane, then with moist silver oxide, and heated. Ethene is the only alkene formed. Identify the amine and the other product. Pause the video and try it first. Done? Okay, let's check. The amine is tertiary, so it has three groups on nitrogen, with five carbons in all, and iodomethane adds a fourth group, a methyl. Remember the special case: methyl groups have no beta hydrogen, so they cannot eliminate. Okay? And ethene is the only alkene. So every group that is able to eliminate must be an ethyl group, because only an ethyl group gives ethene. Right? Now, a propyl group and two methyl groups also make five carbons, three plus one plus one. So could the amine be N, N dimethylpropan one amine? It cannot be, and that is the trap, because a propyl group would eliminate to give propene, not ethene. So the five carbons must be two ethyl groups and one methyl group: two plus two plus one gives five. Clear? So the amine has two ethyl groups, C two H five in brackets with a small two, and a methyl group, all on nitrogen. Its name is N ethyl N methylethanamine; the italic N says the group sits on nitrogen. Got it? Now heat the quaternary hydroxide, with its two ethyls and two methyls on nitrogen. One ethyl group leaves as ethene, and nitrogen keeps the other ethyl and the two methyls. So the other product is N, N dimethylethanamine. With me?

7:48Quick revision: steps, CH3I moles and the Hoffmann rule

So, quick revision. Excess iodomethane, then moist silver oxide, then heat, gives an alkene, a tertiary amine and water. Iodomethane used: three moles for a primary amine, two for a secondary, and one for a tertiary amine. And the Hoffmann rule gives the less substituted alkene, the opposite of Saytzeff. Now go and try the practice problems on this page. All the best!