The Hoffmann Bromamide Degradation

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Most ways of making an amine keep the number of carbons the same, and the cyanide route even adds one. But take an amide, add bromine and sodium hydroxide, and out comes an amine with one carbon fewer! So where did that carbon go? By the end of this video you will know. This reaction comes up often in JEE, so let's begin.

0:27The overall reaction

So, here is the reaction, and the figure shows its steps. We start with an unsubstituted amide, RCONH two, where R is an alkyl or aryl group and C O is the carbonyl group. We add bromine, B r two, and four N a O H, sodium hydroxide, aqueous or alcoholic, as the short forms in brackets say. The small numbers written low are subscripts; they count atoms. Okay? Now count the carbons. The amide has the carbons of R, plus one more, the carbon of the C O group. But the amine we get is only RNH two. So let me ask you this: where does that carbonyl carbon end up? It ends up in sodium carbonate, yes, exactly! The arrow means gives: RNH two, plus N a two C O three, sodium carbonate, plus two N a B r, sodium bromide, plus two water molecules. So the lost carbon leaves as carbonate. That solves our puzzle, clear? And the product is a primary amine; the small raised circle is read as degree, and one degree means primary. It has one carbon fewer than the amide, so this is called a degradation: it steps a chain down by one carbon. Got it?

1:48Inside the reaction: a group moves to nitrogen

Now, what happens inside? Follow the figure with me. First, bromine replaces one hydrogen on nitrogen, giving the N bromoamide; the italic N says the bromine is on nitrogen. Then hydroxide pulls off the second hydrogen, giving its anion, with a negative charge on nitrogen. Okay? Next, the anion loses a bromide ion, and as it goes, the nitrogen becomes short of electrons. Something must move onto that electron deficient nitrogen. Look at the figure: which group moves? It is the group R that moves, yes, exactly! As B r minus leaves, the raised minus being its charge, the group R shifts from the carbon to the nitrogen. This shift is the heart of the whole reaction. With me? What we get is an isocyanate, RNCO, where each equals sign is a double bond and the small dash is a single bond. Now the group R sits on nitrogen, and the old carbonyl carbon sits alone between nitrogen and oxygen. Right? Finally, as the figure shows, water adds to give a carbamic acid, and then O H minus, hydroxide, breaks it into RNH two and carbonate, C O three two minus; the raised two minus is its charge. So that lonely carbon leaves as carbonate. Clear?

3:10Four facts the questions test

Now, four facts that questions like to test. Number one: the migration is intramolecular. That means the group R moves within the same molecule; it never breaks away. Okay? Number two: the migrating group keeps its configuration. So if the moving carbon is a chiral centre, its arrangement in space stays the same in the amine. Got it? Now go back to the mechanism. Bromine replaced one hydrogen on the nitrogen, and hydroxide removed the other. So tell me, how many hydrogens must the amide have on its nitrogen? It needs two hydrogens, exactly! So only RCONH two reacts. An N substituted amide, with a carbon group on nitrogen in place of a hydrogen, does not react. This is a common trap, so look at the nitrogen first. Clear? And number four: R may be an aryl group, such as the phenyl group from benzene. So benzamide gives aniline, a neat way to make it. Right?

4:11Choosing a route: count the carbons

So, when should you choose this reaction? Count the carbons. If the amine needs the same number of carbons as the alkyl halide, use the Gabriel synthesis, which gives only primary aliphatic amines, or ammonolysis, heating the halide with ammonia. Okay? One carbon more than the halide? Then use KCN, potassium cyanide, and reduce the nitrile you get. The cyanide brings its own carbon. Right? The same number of carbons as the amide? Reduce the amide with lithium aluminium hydride, L i A l H four. Every carbon stays. With me? Now suppose the amine must have one carbon fewer than the amide. Which route will you pick? You pick the Hoffmann bromamide degradation, of course! That is B r two slash N a O H, where the slash means together with. So always count the carbons before you choose. Got it?

5:07Example 1: amines from butanamide and benzamide

Now Example one, an easy one. Name the amine formed when, part a, butanamide, and part b, benzamide, are treated with bromine and sodium hydroxide. Pause the video and try it first. Done? Okay, let's check. Butanamide, on screen, is a straight chain of three carbons joined to the CONH two group: four carbons in all. Hoffmann removes one, so four carbons become three. Okay? Now, many students simply change amide to amine and write butan one amine. Is that right? That is exactly the trap, and it is wrong! The C O carbon has left as carbonate. So the amine has a straight chain of three carbons, with N H two at the end: propan one amine. Clear? Part b is quick. Benzamide, C six H five CONH two, with C six H five the phenyl group, has seven carbons. Remove the C O carbon, and C six H five N H two is left, with six carbons: aniline! Got it?

6:17Example 2: identify amide C4H9NO and its amine B

And now Example two, a harder one. An amide, called A, with formula C four H nine N O, gives with bromine and KOH, potassium hydroxide, a primary amine, called B, whose nitrogen is attached to a secondary carbon. Identify the amide and the amine. Pause the video and try it first. Done? Okay, let's check, and start from the amine B, okay? Hoffmann removes one carbon, so the amine has three carbons. And its nitrogen sits on a secondary carbon, two degree, which means a carbon joined to two other carbons. Right? Now, the amide has four carbons, and we just met butanamide in Example one. So the amide is butanamide, right? Careful, that cannot be right, because butanamide gives propan one amine, with nitrogen on an end carbon. With nitrogen on the middle carbon, that carbon carries N H two and two methyl groups; the brackets with a two mean two methyl groups. That is propan two amine, our amine B, clear? Now go backwards to the amide. When the group R moves, the carbon of R that was bonded to the C O group is the one that becomes bonded to nitrogen. So in the amide, the CONH two group sat on that same middle carbon. With me? Two methyl groups on a carbon carrying CONH two: that is two methylpropanamide, and its formula is C four H nine N O, as given. So the amide is two methylpropanamide, and the amine is propan two amine. Butanamide has the same formula, but it fails the clue. Got it?

7:56Quick revision: one carbon fewer, isocyanate, RCONH2 only

So, quick revision. An amide, RCONH two, with bromine and sodium hydroxide gives RNH two, a primary amine with one carbon fewer. It goes through the isocyanate, and R migrates, keeping its configuration. Only RCONH two reacts, and benzamide gives aniline. Now try the practice problems on this page. Count the carbons, and they will feel easy. All the best!