Conversions Through Nitrogen Compounds

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Two students want three bromoaniline from benzene, and both use the same three steps: nitrate, brominate, and reduce. The first student does them in exactly that order. The second one does nitrate, reduce, and then brominate. Only one of them gets three bromoaniline. Which one, and why? Keep that question in mind, we will answer it later. Today we plan conversions through nitrogen compounds: count the carbons, pick the reagent, and on a ring, choose the order. These questions come up often in JEE, so let's begin.

0:38Count the carbons: one carbon more

So, the first rule of every conversion: count the carbons. One carbon more, the same number, or one fewer? Each case has its own reagent. First, one carbon more. On screen, R X is an alkyl halide, with R an alkyl group and X a halogen. Potassium cyanide, KCN, above the arrow, turns it into RCN, a nitrile, and the C N brings in one new carbon. Okay? Now reduce the nitrile with L i A l H four, lithium aluminium hydride; the small numbers written low are subscripts. We get RCH two N H two, a primary amine, one carbon longer than the halide. Right? Or hydrolyse it with H three O plus, aqueous acid, the raised plus being its charge. That gives RCOOH, a carboxylic acid, again one carbon longer. So potassium cyanide is your step up. Got it?

1:37Count the carbons: the same number

Next, the same number of carbons. An acid, RCOOH, with ammonia, N H three, and heat, written above the arrow, gives an amide, RCONH two. No carbon is added, and none is lost. With me? Now reduce this amide with L i A l H four. Look at its C double bond O, the carbonyl group, written C equals O in the box on screen. Let me ask you this: does that carbon stay in the amine, or is it lost? Yes, exactly, it stays. The C O just becomes C H two, giving RCH two N H two, with no carbon lost. Clear? Also on this rung, the Gabriel phthalimide synthesis turns R X into RNH two, but it gives only primary aliphatic amines, never aniline. Okay? And nitrous acid, HNO two, turns a primary aliphatic amine into an alcohol, ROH, with the same carbons. Right?

2:43Count the carbons: one carbon fewer

Now the ladder going down. The amine must have one carbon fewer than the amide. Which reagent will you use? Yes, it is bromine with sodium hydroxide, written B r two slash N a O H, where the slash means together with. The amide gives RNH two: R is kept, but the C O carbon is gone. Okay? This is the Hofmann bromamide reaction, and it removes exactly one carbon. So remember the pair: L i A l H four on an amide keeps every carbon, bromine with sodium hydroxide removes one. Mixing them up is a common trap. Got it?

3:21On a benzene ring: in through nitrogen

Now the benzene ring. Here, A r means an aryl group, so A r H is an arene such as benzene. Nitration with HNO three slash H two S O four gives A r N O two. Then tin and hydrochloric acid, S n slash H C l, reduce it to A r N H two, an aromatic primary amine like aniline. With me? Next we turn this amine into a diazonium salt. So tell me, should this be done warm, or cold? Yes, it must be cold, two seventy three to two seventy eight kelvin, where the K stands for kelvin. Sodium nitrite, N a N O two, and H C l give the diazonium salt, A r N two plus; the raised plus is its charge. Then its N two plus group can be replaced by another group. Okay? And one smart idea: the N H two group can direct a new group into the ring, and then be removed through the diazonium salt. So on a ring, plan which group does the directing. Right?

4:26A primary amine to its N-methyl derivative

One more conversion: exactly one methyl group on the nitrogen of a primary amine. The italic N means the methyl sits on nitrogen. So why not just methylate the amine directly? Because direct methylation gives a mixture: the product reacts again, and the tertiary amine forms too. Clear? So, chloroform, CHC l three, with alcoholic KOH, a l c being short for alcoholic. This carbylamine reaction turns RNH two into RNC, an isocyanide. With me? Then L i A l H four reduces the isocyanide to RNHCH three, the N methyl amine, with no tertiary amine mixed in. Got it?

5:11Example 1: ethanoic acid to methanamine, one carbon fewer

Now Example one, an easy one. How is ethanoic acid converted into methanamine? Count the carbons, then pause the video and try it. Done? Okay, let's check. Ethanoic acid, C H three COOH, has two carbons, written two C in brackets, and methanamine, C H three N H two, has one. So one carbon must be lost. Right? Many students write this: make the amide, then reduce it with L i A l H four. Will that give methanamine? No, and that is exactly the trap. L i A l H four keeps every carbon, so it gives ethanamine, with two carbons. Clear? So first, ammonia and heat turn ethanoic acid into ethanamide, C H three CONH two, still two carbons. Okay? Then the Hofmann bromamide reaction, bromine with sodium hydroxide, removes the carbonyl carbon and gives methanamine, C H three N H two. That is two steps, and we are done. With me?

6:20Example 2: benzene to 3-bromoaniline, nitro group first

And now Example two, a harder one, and it answers our puzzle. How is benzene converted into three bromoaniline? Here is a hint: which group should direct the bromine? Pause the video and try it first. Done? Okay, let's check. In three bromoaniline, N H two is on carbon one and bromine on carbon three, so they are meta to each other. Okay? Now the easy plan: nitrate, reduce to aniline, then brominate, the order from our puzzle. Will that give the meta product? No, and that is the trap. In aniline, the N H two group directs ortho and para, so the bromine goes ortho and para, not meta. Right? So first, nitrate benzene, C six H six, with HNO three slash H two S O four, to get nitrobenzene, C six H five N O two. Clear? Now brominate straight away, with B r two slash F e B r three, bromine with iron three bromide. The nitro group directs meta, so bromine enters at the meta position, giving three bromonitrobenzene. Got it? Finally, tin and hydrochloric acid, S n slash H C l, reduce the nitro group to N H two, giving three bromoaniline. So our puzzle is solved: brominate before you reduce, while the meta directing nitro group is still there. The order decides the product. Okay?

7:52Quick revision: one carbon more, same, fewer; ring order

So, quick revision. One carbon more: potassium cyanide, then L i A l H four or acid. Same carbons: L i A l H four on the amide; one carbon fewer: bromine with sodium hydroxide. On a ring, go through nitro, amine and diazonium, and plan the order. Now try the practice problems on this page. Count the carbons first, and they will feel easy. All the best!