Replacement of the Diazonium Group

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. In toluene, the methyl group sends an incoming bromine atom to the ortho and para positions. So how would you make a toluene with two bromine atoms, both meta to the methyl group? Looks impossible, right? By the end of this video, you will be able to do it. Today we see how the diazonium group is replaced by halogens, the cyano group, the hydroxyl group, the nitro group and even a plain hydrogen atom. Questions on this come up often in JEE, so let's begin.

0:35Copper reagents: Sandmeyer and Gattermann

So, here is a diazonium salt. The letters A r stand for an aryl group, such as a benzene ring, and the small numbers written low are subscripts. The plus written high is the positive charge on the N two group, and X minus is the negative ion, for example chloride. In almost every reaction today, this N two plus group is replaced, and it leaves as nitrogen gas, N two. Okay? First, the Sandmeyer reaction. Treat the salt with cuprous chloride, C u C l, together with H C l; the slash means together with, and the arrow means gives. We get the aryl chloride. In the same way, cuprous bromide with H B r gives the aryl bromide. Right? And with cuprous cyanide, C u C N, and potassium cyanide, KCN, we get the aryl cyanide. That is also called a Sandmeyer reaction. With me? Now let me ask you this. What if we use plain copper powder with hydrochloric acid, instead of cuprous chloride? Is that still the Sandmeyer reaction? No, it has its own name, the Gattermann reaction. Copper powder, C u, with H C l or H B r, gives the aryl chloride or the aryl bromide. But the yield is lower than in the Sandmeyer reaction. So in naming questions, remember this: a cuprous salt means Sandmeyer, and copper powder means Gattermann. Clear?

2:04Iodine and fluorine

Next, iodine and fluorine. Let me ask you this: can we make fluorobenzene or iodobenzene conveniently, just by direct halogenation of benzene? Not conveniently, and that is exactly where the diazonium salt helps. The aryl iodides and the aryl fluorides are not made conveniently by direct halogenation, so we make them from the diazonium salt. Okay? For iodine, it is really simple. Warm the salt with aqueous K I, potassium iodide, and you get the aryl iodide. Got it? For fluorine, we use the Balz Schiemann reaction. Add HBF four, fluoroboric acid, and the diazonium fluoroborate comes out as a solid. Then heat the solid, and it gives the aryl fluoride, plus B F three, boron trifluoride, plus nitrogen gas. Clear?

2:55Hydroxyl, nitro and hydrogen

Now, the other groups. Warm the diazonium salt with water, H two O, and the hydroxyl group, O H, takes its place. We get a phenol. Right? For the nitro group, we first make the solid diazonium fluoroborate with HBF four, just as before. Then we heat it with sodium nitrite, N a N O two, and copper. We get the nitro compound, A r N O two. With me? Now here is something surprising. Hypophosphorous acid, H three P O two, with water, or simply ethanol, replaces the diazonium group by a plain hydrogen atom. Okay? Here is the full equation with hypophosphorous acid. The diazonium chloride, plus H three P O two, plus water, gives the arene, plus nitrogen gas, plus H three P O three, phosphorous acid, plus H C l. Got it? Now, with ethanol, the diazonium salt is reduced, so the ethanol itself must be oxidised. What does ethanol become? Yes, exactly, the ethanol is oxidised to ethanal, while the diazonium group is replaced by hydrogen. Okay?

4:01Nitrogen kept, and why this route matters

So far, the nitrogen always left as a gas. Now tell me, does it leave in every reaction of a diazonium salt? No, not always, and this is a common trap. With stannous chloride, S n C l two, and H C l, the salt is reduced, and both nitrogen atoms stay on. We get A r NHNH two, and when the aryl group is C six H five, this is phenylhydrazine. Got it? Now, here is why chemists like this route so much. It brings in fluorine, iodine, the cyano group, the hydroxyl group and the nitro group, where direct substitution fails or is not convenient. Right? Or where direct substitution does work, but puts the group in the wrong position, giving the wrong isomer. Clear? And here is a topper's trick. Since the amino group can be removed, by diazotising it and then using hypophosphorous acid, we can use it to direct other groups first, and then take it away. Hold on to this idea, because our puzzle needs it. With me?

5:03Example 1: benzenediazonium with CuBr, KI, warm water, H3PO2

Time for Example one. Name the product when benzenediazonium chloride reacts with, first, cuprous bromide and H B r; second, K I; third, warm water; and fourth, hypophosphorous acid. Pause the video and try it first. Done? Okay, let's check. In benzenediazonium chloride, the aryl group is C six H five. Cuprous bromide with H B r is the Sandmeyer reaction, so bromine replaces the diazonium group. We get bromobenzene, C six H five B r. Okay? With potassium iodide, iodine takes the place of the diazonium group. So we get iodobenzene, C six H five I, where I stands for iodine. Right? With warm water, the hydroxyl group comes in, so we get phenol, C six H five O H, clear? And finally, hypophosphorous acid replaces the diazonium group by hydrogen. So we are back to benzene itself, C six H six. Got it?

6:10Example 2: 4-methylaniline to 3,5-dibromotoluene

Now Example two, and this is our puzzle from the start. How is four methylaniline converted into three five dibromotoluene? The numbers three and five are the positions of the two bromine atoms, counted from the methyl group on carbon one. Pause the video and try it first. Done? Okay, let's see. Many students write this: diazotise, remove the amino group to get toluene, and then brominate the toluene twice. Does that give our product? That is exactly the trap, so be careful here. Direct bromination of toluene gives the ortho and para isomers, because the methyl group directs ortho and para. We need the bromines meta, so we must brominate while the amino group is still on the ring. Okay? So, first we add bromine water. The amino group sends two bromine atoms to the two positions ortho to itself, since para is already taken by the methyl group. We get two six dibromo four methylaniline. Right? Next, we diazotise with sodium nitrite and H C l, at two hundred and seventy three to two hundred and seventy eight kelvin; the capital K on screen stands for kelvin. The amino group becomes the diazonium group. With me? Now our topper's trick. Hypophosphorous acid replaces the diazonium group by hydrogen. The amino group has done its job of steering the bromines, and now it is gone. Clear? Last, let us check the positions, where C two means carbon two. In the amine, the amino group was on carbon one, the methyl group on carbon four, and the bromines on carbons two and six. Now number from the methyl group, and those bromines sit on carbons three and five, both meta to the methyl. So we get three five dibromotoluene, and our puzzle is solved. Got it?

8:03Quick revision: replacing N2+ by Cl, Br, CN, I, F, OH, H

So, quick revision. Sandmeyer uses cuprous chloride, bromide or cyanide, and Gattermann uses copper powder, with a lower yield. Potassium iodide gives the aryl iodide, fluoroboric acid and heat give the aryl fluoride, and warm water gives the phenol. Hypophosphorous acid or ethanol puts back a hydrogen, and stannous chloride with H C l keeps the nitrogen. Now go and try the practice problems on this page. All the best!