Aryl Halides from Diazonium Salts

Transcript

0:00Introduction

Hello everyone! Let us start with a plan that looks strange. You want a chlorine atom on a benzene ring. But instead of adding chlorine directly, the chemist first puts an amino group on the ring, and then throws that group away. Why take such a long road? Sounds like wasted effort, right? By the end of this video, you will see that the long road is the smart one. Today we make aryl halides from diazonium salts, and questions on this come up often in JEE, so let's begin.

0:32Step 1: Diazotisation

So, step one. We start with a primary aromatic amine. On screen, the letters A r stand for an aryl group, such as a benzene ring, and N H two is the amino group. The simplest example is aniline, C six H five N H two; the small numbers written low count the atoms. Okay? Now we treat the amine with N a N O two, sodium nitrite, and two moles of a mineral acid, H X, for example hydrochloric acid. The arrow means gives. We get the diazonium salt, A r N two plus X minus, then N a X, a sodium halide, and two molecules of water. With me? Here it is, the diazonium salt. The small plus sign written high is a positive charge on the N two group, and the minus sign is the negative charge on the halide ion. Making this salt is called diazotisation. Clear? And look at the temperature: two hundred and seventy three to two hundred and seventy eight kelvin, where capital K is kelvin. Very cold, right? Now let me ask you this: why must diazotisation be done this cold? Yes, exactly, because diazonium salts are unstable. Above about two hundred and seventy eight kelvin, the salt decomposes in water to a phenol, and nitrogen gas, N two, escapes. So, keep it cold. Got it?

2:01Step 2: Sandmeyer, Gattermann and KI replace the N2+ group

Now step two. The N two plus group is replaced, and it leaves the ring as nitrogen gas, N two. Every reaction on this screen works this way. Okay? First, the Sandmeyer reaction. Treat the diazonium salt with C u C l, cuprous chloride, together with H C l, and you get an aryl chloride, A r C l. The slash simply means together with. The same way, C u B r, cuprous bromide, with H B r gives an aryl bromide. Right? And with C u C N, cuprous cyanide, and KCN, potassium cyanide, the same reaction gives A r C N, an aryl cyanide. Not a halide, but remember it too. With me? Now a question for you. What if we use plain copper powder instead of the cuprous halide? Does it still work? It works, and 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. Clear? And for iodine, it is really simple. Just warm the diazonium salt with aqueous K I, potassium iodide, and you get the aryl iodide. No copper is needed at all.

3:20Fluorides: the Balz–Schiemann reaction

Got it? Now fluorine, which has its own reaction, the Balz Schiemann reaction. To the diazonium salt we add HBF four, fluoroboric acid. The salt with B F four minus as the negative ion comes out as a solid; it precipitates. Okay? Then we dry this salt and heat it. It breaks up into A r F, the aryl fluoride, plus B F three, boron trifluoride, plus nitrogen gas. Right? Now, back to our puzzle. Why does the chemist take the long road through the amino group, instead of adding the halogen directly? Because the halogen goes exactly where the amino group was, yes! Direct halogenation of a ring that already carries a group can give a mixture of ortho and para isomers. This route gives one definite position. With me? And this route is also the usual way to make aryl iodides and aryl fluorides. So the long road is the smart road. Clear?

4:17Example 1: volume of N2 at STP from 9.3 g of aniline

Time for Example one. Which gas is evolved in the Sandmeyer reaction? And find its volume at STP, standard temperature and pressure, from nine point three grams of aniline, converted completely into chlorobenzene. Pause the video and try it first. Done? Okay, let's check. The gas is the easy part. When the N two plus group is replaced by chlorine, it leaves as nitrogen gas, N two. Right? Now the moles of aniline. Capital M is the molar mass. Aniline has six carbons, seven hydrogens and one nitrogen. Six times twelve is seventy two, plus seven, plus fourteen, which gives ninety three grams per mole; the minus one written high means per mole. Okay? So, n, the number of moles, is nine point three over ninety three, which gives zero point one mole. With me? Now think carefully. Nitrogen gas has two nitrogen atoms in each molecule, and aniline has only one. So zero point one mole of aniline gives only zero point zero five mole of nitrogen gas, right? No, no, wait, that is exactly the trap! In diazotisation, one nitrogen atom came from aniline, and the other came from sodium nitrite, N a N O two. So one mole of aniline gives one mole of nitrogen gas. Clear? And one mole of a gas at STP occupies twenty two point four litres. So V, the volume, is zero point one times twenty two point four, which gives two point two four litres of nitrogen gas. Got it?

5:57Example 2: m-bromochlorobenzene from benzene

Now a harder one, Example two. How would you prepare meta bromochlorobenzene from benzene? The small italic m stands for meta: the bromine and the chlorine sit on carbons one and three. Pause the video and try it first. Done? Okay, let's see. Many students write this: chlorinate benzene first, then brominate it. Two steps and finished, right? Oh no, careful, that does not work! Chlorine on a ring directs the next group to the ortho and para positions, so you would get ortho and para bromochlorobenzene, not meta. But our bromine and chlorine must be meta, on carbons one and three, so a meta directing group must be present when one of the halogens goes in. Okay? So, first we nitrate the benzene with concentrated nitric and sulphuric acids, HNO three and H two S O four; the short form c o n c means concentrated. We get nitrobenzene, and the nitro group directs meta. Right? Next, we brominate with B r two and F e B r three, that is, bromine with ferric bromide. The nitro group sends the bromine to the meta position, giving meta bromonitrobenzene. With me? Then we reduce with S n and H C l, tin and hydrochloric acid. The nitro group becomes an amino group, so we get meta bromoaniline. Clear? Now our diazonium route begins. We diazotise with sodium nitrite and H C l, at two hundred and seventy three to two hundred and seventy eight kelvin, cold as always. With me? Finally, the Sandmeyer reaction with cuprous chloride and H C l. The chlorine goes exactly where the amino group was, meta to the bromine, so we get meta bromochlorobenzene. See, this is our puzzle again: the long road through the amino group put the chlorine exactly where we wanted it.

7:54Quick revision: diazotisation, Sandmeyer, Balz-Schiemann

Got it? So, quick revision. First, diazotise the aromatic amine with sodium nitrite and a mineral acid, at two hundred and seventy three to two hundred and seventy eight kelvin. For chlorides and bromides, use Sandmeyer with a cuprous halide, or Gattermann with copper powder, which gives a lower yield. For iodides, warm with potassium iodide; for fluorides, add fluoroboric acid, then heat the dry salt. Go and try the practice problems on this page. All the best!