AIEEE and Earlier JEE Main Questions: Diazonium Salts

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. One old exam question says, keep the reaction at two hundred and seventy eight kelvin. Another says, keep it at zero to five degrees Celsius. So, are these two different conditions, or the same condition written in two ways? Keep that puzzle in mind; we will settle it today. We solve two past questions, from A I triple E two thousand and ten and JEE Main two thousand and fifteen, where the products hide behind letters, and you must name each one. Questions like these do appear in JEE Main, and each of these two needs just two moves. So, let's begin.

0:42Move one: diazotisation, always cold

So, here is the map of this chapter. The box on the far left, aniline with sodium nitrite and hydrochloric acid, is diazotisation, move one. On screen, A r N H two is the general starting compound. Here, A r stands for an aryl group, such as a benzene ring, and the small numbers written low are subscripts. It must be a primary aromatic amine, with the N H two group directly on the ring. Okay? The reagents are sodium nitrite, N a N O two, and H C l; the slash means together with. The temperature is two seventy three to two seventy eight kelvin, where capital K is kelvin and the dash means to. Right? Now, back to our puzzle. One question said two hundred and seventy eight kelvin, and the other said zero to five degrees Celsius. So tell me, is that the same condition, or a different one? Yes, exactly, it is the same range. The small raised circle with capital C means degrees Celsius, and the wavy equals sign means approximately equal to. To change Celsius into kelvin, add about two hundred and seventy three. So zero degrees Celsius is about two hundred and seventy three kelvin, and five degrees Celsius is about two hundred and seventy eight kelvin. Same ice-cold condition, two different units. Clear? The product is the diazonium salt, A r N two plus, C l minus. The plus written high is the positive charge on the N two group, and C l minus is the chloride ion. From aniline itself, this is benzenediazonium chloride, the big box in the middle of the map. With me? And here is why we work so cold. The salt is stable only in cold solution, as the map says under the first box. So we never store it; we make it, and we use it at once in the next step. Got it?

2:31Move two: replace the N2+ group by CN or F

Now move two, on the right of the map. The top eight arrows are replacement reactions. In each of them, the diazonium group leaves the ring as nitrogen gas, N two, and a new group takes its place. Okay? Our second question needs the cyanide box, second from the top. Cuprous cyanide, C u C N, with potassium cyanide, KCN, gives the aryl cyanide, A r C N; the arrow means gives. This is a Sandmeyer reaction. Right? Our first question needs the fluorine box, fifth from the top. Add HBF four, fluoroboric acid, and we get the diazonium fluoroborate, A r N two plus, B F four minus. Unlike the chloride, this solid is stable at room temperature, as the line on screen says. With me? Then we heat the solid. It gives the aryl fluoride, A r F, plus boron trifluoride, B F three, plus nitrogen gas. This is the Balz Schiemann reaction. Clear? Now let me ask you this. Suppose the ring also carries a methyl group, C H three. When the diazonium group is replaced, what happens to that methyl group? Nothing happens to it at all; it stays exactly where it was. Only the N two plus group is replaced, and the rest of the aryl group, A r, comes through unchanged. So a group at position four stays at position four. Got it?

3:57Example 1: aniline to fluorobenzene by Balz-Schiemann

Time for Example one, the easier one, from A I triple E two thousand and ten. Aniline is treated with sodium nitrite and H C l at two hundred and seventy eight kelvin to give A, and A with HBF four gives B, on heating. So, what are A and B, the two products? Pause the video and try it first. Done? Okay, let's check. Two hundred and seventy eight kelvin is about five degrees Celsius, our cold range from the puzzle. Aniline is a primary aromatic amine, and with sodium nitrite and H C l in the cold, that means diazotisation. Right? So compound A is benzenediazonium chloride, C six H five N two plus, C l minus. Here C six H five is the phenyl group, the A r of our general formula. Okay? Next comes fluoroboric acid. It turns compound A into the solid fluoroborate, C six H five N two plus, B F four minus, which is stable at room temperature. With me? So the fluoroborate must be compound B, right? Not quite, and this is the trap, because the question says that B forms on heating. When we heat the solid fluoroborate, it breaks up into fluorobenzene, C six H five F, plus boron trifluoride and nitrogen gas. Clear? So compound B is fluorobenzene, made by the Balz Schiemann reaction, exactly as on our map. If you named both products correctly, very good!

5:28Example 2: 4-methylaniline to 4-methylbenzonitrile (CuCN)

Now Example two, a little harder, from JEE Main two thousand and fifteen. Four methylaniline is treated with sodium nitrite and H C l, at zero to five degrees Celsius, to give D, and D is then heated with C u C N and KCN to give E and nitrogen. The four in the name is a position: the methyl group is on carbon four, para to the amino group on carbon one. So, can you name E, the final product? Pause the video and try it first. Done? Okay, let's check. Zero to five degrees Celsius is two seventy three to two seventy eight kelvin, the same cold range as in Example one. So our puzzle is settled: both questions use the same condition. Four methylaniline is a primary aromatic amine, so this is diazotisation. Okay? So D is four methylbenzenediazonium chloride. The formula on screen shows the benzene ring carrying the methyl group, with the N two plus group where the amino group used to be. Right? Next, cuprous cyanide with potassium cyanide. That is the Sandmeyer reaction from our map, so the N two plus group is replaced by C N, the cyano group. With me? Now here is a doubt many students have. Compound E still contains a nitrogen atom. So did one nitrogen of the diazonium group stay behind on the ring? No, it did not, and the question itself tells you so, because it says E and nitrogen. Both nitrogen atoms of the diazonium group leave together as nitrogen gas, N two. The C N group comes from the cyanide reagent. Clear? And the methyl group stays exactly where it was, at position four. So E is four methylbenzonitrile, with the methyl group still para to the cyano group. Two moves, cold diazotisation and then the replacement, and the question is done. Got it?

7:25Quick revision: diazotisation at 0-5 C, then CuCN or HBF4

So, quick revision. Move one: a primary aromatic amine with sodium nitrite and H C l, at two seventy three to two seventy eight kelvin, which is zero to five degrees Celsius, gives the diazonium salt. Move two: cuprous cyanide with potassium cyanide gives the aryl cyanide, and fluoroboric acid, then heat, gives the aryl fluoride. Nitrogen gas leaves, and every other group on the ring keeps its place. Now go back to these past-paper questions on this page and solve them yourself. All the best!