JEE Advanced and IIT-JEE Questions (2000–2016): Diazonium Salts

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle from JEE Advanced, twenty sixteen. Aniline is treated with five reagents, one after another, like runners in a relay race, and the last runner is copper with H B r. So what comes out at the end of the race? Today we solve it, along with other past JEE questions on diazonium salts. They come up often, and all of them use one small toolkit. So, let's begin.

0:30Every route starts here: diazotisation

So, every route starts on the left of the course's map. We take a primary aromatic amine, A r N H two, where A r is an aryl group, like phenyl, and add sodium nitrite, N a N O two, with hydrochloric acid, at two seventy three to two seventy eight kelvin. Small numbers written low count atoms, the slash means with, and capital K is kelvin. Okay? This gives the diazonium salt, A r N two plus, C l minus, where the raised signs are the charges on the ions. It is stable only in cold solution, so we use it at once. Right? Some papers write zero degrees Celsius instead, where the small raised circle means degrees. That is about two seventy three kelvin, and the wavy equals sign means about, so it is the same ice-cold condition. Got it? And one rule you must never forget. Diazotisation needs a primary aromatic amine, with the N H two directly on the benzene ring. The ring stabilises the diazonium ion by resonance; an alkanediazonium ion has no such help and loses nitrogen at once. With me?

1:36Replacement: nitrogen gas leaves the ring

Now the right side of the map. Nitrogen gas, N two, leaves the ring, and a new group takes its place. With copper salts, C u C l with H C l, C u B r with H B r, or C u C N with KCN, this is the Sandmeyer reaction. Clear? Now, the twenty sixteen relay ends with copper metal and hydrobromic acid, C u and H B r. What will that do to a diazonium salt? It puts a bromine atom where the nitrogen was, very good! But be careful with the name, because copper powder with H C l or H B r is the Gattermann reaction, not Sandmeyer. Same product, different reagent. Okay? Next, hypophosphorous acid, H three P O two, or ethanol, replaces the diazonium group, N two plus, by plain hydrogen, giving A r H, so the old amino group is simply removed from the ring. Got it? This is the big idea behind most past questions. The amino group first directs a new group where we want it, and then it is replaced, or removed completely. Clear?

2:41Coupling: nitrogen is kept

Now the other family, coupling, shown in the figure. The diazonium ion is a weak electrophile, so it attacks only strongly activated rings, like phenols and arylamines, usually at the para position. Phenol couples in weakly alkaline solution, the O H minus above the arrow, and aniline in weakly acidic solution, the H plus. With me? The product joins the two rings through N double bond N, the dashes being bonds. It is an azo compound, and the prime on the second A r means a different aryl group. So in coupling, the nitrogen is kept. Okay? Now a special partner, two naphthol, a phenol with two fused rings. In alkali, it couples at C one, the carbon numbered one, right beside the carbon with the O H group. Clear? Here is JEE Advanced twenty fifteen. Aniline is diazotised to give V, and the arrow means gives. Then V, with two naphthol and sodium hydroxide, gives W, the product we want. So, what is W here? Yes, exactly, W is one phenylazo two naphthol. The brackets simply keep the phenylazo part of the name together. The phenylazo group, a benzene ring with its N double bond N, sits on carbon one, and the compound is an orange red dye. Got it? Now, IITJEE two thousand and eight said this coupling gives a dark blue precipitate, coloured because of extended conjugation. Is the dark blue part true or false? It is false, because the dye is orange red, not dark blue. But the reason is true: N double bond N joins both rings into one long conjugated system, and that extended conjugation gives the colour. Right?

4:27Example 1: which amine gives the azo dye test (2011)

Time for Example one, from IITJEE twenty eleven. There are four amines: N N dimethylaniline, N methylaniline, four methylaniline, and benzylamine. Which one forms a brilliantly coloured dye with sodium nitrite in dilute H C l, then alkaline two naphthol? Pause the video and try it first. Done? Okay, let's check. This is the azo dye test, and only a primary aromatic amine gives it, with the N H two directly on the ring. Okay? First, N N dimethylaniline, where the italic letters N say both methyl groups sit on the nitrogen. With the ring, that makes it tertiary, with no N H bond at all, so it cannot be diazotised. Right? Next, N methylaniline has one methyl group and the ring on its nitrogen, so it is secondary, and with nitrous acid it gives a yellow oily nitrosoamine, not a dye. With me? Now benzylamine has an N H two group, so it is a primary amine. So it gives the dye too, right? No, it does not, and that is exactly the trap here. Look at the formula on screen: its N H two sits on a C H two group, not on the ring. So its diazonium ion has no ring to stabilise it, and loses nitrogen at once. Clear? That leaves four methylaniline, with its N H two directly on the ring. It is diazotised in the cold and couples with two naphthol, so four methylaniline forms the dye. Got it?

6:02Example 2: aniline to 1,4-dibromobenzene in five steps

And now, our relay race from JEE Advanced twenty sixteen. The small Roman numerals number the five reagents. They are ethanoic anhydride with pyridine; potassium bromate, K B r O three, with H B r; then H three O plus, which is aqueous acid, with heat; then sodium nitrite and H C l at two seventy three to two seventy eight kelvin; and finally copper with H B r. What is the major product? Pause the video and try it first. Done? Okay, let's check. But first, why does the relay begin with ethanoic anhydride? Why not brominate aniline directly? Because the N H two group activates the ring so strongly that bromine water gives two four six tribromoaniline at once. So reagent one acetylates it, with pyridine as the base, to give acetanilide, the formula on screen, with an acetyl group, COCH three, on the nitrogen. Its nitrogen's lone pair is now shared with the C O group, so the ring is only moderately activated. With me? Reagent two now brominates acetanilide, and the group on the ring still directs ortho and para. So where does the bromine mainly go? It goes mainly to the para position, opposite the nitrogen, yes, exactly! We get para bromoacetanilide, where the italic p means para. Okay? Reagent three hydrolyses the amide and gives back the free amine, four bromoaniline. The acetyl group has done its job. Right? Reagent four is our cold diazotisation, and this primary aromatic amine gives four bromobenzenediazonium chloride. Got it? Reagent five, copper with H B r, is the Gattermann reaction, so nitrogen gas leaves, and a second bromine atom takes the place of the diazonium group. Clear? So the two bromines sit at positions one and four, opposite each other, and the major product is one four dibromobenzene. Five reagents, and every one came from our small toolkit. Okay?

8:09Quick revision: diazotisation, replacement, coupling

So, quick revision. Diazotisation works for primary aromatic amines only, at two seventy three to two seventy eight kelvin. Replacement reactions, like Sandmeyer and Gattermann, lose nitrogen, while coupling keeps it. And protect the amino group as acetanilide for clean para substitution. Now try the other past questions on this page. All the best!