Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Warm chloroethane with aqueous potassium hydroxide, and the chlorine is replaced: we get ethanol. Now take chlorobenzene, a chlorine on a benzene ring, and try the same thing, and nothing happens at all! To swap that chlorine for O H, it takes sodium hydroxide at six hundred and twenty three kelvin, and three hundred atmospheres. Same carbon chlorine bond, so why such a huge difference? Strange, right? Today, one map takes us through the whole chapter on haloarenes, and it solves this puzzle too. Questions on haloarenes come up often in JEE papers. So, let's begin.

0:45The chapter map: making chlorobenzene

So, here is our map. In the centre sits chlorobenzene, C six H five C l, where the small numbers written low count the atoms. Arrows on the left make it, and arrows on the right use it. The direct way first: benzene, C six H six, plus chlorine, C l two, with ferric chloride, F e C l three, a Lewis acid, in the dark. That gives substitution on the ring, and the product after the colon, chlorobenzene. Okay? The second way starts from benzenediazonium chloride, C six H five N two plus C l minus, where the plus and the minus written high are the charges. Treat it with cuprous chloride in hydrochloric acid, C u C l slash H C l, and nitrogen gas leaves while chlorine takes its place. This is the Sandmeyer reaction. With me? Use copper powder instead, C u slash H C l, and it is the Gattermann reaction: the same product, but a lower yield. Don't mix them up: cuprous chloride means Sandmeyer, and copper metal means Gattermann. Clear? Here, A r stands for an aryl group. For an aryl iodide, A r I, simply warm the diazonium salt with potassium iodide, K I, and no copper is needed. For an aryl fluoride, A r F, the map shows the Balz Schiemann reaction: the salt A r N two plus B F four minus is heated. So iodides and fluorides are usually made through the diazonium salt. Got it?

2:14Why the chlorine refuses to leave

Now back to our puzzle. This line is Dow's process: sodium hydroxide, N a O H, at six hundred and twenty three kelvin, written with a capital K, and three hundred atmospheres. Then H plus, an acid, gives phenol. Such harsh conditions for one chlorine! So let me ask you this. What makes the carbon chlorine bond in chlorobenzene so hard to break? The five structures in the figure give a hint. Yes, it is resonance! Chlorine pushes one of its lone pairs into the ring. In structures two, three and four, chlorine carries a plus and is joined to the ring by a double bond, while a minus appears on the ring. The double headed arrows mean resonance. So C, dash, C l, the carbon chlorine bond, is partly a double bond, and much harder to break. Okay? Second reason: the ring carbon is s p two hybridised, with the two written high. It has more s character than the s p three carbon of an alkyl halide, so it holds the bonding pair more tightly. The bond is shorter too: one hundred and sixty nine picometres, p m, against one hundred and seventy seven in an alkyl chloride. Right? Replacement becomes far easier with a nitro group, N O two, at the ortho or para position. But careful: a nitro group at meta gives almost no help. That is a common trap. Clear?

3:37Electrophiles: slower, yet ortho and para

Now the right side of the map, with electrophiles. Chlorine with ferric chloride gives ortho and para dichlorobenzene. On screen, ortho and para are written as an italic o and an italic p. Ortho is the position next to the chlorine, and para is directly opposite. Okay? Next comes nitration. The short form conc means concentrated, so these are concentrated nitric and sulphuric acids, HNO three slash H two S O four. It gives ortho and para nitrochlorobenzene. And the same pattern for sulphonation with concentrated sulphuric acid, and for C H three C l or C H three COC l with aluminium chloride, A l C l three. So again, we get ortho and para products. Right? Now here is the trap. Ortho and para directing sounds like an activating group, doesn't it? So tell me, with electrophiles, does chlorobenzene react faster or slower than benzene? The answer is slower! If you said faster, you just fell into the trap. So chlorine deactivates the ring. Minus I, the inductive effect, pulls electrons out of the whole ring. But plus R, resonance, pushes a lone pair into the ortho and para positions, so the electrophile attacks there. So chlorine is deactivating, and yet ortho and para directing. With me? And of the two, the para product is the major one. Got it?

5:00Metals: Wurtz–Fittig, Fittig and Grignard

Last, the bottom right of the map: metals. Sodium, N a, with an alkyl halide, R X, in dry ether: the aryl group and the alkyl group join, giving an alkylbenzene. This is the Wurtz Fittig reaction. Okay? Now leave out the alkyl halide, and with sodium in dry ether, two aryl groups join each other. C six H five, dash, C six H five is biphenyl, where the dash joins the two rings. This is the Fittig reaction. And magnesium, M g, in THF, the solvent tetrahydrofuran, turns chlorobenzene into C six H five M g C l, a Grignard reagent. Clear? Bromobenzene forms its Grignard reagent with magnesium in dry ether, but chlorobenzene needs THF, the higher boiling solvent. Okay?

5:45Example 1: para products of nitration, sulphonation of PhCl

Now let us use the map. This first example comes from the section on electrophilic substitution. Name the major product of, part a, nitration, and part b, sulphonation, of chlorobenzene. Pause the video and try it first. Done? Okay, let's check. Chlorine directs the electrophile to ortho and para, and the para product is the major one. Right? So in nitration, the nitro group goes para: one chloro four nitrobenzene. Chlorine is on carbon one, and nitro on carbon four, directly opposite. Okay? And in sulphonation, the sulphonic acid group goes para: four chlorobenzenesulphonic acid. Writing a meta product is the trap here, because chlorine directs ortho and para, not meta.

6:33Example 2: PhCl to toluene, benzoic acid, biphenyl, benzene

Clear? Now a harder one, from the section on reactions with metals. How would you convert chlorobenzene into, part a, toluene, part b, benzoic acid, part c, biphenyl, and part d, benzene? Here is a hint: start from the map. Pause the video and try it first. Done? Okay, let's check. Toluene is methylbenzene, so we must attach a methyl group, and the Wurtz Fittig arrow attaches an alkyl group. So chlorobenzene with chloromethane, C H three C l, and sodium in dry ether gives toluene. Okay? Now benzoic acid: no arrow on the map gives it. So which product on the map can we build on? Yes, it is the Grignard reagent! Magnesium in THF gives C six H five M g C l. Then carbon dioxide, C O two, followed by acid, gives benzoic acid. Right? Biphenyl comes straight from the map: sodium in dry ether, the Fittig reaction. And for benzene, make the same Grignard reagent, and then simply add water. The M g C l part is replaced by hydrogen, and we get benzene. See, one map, four conversions!

7:47Quick revision: making, nucleophiles, electrophiles, metals

So, quick revision. To make a haloarene, use the halogen, X two, with a Lewis acid in the dark, or go through the diazonium salt. Nucleophiles find it very hard, because of resonance and the s p two carbon, unless nitro groups sit at ortho or para. Electrophiles react slower than with benzene, at ortho and para, with para major. And metals give the Wurtz Fittig, Fittig and Grignard reactions. You now have the whole chapter on one map, and that is a great start. Now go and try the practice problems in the next sections of this chapter. All the best!