Low Reactivity towards Nucleophiles

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Chloroethane and chlorobenzene both have a carbon to chlorine bond, okay? Treat chloroethane with aqueous KOH, and the chlorine is replaced by O H, giving ethanol. But to do the same with chlorobenzene, you need sodium hydroxide at six hundred and twenty three kelvin, and three hundred atmospheres of pressure! So why is chlorobenzene so stubborn? That is strange, isn't it? This comes up often in JEE questions, so today we find out exactly why, and when the chlorine does leave. Let's begin.

0:37Reason 1: resonance

So, structure one, on the left, is chlorobenzene. The symbol C l is chlorine, and the hexagon with alternate double bonds is the benzene ring. Chlorine has lone pairs, right next to the double bonds of the ring. So tell me, what can one of those lone pairs do? It can spread into the ring, yes, exactly! One lone pair is delocalised into the ring, giving structures two, three and four. In each, chlorine carries a plus sign, written high, with a double bond to the ring, and a minus sign sits on an ortho or a para carbon. Structure five, on the right, is chlorobenzene again, with the ring's double bonds drawn the other way. The Roman numerals below simply number the structures, and the double-headed arrows mean resonance. With me? So in the real molecule, the carbon to chlorine bond, written C dash Cl on screen, has partial double-bond character. And a bond that is partly double is hard to break, right? That is reason number one.

1:37Reason 2: hybridisation

Now reason number two, hybridisation. In chlorobenzene, the carbon carrying the chlorine is a ring carbon, s p two hybridised; the small two is written high. In a chloroalkane, that carbon is s p three. Clear? So which carbon holds the bonding pair more tightly, the s p two or the s p three? It is the s p two carbon, correct, because it has more s character, so it holds the bond pair more tightly. The double arrow on screen means so, or therefore. Okay? And the bond lengths show it. The carbon to chlorine bond is one hundred and sixty nine picometres in chlorobenzene, and one hundred and seventy seven picometres in a chloroalkane; p m stands for picometre. So it is shorter, held more tightly, and harder to break. Got it?

2:27Why neither SN1 nor SN2 works for chlorobenzene

Now the two routes we learnt for alkyl halides, S N one and S N two, with the N written low. Try S N one first. The chlorine would leave on its own, giving a cation on the ring, the phenyl cation. Will that cation be stable enough to form? No, it will not, very good! The phenyl cation is not stabilised by resonance, so it does not form, and S N one is ruled out. Right? Now S N two, where the nucleophile attacks the carbon from the back. There are two problems here. First, the electron-rich ring repels the approaching nucleophile. Okay? And second, the back of that carbon lies inside the ring itself, so the nucleophile cannot get there. So S N two is ruled out too. Four reasons, and both routes are blocked.

3:15When the chlorine does leave

With me? But under harsh conditions, the chlorine does leave, okay? This is Dow's process: chlorobenzene with aqueous sodium hydroxide, N a O H, at six hundred and twenty three kelvin, capital K, and three hundred atmospheres, written a t m. The arrow means gives. We get sodium phenoxide, and then acid gives phenol. Right? Now the beautiful part. Put a nitro group, N O two, with the small two written low, para to the chlorine. Will the replacement get harder, or easier? It gets much easier, yes, exactly! Four nitrochlorobenzene, with the nitro group at position four, needs sodium hydroxide at only four hundred and forty three kelvin. Okay? With two nitro groups, at positions two and four, so ortho and para to the chlorine, the temperature drops to three hundred and sixty eight kelvin. Clear? And with three nitro groups, in two, four, six trinitrochlorobenzene, warm water alone does the job! That is the power of electron-withdrawing groups at ortho and para. With me? So any nitro group helps, and a nitro group at the meta position should help too, right? No, no, careful, that is exactly the trap, because a nitro group at the meta position has almost no effect. The position matters, not just the number of nitro groups. To see why, let us look at the mechanism.

4:36SNAr: addition-elimination via the Meisenheimer complex

Here is the mechanism, addition elimination, or SNA r, where A r stands for aromatic. On the left, four nitrochlorobenzene meets the hydroxide ion, O H minus, with the minus written high. Step one, marked slow over the first arrow: the hydroxide adds to the carbon carrying the chlorine. Okay? This gives a carbanion, called the Meisenheimer complex. The attacked carbon now holds both the chlorine and the O H group, so it is s p three. Right? Now the key question. This carbanion carries a negative charge. On which ring carbons can that charge sit? On the carbons ortho and para to the point of attack, exactly! In the figure, it sits on the para carbon, which holds the nitro group. With me? Now the second structure, after the double-headed arrow. When the nitro group is ortho or para, it takes up the charge. In the figure, the nitrogen gets a plus sign, and its oxygens carry the minus signs. That stabilises the carbanion. From the meta position, as the figure says, it cannot do this, and that is why meta has almost no effect. Clear? Step two, marked fast over the last arrow: the chloride ion, C l minus, leaves, and the ring becomes aromatic again. The product, on the right, is four nitrophenol. Got it?

6:00Example 1: two reasons chlorobenzene resists aqueous KOH

Now an easy example from the course. Give two reasons why chlorobenzene is less reactive than chloroethane towards aqueous KOH, which is potassium hydroxide. Pause the video and try it first. Done? Okay, let's check. Reason one is resonance: one lone pair of chlorine spreads into the ring, so the carbon to chlorine bond gets partial double-bond character. Right? Reason two is hybridisation. The s p two ring carbon of chlorobenzene holds the bond pair more firmly than the s p three carbon of chloroethane. And that solves our opening puzzle: the same bond on paper, but a very different bond in reality. Okay?

6:42Example 2: ease of Cl replacement, nitro vs methyl at para

Now a harder one. Arrange these in order of increasing ease of replacement of chlorine by hydroxide: four chlorotoluene, chlorobenzene, and four nitrochlorobenzene. Here, four chlorotoluene has a methyl group para to the chlorine. Pause the video and try it first. Done? Okay, let's check. In four nitrochlorobenzene, the nitro group at para stabilises the carbanion, so its chlorine is replaced most easily. Okay? Now four chlorotoluene. Its methyl group also sits at the para position, the special position. So it should help as well, right? No, no, wait, it works the other way round, because methyl, C H three, is an electron-releasing group. It pushes electrons into a ring that is already negative, so it destabilises the carbanion. That makes four chlorotoluene the hardest of the three. Clear? So the order is four chlorotoluene, less than chlorobenzene, less than four nitrochlorobenzene; the angled sign is read as less than. The kind of group and its position both matter. Got it?

7:49Quick revision: four reasons, Dow's process, ortho-para NO2

So, quick revision. Aryl halides, like chlorobenzene, resist nucleophilic substitution for four reasons: resonance, the s p two carbon, no S N one, and no S N two. The halogen leaves under harsh conditions, as in Dow's process, or with nitro groups at ortho or para. The mechanism is addition, then elimination, through a carbanion, and a meta nitro group does not help. Now go and try the practice problems on this page. All the best!