Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle from Set two. Take ethylbenzene, a benzene ring carrying an ethyl group, and add chlorine. In the dark, with F e C l three, the chlorine goes into the ring. But in light, it goes into the side chain instead. Same two chemicals, two different products. So how does the chlorine decide? The answer is in the conditions. Questions like this come up often in JEE Main, so let's begin.

0:31Benzene: substitution, not addition

So, first, the big picture. This map from the course has benzene in the centre, C six H six; the small sixes written low are subscripts. On the left are ways of making it, and on the right, its reactions. The top five boxes are bracketed as electrophilic substitution, and the next two as addition. Today we need only the two chlorine boxes. Okay? Now, in benzene, the six pi electrons, pi is the Greek letter, are delocalised over all six carbons of the ring. That makes benzene extra stable. Right? Alkenes add very easily, so let me ask you this: why does benzene avoid addition? Yes, exactly! Addition would destroy this delocalised ring of electrons. So benzene keeps its ring and reacts by substitution instead. Clear? Now, chlorine, C l two. With F e C l three, and the short form before it means anhydrous, completely dry, chlorine substitutes in the ring: one hydrogen is replaced, and we get chlorobenzene, C six H five C l. That is the second box from the top of the map. Got it? But change the conditions to ultraviolet light, U V, at five hundred kelvin, the capital K on screen. Now three molecules of chlorine add, and we get C six H six C l six, benzene hexachloride, or BHC for short. That is the lower addition box, so the conditions decide. With me?

2:04Where does chlorine enter the ring?

Now the next question: when the ring already carries a group, where does the chlorine go? Look at the first drawing, phenol, with O H on top. The two carbons next to the group are ortho, the two marked with a small m are meta, and the carbon opposite is para. Okay? The O H pushes electrons into the ring; the figure calls this plus R, the resonance effect. The delta minus signs, delta the Greek letter, show a small negative charge at ortho and para. Now, an electrophile is hungry for electrons, so where will it go? To the electron-rich carbons, ortho and para, just as the label under the drawing says. Right? So, in our first line, a plus R donor makes ortho and para electron-rich, and it is ortho, para-directing, the small o and small p on screen. The other two drawings, nitro with delta plus and chlorine with minus I, show groups that pull electrons; we skip them today. With me? Now, our ethyl group is an alkyl group, and alkyl groups also release electrons into the ring, so they too are ortho, para-directing, and activating. Got it? Now a small trap. There are two ortho positions, but only one para position. So the ortho product should be the major one, right? Oh no, careful here, because para usually predominates over ortho. Ortho sits right next to the group and is crowded; para is far away. Clear?

3:36Ring or side chain? The conditions decide

Now the real puzzle: ring or side chain? With a Lewis acid such as F e C l three, in the dark, C l two plus F e C l three gives C l plus, the small plus for its positive charge, and F e C l four minus, in square brackets; the arrow means gives. C l plus is an electrophile, so it attacks the ring. Okay? Now switch on the light instead. What do you think light does to a chlorine molecule? Yes, it splits it into chlorine atoms, free radicals, and heat does the same. With a side chain present, the halogen goes to the benzylic position, the side-chain carbon joined directly to the ring. Right? For example, toluene, with a methyl side chain, gives benzyl chloride, then benzal chloride, then benzotrichloride. Every chlorine goes to the side chain. With me? And why the benzylic carbon? Because the benzylic radical is stabilised by the ring. Clear? So, one line to remember, with X two standing for any halogen. X two with light goes to the side chain, and with a Lewis acid, to the ring. Got it?

4:43Example 1: mass of BHC from 7.8 g of benzene in UV light

Now, Example one, from Set two. What mass of BHC is formed from seven point eight grams of benzene and excess chlorine in ultraviolet light? Pause the video and try it first. Done? Okay, let's check. First, the product. Benzene prefers substitution, so benzene and chlorine give chlorobenzene, right? Oh no, careful, look at the conditions: ultraviolet light! In U V light, three molecules of chlorine add. So C six H six plus three C l two gives C six H six C l six, benzene hexachloride, and all six hydrogens stay. Okay? Now the moles of benzene. Its molar mass, capital M, is six times twelve, plus six times one, which gives seventy eight; g mol to the power minus one means grams per mole. So seven point eight over seventy eight gives zero point one mole. Right? Next, the molar mass of BHC, which is benzene plus six chlorine atoms: seventy eight, plus six times thirty five point five. Six times thirty five point five is two hundred and thirteen, and seventy eight plus two hundred and thirteen gives two hundred and ninety one grams per mole. With me? And one mole of benzene gives one mole of BHC, one to one. So zero point one times two hundred and ninety one gives twenty nine point one grams, and that is our answer!

6:15Example 2: ethylbenzene and Cl2, with FeCl3 or in light

And now, Example two, our puzzle from the start. Ethylbenzene reacts with chlorine, part A, with F e C l three in the dark, and part B, in light; the letters in brackets mark the parts. Name the major product in each case. Pause the video and try it first. Done? Okay, let's check. Ethylbenzene is C six H five, C H two, C H three; the dashes are single bonds. So the ring carries an ethyl group, an alkyl group. Okay? Part A first. F e C l three in the dark gives C l plus, an electrophile, so it attacks the ring. And the ethyl group sends it ortho or para. Right? So which product is the major one, ortho or para? Para, yes, exactly, because para is less crowded. The chlorine and the ethyl group sit opposite each other, on carbons one and four, so the name is one chloro four ethylbenzene. Got it? Now part B, in light. The side chain has two carbons: the C H two joined to the ring, and the C H three at the end. Which carbon gets the chlorine? The C H two, the benzylic carbon, that is correct! Light gives chlorine atoms, and the benzylic radical is stabilised by the ring. Clear? So the product is C six H five, CHC l, C H three, named one chloroethyl benzene, with one chloroethyl in brackets. And see, our puzzle is solved: with a Lewis acid in the dark, the ring; in light, the side chain!

7:58Quick revision: substitution, addition, ring or side chain

So, quick revision. First, benzene substitutes with chlorine and F e C l three, but in U V light it adds, giving BHC, benzene hexachloride. Second, alkyl groups direct ortho and para, and para usually predominates. Third, a halogen with light goes to the side chain, at the benzylic carbon; with a Lewis acid, it goes to the ring. Now open Set two, give yourself twenty five minutes, and try all fifteen problems, then check the solutions. All the best!