From Hydrocarbons

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Pass chlorine into toluene, which is benzene with a methyl group, in sunlight, and the chlorine goes onto the methyl group. Now do it in the dark, with a Lewis acid like ferric chloride, and the chlorine goes onto the ring! Same two chemicals, two different places. So how does chlorine decide where to go? By the end of this video, you will predict it every time. Such questions come up often in JEE, so let's begin.

0:31Alkanes: free-radical substitution

So, let us start with alkanes. On screen, R H is an alkane, R being the alkyl group, and X two is a halogen; the small two written low means two atoms, and the small dash is a bond. With light or heat, written over the arrow, a halogen atom replaces one hydrogen, giving R X, the alkyl halide, along with H X, a hydrogen halide. It goes through free radicals, so it is called free radical substitution. With me? Now let me ask you this. Is this a good way to make one pure alkyl halide? Not really, and here is why. The halogen does not stop after one hydrogen, so we get monohalides and polyhalides, and it attacks different carbons, so we get position isomers too. Their boiling points are close, so they are hard to separate. Right? So this method is of limited use. One fix helps: a large excess of the alkane favours the monohalide. Clear?

1:31Allylic and benzylic positions are selective

Now, some positions are halogenated selectively. Pass propene, which has a double bond, with chlorine, C l two, at seven hundred and seventy three kelvin; capital K is kelvin, and the plus sign means together with. With a double bond present, do we get addition, or substitution? It is substitution, very good! At this high temperature, substitution at the allylic carbon, the one next to the double bond, is preferred to addition. We get C H two, double bond, C H, single bond, C H two C l; the double line is the double bond. This is allyl chloride. Okay? For bromine, there is a special reagent, NBS, or N bromosuccinimide; the italic capital N means the bromine sits on the nitrogen. It brominates the allylic carbon, or the benzylic carbon, the one next to a benzene ring. Got it? And sulphuryl chloride, S O two C l two, also chlorinates by a radical path. Right?

2:35Toluene in light: the side chain

Now, back to toluene in light. Its methyl group is benzylic, so chlorine attacks there, giving C six H five C H two C l, benzyl chloride; C six H five is the phenyl group. If more chlorine keeps coming, what forms next? Another side chain hydrogen is replaced, yes! That gives C six H five CHC l two, benzal chloride. Okay? And then the last one: C six H five C C l three, benzotrichloride. In light, the chlorine stays on the side chain. With me?

3:13Alkenes: adding HX and halogens

Now, let us move to alkenes. H X, a hydrogen halide, adds across the double bond by Markovnikov's rule: the proton goes to the carbon with more hydrogens, so the more stable carbocation forms, a carbon with a positive charge. The halide then joins it. Okay? But H B r with a peroxide adds the other way round, because the mechanism is now a radical one, with no carbocation. Keep this in mind for Example one. Right? Next, an alkene with X two in carbon tetrachloride, C C l four, gives a vicinal dihalide, with the two halogen atoms on neighbouring carbons. Now, bromine is red-brown. What do you see when it adds to an alkene? The colour disappears, exactly! Bromine is used up as it adds, so this loss of colour is a test for unsaturation.

4:05Arenes: ring substitution in the dark

Got it? Now, let us take arenes, like benzene. We use chlorine or bromine with a Lewis acid, ferric chloride, F e C l three, ferric bromide, F e B r three, in the dark. Iron, F e, also works, because it reacts with the halogen to form the ferric halide. Okay? The Lewis acid supplies C l plus or B r plus; the small plus written high is a positive charge. This ion attacks the ring. So, toluene in the dark with a Lewis acid. Which ring isomers form? Ortho and para, yes! They separate easily, because their melting points differ widely. And our puzzle is solved: light sends chlorine to the side chain, and a Lewis acid in the dark sends it to the ring. Clear? Now, what about iodine? Benzene with iodine alone gives a poor yield of iodobenzene. Why is that? Because iodination is reversible, and the H I formed pushes it back! So we add an oxidising agent, nitric acid, HNO three, or iodic acid, HIO three, to destroy the H I as it forms. With me? And what about fluorine? Fluorination is too violent to control, so fluoro compounds are made indirectly. Got it?

5:27Example 1: HBr adds to propene with and without peroxide

Now, Example one, an easy one. Write the product of addition of H B r to propene, part a, without a peroxide, and part b, with a peroxide. Pause the video and try it first. Done? Okay, let's check. Here is propene: C H three, C H, double bond, C H two. Part a has no peroxide, so by Markovnikov's rule the proton adds to the carbon with more hydrogens. So which carbon gets the bromine? The middle carbon, yes, exactly! The proton goes to the end C H two, the more stable carbocation forms on the middle carbon, and bromide joins it. So we get C H three, CHB r, C H three, which is two bromopropane. Okay? Now part b, with a peroxide. Same H B r, same propene, so two bromopropane again, right? No, no, wait, that is the trap! With a peroxide, H B r adds by a radical mechanism, the other way round. So the bromine goes to the end carbon: C H three, C H two, C H two B r, which is one bromopropane. Clear?

6:45Example 2: chlorinating p-xylene in sunlight and with FeCl3

Now, Example two, a little harder. One four dimethylbenzene, or para xylene, reacts with one equivalent of chlorine, part a, in sunlight, and part b, in the dark with ferric chloride. Name the product in each case. Pause the video and try it first. Done? Okay, let's check. The small italic p on screen means para: two methyl side chains on the ring, at positions one and four. Part a is in sunlight, which gives chlorine atoms. Do they attack the ring, or a side chain? It is a side chain, exactly, just like toluene! With one equivalent of chlorine, one methyl hydrogen is replaced, giving C H three, C six H four, C H two C l, which is four methylbenzyl chloride. Okay? Now part b, in the dark with ferric chloride. The Lewis acid gives C l plus, and this ion attacks the ring, not the side chain. Right? All four free ring positions are alike, so the chlorine enters at position two, next to a methyl group, giving two chloro one four dimethylbenzene. So light gave the side chain, and the Lewis acid gave the ring. Got it?

8:01Quick revision: halogenating alkanes, alkenes and arenes

So, quick revision. Alkanes in light give radical mixtures, but allylic and benzylic positions are halogenated selectively. Alkenes add H X by Markovnikov's rule, H B r with a peroxide adds the other way, and halogens give vicinal dihalides. And arenes with a Lewis acid in the dark are substituted on the ring, while light attacks the side chain. Now go and try the practice problems on this page. All the best!