AIEEE and Earlier JEE Main Questions: Haloarenes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two bottles in a lab have lost their labels. One holds iodobenzene, and the other holds benzyl iodide. Both are iodides with a benzene ring. With sodium hydroxide, nitric acid and silver nitrate, can you tell which is which? Surely two iodides behave the same way, right? Hold that thought, and we will come back to it. Today we solve older A I triple E and JEE Main questions on haloarenes, and questions like these come up often. Let's begin.

0:36Why a ring halogen stays put

So, first, a quick recap from this chapter. In an aryl halide, the halogen, X, sits on a ring carbon, which is s p two hybridised, with the small two written high. The figure on the right shows resonance in chlorobenzene, and we will read it in a moment. Okay? Now let me ask you this. Will this ring halogen leave as easily as the chlorine of chloroethane? No, it holds on tightly! Look at the figure. Structure one, marked with a Roman numeral, is chlorobenzene. One lone pair of chlorine spreads into the ring, giving structures two, three and four: a plus sign on chlorine, a double bond from chlorine to the ring, and a minus sign on an ortho or para carbon. Structure five is chlorobenzene again, and the double-headed arrows mean resonance. So the C X bond, written C dash X, is partly double and hard to break. Right? Next, S N one, with the N written low. It would need a phenyl cation, which is not stabilised by resonance, so S N one is ruled out. Clear? And S N two? The electron-rich ring repels N u minus, the nucleophile, with the minus sign written high, and the back of the carbon lies inside the ring, so the back side is blocked. So S N two is ruled out too. With me?

1:54Ring halogen versus side-chain halogen

Now the idea behind every question on this page. In chlorobenzene, C six H five, dash, C l, the chlorine sits on the ring itself: an aryl halogen. The small numbers written low count the atoms. Okay? Now benzyl chloride, C six H five, C H two, C l. One extra carbon, the C H two, sits between the ring and the chlorine. So this chlorine is on an s p three carbon next to the ring, and we call it a benzylic halogen. Right? So tell me this. Warm each one with alcoholic silver nitrate. Which one gives a precipitate? Benzyl chloride, yes, exactly! It gives a precipitate at once, and chlorobenzene gives none. On screen, a l c is short for alcoholic, and A g N O three is silver nitrate. Clear? And why? The benzyl cation left behind is stabilised by resonance with the ring, so it forms easily. The phenyl cation is not, so it does not form. That one difference answers every question on this page. The same holds with potassium cyanide, KCN, ammonia, N H three, or a sodium alkoxide, N a O R, as reagents: only the side-chain halogen is replaced. Got it?

3:05Two tools for these questions

Now two tools that these past questions keep using. Tool number one is a test. Boil with aqueous sodium hydroxide, N a O H, then acidify with dilute nitric acid, HNO three, and then add silver nitrate. The short forms a q and d i l mean aqueous and dilute. Okay? If the halogen is on a side chain, it comes off as a halide ion, X minus. Then the silver ions give a silver halide precipitate. Right? And if the halogen is on the ring, it stays on, so there is no halide ion and no precipitate. Clear? Tool number two is oxidation. Hot potassium permanganate, K M n O four, turns each side chain with a benzylic hydrogen into COOH, the acid group; the arrow means gives. Here, a benzylic hydrogen is one on the carbon joined to the ring. With me? Now tell me this. If the ring carries two such side chains, how many COOH groups will you get? Two, one for each side chain, correct! And a ring halogen survives the oxidation. Example one uses tool number one, and Example two uses a silver nitrate test together with this oxidation. Got it?

4:17Example 1: iodobenzene or benzyl iodide, AgNO3 test

Now Example one, from A I triple E two thousand and three, our puzzle from the start. Bottles containing C six H five I, and C six H five C H two I, lost their labels and were marked A and B, okay? Each was boiled with sodium hydroxide solution, acidified with dilute nitric acid, and silver nitrate was added. Bottle B gave a yellow precipitate. Which bottle holds which? Pause the video and try it first. Done? Okay, let's check. C six H five I is iodobenzene, where the letter I stands for iodine, and that iodine sits on the ring. Right? And C six H five, C H two, I is benzyl iodide. Its iodine is on the C H two, the side chain, so it is a benzylic halogen. Clear? Now, both are iodides. So both bottles give the yellow precipitate, right? No, no, careful, that is the trap! The iodine of iodobenzene is on the ring, and boiling sodium hydroxide cannot replace it. So no iodide ion, I minus, forms, and that bottle gives no precipitate. With me? Benzyl iodide is different. Its side-chain iodine is hydrolysed, so iodide ions go into the solution. Then silver ions, A g plus, with the plus written high, meet the iodide and give A g I, silver iodide, the yellow precipitate. Okay? So bottle B, with the yellow precipitate, is benzyl iodide, and bottle A is iodobenzene. Puzzle solved: both are iodides, but only the side-chain iodine comes off. Got it?

5:54Example 2: identify C8H9Br that oxidises to phthalic acid

Now Example two, a harder one, from JEE Main two thousand and thirteen. Compound A, C eight H nine B r, gives a white precipitate when warmed with alcoholic silver nitrate. Oxidation of A gives an acid B, C eight H six O four, which easily forms an anhydride on heating. Identify compound A, okay? Pause the video and try it first. Done? Okay, let's check. Many students put this bromine straight onto a benzene ring. Does that fit the first clue? No, it does not! Bromine on the ring gives no precipitate. So the precipitate tells us this bromine is on a side chain. One small point: the paper calls it white, but silver bromide, A g B r, is really pale yellow. Either way, a precipitate forms. Right? Now look at acid B, C eight H six O four. It is C six H four, with COOH in brackets and a small two outside: a benzene ring with two COOH groups. Each one came from a side chain, so compound A has a ring with two side chains. With me? Now, are the two COOH groups ortho, meta or para? Which clue decides it? The anhydride clue, exactly! Easy anhydride formation on heating means the two groups are ortho, next to each other. So acid B is phthalic acid. Clear? Now count the carbons. The ring has six, and compound A has eight, so each side chain has one carbon: C H two B r, which carries the bromine, and a methyl group, C H three. And the hydrogens: four on the ring, plus two, plus three, gives nine. So C eight H nine B r fits perfectly. Okay? So compound A is two methylbenzyl bromide, with the methyl and the C H two B r groups ortho to each other. Its side-chain bromine gives the precipitate, and oxidation turns both side chains into COOH, giving phthalic acid. If you got this one on your own, very good, that was a proper JEE Main puzzle!

8:05Quick revision: ring vs side-chain halogen, AgNO3, oxidation

So, quick revision. The ring halogen is hard to replace, because of resonance and the s p two carbon, while a side-chain halogen is reactive. Next, a precipitate with silver nitrate means a side-chain halogen; the double arrow means so. And oxidation turns each side chain with a benzylic hydrogen into COOH, while easy anhydride formation means ortho. One more past question waits on this page, with three iodobenzyl chloride and sodium cyanide. Try it yourself. All the best!