AIEEE and Recent JEE Main Questions: Ethers

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You want methyl phenyl ether: a benzene ring and a methyl group, joined by one oxygen. The first plan is bromobenzene with sodium methoxide. The second plan is sodium phenoxide with bromomethane. Both bring a phenyl and a methyl, right? But only one of them works, so which one is it? Hold that thought; we will settle it in Example one. Questions like these appear in JEE Main, and they need just three ideas. So, let's begin.

0:33Idea 1: making ethers, the Williamson synthesis

So, idea number one: the Williamson synthesis. RON a is a sodium alkoxide, and R prime X an alkyl halide, X being a halogen; prime, the small tick mark, marks a second group that may differ, and each short dash is a bond. The arrow means gives: the ether ROR prime, plus N a X, the salt. The figure's top line shows it with O minus and N a plus. Okay? Above the arrow in the figure, S N two, with a small N written low, means bimolecular nucleophilic substitution. So which alkyl halides will work? Only methyl or primary halides, yes, exactly! It is S N two, so R prime X must be methyl or primary, while the alkoxide can be primary, secondary or tertiary. Right? But a secondary or tertiary halide will not do. The alkoxide is a strong base, so that halide undergoes elimination, giving an alkene, not an ether. Clear? Now aryl ethers. The benzene ring must come from sodium phenoxide, and the alkyl group from an alkyl halide. Got it? Never the other way round: aryl halides do not undergo nucleophilic substitution under these conditions. Remember this; it decides our puzzle. With me?

1:53Idea 2: breaking ethers with HI

Now idea number two: breaking ethers. Ethers are quite unreactive, but concentrated H I or H B r cleaves them; the greater than signs say H I works best, then H B r, then H C l. Okay? In the figure's bottom line, H I first protonates the oxygen, shown by the plus sign; the two half arrows mean this step is reversible. Then iodide, I minus, attacks a carbon. In methoxyethane, methyl on one side and ethyl on the other, which side does iodide attack? The smaller methyl group, very good! It is S N two, so iodide attacks the less hindered group. C H three O C two H five, plus H I, gives iodomethane, C H three I, plus ethanol, C two H five O H; the small numbers written low count atoms. The smaller group becomes the iodide. Right? But if one group is tertiary, benzylic or allylic, it leaves as a carbocation, by S N one, and that group becomes the iodide. Clear? With excess hot H I, the alcohol formed also becomes an iodide. ROR prime, plus two H I, gives R I, plus R prime I, plus H two O, water. So we get two iodides; keep this for Example two. Got it? Now a classic trap. Anisole is methoxybenzene. With H I, does it give iodobenzene, or phenol? It gives phenol, not iodobenzene; if you said iodobenzene, that was the trap! Alkyl aryl ethers always give the phenol and the alkyl iodide, here iodomethane, because the bond between the ring and oxygen has partial double bond character, so it stays. With me?

3:47Idea 3: the iodoform test

Idea number three, from the alcohols chapter, is the iodoform test. Its reagent is iodine, I two, with sodium hydroxide, N a O H, sometimes written N a O I, sodium hypoiodite. If positive, you see a yellow precipitate of iodoform, CHI three. Okay? Now, the alcohols that pass are those with C H three CHOH joined to a carbon or a hydrogen; the brackets put the O H on the second carbon, and the end dash is the free bond. Right? Now ethanol, C H three C H two O H, has nothing after that second carbon. So ethanol fails the test, doesn't it? Oh no, careful, that is the trap! Write ethanol as C H three CHOH, joined to H, a hydrogen, so it passes the iodoform test. Got it?

4:45Example 1: which reactant pair makes methyl phenyl ether

Time for Example one, from JEE Main, April two thousand and twenty three. Which is the suitable combination for preparing methyl phenyl ether? Benzene with M e B r; sodium phenoxide with M e O H; bromobenzene with sodium methoxide; or sodium phenoxide with M e B r. M e means methyl, so M e B r is bromomethane, and M e O H is methanol. Pause the video and try it first. Done? Okay, let's check. Benzene with bromomethane: benzene has no oxygen at all, so it cannot give an ether. Okay? Next, sodium phenoxide with methanol. The Williamson synthesis needs an alkyl halide, and methanol is an alcohol, not an alkyl halide. With me? Now our puzzle: bromobenzene with sodium methoxide. Phenyl from one, methyl from the other, so it looks perfect, right? No, no, careful, this is exactly the trap from our puzzle! Bromobenzene is an aryl halide, and aryl halides do not undergo nucleophilic substitution here, so the first plan fails. Clear? So the ring comes from the phenoxide, and the methyl from the halide. C six H five O N a, plus C H three B r, gives methyl phenyl ether, C six H five OCH three, plus N a B r. The answer is sodium phenoxide with M e B r, our second plan. Easy, isn't it?

6:17Example 2: which ether gives two iodoform-positive alcohols

Example two is harder, from JEE Main, January two thousand and twenty six. Some mixed ether, P, heated with excess hot concentrated H I, gives two different alkyl iodides. With aqueous N a O H these give Q and R, and both give a yellow precipitate with N a O I, the hypoiodite. Which is P? Benzyl sec butyl ether, ethyl one phenylpropyl ether, two ethoxy two methylbutane, or two ethoxybutane? Pause the video and try it first. Done? Okay, let's decode it. Excess hot H I, then aqueous sodium hydroxide. So what are Q and R, really? They are just the alcohols on the two sides of the oxygen! Excess hot H I turns both sides into iodides, and aqueous N a O H puts an O H back in place of each iodine. With me? And a yellow precipitate with N a O I is the iodoform test, idea three. So both alcohols must contain C H three CHOH, okay? Now the options. Benzyl sec butyl ether, sec meaning secondary, gives benzyl alcohol, C six H five C H two O H, on one side. It has no C H three CHOH group, so it fails. Right? Next, ethyl one phenylpropyl ether. Its phenylpropyl side gives one phenylpropan one ol, which has a C H carrying an O H group. Does it pass? No, it fails; that is a common slip. In C six H five CHOHCH two C H three, the carbon with the O H carries a phenyl and an ethyl, but no methyl. So the group is missing. Clear? Then two ethoxy two methylbutane gives two methylbutan two ol, a tertiary alcohol. The carbon carrying the O H has no hydrogen, so it fails too. Got it? Finally, two ethoxybutane. The ethyl side gives ethanol, which passes. The other side gives butan two ol, with the O H on the second carbon, next to a methyl group, so it passes too. So P is two ethoxybutane!

8:39Quick revision: Williamson, HI cleavage, iodoform test

So, quick revision. Williamson synthesis: an alkoxide or phenoxide, with a methyl or primary halide. Cleavage by H I: the smaller group, or a tertiary or benzylic one, becomes the iodide, and the ring to oxygen bond stays. And for the iodoform test, the alcohol needs a C H three CHOH group. Three more past questions are solved on this page; try each one yourself first. All the best!