Reactions of Ethers

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle before we start. Ethers are lazy molecules: bases, sodium, even oxidising agents leave them alone. But add concentrated hydrogen iodide, and the ether splits into two pieces. So when the two sides are different, which side becomes the iodide? Chemists even use this split to count the methoxy groups in a molecule. Questions on ethers come up often in JEE, so let's begin.

0:30Ethers ignore most reagents, but not HI

So, first the lazy side of ethers. They are unreactive towards bases, towards sodium, the N a on screen, and towards oxidising and reducing agents. Okay? But concentrated hydrogen iodide, H I, or hydrogen bromide, H B r, can split an ether apart. We call this cleavage. H I is the most reactive, then H B r, then hydrogen chloride, H C l; the sign between them means greater than. So in most questions, the reagent is hydrogen iodide. Clear?

1:07How HI cuts an ether

In step one, the H plus from the acid attaches to the oxygen; the plus sign written high is its charge. We say the oxygen is protonated. Okay? In step two, the iodide ion, I minus, attacks a carbon attached to that oxygen; the minus written high is its charge. So the ether splits in two. With me? When both groups are primary or secondary, this attack is S N two, written with a small N low down. It happens at the less hindered, less crowded group. Right? Let us try one. Methoxyethane is C H three, O, C two H five; the small numbers written low count the atoms, and the dashes are single bonds. The arrow means gives, and the question mark is the product. So tell me, which group becomes the iodide, the methyl or the ethyl? The methyl group, yes, exactly! Methyl is smaller and less hindered, so the iodide attacks there. We get C H three I, iodomethane, and C two H five O H, ethanol. So the rule is: the smaller group becomes the iodide, and the larger one the alcohol. Got it?

2:20The trap: a tertiary group

Now look at this ether. C H three in brackets, with a small three after the bracket, means three methyl groups on one carbon: a tertiary group. The other side is a methyl. By our rule, the smaller methyl becomes the iodide, so iodomethane again, right? Oh no, careful, that is the trap here! When one group is tertiary, or benzylic, or allylic, it leaves as a carbocation, a carbon carrying a positive charge. This path is S N one. Clear? So now the tertiary group becomes the iodide, C H three in brackets, three, then C I, and the methyl gives methanol, C H three O H, okay? So before you use the smaller group rule, always check for a tertiary, benzylic or allylic group. Right?

3:09Excess HI, and aryl ethers

Next, suppose the H I is hot and in excess. Then the alcohol formed is also converted into its iodide. R and R prime stand for two groups; the small mark is called prime. ROR prime, plus two H I, gives R I, plus R prime I, plus H two O, water. Both groups end up as iodides. Got it? Now an alkyl aryl ether, where one group is a benzene ring, like anisole. With H I, can the iodide attack the ring carbon and give iodobenzene? No, the iodide cannot attack the ring carbon. Alkyl aryl ethers always give the phenol and the alkyl halide. Right? The reason is the bond between the ring and the oxygen, written aryl, dash, O, on screen. It has partial double bond character, so it is not broken. And when both groups are rings, a diaryl ether, nothing is cleaved at all. Clear?

4:09Zeisel's method: counting methoxy groups

Now, a smart use of this cleavage, called Zeisel's method. The methoxy group is OCH three, with a dash for its bond to the molecule. With H I, each methoxy group gives exactly one iodomethane. Okay? And the iodomethane is estimated as silver iodide, which is the A g I on screen. So the moles of silver iodide, divided by the moles of the compound, tell us the number of methoxy groups in each molecule. We will use this in example two. Right?

4:41Anisole: an activated ring

Now the ring of an aryl ether. The alkoxy group, like OCH three in anisole, is activating and ortho, para directing, so electrophilic substitution on the ring is easy. Okay? For bromination, benzene needs a Lewis acid catalyst. So, does anisole, methoxybenzene, also need one? It needs no catalyst at all, because its ring is strongly activated! The methoxy group donates a lone pair into the ring. Bromine, B r two, in ethanoic acid gives four bromoanisole; the four is the para position. Right? Nitration and Friedel-Crafts reactions also give the four and two substituted products, para and ortho, with the para product as the major one. Got it?

5:26More reactions to remember

Now some quick facts. On standing in air and light, ethers form peroxides, and these are explosive. To remove them, the ether is shaken with ferrous sulphate solution. Okay? Next, the oxygen of an ether has lone pairs, so ethers are Lewis bases. They dissolve in cold concentrated sulphuric acid, H two S O four, as oxonium salts, and they form complexes with B F three. They also solvate Grignard reagents. With me? Last, epoxides, also called oxiranes, are strained ring ethers, so they open easily. In acid, the nucleophile attacks the more substituted carbon; with a basic nucleophile, the less substituted one. Do not mix them up. Clear?

6:14Example 1: benzyl ethyl ether with one equivalent of HI

Time for example one. Benzyl ethyl ether is treated with one equivalent of H I, so no excess this time. Name the products. There is a trap in it, so pause the video and try it first. Done? Okay, let's check. The ether is C six H five C H two, the benzyl group, then O, then C two H five, the ethyl group. Ethyl is smaller, so we get iodoethane, right? No, no, wait, that is the trap we just learnt! Benzyl is benzylic, so it leaves as the benzyl cation, which is resonance stabilised: its positive charge spreads into the ring. This is S N one. Okay? So the benzyl group becomes the iodide. The products are benzyl iodide, C six H five C H two I, and ethanol. Got it?

7:08Example 2: methoxy groups by Zeisel method from AgI mass

Now example two, our counting puzzle. Zero point one three eight grams of a compound, of molar mass one hundred and thirty eight grams per mole, is heated with hydrogen iodide. The g is grams, and mol to the power minus one means per mole. The iodomethane gives zero point four seven zero grams of silver iodide, molar mass, capital M, two hundred and thirty five. How many methoxy groups does one molecule contain? Pause the video and try it first. Done? Okay, let's check. Moles of the compound: zero point one three eight over one hundred and thirty eight gives zero point zero zero one mole. Okay? Now the silver iodide: zero point four seven zero over two hundred and thirty five gives zero point zero zero two mole. Each silver iodide came from one iodomethane. So, how many methoxy groups per molecule? The answer is two, yes, exactly! Zero point zero zero two over zero point zero zero one gives two. Each methoxy group gave one iodomethane, so each molecule carries two methoxy groups. And that solves our counting puzzle from the start!

8:28Quick revision: HI cleavage, SN1 vs SN2, aryl ethers, Zeisel

So, quick revision. With H I, when both groups are primary or secondary, the smaller group becomes the iodide and the larger group the alcohol, by S N two. But a tertiary, benzylic or allylic group leaves as a carbocation and becomes the iodide, by S N one. And alkyl aryl ethers give the phenol and the alkyl halide, while each methoxy group gives one iodomethane. Now go and try the practice problems on this page. All the best!