Preparation of Ethers

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle before we start. Two students want to make the same ether, two methoxy two methylpropane. The first mixes sodium methoxide with two bromo two methylpropane. The second mixes sodium tert butoxide with bromomethane. On paper, both join the same two groups through an oxygen. But one flask gives no ether at all, only an alkene. Which flask fails, and why? Today we learn the two ways to prepare ethers, and exactly where each one fails. Questions on these come up often in JEE, so let's begin.

0:39Method 1: dehydration of alcohols

So, method one is the dehydration of alcohols. We heat a primary alcohol with concentrated sulphuric acid, H two S O four, at four hundred and thirteen kelvin, the capital K, with the alcohol in excess. The small circle written high means degree, so one degree is primary, and the small numbers written low count the atoms. Now, two ethanol molecules join through one oxygen to make one ether. So what small molecule must come out? Water, yes, exactly! That is why we call it dehydration. Two C two H five O H give C two H five O C two H five, which is ethoxyethane, plus H two O; the arrow means gives. Right? Here is how it happens. The acid protonates one alcohol molecule, a second alcohol molecule attacks its carbon, and water leaves. Then the protonated ether loses H plus, and we have our ether. That is an S N two reaction, written S with a small N below and a two, the same S N two you met with alkyl halides. Clear?

1:47Dehydration: where it fails

Now, here comes a trap. Same alcohol, same acid, but we heat a little more, to four hundred and forty three kelvin. Do we still get the ether? No, and this is where students slip! At four hundred and forty three kelvin, even ethanol gives ethene, an alkene, not the ether. So keep the two numbers apart: four hundred and thirteen kelvin for the ether, and four hundred and forty three kelvin for ethene. Got it? Next, suppose we want ethyl methyl ether, so we heat methanol and ethanol together with the acid. Tell me, how many different ethers will we get? Three, yes, exactly! Two methanols give methoxymethane, two ethanols give ethoxyethane, and one of each gives methoxyethane, the one we wanted. So we get a mixture of three ethers, and that is no good. With me? So dehydration suits only symmetrical ethers, the same group on both sides of the oxygen, from primary alcohols. Secondary and tertiary alcohols, two degree and three degree on screen, give alkenes instead. For everything else, we need method two. Okay?

2:59Method 2: the Williamson synthesis

Method two is the Williamson synthesis. We take a sodium alkoxide, or a sodium phenoxide, and treat it with an alkyl halide. Right? Here is the equation: RON a, plus R prime X, gives ROR prime, plus N a X, okay? R and R prime stand for the two groups joined to the oxygen, the small dash called prime marks the second one, and X is the halogen; the short dashes are bonds. The top line of the figure shows the same reaction; its bottom line, cleavage by hydrogen iodide, is for our next lesson. Clear? Now, the Williamson synthesis is an S N two reaction. Knowing that, which alkyl halides will work well here? Methyl and primary halides, exactly! Because it is S N two, the halide must be methyl or primary. Got it? But now take a secondary or tertiary halide. Remember, an alkoxide is a strong base. So what will that halide do, instead of giving an ether? It undergoes elimination, yes! The strong base gives an alkene, and we lose our ether. This is a common trap in Williamson questions. Okay? Now the good news. The alkoxide itself may be primary, secondary or tertiary. So here is a topper's trick: put the crowded group on the alkoxide, and keep the halide methyl or primary. With me?

4:27Aryl ethers: the ring comes from the phenoxide

Now ethers with a benzene ring on the oxygen, called aryl ethers. Here we use sodium phenoxide with an alkyl halide. Clear? For example, ethoxybenzene is made from sodium phenoxide and bromoethane. The ring comes from the phenoxide. Right? Now, anisole is methoxybenzene. Why can we not make it from bromobenzene and sodium methoxide? Because bromobenzene is an aryl halide, and an aryl halide cannot be used. Aryl halides do not undergo nucleophilic substitution under these conditions. So the ring must always come from the phenoxide. Got it?

5:06Example 1: choosing reactants for methyl tert-butyl ether

Now, Example one, and it is our puzzle from the start. Which pair of reactants should be used to make two methoxy two methylpropane: sodium methoxide with two bromo two methylpropane, or sodium tert butoxide with bromomethane? And why? Pause the video and try it first. Done? Okay, let's check. First, the target. It is C H three in brackets, with a small three after the bracket, which means three methyl groups on one carbon; then O, then C H three. So one side of the oxygen is a tertiary group, and the other side is a methyl group. Clear? Now look at the first pair: sodium methoxide, with two bromo two methylpropane. Which of our rules does this pair break? The halide rule, exactly! On screen, C H three O N a is sodium methoxide, and the bromide next to it is a tertiary halide. Methoxide is a strong base, so we get elimination: only two methylpropene, an alkene, and no ether at all. So that is the flask that failed in our puzzle. Right? Now the second pair: sodium tert butoxide, tert being short for tertiary, with C H three B r, which is bromomethane. The halide is methyl, perfect for S N two, and the tertiary group sits on the alkoxide, which is allowed. So we get our ether, plus N a B r, sodium bromide. Got it? So the answer is sodium tert butoxide with bromomethane. Crowded group on the alkoxide, small group on the halide, and our puzzle is solved. With me?

6:46Example 2: tetrahydrofuran from 4-bromobutan-1-ol and NaH

Now a harder one, Example two. Four bromobutan one ol is treated with sodium hydride. Name the cyclic product, and the type of reaction. Cyclic means the product is a ring. Pause the video and try it first. Done? Okay, let's check. Here is the molecule: H O, then four C H two groups in a chain, and B r at the end. So there is an O H group on carbon one, and a bromine on carbon four: one molecule, two ends. Okay? Sodium hydride, N a H, turns the O H end into an alkoxide, O minus with N a plus; the minus and plus written high are the charges. Now we have an alkoxide end, and a primary bromide end, in one molecule. Right? So, this looks just like a Williamson synthesis. But no separate alkyl halide is added. So what can the alkoxide end attack? Its own other end, yes, exactly! The alkoxide end displaces bromide, B r minus, from the other end of the same molecule. Clear? So the chain closes into a ring of one oxygen and four carbons. This ring is tetrahydrofuran, also called oxolane. Got it? So the reaction is a Williamson synthesis inside one molecule: an intramolecular Williamson synthesis. That was the hardest one today, so well done if you got it. With me?

8:16Quick revision: dehydration at 413 K and Williamson rules

So, quick revision. Dehydration: a primary alcohol with concentrated sulphuric acid at four hundred and thirteen kelvin gives symmetrical ethers only. Williamson: sodium alkoxide plus R prime X, by S N two, where R prime X must be methyl or primary. And put the crowded group on the alkoxide, and take the aryl group from the phenoxide. Now go and try the practice problems on this page. All the best!