Reactions with Metals

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. In bromoethane, the carbon carries a small positive charge, so nucleophiles attack it. Now add magnesium, in dry ether, and that same carbon turns negative! The target becomes the attacker. But one drop of water destroys this new reagent. That sounds strange, right? Today, the reactions of alkyl halides with metals. They come up often in JEE, so let's begin.

0:30Making a Grignard reagent

So, let's make it. R X is an alkyl or aryl halide: R is the alkyl or aryl group, X is the halogen, and the short dash is a single bond. With magnesium, written M g, and with dry ether written above the arrow as the condition, the arrow meaning gives, we get R Mg X, an alkylmagnesium halide, the Grignard reagent. Okay? Now, the ease of formation. The iodide, R I, forms it most easily, then the bromide, R B r, then the chloride, R C l; the greater than sign means more easily. It is the same order as the reactivity of alkyl halides in chapter one. Right?

1:14Inside the reagent: the charge flips

Now, inside the reagent. Remember R X from chapter one: the halogen is more electronegative than carbon, so delta plus sits on the carbon and delta minus on the halogen. Delta, the small Greek letter written high, marks a partial charge, and nucleophiles attack that positive carbon. With me? Now the carbon is bonded to magnesium, a metal. So which end of this new bond carries the negative charge? Yes, this time it is the carbon! Carbon is delta minus, and magnesium is delta plus. The carbon magnesium bond is covalent, but highly polar, and the magnesium halogen bond is ionic. The charge has flipped, and our puzzle is solved. Clear? So the carbon behaves as a carbanion, a negatively charged carbon. That makes it a strong base and a strong nucleophile. Remember the word base. Got it?

2:09Any acidic hydrogen destroys it

Now, this strong base meets water, which can give away a proton. So what happens? The base grabs the proton, exactly! R Mg X plus H A gives R H, plus Mg A X; H A is any compound with an acidic hydrogen, and the round brackets keep A on the magnesium. So the reagent becomes a plain hydrocarbon. Okay? Such compounds include water, alcohols, amines, carboxylic acids, and terminal alkynes, with a hydrogen on the triple bonded carbon. Each one destroys the reagent. Right? So the apparatus and the ether must be perfectly dry; that is why water was so deadly in our puzzle. And careful, the acidic hydrogen can sit inside the halide itself. Two bromoethanol has its own O H group, which would destroy its Grignard reagent as it formed. Clear?

3:03Heavy water and carbon dioxide

Now, two useful reactions. First, heavy water, D two O, where D is deuterium, and the small two written low means two atoms. It acts like water, so R Mg X gives R D, with deuterium exactly where the halogen was. Okay? Second, carbon dioxide, C O two, and then acid. The product is RCOOH, a carboxylic acid. Right? So tell me, how many carbons does the acid have, compared with the R group? It has one carbon more, yes! The carbon of carbon dioxide joins on. Propylmagnesium bromide, C three H seven Mg B r, on dry ice, which is solid carbon dioxide, and then acid, gives C three H seven COOH, butanoic acid. Got it?

3:55Wurtz reaction: sodium joins two R groups

Now, sodium. Two R X with two sodium, written N a, in dry ether, give R R plus two N a X, the sodium halide. The two R groups join. This is the Wurtz reaction. Okay? So think about this: R R has twice the carbons of the R group. Is that number odd, or even? It is always even, that is correct! So Wurtz suits symmetrical alkanes with an even number of carbons: bromoethane gives butane. Right? But with two different halides, sodium joins every possible pair, giving a mixture of three alkanes, hard to separate. And tertiary halides give mainly alkenes. With me?

4:38Corey–House: joining two different groups

For a clean join of two different groups, we use the Corey House synthesis. R two C u L i is a lithium dialkyl cuprate, with copper, C u, and lithium, L i. It reacts with R prime X; the small mark called a prime means a second, different group. We get R R prime, R C u, and L i X, the lithium halide. Okay? So the two different groups join with no mixture, giving unsymmetrical alkanes, cleanly. It even works with vinyl and aryl halides. Clear?

5:11More reactions with metals

Now, three more reactions, quickly. First, a dihalide with sodium or zinc can close a ring. One three dibromopropane with zinc, Z n, joins its two ends to give cyclopropane, a ring of three carbons. Got it? Second, reduction to alkanes, using zinc with hydrochloric acid, the zinc copper couple in ethanol, lithium aluminium hydride, L i Al H four, or hydrogen with palladium, P d. The slash means together with. Okay? And third, four molecules of chloroethane, C two H five C l, with four N a P b, a sodium lead alloy, give tetraethyl lead: C two H five in brackets, four times, then P b for lead. Four sodium chloride and three lead also form. Right?

6:00Example 1: Wurtz mixture from CH3Cl and 2-chloropropane

Now, Example one. Here, a mixture of chloromethane and two chloropropane is treated with sodium in dry ether. Which alkanes are formed? Pause the video and try it first. Done? Okay, let's check. R is C H three, methyl, and R prime is C H three twice in brackets, then C H; the two after the bracket means two methyl groups. Join R and R prime, and the answer is just two methylpropane, right? Oh no, careful, that is only one of the products! Sodium joins any two groups, so two methyls give C H three C H three, ethane. Okay? Two of the R prime groups join, giving a six carbon alkane, two three dimethylbutane. Right? And one of each gives two methylpropane. So we get three alkanes in all, exactly the mixture we warned about. Got it?

6:58Example 2: methane volume from ethane-1,2-diol and CH3MgI

Now, Example two, a harder one. Zero point six two grams of ethane one two diol, of molar mass sixty two, is treated with excess methylmagnesium iodide. Find the volume of methane evolved at STP, standard temperature and pressure. Capital M is the molar mass, and the small g stands for grams. Pause the video and try it first. Done? Okay, let's check. Moles of diol equals zero point six two over sixty two, the slash meaning over, which gives zero point zero one mole, written m o l. Okay? Each acidic hydrogen gives one methane. So zero point zero one mole of methane, and that is two hundred and twenty four millilitres, right? No, no, wait, look at the molecule! Ethane one two diol, HOCH two C H two O H, has two O H groups, each with an acidic hydrogen. So each molecule gives two methanes. With me? So methane is two times zero point zero one, giving zero point zero two mole; the cross means times. One mole of gas at STP occupies twenty two thousand four hundred millilitres. Zero point zero two times twenty two thousand four hundred gives four hundred and forty eight millilitres, written m L on screen. Clear?

8:26Quick revision: Grignard reactions, Wurtz and Corey-House

So, quick revision. Magnesium in dry ether turns R X into the Grignard reagent, whose carbon acts as a carbanion, so keep it dry. Acidic hydrogen gives R H, heavy water gives R D, and carbon dioxide, then acid, gives an acid with one carbon more. Wurtz joins R to R, and Corey House joins R to R prime cleanly. Now go and try the practice problems on this page. You can do them. All the best!