Preparation of Alkenes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Heat three, three dimethyl butan two ol with concentrated sulphuric acid, and it loses water to give an alkene. The O H group sits on carbon two, so the double bond should form right there, with the carbon skeleton unchanged, right? Well, the major product is two, three dimethyl but two ene, and a methyl group has actually moved! Let's find out why. This comes up often in JEE, so let's begin.

0:31The ways to make an alkene

Here is the full map from the course. In the centre is the alkene: the equals sign between the two carbons is the double bond, and the pointed signs stand for the other two bonds of each carbon, okay? The boxes on the left, under Preparation, are the ways to make it, and the orange ones on the right, under Reactions, are its reactions, for later lessons. With me?

0:52Method one: partial reduction of alkynes

Let us start with alkynes, which have a triple bond. Alkyne plus H two, with Lindlar's catalyst, gives the cis alkene. The small two written low is a subscript, and the arrow means gives. Cis means the two groups sit on the same side of the double bond. Okay? So what is Lindlar's catalyst? Palladised charcoal, or palladium, P d, on calcium carbonate, C a C O three, or on barium sulphate, B a S O four. Right? Here is the clever part: this catalyst is weakened on purpose. Why would a chemist want a weaker catalyst? Yes, exactly, so that the reaction stops halfway! It is partially deactivated with sulphur compounds or quinoline, so the alkyne stops at the alkene, not the alkane. Clear? Now change the reagent to sodium in liquid ammonia, N H three. Same alkyne, but now we get the trans alkene, with the two groups on opposite sides. So remember: Lindlar gives cis, and sodium gives trans. Got it?

1:54Method two: dehydrohalogenation

Next, dehydrohalogenation. Heat an alkyl halide with alcoholic KOH, potassium hydroxide in alcohol, and it loses H X, a hydrogen halide, with X for the halogen. Okay? The hydrogen and the halogen leave from adjacent carbons. The carbon next to the one carrying the halogen is the beta carbon, the Greek letter beta on screen, so this is beta elimination. Right? Now, reactivity. R is the alkyl group, and each pointed sign means greater than. So R I, the iodide, is greater than R B r, greater than R C l. And the small circle up high is read as degree, so three degree is greater than two degree, greater than one degree, which means tertiary halides react fastest. Clear? Let me ask you this: what if the beta carbon carries no hydrogen at all? Then no alkene forms at all. Look at this one: C H three in brackets with a small three after it, meaning three methyl groups on one carbon, then C H two B r. That is one bromo two, two dimethyl propane. Its beta carbon has no hydrogen. That is a classic trap. Got it?

3:02Method three: dehydration of alcohols

Method three is dehydration of alcohols. Heat an alcohol with concentrated sulphuric acid, H two S O four, at four hundred and forty three kelvin; capital K is kelvin. It loses H two O, a water molecule, giving an alkene. Okay? Other reagents work too: phosphoric acid, H three P O four, or alumina, A l two O three, at six hundred and twenty three kelvin. Right? Now tell me, which alcohol loses water most easily: primary, secondary or tertiary? Yes, tertiary alcohols lose water most easily. Three degree greater than two degree greater than one degree, the same order as for the halides. Clear? And why is that? The reaction goes through a carbocation, a carbon with a positive charge, and the rate follows the stability of that carbocation. And here is the key to our puzzle: a carbocation can rearrange into a more stable one. Got it?

3:59Two more methods

Two more methods, quickly. Dehalogenation: a vicinal dihalide has two halogen atoms on adjacent carbons; here both are bromine, and each dash is a single bond. With zinc, Z n, both bromines leave as Z n B r two, zinc bromide, and a double bond forms. Okay? Second, Kolbe's electrolysis of potassium butanedioate, also called potassium succinate. It gives ethene and C O two, carbon dioxide, at the anode. With me?

4:29Which alkene? Saytzeff and Hofmann

Now the big question. Often there are two different beta carbons, so two alkenes are possible. Which one is the major product? Saytzeff's rule answers it. In dehydrohalogenation and dehydration, the major product is the more substituted alkene, with more alkyl groups on its double bond carbons, which is also more stable. Okay? In practice, the hydrogen is lost from the beta carbon that has fewer hydrogens. Example one tests exactly this. Right? But change the base. With a bulky base such as potassium tert butoxide, the less substituted alkene, the Hofmann product, predominates. Clear? The same happens when the leaving group is N R three plus: a nitrogen carrying three R groups, with the small plus on top for its positive charge. So, normal base, Saytzeff; bulky base, Hofmann. Got it?

5:22Example 1: 2-bromobutane with alcoholic KOH, major product

Alright, Example one. Two bromobutane is heated with alcoholic KOH, okay? Name the major and the minor products. Pause the video and try it first. Done? Okay, let's check. Two bromobutane is C H three, CHB r, C H two, C H three, so the beta carbons are carbon one and carbon three, written C dash one and C dash three. Carbon one has three hydrogens, and carbon three only two. More hydrogens, more chances, so but one ene is major, right? No, no, careful, that is the trap! Saytzeff says the hydrogen is lost from the beta carbon with fewer hydrogens, so from carbon three. That gives C H three, C H double bond C H, C H three, which is but two ene, the more substituted alkene, and it is major. Clear? And losing a hydrogen from carbon one gives C H two double bond C H, C H two, C H three, which is but one ene, the minor product. Got it?

6:21Example 2: 3,3-dimethylbutan-2-ol dehydration, methyl shift

Now the harder one, and it is our puzzle from the start. Three, three dimethyl butan two ol is heated with concentrated sulphuric acid. Give the major alkene, and explain. Pause the video and try it first. Done? Okay, let's check. Here is the alcohol: three methyl groups on carbon three, then C H with the O H group, then C H three. Notice that carbon three carries no hydrogen. With me? The acid removes the O H group as water, leaving a secondary carbocation on carbon two; the small plus sign over that carbon is its positive charge. Its neighbour carries three methyl groups. Can this cation become more stable? Yes, it can, because one methyl group shifts over to the positive carbon, and the positive charge moves to the carbon the methyl left. That carbon now has two methyl groups and is joined to a C H with two more, so it is a tertiary cation, more stable. Clear? Finally, the cation loses a proton, H plus, with the small plus up high for its charge, from the neighbouring C H, the beta carbon with fewer hydrogens. We get two methyl groups, C double bond C, and two more methyl groups: two, three dimethyl but two ene, the tetrasubstituted alkene. Puzzle solved: the methyl group moved to make a more stable carbocation. Got it?

7:46Quick revision: Lindlar, Saytzeff, Hofmann, rearrangement

So, quick revision. From alkynes, Lindlar's catalyst gives cis, and sodium in liquid ammonia gives trans. Dehydrohalogenation needs a beta hydrogen, and the more substituted alkene, the Saytzeff product, is major. With a bulky base the Hofmann product wins, and in dehydration the carbocation can rearrange. Now go and try the practice problems on this page. All the best!