Addition Reactions of Alkynes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Add water across the triple bond of ethyne, with the right catalyst. Water is H and O H, so we should get an alcohol, right? But what we collect is ethanal, an aldehyde, with no O H group at all! So where did the alcohol go? Strange, isn't it? Hold on to that question. Questions on these additions come up often in JEE, so let's begin!

0:30Adding hydrogen: three reagents, three results

So, let us start with hydrogen. H two, the H with a small two, is hydrogen gas. With nickel, platinum or palladium, written N i, P t or P d, it adds all the way, and the alkyne becomes the alkane. Okay? But often we want only the alkene. So tell me, can we stop halfway? Yes, we can, with Lindlar's catalyst. It stops at the alkene and gives the cis alkene. This is a syn addition: both hydrogens add from the same side. Clear? And for the trans alkene, use sodium or lithium, written N a or L i, in liquid ammonia, N H three. This is an anti addition: the hydrogens add from opposite sides. Same alkyne, two different alkenes! Got it?

1:19Halogens, then HX by Markovnikov's rule

Now let us add halogens. X two stands for a halogen molecule. One mole of it gives the trans dihalo alkene, and two moles go all the way to the tetrahalide. Okay? And bromine water is decolourised by an alkyne: its colour disappears as the bromine adds. With me? Next, hydrogen halides, H X, which add by Markovnikov's rule, just as with alkenes. Take propyne and H B r. Which carbon of the triple bond takes the hydrogen? Yes, the end carbon, which already has a hydrogen! On screen: C H three, C, three short lines for the triple bond, then C H; the arrow with H B r over it means add H B r. The bromine goes to the middle carbon, giving C H three, C B r, two short lines for the double bond, and C H two. Bromine on a doubly bonded carbon: that is a vinylic halide. Where will a second H B r go? By Markovnikov's rule again, the hydrogen goes to the C H two end, and the bromine joins the carbon that already has one. C H three, C B r two, C H three: the small two means two bromines, both on the same carbon. That is a geminal dihalide. Clear?

2:42Adding water: what do we really get?

Now, back to our puzzle: adding water. It needs mercuric sulphate, H g S O four, and dilute sulphuric acid, H two S O four, at three hundred and thirty three kelvin, the capital K on screen. Then water adds the Markovnikov way. Okay? So, the H goes to one carbon, and the O H to the other. That makes an alcohol, right? Not quite, and this is the key to our puzzle. The O H sits on a doubly bonded carbon: ene plus ol, so it is an enol. And an enol changes at once into its keto tautomer, the form with C double bond O, a carbonyl group. Got it? So, for ethyne: H C, triple bond, C H, plus water, and each arrow means gives, so first the enol, C H two, double bond, C H, dash, O H, in square brackets because it is only an intermediate. At once it becomes C H three CHO, which is ethanal. Puzzle solved: the alcohol was there, but only for a moment! Now take propyne instead: aldehyde or ketone? It is a ketone, yes! Markovnikov's rule puts the O H on the inner carbon. So every alkyne other than ethyne gives a ketone, and a terminal alkyne gives a methyl ketone: propyne gives propanone. Right? And if you want the aldehyde from a terminal alkyne, use hydroboration oxidation, which works the anti-Markovnikov way. Ketone from mercury, aldehyde from hydroboration: this pair comes up often. Okay?

4:26Alkynes or alkenes: which adds faster?

Now, one last idea before the examples. An alkyne has two pi bonds, pi being the Greek letter on screen, while an alkene has only one. So alkynes should be more reactive towards electrophiles, right? No, no, careful, it is the other way round! Towards electrophiles, alkynes are less reactive than alkenes, and there are two reasons. With me? Reason one: the carbons of the triple bond are s p hybridised, and an s p carbon holds the pi electrons more tightly. Right? Reason two: when an electrophile adds, the positive charge sits on a doubly bonded carbon. That is a vinyl cation, and it is unstable. Clear?

5:07Example 1: but-2-yne with Lindlar, Na in NH3 and excess H2

Time for the first example. Name the products when but two yne reacts with: part A, H two and Lindlar's catalyst; part B, sodium in liquid ammonia; and part C, excess H two over nickel. Pause the video and try it first. Done? Okay, let's check. But two yne is C H three, dash, C, triple bond, C, dash, C H three, with the triple bond in the middle. Now part A, Lindlar's catalyst: will it give cis or trans? It is cis, yes, exactly! Syn addition puts both hydrogens on the same side, so the methyl groups stay on the same side: cis but two ene. Okay? Part B, sodium in liquid ammonia, is an anti addition: the hydrogens come from opposite sides, so the methyl groups end up on opposite sides, giving trans but two ene. The reagent decides the shape. Right? And part C, excess H two over nickel. Many students write but two ene here too. But nothing stops this reaction: two moles of hydrogen add, and we get butane. Got it?

6:22Example 2: which terminal C5 alkyne gives a branched ketone

Now let us try a harder one. An alkyne, C five H eight, gives a white precipitate with ammoniacal silver nitrate. On hydration with H g S O four and H two S O four, where the slash means the two together, it gives a ketone with a branched chain. Identify the alkyne. Pause the video and try it first. Done? Okay, let's check. Clue one is the white precipitate with ammoniacal silver nitrate, written A g N O three. From the lesson on acidity, only a terminal alkyne gives it. So, which C five H eight alkynes have the triple bond at the end? The first is pent one yne, a straight chain of five carbons with the triple bond at the end. It passes clue one. But what ketone does it give? Water adds the Markovnikov way, so we get the methyl ketone, pentan two one: C H three, C O, then three carbons in a straight line. No branch, so pent one yne fails clue two. Careful here: many students stop at the first terminal alkyne! Okay? The second is three methyl but one yne. C H three in brackets with a small two outside means two methyl groups on one carbon; then C H, dash, C, triple bond, and C H at the end. What ketone does this one give? By Markovnikov's rule, the O H goes to the inner carbon, and the enol becomes a methyl ketone, with C O next to the end methyl: three methyl butan two one. Those two methyl groups make its chain branched, exactly what the question wanted! With me? So the answer is three methyl but one yne. It is terminal for the silver test, and its ketone is branched. If you got this one on your own, very well done! Got it?

8:20Quick revision: Lindlar, Markovnikov HX, hydration to ketone

So, quick revision. Hydrogen: nickel gives the alkane, Lindlar's catalyst the cis alkene, and sodium in liquid ammonia the trans alkene. H X and water add the Markovnikov way, giving a geminal dihalide, and an enol that becomes a ketone. Only ethyne gives an aldehyde with mercuric sulphate, and towards electrophiles, alkynes are less reactive than alkenes. Now go and try the practice problems on this page. All the best!