Set 3

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take but two yne, four carbons with a triple bond in the middle. In one flask, we add hydrogen over Lindlar's catalyst; in another flask, we use sodium in liquid ammonia. Both add one mole of hydrogen and give but two ene. Then we add bromine to each flask. Same alkyne, same bromine, so the same final product, right? Hmm, are you sure about that? We will settle it in our second example. This is Set three of the mixed JEE Main pattern problems on hydrocarbons, and we take two of its alkyne questions. So, let's begin.

0:42Tool 1: water on a triple bond

So, first tool: adding water to a triple bond. On the right is the map of alkynes: grey boxes make them, orange boxes are their reactions. Today we use the hydrogen box at the top, and the water box, fifth from the top, okay? The conditions: H g S O four, mercuric sulphate, with dilute H two S O four, sulphuric acid, at three hundred and thirty three kelvin, the capital K on screen. Small numbers written low count atoms. Clear? Now let me ask you this. We add water, an H and an O H, across the triple bond of ethyne. So the final product is an alcohol, right? Not quite, and this is where students slip. Water adds by Markovnikov's rule, and what forms first is an enol: ene for the double bond, and ol for the O H group, so an alcohol on a double bond carbon. But the enol does not stay. With me? Look at ethyne. H C, three lines for the triple bond, C H, plus H two O, water, gives C H two, two lines for the double bond, C H, a dash for the single bond, and O H, in square brackets, which mean an intermediate: ethenol, or vinyl alcohol. Then the arrow takes it to C H three CHO, ethanal. Okay? Why does it change? Because the enol is much less stable than its keto tautomer, the form with a carbon oxygen double bond, so it rearranges at once, and we get a carbonyl compound. Right? And the rule to remember: ethyne gives ethanal, an aldehyde, and every other alkyne gives a ketone. Only ethyne gives an aldehyde here; that is a common trap. Got it?

2:33Tool 2: adding hydrogen, stop at cis or trans

Now, second tool: adding hydrogen. H two with nickel, platinum or palladium, the symbols N i, P t and P d, goes all the way to the alkane. But often we want to stop at the alkene, okay? For that, we use Lindlar's catalyst: palladium on charcoal, or on calcium carbonate, C a C O three, or on barium sulphate, B a S O four, partly deactivated. This poisoned catalyst stops at the alkene. Clear? On Lindlar's catalyst, both hydrogen atoms add to the same face of the triple bond, which is syn addition. So tell me, do we get the cis alkene or the trans alkene? The cis alkene, yes, exactly! Same face, so the two new hydrogens end up on the same side of the double bond. Right? And for the trans alkene, we use sodium or lithium, N a or L i, in liquid ammonia, N H three. Here the hydrogens add from opposite sides, which is anti addition, so we get the trans alkene. Same alkyne, but two different alkenes: that is the first half of our puzzle. Got it?

3:42Tool 3: bromine adds anti, racemic or meso?

Now, third tool: bromine. B r two, in carbon tetrachloride, C C l four, gives a vicinal dibromide, which means the two bromine atoms sit on neighbouring carbons. Okay? It goes through a cyclic bromonium ion, which the bromide ion opens from the opposite side. So the two bromines add to opposite faces of the double bond: anti addition. With me? Now the stereochemistry. Read the plus sign as and, and the arrow as gives. So a cis alkene with anti addition gives the racemic dibromide: two mirror image forms in equal amounts, shown by the plus minus sign in brackets. Clear? And a trans alkene with anti addition gives the meso dibromide, when the two ends of the double bond are alike, as in but two ene. Meso means the product has a plane of symmetry: one half is the mirror image of the other. Right? Now, Baeyer's reagent adds syn, to the same face, and for syn addition the result is the reverse. Okay? So tell me, a cis alkene with syn addition: racemic, or meso? Meso, yes, very good! Here is a topper's trick. Cis, same side, and syn, same face, match. Trans, opposite sides, and anti, opposite faces, also match. Matching pairs give meso, and mixed pairs give racemic, but only when the two ends are alike. Got it?

5:13Example 1: mass of ethanal from 2.6 g of ethyne

Now, Example one, an easy one. What mass of ethanal is obtained by complete hydration of two point six grams of ethyne? The small g is grams. Pause the video, and try it first. Done? Okay, let's check. H C triple bond C H, plus H two O, gives C H three CHO, ethanal. So one mole of ethyne gives one mole of ethanal; m o l on screen is mole. Clear? Capital M, with the formula in brackets, is the molar mass. Two carbons at twelve each, and two hydrogens at one each, so twenty four plus two, which is twenty six grams per mole; the minus one written high means per mole. Okay? So the number of moles, small n, is two point six over twenty six, which gives zero point one mole. And so we get zero point one mole of ethanal. Right? Now, careful here. For the product, which molar mass do we use: forty four, or forty six? Forty four is correct, because the product is ethanal, not ethanol. Twenty four for carbon, plus four for hydrogen, plus sixteen for oxygen, which is forty four grams per mole. Forty six belongs to ethanol, the trap from our first tool. Got it? So the mass of ethanal is zero point one times forty four, which is four point four grams. The mass went up because each ethyne molecule picked up a whole water molecule, okay?

6:45Example 2: but-2-yne, Na in liquid NH3, then Br2 in CCl4

Now, Example two, a harder one. But two yne is reduced with sodium in liquid ammonia, and the product is treated with bromine in carbon tetrachloride. Describe the final product. Pause the video, and try it first. Done? Okay, let's check. But two yne is C H three, C, triple bond, C, C H three. Above the arrow is sodium in liquid ammonia; l i q with a dot means liquid. That is anti addition of hydrogen, so we get trans but two ene. Right? Step two: trans but two ene, plus bromine in carbon tetrachloride, and bromine adds anti. So tell me, a trans alkene with anti addition: racemic, or meso? Meso, yes, perfect! Trans and anti are a matching pair. So the final product is meso two, three dibromobutane: C H three, CHB r, CHB r, C H three. Both ends of but two ene are alike, so the meso result holds. Clear? And now our puzzle from the start. With Lindlar's catalyst instead, we get cis but two ene, and anti addition of bromine then gives the racemic dibromide. So the two flasks give different products: one meso, one racemic. With me?

8:09Quick revision: alkyne hydration, Lindlar vs Na, Br2 anti

So, quick revision. Hydration of an alkyne with mercuric sulphate and dilute sulphuric acid gives an enol, which turns at once into a carbonyl compound, and only ethyne gives an aldehyde. Lindlar's catalyst gives the cis alkene, while sodium or lithium in liquid ammonia gives the trans alkene. And bromine adds anti: cis gives racemic, and trans gives meso. Now go and try the other problems of Set three on this page. You can do this. All the best!