AIEEE Questions: Alkynes and Dienes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. When H B r adds to a double bond, the H and the B r go onto its two carbons, side by side, right? Now add one molecule of H B r to buta one three diene at forty degrees Celsius. In the main product, the H and the B r sit at the two ends of a four carbon chain, and the double bond has moved to the middle! Strange, isn't it? Today we solve five older entrance-paper questions on alkynes and dienes. The same ideas are still tested in JEE Main, so let's begin.

0:38Idea 1: the acidic hydrogen of a terminal alkyne

Idea number one is the acidic hydrogen. R C triple bond C H is a terminal alkyne: R is an alkyl group, the three short lines are the triple bond, and the dash is a single bond. Its hydrogen sits on an s p carbon, with fifty percent s-character; that sign is percent. That carbon holds the electron pair tightly, so the hydrogen can leave as a proton. Okay? Now the order of acidity, strongest first. H two O, water, then ROH, an alcohol, then ethyne, then N H three, ammonia, and last, R H, an alkane. The small numbers written low count the atoms, and the pointed signs mean greater than. Right? So which bases can take that proton? N H two minus, the amide ion, can; the raised minus is its charge. Grignard reagents can too, because they end up as R H, at the bottom of the list. Clear? Now a question from the two thousand eight paper. C H three M g X is a Grignard reagent; M g is magnesium, and X stands for a halogen. It is treated with propyne. What is produced? The answer is methane, yes, exactly! C H three M g X plus propyne gives C H four, because the acidic hydrogen of propyne protonates the Grignard reagent. Got it?

2:03Idea 2: an alkyne adds two molecules of HBr

Idea number two: an alkyne can add two molecules of a hydrogen halide, H X, where X is again a halogen. Both follow Markovnikov's rule. The first gives a vinylic halide, with the halogen on a doubly bonded carbon, and the second a geminal dihalide, with both halogens on one carbon. Okay? So, from the two thousand seven paper: which reaction gives two, two dibromopropane? Option A is propene plus H B r, option B is propyne plus two H B r, option C is one bromopropene plus H B r, and option D is ethyne plus two H B r. Let's build it. Propyne, C H three C triple bond C H, takes H B r, written over the arrow, and gives C H three C B r double bond C H two. Markovnikov again, and a second H B r gives C H three C B r two C H three. With me? So option B is correct! Both bromines sit on the middle carbon, C dash two on screen, so this is two, two dibromopropane. Clear? And the others? Option A adds only one bromine, option C already has a bromine on carbon one, the end carbon, and option D has only two carbons. Right?

3:27Idea 3: HBr and buta-1,3-diene

Now idea number three, and our puzzle. In the figure, buta one three diene takes H plus, a proton, at carbon one, the end carbon. The cation formed is allylic: the square brackets hold its two resonance forms, joined by the double-headed arrow, and the little plus is the positive charge. So the charge is shared by carbon two and carbon four. With me? Now B r minus, bromide, can attack either end. On the left path, it attacks carbon two, so H and B r sit on neighbouring carbons: one two addition. It forms faster, so it is the major product at low temperature: kinetic control. Okay? On the right path, bromide attacks carbon four, the far end, and the double bond shifts to the middle: one four addition. The figure says this product is more stable. Can you tell me why? Yes, exactly! Its double bond is now in the middle of the chain, so it is a more substituted alkene, and more stable. At about forty degrees Celsius, the small circle meaning degree, the additions are reversible, so the more stable product wins: thermodynamic control. Right? That was the two thousand five question: one molecule of H B r and buta one three diene at forty degrees Celsius. The predominant product is one bromobut two ene, under thermodynamic control. Puzzle solved: H and B r at the two ends, and the double bond in the middle! Clear?

4:56Example 1: which reagent does not react with ethyne (2002)

Time for Example one, from the two thousand two paper. Which of these does not react with ethyne: N a O H, sodium hydroxide; ammoniacal A g N O three, silver nitrate in ammonia; N a, sodium metal; or H C l, hydrogen chloride? Pause the video and try it first. Done? Okay, let's check. Ethyne has acidic hydrogens, so sodium metal replaces one, and H two, hydrogen gas, is given off. So sodium reacts. Okay? Ammoniacal silver nitrate is our test for a terminal alkyne, and ethyne gives a white precipitate of silver acetylide. That reacts too. Right? And what about H C l? From idea two, H C l adds across the triple bond, and one mole of it turns ethyne into chloroethene. Clear? Now only N a O H is left, and sodium hydroxide is a strong base. Surely it takes the proton from ethyne, right? No, no, careful, it does not! Look at the order from idea one: water is a stronger acid than ethyne. So O H minus, the hydroxide ion, is too weak a base to remove the proton of ethyne. Got it? So the answer is N a O H, sodium hydroxide, the only one that does not react. With me?

6:22Example 2: ethyne and HOCl give dichloroethanal (2002)

Now the harder one, also from the two thousand two paper. What is the product when ethyne reacts with hypochlorous acid? Pause the video and try it first. Done? Okay, let's check. Hypochlorous acid, HOC l, adds across the triple bond with chlorine on one carbon and O H on the other, as it does with alkenes. So H C triple bond C H gives C l C H double bond C H, with O H in brackets. Okay? Now look, this product still has a double bond. So does the reaction stop here, after one molecule of HOC l? It does not stop there; just like H B r in idea two, a second molecule adds. Chlorine joins the carbon that already has chlorine, and O H joins the other one. We get C l two C H, single bond, C H, O H two in brackets. Right? So, is this the final product, the one with two O H groups on one carbon? Oh no, careful, that is the trap! This compound loses a molecule of water, H two O, so it is not the final product. Clear? So the product is dichloroethanal: C l two C H, single bond, CHO, where CHO is the aldehyde group. Stop after one molecule, or keep both O H groups, and you lose the marks. Got it?

7:52Quick revision: alkyne acidity, two HX, 1,2 vs 1,4 addition

So, quick revision. The hydrogen of a terminal alkyne is acidic: the amide ion and Grignard reagents remove it, but hydroxide cannot. Alkynes add two H X by Markovnikov's rule, and ethyne adds two HOC l. And for a diene with H B r, one two addition is kinetic, and one four addition is thermodynamic, at about forty degrees Celsius. Now solve the five questions on this page again, without looking. All the best!