Addition of Halogens, Water and Hydrogen

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take one alkene, three methyl but one ene, and add water to it: overall, one hydrogen and one O H group. Simple, right? But change the reagent, and the same alkene gives three different alcohols! Same alkene, same water, three products? Strange, isn't it? By the end of this lesson, you will predict all three. Questions on these additions come up often in JEE, so let's begin!

0:34Halogens add through a halonium ion

So, first the halogens. Bromine, B r two, or chlorine, C l two, the small two meaning two atoms, in C C l four, carbon tetrachloride. The arrow means gives, and we get a vicinal dihalide, with the two halogens on neighbouring carbons. Okay? But the halogen does not add in one go. The reaction goes through a cyclic halonium ion: a small ring in which the positive halogen is joined to both carbons, on one face of the molecule. With me? Now the halide ion comes in to open this ring. The ring covers one face. So from which side can the halide ion attack? From the opposite side, yes, exactly! So the two halogen atoms end up on opposite faces, and we call this anti addition. Remember this word anti, because it decides the next scene. Clear?

1:25Anti addition decides the stereochemistry

Now the stereochemistry with bromine, and here is a trap. Take a cis alkene. It looks symmetric, so it should give the symmetric product, the meso compound, right? No, no, careful, it is the other way round! Anti addition to a cis alkene gives the racemic dibromide: the two mirror image forms in equal amounts, marked by the plus or minus sign in brackets. Okay? And the trans alkene gives the meso dibromide. But note the condition in brackets: the two ends of the double bond must be alike. Got it? And a cycloalkene? Anti addition puts the two halogens on opposite faces of the ring, so we get the trans one two dihalide, with the halogens on carbons one and two. Right?

2:09Halogen in water: the halohydrin

Now let us change the solvent to water. X two plus H two O, where X stands for a halogen, acts as HOX, a hypohalous acid. The product is a halohydrin, with a halogen and an O H group on neighbouring carbons. Okay? X plus, the positive halogen, adds first, to the carbon with more hydrogens; the small plus sign shows its charge. With me? Then water attacks the more substituted carbon. So the O H group ends up there, and the halogen on the carbon with more hydrogens. Right? So, propene with bromine water gives one bromo propan two ol: bromine on carbon one, O H on carbon two. Notice, the halogen, not a hydrogen, goes to the carbon with more hydrogens. Clear?

2:55Three ways to add water

Now water itself, and there are three ways to add it. First, dilute H two S O four, sulphuric acid. The orientation is Markovnikov, but it goes via a carbocation, so a rearrangement is possible. Keep this in mind for Example two. Okay? Second, oxymercuration, as you see on screen. H g, then O A c in brackets with a small two, is mercuric acetate, used with water; then, as a second step, N a B H four, sodium borohydride. Again Markovnikov, but with no rearrangement. Got it? Now let me ask you something. Both methods put the O H group on the more substituted carbon. What if we want it on the carbon with more hydrogens? Hydroboration oxidation, that is the answer! B two H six, diborane; then H two O two, hydrogen peroxide, with O H minus, the hydroxide ion; the small minus sign shows its charge. Anti-Markovnikov, syn addition, so H and O H go to the same face, and no rearrangement. With me? And there is one more route. Cold, concentrated sulphuric acid adds to give an alkyl hydrogen sulphate, with Markovnikov orientation, and water then hydrolyses it to the alcohol.

4:13Hydrogenation: syn addition, and a way to count

Clear? Finally, hydrogen. H two with nickel, platinum or palladium, written N i, P t or P d on screen. Both hydrogens add to the same face on the metal surface: syn addition, the opposite of bromine's anti addition. Okay? And the less substituted, less crowded the double bond, the faster it is hydrogenated, because it reaches the catalyst surface more easily. Right? Now a topper's tool. One mole of H two, or of B r two, is taken up per double bond. So the amount taken up lets us count double bonds, or find a molar mass. Let's use it right now!

4:48Example 1: alkene of 2.8 g that decolourises 8 g of bromine

Time for the first example. Two point eight grams of an alkene decolourises eight grams of bromine. Identify the possible alkenes. Pause the video and try it first. Done? Okay, let's check. First, the moles of bromine. The atomic mass of bromine is eighty, so we divide eight by eighty, right? Oh no, careful, bromine adds as the molecule B r two! One double bond, the C, double bond, C you see on screen, takes one mole of B r two, of molar mass one hundred and sixty grams per mole; g mol to the power minus one means grams per mole. With me? So n of B r two, the number of moles of bromine, is eight over one hundred and sixty, which gives zero point zero five moles. One mole of bromine per mole of alkene, so n of alkene is also zero point zero five moles. Okay? Now the molar mass, capital M, is two point eight over zero point zero five, which gives fifty six grams per mole. Right? Which alkene is that? An alkene is C n H two n, with n carbons and two n hydrogens. Each C H two unit weighs fourteen, so fourteen times n equals fifty six, and n equals four. The formula is C four H eight. Clear? So the possible alkenes are but one ene, but two ene, or two methyl propene. All three are C four H eight, so these data cannot tell them apart. Got it?

6:18Example 2: three alcohols from hydrating 3-methylbut-1-ene

Now the harder one, and it is our puzzle from the start. Three methyl but one ene is hydrated by dilute sulphuric acid in part a, by oxymercuration demercuration in part b, and by hydroboration oxidation in part c, okay? Name the major alcohol in each case. Pause the video and try it first. Done? Okay, let's check. Here is the structure: two C H three groups, the brackets with a small two, on a C H, then C H, a double bond, and C H two at the end. Part a, dilute acid, Markovnikov: hydrogen to the end carbon, O H to carbon two. So three methyl butan two ol, right? No, no, careful, this goes via a carbocation! The secondary cation on carbon two sits next to a tertiary carbon, so a one two hydride shift gives the more stable tertiary cation. Water attacks it, and part a gives two methyl butan two ol. With me? Part b, oxymercuration demercuration. Markovnikov again, but with no rearrangement, so the O H stays on carbon two: three methyl butan two ol. Got it? And part c, hydroboration oxidation, is anti-Markovnikov: the O H goes to the end carbon, giving three methyl butan one ol. And there it is, our puzzle solved: one alkene, three reagents, three different alcohols! Okay?

7:46Quick revision: anti, halohydrin, three hydrations, syn H2

So, quick revision. Halogens add through a halonium ion, anti; with B r two, cis gives racemic and trans gives meso. Halogen in water gives a halohydrin, with O H on the more substituted carbon. Water adds with dilute acid, which may rearrange, by oxymercuration, or by hydroboration, which is anti-Markovnikov. And hydrogen with nickel, platinum or palladium adds syn, one mole per double bond. Now go and try the practice problems on this page. All the best!