Addition of Hydrogen Halides

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Add hydrogen bromide to propene, and the bromine ends up mainly on the middle carbon, giving two bromo propane. Now add a little peroxide to the flask, and the bromine goes to the end carbon, giving one bromo propane! So how does the peroxide change the product? Strange, right? Hold on to that question. In JEE, questions on this come up often. So, let's begin!

0:29Markovnikov's rule: where does the H go?

So, look at the picture. On the left is propene: C H three, C H, two short lines for the double bond, and C H two; the small numbers count hydrogen atoms. Hydrogen bromide adds across the double bond, hydrogen to one carbon, bromine to the other. So which carbon takes the hydrogen? Yes, exactly, the C H two carbon, which already has more hydrogens! This is Markovnikov's rule. When an unsymmetrical reagent like H X adds to an unsymmetrical alkene, the hydrogen goes to the doubly bonded carbon with more hydrogen atoms. Okay? And X, the other part of the reagent, goes to the other carbon. But a rule is not a reason, right? Why should the hydrogen go to that carbon? It is because of the carbocation. H plus adds first; the small raised plus is its positive charge. Follow the arrows: on the top path, H plus goes to the C H two end, and the plus sign sits on the middle carbon. The two with a small circle means secondary: the more stable secondary carbocation. The lower path gives a less stable primary one. In modern form, the electrophile adds to give the more stable carbocation. Got it? Then B r minus, the bromide ion with a small raised minus, attacks the cation. The secondary cation forms faster, so propene plus H B r gives two bromo propane, the major product; the arrow means gives. One bromo propane is the minor product.

2:02How HX adds: two steps through a carbocation

Clear? So, let us write the mechanism. Step one is the slow step. The pi electrons of the double bond, written with the Greek letter pi, take H plus, and a carbocation is formed. Okay? Step two is fast. X minus, the halide ion, attacks the carbocation from either face. Remember the words either face; we will need them in a minute. With me? Now, which hydrogen halide adds the fastest: H I, H B r, or H C l? Here is a hint: in the slow step, H X has to give away its H plus. Yes, exactly, H I is the fastest! H I, greater than H B r, greater than H C l; the pointed signs mean greater than. That is the order of acid strength. Right? And what about the alkenes? C H three in brackets with a small two outside means two methyl groups on one carbon: two methyl propene, the fastest, then propene, then ethene. They give tertiary, secondary and primary cations. So the more stable the carbocation formed, the faster the addition. Clear?

3:09A carbocation can rearrange, and it is planar

Now, because a carbocation is formed, it can rearrange. In a one, two hydride shift, a hydrogen moves with its electron pair from the next carbon over to the positive carbon. In an alkyl shift, an alkyl group moves instead. Okay? And when does it happen? Whenever the shift gives a more stable cation, say secondary turning into tertiary. This is a common trap, and our second example is built on it. Got it? One more question. Suppose the addition creates a new chiral centre. Will the product be optically active? No, and here is why. The carbocation is planar, so X minus attacks it from either face, equally. Both mirror image forms are made in equal amounts, a racemic mixture, which is optically inactive. Clear?

3:58The peroxide effect: anti-Markovnikov, HBr only

Now, back to our puzzle. When a peroxide is present, H B r, and only H B r, adds by a different path, a free-radical chain. This is the peroxide effect. Okay? Here is how the chain starts. R O, a dash, O R is the peroxide; it breaks into two R O radicals, and the dot marks the unpaired electron. R O dot then takes the hydrogen of H B r, giving ROH and B r dot, a bromine atom. Right? Now the bromine atom meets propene. Which carbon will it attack? Here is a hint: a secondary radical is more stable than a primary one. Yes, exactly, it attacks the C H two end! The dot above the C shows the radical on the middle carbon, the more stable secondary radical. And the bromine is already on the end carbon. With me? This radical then takes a hydrogen from another H B r. We get C H three, C H two, C H two B r, which is one bromo propane, plus a fresh bromine atom to carry the chain on. Okay? So that is our puzzle solved! Without peroxide, H plus adds first; with peroxide, the bromine atom adds first, to the carbon with more hydrogens. That gives the anti-Markovnikov product. Got it? Now a common question. Will H C l or H I show this peroxide effect too? No, neither of them does. With H C l, the hydrogen chlorine bond is too strong to be broken by the carbon radical. With H I, the iodine atoms combine with each other instead of adding to the double bond. Clear?

5:42Example 1: HCl adds to 2-methylpropene

Time for the first example. What is formed when H C l, hydrogen chloride, adds to two methyl propene? Pause the video and try it first. Done? Okay, let's check. Two methyl propene is C H three in brackets with a small two, then C, the double bond, and C H two. One carbon of the double bond has two hydrogens, the other has none. So which one takes the H plus? That is correct, it is the C H two carbon, by Markovnikov's rule. The plus charge then sits on the central carbon, with three methyl groups: C H three in brackets with a small three, then C plus. That is a tertiary cation. Okay? Finally, C l minus attacks the cation, and the chlorine sits on the central carbon. The product is two chloro, two methyl propane. See, the hydrogen went to the carbon that already had more hydrogens. Clear?

6:39Example 2: 3-methylbut-1-ene with HCl and a hydride shift

Now the harder one. Three methyl but one ene reacts with H C l. Name the two products, and say which one is formed by rearrangement. Pause the video and try it first. Done? Okay, let's check. Three methyl but one ene is C H three in brackets with a small two, then C H, then C H, the double bond, and C H two. H plus adds to the C H two end, and the plus sign sits on the next carbon, a secondary carbocation. With me? If C l minus attacks this cation, we get two chloro, three methyl butane. Many students stop right here, with just this one product. But the question asks for two, so where does the second one come from? Look at the carbon next door, which has one hydrogen and two methyl groups. This is the rearrangement we just learnt. That hydrogen moves over with its electron pair, a one, two hydride shift, and the plus sign moves to that carbon. That is a more stable tertiary cation. Clear? When C l minus attacks this tertiary cation, we get two chloro, two methyl butane. So both products form, and two chloro, two methyl butane is the rearranged one. Got it?

7:55Quick revision: Markovnikov, rearrangement, peroxide effect

So, quick revision. Markovnikov's rule: the hydrogen goes to the carbon with more hydrogens, giving the more stable carbocation. The carbocation may rearrange, and a new chiral centre comes out racemic. And with a peroxide, H B r alone adds by free radicals, the anti-Markovnikov way. Now go and try the practice problems on this page. All the best!