Halogenation

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Propane has six hydrogens on its two end carbons, and only two on the middle carbon. In light, chlorine replaces one of them. Six against two, so the end carbon product should win easily, right? Well, it does not, and the middle product actually wins. Let us see why, and predict the percentages. This comes up often in JEE, so let's begin.

0:27Chlorination of methane in light

So, start with methane, C H four, the simplest alkane. The small four written low is a subscript, meaning four hydrogen atoms. Mix it with chlorine, C l two, in light. The h nu over the arrow, the letter h and the Greek letter nu, means light, and the arrow means gives, okay? We get C H three C l, chloromethane, plus H C l, hydrogen chloride. This is halogenation. With me? And it does not stop there. Chlorine keeps replacing hydrogens one at a time, giving C H two C l two, then CHC l three, and finally C C l four, tetrachloromethane. So we can get a mixture. Clear?

1:15The free-radical chain mechanism

Alright, how does this happen? Light brings energy, so some bond breaks first. Is it a carbon hydrogen bond of methane, or the bond between the two chlorine atoms? The bond between the two chlorine atoms, yes, exactly! It is the weakest bond here, the dash on screen. Each chlorine atom leaves with one electron, the dot, and an atom with an unpaired electron is called a free radical. So one chlorine molecule gives two chlorine radicals, and this first step is initiation, okay? Next comes propagation. Now a chlorine radical pulls a hydrogen atom off methane. That gives H C l, and C H three with a dot, a methyl radical. Right? This methyl radical now meets a chlorine molecule. Tell me, what does it make, and what is left over? It takes one chlorine atom to become C H three C l, our product, and the other comes off as a new chlorine radical! That radical attacks another methane, so these two steps repeat thousands of times, as a chain reaction. Got it? The chain stops only when two radicals meet, and this is termination. Two chlorine radicals give C l two, two methyl radicals give C two H six, ethane, and a methyl and a chlorine radical give chloromethane. That is why a little ethane turns up in the products. Interesting, isn't it?

2:40Which halogen? The reactivity order

Now, the other halogens. In this order, F two, C l two, B r two, I two, each pointed sign means greater than. So fluorine is the most reactive, and iodine the least. Okay? Fluorine is so reactive that fluorination is explosive. Not something for your lab bench, right? At the other end, iodination is reversible. The H I formed can turn the products back into methane and iodine. So how do chemists push this reaction forward? Get rid of the H I, exactly! We add an oxidising agent, HIO three, iodic acid, or HNO three, nitric acid, to destroy the H I formed. So the reaction goes forward. With me?

3:24Which hydrogen is replaced first?

Now, bigger alkanes have different kinds of hydrogen. The small circle written up high is read as degree. One degree, or primary, is a hydrogen on a carbon joined to one other carbon; two degree, secondary, to two carbons; and three degree, tertiary, to three. Okay? Let me ask you this: which hydrogen does a halogen radical pull off most easily, primary, secondary or tertiary? Think about the radical left behind. Tertiary first, then secondary, then primary, very good if you said that! Three degree greater than two degree greater than one degree. And the reason is the stability of the radical formed: tertiary radicals are the most stable. Right?

4:08Reactivity per hydrogen decides the mix

Now let us put numbers on it. In chlorination at two hundred and ninety eight kelvin, primary to secondary to tertiary is one to three point eight to five per hydrogen, and the colons mean a ratio. So a tertiary hydrogen is only five times as reactive as a primary one. Okay? Now bromination at four hundred kelvin: one to eighty two to sixteen hundred! Here a tertiary hydrogen is sixteen hundred times as reactive as a primary one. Huge, right? The reason for such a big gap is that bromine is less reactive than chlorine, and therefore far more selective. It gives almost only the product from the most stable radical. Got it? So now we can work out how much of each product we get. The amount is proportional to the number of equivalent hydrogens, times the reactivity per hydrogen, and the curly sign means proportional to. Equivalent hydrogens give the same product. Write this rule down. Clear?

5:05Example 1: 1-chloropropane vs 2-chloropropane percentages

Now, remember our puzzle? Let's finish it. Propane is monochlorinated at two hundred and ninety eight kelvin. Find the percentages of one chloro propane and two chloro propane. Pause the video and try it first. Done? Okay, let's check. Propane has six primary hydrogens and two secondary ones. So, many students say six out of eight, that is seventy five percent one chloro propane, right? Oh no, careful, that counts the hydrogens but forgets their reactivity! Use our rule. Six primary hydrogens times one gives six, and the arrow shows they lead to one chloro propane. Okay? Two secondary hydrogens times three point eight gives seven point six. These lead to two chloro propane, with chlorine on the middle carbon. Right? The total is thirteen point six. So one chloro propane is six over thirteen point six, times a hundred, about forty four percent. The wavy equals sign means approximately. With me? And two chloro propane is seven point six over thirteen point six, times a hundred, about fifty six percent. Only two hydrogens, yet the bigger share, because each is three point eight times as reactive. That solves our puzzle!

6:23Example 2: monochlorination percentages for 2-methylbutane

Now a harder one. Find the percentage composition of the four monochloro products of two methyl butane, as structural isomers, using one to three point eight to five per hydrogen. Pause the video and try it first. Done? First the structure: C H three, then C H with a C H three branch in brackets, then C H two, then C H three, and the dashes are bonds. Carbons are numbered one to four from the left, so C dash two is carbon two. Now, which hydrogens are equivalent? Look at carbon two. Its two methyl groups, carbon one and the branch, are exactly alike, so their six primary hydrogens are equivalent. Six times one gives six, for one chloro two methyl butane. Okay? Carbon two itself has one hydrogen, and it is tertiary. One times five gives five, for two chloro two methyl butane. Right? Carbon three is a C H two, with two secondary hydrogens. Two times three point eight gives seven point six. Numbered from the other end, this product is two chloro three methyl butane. Clear? And carbon four, the last C H three, has three primary hydrogens. Three times one gives three, for one chloro three methyl butane. With me? The total is six plus five plus seven point six plus three, which is twenty one point six. As percentages, in the same order, we get twenty seven point eight percent, twenty three point one percent, thirty five point two percent, and thirteen point nine percent. See, the tertiary product is not the largest: there is only one tertiary hydrogen. Both factors matter!

8:12Quick revision: radical chain, reactivity order, product mix

So, quick revision. Halogenation is a free radical chain: initiation, propagation that repeats, and termination. Halogens react from fluorine down to iodine, and hydrogens tertiary, then secondary, then primary. The amount of a product is equivalent hydrogens times reactivity per hydrogen, and bromine is more selective. Now go and try the practice problems on this page. All the best!