Isomerism in Halogen Compounds

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take one formula, C five H eleven C l. One student draws every possible structure, very carefully, and gets eight. The answer key says eleven. So who made the mistake? Maybe nobody made a mistake at all. By the end of this video, you will know exactly where those three extra isomers come from. Counting isomers comes up often in JEE, and with a clear method it is easy marks. So, let's begin.

0:34Monohalides: one isomer per kind of hydrogen

So, first, monohalides, with just one halogen atom. On screen, the letter X stands for the halogen: fluorine, chlorine, bromine or iodine. The general formula is C n, H two n plus one, X, where the small letters and numbers written low are subscripts that count the atoms. Okay? Now the rule. Take an alkane skeleton and replace one hydrogen by the halogen. Hydrogens that sit in identical positions give the same compound, so they count as one kind. So, one isomer for each kind of hydrogen, in each skeleton. Clear? Let me ask you this. Propane has three carbons. So, the halogen on carbon one, carbon two or carbon three gives three isomers of C three H seven X, right? No, no, wait, that is the trap! Carbon one and carbon three are both end carbons, so their hydrogens are the same kind. Propane has only two kinds of hydrogen, on an end carbon or on the middle carbon. So C three H seven X has just two isomers. Got it? In the same way, C four H nine X has four isomers, and C five H eleven X has eight. We will build both of these ourselves in the examples. With me? And here is the trick that toppers use. First list every carbon skeleton, straight and branched, then count the kinds of hydrogen in each one. That way you never miss an isomer, or count one twice. Right?

2:07Chiral centres add enantiomers

Now, the second idea. Look at each carbon in the molecule. If a carbon carries four different groups, we call it a chiral centre. Okay? So, take two bromobutane, with the bromine on carbon two. Count the groups on that carbon. Are all four of them different? Yes, exactly, all four are different! On screen, C dash two means carbon number two. It carries a hydrogen, a bromine, a methyl group, C H three, and an ethyl group, C two H five. So carbon two is a chiral centre. Right? Such a molecule cannot be placed exactly on its own mirror image, just like your left and right hands. So it exists in two forms, the molecule and its mirror image, and the two are called a pair of enantiomers. Clear? So here is the counting rule. When the question asks for stereoisomers too, each compound with one chiral carbon adds one more, its enantiomer. When it asks only for structural isomers, add nothing. Read the question carefully, because this is where marks are lost.

3:15Double bonds and benzene rings

Got it? Now, a third way to get isomers: the double bond. Take one two dihaloethenes, like one two dichloroethene, with one halogen on each carbon of the double bond. There is no free rotation about the double bond, so these compounds show cis trans isomerism. With me? In the cis form, both halogens sit on the same side of the double bond. In the trans form, they sit on opposite sides. Same formula, same bonds, but a different arrangement in space. Okay? And the last idea, the benzene ring. Put one group on carbon one of the ring. Where can the second group go, and how many different compounds do you get? Very good if you said three! Two groups on a benzene ring give three position isomers. Right? Next door, at positions one and two, is ortho, written as a small o. One carbon apart, at one and three, is meta, a small m. Directly opposite, at one and four, is para, a small p. And positions one five and one six just repeat meta and ortho. Clear?

4:25Example 1: the 4 structural isomers of C4H9Br and classes

Alright, Example one, and it is your turn. Write all the structural isomers of C four H nine B r, with their IUPAC names. And for each one, say whether it is primary, secondary or tertiary. Pause the video and try it first. Done? Okay, let's check. Four carbons give two skeletons: the straight chain, butane, and the branched one, two methylpropane. Each skeleton has two kinds of hydrogen. Right? On butane, bromine at an end carbon gives one bromobutane. The carbon carrying bromine is joined to just one other carbon, so it is primary. On screen, that is a one with a small circle written high, read as one degree. Bromine on an inner carbon gives two bromobutane, joined to two carbons, so it is secondary, two degree. Okay? Now, many students write one more structure here, with the bromine on carbon three. Is that a new isomer? No, it is not, so be careful here! Number the chain from the other end, and carbon three becomes carbon two. So it is just two bromobutane again, the same kind of hydrogen. Counting it twice is a very common mistake. Got it? Now the branched skeleton. Bromine on one of the C H three groups gives one bromo two methylpropane, joined to one carbon, so primary. Bromine on the central carbon gives two bromo two methylpropane, joined to three carbons, so tertiary, three degree. With me? So C four H nine B r has four structural isomers, exactly the four that our rule promised. That's it, all four with names and classes!

6:11Example 2: the 3 chiral isomers of C5H11Cl, 11 in total

Now the harder one, and it brings back our puzzle. Example two. Which of the eight isomers of C five H eleven C l are chiral, and what is the total number of isomers, including stereoisomers? Pause the video and try it first. Here is a hint: hunt for carbons with four different groups. Done? Okay, let's check. Start with the straight chain, pentane. It has three kinds of hydrogen, giving one chloropentane, two chloropentane and three chloropentane. Okay? Now let me ask you this. In three chloropentane, carbon three carries a hydrogen and a chlorine. Is it a chiral centre? Oh no, careful, it is not chiral! Its other two groups are both ethyl, C two H five, so they are the same. But in two chloropentane, carbon two carries hydrogen, chlorine, methyl and propyl, four different groups, so it is chiral. And one chloropentane has no chiral carbon at all, since every one of its carbons carries at least two hydrogens. So from pentane, only two chloropentane is chiral. Right? Next, two methylbutane has four kinds of hydrogen, so four isomers. And two two dimethylpropane has just one kind, so one isomer. Three plus four plus one gives eight, the eight in the question. With me? Among the four from two methylbutane, two are chiral: one chloro two methylbutane and two chloro three methylbutane. In both, carbon two carries four different groups. The other two have no such carbon, and neither does the one from two two dimethylpropane. Clear? So three chiral compounds, and each one adds its enantiomer. Eight plus three gives eleven isomers in all. And there is our puzzle, solved! The student's eight counts only structures, and the answer key's eleven counts the stereoisomers too. Both were right, for different questions.

8:13Quick revision: kinds of H, enantiomers, cis-trans, o m p

Okay? So, quick revision. For monohalides, count one isomer for each kind of hydrogen, in each skeleton. Each compound with one chiral carbon adds its enantiomer, but only when stereoisomers are asked. One two dihaloethenes show cis and trans forms, and disubstituted benzenes have three, ortho, meta and para. Now you have a clear method, and that is real progress. Now go and try the practice problems on this page. All the best!