Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small counting puzzle. Take three carbon atoms, five hydrogen atoms and one chlorine atom, the formula C three H five C l. How many different compounds can you build from it? Count them in your head: chains, rings, everything. Did you get three, or maybe four? The answer is five. So where are the missing ones hiding? We will find them today, in this unit problem bank. Questions like this come up often in JEE, so let's begin.

0:35How to use this chapter

So, first, how to use this chapter. It has three parts. Part A is at the JEE Main level, so begin there. Okay? Part B is at the JEE Advanced level, so expect harder questions here. Right? And Part C is a set of concept checks. They test the ideas behind the rules: not only what a rule says, but why it is true. Clear? Now, the most important line. Attempt each part as a test first, and read the solutions only after that. That is how you find out what you really know. Got it?

1:10Part C warm-up: two concept checks

Now let us warm up with two questions from Part C, the concept checks. The first one is on naming: a trimethyldecane, which is a ten carbon chain with three methyl groups. Numbering from one end gives the locants two, seven and eight; from the other end, three, four and nine. The curly brackets on screen just hold each set. With me? Now add each set. Two plus seven plus eight gives seventeen, and three plus four plus nine gives sixteen. Lowest locants means the smaller sum, so three, four, nine is correct, right? No, no, careful, the sum is the trap here! The IUPAC rule compares the two sets term by term, and picks the lower set at the first point of difference. Two is lower than three, so two, seven, eight wins, and the rest does not matter. Okay? The second check is about optical activity. Here, a meso compound has chiral carbons, and yet it is optically inactive. So why does it not rotate plane polarised light at all? Because it has a plane of symmetry, or, as the round brackets add, a centre of symmetry. The rotation due to one half of the molecule is cancelled by the other half. So chiral carbons alone are not enough: a molecule is optically active only if it is chiral. Keep this in mind for Example one.

2:34Tools for counting isomers

Right? Now, the tools for our examples. First, the degree of unsaturation, or D U for short. D U equals two c, plus two, plus n, minus h, minus x, all over two, where the small letters count the carbons, nitrogens, hydrogens and halogens. And D U is the number of rings plus the number of pi bonds; pi is the Greek letter on screen. Clear? Second, geometrical isomerism. Here a, b, d and e are the four groups on a C C double bond, shown by the two lines between the carbons: a and b on one carbon, d and e on the other. The equals sign with a slash through it means not equal to. So cis and trans, or E and Z, isomers exist only if a is not equal to b, and d is not equal to e. Got it? So let me ask you this. If one carbon of the double bond is a C H two group, can the molecule show geometrical isomerism? No, it cannot, very good! Its two hydrogens, shown by the small two written low, are identical groups, so there is only one arrangement, and no geometrical isomers. We need this in Example two. Okay? Third, counting stereoisomers. Let n be the number of stereogenic units: the chiral carbons, and the double bonds that show geometrical isomerism. If the molecule cannot be divided into two identical halves, it has two to the power n stereoisomers; the small n written high is the power. With me?

4:07Example 1: stereoisomers of 2-bromo-3-chlorobutane

Now Example one, a JEE Main level question from Part A, okay? How many stereoisomers does two bromo three chloro butane have? And is any of them meso? Pause the video and try it first. Done? Okay, let's check. The chain is C H three, CHB r, CHC l, C H three; each dash is a single bond. C dash two means carbon number two. Carbon two carries hydrogen, bromine, a methyl group and the rest of the chain: four different groups, so it is chiral. Carbon three is chiral in the same way, with chlorine, so n is two. With me? Now look at the two ends: both are C H three groups. Two chiral carbons and two identical ends, so there must be a meso form, right? Oh no, careful, look at the middle of the chain! One chiral carbon carries bromine, and the other carries chlorine. So the two halves are different, and the molecule has no plane of symmetry. Clear? So we use two to the power n: two squared, which is four stereoisomers. And with no plane of symmetry, none of them is meso. Got it?

5:18Example 2: the five isomers of C3H5Cl, E and Z plus a ring

Now Example two, a JEE Advanced level question from Part B, and this is our opening puzzle. How many isomers, cyclic and acyclic, and including stereoisomers, have the formula C three H five C l? Here is a hint: find the D U first. Pause the video and try it first. Done? Okay, let's check. Three carbons, five hydrogens, one chlorine, and no nitrogen. So D U equals six plus two, minus five, minus one, all over two. That is two over two, which gives one. So every isomer has either one C C double bond, or one ring. Okay? First, the open chains: propene with one chlorine. Put the chlorine on carbon three, away from the double bond: three chloro propene, C H two double bond C H, then C H two C l. It has a C H two end on the double bond, so just one form. Right? Now put the chlorine on the middle carbon: two chloro propene. Again there is a C H two end on the double bond, so there is only one form. Clear? Next, put the chlorine on carbon one, which gives one chloro propene. Just one more compound, right, or could it give more? Look at both carbons of its double bond. It gives two forms, and this is the one that is easy to miss! Carbon one carries chlorine and hydrogen, two different groups, and carbon two carries a methyl group and hydrogen, also different. So both conditions hold, and we get two geometrical isomers, the E form and the Z form. Got it? And now, do not forget the ring! Our D U of one can also mean one ring: a three carbon ring carrying the chlorine, chlorocyclopropane. That is one more compound, with one form. With me? So the total is one plus one plus two plus one, which gives five. There are our missing compounds: the second geometrical form of one chloro propene, and the ring. And that solves our opening puzzle. Okay?

7:22Quick revision: locants, meso forms, counting isomers

So, quick revision. Part A is Main level, Part B is Advanced level, and Part C checks the ideas; take each part as a test first, then read the solutions. Locants are compared at the first point of difference, not by their sum, and a meso compound needs a plane or centre of symmetry. To count isomers, find the D U first, then check the chains, the rings, the E and Z forms, and the chiral carbons. Now open Part A, take it as a test, and start. All the best!