Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Two three dibromo butanedioic acid has two chiral carbons. One chiral carbon gives two mirror image forms, so two chiral carbons give two times two, four stereoisomers, all optically active, right? Hold that thought for a few minutes. This chapter gives you three timed sets of mixed problems from the whole unit. So let us see how to use them, and pick up the quick tools you will need on the way. Let's begin.

0:33How this chapter works

So, first, what is in this chapter? Three sets, with fifteen original problems in each, written in the style of recent JEE Main papers. Okay? And here is the important part. Every set mixes all seven topics of the unit, so the problems do not come chapter by chapter. For each one, you first decide which idea it needs. Right? Now, the timing. Give yourself twenty five minutes per set, and only then check the solutions. So tell me, twenty five minutes for fifteen problems: how much time is that for each problem? Twenty five times sixty, over fifteen; the cross means times, and the long line is a fraction bar. That is one thousand five hundred seconds over fifteen, which gives one hundred seconds per problem, a little under two minutes. So keep an eye on the clock. With me?

1:24Seven topics, mixed in every set

So, which seven topics? Chapters one and two: structure and bonding, and IUPAC nomenclature, the naming rules. Okay? Chapters three and four: structural, geometrical and optical isomerism, and then conformations. Right? Chapters five and six: electronic effects and aromaticity, and then acids and bases. Clear? And chapter seven: intermediates, reagents and reaction types. In each set, all of these come mixed up, in any order. Got it?

1:59Tools for counting: bonds and unsaturation

Now, our first tools, for counting problems. Every single bond is one sigma bond, sigma being the Greek letter on screen, and every double or triple bond also contains exactly one sigma bond. For a molecule with capital N atoms and small r rings, as the brackets say, the number of sigma bonds is N minus one, plus r. Okay? Pi, the next Greek letter, counts the extra bonds. Each double bond has one pi bond, and each triple bond has two. So pi bonds equal the double bonds, plus two times the triple bonds. Right? Let us try one from Set one. The compound on screen is C H three, C O, C H, double bond, C H two. Each dash is a single bond, the two lines are a double bond, and the small numbers written low are subscripts. How many sigma bonds and how many pi bonds does it have? Its formula is C four H six O, so four plus six plus one gives eleven atoms, and there is no ring. So the sigma bonds are eleven minus one, which is ten. And one C double bond O, plus one C double bond C, gives two pi bonds. Ten sigma and two pi, counting every C H bond, even the ones the formula does not draw. Clear? Next, the degree of unsaturation, D U, which is the number of rings plus pi bonds. Write the formula with a small letter written low after each atom, giving its count, and capital X for any halogen. Then D U equals two times c, plus two, plus n, minus h, minus x, all over two. Got it? And two things to remember. Oxygen and sulphur, O and S on screen, do not enter the formula at all. And to read the answer: a ring or a double bond counts one, a triple bond counts two, and a benzene ring counts four. With me?

4:00Tools for stereoisomers

Now, the tools for stereoisomers. Remember, a molecule is optically active only if it is chiral, that is, it has no plane of symmetry and no centre of symmetry. Okay? Then count n, the stereogenic units: the chiral carbons, each bonded to four different groups, plus the double bonds, C double bond C on screen, that show geometrical isomerism. Right? If the molecule cannot be divided into two identical halves, there are two to the power n stereoisomers; the small n written high is the power. Clear? But what if the molecule is symmetrical, with two identical halves? Will it have more forms than two to the power n, or fewer? Yes, exactly, fewer! Some of the forms are meso: they have chiral carbons, but a plane of symmetry makes them inactive. For a symmetrical molecule with n chiral carbons, and n even, there are two to the power, n minus one, active forms, and two to the power, n over two, minus one, meso forms. Got it?

5:06Example 1: degree of unsaturation of C9H10O2

Now Example one, an easy one from Set one. Find the degree of unsaturation of C nine H ten O two. Pause the video and try it first. Done? Okay, let's check. Some students subtract the two oxygens as well, like hydrogens, and get four. Is that right? No, that is the trap: oxygen does not enter the formula at all. So the carbons, c, are nine, the hydrogens, h, are ten, and the two oxygen atoms are simply ignored. Okay? Two times nine is eighteen. Eighteen plus two, minus ten, gives ten, and ten over two gives five. Right? So this molecule has five rings plus pi bonds in all. If it contains a benzene ring, that alone counts four, leaving room for one more ring or pi bond.

6:02Example 2: stereoisomers of 2,3-dibromobutanedioic acid

Clear? And now Example two, a medium one from Set one, and it is our opening question. How many stereoisomers does two three dibromo butanedioic acid have, and how many are optically active? Here is a hint: look for two identical halves. Pause the video and try it first. Done? Okay, let's check. Butanedioic acid is a four carbon chain with a COOH group at each end, and the bromines sit on carbon two and carbon three; C dash two on screen means carbon number two. Each of these carbons carries hydrogen, bromine, a COOH group and the rest of the chain, which makes four different groups. So both are chiral, and n is two. Okay? So, two to the power two, four stereoisomers, all optically active, just as we guessed at the start, right? Oh no, careful, look at the two halves! Each half is CHB r joined to COOH, so the two halves are identical. This is a symmetrical molecule, and n is two, which is even. Clear? So the active forms number two to the power, n minus one, that is two to the power one, which is two. Those two are a pair of enantiomers, mirror images of each other. Right? And the meso forms number two to the power, n over two, minus one, that is two to the power zero, which is one. So three stereoisomers in all, and only two of them are optically active. That settles our opening question. Got it?

7:39Quick revision: timed sets, DU, sigma bonds, meso forms

So, quick revision. Three sets of fifteen, all seven topics mixed: twenty five minutes each, then check the solutions. For counting: sigma bonds are N minus one plus r, D U ignores oxygen, and a benzene ring counts four. For stereoisomers: two to the power n, but fewer for symmetrical molecules, because of meso forms. Now open Set one, start your twenty five minute timer, and then check the solutions. All the best!