Set 1

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a molecule with two chiral carbons. Each chiral carbon can be arranged in two ways, so two times two gives four stereoisomers, right? Well, for one acid in today's set, that answer is wrong. So where does the missing one go? This is Set one of the mixed JEE Main pattern problems: fifteen questions that mix all seven topics of this unit, with twenty five minutes for the whole set. We will sharpen three tools from the set, then solve two of its questions. So, let's begin.

0:37Tool 1: hybridisation from pi bonds

So, tool number one is hybridisation. For a carbon with no charge, the course gives a one-line rule: just count its pi bonds, where pi is the Greek letter you see on screen. Zero pi bonds gives s p three, one pi bond gives s p two, and two pi bonds give s p. The small raised numbers are part of the name, and the double arrow means gives. Okay? Now, question one of the set. C H two, double bond, C H, single bond, C H two, single bond, C, triple bond, N; the small numbers count hydrogens, and the three lines are the triple bond. Give the hybridisation of each carbon, from left to right. Okay, let us check. The first two carbons share the double bond, so one pi bond each, and both are s p two. The C H two in the middle has only single bonds, so it is s p three. And the last carbon is triple bonded to nitrogen, not to carbon, but it still has two pi bonds, so it is s p. With me?

1:43Tool 2: carbocation stability

Now, tool number two: carbocations, where a carbon carries a positive charge, the small raised plus sign. Anything that releases electrons to that carbon stabilises it. So, by the plus I effect and hyperconjugation, tertiary beats secondary, secondary beats primary, and primary beats the methyl cation, C H three plus. The small circle means degree, and the sign between them means greater than. Okay? Question twelve: arrange these four in order of decreasing stability. The first three are tertiary, secondary and primary; a bracket with a small number after it repeats the C H three group. The last one is C six H five, a phenyl ring, on a carbon with the plus written above it and two methyl groups. So the tertiary one on the left is the most stable, right? No, no, careful, that is the trap! Look at the last one again. Its charged carbon holds two methyls and a ring, so it is tertiary, and benzylic too; the ring spreads the charge by resonance. Tertiary and benzylic together beats tertiary alone. Right? So the order is: the phenyl one first, then tertiary, then secondary, and the primary cation last. Always check for resonance before you count methyl groups. Clear?

3:03Tool 3: chiral carbons and meso forms

Now, tool number three, and this one solves our puzzle. Look at the figure: two lactic acids, mirror images across the dashed line, like your left and right hands. The solid wedge points towards you, and the dashed bond points away. The central carbon carries four different groups, H, O H, C H three and COOH, so it is a chiral carbon. Okay? So let me ask you this: if a molecule has chiral carbons, is it always optically active? Not always, and this is the key idea! If the molecule has a plane or a centre of symmetry, it can be placed exactly on its own mirror image. Then it is achiral and optically inactive, even with chiral carbons inside, and we call it a meso compound. Got it? Now, counting. Let n be the number of stereogenic units, that is, chiral carbons and double bonds that show geometrical isomerism. If the molecule cannot be split into two identical halves, it has two to the power n stereoisomers. Right? But if it splits into two identical halves, and n is even, there are two to the power, n minus one, optically active forms. Clear? And two to the power, n over two, minus one, meso forms. This needs n to be even; for odd n, the course gives a separate rule. With me?

4:31Example 1: 3 chiral carbons, no symmetry, 8 stereoisomers

Now, example one, which is question nine of the set. C H three, CHC l, C H with O H in brackets, CHB r, C H three. How many chiral carbons does it have, and how many stereoisomers? Pause the video and try it first. Done? Okay, let's check. Number the chain from one to five. The two end carbons are methyl groups with three hydrogens each, so they are not chiral. Carbon two, written C dash two, has H, chlorine, a methyl and the rest of the chain: four different groups. Carbon four is the same story, with bromine. Okay? Now carbon three, in the middle. It has H and O H, and two arms, each starting with a C H group. So is carbon three chiral? Yes, carbon three is chiral too, because one arm carries chlorine and the other carries bromine, so the two arms are different, and carbon three also has four different groups. Clear? So n is three. And this molecule cannot be cut into two identical halves, because one end has chlorine and the other has bromine, so there is no meso form here. Right? So the first rule applies: two to the power three, which gives eight stereoisomers. So the answer is three chiral carbons and eight stereoisomers. Now watch what changes in the next one.

6:00Example 2: 2,3-dibromobutanedioic acid and its meso form

And now, example two, which is question ten of the set, and it is our opening puzzle. How many stereoisomers does two, three dibromo butane dioic acid have, and how many of them are optically active? Pause the video and try it first. Done? Okay, let's check. Read the formula on screen: HOOC, CHB r, CHB r, COOH, and look at carbons two and three. Each carries H, bromine, a COOH group and the rest of the chain, so both are chiral, and n is two. Okay? Two chiral carbons, two arrangements each. So two to the power two, four stereoisomers, right? No, no, careful, this is exactly the trap from our puzzle! Cut the molecule between carbons two and three, and each half is CHB r, COOH, identical. So it is symmetrical, n is even, and we must use the second rule. Clear? Optically active forms: two to the power, two minus one, that is two to the power one, which is two. These two are a pair of enantiomers, mirror images of each other. Right? And meso forms: two to the power, two over two, minus one, that is two to the power zero, which is one. Draw this form as a Fischer projection, and its top half is the mirror image of its bottom half, so it is optically inactive. With me? So the answer is three stereoisomers, and two of them are optically active. And our puzzle is solved: the fourth one never existed, because the meso form is its own mirror image, so that would-be pair is just one compound. If you got this one, really good work!

7:54Quick revision: pi bonds, benzylic cations, meso check

So, quick revision. For a carbon with no charge, zero, one or two pi bonds mean s p three, s p two or s p. Tertiary and benzylic together beats plain tertiary. And when a molecule splits into two identical halves, two to the power n fails, so look for the meso form. Now go and try the other questions of this set, timed, and then check the solutions on this page. All the best!