Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a trimethyldecane, a ten carbon chain with three methyl groups, and number it from either end. One way gives the locants two, seven and eight; the other way gives three, four and nine. Two plus seven plus eight is seventeen, but three plus four plus nine is only sixteen. So the smaller sum wins, and the second numbering is correct, right? Hold your answer, because Example two settles it. This is Part C of the unit problem bank: short concept checks, where every answer needs a reason. Questions like these come up often in JEE, so let's begin.

0:43Check 1: conformers versus cis and trans

Let us begin with check number one. Butane twists about its central single bond, and each arrangement is a conformer, like the anti and the gauche forms. Yet nobody can fill one bottle with anti butane and another with gauche butane. With me? The reason is the barrier. Rotation about a single bond has only a low energy barrier: about twelve point five kilojoules per mole for ethane, and not much more for butane. On screen, the minus one written high on mol means per mole. So at room temperature the conformers interconvert rapidly, and cannot be isolated. Okay? Now take cis and trans but two ene, where the two is the position of the double bond. These two isomers can be separated. So what is different about them? Yes, exactly, it is the double bond! To turn one carbon of a double bond, you must break the pi bond, pi being the Greek letter on screen. That does not happen at room temperature. Right? So conformers cannot be isolated, while cis and trans isomers can be separated. Clear?

1:53Check 2: why planar, and why pyramidal?

Check number two is about shape. Here are the carbon intermediates, where each R is an alkyl group. From the left, we have the carbocation with its plus charge, the free radical, the carbanion with its minus charge, and the singlet carbene. Today we need the first and the third. So why is a carbocation planar, but a simple carbanion pyramidal? Let us count the electron pairs. The carbocation carbon has just three bond pairs, and nothing else. Three pairs repel and spread as far apart as possible, one hundred and twenty degrees apart; the small circle means degrees. Okay? Three pairs at one hundred and twenty degrees means s p two, with the two written high, and a flat, planar shape. The leftover p orbital stays empty, the lobes marked empty p in the figure. Right? Now the carbanion. It also has three bond pairs, but look at the two dots on top of the carbon: a lone pair. That lone pair is a fourth electron pair. Got it? Four pairs means s p three, and with one corner taken by the lone pair, the shape is pyramidal. But note the condition: a carbanion that is conjugated, like the allyl, benzyl or enolate ion, is s p two and planar. Clear?

3:11Check 3: one alkyl group, two opposite effects

Next comes check number three. Alkyl groups release electrons, and this is the plus I effect, with I for inductive. Okay? Alkyl groups stabilise carbocations. So they should stabilise carbanions too, right? No, no, wait, look at the charge first. On a carbocation, the electrons pushed in reduce the positive charge. So tertiary is more stable than secondary, then primary, then the methyl cation, C H three plus, with the small three written low; the greater than signs on screen mean more stable than. Three with a small circle written high is read as tertiary, two as secondary, and one as primary. With me? On a carbanion, the same electrons add to a negative charge that is already there. So the order turns upside down: the methyl anion, C H three minus, is the most stable, then primary, then secondary, and tertiary is the least. Got it?

4:05Check 4: benzene and bromine water

And here is check number four. Draw benzene the Kekule way, and you see three C double bond C, the two short lines being a double bond. Yet benzene does not decolourise bromine water. So why does it not add bromine? The answer is aromaticity. By Huckel's rule, a ring is aromatic if it is cyclic, planar and fully conjugated, with four n plus two pi electrons, where n is zero, one, two, and so on. Okay? Benzene has six pi electrons. Four times one, plus two, gives six, so n equals one, and these six electrons are the aromatic sextet. Right? Adding bromine would destroy that sextet, and with it the resonance energy of about one hundred and fifty kilojoules per mole. So benzene does not decolourise bromine water. Clear?

4:55Example 1: why a meso compound is optically inactive

Now the two examples. Example one says: a meso compound has chiral carbons, so why is it optically inactive? Pause the video and try it first. Here is a hint: look for symmetry. Done? Okay, let's check with tartaric acid, drawn as Fischer projections, where each crossing point is a chiral carbon. The first drawing, D glyceraldehyde, is not needed today. Next come the two active forms, two R, three R, and two S, three S, mirror images across the grey dashed line. On the right is the meso form, two R, three S; capital R and S give the configurations of carbons two and three. With me? Now, the meso form has two chiral carbons. So it must rotate plane polarised light, right? No, careful here, because this is where students slip! Look at the dashed line marked plane. The top half, with its COOH, H and O H, is the mirror image of the bottom half: this is a plane of symmetry. Okay? So the top half rotates the light one way, and the bottom half by the same amount the other way. The two rotations cancel, and this is called internal compensation. Right? So the meso compound is achiral and optically inactive; a centre of symmetry, shown in brackets, does the same job as a plane. Remember, chirality of the whole molecule, not just a chiral centre, decides optical activity. Clear? And one more trap to remember. The racemic mixture is also inactive, but it is a mixture of two compounds and can be resolved, while a meso compound is one compound and cannot. Got it?

6:35Example 2: locants 2,7,8 vs 3,4,9, first point of difference

And now Example two, a step harder: our opening puzzle. For a trimethyldecane, the two directions of numbering give the locants two, seven, eight, and three, four, nine; the curly brackets just hold each set. The second has the smaller sum. Which is correct, and why? Pause the video and try it first. Done? Okay, let's check. Two plus seven plus eight gives seventeen, and three plus four plus nine gives sixteen. So three, four, nine wins, right? No, no, careful, because this is exactly the trap! The IUPAC rule never adds the locants. It compares the two sets term by term, starting from the first locant. With me? So compare the first terms, two and three. They already differ, and two is lower than three; the sign on screen means less than. This is the first point of difference, so we stop right here. Okay? So the set two, seven, eight is correct, even though its sum is larger. Puzzle solved, the smaller sum was the trap! If you spotted that, very well done.

7:45Quick revision: conformers, carbanions, meso, lowest locants

So, quick revision. Conformers interconvert, but turning cis into trans means breaking the pi bond. Carbocations are s p two and planar, alkyl carbanions s p three and pyramidal. Alkyl groups stabilise carbocations but destabilise carbanions. Benzene keeps its aromatic sextet, and a meso compound has a plane of symmetry. And locants are compared at the first point of difference, never by the sum. Now go and try the practice problems on this page. All the best!