Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a chain with two double bonds, where each double bond can be cis or trans. Two choices for the first one, two choices for the second. So two times two gives four geometrical isomers, right? Well, for one molecule in this problem bank, the answer is only three. So where did the fourth one go? We will catch it today. Questions like this, where you rebuild a molecule from its pieces and then count its isomers, come up often in JEE, so let's begin.

0:34How to use this chapter

So, first, how to use this chapter. This unit problem bank on hydrocarbons has three parts. Part A is at the JEE Main level, so start 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 on the reasons behind the reactions. Not only what happens, but why it happens. 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?

1:10Ozonolysis: cut at the double bond, then join back

Now, both our examples today use one tool, ozonolysis, so let us set it up. O three, with the small three written low, is ozone. It adds to the C C double bond, the two lines between the carbons, and forms an ozonide. Okay? Then we break the ozonide with zinc and water, written Z n and H two O on screen. The result is simple: the molecule is cut at the double bond, and each end gets an oxygen atom. So we get aldehydes and ketones. With me? So, the two lines in front of each group mark the double bond end. An end with C H two gives HCHO, which is methanal. An end with CHR, where the letter R stands for the rest of the chain, gives RCHO, an aldehyde. And an end with C R two gives R two C O, a ketone. Right? Now reverse it. Given the products, how do we get the alkene back? Remove the two oxygen atoms, and join the two carbonyl carbons by a double bond. That's it! Clear? Now here is a question straight from Part C, the concept checks. Why is ozonolysis so useful in finding the structure of an alkene? Because it cuts the molecule exactly at the double bond, and nowhere else. So the carbonyl pieces show you exactly where the double bond was. Very good if you said that! Got it? Two more facts that we need today. When the double bond sits inside a ring, the alkene gives one dicarbonyl compound, a single molecule with two carbonyl groups. And a diene, with two double bonds, uses two moles of ozone. With me?

2:48Example 1: alkene that gives cyclohexanone and methanal

Now Example one, a JEE Main level question from Part A, okay? An alkene gives cyclohexanone and methanal on ozonolysis. Name the alkene. Pause the video and try it first. Done? Okay, let's check. Cyclohexanone has a ring of six carbons. So the alkene must be cyclohexene, right? Oh no, careful, that is the trap here! With its double bond inside the ring, cyclohexene would give one dicarbonyl compound, remember? Here the ring survives, so the double bond was outside it. Now look at the pieces. In cyclohexanone, the round brackets with a small five mean five C H two groups, closed into a ring with one C O carbon. And methanal is O, double bond, C H two. With me? Now join them. Remove the two oxygen atoms, and connect the ring carbon to the C H two carbon by a double bond. Six carbons from the ring and one from methanal, so seven in all. The C H two group now hangs outside the ring by a double bond. Right? Its name is methylidene cyclohexane, also called methylene cyclohexane, the name in the round brackets. Ozonolysis cuts it just outside the ring, so the ring stays whole. Got it?

4:09Example 2: C6H10 to ethanal and ethanedial, count isomers

Now Example two, a JEE Advanced level question from Part B, and it holds our opening puzzle. Here, a hydrocarbon, C six H ten, takes up two moles of hydrogen. On ozonolysis, with zinc and water, it gives two moles of ethanal and one mole of ethanedial. Identify it, and find how many geometrical isomers it has. Pause the video and try it first. Done? Okay, let's check. It takes up two moles of hydrogen, H two on screen, and ozonolysis gives three pieces. Three pieces need two cuts, so the molecule has two C C double bonds. It is a diene. Okay? Now look at the pieces. Ethanal is C H three, C H, double bond O, and we have two of them, as the round brackets say. Ethanedial is O, double bond, C H, then a short dash for a single bond, then C H, double bond, O, so it has a carbonyl group at each end. Right? So let me ask you this: which piece goes in the middle of the chain? Ethanedial, yes, exactly! It has two carbonyl groups, so it joins on both sides. The two ethanals sit at the ends, each with one carbonyl group. So we get C H three C H, double bond, C H, a single bond, then C H, double bond, CHCH three. That is hexa two, four diene, with double bonds starting at carbons two and four. Clear? Let us do a quick check. Two plus two plus two carbons gives six. For hydrogens, there are four on each end piece and two in the middle, so four plus two plus four gives ten. That is C six H ten, perfect! With me? Now let us count the isomers. Look at each carbon of the two double bonds. Every one carries a hydrogen and a different group, so each double bond can be cis or trans. Got it? Two double bonds, each cis or trans. So two times two gives four isomers, right? Oh no, careful, this is exactly our opening puzzle! The two double bonds are similar, and the two ends of the chain are the same. So cis, trans is the same molecule as trans, cis: just turn it end to end, and the equals sign on screen says so. That leaves trans trans, cis cis, and cis trans, only three isomers. And that is where the fourth one went!

6:33Quick revision: ozonolysis back to alkene, cis,trans trap

So, quick revision. Part A is Main level, Part B is Advanced level, and Part C checks the reasons; take each part as a test first, then read the solutions. Ozonolysis cuts at each double bond and puts an oxygen on each end; to go back, remove the oxygens and join the carbonyl carbons. And when the double bonds are similar and the ends are the same, cis trans and trans cis are one molecule, so count carefully. Now open Part A, take it as a test, and start. All the best!