Counting Structural Isomers

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. The formula C five H twelve has just three structural isomers, three different carbon skeletons. Now take out one hydrogen, and put a chlorine atom in its place. That gives C five H eleven C l. How many structural isomers does that have? Three skeletons, so three compounds, right? Well, the real answer is eight! So where do the other five come from? And how do we count them without missing one, or counting one twice? Counting questions like this do come up in JEE, so let's begin.

0:37A method that does not miss any

So, how do we count isomers without missing any? If you draw structures at random, you will surely miss one, or draw the same one twice. So we follow a fixed method, in four steps. Step one: find the degree of unsaturation, D U for short. It tells you how many rings or pi bonds you have to use, and pi is the Greek letter on screen. Okay? Step two: write all the carbon skeletons. First the straight chain. Then shorten it by one carbon, and put that carbon back as a methyl branch. Then shorten it by two carbons, and so on. This fixed order is how you make sure that no skeleton is missed. Right? Now step three, and here comes a trap. Take the straight chain of pentane, five carbons in a row. Put a chlorine on carbon one. Then, instead, put it on carbon five. Are these two different compounds? No, no, wait, they are the same! Just turn the molecule around, and carbon five becomes carbon one. So step three says: place the group, or the double bond, on every different kind of position, and use symmetry to reject the repeats. Got it? So in pentane, where C one means carbon one, carbons one and five are alike, carbons two and four are alike, and carbon three stands alone. Five carbons, but only three kinds of position, so pentane gives just three monochloro products. With me? And step four is the safety check: name every structure you draw. If two structures get the same IUPAC name, they are the same compound, so you count it only once. Okay?

2:08Numbers worth remembering

Now, some numbers worth remembering. The small numbers written low are subscripts; they count the atoms. The formula C four H ten has two structural isomers. C five H twelve has three, C six H fourteen has five, and C seven H sixteen has nine. Clear? Next, the alkyl groups. Remove one hydrogen from an alkane and you get an alkyl group, and the short dash after the formula is its free bond. There are two propyl groups, four butyl groups, C four H nine, and eight pentyl groups, C five H eleven. Right? And here is the shortcut. Take C n H two n plus one X, a monosubstituted alkane, where X stands for the one group that took the place of a hydrogen. The number of such compounds, with X on the carbon chain, is simply the number of alkyl groups. Okay? So now, back to our puzzle from the start. Chlorine on the pentanes, C five H eleven C l. Using this shortcut, how many isomers does it have? Yes, it has eight, one for each pentyl group! And here is where they come from. Count the different kinds of hydrogen on each skeleton. Pentane has three kinds, two methyl butane has four, and two, two dimethyl propane has just one kind. Got it? So three plus four plus one gives eight. That is our puzzle solved. Each skeleton gives one compound for every different kind of hydrogen it has, not just one. With me?

3:31Example 1: the 4 structural isomers of C4H9Cl

Now Example one, the easier one. Write all the structural isomers of C four H nine C l, where C l is the symbol for chlorine. Use the four steps. Pause the video and try it first. Done? Okay, let's check. In step one, C four H nine C l is just butane, C four H ten, with one hydrogen replaced by chlorine. So the D U is zero, which means no rings and no pi bonds, only open chains with single bonds. Okay? Now step two, the skeletons. With four carbons, how many different skeletons can you make? Yes, there are just two skeletons. The straight chain, butane, and the chain shortened by one carbon with a methyl branch, two methyl propane. Right? Now, butane has four carbons, so the chlorine can go on carbon one, two, three or four. So that makes four isomers from butane alone. Is that correct? Oh no, careful, that is the trap we just saw! Carbon four is the same kind as carbon one, and carbon three is the same kind as carbon two. Name them and you will see it: chlorine on carbon three is still two chloro butane, by the lowest locant rule. So butane gives two, one chloro butane and two chloro butane. With me? Now two methyl propane. Its three end carbons are all alike, and the centre carbon is the other kind. So we get one chloro two methyl propane, and two chloro two methyl propane. Got it? So the answer is four isomers, two plus two. And notice, that is exactly the number of butyl groups in our table. Clear?

5:04Example 2: isomers of C4H10O, 4 alcohols and 3 ethers

Now Example two, a little harder. How many structural isomers does C four H ten O have? Here the O is oxygen. Pause the video and try it first. Done? Okay, let's check. C four H ten O has the same number of hydrogens as butane, so again the D U is zero. Only single bonds, and no rings. Right? Now, where can the oxygen go? First, it can carry a hydrogen, as an O H group on a carbon. That gives an alcohol, C four H nine O H, okay? And here our shortcut works again: one alcohol for each butyl group, so four alcohols. With me? Here they are. Butan one ol and butan two ol from the straight chain, and two methyl propan one ol and two methyl propan two ol from the branched one. Clear? Now, a lot of students stop right here, and write four. Is that the full answer? No, no, wait, that is the classic trap! The oxygen can also sit inside the chain, between two carbons. That gives an ether. Methyl with the straight propyl gives one methoxy propane, methyl with the branched propyl gives two methoxy propane, and ethyl with ethyl gives ethoxy ethane. So that makes three ethers. Got it? So the total is four plus three, which is seven: four alcohols and three ethers. So the shortcut counts only the alcohols; the ethers you must add yourself. See, the method found every one, and none of them twice. With me?

6:33Quick revision: skeletons, symmetry, one per alkyl group

So, quick revision. Write the skeletons in order: the straight chain first, then shorter chains with branches. Put the group on every different kind of position, and let symmetry reject the repeats. And for C n H two n plus one X, there is one compound for each alkyl group: four for butyl, and eight for pentyl. Now go and try the practice problems on this page. All the best!