Drawing Structures; Degree of Unsaturation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Benzene has the formula C six H six: six carbon atoms, but only six hydrogen atoms. Now, six carbons in a fully saturated chain would carry fourteen hydrogens. So where did the other eight hydrogens go? Strange, right? By the end of this video, you will read the answer straight from the formula. And on the way, we will learn the quick ways chemists draw molecules. Questions on both come up often in JEE, so let's begin.

0:36Ways of writing a structure

So, how do we draw a molecule on paper? The first way is the complete structure, also called the Lewis or dash structure. Every bond is drawn as a line, even every bond to hydrogen. Clear, but slow for a big molecule. Okay? The second way is the condensed structure. Here the bonds to hydrogen, and often all the bonds, are left out. The first example reads C H three, C H two, C H two, then O H, all in one line. The small numbers written low are subscripts, and each one counts the atoms just before it. In the second, the bracket with a small four outside means four C H two groups in a row. Right? The third way, and the most useful, is the bond-line structure. Only the carbon skeleton is drawn, as a zig-zag, and every corner and every line end is a carbon. The figure shows five molecules drawn this way: pentane, three methyl but one ene, propanoic acid, benzene and cyclohexanol. Look at pentane, the plain zig-zag on the left. How many carbons does it have? It has five, yes exactly! Two line ends, plus three corners, gives five carbon atoms. So count the corners and the ends, every time. Clear? But where are the hydrogens? They are not drawn at all. Each carbon simply carries enough hydrogens to make four bonds. Atoms other than carbon and hydrogen are always written, like the O H in propanoic acid and in cyclohexanol. Got it?

2:03Wedge–dash: bonds in three dimensions

The fourth way is the wedge dash structure, which shows a molecule in three dimensions on flat paper. Here, a plain line is a bond lying in the plane of the paper. With me? The solid wedge, a filled triangle, is a bond coming out of the paper, towards you. So what do you think the dashed wedge shows? Yes, just the opposite! The dashed wedge is a bond going away from you, behind the paper. Okay?

2:30Degree of unsaturation (DU)

Now, back to our puzzle. For n carbon atoms, the saturated formula is C n H two n plus two: n carbons, and two n plus two hydrogens. Right? So six carbons would carry two times six plus two, which is fourteen hydrogens: C six H fourteen. Benzene, C six H six, is eight hydrogens short of that. Clear? So let me ask you this: what in a molecule can use up hydrogens like that? Think about multiple bonds, and about rings. Yes, exactly, both of them! Each ring, and each pi bond, removes two hydrogens from the saturated formula. Pi is the Greek letter you see on screen. The pi bond is the extra bond in a multiple bond, so a double bond has one pi bond and a triple bond has two. Okay? So we define the degree of unsaturation, D U for short, as the number of rings plus the number of pi bonds. Got it? You will also meet it as the double bond equivalent, or the index of hydrogen deficiency. Three names for one idea.

3:31DU straight from the molecular formula

With me? Now, the shortcut that works for any formula. Write the molecule with capital letters for the atoms and a small letter written low after each one, counting those atoms. Capital X here stands for any halogen. Okay? Then D U equals two c, plus two, plus n, minus h, minus x, all over two. So carbons count double, nitrogens add, and hydrogens and halogens subtract. Right? Now look carefully. The molecule may contain oxygen, but where is oxygen in the D U formula? It is nowhere, and that is the point! Oxygen and sulphur do not enter the formula at all, so just ignore them. And the halogen is subtracted exactly like a hydrogen. Many students add it by mistake, so don't do that. Clear? And here is how to read the number. Each double bond or ring counts one, each triple bond counts two, since it has two pi bonds, and a benzene ring counts four. Got it? So let us settle our puzzle. Benzene is C six H six: twelve plus two minus six is eight, and eight over two gives four. Four means one ring plus three C C double bonds, the equals sign inside showing each double bond. Two hydrogens for each of the four, and there go our eight missing hydrogens. So the mystery is solved!

4:51Example 1: cyclohexanol formula from its bond-line structure

Alright, now Example one. Write the molecular formula of the compound whose bond-line structure is a regular hexagon with one O H group attached. That is cyclohexanol, on the right of the figure. Remember the hidden hydrogens. Pause the video and try it first. Done? Okay, let's check. The hexagon has six corners, and every corner is a carbon. The double arrow on screen means therefore. So, six carbon atoms. Right? Now the hydrogens. Each corner has two ring bonds, so it needs two hydrogens to make four bonds. So every corner gets two hydrogens, right? No, no, wait, look at the carbon carrying the O H group. It already has three bonds, two in the ring and one to oxygen, so it has room for only one hydrogen. The other five carbons have just their two ring bonds, so two hydrogens each. Clear? Now add them up. The cross on screen means times. One, plus five times two, which is ten, plus the hydrogen inside the O H group, written in brackets. One plus ten plus one gives twelve hydrogens. Okay? So the formula is C six H twelve O, for cyclohexanol. If you got thirteen hydrogens, you gave the O H carbon two hydrogens, and that is the trap. Got it?

6:05Example 2: rings in C10H16 that takes up 2 mol H2

Now a harder one. Here is a hydrocarbon, C ten H sixteen. On complete catalytic hydrogenation, it absorbs two moles of hydrogen gas, H two, per mole. How many rings does it have? Here is a hint: find the D U first. Pause the video and try it first. Done? Okay, let's check. Ten carbons, sixteen hydrogens, and no nitrogen or halogen. So D U equals twenty plus two minus sixteen, all over two. That is six over two, which gives three. Right? So the compound has three rings, right? Or is it three double bonds? Which one? Careful, we cannot say yet, and this is the trap! D U counts rings and pi bonds together, so on its own it cannot tell them apart. To split them, use the hydrogenation. In this kind of question, hydrogen adds only to the pi bonds, one mole of hydrogen for each pi bond, and the rings stay as they are. So two moles of hydrogen means two pi bonds. Clear? So the number of rings is three minus two, which is one. One ring and two pi bonds, so the answer is one ring. Got it?

7:14Quick revision: hidden H, DU formula, rings vs pi bonds

So, quick revision. In a bond-line structure, every corner and line end is a carbon, and you add the hidden hydrogens yourself. D U is rings plus pi bonds, and it equals two c, plus two, plus n, minus h, minus x, all over two. Oxygen and sulphur are ignored. And hydrogen uptake counts the pi bonds; the rest of the D U is rings. Now go and try the practice problems on this page. You will find them easy. All the best!