Counting σ and π Bonds

Transcript

0:00Introduction

Hello everyone! Let us begin with a small puzzle. Ethene, the simplest alkene, has five sigma bonds and only one pi bond. Sigma and pi are the two Greek letters you see in the title. So here is the puzzle: can a molecule ever have as many pi bonds as sigma bonds? Sounds strange, right? Keep that question in your head; we will meet such a molecule at the end. And counting these bonds comes up often in JEE questions. So, let's begin.

0:32Every bond has exactly one sigma bond

So, two kinds of bond today. The Greek letter like a small o with a tail is sigma, and the one like a little table is pi. Here pi is just the name of a bond, not three point one four, okay? First, a single bond: one line between two atoms, like the C, dash, C on screen. Every single bond is one sigma bond. Clear? Now, a double bond. Let me ask you this: is a double bond just two sigma bonds? No, it is not, and this is the key idea. In organic molecules, the first bond between two atoms is always a sigma bond, and every extra bond is a pi bond. So the double bond, the two lines on screen, is one sigma plus one pi. Okay? The same rule works for the triple bond, the three lines you see here. One sigma, and the two extra bonds are both pi bonds. So one sigma plus two pi. Got it?

1:23Counting pi bonds and pi electrons

Alright, counting pi bonds is really simple. The number of pi bonds equals the number of double bonds, plus two times the number of triple bonds. Each double bond gives one, and each triple bond gives two. With me? Let me test you. Here is propyne, written as C H three, single bond, C, triple bond, C H, on screen, with one triple bond. The small three written low after the H is a subscript: three hydrogen atoms on that carbon. So tell me, how many pi bonds does propyne have? Yes, exactly, two pi bonds! There is one triple bond, and a triple bond carries two pi bonds. The double arrow on screen simply means therefore, okay? And notice that the triple bond still has its one sigma bond as well. Now one more fact. Each pi bond holds two pi electrons. So for pi electrons, count the pi bonds and double the number. Clear? For example, buta one three diene. The name diene tells you there are two double bonds, at positions one and three. So two pi bonds, and that means four pi electrons. Easy, isn't it?

2:32Counting sigma bonds: the N minus 1 plus r shortcut

Now the sigma bonds. You could draw every bond and count, but there is a faster way. Let capital N be the number of atoms in the molecule, and small r the number of rings. Then the number of sigma bonds equals N, minus one, plus r. Here is the reason: joining N atoms without any ring needs N minus one links, each with exactly one sigma bond. Each ring closes one extra link, so we add r, okay? Let us try it on ethene, C H two, double bond, C H two, as you see here. Count the atoms first, and then tell me the number of sigma bonds. Two carbons plus four hydrogens, that is six atoms, and no ring. So sigma equals six minus one plus zero, which gives five. If you got five, very good! Check by drawing it: four C H bonds and one C C sigma bond, so five sigma bonds. The shortcut works! And the double bond gives one pi bond. Five sigma and one pi, just as in our puzzle.

3:33The trap: hidden hydrogens

Right? Now, the most common mistake. Benzene, C six H six, is usually drawn in bond-line form: a hexagon with three double bonds inside, and no hydrogen drawn at all. Okay? So let us count the lines. Six C C bonds around the ring, six sigma, plus three pi bonds from the double bonds. So benzene has six sigma bonds, right? Oh no, careful, this is exactly the trap! Each carbon of the ring also carries one hydrogen that the drawing does not show. Write them out: six C C bonds plus six C H bonds, which gives twelve sigma bonds. Clear? And the shortcut agrees. Twelve atoms and one ring, so twelve minus one plus one, equals twelve. So benzene has twelve sigma and three pi bonds. With me? So here is the rule. Bond-line and condensed formulas hide the C H bonds, so always write out the hydrogens before you count.

4:27Example 1: sigma and pi bonds in acetic acid

Got it? Alright, now Example one. Count the sigma and pi bonds in acetic acid, written as C H three COOH on screen. Remember the trap we just saw. Pause the video and try it first. Done? Okay, let's check. First, write it out fully. In the COOH group, the carbon has a double bond to one oxygen and a single bond to an O H group. The equals sign inside the bracket shows that double-bonded oxygen. Okay? Now count the atoms: two carbons, four hydrogens and two oxygens, so eight atoms in all, and no ring. So sigma equals eight minus one, which is seven. Right? Let us check by listing them: three C H bonds, one C C bond, the sigma part of the C O double bond, the C O single bond, and the O H bond. That makes seven. Many students miss the O H bond, because the condensed formula does not draw it, so do not fall for that. Now the pi bonds. Look at the structure again: how many are there? Yes, just one, from the C O double bond, so pi equals one. So acetic acid has seven sigma bonds and one pi bond. Clear?

5:40Example 2: sigma and pi bonds in tetracyanoethene

Now a harder one. Count the sigma and pi bonds in tetra cyano ethene. On screen, N C sits inside a bracket with a small two outside, then C double bond C, then C N in a bracket with a small two again. The small two after a bracket means two such groups, so there are four C N groups in all. Pause the video and try it first. Done? Okay, let's check. Each C N here is a cyano group: a carbon joined to a nitrogen by a triple bond. Keep that triple bond in mind, okay? Now count the atoms. Six carbons, two in the middle and four in the cyano groups, plus four nitrogens, so ten atoms. No ring, and look, no hydrogen at all, so nothing is hidden this time. Right? So sigma equals ten minus one, which gives nine sigma bonds: one in the central double bond, four C C single bonds to the cyano groups, and four inside the triple bonds. Clear? Now the pi bonds. There is one double bond in the middle and there are four triple bonds. Tell me the total. One from the C C double bond, plus four times two from the four triple bonds. One plus eight gives nine. If you said five, you counted each triple bond as only one pi bond. That is the trap, okay? So the answer is nine sigma and nine pi. Remember our puzzle from the start? Here is the answer: tetra cyano ethene has exactly as many pi bonds as sigma bonds! Got it?

7:10Quick revision: bond rules, N minus 1 plus r, hidden H

So, quick revision. Every bond has one sigma; a double bond adds one pi, and a triple bond adds two. Sigma equals N minus one plus r, and pi equals the doubles plus two times the triples. And always write out the hidden hydrogens before you count. Now go and try the practice problems on this page. You will find them easy. All the best!