Conformations of Ethane and Butane

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Ethane has two carbon atoms joined by a single bond, and the two ends can turn about that bond, like wheels on an axle. Every small turn gives a new arrangement in space. So, how many shapes does ethane have? The answer is infinitely many! And yet, you can never bottle even one of them. Strange, right? Today we draw these shapes and compare their energies. JEE questions do test these ideas, so let's begin.

0:33Conformations and the Newman projection

So, look at the top right of the figure: ethane, in a sawhorse drawing. Hold the front carbon still, and rotate the back carbon about the carbon carbon single bond. Every arrangement you get from this rotation is called a conformation. Okay? These are also called conformers, or rotamers. And here is why we can never bottle one of them. At room temperature they interconvert rapidly, so they cannot be isolated. Right? To draw them, we use a Newman projection, like the drawings on the top left. We look straight down the carbon carbon bond. The front carbon is just a point, where the three black bonds meet. The back carbon is the circle, and its blue bonds start from its edge. Clear? And the angle between a front bond and a back bond, seen this way, is the dihedral angle, also called the torsional angle. With me?

1:23Ethane: staggered and eclipsed

Now, the first drawing on the top left is staggered ethane. Each blue hydrogen at the back sits exactly between two black ones at the front, so the dihedral angle is sixty degrees; the small circle on screen means degrees. Now, if we turn the back carbon by sixty degrees more, where do the blue hydrogens go? Yes, exactly, right behind the front hydrogens! This is eclipsed ethane, the second drawing, with a dihedral angle of zero degrees. The figure turns the back bonds slightly so you can see them. So tell me, which form is more stable? Yes, the staggered form, by about twelve point five kilojoules per mole. The k J mol with a small minus one, a superscript, means kilojoules per mole. Got it? This extra energy of the eclipsed form is called torsional strain. Okay? And in between, there are infinitely many skew forms. So that is our puzzle solved. Clear?

2:23Butane: rotating about the C2-C3 bond

Now, let us move on to butane. We rotate about the bond between carbon two and carbon three; in the heading, the small numbers are subscripts, and the dash is the bond. Each of these two carbons carries a C H three, a methyl group. So tell me, where would the two methyl groups like to be? As far apart as possible, yes, exactly opposite! That is the anti form, the first drawing in the second row, with a dihedral angle of one hundred and eighty degrees. Okay? In the gauche form, the methyls are sixty degrees apart, and it is still staggered. In the eclipsed form, at one hundred and twenty degrees, the back methyl hides behind a front hydrogen. And in the fully eclipsed form, at zero degrees, methyl sits right behind methyl. Right?

3:12Butane: the stability order

Now look at this graph: the energy of butane, in kilojoules per mole, against the dihedral angle between the two methyls. Anti is the lowest point, at one hundred and eighty degrees, and the valleys at sixty and three hundred are gauche. The peaks at one hundred and twenty and two hundred and forty are eclipsed, and the highest, at zero, is fully eclipsed. So that is the order on screen; the greater than signs mean more stable than. Okay? Gauche is three point eight kilojoules per mole less stable than anti. But both of them are staggered forms. So why is gauche higher? Because in gauche, the two methyl groups are close enough to crowd each other. This crowding is called steric strain. Got it? Now, two quick rules for moving between forms. Turning by sixty degrees takes a staggered form to an eclipsed one; the arrow on screen means turns into. With me? And turning by one hundred and twenty degrees takes anti to gauche. On the graph, start at one hundred and eighty, go back by one hundred and twenty, and you land at sixty, a gauche valley. Clear?

4:21The exception: hydrogen bonding

Now, careful, here is a classic trap. Take ethane one two diol, with an O H group on each carbon. These two groups can form a hydrogen bond with each other. So which form is more stable here, anti or gauche? If you said anti, then no, wait a moment. Ethane one two diol, two fluoro ethanol, two chloro ethanol and two amino ethanol are the course's examples. The numbers in the names are locants, the carbons carrying each group. Right? For all of these, gauche is more stable than anti, because to form a hydrogen bond the two groups must be close together. So the usual order flips. Okay?

5:03Example 1: fully eclipsed butane turned 60, 120, 180 degrees

Now, let's try one. Start from the fully eclipsed form of butane, and rotate the back carbon by sixty, by one hundred and twenty, and by one hundred and eighty degrees. Name the conformation you reach each time. Pause the video and try it first. Done? Okay, let's check. In the fully eclipsed form, the dihedral angle between the methyls is zero. Turn by sixty degrees, and the angle becomes sixty, the first valley on the graph, which is gauche. Right? Now one hundred and twenty degrees. That turn takes anti to gauche, staggered to staggered. So we get a staggered form here too, don't we? Oh no, careful, look at where we started! From an eclipsed form, a turn of one hundred and twenty degrees lands on another eclipsed form. The angle becomes one hundred and twenty, the next peak on the graph, so it is the eclipsed form. Got it? And at one hundred and eighty degrees, we reach the lowest point, the anti form. So the answers are gauche, eclipsed and anti. Clear?

6:12Example 2: energy of one H/CH3 eclipsing from propane

Now a harder one, with numbers. The barrier to rotation in propane is fourteen kilojoules per mole, and in ethane it is twelve point five. Estimate the cost of one H slash C H three eclipsing interaction; the slash means a hydrogen lined up with a methyl group. Pause the video and try it first. Hint: start with ethane. Done? Okay, let's check. The barrier is the energy gap between the eclipsed and staggered forms. In eclipsed ethane, every front hydrogen has a back hydrogen right behind it, so there are three H slash H eclipsings. With me? So the twelve point five is shared by these three. Twelve point five over three is about four point two kilojoules per mole for each; the wavy equals sign means approximately. Right? Now propane, with a barrier of fourteen kilojoules per mole. Can we simply divide fourteen by three as well? No, and that is the trap here. Looking along carbon one to carbon two of propane, the back carbon carries one methyl group and two hydrogens. So eclipsed propane has two H slash H eclipsings and one H slash C H three eclipsing, which are not the same. Clear? So take the two hydrogen hydrogen costs away from fourteen. Fourteen minus two times four point two; the brackets mean multiply. That is fourteen minus eight point four, which gives about five point six kilojoules per mole for one hydrogen methyl eclipsing. Got it?

7:46Quick revision: ethane energy gap, butane order, H bonding

So, quick revision. Ethane: staggered, at sixty degrees, beats eclipsed, at zero degrees, by about twelve point five kilojoules per mole. Butane: anti, then gauche, then eclipsed, then fully eclipsed, in order of stability. And if the two groups can form a hydrogen bond, gauche beats anti. Now go and try the practice problems on this page. All the best!