The Double Bond
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The carbon-carbon double bond of alkenes for JEE Main and Advanced: sp2 carbons, one sigma and one weaker pi bond, why alkenes are attacked by electrophiles and undergo addition, bond length and bond enthalpy, planarity and geometrical isomers, tests with bromine in CCl4 and Baeyer's reagent, stability from heats of hydrogenation, and two solved problems: the strength of the pi bond and the most stable butene.
Chapters
- 0:00Introduction
- 0:32Inside the double bond
- 1:33Electron-rich, so attacked by electrophiles
- 2:33The numbers: bond length and bond enthalpy
- 3:24Planar, with restricted rotation
- 3:47Tests for unsaturation
- 4:11Stability and heat of hydrogenation
- 4:58Example 1: strength of the pi bond from bond enthalpies
- 5:57Example 2: most stable butene from heats of hydrogenation
- 7:29Quick revision: pi bond, addition, tests, stability
Transcript
0:00Introduction
Hello everyone! Let us start with a small puzzle. The carbon carbon double bond is stronger than the single bond: six hundred and eighty one kilojoules per mole, against three hundred and forty eight. Yet hex one ene takes away the orange-red colour of a bromine solution at room temperature, in the dark, while hexane, an alkane, does not. So the stronger bond is the more reactive one? That sounds strange, right? In JEE, questions on this bond come up often, so let's find out!
0:32Inside the double bond
So, what is inside the double bond? On screen, the two short lines between the two C symbols mean a double bond between two carbon atoms. Both carbons are s p two hybridised; the small raised two means one s orbital mixes with two p orbitals. Okay? Now let me ask you something. There are two bonds between these carbons. Are they two identical bonds, of the same strength? No, and this is the key point. One is a sigma bond, shown by the Greek letter sigma, and the other is a pi bond, shown by the Greek letter pi. And the pi bond is the weaker one. Clear? Now, how is the pi bond formed? By the sideways overlap of p orbitals: the p orbitals of the two carbons meet side by side. Got it? And where do the pi electrons sit? They lie above and below the plane of the molecule. They are held loosely, and they are exposed to attack. With me?
1:33Electron-rich, so attacked by electrophiles
Alright, put these facts together. Exposed, loosely held pi electrons make the alkene a source of electrons. So tell me: an electrophile seeks electrons, and a nucleophile brings its own. Which one attacks an alkene? Yes, exactly, the electrophile! So alkenes react with electrophiles, not with nucleophiles. Right? And what kind of reaction follows? It is addition, in which the pi bond is replaced by two sigma bonds. Okay? Here is a quick check for you. When ethene is hydrogenated, hydrogen adds across the double bond. How does the hybridisation of the two carbons change? From s p two to s p three, correct! The arrow means changes to, and the raised three means three p orbitals now mix with the s orbital. The pi bond is gone, so each carbon is s p three, like in an alkane. Got it?
2:33The numbers: bond length and bond enthalpy
Look at the numbers first. The double bond is one hundred and thirty four picometres long, p m for picometre, and its bond enthalpy is six hundred and eighty one kilojoules per mole; k J is kilojoules, and mol to the power minus one means per mole. Clear? Now the single bond, with one dash between the carbons: one hundred and fifty four picometres, and three hundred and forty eight kilojoules per mole. So the double bond is shorter, and stronger. Okay? So here is a trap. The double bond is stronger, so an alkene should be less reactive than an alkane, right? No, no, careful, the double bond is not one strong bond! It is a sigma bond plus a pi bond, and the pi bond is weaker than the sigma bond. Example one puts a number on it. With me?
3:24Planar, with restricted rotation
Next, let us look at the shape. The molecule is planar at the double bond, and rotation about the double bond is restricted. Right? This restricted rotation can give geometrical isomers, the cis and the trans forms, when each carbon of the double bond carries two different groups. You will meet one such pair in Example two. Clear?
3:47Tests for unsaturation
So, how do we test for a double bond? First, bromine in C C l four, carbon tetrachloride, with a small four for four chlorine atoms. An alkene takes away its orange-red colour. Got it? Second, cold, dilute, alkaline K M n O four, potassium permanganate, named in brackets as Baeyer's reagent. An alkene decolourises it too.
4:11Stability and heat of hydrogenation
Okay? Alright, the last idea is stability. The more alkyl groups on the double bond, the more stable the alkene. With me? And for the geometrical isomers, trans is more stable than cis. Right? Now, how do we compare stability? We hydrogenate the alkene and look at the heat released, the heat of hydrogenation. So tell me, does a more stable alkene release more heat, or less? Less heat, yes, exactly! The more stable alkene has less energy to start with, so it gives out less. So the more stable the alkene, the smaller its heat of hydrogenation. And the note in brackets is an important condition: compare only alkenes that give the same alkane. Clear?
4:58Example 1: strength of the pi bond from bond enthalpies
Time for the first example, and it answers our puzzle. Using the bond enthalpies, six hundred and eighty one for the double bond and three hundred and forty eight kilojoules per mole for the single bond, estimate the strength of the pi bond. Pause the video and try it first. Done? Okay, let's check. The double bond is one sigma bond plus one pi bond, and the single bond is just one sigma bond. With me? So the pi bond is about six hundred and eighty one minus three hundred and forty eight, which gives three hundred and thirty three kilojoules per mole. The wavy equals sign means about, because this is only an estimate. Clear? Three hundred and thirty three is less than three hundred and forty eight; the pointed sign means less than. So the pi bond is weaker than the sigma bond. And that solves our puzzle! The double bond as a whole is stronger, but its exposed pi bond is the weak link, and it gives way in addition. Got it?
5:57Example 2: most stable butene from heats of hydrogenation
Let's go to the harder one. The heats of hydrogenation of but one ene, cis but two ene and trans but two ene are minus one hundred and twenty six point eight, minus one hundred and nineteen point seven, and minus one hundred and fifteen point five kilojoules per mole. Which is the most stable, and by how much is trans but two ene more stable than cis? Pause the video and try it first. Done? Okay, let's check. All three add H two, hydrogen, and give the same product, butane; the arrow means gives. Same product, so the heats can be compared. With me? So, which one is the most stable? The biggest heat, one hundred and twenty six point eight, comes from but one ene. So the biggest number means the most stable, right? No, no, careful, it is the other way round! The less heat an alkene releases, the less energy it had, and the less energy, the more stable it is. Each double arrow on screen reads as therefore. Okay? The least heat is one hundred and fifteen point five, for trans but two ene. So trans but two ene is the most stable, which fits our rule: trans is more stable than cis. Clear? And by how much is trans more stable than cis? Cis releases one hundred and nineteen point seven, and trans releases one hundred and fifteen point five. Subtract them, and the difference is four point two kilojoules per mole. So trans is more stable by four point two. Got it?
7:29Quick revision: pi bond, addition, tests, stability
So, quick revision. The double bond joins two s p two carbons by one sigma bond and one weaker pi bond. Its exposed pi electrons are attacked by electrophiles, and the alkene undergoes addition. It is planar with restricted rotation, and we test for it with bromine in C C l four, or Baeyer's reagent. More alkyl groups, or the trans form, mean a more stable alkene and a smaller heat of hydrogenation. You handled this really well. Now go and try the practice problems on this page. All the best!
More videos in Unit 15 · Hydrocarbons
- Structure and Physical Properties
- Preparation of Alkanes
- Halogenation
- Combustion and Other Reactions
- Preparation of Alkenes
- Addition of Hydrogen Halides
- Addition of Halogens, Water and Hydrogen
- Oxidation, Ozonolysis and Polymerisation
- AIEEE and Earlier JEE Main Questions: Alkenes
- The Triple Bond and its Acidity
- Preparation of Alkynes
- Addition Reactions of Alkynes
- Oxidation and Polymerisation
- Conjugated Dienes
- AIEEE Questions: Alkynes and Dienes
- Benzene: Structure and Preparation
- Electrophilic Substitution
- Directive Influence of Substituents
- Addition, Oxidation and Side-Chain Reactions
- Overview
- Set 1
- Set 2
- Set 3
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks