1. Unit 3 – Organic Chemistry
Advanced Higher
Unit 3 – Organic Chemistry

2. 1. Cobalt Complexes 2. Adsorption of Charcoal 3. Wine Analysis 4
1. Cobalt Complexes  2. Adsorption of Charcoal  3. Wine Analysis  4. Harcourt- Essen Reaction  5. Conductrimetric Titrations  6. Ester Hydrolysis  7. Vitamin C Analysis 8. Aspirin Analysis  9. Buffer Solution 10. Wine Analysis

3. By the end of this course:
You’ll be able to tell me What this is Some its made The problems of this drug The controversy behind the drug

4. Organic chemistry looks at reactions in chemistry, the main ones are: Addition, condensation, hydrolysis, oxidation, reduction, substitution, elimination, acid/base.

5. All chemical reactions involve breaking bonds in order to make new bonds/molecules.
There are two main ways of doing this
Homolytic fission
Eg. H:Cl → H• + Cl•
The dot means that there is an unpaired electron. The atoms are neutral (even protons and electrons) but it is v. reactive because the unpaired electron. An atom with an unpaired electron is called a free radical.

6. 2. Hetrolytic fission Eg. H:Br → H⁺ + Br⁻ This is more likely when the bond is already polar (difference in electronegativity is about 2.5) eg. CH₃Br This example shows curly arrows, nucleophiles and electrophiles

7. Electrophiles are electron seekers and want to get electrons from other atoms. Eg. CH₃⁺(carbocation), Li⁺
Nucleophiles are nucleus seekers, there are electron rich and look to bond to positive centres. Eg. Cl⁻, Br⁻, F⁻, I⁻, CN⁻, OH⁻
NOTE: there are also positive and negative centres(through polarity)

8. Have you thought about projects?
You should!

9. Last day we mentioned:
Functional groups Types of reactions Why methane didn’t make sense looking at its structure and the orbitals it has

10. Methane, CH₄
For this we look at the carbon atom and its bonding orbitals. Carbon: s1, s2, p2

11. Energy levels of carbon
Increasing energy 2s 2p
hybridised orbitals
We know that this doesn’t make sense, because this would mean that carbon can only make 2 bonds.

12. Hybridisation The 4 orbitals below combine to give 4 new oritalsx z y x z y
2s orbital
2pz orbital x z y x z y
New Hybrid sp³ orbitals
2px orbital
2py orbital

13. When we add the hydrogen atoms (1s orbitals) in we get: These bonds are called sigma bonds σ - bonds

14. This works in a similar way for other alkanes
Eg. ethane
Etc.

15. Reactions of alkanes
Alkanes are relatively unreactive compared to other organic compounds. To make the reaction work sunlight is often used to start (initiate) the reaction, this is a way of creating a free radical which is very reactive and can force the unreactive alkane to react. The reaction has three steps

16. Step one- initiation
The initiation step is where a free radical is made which will react with the alkane.
Insert homolytic fission mechanism here

17. Step 2- propagation
Radicals are very reactive and will react with anything to get a full shell. Add mechanisms in here.

18. Step 3- termination
The reaction will continue until all the reactants are used up. They can be used in a few different ways. Add mechanisms

19. Notes about the reaction:
The reaction will continue in the dark after it has started
Depending on the proportions of alkane and the bromide multiple substitutions can occur. (which isn’t always a good thing)
To ensure monosubstitution (one addition of bromine) large excess of alkane is used
3. To remove any unwanted products, alkanes multisubstitutions etc distillation is used.

20. Bonding in Alkenes
How can we explain the existence of double bonds as observed in alkenes?
C C H
C C H H
As with alkanes, bonding in alkenes is due to hybridisation.

21. single unhybridised 2p orbital
As with alkanes, an electron from the 2s shell is promoted to the empty 2p orbital.
This results in the formation of three hybrid orbitals, with one remaining unhybridised 2p orbital.
single unhybridised 2p orbital
Increasing energy 2s 2p
hybridised orbitals
The hybrid orbitals formed from one s orbital and two p orbitals are called sp2 orbitals.

22. The three sp2 orbitals repel each other, resulting in a bond angle of 120° between them.
The hybrid orbitals are responsible for overlapping to form σ bonds joining their central carbon atoms to both carbon and hydrogen. C H
sp2 orbitals

23. unhybridised 2p orbitals
The unhybridised p orbitals are perpendicular to the plane of the molecule.
unhybridised 2p orbitals C H
σ bonds
σ bonds
σ bond C H
The p orbitals of the carbon atoms are parallel and close enough to overlap sideways.

24. This new orbital is called a pi (π) orbital or more commonly a π bond.
This sideways overlap between the 2p orbitals produces a new molecular orbital between the two carbon atoms. C H
A π bond is a covalent bond formed by the sideways overlap of two parallel atomic orbitals.
This new orbital is called a pi (π) orbital or more commonly a π bond.

25. H C
σ and π bonds
Looking at information comparing σ and π bonds, we can see that double bonds are stronger than single bonds, but not twice as strong. This is because the sideways overlap (π bond) is weaker than the end-on overlap (σ bond).
C C
C C
C C

26. Synthesis of Alkenes There are two main routes used in the laboratory:
Acid-catalysed dehydration of the appropriate alcohol
From monohalogenoalkane

27. Acid-catalysed dehydration of the appropriate alcohol
This can be accomplished by mixing the alcohol with sulphuric acid or phosphoric acid and heating, water is eliminated to form the alkene. In general: Alcohol + sulphuric acid → water + alkene NOTE: an elimination reaction is where any small molecule (eg. H₂O or HBr) is removed or eliminated from a larger molecule. Example, butan-2-ol
Heat

28. Mechanism
Insert mechanism here

29. Monohalogenoalkanes
A monohalogenoalkane is an alkane but instead of all hydrogen’s there is a halogen. If a monohalogenoalkane is reacted with potassium hydroxide in ethanol solution, the hydrogen halide is eliminated to leave appropriate alkene. In general: Monohalogenoalkane + potassium hydroxide → hydrogen halide + alkene

30. This can happen through two possible routes:
Elimination 1 (E1)
Insert mechanism here
Elimination 2 (E2)
Insert mechanism here (mention LGs)

31. Reactions of Alkenes
Almost all of the reactions involve addition across the double bond, eg HCl, H₂, Br₂, to form saturated products.

32. Addition of Hydrogen (Hydrogenation)
The hydrogenation reaction is slightly exothermic but so slow at room temperature that it is unobservable. However, the reaction can be carried out successfully at higher temperatures with a suitable catalyst (eg. Ni, Pt, Pd). The catalyst absorbs hydrogen onto its surface, which lowers bond energy and makes an easier transition phase.

34. Addition of Halogens
As we know from standard grade/higher that bromine water rapidly decolourises in the presence of an alkene,(carbon-carbon double bond). This reaction explains why. Note: this could work for different halogens as well. The reaction has two steps.

35. Step 1
Electron rich C=C is approaches a non-polar Br₂, the Br₂ then becomes a temporary dipole due to electron repulsion and a reaction can take place. Add mechanism here The intermediate is called a bromonium ion, the ion is positive and is stabilised by delocalisation.

36. Step 2
Rapid nucleophilic attack of the bromine ion on one of the carbons from the brominium ion to give the product 1,2-dibromomethane. Add mechanism

37. Notes:
If the reaction is carried out in the presence of NaCl some 1-bromo, 2-chloroethane is formed. This means that the Cl⁻ion completes and that there must be a positive ion and that it is not a radical reaction.
This reaction could be carried out with Cl₂ via a similar mechanism.

38. Addition of hydrogen halides
Hydrogen halides react quickly fairly readily with alkenes, in a similar way to before, in the gas phase and in non-polar liquids (eg hexane) Eg. CH₂= CH₂ + H-I → CH₃-CH₂I If you were to do this with an unsymmetrical alkene you would get a mix of halogenoalkanes. Eg propene and HBr

39. Why?
Again we look at the mechanism.

40. Note:
Just like 2⁰ alcohols are more stable that 1⁰, the positive charge on the 2⁰ carbon is more stable. More stable is favourable=>major products This is the Markovnikov rule, the more stable product is favoured 3⁰>2⁰>1⁰

41. Why is it more stable?
The positive carbon pulls electrons from its neighbouring carbons which stabilises it. More neighbours, more stable. There are variations in this rule, like the questions from the other day is other things can have an effect. Eg carbonyl groups would unstablise the carbon because of its pull on the electons.

42. Variations
In the last reaction the conditions were in gas phase or in a non-polar solution, if we change this we get a very different result. Instead of the previous the following happens. Insert mechanism here

43. This could also be done using cold sulphuric acid and then heating with water to hydrolyse. Insert reaction scheme here

44. Halogenoalkanes
We know how to make them, now well learn how to name them:
Rules
The basic name comes from the longest unbranched chain
Depending on the halogen use the appropriate prefix, flouro-, chloro-, bromo-, iodo-

45. 3. If there is more tha one atom of the same halogen use the prefixes di-, tri-, etc 4. The number of the substituents is shown by a number in front of the prefix.

46. Examples

47. Reactions of halogenoalkanes
The reaction of halogenoalkanes depeds upon two factors:
The halogen
R-I < R-Br < R-Cl < R-F
2. The position (eg 1⁰,2⁰,3⁰)
3⁰ > 2⁰ > 1⁰
Weakest bond
Strongest Bond
Most reactive
Least reactive
Most stable
intermediate
Least stable intermediate
Slowest made= worst at reacting
Quickest made= quickest at reacting

48. Lets see this in action- nucleophilic substitution reaction
In a halogenoalkane, there will be a polarised bond thanks to the carbon-chlorine bond. This means that the carbon is susceptible to attack by nucleophiles (negative things). Halogens are stable ions and are called GOOD LEAVING GROUPS represented as LG⁻ or Y⁻.

49. In fact-there are two possible mechanisms
SN2 reaction
Consider the following:
C2H5Br + OH- → C2H5OH + Br-
Examining the rate (which you’ll do in unit 2)
We deduce that in this instance that the transition state has the leaving and joining atoms in situ.
Hopefully this makes sense when you see the mechanism

51. SN1
(CH3) 3CBr + H₂O → (CH3) 3COH + HBr
Again looking at the kinetics we could work out that the transition state has only one particle in it, as the mechanism shows.