1. PHYSICAL PROPERTIES OF ALKANES and ALKENES
SAME TREND AND REASON FOR BOTH “ANES” AND “ENES”
ALL NON POLAR AND HAVE WEAK VAN DER WALLS FORCES
Melting point general increase with molecular mass
the trend is not as regular as that for boiling point.
Solubility alkanes are non-polar so are immiscible with water
they are soluble in most organic solvents.

2. CHEMICAL PROPERTIES OF ALKANES
Introduction - fairly unreactive
2 REACTIONS:
Combustion CH4(g) O2(g) ——> CO2(g) H2O(l)
2) Substitution (FREE RADICAL)
Equation: CH4(g) Cl2(g) (uv light)——> HCl(g) CH3Cl(g) chloromethane
_______________________Free Radical Substitution__________________
Initiation Cl2 ——> 2Cl• RADICALS CREATED
Propagation Cl• + CH4 ——> CH3• HCl RADICALS USED
Cl2 + CH3• ——> CH3Cl Cl•
Termination Cl• + Cl• ——> Cl RADICALS REMOVED
Cl• + CH3• ——> CH3Cl
CH3• + CH3• ——> C2H6

3. CHEMICAL PROPERTIES OF ALKENES ELECTROPHILIC ADDITION
Equation C2H4(g) HBr(g) ———> C2H5Br(l) bromoethane
Mechanism
Double bond breaks and the
2e- instead bond to Hydrogens
That leaves a carbon devoid
Of an e-

4. ANIMATED MECHANISM CHEMICAL PROPERTIES OF ALKENES
ELECTROPHILIC ADDITION OF HYDROGEN BROMIDE
ANIMATED MECHANISM
Animation repeats continuously after every 10 seconds

5. 5 ADDITION REACTIONS (ALL GO BACKWARDS)
Addition Hydrogenation (H2)
C2H4 (g) H2 (g) —Ni—> C2H6 (g)
2) Addition Halogenation (Cl2, Br2, I2)
C2H4 (g) Cl2 (g) ——> C2H4Cl2 (g)
Addition Hydrogenhalides
C3H6(g) HBr(g) ———> C3H7Br(l)
4) Addition Hydration (H2O or HOH) Le Chatieler (*BOTH WAYS)
C2H4 (g) + H2O(g) (or HOH) H2SO4  C2H5OH(g) ΔH KJ/mol
5) Addition Polymerization

6. Markovnikov's Rule The hydrogen adds to carbon with most H’s already The rich get richer. Cambridge Exams want to see these words: “Secondary carbcations are more stable”
MINOR
MAJOR

7. GEOMETRICAL ISOMERISM IN ALKENES
Introduction
occurs due to the RESTRICTED ROTATION OF C=C bonds
CIS
Groups/atoms are on the
SAME SIDE of the double bond
TRANS
Groups/atoms are on OPPOSITE SIDES across the double bond

8. GEOMETRICAL ISOMERISM IN ALKENES RESTRICTED ROTATION OF C=C BONDS
Single covalent bonds can easily rotate..
ALL THESE STRUCTURES ARE THE SAME BECAUSE C-C BONDS HAVE ‘FREE’ ROTATION.
DOUBLR BONDS DO NOT

9. FEEDSTOCK for POLYMERS Polymers (next slide)
DEHYDROGENATION OF ALKANES TO ALKENES
Reduction Reaction (lose H2)
900 0C
ALKANES  ALKENES ∆
C2H6  C2H H2
Conditions: High heat , high pressure and a catalyst

10. POLYMERISATION OF ALKENES LOTS OF HEAT AND PRESSURE∆
TiCl4.
Uses
POLY(ETHENE)
Shopping Bags
ETHENE ∆
TiCl4
PROPENE
POLY(PROPENE)
Hard Plastics ∆
TiCl4
CHLOROETHENE
POLY(CHLOROETHENE)
POLYVINYLCHLORIDE PVC
Windows (building Materials) ∆
TETRAFLUOROETHENE
POLY(TETRAFLUOROETHENE)
PTFE “Teflon”
Slippery, non-stick frying pans

11. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Addition
Addition
ALKANES
ALKENES
ALCOHOLS
Addition
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

12. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Addition
Addition
ALKANES
ALKENES
ALCOHOLS
Dehydrogenation
Addition
Free Radical
Substitution
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

13. Next Functional Group ALCOHOLS (R-OH)
Reactions
Destroying Alcohols
reflux
Elimination (DEHYDRATION) of Water  C=C (Alkenes)
H2SO4 Conc
2) Oxidation A)  Aldehydes  Carboxylic Acids
B)  Ketones
C)  NO GO

14. INDUSTRIAL MANUFACTURE OF ETHANOL
Making Alcohols
Fermentation: (No O2 ), body temp, yeast
C6H12O6 — NO O2  2 C2H5OH CO2
HYDRATION OF ETHENE (high heat and high pressure)
∆ (STEAM) 300 C
C2H H2O —H2SO4--> C2H5OH
Ex2:
=>

15. Boiling points are high for alcohols due hydrogen bonding.
Physical Properties
Boiling points are high for alcohols due hydrogen bonding.
Melting points
B) Solubility Low molecular mass alcohols are miscible with water
Due to hydrogen bonding between the two molecules
Heavier alcohols are less miscible
PRIMARY 1° SECONDARY 2° TERTIARY 3°

16. CHEMICAL REACTIONS ELIMINATION REACTION (DEHYDRATION)
Equation e.g. C2H5OH(l) —H2SO4—> CH2 = CH2(g) + H2O(l)
Mechanism
Step 1 protonation of the alcohol using a lone pair on oxygen
Step 2 loss of a water molecule to generate a carbocation
Step 3 loss of a proton (H+) to give the alkene

17. Zaitsev’s Rule
The hydrogen is removed from the carbon atom in the chain with the smaller number of H atoms “ the poor get poorer “ 17

18. Another example of Zaitsev’s Rule

19. Answer…
Major
minor
Remember that the more substituted product is preferred!
(1-butene also forms, but it is the minor product).

20. This one, then one more… 20

21. The more substituted product is preferred…21

22. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Hydration
ALKANES
ALKENES
ALCOHOLS
Elimination
Dehydration
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

23. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Addition
Addition
ALKANES
ALKENES
ALCOHOLS
Elimination
Dehydration
Dehydrogenation
Addition
Free Radical
Substitution
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

24. Next Reaction Type OXIDATION OF PRIMARY ALCOHOLS
K2Cr2O7 AND H+ ( H2SO4 ) Conc
Controlling the products
e.g. CH3CH2OH(l) [O] ——> CH3CHO(l) H2O(l)
then CH3CHO(l) [O] ——> CH3COOH(l)
OXIDATION TO CARBOXYLIC ACIDS
REFLUX
OXIDATION TO ALDEHYDES
DISTILLATION
Aldehyde has a lower boiling point so distils off before being oxidised further
Aldehyde condenses back into the mixture and gets oxidised to the acid

25. OXIDATION OF ALCOHOLS
10 Primary alcohols  aldehydes (IN acidified K2Cr2O7) Conc
STEP e.g. CH3CH2OH(l) [O] ——> CH3CHO(l) H2O(l)
ethanol ethanal
Aldehydes have low boiling points - no hydrogen bonding - they distil off immediately
STEP CH3CHO(l) [O] ——> CH3COOH(l)
ethanal ethanoic acid
to oxidise an alcohol straight to the acid, reflux the mixture
_____________________________________________________________
Secondary ketones - more difficult than 10 must be refluxed
(in acidified Conc K2Cr2O7)
e.g CH3CHOHCH3(l) [O] ——> CH3COCH3(l) H2O(l)
propan-2-ol propanone
____________________________________________________________
30 Tertiary Not oxidised

26. 1° 2° 3° OXIDATION OF ALCOHOLS H H R C O + [O] R C O + H2O H H H H
Why 1° and 2° alcohols are easily oxidised and 3° alcohols are not
For oxidation you must have 2 hydrogen atoms on adjacent C and O atoms.
H H
R C O [O] R C O H2O
H H 1° 2° 3°
H H
R C O [O] R C O H2O
R R
This is possible in 1° and 2° alcohols but not in 3° alcohols.
R H
R C O [O] not enough H’s to remove R

27. Test Results for oxidation with Orange/Yellow is H+/Cr2O7 2- GREEN (GOES) is Cr +3
Green means it successfully reacted
Thus, 1o and 20 DO React
3o color stayed so did NOT react
Chapter 14

28. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Oxidation
Oxidation
ALKANES
ALKENES
ALCOHOLS
ALDEHYDES
HALOGENOALKANES
Oxidation
AMINES
NITRILES
CARBOXYLIC ACIDS

29. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Oxidation
Addition
Addition
Oxidation
ALKANES
ALKENES
ALCOHOLS
Elimination
Dehydration
Dehydrogenation
Free Radical
Substitution
Addition
ALDEHYDES
HALOGENOALKANES
Oxidation
AMINES
NITRILES
CARBOXYLIC ACIDS

30. HALOALKANES

31. Nucleophilic Substitution Swapping
Generic Equation: Swap R-X + Nu:  R-Nu + :X- The problem lies in the mechanism.

32. There are 4 Substitutions :NU :OH- :CN- :NH 3 H2O
1) WITH HYDROXIDE
C2H5Br(l) + OH- (aq) -->C2H5OH(l) +Br-(aq)
2) WITH A CYANIDE
C2H5Br(l) + CN- (aq/alc) —>C2H5CN + Br-(aq)
3) WITH 2 AMMONIAS
C2H5Br(l) + 2NH3(aq/alc)—> C2H5NH2 + NH4Br
4) WITH H2O (solvolysis) solvent and reactant at same time
C2H5Br(l) + H2O (l) —> C2H5OH(l) + HBr (aq)

33. NUCLEOPHILIC SUBSTITUTION AQUEOUS SODIUM HYDROXIDE
Equation e.g. C2H5Br(l) NaOH(aq) ——> C2H5OH(l) NaBr(aq)
Mechanism
* WARNING It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent - SEE LATER

34. NUCLEOPHILIC SUBSTITUTION AQUEOUS SODIUM HYDROXIDE
ANIMATED MECHANISM

35. Mechanism for 3o Halides SN1
SLOW (RDS)
STEP
Carbocation Intermediate
FAST STEP

36. SN1 Top pic 3o SN2 Bottom

37. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
ALKANES
ALKENES
ALCOHOLS
Nucleophilic
Substitution
ALDEHYDES
HALOGENOALKANES
Nucleophilic
Substitution
Nucleophilic
Substitution
AMINES
NITRILES
CARBOXYLIC ACIDS

38. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Oxidation
Addition
Addition
Oxidation
ALKANES
ALKENES
ALCOHOLS
Elimination
Dehydration
Dehydrogenation
Free Radical
Substitution
???
Addition
Nucleophilic
Substitution
ALDEHYDES
HALOGENOALKANES
Oxidation
Nucleophilic
Substitution
Nucleophilic
Substitution
AMINES
NITRILES
CARBOXYLIC ACIDS

39. ELIMINATION v. SUBSTITUTION
The products of reactions between haloalkanes and OH¯ are influenced by the solvent
Modes of attack
:Nu
Aqueous soln OH¯ attacks the slightly positive carbon bonded to the halogen.
OH¯ acts as a nucleophile
BASE
Alcoholic soln OH¯ attacks one of the hydrogen atoms on a carbon atom adjacent
the carbon bonded to the halogen.
OH¯ acts as a base (A BASE IS A PROTON ACCEPTOR)
Both reactions take place at the same time but by varying
the solvent you can influence which mechanism dominates.
SOLVENT
ROLE OF OH–
MECHANISM
PRODUCT
WATER
NUCLEOPHILE
SUBSTITUTION
ALCOHOL
BASE
ELIMINATION
ALKENE

40. ELIMINATION Reagent Alcoholic REFLUX with KOH or NaOH
Equation C3H7Br NaOH(alc) ——> C3H H2O NaBr
Mechanism
the OH¯ ion acts as a base and picks up a proton

41. ELIMINATION
ANIMATED MECHANISM

42. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
Addition
Addition
Oxidation
Addition
Addition
Oxidation
ALKANES
ALKENES
ALCOHOLS
Elimination
Dehydration
Dehydrogenation
Free Radical
Substitution
Elimination
Addition
Nucleophilic
Substitution
ALDEHYDES
HALOGENOALKANES
Oxidation
Nucleophilic
Substitution
Nucleophilic
Substitution
AMINES
NITRILES
CARBOXYLIC ACIDS

43. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
ALKANES
ALKENES
ALCOHOLS
Elimination
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

44. REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
ALKANES
ALKENES
ALCOHOLS
ALDEHYDES
HALOGENOALKANES
AMINES
NITRILES
CARBOXYLIC ACIDS

45. 7 Reaction Mechanisms Free Radical Substitution (UV Light)
Electrophilic Addition (5 types)
Oxidation (K2Cr2O7 / H+ and distil or reflux (from H2SO4)
Polymerization (High Heat, High Temp, TiCl4 (catalyst)
Elimination (Alcohol dehydration) (H+ from H2SO4)  C=C
Elimination Halogen ( :OH- Base in Alcohol)  C=C
:Nu Substitution (:OH- (aq) ) OR R-X  R-OH

46. LAST TOPIC ANALYSIS IR MS H-NMR C-NMR
END OF REACTIONS
LAST TOPIC ANALYSIS IR MS H-NMR C-NMR

47. FINGERPRINT REGION
• organic molecules have a lot of C-C and C-H bonds within their structure
• spectra obtained will have peaks in the 1400 cm-1 to 800 cm-1 range
• this is referred to as the “fingerprint” region
• the pattern obtained is characteristic of a particular compound the frequency
of any absorption is also affected by adjoining atoms or groups.

48. Elimination (BASE) :OH- (alcohol) Cascade
7 Mechanisms
Elimination (BASE) :OH- (alcohol) Cascade
Elimination (Acid) : H+ ( from H2SO4)
Free Radical Substitution (UV light)
Nucleophilic Substitution :Nu
Electrophilic Addition
Oxidation: [O] (H+ , K2Cr2O7)
Polymerization (High temp, high heat, TiCl4)

49. IR SPECTRUM OF A CARBONYL COMPOUND
• carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm-1
• this is due to the presence of the C=O bond

50. IR SPECTRUM OF AN ESTER
• esters show a strong absorption between 1750 cm-1 and 1730 cm-1
• this is due to the presence of the C=O bond

51. IR SPECTRUM OF AN ALCOHOL
• alcohols show a broad absorption between 3200 and 3600 cm-1
• this is due to the presence of the O-H bond

52. IR SPECTRUM OF A CARBOXYLIC ACID
• carboxylic acids show a broad absorption between 3200 and 3600 cm-1
• this is due to the presence of the O-H bond
• they also show a strong absorption around 1700 cm-1
• this is due to the presence of the C=O bond

53. MASS SPECTROMETRY

54. . THE MASS SPECTRUM - THE MOLECULAR ION Abundance % m/z
The small peak (M+1) at 115 due to the natural abundance (about 1%) of carbon-13. The height of this peak relative to that for the molecular ion depends on the number of carbon atoms in the molecule. The more carbons present, the larger the M+1 peak.
Abundance %
114 .
m/z

55. FRAGMENTATION PATTERNS
ALKANES
The mass spectra of simple hydrocarbons have peaks at m/z values corresponding to the ions produced by breaking C-C bonds. Peaks can occur at ...
m/z etc CH3+ C2H5+ C3H C4H C5H C6H13+
• the stability of the carbocation formed affects its abundance
• the more stable the cation the higher the peak
• the more alkyl groups attached to the carbocation the more stable it is
most stable tertiary 3° > secondary 2° > primary 1° least stable
alkyl groups are electron releasing and stabilise the cation

56. Tetramethylsilane Downfield from what?????
All the protons are chemically equivalent - PRODUCES A SINGLE PEAK
Non-toxic liquid - SAFE TO USE
Inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED
THE BEST SHEILDING ANYWHERE.
WHY?
Carbon actually is MORE electronegative than Silicon, thus all the electrons are pulled into the protons (H+) and it is well shielded from external magnetic fields.
The molecule contains four methyl groups attached to a silicon atom in a tetrahedral arrangement. All the hydrogen atoms are chemically equivalent. THUS only 1 signal is given off.

57. Chemical Shift (look up on the chart)
1) Ratio of shift downfield from TMS (Hz) Called the delta scale (𝛅 ppm).
2) Based on polarity, the more polar atoms strip away the H+ shielding electrons, leaving them “naked” and frozen in the external magnetic field
3) Unprotected the H+ are held well, thus we need more RF energy to make them resonate.
4) The more polar the more downfield shifted from TMS a molecule is (R- COOH) 𝛅 ppm, while CH3 is 0.5 ppm

58. Signals What is the patttern?
More electronegative atoms deshield more and give larger shift values.
Additional electronegative atoms cause increase in chemical shift.

59. Spin-Spin Splitting
Protons (H+) on adjacent (neighboring) carbons have magnetic fields that may align with other adjacent protons and this increases their signal.
Hydrogen's in similar environments will have the same magnetic spin
This magnetic coupling causes the proton peaks to split into doublets, triplets, quintet. (n+1)
EXAMPLES of Proton Coupling or peak splitting
Singlet has 0 Proton neighbors or n=0 so n+1 = 1 peak
Doublet has 1 Proton neighbor or n=1 so n+1 = 2 peaks
Triplet has 2 Proton neighbors or n=2 so n+1 = 3 peaks
Quartet has 3 Proton neighbors or n=3 so n+1 = 4 peaks

60. n+1 rule
Hydrogen's in similar environments will have the same magnetic spin. Ex CH3 here: CH3CH2CH3. The CH3 will send out the SAME signals they are equivalent. SO, we say we have only 2 environments for hydrogen. The 2 CH3 have symmetry Different environments in neighboring hydrogen's often do not have the same signal, so they, being different, have slightly different signals. Thus different neighbors interfere slightly with each others signal (called coupling) and cause them to be split (called splitting, note splitting = coupling). They are split in a specific pattern: (splits) = n+1 where n=protons neighbors (non-equivalent)

61. Integration Ex #2
The vertical displacement of the peaks (called integrals) gives the relative number of protons
36.2/15.2 = LOWEST INTEGRAL (15.2) 15.2/15.2 =  22.8/15.2 = NOW all X 2
=4.76/2.0/ rounding 5/2/3