University of Sydney
Chemistry 1B (CHEM1102)
Organic Chemistry Lecture Notes
Topic 1
Introduction & isomers
2
Topic 2
Alkenes, alkynes, arenes
23
Topic 3
Structure determination
32
Topic 4
Alcohols and amines
43
Topic 5
Stereochemistry
49
Topic 6
Alkyl halides, carbonyls
55
Topic 7
Carboxylic acids & derivatives
69
Summary of Reactions
79
Inorganic & General Chemistry Lecture Notes
Topic 8
Acids and Bases
Topic 9
Solubility
Topic 10
Coordination Chemistry
Topic 11
Kinetics
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2012 (John Wiley)
ISBN: 978 1 74246 707 8
1 85
Topic 1 – Introduction, representation of structures,
functional groups, nomenclature, isomers
Introduction - Definition of Organic Chemistry
Organic chemistry deals with compounds in which carbon is the principal element • Organic substances arise in all sorts of places -­‐ plants, animals, food, medicines, industry, research laboratories • Every living organism is made of organic chemicals • 15 million organic substances -­‐ number increases by 10,000 per week! The Unique Nature of Carbon – Bonding in Organic Carbon
•
•
•
•
•
•
Catenation -­‐ Carbon atoms can bond together to form • Stable extended chains of atoms • Rings • Multiple bonds Carbon has atomic number = 6 • Electronic configuration: 1s2 2s2 2p2 • Valency of 4 – four bonds to completely fill outer electron shell of carbon Carbon forms bonds to • Itself – single, double, triple • Metals – Na, K, Fe, Cu, Mg … • Heteroatoms – N, P, O, S, X C-­‐C C-­‐H Si-­‐Si N-­‐N O-­‐O I-­‐I C-­‐C and C-­‐H bonds are strong C-­‐C and C-­‐H bonds are non-­‐polar 348 kJ mol-­‐1 412 176 163 146 152 A combination of the strength and non-­‐polar nature make C-­‐C and C-­‐H bonds unreactive 1. Propane is CH3CH2CH3 (a chain of 3 carbons, with the number of hydrogens indicated attached to each carbon). Draw a picture to show the structure of this molecule. Could you draw your picture quickly and efficiently? How well would this drawing method work for a bigger molecule (e.g. octane (C8H18), octadecane (C18H38) or taxol (C47H51NO14))? How well does the picture above represent the actual structure of propane? 2 Representation of Organic Molecules
Organic compounds may be represented in several ways as illustrated in the examples below. H
H
H
C
C
H
H
O
H
CH 3CH2OH
C2H 6O
H
H
H
H
O
H
H
C
C
O
H
Concept § Catenation – chains and rings of carbon atoms § Bonding – single, double, triple bonds possible § Geometry – depends on hybridisation of carbon atom § Valence – 4 for carbon (3 – N, 2 – O, 1 – H, Cl, Br, I) § Reactivity – often associated with heteroatoms (atoms other than C or H) In the stick representation: Ø C is basis of structure Ø Carbon atoms are not usually shown but assumed to be at the intersection of 2 or more bonds and at the end of each line Ø 1, 2 or 3 lines to represent single, double and triple bonds Ø Indicate bond angles (~ 1200 or 1800) Ø Omit all C-­‐H bonds, hydrogen count assumed Ø Specify all heteroatoms (O, N, F, Cl, B, I, S etc) and H – heteroatom bond Ø When drawing neutral organic molecules: Ø C always has a valence of 4 Ø N always has a valence of 3 Ø O always has a valence of 2 Ø X (halogen) always has a valence of 1 3 Example: From a condensed structural formula, first draw the structure out in full then remove all C atoms with the H attached to them and replace them with a kinked chain showing all C-­‐C bonds. Show all heteroatoms and H attached to heteroatoms. With practice you will be able to convert condensed structural formula to stick structures directly. Condensed Structural Formula CH3CH2CO2H or CH3CH2COOH CH3CHClCH=CH2 or CH3CHClCHCH2 CH3COCH2CH3 Structural Formula H H O
H C C C O H
H H
H H H
H
H C C C C
H
H Cl
H O H H
H C C C C H
CH3OCH2CH3 H
H
H H
H
H H
C O C C H
H
Stick representation O
OH
Cl
O
O
H H
2. Now draw a line structure for octane , C8H18 (remember lines represent bonds, C atoms aren’t shown (they’re assumed to be at the intersections and ends of lines) H atoms aren’t shown, and angles are drawn close to the actual bond angles. How quickly/ efficiently could you draw that picture? How well does it represent the actual structure of the molecule? 4 3. Next try drawing line structures for 1-­‐pentanol, 2-­‐bromopropane and 2-­‐aminobutane. 1-­‐pentanol (CH3CH2CH2CH2CH2OH) 2-­‐bromopropane (CH3CHBrCH3) 2-­‐aminobutane (CH3CH(NH2)CH2CH3) 4. And now work backwards and draw condensed formulae (e.g. CH3CH2CH3) for these structures. ________________ ________________ ________________ Can you name any of these three compounds? ________________ ________________ ________________ Where it is important to represent the three-­‐dimensional shape of a molecule, the following convention is adopted: Bonds in the plane of the paper
A B
A B
Bonds coming towards the observer
Bonds going away from the observer
A B or A B
H
For example, CH 4
C
H
5 H
H
5. Try drawing 3D structures for these compounds, using wedges and dashes to show bonds coming out of/ going into the page. dichloromethane (CH2Cl2) methanol (CH3OH) For more practice, try these extra questions … 6. Try converting the following to stick structures. 6 dimethylpropane (CH3C(CH3)3) Reverse the process to go from a stick structure to a condensed molecular formula by first drawing in the C atoms then add H atoms to ensure the valence of each C is four. Example: 7. Try converting the following stick structures to condenser structural formula. 7 Hybridisation
8. Methane (CH4) has a tetrahedral shape. A tetrahedron is symmetrical: What does this tell you about the four C–H bonds in methane? What does this tell you about the four atomic orbitals at carbon that are involved in forming these four bonds? There are four experimentally observed bonding arrangements for carbon: Type of bonds Hybridisation Geometry Examples 4 single bonds sp3 Tetrahedral CH4, CH3CH3 2 single, 1 double bond sp2 Trigonal planar CH2=CH2, CH2=O 1 single, 1 triple bond sp Linear H-­‐C≡C-­‐H 2 double bonds CH2=C=O 9. Let’s look at this another way. What is the electronic configuration of carbon? So which of carbon’s atomic orbitals are involved in making these bonds to the hydrogens? What shape would methane have if these orbitals behaved as separate, independent orbitals (i.e. one 2s orbital + three 2p orbitals)? 10. Now imagine you run a juice bar. You have one apple and three carrots, a blender but no knife. (It’s a poorly equipped juice bar.) How many different juice blends can you make? What is the apple : carrot ratio in each of them? How many carrots are left out in each case? Juice Blend Apple : Carrot Ratio 8 Left Over Carrots 11. Now it’s orbitals you are mixing. You have one s orbital and three p orbitals, and again a blender (hybridizer) but no knife. How many different orbital blends can you make? What is the s : p ratio in each of them? How many p orbitals are left out in each case? Orbital Blend s : p Ratio Left Over p orbitals Hybridisation is a convenient way of describing the sigma-­‐bonding orbitals and lone pair orbitals (if present) of an atom. In alkanes, hybridisation gives four identical sp3 orbitals, which form a tetrahedral shape to minimise repulsion between the orbitals. Energy
Hybridisation
Carbon Carbon Carbon ground
state
bonding
state
four sp 3 orbitals
Note: A quick way of determining the hybridisation of an atom is to count the σ bonds and lone pairs around that atom and assign one hybrid orbital to each. 2 electron pairs ⇒ sp, 3 electron pairs ⇒ sp2, 4 electron pairs ⇒ sp3 Example: Work out the hybridization of nitrogen in the following molecules:
9 12. So what is the hybridization at each carbon in propane, propene and propyne shown below? Once you’ve answered that, match the correct name to each compound. ________________ ________________ ________________ What is the bond angle around the central carbon in each of these compounds? 0
0
0
13. Now draw structures for 1-­‐pentyne (CHCCH2CH2CH3) and 2-­‐pentyne (CH3CCCH2CH3), taking care that your bond angles reflect the hybridization. 1-­‐pentyne (CHCCH2CH2CH3) 2-­‐pentyne (CH3CCCH2CH3) Functional Groups
Organic compounds are classified by Functional Groups, which are responsible for chemical behaviour. Functional groups are involved in naming organic compounds. • C=C, C≡C and the polar bonds from carbon to heteroatoms are more reactive than C-­‐C or C-­‐
H bonds and hence where the chemistry takes place. This part of the molecule is called the Functional Group. • An organic compound can be viewed as a backbone (skeleton) of carbon-­‐carbon single bonds with other groups of atoms, functional groups, attached at various points. 10 •
•
•
•
Functional group
The combination of the ability to form a vast range of unreactive carbon frameworks to which can be added special reactive sites gives the diversity of organic chemistry. Functional groups confer the characteristic chemical and physical properties of the compounds that contain them. Functional groups undergo the same chemical reactions irrespective of the type of molecule that contains them. A molecule containing several functional groups displays reactions that represents the sum of the reactions of each functional group. O
OH
O
O
CH3
Acetyl salicylic acid (aspirin)
shows the properties of
- a carboxylic acid
- an ester
- an aromatic compound
14. Circle or highlight all the functional groups in the structure of taxol below. How many are there? Taxol (paclitaxel) 11 Summary Table of Functional Groups
Note: R is the general abbreviation for the “rest of the molecule”. Class General formula C & H alkane RCH3
alkene alkyne aromatic compound C, H, O -­‐O-­‐ C, H, O =O alcohol ether aldehyde ketone carboxylic acid ester C, H, N amine C, H, N, O amide R2C=CR2
C, H, X acid chloride Alkyl halide CH4 CH3CH2CH2CH3 CH
3
CH3
C C
CH3
CH3
RC≡CR
R-OH
R-O-R
O R C H
O R
R
R
R
C, H, Cl, O Examples C R
O C OH
O C OR
R3N O C NR2
R C Cl
R-X
(X = Cl, Br, I) O
CH3C CH
CH3
OH
CH3CH2OH
O
CH3CH2OCH3
O
O
C
CH3 C
H
H
O
O
CH3CCH3
O
CH3COOH
CH3COCH3
(CH3)3N
O
OH
O
O
O
NH2
CH3
O
C N
H
C
CH3
NH2
O
CH3COCl
C
Cl
CH3CH2Br
12 Cl
15. Name the functional groups in the following molecules: OH
Vitamin A
O
Acetyl salicylic acid
OH
O
O
CH3
HO
Morphine
O
NCH3
HO
O
(CH3CH2)2N
NCH3
O
Lysergic acid diethyl amide
NH
13 Nomenclature
The name of an organic molecule consists of several parts: Number(s)
Substituent
Stem
Ending
The stem indicates the number of carbon atoms in the longest chain containing the functional group. The ending indicates the nature of the functional group present. Substituents are indicated by prefixes and a number is used to locate any substituent or functional group unambiguously in a molecule. Rules: • Find and name the longest carbon chain → stem alkane
• Identify substituents stemane
• Number longest carbon chain to give lowest numbering for number of carbons
substituents • Allocate a number to every substituent • List substituents in alphabetical order • Identical side chains are indicated by using prefixes: di (2), tri (3), tetra (4) etc Methane CH4 Ethane CH3CH3 Propane CH3CH2CH3 Butane CH3CH2CH2CH3 Pentane CH3CH2CH2CH2CH3 Hexane CH3CH2CH2CH2CH2CH3 Heptane CH3CH2CH2CH2CH2CH2CH3 Octane CH3CH2CH2CH2CH2CH2CH2CH3 Nonane CH3CH2CH2CH2CH2CH2CH2CH2CH3 Decane CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 Alkane Alkyl group substituent propane CH3CH2CH3 propyl CH3CH2CH2-­‐ butane CH3CH2CH2CH3 butyl CH3CH2CH2CH2-­‐ 14 16. Give the name of the following molecules: Cl
Cl
O
Isomerism
The problem of isomers… § How many different compounds are there with the formula C2H2BrClO? (Excluding those with charged atoms and O–Cl or O–Br bonds) § How do we tell them apart? § How do we name them? § How are they different? § Do they have different properties? There are 16 different possible compounds O
O
O
O
Br
Cl
Cl
Cl
H C
C
C
C
C
C
H
H
ClCH2 C Br
H
Br
Br
BrCH2 C
Cl
H
C
Br
Cl
O
O
O
Cl
H
C
C
H
Br
H
C H
O
Br
H
OH
C
C
Br
O
C
Cl
H
H
C
H
H
Br
Cl
OH
C
C
C
Br
C
OH
C
Cl
Cl
Br
Cl
OH
Cl
OH
Cl
H
H
C
H
H
C
Cl
Br
Br
C H
C
H
C
C
H
OH
C
C
Br
C
H Isomers are § Compounds that have the same molecular formula but different structures. § Very important in chemistry but especially important in the chemical reactions that take place in living organisms -­‐ the shape is as important as the functional groups present! § Classified according to type. O
Cl
15 Br
17. In the ancient Greek, isos = equal and meros = part, so what are isomers? Here are three to help you: ________________ ________________ ________________ (While we’re looking at them, can you name these three compounds?) Classification of isomers
Isomers
same molecular formula
Constitutional Isomers
Stereoisomers
Different nature/sequence of bonds
Different arrangement of groups in space
Conformational
Isomers
Configurational
Isomers
Differ by rotation about a single bond
Interconversion requires breaking bonds
Enantiomers
Diastereoisomers
Non-superposable mirror images
Not mirror images
18. Match these three words beginning with C with their dictionary.com definitions. constitution the way in which a thing is composed or made up the relative disposition or arrangement of the parts or elements of a thing manner of a formation, structure or form as a physical entity configuration conformation 16 Constitutional isomers
Isomers differ in the nature and/or sequence of bonding § Within a homologous sequence of alkanes, the number of constitutional isomers increases rapidly No. of Constitutional
Constitutional formula
n
iosmers
CnH2n+2
1
CH4
1
CH4
2 CH3CH3
1
C2H6
CH3
3 CH3CH2CH3
1
C3H8
4 CH3CHCH3
CH3CH2CH2CH3
2
C4H10
AND
5
3
C5H12
6
5
C6H14
: :
:
10 75
C10H22
20 366 319
C20H42
19. How many constitutional isomers are there of molecular formula C6H14? Draw them. §
§
The physical and chemical properties of constitutional isomers may be very different, particularly when different functional groups are present. For example the molecule with formula C4H8O may be a ketone, aldehyde, alkene/ether or alkene/alcohol. O
O
17 O
OH
Stereoisomers
Stereoisomers have the same nature and/or sequence of bonding but differ in the arrangement of groups in space. § There are two groups of stereoisomers § Conformational isomers or conformers differ by rotation about a single bond and can not normally be separated from each other at room temperature. § Configurational isomers may be interconverted only by breaking and remaking bonds. This process normally requires considerable energy and does not happen at room temperature. H
Conformational isomers
H
H
H
Straight chain alkanes C
C
H
§ Rotation around each C-­‐C bond readily occurs. C
C
H
§ Conformational isomers result. H H
H H
Use ethane as an example (CH3CH3) H
H
H
H
H
H
HH
Sawhorse representation
H
H
H
H
rotate back carbon 60°
H
HH
H
H
Newman projection
H
H
H
H
H
H
H
eclipsed
staggered
Conformers differ in energy § In ‘straight chain’ alkanes rotation about a C-­‐C single bond occurs rapidly at room temperature § The difference in energy between conformers arises from steric interaction 20. Two conformational isomers of ethane are shown below. One of these is called the staggered conformation, the other is an eclipsed conformation. Which do you think is which? 18 Configurational isomers – ‘Diastereoisomers”
Inn cyclic alkanes § Rotation is restricted within a ring of carbon atoms. § This is because rotation would require the atoms attached to carbon to pass through the centre of the ring – this has a high energy barrier. H
H
C
H
C
H
Disubstituted cycloalkanes
A number of types of isomers are possible. § Constitutional: eg 1,1-­‐dichlorocyclopentane, 1,2-­‐dichlorocyclopentane and 1,3-­‐dichlorocyclopentane. § Configurational: eg two forms of 1,2-­‐dichlorocyclopentane: Nomenclature: cis and trans
§
§
§
These structures are diastereoisomers. They have different physical and chemical properties. The terms cis and trans are used to distinguish them. 21. Three pairs of cyclobutanes are shown below. What type of isomerism is shown in each? __________________________ __________________________ 19 __________________________ Alkene ‘diastereoisomers’
22. Are these the same compound? Why so/ why not? What about these two? Why so/ why not? What type of isomers are these? And what is required for this type of isomerism to occur? A double bond is constructed of a σ-­‐ and π-­‐bond. § Pi-­‐bonds result from p-­‐orbital overlap and are directional with a fixed geometry. § Electron density concentrated above and below plane. § Rotation around the C-­‐C axis would require breaking the pi-­‐bond (~128 kJ mol–1) and does not occur at room temperature. H
CH3
H3C
C
Different
H3C C C H
C
CH3
compounds!
H
H
melting point = -106 ºC
melting point = -139 ºC
boiling point = 4 ºC
boiling point = 1 ºC
π - bond
σ - bond
(E)- 2-butene
(Z)- 2-butene
Both ends of the C=C bond must have two different groups. § If these conditions are met, TWO diastereoisomers result: H
H
H
Cl
A
X
The same
C C
C C
compound!
A≠ B
X≠Y
C C
H
Cl
H
H
B
Y
H
H
H
Different
C
C
C
But A and B can be the same as X and Y
compounds!
HOOC
20 COOH
HOOC
COOH
C
H
21 Nomenclature: Z and E
The rules § Z/E determined by assigning a priority to each of the pairs of groups on each carbon of the double bond. § The higher the atomic number of the atom attached, the higher the priority. § If identical atoms are attached to each carbon of C=C, work outwards along the chain until the first point of difference is reached. § If groups of high priority are on the same side of the double bond the alkene is denoted (Z). § If groups of high priority are on the opposite side of the double bond the alkene is denoted (E). Higher priority groups
Higher priority groups
alkene is
alkene is
on the same side
on opposite sides
denoted (Z)
denoted (E)
of double bond
of double bond
Example: Br
CH3
H
H
H
CH3
C C
C C
C C
H
Cl
H3C
CH3
H3C
H
(Z)- 2-butene
(E)- 2-butene
(E)-1-bromo-2-chloropropene
higher priority groups
higher priority groups
higher priority groups
on opposite sides
on opposite sides
on
same
side
23. Question: name these molecules, including stereochemical descriptor where appropriate A.......................................
B.......................................
(cis/trans, Z/E) C.......................................
D.......................................
24. What is the isomeric relationship between the pairs: A and B…………………..…… C and D……………………….. 22 Note: No diastereoisomers of alkynes
§
§
§
Two pi bonds and two sigma bonds associated with each carbon of the triple bond. Geometry of both carbon atoms is linear. There is only one way to attach two substituents in a straight line. H3C
π - bonds
23 C
C
H