Organic Chemistry Dr V.O. Nyamori
Textbook by i) Bruice “Organic Chemistry” 5th Editionii) Hart et al. “Organic Chemistry: A Short
Course” 12th Edition• The study of carbon‐containing compounds and their properties.
• The vast majority of organic compounds contain chains or ringsof carbon atoms.
• They form the basis of, or are important constituents of manyThey form the basis of, or are important constituents of manyproducts (plastics, drugs, petrochemicals, food, explosives,paints, to name but a few) and, with very few exceptions, theyform the basis of all earthly life processes.
Soap/detergent ‐surfactant C17H35COO‐e g
1
Sugar ‐glucose C6H12O6Medicine ‐ ascorbic acid HC6H7O6
e.g.
Carbon: group 14, atomic no. 6Recall: Electronic configuration for carbon?Recall: Electronic configuration for carbon?
Periodic Table
Structure of Carbon CompoundsStructure of Carbon CompoundsThree hybridization statesThree hybridization states
C C C C C C
each satisfies the octet rule for each carbon!!1.54 Å 1.20 Å1.33 Å
each satisfies the octet rule for each carbon!!
Hart et al. “Organic Chemistry: A Short Course”, 12th Edition,Chapter 1.14 ‐ 1.18
Geometries of carbon compounds
sp3 Tetrahedral 108 5°sp Tetrahedral 108.5
sp2 Trigonal planar 120°sp Trigonal planar 120
sp Linear 180°sp Linear 180
Methane: CH4 Hydrogens are in a tetrahedral
arrangement around the sp3
4
arrangement around the sp3
hybridized carbon atom. Hydrogens bond to the carbon sp3
orbitals with 1s orbitals.
sp3 Hybridizationcarbon
2p
Energy
sp32s Hybridization
E sp1s
sp2 Hybridizationcarbony
2p
Energy
22s Hybridization
sp21s
Ethene: C HEthene: C2H4sp Hybridizationcarbon
2p
gy
sp2s Hybridization
Energ
1sEthyne: C2H2
HYDROCARBONS• compounds composed of only carbon and hydrogen
h i f b b d d h h d• chain of carbon atoms bonded to enough hydrogen atoms to satisfy the octet rule for each carbon
• chain is bent because of the 109.5° C–C–C tetrahedral angle
C H
HH CHHC
CH2C
C
e.g.
C
CC
CC
CC H
HH
HH
CH3H3C
HHH
H
Line notation
HYDROCARBONS
H d b i h ll i l b b b d
Alkanes• Hydrocarbons with all single carbon‐carbon bonds
(no double or triple bonds)
• They contain the maximum number of hydrogen atoms• Alkanes are SATURATED
y y g
Alkenes, alkynes and aromatic compounds
• UNSATURATED hydrocarbons
‐ they ARE NOT ALKANES
• contain carbon‐carbon multiple bonds
8
‐ they ARE NOT ALKANES
Past exam QuestionPast exam Question
C‐1: _____ C‐3: _____
C‐2: _____ C‐4: _____
Example 1. Indicate the hybridization for carbons 1 – 10 and their
respective geometry. Include bond angles in your answer.
CC
OC
HBr
H1 C
C
C CC
CC
N
ClH
H
12 3
89
10CC
C N
F H
O
H
46
78 10
H
Hydrocarbons• Four basic types:
‐ Alkanes‐ AlkanesCnH2n+2
C H Ethane
‐ Alkenes
C2H6 Ethane
CnH2n
C2H4 Ethene 120°
‐ AlkynesCnHn
‐ Aromatic hydrocarbonsC2H2 Ethyne
HH
yCnHn
H
H H
H
C6H6 Benzeneor
Organic Nomenclature
• Three parts to a compound name:
1 2 31 2 3
1. Prefix
2. Base/parent
3. Suffix
Chapter 2: “Organic Chemistry” 5th Edition , Bruice P. Y.
Organic Nomenclature ‐ IUPAC Rules
S ffi T ll h t t f d it i
Base/parent: Tells how many carbons are in the longest continuous
Suffix: Tells what type of compound it is.
Prefix : Tells what substituent(s) are attached, if any.
/p y gchain.
prefix base suffix
What b i ? How many
What substituent? How many
carbons?family?
Alkanes
• Only van der Waals force: London force.• Boiling point increases with length of chain• Boiling point increases with length of chain.• Nomenclature suffix “‐ane”
To Name a Compound…1. Determine what type of
compound it is.2. Find the longest chain in the
molecule.3. Number the chain from the
end nearest the first substituent encountered.
4. List the substituents as a prefix in alphabetical order along with the number(s) of the carbon(s) to which they are attached.
CH2CH
CH2H2C
CH2H3C CH2CH2
H3C
3
CH3Name??
ExampleCH3
CH2CH
CH CH CH2 CH3
CH3
CHCH3 CH3
If there is more than one type of substituent in the molecule
CH3
If there is more than one type of substituent in the molecule,list them alphabetically i.e. name of substituent, not prefix forfrequency e g di tri tetra etc are not consideredfrequency e.g. di, tri, tetra, etc...are not considered.
Cycloalkanes• Carbon can also form ringed structures.
b bl• Five‐ and six‐membered rings are most stable.
– Can take on conformation in which angles aregvery close to tetrahedral angle.
ll i i i d– Smaller rings are quite strained.
General formulaC HCnH2n
cyclohexane cyclopentane cyclopropane
CYCLIC ALKANESHow do we name….
CH33
CH2CH32 3
Alkenes
VSEPR Theory
120°
• Contain at least one carbon–carbon double bond
• Unsaturated
– Have fewer than maximum number of hydrogens
Th C t d bl b d 2 h b idi d– The C atoms on double bond are sp2 hybridized
Structure of Alkenes
• Unlike alkanes alkenes cannot rotate• Unlike alkanes, alkenes cannot rotate freely about the double bond.
– Side‐to‐side overlap makes this impossible without breaking ‐bond.
C C
impossible without breaking bond.
Structure of Alkenes
This creates geometric isomers
difference in the spatial parrangement of groups about the double bond Z‐2‐Pentene
Cis isomer “Z” isomer
Z‐2‐Pentene
Cis‐ isomer “Z”‐isomer
Trans‐ isomer “E”‐isomer
E‐2‐Pentene
NAMING ALKENES
1. Find the longest unbranched carbon chain containing the double bond. • Name chain according to number of carbon atoms. add ‐ene as a suffix
2. Number the carbon atoms in the main chain.
add ene as a suffix
• Start from the end of the chain that is closest to thedouble bond.
location of the double bond is numbered with the
22
lowest‐numbered carbon in the double bond.
Example
Name this alkene
HCH
C CHCH3 C
CH CH2 CH3
CH3
H
CHCH
H C CH2H2C
23
Example
CH3 CH
CH3
Name this alkene CH
CHCH3 C
CH CH2
CHCH3
H
CH2
CHH2C
1. The longest unbranched chain containing the double bond which is the functional group (suffix ‐ ene)
2 The chain numbering starts closest to the double bond2. The chain numbering starts closest to the double bond.
24
3. There are two substituent groups on this alkene:
CH3 CH
CH3
CH
CHCH3 C
CH CH2
CHCH3
H
CH2
CHH2C
4. Compose the name…..• Add the substituent groups alphabetically to the alkene nameAdd the substituent groups alphabetically to the alkene name
• Specify the position of each group on the main chain
name of the alkene is?
Properties of AlkenesExample: C4H8
2‐Methyl‐1‐propenebp ‐7 ⁰C
1‐Butenebp ‐6 ⁰C
Cis‐2‐Butenebp +4 ⁰C
Trans‐2‐Butenebp +1 ⁰C
Structure also affects physical properties of alkenes
bp. 7 C bp 6 C bp +4 C bp +1 C
Can we have more than one double bond?
GEOMETRICAL ISOMERSAlkenes exhibit cis‐trans isomerism.
Trans‐ isomer “E‐” Cis‐isomer “Z‐”• Identical substituents onopposite sides of the double
• Identical substituents on
Trans isomer E Cis isomer Z
opposite sides of the doublebond
H CH
same side of the doublebond
trans cis
CCH
HC
H
CH3
CCH3
CCH3
trans‐ cis‐
CH3 HH H
i k di“Z‐”
“E‐”
Stick diagram
27
Priorities are assigned by the atomic numbers of the atomsbonded to the carbon in the double bond.
Geometric isomers of Alkenes• Cis‐alkenes have similar higher priority elements or group in
the chain on the same side of the molecule (or Z‐isomer i.e.have higher priority elements but not necessarily the same onthe same side of the molecule)
• Trans‐alkenes have similar higher priority elements or group inthe chain on opposite sides of the molecule (or E‐isomer i.e.have higher priority elements but not necessarily the same onopposite sides of the molecule).
28
Examples1. Name the following alkenes and determine whether there
are geometric isomers i.e. either Trans‐ (E‐) or Cis‐ (Z‐)g ( ) ( )isomers.
a) Br
Hb)
H
Br
c)
F
FUNCTIONAL GROUPS
Certain groups of atoms give a molecule a....FUNCTION
Acidic, basic, alcohol, etc…The GROUPS are called functional groups.
Each functional group is specified by a suffix or prefix
The GROUPS are called functional groups.
depicted on the nomenclature of the organic molecule
Functional groups are given an order of priority to Functional groups are given an order of priority to decide on which is the suffix.
Please refer to your textbooks on priority preference by:HOMEWORK!
30
i) Bruice “Organic Chemistry” 5th Editionii) Hart et al. “Organic Chemistry: A Short Course” 12th Edition
Group / Family Formula Structural
Formula Prefix Suffix ExampleFamily Formula
Alkane RH alkyl aneAlkane RH alkyl- -ane
Ethane
Alkene R2C=CR2 alkenyl- -ene
Ethene
Alkyne RC≡CR' alkynyl- -yneEthyne
Benzene derivative
RC6H5RPh phenyl- -benzenederivative RPh p y
2-phenylpropaneisopropylbenzene
Group / Family Formula Structural
Formula Prefix Suffix Example
Haloalkane RX halo- alkyl halide
ChloroethaneEthyl chloride
Fluoroalkane RF fluoro- alkyl fluoride
y
FluoromethaneMethyl fluoride
Chloroalkane RCl chloro- alkyl chloride
y
ChloromethaneM th l hl idchloride
alkyl
Methyl chloride
BromomethaneBromoalkane RBr bromo- alkyl bromide
BromomethaneMethyl bromide
Iodoalkane RI iodo- alkyl iodide
IodomethaneMethyl iodideiodide
Group / Family Group Formula Structural
Formula Prefix Suffix Example
Primary amine
RNH2 amino- -amine
Secondary R NH amino amine
Methylamine
yamine
R2NH amino- -amine
Dimethylamine
Amines Tertiary amine
R3N amino- -amine
T i th l i
H3C CH3
Trimethylamine
Quaternaryammonium
ionR4N+X-
ammonio--ammonium
N
H CH3
Cl
T i th lion ammonio Trimethyl-ammonium
chloride
Group / Family Formula Structural
Formula Prefix Suffix Example
Alcohol ROH hydroxy- -ol
Methanol
Ketone RCOR' keto-, oxo- -oneButanoneMethyl-
ethyl ketoneethyl ketone
Aldehyde RCHO aldo- -al
EthanalAcetaldehyde
Group / Family Formula Structural
Formula Prefix Suffix Example
Carboxylic acid
RCOOH carboxy- -oic acidacid
yEthanoic acid
Acetic acid
Acyl halide RCOX haloformyl- -oyl halideEth l hl idEthanoyl chloride
Acetyl chloride
Ether ROR' alkoxy-alkyl alkyl
ether EthoxyethaneDiethyl ether
Ester RCOOR'alkyl
Diethyl ether
Ester RCOORalkanoate Ethyl butanoate
Ethyl butyrate
Primary (1°) alcohols and amines
General structure Example
AlcoholR1 C
HOH
HR1 C
HOH
°
CH
OHCH3CH2
1°
Amine N
H
CH CHR1 N
H
Amine
1°
NH
CH3CH2R1 NH
1
Secondary (2°) alcohols and amines
General structure Example
Alcohol1
H HR1 C
R2OH C
CH3
OHCH3CH2
2° Alcohol3
Amine N
H
CH CHR1 N
HAmine N
CH3
CH3CH2R1 NR2
2° Amine
Tertiary (3°) alcohols and aminesGeneral structure
3
Example
AlcoholR1 C
R3
OH CCH3
OHCH3CH2R CR2
OH
° l h l
CCH3
OHCH3CH2
3° alcohols
Amine N
CH3
CH CHR1 NR3
Amine N
CH3
CH3CH2R1 NR2
3° Amine
Quaternary amines
General structure Example
R2 CH3Amine
R4 N R3+ NCH3
CH3CH3CH2+
R13
Exercise
1. Draw the structures of the following alcohols and amines and classify them as either 1°, 2°, 3° or quaternary
a) Pentan 1 ola) Pentan‐1‐ol
b) Dimethylamine
c) 3‐Ethyl‐hexan‐3‐ol
d) Diethyl‐methyl‐amine
e) Butan‐2‐ole) Butan‐2‐ol
f) Triethylmethly ammonium ion
a) Pentan‐1‐olSolutions
b) Dimethylamine c) 3‐Ethyl‐hexan‐3‐ola) Pentan 1 ol b) Dimethylamine c) 3 Ethyl hexan 3 ol
d) Diethyl‐methyl‐amine e) Butan‐2‐ol f) Triethylmethly‐ammonium ion
Naming Hydrocarbons with Functional Groups
Name the other substituent groups, using theprefixes for alkyl groups and the prefixes for anyother functional groupsother functional groups
Specify the position of each group on the main chain.p y p g p
Add the substituent groups alphabetically to the name g p p yof the alkane (or alkene or alkyne) along with the frequency of each groupfrequency of each group.
Example42
Example…..
Example: Name this organic molecule:
OH NO2
CH3 CHCH CH3
43
Example: Name this organic molecule:
OH NO2
CH3 CHCH CH3
1. This molecule contains a ……………. and …………. group
Only the …………… group has priority.
So this is an ................
44
Example: Name this organic molecule:
OH NO
CH3 CHCH
OH
CH3
NO2
3 3
l b h i i i h h d l longest carbon chain containing the hydroxyl group has ………… carbons.
Th f i “ ” An alcohol suffix is ‐OL
45
Therefore its a “….……….. ”
OH NO2
CH3 CHCH CH3
3. Number the carbons, starting NEAREST the3. Number the carbons, starting NEAREST the functional group.
The hydroxyl group is on position …
so this is a or…so this is a ………… or .…………
4. This molecule has one substituent4. This molecule has one substituentA nitro group on position …
46
………………………… …………………….or
Example: CH3p
Name this organic CH
CH2C CH3H2N
Name this organic molecule ?? CH3 CH CH CH2 C
CH2
OH
O
CH3 CH2
CH3
OH
C 3
1 Identify functional groups1. Identify functional groups
2. Determine longest carbon chain that contains the highest priority functional group.
47
Example: Name this organic molecule:
CHC
CH3
CHH N
CH2
CH2C CH3
O
H2N
CH3 CH CH CH2
CH3
C
CH2OHCH3 CH2
CH3
The longest carbon chain has 8 carbons.
BUT………………...
THIS CHAIN DOES NOT CONTAIN THE CO H group48
THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group
Example: Name this organic molecule
CHC
CH3
CHH N
CH2
CH2C CH3
O
H2N
CH3 CH CH CH2
CH3
C
CH2OHCH3 CH2
CH3
The longest carbon CONTAINING the ‐CO2H
chain has carbonschain has …. carbons.
so this molecule is based on a ..………… ..
49
CH3
CH2
CH2C CH3
O
H2N
CH3 CH CH CH2 C2
OH
O
CH3 CH2
CH3
OH
Highest priority functional group is a carboxyl group
Suffix ‐OIC ACIDHence… Heptanoic acidHence… Heptanoic acid
Now number chain50
Now number chain…….
CH3Heptanoic acid
CH2
CH2C CH3
O
H2N
CH3 CH CH CH2 C2
OH
O
CH3 CH2
CH3
OH
3. Number the carbons starting with the functional group
The carboxyl group is on position 1,
do not include in the name because
the carboxyl group is always a terminal groupthe carboxyl group is always a terminal group.
CH35 6 7
Heptanoic acid
CH2
CH2C CH3
O4
5 6 7H2N
CH3 CH CH CH2 C2
OH
O123
CH3 CH2
CH3
OH
4. This molecule has four substituents
Compose the name
ALPHABETICAL LIST
52INTERPETING AN IUPAC NAME…...
WHAT IS THE STRUCTURAL FORMULA OF
BUTANONE ?
CH CHC
O
CHCH3 CH2C CH3
C b l CANNOT B t th END!! WHY?????Carbonyl CANNOT Be at the END!! WHY?????O BUTANALC CH2CH2 CH3H
53
ExamplesGive the correct IUPAC name for the following compounds.
Bra) b)
Hc) d)
H
OHOH
Examples
O
Give the correct IUPAC name for the following compounds.
OH CH3
f) g)
O
ClCl
OHh) i) O
OO
ExamplesDraw structural formulae for the following compounds:
) 2 3 5 i h lha) 2,3,5‐trimethylhexane
b) (Z)‐3‐chloro‐hept‐2‐eneb) (Z) 3 chloro hept 2 ene
c) 3‐ethylnonanol
d) 2,3‐dimethylpentanoic acid
e) methylhexanoate
f) 3 iodohexanalf) 3‐iodohexanal
g) pentan‐2‐one
h) 3‐aminopentane T1
Solutions
2,3,5‐trimethylhexane (Z)‐3‐chloro‐hept‐2‐enea) b)) b)
3‐ethylnonanol 2,3‐dimethylpentanoic acidc) d)
Solutions
methylhexanoate 3‐iodohexanale) f)
2) h) 3 ipentan‐2‐oneg) h) 3‐aminopentane
ISOMERS
(a) Structural isomers
Two types: (a) Structural isomers (b) Stereoisomers
(a) Structural isomers Molecules with the same chemical formula but different
CONSTITUTIONAL ISOMERSbonds between the atoms
Now called...
Constitutional isomers have different properties:e g butane (C H ) has 2 structural isomers
These are…..?
e.g. butane (C4H10) has 2 structural isomers
B C H 2 h l C HCH3 CH2 CH2 CH3 CH3 CH CH3
n‐Butane: C4H10 2‐methylpropane: C4H10
bp = ‐12 °C
CH3
pmp = ‐159 °Cbp = 0 °C mp = ‐138 °C
ExampleHow many constitutional isomers are formed from C5H12? Draw their structures.
Solution:
60
CONSTITUTIONAL (STRUCTURAL) ISOMERS
The general formula for ALKANES is…... 22 nnHCg
The number of ISOMERSn = 1, 2 and 3 1 ISOMER
The number of ISOMERS increases with n….. n = 4 2 ISOMERS
n = 5 3 ISOMERS
n = 6 5 ISOMERS
7 9 ISOMERSn = 7 9 ISOMERS
n = 8 18 ISOMERSn = 40
62,491,178,805,831 ISOMERSn = 9 35 ISOMERS
n = 10 75 ISOMERS
, , , ,
HOMEWORK: n = 10 75 ISOMERS
n = 20 366,319 ISOMERS
HOMEWORK: DRAW THE ISOMERS OF C40H82 !!!
Constitutional isomers for multibonds
C HExample: Alkene
C6H10
Constitutional isomers for multibondsExample: Alkyne
C HC6H6
H3C C C C C CH3H3C C C C C CH3
2,4‐Hexadiyne
CH3Hexa‐2,4‐diyne
C C C C CH2H
1 3 H di1,3‐HexadiyneHexa‐1,3‐diyne
63
OPTICAL ISOMERISM
O i l i i i h l l h Optical isomerism arises when molecules have astructure such that the mirror image is notsuperimposable on the original molecule.
occurs whenever there are four different groups bound to the same tetrahedral carbon atombound to the same tetrahedral carbon atom.
Some terminology…….
64
OPTICAL ACTIVITY IN ORGANIC COMPOUNDSCOMPOUNDS
Stereogenic centre has four different groups attachedto a tetrahedral carbon atom
W Chi l bWStereogenic centre
Chiral carbon atom
ZX
C*
The carbon involved is called a chiral carbon or
XY
The carbon involved is called a chiral carbon orstereogenic carbon and the molecule is known as a chiralmolecule
65
molecule.Example: 2‐butanol
Perspective formula
2‐Butanol
Dash shows bond going backwards from the viewer
Bonds aligned to the asymmetric
OHbackwards from the viewer the asymmetric
center in the plane
CH C
*CH2CH3
HH3C
Solid wedge represents a bond extending out towards
the viewerChiral centre
the viewer
Fischer projection
2‐Butanol
Fischer projection
CH3Bonds aligned to the asymmetric
Bond going backwards from the CH3 the asymmetric
center in the planebackwards from the
viewer
CH2CH3HO *Bond extending
H
Bond extending out towards the
viewer HChiral centre
67
ENANTIOMERISM in ORGANIC CHEMISTRY
2‐butanol has two optical isomers.
non‐superimposable mirror images of each otherA pair of isomers called enantiomers ‐
non‐superimposable mirror images of each other.
Vi i 3 DView in 3‐D“mirror”
View in 3‐D
OH OH
CCH CHH3C
C
CHCH CHCH2CH3H
3
HCH3CH3CH2
68
Identify the chiral carbon (stereogenic centre) and draw theQuestion
Identify the chiral carbon (stereogenic centre) and draw thestructural formula for each of the following molecules:
(a) 1‐chloroethanol (b) 2,3,5‐Trimethyl‐hexane
(c) Methylcyclohexane (d) 1,3‐dimethlycyclopentaneSolution(a) 1 chloroethanol (b) 2,3,5‐Trimethyl‐hexane(a) 1‐chloroethanol ( ) , , y
(d) 1 3 di hl l(d) 1,3‐dimethlycyclopentane(c) Methylcyclohexane
NAMING OPTICAL ISOMERS
Stereogenic centre creates twomolecular optical isomers
Cl ClTwo
configurations
CH OH
CH
HO
Enantiomers
How do we name these isomers??
H3C HO CH3
Solution: Use R‐S nomenclature system for designating
How do we name these isomers??
y g gthe configuration
h70
We assign priorities as in the E, Z system……...
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
1. Assign relative priorities to each of the four groups on the stereogenic carbon to describe the configuration.
The priorities are given by rules:
• Higher atomic numbers are given higher priorities.
• If necessary, the second atom in each substituent is used to determine the priorities.p
71
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
2. Draw the molecule with the lowest‐priority group pointing directly into the page….
and the other three groups pointing out of the page inand the other three groups pointing out of the page in an arrangement like a steering wheel.
Example: 2‐ Butanol….. OHpCH2CCH3 CH3
HH
Draw molecule as a wedge and dashed line diagram
72
Draw molecule as a wedge and dashed line diagram
2‐ butanol OH
Hart et al. “Organic Chemistry: A Short Course”
th d h
CH2CCH3
OH
CH3
p12th Edition. Chapter 5
OH OH
23 3
H
C C
H3CCH2CH3
CHH3CH2C
C
CH3
H
3 3
DO NOT FORGET THE OTHER ISOMERDO NOT FORGET THE OTHER ISOMER……….
THESE ARE THE TWO…. ENANTIOMERS
73...or ....OPTICAL ISOMERS
1. Assign priorities:2‐ butanol OH 1. Assign priorities:
O > CMe = CEt > HC
C,H,H > H,H,HC
CH3
CH2CH3H
CEt > CMe
3
74
OH1 2‐ Butanol
CH 241. Assign priorities to each group:
CH3
CH2CH3
3O > CEt > CMe > H
OH1
C
OH1
2
2. Redraw the molecule with the lowest priority group facing in.
H3C CH2CH3
C 2
3
lowest priority group facing in.
Now what?????
75
12 ButanolOH 2‐ Butanol
H3C CH2CH3
2
3
C
3 (R)‐2‐Butanol
3. Look at the direction in which the priorities decrease.
If h d i l k i di i h If they decrease in a clockwise direction, the stereogenic centre is called “R” or rectus
which is Latin for “right.” Or….
76
12 ButanolOH 2‐ Butanol
CH3CH3CH2
2 3C
(S)‐2‐Butanol
If the priorities decrease in a counter‐clockwise
direction the stereogenic centre is called “S” or
sinister, which is Latin for “left.”
Example: What are the configurations of the following chiral molecules?
NH NH
chiral molecules?
C
NH2
C
NH2
CHCH2CH3
CH
CH3CCH2CH3
HCH3 H
H3C CH2CH3CH2CH3H2NC
2 3C
2
78 NH2CH3
Example: Give the configuration of the stereogeniccentre in each of the following molecules:
ClCH2OH
C
Cl
CHC OH
CHCH2OHHO CCl O
HO CO
(A) (B)(A) (B)
2‐chloro‐3‐hydroxy‐propanoic acid
THEY LOOK DIFFERENT BUT ARE THEY???
y y p p
79
THEY LOOK DIFFERENT BUT ARE THEY???
A) CH2OH 1. Assign priorities to each group:
CH
ClC OH
O
CH2OH2. Redraw the molecule with the
lowest priority group facing in
C
lowest priority group facing in.
3. Look which way the priorities Cl C OH
O
3. ook which way the prioritiesdecrease.
O
The priorities decrease anti‐clockwise, so this centre is “.…”
8080
1. Assign priorities to each groupClB)
C
B)
CHCH2OHHO C
2. Redraw the molecule with the lowest priority groupO
Othe lowest priority group facing in.
Cl 3. Direction of the priorities decrease?
CCH2OHHO C
decrease?
The priorities decrease 2O C
O anti‐clockwise, so this centre is “…”
Example: Give the configuration of the stereogenic h f h f ll l lcentre in each of the following molecules:
ClCH OH B)A)
CC
CH2OH B)A)
CHCH2OHHO C
CH
ClC OH
HO CO
ClO
2‐chloro‐3‐hydroxy‐propanoic acid
THEY LOOK DIFFERENT BUT ARE THEY???
82 WHAT IF?????
If we flip the COOH and H ???????
CH2OHA) CH2OHA’)
C
2
C
CH2OH
CHCl
C OHO
C H
Cl
C
O
HO
O ClO
CH2OH CH2OH
C C
2
Cl C OH
O
CC
O
HO ClO O
EXAM QUESTION
What is the structural formula of (R)‐2‐chloro‐2‐butanol?
DRAW MOLECULAR STRUCTURESolution:
ASSIGN PRIORITIES
84
(R) 2 chloro 2 butanolAssign priorities
(R)‐2‐chloro‐2‐butanol
Draw steering wheel with lowest priority group pointingDraw steering wheel with lowest priority group pointing in…….
MAKE SURE PRIORITY GOES CLOCKWISE FOR R
DRAWMOLECULEC
85
DRAW MOLECULE…..
DRAWMOLECULE TO SEEDRAW MOLECULE TO SEE ALL GROUPS…..C
Remember lowest priority group has a “dashed” bond
Make sure priority goes clockwise for “R”
C
(R)‐2‐chloro‐2‐butanol
CHCH3 CH CH3* *
[2R,3R] [2S,3S] [2R,3S] [2S,3R]
3
Br Cl3
[2R,3R] [2S,3S] [ , ] [ , ]
a b c d
RCH3
SCH3 R
C
CH3
HBS
C
CH3
BHRCC
HCl
BrH
SCC
BrH
HCl S
CC
HH
BrCl
R
CC
BrCl
HH
RC
CH3
ClHS
C
CH3
HCl SCH3
RCH3
enantiomeric pairs enantiomeric pairs enantiomeric pairs e a t o e c pa s
• Enantiomeric pair differ only in optical activity
Diastereomers b & ca & ca & c
b & da & d
RC
CH3
HBrS
C
CH3
BrHR
C
CH3
HBrS
C
CH3
BrHa b c d
S
CC
HH
BrCl
R
CC
BrCl
HH
R
CC
HCl
BrH
S
CC
BrH
HCl
SCH3
RCH3
RCH3
SCH3
i di i h i l d Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp.,
sol bilit etc, solubility etc…molecule with n stereogenic centres may exist in
i f 2n t i i f ith i fmaximum of 2n stereisomeric forms, with maximum of2n/2 enantiomeric pairs
Meso compoundCHHOC CH COH
OO * *CHHOC CH COHOH OH
RC
CO2H
OHHS
C
CO2H
HHOR
C
CO2H
OHHS
C
CO2H
HHO
R
CC
CO2H
OHHO
S
CC
CO2H
OHOH
S
CC
CO2H
OOHH
R
CC
CO2H
HO
HO
CO2H CO2H CO2H CO2H
170 °C 170 °C 140 °C+12° ‐12° 0°
meso compound ‐ an achiral (optically inactive)
+12 ‐12 0
diastereomer of compound with stereogenic centres arises because 4 different groups making each of C‐2 & C‐3 g p g
stereogenic are same 4 different groups…(!)
Meso compound
CO2H CO2H CO2H CO2HR
C
C
CO2H
OH
H
H
HO
SC
C
CO2H
H
OH
HO
H
RC
C
CO2H
OH
OH
H
H
SC
C
CO2H
H
H
HO
HO-------------------- -------------------- -------------------- --------------------
RC
CO2H
HHOS
C
CO2H
OHHS
C
CO2H
OHHR
C
CO2H
HHO
Enantiomers, Chiral Identical, achiral, Meso form
possess plane of symmetry bisecting central C‐C bond
Newman projections
Shows arrangements in space e.g. Ethane C2H660°
H HH
HH
H
60° Staggered
conformationH H
H
HH
H
HH
HH
conformationH
HH
H
“dash‐wedge” “sawhorse” NewmanH
H HH
H0° Eclipsed
conformationHH
HHH H
HH
H
conformationHHH
HH
“dash‐wedge”H H“sawhorse” Newman
H
Cyclohexane conformations
ae a
a"fli "1 5
aea
ae 1
65
e
e
ae
ae
e
"flip"1 65
43
2
ea ee
e4
32
a ae aa
Ch i f i
H H
Chair conformations
H
H HH
HH
HH
H
HH
HCyclohexane
H HBoat conformation
Examples1 2‐Dimethylcyclopentane1,2‐Dimethylcyclopentane
CH3 CH3 HH
H
H CH3 H HH
H
H
H
H HH
H H
H H CH3H
Summary of i i
Hart et al. “Organic Chemistry: A Short Course” 12th Edition page 52‐54
Structural
isomerism Course 12 Edition, page 52 54
Different bond pattern
Structural (constitutional)
isomerisomers
isomer
Same bond
Stereoisomer
Same bond pattern
Stereoisomer
Interconvertible by Not interconvertiblesingle bond rotation by bond rotation
Conformers ConfigurationalConformers (rotamers)
Configurational isomers
Exercise1. Draw the structures of
a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐Heptadiene
b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐Bromobutan‐2‐olb) (S) 2 bromopropan 1 ol d) (2S,3R) 3 Bromobutan 2 ol
2 Using the Newman projection draw the structure a2. Using the Newman projection draw the structure astaggered conformation of butane.
3. Using the Fischer projection draw a structure of(S)‐2‐methylpentanoic acid.
4 Draw the structure for the cis and trans isomers of4. Draw the structure for the cis and trans isomers of1‐bromo‐2‐chlorocyclopropane
Solutions
1. a) (Z)‐3‐methyl‐2‐pentene b) (S)‐2‐bromopropanol
c) (E Z)‐2 4‐Heptadiene d) (2S 3R)‐3‐Bromobutan‐2‐olc) (E,Z)‐2,4‐Heptadiene d) (2S,3R)‐3‐Bromobutan‐2‐ol
Solutions
Staggered conformation of butane2.
Newman projection
3. (S)‐2‐methylpentanoic acid
Fischer projection
3. (S)‐2‐methylpentanoic acid
(4) 1‐bromo‐2‐chlorocyclopropane