organic chemistry - university of kwazulu-natalcheminnerweb.ukzn.ac.za/files/students organic...

Organic Chemistry Dr V.O. Nyamori

• The study of carbon‐containing compounds and theirproperties.

• The vast majority of organic compounds contain chains orrings of carbon atoms.

• They form the basis of, or are important constituents ofmany products (plastics, drugs, petrochemicals, food,explosives, paints, to name but a few) and, with very fewexceptions, they form the basis of all earthly life processes.

Soap/detergent ‐surfactant C17H35COO‐

Sugar ‐glucose C6H12O6

Medicine ‐ ascorbic acid HC6H7O6 1

Examples

Carbon: group 14, atomic no. 6Recall: Electronic configuration for carbon?Recall: Electronic configuration for carbon?

Periodic Table

Structure of Carbon Compounds

C C C C C C

Three hybridization states

each satisfies the octet rule for each carbon!1.54 Å

1.20 Å1.33 Å

“Organic Chemistry: A Short Course” 12th Edition, by Hart etal. Chapter 1.14 ‐ 1.18

“Chemistry: The Central Science: A Broad Perspective”, 2nd

Edition, by LeMay et al. Chapter 21.1 & Chapter 23.1 pg 852 1

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 3

1. Prefix

2. Base

3. Suffix

Chapter 2: “Organic Chemistry”, 5th Edition , Bruice P. Y.

Chapter 21: pg. 807‐9: “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, LeMay et al.

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.• Combust to give mainly CO2 and H2O• Nomenclature suffix “‐ane” 1

PropertiesProperties

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

Unsaturated Hydrocarbons

Carbon can form multiple bonds with itself or other atoms,e.g. N or O.

Multiple bonds affect, physical and chemical properties(reactivity of organic molecules).

Hydrocarbons that contain one or more double or triplebonds are called unsaturated hydrocarbons.

C

CC

C

CC

C C

C C

H

H

H

H

H H

H

H H

H

H

H

CH3C

O

CH3

CH3C

N

CH3C

H3C

S

CH3

H

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

34

Each functional group is specified by a suffix or prefix

Certain groups of atoms give a molecule a.... FUNCTIONAcidic, basic, alcohol, etc…

The GROUPS are called functional groups.

depicted on the nomenclature of the organic molecule

FUNCTIONAL GROUPS

34

Functional groups are given an order of priority to decide on which is the suffix.

HOMEWORK!

• “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay etal. Chapter 24.1, 25.1, 26.1 & 26.3

• “Organic Chemistry” 5th Edition, by Bruice• Hart et al. “Organic Chemistry: A Short Course” 12th Edition

Hint: Please refer to your textbooks !!

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


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


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



Alcohol ROH hydroxy- -ol

Methanol

Ketone RCOR' keto-, oxo- -oneButanoneMethyl-

ethyl ketoneethyl ketone

Aldehyde RCHO aldo- -al

EthanalAcetaldehyde



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

AlcoholR1 C

H

HOH

Amine

Example

CH

HOHCH3CH2

N

H

HCH3CH2R1 N

H

H

1

Secondary (2°) alcohols and amines

General structure

AlcoholR1 C

H

R2OH

Amine

Example

CH

CH3

OHCH3CH2

N

H

CH3

CH3CH2R1 N

H

R2

2

Tertiary (3°) alcohols and aminesGeneral structure

AlcoholR1 C

R3

R2OH

Amine

Example

CCH3

CH3

OHCH3CH2

N

CH3

CH3

CH3CH2R1 NR3

R2

3

Quaternary amines

General structure

AmineR4 N

R2

R1R3+ N

CH3

CH3

CH3CH3CH2+

Example

4

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‐ol

b) Dimethylamine

c) 3‐Ethylhexan‐3‐ol

d) Diethylmethylamine

e) Butan‐2‐ol

f) Triethylmethyl ammonium ion

1

a) Pentan‐1‐olSolutions

b) Dimethylamine c) 3‐Ethylhexan‐3‐ol

d) Diethylmethylamine e) Butan‐2‐ol f) Triethylmethylammonium ion

2

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…..

1

Example: Name this organic molecule:

CH3 CHCH

OH

CH3

NO2

2


1. This molecule contains a …………… and a ..….…. group

CH3 CHCH

OH

CH3

NO2

Only the hydroxyl group has priority.

So this is an ………………..…..

3


longest carbon chain containing the hydroxyl group has …………..

Therefore its a “……………..”.

CH3 CHCH

OH

CH3

NO2

this is an ALCOHOL.

An alcohol suffix is ‐OL

4

3. Number the carbons, starting NEAREST the functional group.

CH3 CHCH

OH

CH3

NO2

The hydroxyl group is on …………..

…so this is a ………….... or ……….………..

4. This molecule has one substituentA nitro group on …………..

or

5

Example: Name this organic molecule

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

1. Identify functional groups

2. Determine longest carbon chain that contains the highest priority functional group.

6


CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

1. Molecule contains a ……………. and an ......………..

Now determine longest carbon chain

The …………….. group has highest priority

7


C

CH3 CH CH CH

CH3

C

CH2

CH2

CH2

CH3

CH3

CH3

OH

O

H2N

The longest carbon chain has 8 carbons.

BUT………………...

THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group

8

Example: Name this organic molecule

The longest carbon CONTAINING the ‐CO2H

chain has ……… carbons.

so this molecule is based on a ………….

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

9

Highest priority functional group is a carboxyl group

Suffix ‐OIC ACIDHence… Heptanoic acid

Now number chain…….

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

3. Number the carbons starting with the functional group

Heptanoic acid

The carboxyl group is on position 1,

do not include in the name because

the carboxyl group is always a terminal group.10

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

11

Compose the nameALPHABETICAL LIST

INTERPETING AN IUPAC NAME…...

Heptanoic acid

4. This molecule has four substituents

CH3 CH CH CH

CH3

C

CH2

CH2C

CH2

CH3

CH3

CH3

OH

O

H2N

12

WHAT IS THE STRUCTURAL FORMULA OF

BUTANONE ?KETONE GROUP

CCH3 CH2

O

CH3

Carbonyl CANNOT BE AT END!!!!WHY?????

CH2C CH2

O

CH3HBUTANAL

ExamplesGive the correct IUPAC name for the following compounds.

Br

H

OH

a)

c)

b)

d)

13

OH

O

CH3

O

Cl

OH

ExamplesGive the correct IUPAC name for the following compounds.

f) g)

h) i) O

O

14

ExamplesDraw structural formulae for the following compounds:

a) 2,3,5‐trimethylhexane

b) (Z)‐3‐chlorohept‐2‐ene

c) 3‐ethylnonanol

d) 2,3‐dimethylpentanoic acid

e) Methyl hexanoate

f) 3‐iodohexanal

g) pentan‐2‐one

h) 3‐aminopentane T115

Solutions

2,3,5‐trimethylhexane (Z)‐3‐chlorohept‐2‐ene

3‐ethylnonanol 2,3‐dimethylpentanoic acid

a) b)

c) d)

16

Solutions

Methyl hexanoate 3‐iodohexanal

pentan‐2‐one

e) f)

g) h) 3‐aminopentane

17

ISOMERS

(a) Structural isomers

CONSTITUTIONAL ISOMERS

Constitutional isomers have different properties:

These are…..?

e.g. butane (C4H10) has 2 structural isomers

Molecules with the same chemical formula but different bonds between the atoms

Now called...

n‐Butane: C4H10 2‐methylpropane: C4H10

bp = ‐12 °C mp = ‐159 °Cbp = 0 °C mp = ‐138 °C

Two types: (a) Structural isomers (b) Stereoisomers

1

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

1

H3CCH2CH3

C

OH

HH3CH2C

C

OH

CH3

H

2‐ butanol

DO NOT FORGET THE OTHER ISOMER……….

CH2CCH3

OH

CH3

H

THESE ARE THE TWO…. ENANTIOMERS...or ....OPTICAL ISOMERS

p

Chapter 22; pg. 829‐839. “Chemistry: The Central Science: A Broad Perspective” 2th Edition.

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


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


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

organic chemistry - university of kwazulu-natalcheminnerweb.ukzn.ac.za/files/students organic...

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