stereochemistry - sorbonne-universite.fr

1

GENERAL CHEMISTRYCHEM-110

1

STEREOCHEMISTRY

Same molecular formulas but different connectivity.

STRUCTURAL ISOMERS

CCl2 CH2 CHCl CHCl1,1-dichloroethene 1,2-dichloroethene

C2H2Cl2

Same molecular formulas but different connectivity.

STRUCTURAL ISOMERS

STEREOISOMERS

Same molecular formulas and same connectivity but different orientation in space.

What is the relation between these two What is the relation between these two compounds?compounds?

C CHH

Cl ClC C

Cl

Cl

H

H

1.1. IdenticalIdentical

Iden

tical

Con

stitut

ional is

omer

s

Ster

eoiso

mers

Stru

ctur

al iso

mer

s

0% 0%0%0%

2.2. Constitutional isomersConstitutional isomers

3.3. StereoisomersStereoisomers

4.4. Structural isomersStructural isomers

STEREOISOMERSCHCl CHCl

1,2-dichloroethene

C CHH

Cl ClC C

Cl

Cl

H

H

cis-1,2-dichloroethene trans-1,2-dichloroethene

STEREOISOMERS

CONFORMERS: stereoisomers resulting from free rotation (or a flip) around carbon-carbon bonds.

ENANTIOMERS: stereoisomers that are non

DIASTEREOMERS: stereoisomers that do not result from free rotation and are not mirror image of another.

ENANTIOMERS: stereoisomers that are non-superimposable mirror image of one another.

2

ETHANE

STAGGERED

ECLIPSED ECLIPSED

3

STAGGERED

Newman Projection

STAGGERED ECLIPSED

4

STEREOISOMERS





These two molecules have differentconfigurations

They are called enantiomers from the Greek enantio or opposite.

They are also called chiral

STEROISOMERS

ENANTIOMERS: NON-SUPERIMPOSABLE MIRROR IMAGES

STEREOISOMERS





5

STEREOISOMERS

DIASTEREOMERS

C CHH

Cl ClC C

Cl

Cl

H

H

cis-1,2-dichloroethene trans-1,2-dichloroethene


C

CH3

HCl

ClC

CH3

HCl

Cl


Iden

tical

0% 0% 0%0%0%


3.3. ConformersConformers

4.4. DiastereomersDiastereomers

5.5. EnantiomersEnantiomers5


C

CH3

HBr Cl

C

CH3

HBrCl


Iden

tical

Con

stitut

ional is

omer

s

Con

form

ers

Dias

tereo

mer

s

Ena

ntiom

ers

0% 0% 0%0%0%




5.5. EnantiomersEnantiomers5


CH3

CH3

CH3

CH3


Iden

tical

Con

stitut

ional is

omer

s

Con

form

ers

Dias

tereo

mer

s

Ena

ntiom

ers

0% 0% 0%0%0%

:05




5.5. EnantiomersEnantiomers

O

CCH2CH3

Carvone

O O

C

CH2CH3

C

H2C CH3

6

d-carvonel-carvone

7

DEXFENFLURAMINE LEVFENFLURAMINEDrowsiness

DEXFENFLURAMINE

8

1996Wyeth-Ayerst

DRY MOUTH

DIARRHEA

FATIGUE

PRIMARY PULMONARY HYPERTENSION

FALL 1997

RULES1. Assign priority to the 4 groups attached.

This is based on atomic NUMBER of the atom.

2 Visualize the molecule with the2. Visualize the molecule with the LOWEST priority group AWAY from you.

3. Trace Clockwise orCounterclockwise.

R

RECTUS = RIGHT

S

SINISTER = LEFT

9

R/S

The absolute configuration for the structure The absolute configuration for the structure below isbelow is

R S

Neit

her

0% 0%0%

5

1. R

2. S

3. Neither

Rules1. Assign priority to the 4 groups attached.

This is based on atomic NUMBER of the atom.

2 Visualize the molecule with the2. Visualize the molecule with the LOWEST priority group AWAY from you.

Double or triple bonds are evaluated as – C holding 2, 3 carbons (Solomons p 192)

C C

H H

H

C C

H H

C

HC

Double or triple bonds are evaluated as – C holding 2, 3 carbons (Solomons p 205)

H HC

C

H

O C

H

O

O C

Same for other species

The absolute configuration for the structure The absolute configuration for the structure below isbelow is

C

CH3

H

CH CH2Cl

R S

0%0%

5

1. R

2. S

10

The absolute configuration for the carvone The absolute configuration for the carvone structure belowstructure below O

CH

R S

0%0%

5

CH3 CH21. R2. S

What relationship exists in the following pair?What relationship exists in the following pair?

H

CH3

H

CH3 H

CH3

H

CH3

Iden

tical

Ena

ntiom

ers

Dias

tereo

mers

0% 0%0%

5

1. Identical

2. Enantiomers

3. Diastereomers

RADIATIONThe transfer of energy throughThe transfer of energy through waves or particles.

Energy

Gamma X rays Ultraviolet rays

Infrared microwave Radio wavesrays

λ

400 nm 600 nm500 nm 700 nm

Hand of Frau Röentgen

November 8, 1895

11

RADIOACTIVITYCan fluorescent materials produce X-rays ?

Antoine HenriBecquerel

1896

The spontaneous emission f di t d

RADIOACTIVITY

of radiant energy, and or, particles from the nucleus of the atom.

Nobel Prize for Physics1903

Ernest Rutherford1899

12

β particles -1 charge

α particles +2 charge

β particles -1 charge

γ radiation

α particlespaper

aluminiumβ particles

leadγ rays

β-

α He24

Nuclei

Electrons: -1e0

γ Electromagnetic radiation

He3 2+

2 protons2 neutrons

2 protons1 neutrons

He4 2+ He2

ISOTOPES

He2

αHe2+ nuclei

24

?Rn86

222He

24

+

13

Which nucleus is produced from the alpha Which nucleus is produced from the alpha decay of radondecay of radon--222?222?

?Rn

86

222He

2

4+

Which nucleus is produced from the Which nucleus is produced from the alpha decay of Radonalpha decay of Radon--222?222?

1.1. RadiumRadium--2262262.2. PoloniumPolonium--22222233 P l iP l i 218218

Rad

ium-226

Polonium-222

Polonium-218

Lead-22

2

None o

f the l

isted

nuclei

0% 0% 0%0%0%

:05

3.3. PoloniumPolonium--2182184.4. LeadLead--2222225.5. None of the listed None of the listed

nucleinuclei

αHe2+ nuclei

24

Rn86

222He

24

+ Po84

218

Half-Life

Time required for half of the number of nuclides to decay

Rn86

222He

24

+ Po84

218

Half-Life = 4 days

NUCLEAR REACTORSSpent FuelPu-239

The halfThe half--life of plutonium is life of plutonium is

1.1. 241 years241 years2.2. 2,410 years2,410 years

241 y

ears

2,41

0 yea

rs

24,10

0 yea

rs

241,0

00 ye

ars

2,41

0,000

years

0% 0% 0%0%0%

:05

3.3. 24,100 years24,100 years4.4. 241,000 years241,000 years5.5. 2,410,000 years2,410,000 years

14

0 50 g

1.0 g

10 Half-Life = 241,000 years

24,100 48,200 72,300 96,400 120,500

0.50 g

0.25 g

0.125 g

40,000 Warheads

150,000 kg of plutonium

Critical mass5kg

If 1.0 grams of plutonium are present If 1.0 grams of plutonium are present initially how much approximately will be left initially how much approximately will be left after 241,000 years?after 241,000 years?

1.1. 100 mg100 mg2.2. 10 mg10 mg

100 m

g

10 m

g 1

mg

0.1 m

g

0.01

mg

0% 0% 0%0%0%

:05

3.3. 1 mg1 mg4.4. 0.1 mg0.1 mg5.5. 0.01 mg0.01 mg

226Ra 218Po

238U

222Rn 214Po

234Th 234U 230Th

Ra Po Rn Po206Pb 210Po

Uranium-238 naturalradioactive decay series

RADIUM WATCHESRadioluminescence

The Radium GirlsOttawa (Illinois)

1920's33 DEATHS

Ra226

Rn222

88 84 He42+

15

226Ra 218Po

238U

222Rn 214Po

234Th 234U 230Th

*Ra PoRn Po206Pb210Po

* gas

16

226Ra 218Po

238U

222Rn 214Po

234Th 234U 230Th

*Ra Po Rn Po206Pb 210Po

Polonium-210

Alexander Litvinenko

Irene Joliot-Curie1897-1956

β-

α He24

Nuclei

Electrons: -1e0


β-

e- particles

1n β- + p10

1 0-1

Which nucleus is produced from the beta Which nucleus is produced from the beta decay of iodinedecay of iodine--131?131?

I53

132β- ?+

0-1

17

Which nucleus is produced from the Which nucleus is produced from the beta decay of Iodinebeta decay of Iodine--131?131?

1.1. TelliriumTellirium--1311312.2. CesiumCesium--13113133 XX 131131

Telliriu

m-131

Ces

ium-131

Xenon-1

31

Telliriu

m-127

None o

f the l

isted

nuclei

0% 0% 0%0%0%

:05

3.3. XenonXenon--1311314.4. TelliriumTellirium--1271275.5. None of the listed None of the listed

nucleinuclei

β-

e- particles

β- Xe131+I131 01β Xe5453

1n β- + p10

1 0-1

-1

Iodine 127

Goiter

Iodine-127

The “goiter belt” was associated with The “goiter belt” was associated with which part of the U.S.?which part of the U.S.?

1.1. The Atlantic coastThe Atlantic coast2.2. The MidwestThe Midwest33 Th S th U STh S th U S

The Atla

ntic co

ast

The Midwes

t

The South

ern U

.S.

The Cali

fornia

coas

t

The South

west

0% 0% 0%0%0%

:05

3.3. The Southern U.S.The Southern U.S.4.4. The California coastThe California coast5.5. The Southwest The Southwest

IODINE-131

18

April 26,1986

IODINE-131IODINE-131

Half-Life 8 Days

IODINE-127Hyperthyroidism

IODINE-131

β- He3 H3+0

1β He21 -1

Tritium Watches

19

β- He3 H3

Half-Life = 12.3 years

+01β He21 +-1

Polonium-210

Po210

Pb206

84 82 He42+

Bi + n Bi210

Po210

+ β-209

83 83 84

RADIATION EXPOSURE

Natural Sources 82%

Medical X-rays 11%Medical X rays 11%

Nuclear Medicine 4%

Consumer Products 3%

Electroluminescence

β-

α He24

Nuclei

Electrons: -1e0


20

γ

High energy electromagneticradiations associated withradiations associated with

nuclear changes Tc43

90m+ γTc90

Half-Life 6 Hours

43 + γTc43

NUCLEAR IMAGING

TECHNETIUM(1937)

Τεχνητόςχ η ς

"Artificial"

β-

α He24

Nuclei

Electrons: -1e0

γ

Positron emission: +1eβ+

Electromagnetic radiation

0

Electron capture

β-

α He24

Nuclei

Electrons: -1e0

γ



0

21

e+ particles

np β+ + 111

0

β+ positron emission

0+1

β+ B5

11+C6

11

p β1 0

0+1

+1

PE TScan

6C6H12O6

Glucose

11Normal Depressed

Frontal lobe

Normal Criminal

22

β-

α He24

Nuclei

Electrons: -1e0

γ



0

Electron capture

Electron capture

np -1e+ 111

00

Rb8137 ++ -1e

0 Kr8136

α

Num

ber o

f Neu

trons

Number of Protons

β+(or electron capture)

β-

NATURALRADIOACTIVITY

Ernest Rutherford

1871-1937

N + He O + H

TRANSMUTATION(1919)

414 1177N + 2He O + 1H8

23

ARTIFICIAL RADIOACTIVITY

Irene and Frederic Joliot-Curie

13Al + 2He 15P + 0n427 30 1

30 30 015P 14Si + +1β30 30 0

NATURALRADIOACTIVITY

How Fast?

Rate of Decay = Activity A

A = k x N

A Number of disintegrations per

k decay constant

N Number of nuclei presentat a given time

unit time

Ln(N0/2No

) = -k x t1/2

No/2 Number of nuclei present at time t1/2

t1/2 Half-Life

Ln( 1/21 ) = -k x t1/2

k x t1/2 = 0.693

24

Ln( NNo

) = -k x t

k = 0.693 / t1/2

CARBON 14 DATINGCARBON-14 DATING

12C 13C 14C*12C 13C 14C* β-N14 C14 + 0 βN7C

6+

-1

Half-life = 5730 YEARS

β-N14 C14 + 0

HC614N

714 + 1

1

Atmosphere

βN7C

6+

-1

Carbon Cycle

14C/12C Constant

25

Carbon Cycle14C/12C decreases

Example

An ancient wooden shovel has an activity of 11.6 dis x min-1 x g-1.

Given that the activity of carbon-14 in equilibrium with the environment is 15.3 dis x min-1 x g-1 calculate the age of the object.

Ln( NNo

) = -k x t

k = 0.693 / 5730 yr -1

Ln( NNo

) = -k x t

k = 1.21 x 10 -4 yr -1

Ln(11.715.3

) = -1.21 x 10 -4 x t

t = 2,200 years

CarbonCarbon--14 dating can be used for samples as 14 dating can be used for samples as old as?old as?

1. 500 years

2. 1,000 years

500 y

ears

1,00

0 yea

rs

10,00

0 yea

rs

50,00

0 yea

rs

100,0

00 ye

ars

0% 0% 0%0%0%

:05

3. 10,000 years

4. 50,000 years

5. 100,000 years

26

The term The term sindonologysindonology applies to the study of?applies to the study of?

1. the Shroud of Turin.

2. old paintings.

the S

hroud

of Turin

.

old pain

tings.

the a

ge of th

e eart

h.

the c

ause

of N

apoleo

n..

the a

ge of a

pes.

0% 0% 0%0%0%

:05

3. the age of the earth.

4. the cause of Napoleon’s death.

5. the age of apes.THE SHROUD OF TURIN

CARBON-14 DATINGDECAY MEASUREMENT

CARBON-14 DATINGDECAY MEASUREMENT

ISOTOPE COUNTS (MACS)

1260-1390 October 5, 2009

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