Graphite Structure

Organic Chemistry –

The chemistry of the compounds of carbon.

(Allotropic forms of carbon: diamond, graphite, fullerenes.)

Graphite —

2

Diamond —

Diamond Structure –

3

Buckministerfullerene —

4

5

Inorganic Chemistry:

The chemistry of the other ~100 elements.

Historical reason for division:

The sources of chemicals for early chemical investigations(last quarter of 18th and first quarter of 19th centuries)were: animal, vegetable, mineral.

Organic chemicals, those from living organisms (animal,vegetable) were complex and contained C, H, and often Nand/or O.

Inorganic chemicals (mineral) were simpler, could containa variety of elements, but only rarely carbon, except forcarbonates.

It seemed that inorganic sources of carbon (carbonate,cyanide, carbon dioxide, etc.) could not be converted intoorganic compounds. This led to — Vital Force Theory: only living organisms can convertcarbon containing inorganic compounds to organiccompounds.

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NH4 OCN H2N C

O

NH2heat

ammoniumcyanate(inorganic)

urea(organic)

Letter from Wöhler to Berzelius (1835):"Organic chemistry just now is enough to drive one mad. It gives me an impression of a primeval tropical forest, fullof the most remarkable things, a monstrous and boundlessthicket, with no way to escape, into which one may welldread to enter."

Friedrich Wöhler, 1828 –

Letter from Wöhler to Berzelius: "I must tell you that I canprepare urea without requiring a kidney or an animal ..."

By 1850, general agreement that organic chemistry is thestudy of the compounds of carbon.

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Contemporary reason for division:

Convenience of study — ~20 million known compounds; over 90% contain C.

(Fortunately, there are classes of organic compounds,each of which is characterized by a "functional group".)

Why so many organic compounds?

C has the almost unique ability to join to itself, as well assome "hetero" atoms (eg O, N, S) to form long chains. Si,also, can do this, but usually does not, and other atoms(eg O, N, S) apparently cannot.

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H C C

H

H H

C

C

H

H

HHHH

H C C

H

H H

C

H

H

C

H

H

H

H

H C C

H

H H

C

H

H

H

H

H C C

H

H H

H

H

H C H

H

H

methane

ethane

propane

butane

isobutane

CH4

C2H6

C3H8

C4H10

Structure (2-D, not 3-D)NameMolecularFormula

Consider the alkanes, the simplest organic compounds,hydrocarbons with single bonds, only.

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Same molecular formula, but different molecules:

isomers.

If molecules differ in regard to which atoms are attached towhich: constitutional or structural isomers.

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C5H12

H C C

H

H C

C

C

H

H

HHHH

HH H

pentane

isopentane

neopentane

Molecular formula: --- the pentanes; 3 constitutional isomers.

H C C

H

H H

C

C

H

C

HHHH

H

H

H

H C C

H

H H

C

H

H

C

H

H

C

H

H

H

H

Structure (2-D, not 3-D)Name

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C6H14: the hexanes; 5 constitutional isomers.C7H16: the heptanes; 9 constitutional isomers.C8H18: the octanes.C9H20: the nonanes.C10H22: the decanes; 75 constitutional isomers theoretically possible.

C20H42: the eicosanes; 366,319 constitutional isomerstheoretically possible (not all could be made; in some theatoms would be too close to each other for the molecule tobe stable [steric hindrance]).

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Proteins are macromolecules formed fromamino acid residue units. Twenty different amino acidsare found incorporated into proteins.

AS AN ANALOGY , WE MIGHT IMAGINE THAT A

PROTEIN MOLECULE IS L IKE A CHAIN WHERE THE

L INKS ARE AMINO ACID RESIDUES . EACH OF

THE L INKS IS ONE OF TWENTY COLORS ,CORRESPONDING TO THE TWENTY DIFFERENT

AMINO ACIDS .

A.A.#1 A.A.#2 A.A.#3 A.A.#20.....

Amino Acids:

Amino Acid Residues Incorporated into a Protein:

A.A.#3 A.A.#8 A.A.#15 A.A.#9 A.A.#12 A.A.#7

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N-terminus C-terminusamino acidresidue

N C

H

R7

C

O

OH

H

N C

H

R9

C

O

H H

O

C

R12

H

CN

An amino acid: CH2N

H

Rx

C

O

OH where Rx can be one of 20 possible groupseg H, CH3, CH(CH3)2, CH2OH, CH2SH, etc.

Amino acid residues: CH2N

H

Rx

C

O

CN

H

Rx

C

O

OH

HCN

H

Rx

C

O

H

N-terminal residue Middle residue(s)C-terminal residue

A protein:

N C

H

R15

C

O

H

H2N C

H

R3

C

O

H

O

C

R8

H

CN

Interchange of any two nonequivalent residues results in adifferent (structurally isomeric) protein.

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Consider a protein having 120 amino acid residues; 20different residues, 6 of each.

The total number of different (isomeric) protein moleculeswhich could be formed merely by permutation of the aminoacid residues is 120!/(20!)6 = 3.23x1088.

Assuming an average MW of the residues to be 100Daltons, the MW of each protein molecule would be12,000 Daltons, and a sample of 1 molecule of eachisomer would have a mass of —

~ 6.45 x 1065 kg.

Assuming a density of 1.0 gram/cubic centimeter, thesample would fill a cubic box

~ 9.13 x 104 light years on an edge,

which is approximately the diameter of the Milky Way.

Since the mass of the universe is believed to be on theorder of 1060 kg, an attempt to make all these isomerswould consume all the known matter in the universe whenfewer than 0.001% of them had been made.