Transcript Aromatic Chemistry, Amines & Natural Products Dr. Fawaz Aldabbagh NUI, Galway

Aromatic Chemistry, Amines &
Natural Products
Dr. Fawaz Aldabbagh
NUI, Galway
[email protected]
Recommended Text: Organic Chemistry by Paula Y. Bruice
4th or 5th Edition
Benzene
First isolated by Michael Faraday in 1825
H
H
H
C
C
C
C
C
C
H
H
H
General Formula: C6H6
Aromatic originally referred to the smell, as opposed to aliphatic
Test for Aromatic: High Carbon content – burn with a smoky flame
Structure of Benzene
1. Each Carbon atom is sp2-hybridised : Making three σ-bonds with two adjacent
carbons and one hydrogen atom: bond angles = 120º
This is the σ-Framework
2. The p-orbital on each carbon atom overlaps above and below the σ-framework
with the p-orbital on the adjacent carbon atom
H
H
H
C
C
C
C
H
C
C
H
H
1856: Friedrich Kekulé Resonance Structures
Resonance arrow indicates structures are electronically equivalent
The problem is that C-C single and double bonds would be different lengths
and molecule would be a distorted hexagon
1930s: X-ray crystal structures showed benzene is planar and six carboncarbon bonds have the same length (C-C 1.39 A), which is shorter than C-C
single bond at 1.54 A. But longer than a C=C.
Thus, benzene does not have alternating bonds!
•
•
•
Kekulé resonance structures
Neither canonical form is correct structure
Resonant Hybrid is correct structure
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
C
C
C
C
C
C
H
H
Another Problem with the Kekulé Model:
Extra Stability than 1,3,5-Cyclohexatriene Structure Would Suggest!
+
H2
-29 kcalmol-1
+ 3 H2
-87 kcalmol-1
(Calculated)
Experimental ∆Hº = 50 kcalmol-1
Benzene is more stable than expected, and cannot be cyclohexatriene.
The six π-electrons must be delocalised and not localised
37 kcalmol-1 of resonance stabilization
Delocalisation of the six π-electrons
A resonance hybrid is more stable than the predicted stability of its
resonance contributor, and the more resonance contributors there
are, the greater the stability
p-orbitals must be aligned!
When Is A Molecule Aromatic?
•
For a molecule to be aromatic it must:
Be cyclic
Have a p-orbital on every atom in ring
Be planar
Posses 4n+2 p electrons (n = number of rings)
benzene
naphthalene
Erich Hückel
phenanthrene
Aromatic Hydrocarbons
Aromatic Molecules?
• Are the following structures aromatic?
Cyclic? P-Orbital? Planar? 4n+2 p electrons?
Cycloheptarienyl Cycloheptarienyl
cation H
anion H
+
C
C
[14]-Annulene
Cyclobutadiene
No
Yes
No
Yes
Which is Aromatic?
Draw Resonance Structures if so!
+
_
Homework: Try Problem 4 : page 645 of Bruice (5th Ed)
or Problem 2 : page 597 of Bruice (5th Ed)
Aromatic Hydrocarbons: Nomenclature
•
Substituent plus benzene
However, there are exceptions
CH3
Ph
Methylbenzene
(Toluene)
OH
CH3
Cl
Ph
Chlorobenzene
NH2
NO2
Cl
Ph
NO2
Nitrobenzene
O
Benzoic acid
OH
Phenol
Ph OH
Aniline
Ph NH2
OMe
Ph
CO2H
Ph OMe
Methoxybenzene
(Anisole)
Nomenclature
• Two identical substituents: di- (3: tri- etc.)
• Numbering of C atoms in ring
6
6
CH3
6
CH3
CH3
5
1
5
1
5
1
4
2
4
2
4
2
3
CH3
1,2-Dimethylbenzene
o-Dimethylbenzene
o-Xylene (o: ortho)
3
CH3
1,3-Dimethylbenzene
m-Dimethylbenzene
m-Xylene (m: meta)
H3C
3
1,4-Dimethylbenzene
p-Dimethylbenzene
p-Xylene (p: para)
Electrophilic Aromatic Substitution
H
E
H
H
E
H
X
H
+
H
H
H
H
H
X
H
H
Electrophilic attack – Slow Rate Determining Step
E
H
E
H
E
H
E
sp3 Transition State or Wheland Intermediate
E
H
Delocalised Cyclohexadienyl cation
Fast Step is the loss of a proton
+ E
---rapid re-aromatization
- H+
E
H
E.g. Nitration of benzene
HNO3(c), H2SO4(c)
NO2
+ H+
Sir Christopher Ingold's ideas (1930s), terminology and
nomenclature for reaction mechanisms (e.g. electrophilic,
nucleophilic, inductive, mesomeric, SN1, SN2 etc) were
generally accepted and employed everywhere.
Derive the Lewis Structure of NO2+, which is the
electrophile (E+) for nitration
Generating NO2+
Sulfuric acid is a stronger acid than nitric acid
O
_
H O S O H
O
_
O S O + 2 H+
O
O
H
HO NO2
H+
H
O+ NO2
NO2+
+
H2O
The Nitration of Benzene
_
O
O
+
N
electrophilic attack
O
electrophile
O2 N
+
O
+N
+
slow
H
- H+
fast
O +
N
_
O
O
+N
=
O
NO2
=
+
Label the parts of the Electrophilic Aromatic Substitution Profile
Some Reactions of Benzene
You should be able to write mechanisms for all the transformations below.
All reactions will be described in your lectures
Br2, FeBr3 (cat)
Cl2, FeCl3 (cat)
c. H2SO4
Br
+
HBr
Cl
+
HCl
SO3H
+
H2O
Friedel-Crafts
H3C
CH3
+
+
AlCl3 (cat)
HCl
Cl
O
H3C
Cl
1. AlCl (cat)
3
+
CH3 +
Charles Friedel
HCl
2. H2O
O
O
H3C
O
CH3
+
O
O
1. AlCl3 (cat)
2. H2O
CH3
+
H3C
O
OH
James Mason Crafts
Homework: Give a mechanism and rationalise the substitution pattern
FeCl3
Cl
Mechanism for Acylation
Mechanism for Alkylation
Water is required here
Reactions of Substituted Benzenes
CH3
CH3
CH3
CH3
NO2
HNO3 / H2SO4
+
+
59%
4%
NO2
toluene
NO2
37%
alkyl groups are weakly activating at ortho- & paraToluene reacts 25 times faster than benzene in nitration
NH2
OH
NH2
Br2
aniline
Br
Note – no catalyst
Br2
Br
Br
100%
OH
phenol
Br
Br
Br
100%
-NH2, -OH, -OMe, MeCONH- are all powerful activating groups
Ortho- and para- directing
Alkyl groups (e.g. CH3) inductively donate electrons (+I)
(through σ-bonds)
CH3
+I
H + H
N
NH2
_
+R activate through π-bonds
+R
Activates o & p towards E+ attack
H + H
N
_
OH
OH
Br
Br2
CH3
NHCOCH3
CH3
The strongest activator wins out
NHCOCH3
Br2
Br
Br
NHCOCH3 directs para, but makes benzene less
reactive compared to aniline
Strong –R Deactivators: NO2, CHO, COOH, CO2H, CN
Deactivates o & p towards E+ attack; Thus, strong meta-directors
NO2
NO2
NO2
NO2
NO2
HNO3 / H2SO4
+
+
6%
93%
NO2
Undergoes nitration 10,000 times slower than benzene
O
+
N
O
O
+
O
N
1%
O
+
O
N
- R, -I
+
+
NO2
CH3
CH3
NO2
HNO3 /H2SO4
NO2
directors reinforce each other
NO2
Halogens are weak deactivators – directing ortho- and paraF
Cl
+
I
Br
+
F
F
_
- I, +R
Br
Br
Cl2, FeCl3
Br
Cl
+
_
o-bromochlorobenzene
Cl
p-bromochlorobenzene
Tutorial Questions
HNO3 / H2SO4
NO2
HNO3 / H2SO4
A + B +C
isomeric dinitrobenzenes
85%
1.
Write a mechanism for the generation of the nitronium ion (NO2+) from concentrated nitric
and sulfuric acid.
2.
Write a mechanism for the nitration of benzene.
3.
Draw dinitrobenzenes A-C.
4.
Predict the ratio of isomeric products A-C from the nitration of nitrobenzene, and comment
on the rate of nitration of benzene compared to nitrobenzene. Rationalize your answers with
resonance structures.
5.
Draw the structure of an aromatic that undergoes electrophilic aromatic substitution faster
than benzene. Briefly give reasons for your answer.
Tutorial Questions
1.
What are the four rules of aromaticity?
2.
Draw one aromatic molecule, which does not contain a benzene ring.
3.
Draw the following molecules;
ortho-dibromobenzene, para-nitroaniline, phenol, meta-nitroanisole and 2,4,6-trinitrotoluene (TNT)
4.
Predict the major product for the following reaction, and name the product.
OH
+
major product
Br2
CH3
5. Provide a brief synthesis for TNT starting from toluene
6. Provide a mechanism for the following reaction;
O
AlCl3
Cl
O
Synthesis of Substituted Benzenes Using
Arenediazonium Salts
_
+
N N Cl
+
Nu
_
_
Nu
+
N2
+
Benzenediazonium
chloride
NH2
NaNO2, HCl
0 ºC
HNO2
Nitrous acid made on ice
_
+
N N Cl
Cl
_
+
N N Cl
+
Nu
_
_
Nu
+
NO2
HCl, NaNO2
_
+
N N Br
_
+
N N Cl
Br
+
Cl
_
+
N N Cl
NH2
Sn / HCl
N2
CN
CuCN
CuBr
- N2
- N2
CH3
_
+
N N Cl
CH3
I
KI
- N2
CH3
CH3
Arenediazonium Ion as an Electrophile
Aryl diazonium salts undergo coupling reactions with activated
aromatic rings, such as phenols and anilines to yield brightly
coloured azo compounds (Ar-N=N-Ar)
NH2
_
+
N N Cl
NH2
N N
+
Diazonium coupling is a typical electrophilic aromatic substitution,
where the phenol or aniline is the electron-rich aromatic and the
diazonium salt is the electrophile.
Electrophilic aromatic substitution occurs at the para-position to the
electron-rich aromatic
William Perkin
Examples of Azo-Dyes
+
N2 Cl
_
H3C
N
CH3
CH3
NaOAc, 0 ºC, H2O
+
N
N
N
CH3
Yellow of margarine
+
N2 Cl
_
OH
+
NaOH, 0 ºC, H2O
N
N
OH
Orange solid
The mechanism for formation of an azo-dye will be described in your lectures
Tutorial Questions
1. Draw the chemical structure of A, and give a full reaction mechanism for the formation of
yellow azo-compound B (Scheme 1).
+
N2 Cl
_
+ A
CH3
NaOAc, 0 ºC, H2O
N
N
N
CH3
B
Scheme 1
NO2
HNO3 / H2SO4
NH2
Br2 / FeBr2
reagents X
C
A
Br
reagents & conditions, Y
D
E
Scheme 2
D
Br
CuBr
Br
2. Draw structures A, C and E, and give reagents & brief conditions X and Y.
3. Give a mechanism for the formation of C (Scheme 2).
Amines
Amines are derivatives of Ammonia
Ammonia & amines are bases by virtue of their lone pair of electrons
..
..
N
H
N
H
R
H
107O
CH3
H
H
Ammonia
NH3
R
R = alkyl groups
107O
H
N
R
N
H
H
Methylamine
NH2Me
1º amine
CH3
CH3
N
H
CH3
Dimethylamine
NHMe2
2º amine
N
CH3
CH3
Triethylamine
NMe3
3º amine
Amines are bases because of the lone pair on the
nitrogen atom - red litmus paper to blue
H
H
Cl
NH2
Base
+
Acid
N H Cl
H
Ammonium Salt
=
O
H O
O
O H
O
oxalic acid
+ 2 N(CH2CH3)3
triethylamine
O
O
+
2 HN(CH2CH3)3
O
triethylaminium oxalate
Aniline is a weaker base than aliphatic amines due to the resonance
delocalisation of the lone pair of electrons in the free amine
H
N
H
H + H
N
_
This delocalisation of the lone pair of electrons makes it less available to H+
The delocalisation is lost upon formation of the salt
Reactions of Amines with Nitrous Acid - Revisited
1º aliphatic amines form very unstable diazonium salts
1º aromatic amines form stable diazonium salts at 0 ºC of great synthetic utility
Secondary Amines with nitrous acid
Me2NH + HCl + NaNO2
N-Nitrosoamines
Me
N
N
O
Me
N-Nitrosodimethylamine (yellow oil)
H
N
+ HCl + NaNO2
N
O
N
Me
Me
N-Nitoso-N-methylaniline (yellow oil)
Tertiary Amines with nitrous acid
Nitrous acid reacts to give salts with 3º aliphatic amines (not useful),
with 3º aromatic amines NO+ becomes an electrophile in electrophilic aromatic substitution
Me
Me
+ HCl + NaNO2
N
Me
O
N
N
Me
p-Nitoso-N,N-dimethylaniline
(exclusive product)
Tutorial Questions
1.
Draw the structure of a primary, a secondary and a tertiary aliphatic amine, and name all three
amines.
2.
Give a mechanism for the reaction of triethylamine with HCl.
3.
Explain the origin of amine basicity and how does the basicity of aniline compare with that of a
primary aliphatic amine. Draw resonance structures to explain your answer.
4.
Describe the reactions of nitrous acid (HNO2) with primary, secondary and tertiary aromatic
amines.
There are about 20 naturally occurring amino acids
20n possible combinations of amino acids in a peptide
There are eight essential amino acids
All are chiral apart from glycine, R = H
H
CO2H
H2N
R
All DNA encoded aa are
usually L-
aa are covalently linked by amide bonds
(Peptide Bonds)
The resulting molecules are called
Peptides & Proteins
R'
R'
N
C
O
R
N
C
R
O
Features of a Peptide Bond;
1. Usually inert
2. Planar to allow delocalisation
3. Restricted Rotation about the amide bond
4. Rotation of Groups (R and R’) attached to the amide
bond is relatively free
CH2OH
CH3
O
O
H3N
CH
H3 N
C
O
C
H3N
C
O
O
O
Serine
Alanine
Valine
- 2 H2O
CH3
O
H
N
H3N
C
C
O
CH2OH
O
N
H
C
O
Tripeptide : Ala . Ser. Val
Strong Acid Required to hydrolyse peptide bonds
Primary Structure is the order (or sequence) of amino acid residues
Peptides are always written and named with the amino terminus on
the left and the carboxy terminus on the right
Lys. Cys. Phe
Phe. Ser. Cys
1. RSH
2. 6 M HCl hydrolysis
Lys + 2 Cys
+ 2 Phe + Ser
Ph
Cysteine residues create
Disulfide Bridges between
chains
(CH2)4NH2
O
H
N
H2 N
This does not reveal
Primary Structure
C
C
OH
N
H
O
O
S
S
Ph
O
H
N
H2 N
C
OH
N
H
O
C
O
HO
C
Two Nucleic Acids (Polymers) –
deoxyribonucleic acid (DNA) and
ribonucleic acid (RNA)
Mild degradation yields monomeric units Nucleotides
Complete degradation yields
1. A Heterocyclic Base
Pyrine or
Pyrimidine
2. A five Membered Monosaccharide
3. A Phosphate ion
O
O
P
O
O
D-Ribose or 2-deoxy-D-ribose
The Phosphate ester
can be at C-5’ or C-3’
RNA - Nucleotide
Hydolysis of
Phosphate
Heterocyclic Base
O
5'
HO
P
N
O
O H2C
1'
4'
OH
Nucleoside
N- Glycosidic linkage
2'
3'
OH OH
DNA - Nucleotide
Heterocyclic Base
O
5'
HO
P
O
O H2C
N
1'
4'
OH
3'
2'
OH H
N- Glycosidic linkage
NH2
N
N
N
Adenine
H
(A)
NH2
N
H3C
(C)
N
N
NH2
Guanine
Pyrimidines
O
H
H
N
O
N
H
Cytosine
H
(G)
O
N
O
N
N
N
H
Purines
N
O
H
Thymine (DNA
(T)
N
O
H
only)
Uracil (RNA
(U)
only)
O
NH2
N
N
N
N
N
HO
N
Nucleosides
that can be
obtained
from DNA
N
HO
NH2
O
O
H
NH
H
H
OH
H
H
H
H
H
OH
H
2'-Deoxyadenosine
H
2'-Deoxyguanosine
NH2
O
N
O
HN
N
HO
O
N
HO
O
O
H
H
OH
H
H
H
2'-Deoxycytidine
H
H
OH
H
H
H
2'-Deoxythymidine
NH 2
Adenine
N
N
N
N O
5'
O
O
H
H
O
O
P
HN
H
3'
H
O
Base Sequence is
the Genetic Code
Primary Structure
H
O
N
Thymine
5'
O
O
O
H
H
H 3'
O
H
O
P
O
H
NH
H
N
5'
O
O
N
O
H
H
O 3'
H
P
N
NH 2
Guanine
O
H
5'
A T G
3'
O
Nucleotides are held together by phosphate ester linkages.
Phosphate esters link 3’- OH of one ribose (or deoxyribose)
with the 5’-OH of another. This makes the nucleic acid a long unbranched
chain with a backbone of sugar and phosphate units with heterocyclic
bases protruding from the chain at regular intervals.
Secondary Structure
H
H3C
H
O
N
N
N
E. Chargaff
N
N
H
N
N
Adenine
O
Thymine
H
N
N
O
H
H
N
N
N
N
O
Cytosine
N
H
Guanine
N
H
Two Complementary Chains Result
Watson-Crick
Model of DNA
(1953)
Double Helix is the Secondary Structure of DNA
Two nucleic acid chains are held together by weak
H-bonds between bases of opposite strands
Wound into a helix with a common axis
The base pairs are on the inside of the helix and
the sugar-phosphate backbone is on the outside
34A repeating unit contains 10 successive
nucleotide pairs
Phosphate-Sugar backbone is regular, base
pairs can assume many different permutations
Acetal Formation is REVERSIBLE and acid catalysed.
Anhydrous conditions
H+
R
C O
H
+
R OH
R OH
R
OR
C
- H2O
H
OR
+
H2O , H
O
Hydrolysis
Tetrahydrfuran
O
Tetrahydropyran
HEMIACETAL FORMATION
CH3
OH
H
H
C
O
H3 C
5-Hydroxyhexanal
(6%)
O
C
OH
2-Hydroxy-6-methyltetrahydropyran
(94%)
Explaining That Glucose is a Reducing Sugar
CH2OH
O
OH
O
H
Open form is an
aldehyde
OH
OH
D-Glucopyranose
To be drawn in the lectures as a
Fischer Formula
(shown as a Howarth Projection)
Since, aldehydes are readily oxidised, glucose is called a Reducing Sugar
Sucrose is a non-reducing Sugar (cannot open up to form an aldehyde)
Tutorial Questions
1. Draw the Haworth projection of glucose, and briefly explain why it is a reducing sugar.
2. Draw the DNA-encoded tripeptide, Ala.Gly.Ser.
3. Draw one DNA and one RNA nucleoside.
4. Write notes on the structure of DNA.
Emil Fischer
1902 Noble Prize in Chemistry
W. N. Haworth
1937 Noble Prize in Chemistry