I got a 1st in my first year studying chemistry at the University of Birmingham using these revision notes that I have uploaded. They include detail on structure of benzene, stability of aromatic compounds, criteria for aromaticity, electrophilic substitution, benzene vs olefin reactivity with elec...
Solution Manual For Organic Chemistry Second Edition Jonathan Clayden, Nick Greeves, and Stuart Warren
Mechanisms
Test Bank For Organic Chemistry 2nd Edition Klein 9780199270293 | All Chapters with Answers and Rationals
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The University of Birmingham (UBir)
Chemistry
Organic Chemistry I
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Aromatic Chemistry
16 October 2017 18:02
INTRODUCTION TO AROMATIC CHEMISTRY Nomenclature of aromatic systems: 1. Structure of benzene
1. Describe the structure of benzene • All carbon and hydrogen atoms are coplanar,
2. Rationalise the extra stability associated with aromatic compounds therefore the molecule as a whole is planar
3. State the conditions for aromaticity and apply these rules to a range of • All six carbons are equivalent and sp2-
molecules to predict whether or not they are aromatic or anti-aromatic hybridised
4. Rationalise the appearance of the 1H-NMR spectrum of benzene and
related aromatic compounds
2. Stability of aromatic compounds (R group drawn as such when it doesn't • Each carbon uses 3 sp2 hybrid orbitals to
π-molecular orbitals of benzene matter what position it is in) make σ-bonds to neighbouring atoms
• Planar π-system made up of overlapping 6 AO's to give 6 new MO's • Remaining p-orbital is orthogonal to the plane of the molecule -
• Can then accommodate the 6 electrons from the original p-orbitals these orbitals can overlap to form the π-system of the molecule
• The 6 electrons are then completely delocalised around the six
carbon atoms in the ring
• Although the Kekule structure of benzene is less correct than the hybrid structure, it helps to
understand chemical reactivity and aids drawing mechanisms.
3. Criteria for aromaticity
1. Cyclic - electrons can be delocalised around the whole ring
2. Planar - the p atomic orbitals must be able to overlap to form the π-system, requiring all
atoms to be in the same plane
3. Fully conjugated - allows the complete delocalisation around the ring
Cyclopentadienyl Anion 4. Hückel's rule - 4n+2 π-electrons (n = 0,1,2,3, etc) - estimation if a molecule will be aromatic
• As predicted by (based on quantum mechanical calculations by Erich Hückel in 1931 and can also be seen as
Hückel's rule, a quick estimation of drawing the MO diagram for a molecule)
the
cyclopentadienyl Cyclooctatetraene (COT)
anion is aromatic and so relatively stable, hence explaining the easy • There can be situations in which cyclic delocalisation of
deprotonation (and low pKa). This can also be seen from the MO electrons is destabilising, rather than stabilising.
diagram • These cyclic systems are said to be anti-aromatic and
• The π-molecular orbital diagrams for conjugated cyclic hydrocarbons Hückel's rule predicts that systems with 4n π-electrons
can be predicted relatively easily. If the energy level of the highest filled have these very unstable electronic arrangements.
MO relative is lower than the energy of p-orbital (dashed line) then • Can see from MO diagram that a particularly unstable
there will be aromatic stabilisation (if higher, not aromatic) bonding arrangement would result from the
For monocyclic delocalisation of the p-electrons around the ring.
compounds, MO • This is actually avoided as the system prevents the
diagrams look the delocalisation of the electrons by adopting a non-planar
same as the molecule conformation.
pointing down.
Cyclobutadiene Low energy form of cyclooctatetraene "tub-shaped" conformation
• Fleeting existence and dimerises at Which are aromatic?
35K (-238oC).
• The MO diagram for a square planar Molecule
molecule predicts it to be very
unstable and a diradical species with
unpaired electrons
• However cyclobutadiene is Cyclic? Y N Y Y Y Y
diamagnetic and to be rectangular in
shape, due to distortions that Planar? N N Y Y Y Y
reduce the orbital overlap, resulting Conjugated? N N Y Y Y Y
in a rectangular molecule that
Hückel? N N Y (6π, n= Y (14π, n=1) N (8π so 4n, Y (6π, n=1)
behaves more like it has 2 isolated
1) n=2)
double bonds.
Aromatic? Not Not aromatic Aromatic Aromatic Anti- Aromatic
aromatic aromatic
1H NMR spectra of arenes
• Aromatic protons are
Pyridine Aniline Pyrrole deshielded compared to
alkenic protons e.g.
cyclohexene vs benzene.
• N is sp2 hybridised • N is sp2 hybridised, allowing • N is sp2 hybridised, allowing full conjugation and • The delocalised π-electron
• N has a lone pair the lone pair on nitrogen to aromatic stabilisation. density above and below the
available to bond conjugate in to the ring (this • If the N was sp3 hybridised (as expected from looking plane of the ring sets up a ring
with H+ (so a weak reduces the availability of at the structure), would not be aromatic as not fully current when the molecule is
base) the lone pair) conjugated. placed in an applied magnetic
• pKa = 5.25 • Aniline is a weak base • Lone pair on N is involved in aromaticity (not available field
• pKa = 4.6 to bond with H+, pyrolle is weak base pKa = -3.8) • This results in protons outside
. the ring experiencing a
stronger local field and so are deshielded (results in a
downfield shift (higher ppm) from a typical sp2 shift).
• Protons above or
inside the ring (not
possible in
N has available lone Lone pair is involved in aromaticity, so protonation will benzene)
pair so protonation disrupt the π-electrons and protonated pyrrole is not experience a
does not disrupt π- aromatic. Protonation weaker local field,
Organic Chemistry I Page 1
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