1
So far...
• We have covered a lot of material...
• You should have an understanding of the structure of
•
•
molecules
You should be familiar with IR spectroscopy
Understand how the structure of compounds relates to their
chemistry
Br
Me
Me
Me
Me
H
H
Br
H
H
H
H
• Have a vague comprehension of 'curly arrows'
O
N C
H
R
NC
O
H
R
H
NC
H
O H
R
• Now before we use that to look at reactions you need to
understand a little more about structure...
2
Isomerism
• Isomers - compounds that have the same numbers & same kind
•
•
of atoms, but they differ in the way they are arranged
Constitutional isomers - same molecular formula but atoms
joined in different ways
As a result can be very different compounds...
pentane
bp 36.2˚C
2-methylbutane
bp 28˚C
2,2-dimethylpropane
bp 9.6˚C
• Only attractive forces in alkanes are van der Waals forces
• These are proportional to surface area
• Smaller the surface area smaller the attractive forces & lower
OH
2-methylpropan-2-ol
C4H10O
mp 26˚C
Presence of hydroxyl group
(alcohol) makes a huge
difference!
the bp
O
diethyl ether
C4H10O
mp –116˚C
3
Conformational isomerism
• Different shapes of the same molecule
• Differ by rotation around a single bond
• Two extremes: eclipsed (below) &
staggered (next slide)
Eclipsed conformation
H
H
H
C C
H
H
H
conventional
representation
≡
H
H
C H
H
H
C
H
sawhorse
projection
≡
H
H
H
H
HH
Newman
projection
view along C–C bond
small dot = front carbon
large circle = back carbon
• Substituents of each atom overlap
• Less stable (more energy) than staggered
4
Conformational isomerism II
Staggered conformation
H
H
H
C C
H
H
≡
H
H
H
C
H
H
C
H
H
sawhorse
projection
conventional
representation
•
•
•
•
H
≡
H
H
H
H
H
Newman
projection
Staggered conformation more stable (lower energy)
Reason for stability is NOT due to sterics (size); H too small to interact
Reason is electronics - electrons of bond repel (so don't like overlap)
Possible that a stabilising overlap of C-H σ and C-H σ* exists
σ*
H
H
C
H
σ
H
C
H
H
5
Conformational isomerism III
•
•
•
•
•
•
R1
Substituted ethanes more complex
Staggered conformations have different energies
Anti-periplanar most stable
Hence most stable alkanes are zig-zag (as we draw!)
Barrier to rotation can be steric now (depends on R)
Each conformation can have effect the behaviour of
the compound (reactivity / mechanism / nmr
spectroscopy)
R1
1
H
H
H
H
R2
HR
H
H
H 2
R
R2
H
H
H
antiperiplanar
anticlinal
H
gauche or
synclinal
R1
R1
R1
H
H
H
H
R2
H
H
H
H
R2
H
H
R1
1
R2
H
H
R2R
H
H
H
H
H
R2
H
H
synperiplanar
H
gauche or
synclinal
R1
R1
H
H
R2
H
H
H
H
H
R2
H
HR
H
H
1
R2
H
anticlinal
R1
H
H
H
H
R2
6
Conformational isomerism IV
H H
H
C
H C
C H
H
H C C C H
H
H
H H
H
H
H
1
H
H5 H
H H
H
H
3
H
H
4
H
6
H
•
•
•
•
H
H
H
H
H
H
boat
(strained)
HH
HH
H
H
6
4
1
5
H
4
2
H
H
H H
H
3
H
1
H
H H
H
H
2
1
H
H
H
H
H
1
H
chair
(strain free)
H
H
H
H
H
H
H
H
chair
(strain free)
H
H
H
1
2
6
H
H
4
H
H
H
Cyclohexane can adopt two (extreme) conformations
The chair is the more stable conformation than the boat
Reason: boat has steric interactions between C1 & C4 hydrogens
Interconversion occurs via ring flipping
7
Substituted cyclohexanes
H
R
R
equatorial
H
• Substituent
R
H
H
H
axial
1,3-diaxial
interactions
prefers to be equatorial
as this reduces 1,3-diaxial
interactions (steric hindrance)
• The bigger R the more equatorial
conformer
H
• In
disubstituted cycohexanes both
substituents will try to be equatorial
• If not possible the largest will adopt
equatorial position
R
R
≡
Me
Me
tBu
R
R
H
H
H
bulky group equatorial
favoured
R
R
H
H
≡
R
R
trans-disubstituted
cyclohexane
R
R
H
H
H
H
R
one substituent up & one down
tBu
tert-butyl group
rarely is axial
H
both substituents up (equatorial preferred)
cis-disubstituted
cyclohexane
H
R
8
Configurational isomerism
• The spatial arrangement of atoms / groups in a molecule is its configuration
• Configurational isomers have the same formula & same bonds
• Configurational isomers can only be interconverted by breaking a bond
Alkenes
• Alkenes with two different substituents at
each end can exist as configurational isomers
• Reason - no rotation around double bond
A
C
B
D
≠
A
D
B
C
• E-alkenes are more stable than Z-alkenes
• Reason - substituents further apart so no steric
R
H
R
R
H
R
H
H
less stable
steric crowding
E or trans
Z or cis
hindrance
9
E and Z nomenclature
• E-alkenes have the groups of highest priority on the opposite sides
• Z-alkenes have the groups of highest priority on the same side
Priority rules
1. Rank atoms directly attached to alkene
in order of decreasing atomic number
2. If atoms are the same, then move
along substituent until a difference is
found
atom = Br>Cl>O>N>C>H
Atomic number = 35, 17, 8, 7, 6 & 1
H2
C
CH3
>
.
3. Multiple bonds are assumed to count
as the same number of single bonds
1 Cl
2
H3C
1 CH2CH3
C C
2
CH3
(Z)-2-chloro-3methylpent-2-ene
1CHO
HO2C 1
C C 2
2
Ph
CH2OH
(Z)-3-hydroxymethyl-4-oxo-2phenylbut-2-enoic acid
O
CH3
C N
≡
CH3
OH
N C
C N
N C
N
1CH OH
C2
2
C C 2
1
Br
CH2CH3
(E)-2-bromo-3(hydroxymethyl)
pent-2-enenitrile
>
H3C
1
2
H3C
O
2 CH
C C
3
1
C6H5
(E)-3-methyl-4phenylpent-3-en-2-one
10
Optical isomerism -stereoisomers
• Optical isomers - configurational isomers with the same chemical
•
•
& physical properties, which rotate plane-polarised light
One isomer rotates light clockwise
Other isomer rotates light anti-clockwise
mirror
plane
• The two compounds are mirror images
• They are asymmetric & can not be
superimposed
• They are called CHIRAL molecules
• If two molecules can be superimposed
they are ACHIRAL
• The
tetrahedral asymmetric carbon
is known as a chiral or stereogenic
centre
• The two isomers are called
enantiomers
• They rotate light in opposite
directions
lactic acid
mirror
plane
H
H
C
C
H3C
CO2H
HO
(–)-enantiomer
[α]D25 = –3.82
CH3
HO2C
OH
(+)-enantiomer
[α]D25 = +3.82
11
Chiral & achiral compounds
H
OH
OH
H
H
OH
HO
OH
H
H
H
OH
OH
≡
compounds
are identical
+
H
H
OH
rotate about
this axis
• Molecule is CHIRAL if it has ONE carbon with FOUR different
•
groups on it
ANY molecule with a plane of symmetry is ACHIRAL
H
Me
OH
4
R4
Me
plane of symmetry
running through central
carbon, hydrogen and OH
ACHIRAL
1
R1
R3
R2
3
2
chiral
R1≠R2≠R3≠R4
3
R3
1
R1
R2
R2
2
2
achiral
12
Nomenclature (groan): (R) or (S)
• Need to be able to define enantiomers
• Assign priorities to each substituent of chiral centre
• Highest atomic number = highest priority
H NH2
CO2H
alanine
(move along chain until difference)
Next, point lowest priority away from you...
4 1
H NH2
3
2
H
Me
3
CO2H
2
rotate around axis
until lowest priority
points away from
you
4
H
1
NH2
CO2H
3
2
• Draw
• If line
CO2H
NH2
1
2 4
HO2C H
3
NH2
1
line pointing from 1 to 3
is clockwise (to the Right)
..then (R)
• If line is anti-clockwise then (S)
≡
2
4 CO2H
H
Me
3
NH2
1
H NH2
CO2H
(S)-alanine
(S)-2-aminopropanoic acid
13
Example of nomenclature
H Me
CHO
• First prioritise substituents
• H is easy!
• Keep moving along chain
..until a difference
a = CH3 = 3
b = CH2 = c
c = CH2 = b
CHO
c
1
CHO
b
c
b = CH2C=C = 2
c = CH2C=O = 1
b = CH2C = c
c = CH2C = b
• Point
a
4 3
H Me
b
2
4 a
H Me
lowest priority away
..from you
• Draw line from 1 to 3
• Line is clockwise so...
H Me
CHO
(R)-3,6-dimethylhept-5-enal
H
2
3
Me
CHO
1