Stereochemistry and Conformation

Question 1

Isomers

Answer 1

molecules which posses the same molecular formula but are different

Question 2

Constitutional isomers

Answer 2

molecules which possess the same molecular formula but different constitutions (connectivities)

Question 3

Stereoisomers

Answer 3

molecules which possess the same molecular formula and the same connectivity (but are nonetheless different)

Question 4

Enantiomers

Answer 4

stereoisomers which are mirror images of each other

Question 5

Diastereomers

Answer 5

stereoisomers which are not mirror images of each other

Question 6

Conformations

Answer 6

different shapes possible for a single molecule (usually related by rotations around single bonds)

Question 7

Cahn-Ingold-Prelog (CIP) convention

Answer 7

1) label groups with decreasing priority
determined by:
atomic no. directly attached atoms
atomic no. largest atom in “second sphere”
no. largest atoms in second sphere
multiple bonds considered as 2/3 single bonds
if no difference, heavier isotope has higher priority than lighter isotope
2) position molecule such that group with lowest priority (4) is away from you
3) clockwise = R
anticlockwise = S

Question 8

Optical activity

Answer 8

property of chiral molecules - rotate plane of polarised light and different enantiomers cause opposite directions of rotation
clockwise - dextrorotary (+)
anticlockwise - laevorotary (-)
no correlation to R/S

Question 9

specific rotation

Answer 9

[a] = a/cl
a = degree of rotation
c = conc in g/dm3
l = length of path in dm

Question 10

distinguish between diastereomers (syn/anti)

Answer 10

arrange longest carbon chain in zig-zag conformation
substituents pointing in same direction have sun relationship
substituents pointing in opposite direction have anti relationship

Question 11

no. stereoisomers related to no. chiral centres

chiral entres and double bonds

Answer 11

general:
n chiral centres
2 options each
= 2^n stereoisomers

both stereogenic elements
rule applies the same to double bods

Question 12

axis of symmetry

Answer 12

n-fold

rotate about axis by 360/n degrees, get the same structure

Question 13

plane of symmetry

Answer 13

reflect through plane, get the same structure

every atom in the plane or reflects/matched by an identical atom on the other side of the plane

Question 14

centre of symmetry

Answer 14

invert through centre

every atom is at centre or matched by an identical atom on a line through the centre

Question 15

relation of symmetry and chirality

Answer 15

high symmetry = achiral
low symmetry = chiral
if a molecule has a plane or centre of symmetry, it will be achiral (reverse not quite true)

Question 16

how can you tell if a molecule is chiral? (final answer)

Answer 16

cannot examine molecule in randomly-chosen conformation. look for an accessible conformation with plane or centre of symmetry (i.e. accessible achiral conformation)
if it has one, no enantiomers

Question 17

CIP rules for double bonds

Answer 17

assign priorities at either end of bond
same side high priority group - Z
opposite sides high priority - E

Question 18

separation of enantiomers techniques

Answer 18

1. derivatives with chiral reagent - forms diastereomers. for preparative reasons make salts where possible, if the two diastereomers have different physical properties easy to separate - i.e. salts crystallise well and they will have different solubilities. add base and extract into organic solvent
2. chromatography with chiral column packing. enantiomer with same chirality will interact better with column and will emerge after. typical stationary phase derived from cellulose
3. action of enzyme - enzymes are chiral and only act on one enantiomer
4. Simple crystallisation - curial shapes, distinguishable by handedness. impractical

Question 19

conformational analysis for: ethane, butane, generalised larger acyclic molecules

Answer 19

Ethane: energy maxima when eclipsed - filled CH bonding orbitals overlap slightly which is always destabilising (raises energy)
minima when staggered - CH bonding orbitals overlap with empty sigma*, overlap of filled orbital with empty is always stabilising (lowers energy)
Butane: central carbon has 3 distinguishable energy minima - gauche, anti-periplanar and gauche which interconvert at RT. anti-periplanar more stable by 3 kJ mol-1. populations determined by equilibrium constant K = [anti-pp]/[gauche] = exp(-delta G/RT)
Generally - as K increases the % of less stable conformer decreases and -delta G increases
larger acyclic molecules - several conformers, interconverting, with somewhat different energies. often controlled by sterics. exceptions occur when groups attract each other e.g. H-bonding

Question 20

conformational analysis: special cases
bonds to sp2 carbons and carbonyls
bonds with intermediated order

Answer 20

bonds to sp2 carbons: one CH bond eclipses double bond to avoid pi electrons (?), also applies to carbonyls
bonds with intermediate order:
amides - planar structure with rotation slower than single bonds, can show two peaks in NMR. secondary amides are more stable in Z conformation so only one NMR peak
esters - less double bond character character than for amides, but planar with Z preferred
Steric effects + overlap between lone pair N/O and empty sigma* from carbonyl carbon ( overlap between full and empty orbitals is always stabilising)

Question 21

3-membered rings

Answer 21

planar
no rotation = no conformational variability
strain
internal angle 60, wants to be 109 - raises energy
reflected in heat of combustion

Question 22

Baeyer strain

Answer 22

strain due to distorted bond angle

Question 23

4-membered ring

Answer 23

planar, 90 degree angle
bent, smaller than 90 so baeyer strain increases
planar = eclipsed
bent = slightly staggered, lower energy
planar conformation - Pitzer strain, due to suboptimal dihedral/torsion angle w
bent preferred - lower Pitzer strain compensates increased Baeyer strain

Question 24

5-membered ring

Answer 24

planar = 108, almost no Baeyer strain but bonds eclipsed so max Pitzer strain
envelope and twisted envelope - similar energies
cyclopentanes are flexible

Question 25

6-membered rings

Answer 25

chair: internal angle about 108, almost no Baeyer strain
torsion angle w = 60, almost no Pitzer strain
almost strain-free
heat of combustion nearly the same as for linear hydrocarbons
twist boats are energy minima
10,000 chairs for each twist boat

Question 26

7-membered ring

Answer 26

comfortable chair
2-fold symmetry
most stable. conformation
still some Pitzer strain

Question 27

8-membered ring

Answer 27

boat chair
most stable conformation
avoids eclipsing interaction
significant Pitzer strain

Question 28

special cases for chair substituents

Answer 28

1. t-Bu has high preference for equatorial, axial t-Bu so bad that two opposing t-Bu flips to twist-boat
2. trans-declin - conformational lock with H trans
3. cis-declin - both rings can flip to give a second all-chair structure

Question 29

6-membered rings in nature: carbohydrates

Answer 29

glucose - chair with one carbon replaced with oxygen
‘anomeric effect’ - axial bond at anomeric centre has an empty anti bonding orbital on the far side of the anomeric carbon. this sigma* overlaps nicely with the axial lone pair on the ring oxygen - stabilising effect
stabilisation is most effective if the sigma* is low in energy i.e. if the bad is to an electronegative atom. the most electronegative of substituents should therefore occupy axial position
O is more EN than H, alpha-anomers of carboyhydrate favoured
anomers = diastereomers

Question 30

6-membered rings in nature: steroids

Answer 30

common framework has trans-declin at centre and therefore is rigid
two important examples:
cholesterol
cholic acid - cis-declin, but no ring flip possible as also contains trans-declin which holds everything in place

Question 31

stereoselective reactions

Answer 31

stereoisomers favoured over others

Question 32

stereospecific

Answer 32

stereisomers in a reaction form more stereoisomers

Question 33

stereospecific reaction examples

Answer 33

1. Diels-Alder - single step, so cis alkene creates cis ring etc
2. Creation of a new chiral centre next to pre-existing one e.g. Grignards forming chelate with substrate (interacting with group on current chiral centra) means that conformation is held and R group can only attack one face, forming almost exclusively one diastereomer
3. enantioselective reaction on achiral substrate - enantiomeric face gives equal enantiomer products

Question 34

diastereotopic/enantiotopic

Answer 34

replacement of hydrogen gives diastereomeric products/ enantiomeric products
or addition to face e.g. aldehyde gives enantiomeric products

Question 35

Problem: just one enantiomer wanted from reaction of achiral reagent - asymmetric catalysis

Answer 35

TS for the two reactions are enantiomeric so use a chiral reagent so that TS are diastereomic. Cant make carbon centre chiral instead add chiral ligand which binds to metal (catalytic amount). reagent which is not bound to ligand will react unselectively.
Start with very poor reagent - organozinc reagent (less reactive Grignard) as reaction is slow, then add chiral ligand which accelerates reaction
e.g. amino-alcohol which binds to Et2Zn, TS are diastereomeric so different energies, big difference means reaction is enantioselective
after reaction, amino-alcohol dissociates and repeats - catalyst

Question 36

enantiomeric excess (% e.e.)

Answer 36

e.e = % major enantiomer - % minor enantiomer

Question 37

enantiomeric vs disasteriomeric TS

Answer 37

enantiomers = identical energies, TS will always be mirror images
diastereomers, different energies (can be similar), TS not mirror images

Question 38

CIP for enantiotopic faces of aldehyde

Answer 38

CIP: 1, 2, 3
anti-clockwise = Si face
clockwise = Re face

Question 39

chiral molecules were obtained as single enantiomers from achiral materials examples

Answer 39

1. Resolution, mandolin acid
2. Kinetic resolution with enzyme
3. Enantioselective organozinc addition with naturally-derived amino-alcohol
4. Enantioselective reduction with enzyme - baker’s yeast

Question 40

how did biological molecules become enantiopure

Answer 40

1. Asymmetric physical process followed by chiral amlification e.g. polarised light destroyed one enantiomer more quickly than other
2. Spontaneous symmetry breaking e.g. saturated solution forms single crystal which acts as a seed

Question 41

acylation of cyclic compounds

Answer 41

Ac2O and pyridine - less aggressive than acyl chloride

equatorial OH reacts faster , monoacetate isolated

Question 42

nucleophilic substitution of cyclic compounds

Answer 42

axial l.g. - no special problem, easy

equatorial l.g. - major steric hindrance of Hs, 30 times slower

Question 43

elimination in cyclic compounds

Answer 43

equatorial l.g. - No anti-periplanar H- E2 cannot happen
axial l.g. - axial H is antiperiplanar - E2 occurs
C-H and C-X have to be anti-periplanar (like trans)

Question 44

hydride reductions in cyclic compounds

Answer 44

subject to steric hindrance

differences can be magnified by using bulky reagents, LisBu3BH - attacks less-hindered face

Question 45

cyclohexanones

Answer 45

axial attack is hindered - usually most important effect
nucleophiles approach carbonyl groups at 107 degrees -not hindered below
cyclohexanones flattened because C-CO are shorter. attack from the top is easier for small reducing agents for electronic reasons (O is already bless towards equatorial orientation)

Question 46

cyclohexanes - subtle effects in additions

Answer 46

Br2 - biaxial product formed through chair-like TS, no problems
alternative attack goes through boat-like TS to give twist boat which flips to diequatorial products - more stable but slower
biaxial product favoured
also applies to epoxides - very reactive, diaxial product formed