Organic Chemistry Flashcards
Working out the equilibrium constant: how to do it and why do we do it?
Use the Ka equation:
Ka = ([produtc 1] x [product 2]) / ([reactant 1] x [reactant 2])
Then use pKa = -log(Ka) to work out the pKa number which is converted into a logarithmic number
This tells us which side of equilibrium is preferred
Stabilising one side of the equilibrium
If a molecule in equilibrium becomes more stable, it is less likely to react to form the other molecule so the equilibrium will be shifted towards its side
Which Ka values show a stronger acid?
Higher Ka values
Bronsted-lowry acid/base
Hydrogen donator/hydrogen acceptor
Lewis acid/base
Electron pair acceptor/donator
Positive ions are normally acids, negative ions are normally bases
Why can carboxylic acids act as acids so easily?
They lose hydrogens way more easily than those without double-bonded oxygen. This is because of resonance (<-> (not same as equilibrium arrow))
Most stable resonance structure
The one with the lowest energy, usually a middle form between two resonance structures
The effect of distributing electron density
As there are more possible resonance structures, electron density (conjugation) is distributed among atoms, and stability is increased
Allyl cation
Carbon one and carbon three will react to electron-rich species while carbon two will not
Allyl: C=C-C / C-C=C
Allyl anion
Carbon one and carbon three will react to electron-poor species while carbon two will not
Benzyl cation
More stable than unconjugated (electron-dense) benzene
Increase in bonding nodes
As the number of bonding nodes increases, the energy increases, and the stability decreases
Carbonyl carbon and carbonyl oxygen
The carbonyl carbon is slightly positive and is electron-poor and so will react with electron-rich species
The carbonyl oxygen is slightly negative and is electron-rich and so will react with electron-poor species
Carbonyl: R-C=O
Why are electron-rich species less likely to react with esters and amides compared with ketones and aldehydes
Aldehydes/ketones have a high electron density and so can easily and quickly react with electron-deficient species
However, esters/amides have a high electron density that is shared due to resonance among atoms so they won’t react as easily/speedily with electron-deficient species
The peptide-amide bond: four key factors
The bond is polar, the polarity of the molecule allows for hydrogen bonding
The N-C bond is shorter than usual N-C bonds (1.33 A in contrast to 1.47 A)
It has planar geometry and allows for resonance stabilisation, causing the strong stability of amino acids in proteins
They display hindered rotation
Huckel’s rule
Resonance in a cyclic array isn’t enough: cyclobutadiene and cyclooctatetraene are not as stable as benzene
Stability is only present when there are (4n + 2) π electrons
Huckel’s rule: why does it work?
When the electrons are 4n + 2, all the bonding orbitals are filled with two bonding electrons
For example, if 4 electrons are involved in the aromatic bonding, two out of the four orbitals will be partially filled, leaving the molecule as unstable
Pyridine
C₅H₅N - relatively unreactive but can act as a weak base
Alkaloids
Nicotine, and quinine are examples
Toxic, form part of some medicines, addictive
σ bonds when dealing with NH, O, and S in five-membered rings compounds
C₄H₄NH - Pyrrole - the nitrogen forms three σ bonds: one with each carbon attached and one with the hydrogen
C₄H₄O - Furan - the oxygen forms two σ bonds: one with each carbon
C₄H₄S - Thiophene - the sulphur forms two σ bonds: one with each carbon
Indole
Pyrrole fused with a benzene ring - found in tryptophan
Basicity of Pyrrole
Can act as a base as the valence electrons can be used to form a bond with hydrogen atoms
Basicity of Indole
Does not act as a base
Imidazole
Heterocycle with 5 members including two nitrogens, part of histidine
The nitrogen bonded to hydrogen has three electrons doing σ bonding so it has two sp² electrons that can contribute to the π bonding of the ring structure
The nitrogen not bonded to hydrogen has two electrons doing σ bonding and two sp² electrons acting as valence electrons so it gives one electron to the c bonding of the ring structure
Four atoms give one electron into the π structure and one atom gives two which produces the aromatic structure of the ring since 4n+2 = 6 (when n=1)
Basicity of imidazole
Since one nitrogen has two sp² electrons, it can act as a base or undergo dative covalent bonding with electron-deficient species
Conformations
The different shapes adopted by molecules, accessible by rotations about single bonds given that no bonds are made or broken in conformational changes
Two types of conformations: staggered and eclipse
Staggered conformation
Sawhorse projection and Newman projection, less energy formation as electrons are not repelling too strongly
Eclipsed conformation
Sawhorse and Newman projection display what it looks like, torsional energy/Pitzer strain
Boltzmann equation
Used to calculate population distribution (P1/P2) between the two conformations from energies:
P1/P2 = e⁻Δᴱ/ᴿᵀ
P1 = usually eclipsed
P2 = usually staggered
ΔE = J mol⁻¹
R = 8.31 J/K⁻¹ mol⁻¹
T = in K
If the above units are not correct, conversion is required
Anti and gauche conformations
Two types of conformations:
Anti - two attached methyl groups are as far apart as possible
Gauche- two attached methyl groups are as close as possible
Occurs in both eclipsed and staggered conformations
Why are long hydrocarbons straight chains?
Multiple gauche conformations are unfavoured due to the repulsion between groups within the hydrocarbons (ie hydrogens)
Are halogenoalcohols more likely to do anti or gauche conformations?
With a halogen and hydroxide group present, the halogen and oxygen exhibit δ⁻ behaviour
The oxygen exhibits δ⁻ behaviour while the hydrogen exhibits δ⁺ behaviour. This hydrogen then forms a weak hydrogen bond which increases the overall stability of the molecule, meaning that the gauche formation is preferred
Why do double bonds have hindered rotation?
The double bond would have to be broken as the p orbitals in the molecule become orthogonal (90°) as the group tries to rotate around the double bond
The energy taken to reach the orthogonal state is very high as strong covalent bonds are needed to be broken
Typical barriers of rotation: C-N, C=N, and C=N (in amides)
C-N: -20 kJmol⁻¹
C=N: -250 kJmol⁻¹
C=N (Amides): -75 kJmol⁻¹
Rate of rotation at room temperature
Isomers interconvert too quickly to isolate if the barrier for interconversion is <60 kJmol⁻¹
If the barrier is >100 kJmol⁻¹ separation IS possible
Changing temperature changes this rule of thumb.
What determines the priority of substituents in a double bond
Determined by:
Double bonds (????)
The atomic number of the molecule
Z and E formations
Z is the same side and E is the opposite side
Z is not always cis and E is not always trans
Antipsychotics
Usually, administered as a mixture of isomers to the individual but, usually, only the Z isomer has the desired psychiatric effect and the E isomer has no biological activity
Sphingomyelins: what are they and what do they do?
One of the most abundant phospholipids and important for the formation of ordered lipid domains and rafts in the cell membranes of vertebrates
Constitutional isomers
Constitutional isomers - have the same molecular formula but different sequences of atoms or linkages between atoms.
Interconversion is possible only by breaking and remaking bonds (i.e. by changing the “connectivity” of the atoms) - structural isomers
Stereoisomers
Have the same molecular formula, the same sequence of atoms or linkages between atoms but different arrangements of atoms in space
Interconversion by rotating or “bending” bonds (no change in atom connectivity: E/Z isomers)
Chiral
A molecule is chiral if it has a non-superimposable mirror image
For chiral compounds, the non-superimposable mirror images are called enantiomers, the
Achiral
A molecule with a superimposable mirror image or a molecule which possesses a plane of symmetry
Two stereoisomers that do not mirror each other are called diastereoisomers
Chiral objects
Hands, corkscrews, shells, all “proteinogenic” α-amino acids other than glycine and most naturally occurring objects are chiral
Optical rotation
Occurs as different enantiomers rotate the plane of plane-polarized light
Each enantiomer will rotate the light by a set amount either clockwise (D) or anticlockwise (L)
Racemic mixture
When both enantiomers are equal in prevalence causing plane-polarised light to not be rotated
Stereogenic centre
Chiral centre - when a point has 4 different groups attached to it
Carbon will be sp³ hybridised
Meso molecule
Has chiral centres, but is achiral due to an internal plane (or point of symmetry)
Configuration of stereogenic centres
To each of the 4 different groups, assign a priority:
- based on the atomic number of the first connected atom (highest = 1 - lowest = 4)
- If there is no distinction at the first attached atom, move to the second, etc until a distinction.
- Double-bonded atom counts as two single bonds.
- Rotate the molecular to view along the C to lowest priority (usually a hydrogen), and examine the sequence of the groups
When the 1-2-3 is clockwise: R-configuration
When the 1-2-3 is anticlockwise: S-configuration
Configuration of stereogenic centres
To each of the 4 different groups, assign a priority:
- based on the atomic number of the first connected atom (highest = 1 - lowest = 4)
- If there is no distinction at the first attached atom, move to the second, etc until a distinction.
- Double-bonded atom counts as two single bonds.
- Rotate the molecular to view along the C to lowest priority (usually hydrogen), and examine the sequence of the groups
When the 1-2-3 is clockwise: R-configuration
When the 1-2-3 is anticlockwise: S-configuration
Constitutional isomers of dimethylcyclopropane
C₅H₁₀ has two constitutional isomers:
1,1-dimethylcyclopropane and
1,2-dimethylcyclopropane
