Organic Chemistry

Question 1

Working out the equilibrium constant: how to do it and why do we do it?

Answer 1

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

Question 2

Stabilising one side of the equilibrium

Answer 2

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

Question 3

Which Ka values show a stronger acid?

Answer 3

Higher Ka values

Question 4

Bronsted-lowry acid/base

Answer 4

Hydrogen donator/hydrogen acceptor

Question 5

Lewis acid/base

Answer 5

Electron pair acceptor/donator

Positive ions are normally acids, negative ions are normally bases

Question 6

Why can carboxylic acids act as acids so easily?

Answer 6

They lose hydrogens way more easily than those without double-bonded oxygen. This is because of resonance (<-> (not same as equilibrium arrow))

Question 7

Most stable resonance structure

Answer 7

The one with the lowest energy, usually a middle form between two resonance structures

Question 8

The effect of distributing electron density

Answer 8

As there are more possible resonance structures, electron density (conjugation) is distributed among atoms, and stability is increased

Question 9

Allyl cation

Answer 9

Carbon one and carbon three will react to electron-rich species while carbon two will not

Allyl: C=C-C / C-C=C

Question 10

Allyl anion

Answer 10

Carbon one and carbon three will react to electron-poor species while carbon two will not

Question 11

Benzyl cation

Answer 11

More stable than unconjugated (electron-dense) benzene

Question 12

Increase in bonding nodes

Answer 12

As the number of bonding nodes increases, the energy increases, and the stability decreases

Question 13

Carbonyl carbon and carbonyl oxygen

Answer 13

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

Question 14

Why are electron-rich species less likely to react with esters and amides compared with ketones and aldehydes

Answer 14

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

Question 15

The peptide-amide bond: four key factors

Answer 15

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

Question 16

Huckel’s rule

Answer 16

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

Question 17

Huckel’s rule: why does it work?

Answer 17

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

Question 18

Pyridine

Answer 18

C₅H₅N - relatively unreactive but can act as a weak base

Question 19

Alkaloids

Answer 19

Nicotine, and quinine are examples

Toxic, form part of some medicines, addictive

Question 20

σ bonds when dealing with NH, O, and S in five-membered rings compounds

Answer 20

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

Question 21

Indole

Answer 21

Pyrrole fused with a benzene ring - found in tryptophan

Question 22

Basicity of Pyrrole

Answer 22

Can act as a base as the valence electrons can be used to form a bond with hydrogen atoms

Question 23

Basicity of Indole

Answer 23

Does not act as a base

Question 24

Imidazole

Answer 24

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)

Question 25

Basicity of imidazole

Answer 25

Since one nitrogen has two sp² electrons, it can act as a base or undergo dative covalent bonding with electron-deficient species

Question 26

Conformations

Answer 26

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

Question 27

Staggered conformation

Answer 27

Sawhorse projection and Newman projection, less energy formation as electrons are not repelling too strongly

Question 28

Eclipsed conformation

Answer 28

Sawhorse and Newman projection display what it looks like, torsional energy/Pitzer strain

Question 29

Boltzmann equation

Answer 29

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

Question 30

Anti and gauche conformations

Answer 30

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

Question 31

Why are long hydrocarbons straight chains?

Answer 31

Multiple gauche conformations are unfavoured due to the repulsion between groups within the hydrocarbons (ie hydrogens)

Question 32

Are halogenoalcohols more likely to do anti or gauche conformations?

Answer 32

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

Question 33

Why do double bonds have hindered rotation?

Answer 33

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

Question 34

Typical barriers of rotation: C-N, C=N, and C=N (in amides)

Answer 34

C-N: -20 kJmol⁻¹
C=N: -250 kJmol⁻¹
C=N (Amides): -75 kJmol⁻¹

Question 35

Rate of rotation at room temperature

Answer 35

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.

Question 36

What determines the priority of substituents in a double bond

Answer 36

Determined by:
Double bonds (????)
The atomic number of the molecule

Question 37

Z and E formations

Answer 37

Z is the same side and E is the opposite side

Z is not always cis and E is not always trans

Question 38

Antipsychotics

Answer 38

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

Question 39

Sphingomyelins: what are they and what do they do?

Answer 39

One of the most abundant phospholipids and important for the formation of ordered lipid domains and rafts in the cell membranes of vertebrates

Question 40

Constitutional isomers

Answer 40

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

Question 41

Stereoisomers

Answer 41

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)

Question 42

Chiral

Answer 42

A molecule is chiral if it has a non-superimposable mirror image

For chiral compounds, the non-superimposable mirror images are called enantiomers, the

Question 43

Achiral

Answer 43

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

Question 44

Chiral objects

Answer 44

Hands, corkscrews, shells, all “proteinogenic” α-amino acids other than glycine and most naturally occurring objects are chiral

Question 45

Optical rotation

Answer 45

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)

Question 46

Racemic mixture

Answer 46

When both enantiomers are equal in prevalence causing plane-polarised light to not be rotated

Question 47

Stereogenic centre

Answer 47

Chiral centre - when a point has 4 different groups attached to it

Carbon will be sp³ hybridised

Question 48

Meso molecule

Answer 48

Has chiral centres, but is achiral due to an internal plane (or point of symmetry)

Question 49

Configuration of stereogenic centres

Answer 49

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

Question 50

Configuration of stereogenic centres

Answer 50

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

Question 51

Constitutional isomers of dimethylcyclopropane

Answer 51

C₅H₁₀ has two constitutional isomers:
1,1-dimethylcyclopropane and
1,2-dimethylcyclopropane