U4 AOS2: Analysis of Organic Compounds

Question 1

Testing for C–C double bonds

Answer 1

• Bromine water test
  
  • Br₂(aq) is light brown/orange
  • If it becomes colourless, this implies the presence of a double bond (colourless dibromoalkane has formed)
  • No info about the type/location of the bonds
• Potassium permanganate test
  
  • KMnO₄(aq) is purple
  • If it becomes colourless and a brown precipitate forms, this implies the presence of a double bond

Question 2

Testing for hydroxyl functional group

Answer 2

• Acidified potassium dichromate test
  
  • In the presence of a 1° or 2° alcohol, the solution will turn from orange to green
  • No colour change = tertiary alcohol

Question 3

Testing for carboxyl group (–COOH)

Answer 3

• Esterification test
  
  • Add an alcohol & a strong acid (e.g. conc. sulfuric acid)
  • This can lead to the formation of an ester
  • Aromatic smell of the ester implies the presence of a carboxyl group
• Reaction with metal carbonates/bicarbonates
  
  • Produces carbon dioxide gas
  • Effervescence/fizzing implies presence of carboxylic acid

Question 4

Testing purity

Answer 4

• Melting point
• Pure substances have a sharply defined MP
• Impure compounds have a lower & wider MP due to impurities disrupting/weakening the molecular structure (less energy required to overcome the forces of attraction)

Question 5

Simple vs fractional distillation

Answer 5

• Impurities can introduce additional intermolecular forces, resulting in melting point elevation
• Distillation separates and verifies components of a liquid mixture
  based on their different boiling points
• Simple – separates liquids with significantly diff BPs
• Fractional – uses a fractionating column, which allows separation of liquids with similar boiling points

Question 6

Mass spectrometry

Answer 6

• Measures the mass-to-charge ratio (m/z) positive ions
• Base peak is the most stable & abundant fragment ion
  
  • Assigned a relative intensity of 100%
  • All other peaks are measured relative to this peak
• Molecular (parent) ion peak
  
  • Forms when the entire molecule loses an electron and becomes positively charged ion w/o fragmentation
  • Tells us the relative molar mass of the molecule
• Smaller peaks are isotopes or fragments

Question 7

Infrared (IR) spectroscopy

Answer 7

• Identifies the types of bonds and functional groups present in an organic molecule
• Use data book (page 18)

Question 8

Carbon-13 nuclear magnetic resonance (NMR) spectroscopy

Answer 8

• Provides information about the number of non-equivalent carbon
  environments in an organic compound
• No. peaks (excluding TMS) = no. carbon environments
• Carbon environments are equivalent when they are attached to the same sequence of bonds and atoms

Question 9

Proton nuclear magnetic resonance (NMR) spectroscopy

Answer 9

• Provides info about the no. non-equivalent proton environments in an organic compound AND the no. protons in those environments
• Splitting pattern follows n+1 rule (n = no. protons in the neighbouring environment
  
  • E.g. quartet (4) = 3 protons in the neighbouring environments

Question 10

Chromatography

Answer 10

• Separates components in a mixture based on their solubility
• Stationary phase – a solid (of known polarity) where components of a mixture adsorb/adhere
• Mobile phase – a fluid/solvent (of known polarity) that carries the sample over the stationary phase, causing some components to desorp/detach
• Polar compounds dissolve in polar solvents, whilst non-polar compounds dissolve in non-polar solvents

Question 11

Retention time (Rt)

Answer 11

• Time taken for a component to pass through a HPLC column
• Rt of a component in a sample can be compared to the Rt of a known substance under identical conditions
  
  • Must consider temp, the type of stationary/mobile phase and the pressure in the column
  • If these variable are diff, Rt is likely to change

Question 12

Effect of column length in HPLC

Answer 12

• Affects retention time and peak separation
• Longer column = more interaction between components of the mixture, higher retention times and greater peak separation
• Shorter column = less interaction between components, lower retention times and smaller peak separation

Question 13

Common solvents

Answer 13

• Hexane – very non-polar
• Acetone (propanone) – quite non-polar
• Ethanol – quite polar
• Water – quite polar

Question 14

Chiral vs achiral compounds

Answer 14

• Chiral compounds contain at least 1 chiral centre which is a carbon bonded to 4 different group of atoms
  
  • Mirror image is not superimposable (would not align)
  • Two non-superimposable mirror images are referred to as optical isomers
• Achiral compounds are superimposable

NOTE: Any carbon atom with a double or triple bond cannot be a chiral centre.

Question 15

Competitive vs non-competitive inhibition

Answer 15

• Stops substrate from binding with active site
• Competitive – attaches to active site
  
  • Can add more substrate to decrease effect
• Non-competitive – attaches elsewhere

Question 16

Protein structures

Answer 16

• Primary
  
  • Linear sequence of amino acids
  • Covalent/peptide/amide bonds
• Secondary
  
  • Polypeptide chain fold into different shapes due to hydrogen bonding between C=O and N-H
  • Results in alpha helices/beta pleated sheets
• Tertiary
  
  • 3D shape produced by further folding of 2° structure
  • Properties of R groups interact
  • Shape held by: ionic interactions, disulfide bridges (type of covalent bond), hydrogen bonds & dispersion forces
• Quaternary
  
  • 2 or more tertiary structures interacting

NOTE: While dispersion forces are the weakest in general, they have the largest effect on protein folding as proteins are very large molecules (dispersion strength increases as molecule size increases)

Question 17

Zwitterions

Answer 17

• Ions with both a positive and negative charge and an overall charge of 0

Question 18

Enzymes

Answer 18

• Biological catalysts that provide an alternative pathway with a lower activation energy, increasing rate of reaction
• Enzyme’s active site binds to a substrate
• Denature at high temp (shape of active site is disrupted)
• Activity is minimised at lower temperatures

NOTE: Denaturation causes irreversible structural damage, meaning the enzyme is
unable to function again.