Optical Isomerism

Do Now 1. Butanal and methylpropanal give slightly different mass spectra. Both give a molecular ion peak at m/z = 72, but butanal gives four other peaks whereas methylpropanal only gives three. a)State the species responsible for the four other peaks in the mass spectrum of butanal and write equations to show their formation from the molecular ion. b)Suggest which one of these peaks will not be present in a mass spectrum of methylpropanal, giving a reason for your choice. 2. Suggest how pentan-2-one and pentan-3-one could be distinguished in a mass spectrum. Write equations to show the formation of any important fragment ions. 3. Write equations to show the formation of at least two species giving intense peaks in the mass spectra of each of the following molecules: a) b) c) d) pentane ethyl ethanoate propanoic acid pentanal

Get your answers out for the challenge we finished with last lesson. Challenge What are the compounds? (C3H6O) Challenge What are the compounds? (C3H8O) Core Practical 4 - Questions 1. Write an equation for the reaction of 1bromobutane with water. 2. In these reactions a precipitate forms. Identify the precipitate formed when the halogenoalkane is 1-iodobutane. 3. Explain why ethanol is used in these reactions. 4. Explain why water is able to act as a nucleophile. 5. Explain why water is used as the nucleophile rather than hydroxide ions? 6. Draw skeletal formulae for each of the halogenoalkanes used in this investigation

Answers to Core Practical 4 1. CH3CH2CH2CH2Br + H2O CH3CH2CH2CH2OH + H+ + Br 2. silver iodide 3. The halogenoalkanes are insoluble in water. Using ethanol ensures that the halogenoalkane dissolves so it can react with the water molecules. 4. Water has lone pair(s) of electrons on the oxygen atom. 5. If hydroxide ions were used, a precipitate of silver hydroxide would form instantly. Answers to Core Practical 4 Draw skeletal formulae for each of the halogenoalkanes used in this investigation (there are 5 of them). Classify each halogenoalkane as primary, secondary or

tertiary. Infrared Spectroscopy Infrared (IR) Spectroscopy Technique used by chemists to help to identify compounds. It helps to identify the FUNCTIONAL GROUPS Use this together with other info (e.g., Mr from mass spec) Atoms, molecules and ions can absorb (or emit) electromagnetic radiation of specific frequencies, and this can be used to identify them. Electromagnetic radiation absorbed What the energy is used for

Spectroscopy technique Ultra-violet / visible Movement of electrons to higher energy levels Ultra-violet / visible spectroscopy Infra-red To vibrate bonds Infra-red spectroscopy Microwaves To rotate molecules

Microwave spectroscopy Radio waves To change nuclear spin NMR spectroscopy How IR Spectroscopy works A pair of atoms joined by a chemical bond are always vibrating. Stronger bonds vibrate faster (at higher frequency) and heavier atoms make the bond vibrate more slowly (at lower frequency). Every bond has its own unique frequency that is in the infra-red region of the electromagnetic spectrum. INFRA-RED SPECTROSCOPY All bonds vibrate at a characteristic frequency. There are different types of vibration.

Symmetric stretch Assymmetric stretch Bending The frequency depends on the mass of the atoms in the bond, the bond strength, and the type of vibration. The frequencies at which they vibrate are in the infra-red region of the electromagnetic spectrum. INFRA-RED SPECTROSCOPY If IR light is passed through the compound, it will absorb some or all of the light at the frequencies at which its bonds vibrate. IR light absorbed is in the range 4000 400 cm-1. Above 1500 cm-1 is used to identify functional groups. Below 1500 cm-1 is used for fingerprinting. The IR Spectrum

This is a typical IR spectrum. The dips in the graph (peaks) represent particular bonds. In your exam, you will be given wavenumbers for some bonds commonly found in organic chemistry. IR Spectrum Bond C-O Location Alcohols, esters Wavenumber/cm-1 C=O Aldehydes, ketones, carboxylic acids, esters

1680-1750 O-H Hydrogen bonded 2500-3000 (broad) in carboxylic acids N-H Primary amines 3100-3500 O-H Hydrogen bonded in alcohols, phenols 3230-3550

1000-1300 Draw out the functional groups below Stretch Alcohols Carbonyls Carboxylics Esters O-H C-H C=O C-O 2500-3300

2850-3100 1640-1750 1000-1300 Draw out the functional groups below Stretch Alcohols O-H C-H C=O C-O

2500-3300 2850-3100 1640-1750 1000-1300 3000-3300 Carbonyls Carboxylics Esters Very Broad

BELOW 1500 cm-1 Fingerprinting

Complicated and contains many signals picking out functional group signals difficult. This part of the spectrum is unique for every compound, and so can be used as a "fingerprint". This region can also be used to check if a compound is pure. cyclohexane CH cyclohexene CH

butanal O CH3 CH2 CH2 C CH H butanal O CH3

CH2 CH2 C H C=O ethanoic acid O CH3 C O OH

H ethanoic acid O CH3 C O C=O H ethanol CH3 CH2

O H OH propanone O CH3 C CH3 C=O O methyl ethanoate CH3

C C=O O CH3 CH3 CH2 O H Aldehyde vs Ketone?? Both have prominent C=O stretch 1700cm-1 ALDEHYDES ALSO have peak about

2820-2720 cm-1 Exercise 1 Match the following eight compounds to the following eight IR spectra. hex-2-ene butanal pentane butanoic acid methylpropan-1-ol propyl ethanoate 2-methylpentan-3-one nitrobenzene