Get your answers out for the challenge we finished with last lesson. 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. 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. 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) pentane b) ethyl ethanoate c) propanoic acid d) 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 1-bromobutane 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 Location C-O Alcohols, esters 1000-1300 C=O Wavenumber/cm-1 C-O Alcohols, esters 1000-1300 C=O Aldehydes, ketones, carboxylic acids, esters 1680-1750 O-H Hydrogen bonded in carboxylic acids 2500-3000 (broad) N-H Primary amines 3100-3500 Hydrogen bonded in alcohols, phenols 3230-3550
Draw out the functional groups below Stretch O-H 2500-3300 C-H 2850-3100 C=O 1640-1750 C-O 1000-1300 Alcohols Carbonyls Carboxylics Esters
Draw out the functional groups below Stretch O-H 2500-3300 C-H 2850-3100 C=O 1640-1750 C-O 1000-1300 Alcohols 3000-3300 Carbonyls Carboxylics Very Broad Esters
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 C–H
cyclohexene C–H
butanal C–H
butanal C=O
ethanoic acid O–H
ethanoic acid C=O
ethanol O–H
propanone C=O
methyl ethanoate C=O
Aldehyde vs Ketone?? Both have prominent C=O stretch 1700cm-1 ALDEHYDES ALSO have peak about 2820-2720 cm-1
A2444 15/11/2018 Top Ketone (acetone) Bottom Aldehyde (formaldehyde)
Exercise 1 Match the following eight compounds to the following eight IR spectra. hex-2-ene pentane methylpropan-1-ol 2-methylpentan-3-one butanal butanoic acid propyl ethanoate nitrobenzene
propyl ethanoate C=O C-O
2-methylpentan-3-one C=O
methylpropan-1-ol O-H
nitrobenzene C-H
pentane C-H
butanal C-H C=O
butanoic acid O-H
hex-2-ene C-H C-H C=C
F
G
9
10
11
P
Q
R
S
T
U
A2444 15/11/2018 IR Problems June 2012 Unit 2 Question 8bii (QS12.2.08) June 2011 Unit 2 Question 6e (QS11.1.06) January 2012 Unit 2 Question 10 (QS12.2.10) June 2009 Unit 2 Question 9 (QS09.2.09)
Is this the IR Spectra of Caffeine?
Identifying Impurities Compare spectrum to database – may show unexpected peaks (from impurity)