A2 Chemistry: 4.1.4 (Amines)

Explain the basicity of amines in terms of proton acceptance by the nitrogen lone pair;

Amines are weak bases. A base is a proton acceptor. When an amine accepts a proton, a dative covalent bond forms between the lone pair of the nitrogen atom and the proton.

Describe the reactions of amines with acids to form salts;

Amines are neutralised by acids to make an ammonium salt. For example, ethylamine, reacts with hydrochloric acid to form ethylammonium chloride (CH3CH2NH3+Cl-).

Describe the preperation of: alipahtic amines by substitution of halogenoalkanes with excess ethanolic ammonia and aromatic amines by reduction of nitroarenes using tin and concentrated hydrochloric acid;

Amines can be made by heating a halogenoalkane with an excess ethanolic ammonia. You’ll get a mixture of primary, secondary and tertiary amines, and quaternary ammonium salts, as more than one hydrogen is likely to be substituted. You can seperate the products using fractional distillation. Example:

2NH3 + CH3CH2Br → CH3CH2NH2 + NH4Br

Nitro compounds, such as nitrobenzene, are reduced in two steps: 1) heat a mixture of a nitro compound, tin metal and conc. hydrochloric acid under reflux– this makes a salt. 2) then to get the aromatic amine, you have to add sodium hydroxide. Example:

C6H5NO2 +6[H] → C6H5NH2 + 2H2O

Describe the synthesis of an azo dye by reaction of an aromatic amine with nitrous acid (<10 degrees celsius), with formation of a diazonium ion, followed by coupling with a phenol under alkaline conditions; (Nitrous acid is generated in situ from NaNO2/HCl)

The first step in creating an azo dye is to make a diazonium salt. The azo dye is then made by coupling the salt with an aromatic compound that is susceptible to electrophilic attack, e.g. a phenol.

Below is the method for creating a yellow-orange azo dye:

Part 1- making the diazonium salt

1) Nitrous acid (HNO2) is unstable, so it has to be made in situ from sodium nitrite and hydrchloric acid:

NaNO2 + HCl → HNO2 + NaCl

2) Nitrous acid reacts with phenylamine and hydrochloric acid to form benzenediazonium chloride. The temperature must be below 10 degress celsius to prevent a phenol forming instead.

Part 2- coupling the diazonium salt with a phenol

1) First, the phenol has to be dissolved in sodium hydroxide solution to make sodium phenoxide solution.

2) It’s then stood in ice, and chilled in the benzenediazonium chloride is added

3) The azo dye will precipitate out of solution immediately

4) Phenol is a coupling agent. The lone pairs on its oxygen release the electron density of the benzene ring, especially around carbons 2, 4 and 6. This gives the diazonium ion (a weak electrophile) something to attack.

State the uses of reactions, such as (d) in the formation of dye stuffs;

n.b– structures are in the CGP book

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A2 Chemistry: 4.3.2 (Spectroscopy)

State that NMR spectroscopy involves interaction of materials with the low-energy radiowave region of the electromagnetic spectrum;

Analyse a carbon-13 NMR spectrum of a simple molecule to make predictions about: the different types of carbon present, from chemical shift values and possible structures of the molecule;

First, count the number of peaks in the spectrum (this is the number of carbon environments in the molecule). If there’s a peak at delta= 0, don’t count it: it’s the reference peak from TMS. Then, look up the chemical shifts in a data table. Match up the peaks with the chemical shifts in the table to work out the carbon environments. Thirdly, compare the area under each peak, in order to work out how many carbon nuclei are in each environment. The ratio of the areas under the peaks is often shown by the numbers above them. Finally, try out possible structures.

n.b– C13 NMR concerns chemical shifts experienced by carbon nuclei in different environments.

Analyse a high resolution proton NMR spectrum of a simple molecule to make predictions about: the different types of proton present (from chemical shift values), the relative numbers of each type of proton present from relative peak areas (using intergration if/when required), the number of non-equivalent protons adjacent to a given proton from the spin-spin splitting pattern, using the n+1 rule and possible structures for the molecule;

Interpreting proton NMR spectra starts off like C13 NMR spectra: counting the number of hydrogen environments, looking the chemical shifts up in a data table, comparing the area under each peak. Sometimes, you can use an integration trace; a series of steps. The ratio of step heights is equal to the ratio of each type of proton.

In proton NMR, a peak that represents a hydrogen envrionment can be split. The splitting is caused by the influence of hydrogen atoms that are bonded to the neighbouring carbons. This effect is called spin-spin coupling. Split peaks are always split into one more than the number of hydrogens on the neighbouring carbon atoms. This is the n+1 rule.

n.b- see text book for practice spectra

Describe the use of tetramethysilane, TMS, as the standard for chemical shift measurements;

TMS produces a single absorption peak in both types of NMR becuase all its carbon and hydrogen nuclei are in the same environment. It’s chosen as a standard because the absorption peak is at a lower frequency than just about everything else. This peak is a given a value of zero and all the peaks in other substances are measured as chemical shifts relative to this.

State the need for deuterated solvents, e.g. CDCl3, when running an NMR spectrum

Deuterated solvents contain Deuterium, which has an even number of nucleons and so produces no signal in an NMR spectrum. Solvents such as CDCl3 are commonly used when running both proton and carbon-13 spectra. After runing an NMR spectrum, the sample can be recovered by evaporating off the CDCl3 solvent.

Describe the identification of O-H and N-H protons by proton echange using D2O;

The chemical shift due to protons attached to oxygen (O-H) or nitrogen (N-H) is very variable. To overcome this, you can run two spectra of a molecule- one with some deuterium oxide, D2O. If an O-H or N-H proton is present is will sawp with deuterium, and so the peak will disappear. So, absorptions due to O-H and N-H protons can be easily identified by comparing the two spectra.

Explain that NMR spectroscopy is the same technology as that used in ‘magnetic resonance imaging’ (MRI) to obtain diagnostic information about internal structures in body scanners

MRI uses the same technology as NMR spectroscopy- the patient is placed inside a very large magnet and radio waves are directed at the area of the body being investigated. Hydrogen nuclei in water molecules in the body absorb energy from the radio waves at certain frequencies. The frequency depends on the kind of tissue the water is in, so an image of different tissues can be built up.

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A2 Chemistry: 4.3.1 (Chromatogrpahy)

Describe Chromatography as an analytical technique that seperates components in a mixture between a mobile phase and a stationary phase;

State that: the mobile phase may be a liquid or a gas and the stationary phase may be a solid or either a liquid or solid on a solid support;

TLC uses a solvent, such as ethanol, for the mobile phase. The mobile phase passes over the stationary phase. That stationary phase in TLC is a thin layer of solid, e.g. silica gel or alumina powder, on a glass or plastic plate.

In GC, the mobile phase is an unreactive carrier gas such as nitrogen or helium. The stationary phase is a viscous liquid, such as an oil, or a solid, which coats the inside of the capillary tubing.

n.b- useful diagrams concerning GC and TLC are in the CGP revision book.

State that: a stationary phase seperated by adsortion and a liquid stationary phase seperates by relative solubility;

Explain the term Rf value, and interpret one-way chromatograms in terms of Rf values;

An Rf value shows how far a component has moved compared with the solvent front. It can be calculated by the following equation:

distance moved by components/distance moved by solvent front

You can work out what is in a mixture by comparing this value to a table of known values.

n.b– Rf values are between 0 and 1

Explain the term retention time, and intepret gas chromatograms in terms of retention times and the approximate proportions of the components in a mixture;

Retention time is the time for component to pass from the column inlet to the detector. Each peak on a chromatogram corresponds to a substance with a particular retention time. Retention times are measured from zero to the centre of each peak, and can be looked up in a reference table to identify the substances present.

n.b– the area under each peak of the chromatogram is proportional to the relative amount of each substance in the original mixture.

Explain that analysis by gas chromatogrpahy has limitations, e.g: similar compounds often have similar retention times and unknown compunds have no reference times for comparison;

Gas Chromatography has limitationsSimilar compounds often have similar retention times, so they’re difficult to identify accurately. A mixture of two similar substances may only produce one peak so you can’t tell how much of each there is. In addition, you can only use GC to identify substances that you already have reliable reference retention times for.

Explain that mass spectrometry can be combined with chromatography (GC-MS and HPLC-MS): to provide a far more powerful analytical tool than from chromatography alone, to generate mass spectra which can be anaylsed or compared with a spectral database by computer for positive identification of a component;

GC can seperate components in a mixture, but cannot identify them conclusively. Mass spec can provide detailed structural information on most compounds for exact identification, but it cannot seperate them. Combining them creates a powerful analytical tool.

Mass spectra can be compared with known spectral values for positive identification.

HPLC stands for High Pressure Liquid Chromatography. The staionary phase is a solid that is packed into a glass column, like tiny silica beads. The mobile phase (a solvent) and the mixture are pushed through the column under high pressure, allowing seperation to happen much faster than GC, where the solvent is just dropped through.

State the use of GC-MS in analysis, e.g. in forensics, environmental analysis, airport security and space probes

Forensics– identify unknown substances found on victims or suspects or at crime scenes

Airport security– can be used to look of specific substances, e.g. explosives or illegal drugs

Space probes– used this technique to examine the atmosphere and rocks of other planets and moons

Environmental analysis– used to detect and track levels of pollutants


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A2 Chemistry: 4.1.3 (Carboxylic Acids and Esters)

Explain the water solubility of carboxylic acids in terms of hydrogen bonding and dipole-dipole interaction

Carboxylic acids contain C=O bonds and O-H bonds, which are highly polar. These allow allow them to form hydrogen bonds with water molecules, as the H delta+ and the O delta- on different molecules are attracted to each other.

Describe the reactions of carboxylic acids with metals, carbonates and bases

Carboxylic acids are weak acids; in water they partially dissociate into carboxylate ions and H+ ions (n.b- this is in equilibrium).

A carboxylic acid reacts with reactive metals to form a salt and hydrogen gas– during the reaction there will be effervescence. Example:

CH3COOH (aq) + Na (s) → CH3COO-Na+ (aq) + ½H2 (g)

n.b– check charge of salt formed is balanced

Carboxylic acids are neutralised by bases to form a salt and water. Example:

2CH3COOH (aq) + MgO (s) → (CH3COO-)2Mg2+ (aq) + H2O (l)

An aqueous solution of a carboxylic acid reacts with carbonates to form a salt, carbon dioxide and water. Example:

2CH3COOH (aq) + Na2CO3 (s) → 2CH3COO-Na+ (aq) + H2O (l) + CO2 (g)

Describe esterification of carboxylic acids with alcohols, in the presence of an acid catalyst; of acid anhydrides with alcohols;

If you heat a carboxylic acid with an alcohol in the presence of an acid catalyst, you get an ester. Concentrated sulfuric acid is usually used as the acid catalyst. This, esterification, is a condensation reaction– where the acid and alcohol combine to form the ester, and water is released. The reaction is reversible, so you need to separate out the product as it’s formed. For small esters, you can warm the mixture and distil off the ester, as it is more volatile than the other compounds. For longer esters, you need to heat them under reflux and use fractional distillation to separate the ester from the other compounds.

Acid anhydrides are made from two identical carboxylic acid molecules. To name them, replace the word ‘acid’ (from carboxylic acid) with ‘anhydride’. These can also be reacted with alcohols to form esters, but a carboxylic acid is formed as well, instead of water. The anhydride is warmed with the alcohol. No catalyst is needed. It is a non-reversible reaction.

n.b– check text book for structures; the naming also suggests what part is from the acid and what part is from the alcohol

Describe the hydrolysis of esters in hot aqueous acid to form carboxylic acid and alcohols and in hot aqueous alkali to form carboxylate salts and alcohols;

In acid hydrolysis, the ester is heated under reflux with dilute sulfuric or hydrochloric acid. The ester is broken down by water (from the dilute acid), with the acid acting as a catalyst. It is a reversible reaction, so you need to use lots of water to shift the equilibrium to the right. Forms a carboxylic acid and an alcohol.

In alkali hydrolysis, aqueous sodium (or potassium) hydroxide is refluxed with the ester. This is a non-reversible reaction, and forms a sodium (or potassium) salt and an alcohol.

n.b– again, please check text book for structures

State the uses of esters in perfumes and flavourings

Describe a triglyceride as a triester of glycerol (propane-1,2,3-triol) and fatty acids;

Fatty acids are long hydrocarbon chains with carboxylic acid groups attached at the end. If the hydrocarbon chain contains no double bonds, then the fatty acid is saturated. If it does contain double bonds, then the fatty acid is unsaturated.

n.b– triglycerides are esters; therefore they are formed by esterification, a type of condensation reaction, where water is eliminated. 

Compare the structures of saturated fats, unsaturated fats and fatty acids, including cis and trans isomers, from systemic names and short hand formulae;

Example for naming system: Octadec-9-enoic acid. ‘Octadec’ refers to the 18 carbons, ‘9’ refers to the double bond being between the 9th and 10th carbon. Short hand version: 18:1 (9).

Unsaturated fatty acids can exist geometric isomers (due to the C=C bond), where they have a cis configuration and a trans configuration. In the cis form, molecules can not pack closely together, so they exist as liquids at room temperature. In the trans form, the shape is quite linear, so the fatty acid chains can pack closely together. This means trans-unsaturated fatty acids have a higher melting point than cis-unsaturated fatty acids.

Compare the link between trans fatty acids, the possible increase increase in bad cholesterol and the resultant increased risk of coronary heart disease and strokes;

High-density lipoproteins (HDLs)- transport cholesterol out of the blood.

Low-density lipoproteins (LDLs)- carry cholesterol in the blood. Can deposit lipids onto artery walls, building up fatty deposits that restrict blood flow.

Trans fats behave like saturated fats, raising LDL levels, increasing the risk of heart disease. They can also lower HDL levels.

Describe and explain the increased use of esters of fatty acids as biodiesel;

Biodiesels are renewable fuels, that are mainly a mixture of methyl and ethyl esters of fatty acids. Triglycerides can be reacted with methanol (KOH catalyst) to form glycerol and methyl ester, which can be used to make biodiesels. Biodiesels can be mixed with conventional diesel, so less crude oil-based diesel is used (which will run out). However, to produce huge quantities of biodiesel would mean devoting huge areas of land to growing biodiesel crops (e.g. rapeseed) rather than food crops.

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Will use this as a revision tool. I think the best way to revise is to explain what you are revising to someone else. In addition, I have a talent for losing things such as school notes; this will be quite hard to do if it is all online

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