Capítulo 7. Otras actividades relevantes
7.1 Participaciones en eventos oficiales
eLithium aluminium hydride reduces C=O in aldehydes, ketones and carboxylic acids, esters and C≡N groups. Hydrogen with a nickel or platinum catalyst reduces C=C, C=O in aldehydes and C≡N, but not carboxylic acids or their derivatives.
Ketones are reduced very slowly by hydrogen.
6 a CH3CH2COOH + NaOH → CH3CH2COONa + H2O b 2CH3CH2COOH + Mg → (CH3CH2COO)2Mg + H2
c CH3CH2COOH + CH3OH CH3CH2COOCH3+ H2O
eThe reactions are:
a acid + base → salt + water b acid + metal → salt + hydrogen c acid + alcohol ester + water
Esterification reactions must be carried out using a small amount of concentrated sulfuric acid catalyst and the mixture has to be heated under reflux.
eHeating under reflux is necessary if:
G the reaction is slow and, therefore, needs heating for several minutes
G the reactants are volatile and would boil off when heated
7 Aspirin is an ester. The –OH on the benzene ring in the starting compound reacts with the acid chloride. The formula of aspirin is:
eIt is a common error to think that acid chlorines react with a COOH group.
8 A carboxylic acid can be prepared from an ester by hydrolysis. If an acid catalyst is used, the reaction is slow and reversible and so the yield is small. The best method is to hydrolyse the ester by heating it with aqueous alkali and then acidifying the solution to produce the carboxylic acid from its salt.
The equations are:
C6H5COOC2H5+ NaOH → C6H5COONa + C2H5OH C6H5COONa + HCl → C6H5COOH + NaCl
The method is:
I Place the ester in a round-bottomed flask and add aqueous sodium hydroxide.
I Heat the mixture under reflux for 30 minutes.
I Allow to cool and add excess dilute hydrochloric (or sulfuric) acid.
I Filter off the precipitate of benzoic acid.
I Recrystallise the benzoic acid by dissolving the solid in the minimum of hot water.
I Use a warm funnel to filter off any insoluble impurities; allow the solution to cool in a flask.
I When the solution is cool, filter the benzoic acid by vacuum filtration. Finally, wash the solid with a little cold water and allow it to dry.
O O
C
C CH3 OH
O
Chapter 8 Carboxylic acids and their derivatives
eThe answer is in two distinct parts. First, you must describe how the benzoic acid is prepared from the ester. Second, you must describe how it is purified. You might be told about its solubility in water, which allows the purification to be done by recrystallisation from water. This works because benzoic acid is soluble in hot water and insoluble in cold water.
9 Add some solid phosphorus pentachloride to dry propanoic acid:
C2H5COOH + PCl5→ C2H5COCl + POCl3+ HCl
Any unreacted propanoyl chloride can be removed from the reaction mixture by distillation.
10 a C2H5COCl + 2NH3→ C2H5CONH2+ NH4Cl
eThe organic product is an amide, propanamide. A common error is to give hydrogen chloride as the second product.
However, this reacts with ammonia, so 2NH3are needed on the left-hand side of the equation.
b C2H5COCl + H2O → C2H5COOH + HCl
eThis hydrolysis reaction demonstrates why acid chlorides should be kept dry. When a bottle containing an acid chloride is opened in air, steamy fumes are observed as the water vapour in the air reacts with the acid chloride, producing gaseous hydrogen chloride.
c 2C2H5COCl + NH2CH2CH2OH → C2H5CONHCH2CH2OOCC2H5+ 2HCl
eThe acid chloride reacts with the amine group to form a secondary amide and with the alcohol group to form an ester.
11 There are two types of transesterification reaction:
I the reaction of an ester with an acid to form a different ester and a different acid, for example:
CH3COOC2H5+ C3H7COOH → C3H7COOC2H5+ CH3COOH ester of acid I acid II ester of acid II acid I
I the reaction of an ester with an alcohol to form a different ester and a different alcohol, for example:
C3H7COOC2H5 + CH3OH → C3H7COOCH3 + C2H5OH ester of alcohol I alcohol II ester of alcohol II alcohol I
In both types of reaction, a catalyst is needed. This is usually an acid, such as dilute sulfuric acid.
12 Both animal fats and vegetable oils are esters of propan-1,2,3-triol (glycerol). However, the acid groups in these esters differ. In animal fats, most of the fatty acids are saturated, for example stearic acid, C17H35COOH. In vegetable oils, some of the fatty acids are saturated, some are monounsaturated and some are polyunsatu-rated. Fish oils contain many polyunsaturated acid groups, such as omega-3 and omega-6 acids.
13 The reaction of an acid chloride with an alcohol is a non-reversible reaction with a high atom economy, as only one H atom and one Cl atom present in the starting materials are not found in the product. The reaction of a carboxylic acid with an alcohol is a slow reversible reaction that requires a concentrated sulfuric acid catalyst.
The yield is therefore much lower than with the method using the acid chloride. The equations are:
CH3COCl + C2H5OH → CH3COOC2H5+ HCl CH3COOH + C2H5OH CH3COOC2H5+ H2O
14 The –OH group on one end of the molecule can form an ester with the –COOH group of another molecule and so on, forming a long-chain polyester:
OC
H CH3
C
O
OC
H CH3
C
O one repeat unit
Chapter 8 Carboxylic acids and their derivatives
15 Vegetable oil is a mixture of esters in which many of the fatty acid groups are polyunsaturated. If some of these are replaced by saturated acids, such as stearic acid, the new ester will have a higher melting temperature.
Such a transesterification reaction does not alter the stereochemistry of the remaining unsaturated acids.
This means that the geometry around the C=C groups remains cis-. The partial hydrogenation reaction saturates some of these C=C groups, and the catalyst causes some of them to change their geometry to form trans- isomers. Trans-fats are more harmful than cis-fats because they increase the level of cholesterol in the blood.
16 This requires a transesterification reaction. The ester in the vegetable oil is reacted with a simple alcohol such as methanol or ethanol. The products are three smaller esters and propan-1,2,3-triol. The smaller esters are easier to volatilise and so are more suitable as a fuel. A typical equation is:
17 Both Jatropha trees and soya beans yield an oil that is suitable for conversion to biodiesel. The Jatropha tree will grow on marginal land — land that is not suitable for growing food crops. It will also grow on the edges of roads and railway lines, neither of which are used to grow food crops. To obtain a high yield, soya beans need good fertile soil. Using soya as a source of biofuels will reduce the food production of a country. If this happens on a large scale, the price of food will rise significantly and may result in starvation in poor countries.
CHOOCR′ + 3CH3OH RCOOCH3 + R′COOCH3 + R′′COOCH3 + CH2OHCH(OH)CH2OH CH2OOCR
CH2COOR′′
Chapter 8 Carboxylic acids and their derivatives
Summary worksheet (www.hodderplus.co.uk/philipallan)
1 B The simplest way to answer this question is to realise that the substance is an ester of ethanol, so the answer must be either A or B. The acid residue is:
The methyl, –CH3, group is on the second carbon atom counting from the carbon of the –COO group. Ethyl butanoate is CH3CH2CH2COOC2H5so option A is incorrect. Options C and D are based on an incorrect alcohol and acid. 1-propyl ethanoate is the ester of ethanoic acid and propan-1-ol and has the formula CH3COOCH2CH2CH3, so option C is incorrect. 2-propyl ethanoate is the ester of ethanoic acid and propan-2-ol, CH3COOCH(CH3)2, so option D is also incorrect.
2 C For these oxygen-containing compounds, there must be an –OH group for intermolecular hydrogen bonding.
Alcohols, acids and water all contain an –OH group which results in a δ+H atom and a δ−O atom. Ester C has two δ−oxygen atoms but no δ+hydrogen, so it cannot hydrogen bond with other ester molecules.
e For negative questions such as 2 and 3, it is worth putting a tick or cross beside each option. You should get three ticks and one cross. If you get two ticks and two crosses then you should look again carefully at the options you marked with a cross.
3 B Under suitable conditions, an alcohol, such as methanol, will react with ethanoic acid to form an ester.
Therefore, option A gets a tick. A reactive metal, such as zinc, will react with an acid to form a salt and hydrogen. So, option C gets a tick. Phosphorus pentachloride reacts with compounds containing an –OH group, so option D gets a tick. As options A, C and D all have ticks, none is the answer to this negative question. Hydrogen, in the presence of a platinum catalyst, will reduce non-polar C=C but not the C=O group in acids (or esters) and so will not react with ethanoic acid. Option B gets a cross and is the correct response to this negative question.
4 D The chlorine atom joins with one of the H atoms of the –NH2group, forming an amide. Option A is totally wrong. The structure in B shows confusion with the reaction of an acid chloride with alcohols to form an ester. Option C is wrong because only one hydrogen atom is substituted in the reaction between a primary amine and an acid chloride. This shows confusion with the reaction of a primary amine with a halogenoalkane, where both hydrogen atoms can be replaced.
5 C The four esters are:
There are three possible acids — propanoic, ethanoic and methanoic. The ester of methanoic acid can be formed from two isomeric alcohols, propan-1-ol and propan-2-ol.
e It is essential to sketch out the structures when answering a question about structural isomers.
6 A For a condensation polymer to form, each monomer must have two functional groups. 1,6-hexandioic acid has two –COOH groups and 1,3-propanediol has two –OH groups, so they will form a condensation polymer. 3-hydroxybutanoic acid has a –COOH group at one end of the molecule and an –OH group at the other, so it too can form a condensation polymer. But-2-ene, like all alkenes, can form an addition polymer.
Therefore, options B, C and D can all form polymers. Methanol in option A has only one –OH group and so cannot form a condensation polymer. therefore, option A is the correct response to this negative question.
C2H5C
Chapter 8 Carboxylic acids and their derivatives
1 a Vibration of the O–H bonds in the water molecules will increase.
b The molecules will rotate faster, gaining rotational energy. On colliding with other water molecules, this energy will be converted to kinetic energy and the water will heat up.
2 The poisonous species is soluble Ba2+ions. Both barium carbonate and barium sulfate are insoluble in water.
The stomach contains hydrochloric acid, which reacts with barium carbonate to form a solution of barium chloride, containing Ba2+(aq) ions:
BaCO3(s) + 2HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g)
Barium sulfate does not react with hydrochloric acid, so no soluble barium compound is formed:
BaSO4(s) + HCl(aq) → no reaction
3 The first step (initiation) is the homolytic fission of a chlorine molecule into two chlorine radicals:
hν Cl2 → 2Cl•
The propagation steps are:
(1) the removal of a hydrogen atom by a chlorine radical:
Cl• + CH4→ HCl + CH3•
(2) the removal of a chlorine atom from a Cl2molecule by the CH3• radical to form a second chlorine radical:
CH3• + Cl2→ CH3Cl + Cl•
The second chlorine radical then continues the chain reaction.
There are some chain-termination processes that could stop the chain reaction:
Cl• + Cl• → Cl2
CH3• + Cl• → CH3Cl CH3• + CH3• → C2H6
eThe presence of traces of ethane is strong evidence for this mechanism.
4 Polar molecules absorb energy in the infrared region of the electromagnetic spectrum. It is because oxygen is more electronegative than hydrogen and because the water molecule is not linear that a water molecule is polar. The light coming from the sun contains higher frequencies than the infrared radiated by the Earth back towards space. The water vapour absorbs some of this infrared radiation, trapping energy in the atmosphere and so keeping the Earth warm.
ePolar water molecules also absorb in the microwave region, which is why aqueous liquids heat up in a microwave oven.
5 Spectrum A: the peak at wavenumber 1715 cm–1is caused by a C=O bond and the peak at 1421 cm–1by a C–C bond.
Spectrum B: the peak at wavenumber 1716 cm–1is caused by a C=O bond and the peak at 1416 cm–1by a C–C bond, but the peak at 2986 cm–1is caused by an O–H bond.
Both propanal and propanoic acid have C=O and C–C bonds, but only propanoic acid has an O–H bond. Thus, spectrum B is that of propanoic acid and spectrum A that of propanal.
eThe broadness of the peak at 3000 cm–1is caused by the intermolecular hydrogen bonding of the –OH group on one molecule with the C=O group on another.