4 The structural formula of ethanamide is:
The carbon atom in the functional group has one double bond and two single bonds (and no lone pairs).
The π-bond does not affect the shape and the electron pairs in the three σ-bonds repel each other to a position of maximum separation. This gives a trigonal planar shape with an N–C–C bond angle of 120°.
eRemember that a double bond consists of a σ- and a π-bond in the same direction between two atoms. The shape is caused by the repulsion between the electrons in the σ-bonds.
5 Ethylamine can form hydrogen bonds with water molecules as it has a δ–nitrogen atom and δ+hydrogen atoms. Therefore, it will dissolve in water. The δ–chlorine in chloroethane is too big to form a hydrogen bond with the δ+hydrogen in water and ethane is non-polar. Thus, neither of these compounds dissolves in water.
eSolubility in water is caused either by ions becoming hydrated by water molecules or by hydrogen bonding between the solute and the water. Polarity in an organic molecule does not mean that it is soluble in polar solvents such as water.
6 Both ethanoic acid and propylamine can form intermolecular hydrogen bonds. Induced dipole (dispersion) forces also exist between their molecules, but these are much weaker than the hydrogen bonds. The oxygen atom in ethanoic acid is more δ–than the nitrogen atom in propylamine, so the O---H hydrogen bond is stronger than the N---H hydrogen bond. This means that more energy is required to separate the ethanoic acid molecules and so ethanoic acid has the higher boiling temperature. Butane has no δ+hydrogen atoms (and no δ–F, N or O atoms) and so cannot form hydrogen bonds. Its only intermolecular interactions are the weaker dispersion forces and so butane has the lowest boiling temperature.
eOxygen is more electronegative than nitrogen and so is more δ–.
7 As with question 1, it makes sense to work backwards, asking how the amine can be prepared. Amines can be prepared from halogenoalkanes, which can themselves be prepared from alcohols. Acids can be reduced to alcohols.
Step 1: add lithium aluminium hydride in dry ether to the ethanoic acid and then hydrolyse with dilute sulfuric acid:
CH3COOH → CH3CH2OH
Step 2: add phosphorus pentachloride:
CH3CH2OH → CH3CH2Cl
Step 3: allow the chloroethane to stand with excess concentrated aqueous ethanolic ammonia:
CH3CH2Cl → CH3CH2NH2
eThe critical step is deducing that a halogenoalkane reacts with ammonia to form a primary amine.
If you are asked for an outline of a synthesis, you should give details of reagents, conditions (if any) and the formulae of the intermediates produced at each step. You do not need to give balanced equations.
8 a
The empirical formula is C2H7ON.
Element % Divide by Ar Divide by smallest Carbon 39.3 39.3/12.0 = 3.275 3.275/1.64 = 2.0 Hydrogen 11.5 11.5/1.0 = 11.5 11.5/1.64 = 7.0 Oxygen 26.2 26.2/16.0 = 1.64 1.64/1.64 = 1 Nitrogen 23.0 23.0/14.0 = 1.64 1.64/1.64 = 1
O
C NH2 CH3
120
Chapter 13 Organic nitrogen compounds
b The molecular formula is also C2H7ON. Since the substance produces steamy fumes with PCl5, it must contain an –OH group. It cannot be a carboxylic acid as it has only one oxygen atom, so it must contain an alcohol group. Its structural formula is CH2OHCH2NH2and its name is hydroxyethylamine (or 2-aminoethanol).
9 a NH2CH2CH2OH + HCl → [NH3CH2CH2OH]+Cl–
eThe –NH2group acts as a base and accepts an H+ion from the acid. The reaction is similar to NH3+ HCl → NH4+Cl–. The reaction can also be written as an ionic equation: NH2CH2CH2OH + H+→ [NH3CH2CH2OH]+.
b NH2CH2CH2OH + 2CH3COCl → CH3CONHCH2CH2OOCCH3+ 2HCl
eThe acid chloride reacts with the amine group to form a substituted amide and with the alcohol group to form an ester.
c NH2CH2CH2OH + CH3I →
ePrimary amines react with halogenoalkanes to form secondary (and tertiary) amines. Alcohols do not react, so only one CH3I is needed in the equation.
The equation NH2CH2CH2OH + CH3I → CH3NHCH2CH2OH + HI would be acceptable, even though the secondary amine produced is a base and would react with the HI to form the salt, as shown above.
10 A base must have a lone pair of electrons that is used to form a bond with an H+ion. The nitrogen atoms in ammonia and in methylamine each have a lone pair, but the methyl group ‘pushes’ electrons towards the nitrogen, which becomes more δ–. This makes the nitrogen in methylamine a stronger base, as it is more negative than the nitrogen in ammonia. Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom becomes partially delocalised with the electrons of the benzene ring. This lowers its ability to accept a proton.
eThis is also the reason why the benzene ring in phenylamine is attacked readily by electrophiles.
11 Aminoethanoic acid is an amino acid. It has an acidic –COOH group and a basic –NH2group. The –COOH group loses a proton and becomes a –COO–group. The –NH2group accepts that proton and becomes an –NH3+
group. The molecule now has a full positive charge on one end of the molecule and a full negative charge on the other end. This species is called a zwitterion. Strong forces of attraction exist between zwitterions and so a large amount of energy is required to separate them, causing aminoethanoic acid to be a solid at room temperature. The positive charge on the nitrogen in the zwitterion causes strong attractions with the δ–oxygen atoms in water; the negative charge on the oxygen interacts similarly with the δ+hydrogen atoms in water.
This makes the amino acid water-soluble.
eYou must make it clear that the forces of attraction that cause the amino acid to be a solid act between zwitterions.
12 When acid is added to ethylamine the volatile base is protonated to form a non-volatile ionic compound.
C2H5NH2+ HCl → C2H5NH3+Cl−
When alkali is added, the C2H5NH3+ion is deprotoanted re-forming the volatile base ethylamine. Therefore, the characteristic smell returns.
C2H5NH3++ OH−→ C2H5NH2+ H2O
13 The reagents are tin and concentrated hydrochloric acid.
N+ CH2CH2OH H3C
H H
I–
Chapter 13 Organic nitrogen compounds
14 a The –NH2group in phenylamine contains two δ+hydrogen atoms and a δ−nitrogen atom. This means that phenylamine can form intermolecular hydrogen bonds. Chlorobenzene does not have any δ+hydrogen atoms or any small δ−atoms and so it cannot form intermolecular hydrogen bonds. The hydrogen bonds between phenylamine molecules are stronger than the dipole–dipole forces and the induced dipole forces present between molecules of each substance. Thus more energy is required to separate phenylamine molecules than is required to separate chlorobenzene molecules and so phenylamine has the higher boiling temperature.
b Although phenylamine can form some hydrogen bonds with water, the large hydrophobic benzene ring reduces its solubility. When acid is added, C6H5NH3+ions are formed. These are hydrated by the water molecules , which results in the substance dissolving.
c Chlorobenzene cannot form hydrogen bonds with water as it has neither δ+hydrogen atoms nor a δ−F, N or O atom. It also does not react with water, so despite being polar, it is insoluble. It does not dissolve in acid, because it is not a base, and so it does not react with H+ions to form an ionic compound.
15 a C6H5NH2+ HCl → C6H5NH3+Cl−
b C6H5NH2+ CH3COCl → C6H5NHCOCH3+ HCl
eHere phenylamine is reacting as a primary amine. In a it forms a salt and in b it forms a substituted amide that has the structural formula:
16 a If the temperature falls below 0°C, the reaction between nitrous acid and phenylamine is too slow.
b If the temperature rises above 10°C, the diazonium ion will decompose as soon as it is formed.
17 a
b
eYou must show the two nitrogen atoms separately with the correct bonds between them. Make sure that you put the positive charge on the appropriate nitrogen atom in the formula for the diazonium ion.
18 The oxygen atom in the C=O group is δ−and the two hydrogen atoms of the NH2group are δ+. Because of this, ethanamide forms strong hydrogen bonds with water molecules and so is soluble.
19
20 Terylene®is a polyamide and so has δ−oxygen, δ−nitrogen atoms atoms and δ+hydrogen atoms spaced regularly along the polymer chain. These atoms form hydrogen bonds with other chains. On melting, these nHOOC–(CH2)3–COOH + nH2NCH2CH2NH2 C (CH2)3 C CH2 CH2
O O
N C
H
N + (2n–1)H2O
n H
N+ N OH + OH–
N N OH + H2O
+
N+ N
NH
CH3 C
O
Chapter 13 Organic nitrogen compounds
hydrogen bonds have to be broken and because they are strong, a high temperature is needed to provide sufficient energy. Addition polymers such as poly(propene) do not form hydrogen bonds because they do not contain any δ+hydrogen atoms or δ−F, N or O atoms. This means that their intermolecular forces (really forces between chains) are weaker.
21 Poly(ethenol) has an –OH group on every other carbon atom in the chain. This means that it can form many hydrogen bonds with water and so, despite the extremely large molecules, it is soluble.
22 a The general formula for a zwitterion of an amino acid is +NH3CHRCOO–. b The equation for the reaction of the zwitterion with acid is:
+NH3CHRCOO–+ H+→+NH3CHRCOOH c For the reaction with base, the equation is:
+NH3CHRCOO–+ OH–→ NH2CHRCOO–+ H2O
eRemember that the positive charge is on the nitrogen atom. This charge must be shown there if the full structural formula is drawn:
23 The two optical isomers of aminobutane-1,4-dioic acid are:
The left-hand compound is the natural isomer, aspartic acid, as it has the same configuration as L-alanine (textbook, p. 275).
24
25 Uracil in mRNA forms two hydrogen bonds to adenine in DNA as shown below:
C
The information needed for questions 26–29 is not given in the A2 chemistry textbook. You are advised to do your own research to find answers to these questions. Possible sources are the internet, biology textbooks and other students who are studying biochemistry.
26 A DNA double helix has two strands. The sense strand has its bases in a particular order. The antisense strand (linked by hydrogen bonds between the two strands) has its bases in the complementary order (A opposite T and C opposite G). The double helix unwinds and a strand of messenger RNA (mRNA) is made with the order of bases exactly in the order of the sense strand of the DNA, except that uracil, U, replaces thymine, T. Three-base sections of the mRNA molecule act as a code for the insertion of a particular amino acid in the polypeptide being synthesised. Each three-letter section is called a codon. A codon for leucine is GAC in the antisense strand of the DNA and hence CUG in mRNA.
27 Chargaff’s rules are:
I the amount of adenine = amount of thymine (A = T)
I the amount of cytosine = amount of guanine (C = G)
I A + G = C + T
These rules apply to DNA but not to RNA.
28 Search for Chargaff’s rule, Crick or Watson on Google. Also don’t forget about the important contribution of Wilkins and Franklin.
29 Each human gene has a sequence of bases, repeated many times, that seem to have no particular function.
These repeating sequences are called minisatellites. Each person, except identical twins, has unique repeats.
One method of DNA fingerprinting is as follows:
I Some tissue is sampled — for example, blood, saliva, semen or even hair.
I DNA is extracted from the cells in the sample.
I The DNA is cut up into the minisatellite fragments by enzymes.
I The fragments are then separated by electrophoresis — chromatography under an electric potential.
I A DNA ‘probe’ is added — a fragment of DNA that has been labelled with a radioisotope. This sticks to certain fragments in the sample.
I A photographic plate is placed over the electrophoresis plate and the X-rays from the radioactive isotope darken the photo plate, showing the position of each fragment.
I Several different DNA probes are required to obtain a definite profile of a person’s DNA.
Your only defence against a DNA match is to say that ‘it must have been my identical twin’.
Chapter 13 Organic nitrogen compounds
Summary worksheet (www.hodderplus.co.uk/philipallan)
1 C An amine must have NH2, NH or N attached to one or more carbon atoms by single bonds. CH3NH2
(option A) is a primary amine, as is CH3COCH2NH2(option D). (CH3)2NH (option B) is a secondary amine, but CH3CONH2(option C) is an amide. Note that compound D is also a ketone. The only non-amine is the compound in option C, which is therefore the correct answer to this negative question.
2 B An alkali will dissolve in, or react with, water to produce OH−ions, so the answer is option B.
The reaction in option A shows that 1-butylamine is a base; that in option C shows it to be a primary amine and that in option D shows that it can act as a ligand.
3 D Phenylamine is a base, so it reacts with acids not bases. Therefore, option A is incorrect. The temperature of 35°C is too high for a diazonium salt to be formed, so option B is wrong. The benzene ring in phenylamine is activated and so it reacts towards electrophiles in the same way as phenol. So the ‘standard’ benzene conditions (option C) are not required and it reacts with bromine water to form a white precipitate of the tri-substituted compound.
4 B The metal can be either tin or iron, but the acid must be concentrated, so option A is wrong. Potassium dichromate is an oxidising agent not a reducing agent, so option C is incorrect. The correct catalyst with hydrogen is either platinum or nickel, not copper. Therefore, option D is incorrect.
5 A There is an –OH group on every other carbon atom in the polymer chain and so it can form thousands of hydrogen bonds with water. It is this that makes it soluble. Polarity (B) is not a guide to solubility and it neither reacts (C) nor breaks down into its monomers in water (D).
6 C The monomer must have two functional groups that can react with each other to form a new covalent bond.
The amino acid in option A has an –NH2group and a –COOH groups that can form amide links. Option B has an alcohol and an acid group and can, therefore, form a polyester. Option D has an alcohol group and an acid chloride group, so it can also form a polyester. Since these three compounds can all form polymers, none is the correct answer. The compound in option C has an alcohol group and an amine group. These groups do not react, so this compound cannot form a polymer. Therefore, option C is the correct answer to this negative question.
7 B The reason for a high melting temperature is strong forces between the particles. Glycine exists as zwitterions that have strong ionic bonds with neighbouring zwitterions.
8 D Amino acids react with acids because of the amine group and with bases because of the carboxylic acid group, thus neither option A nor option B is correct. The acid group will form an ester, so option C is not the correct response. Bromine water does not react with either an –NH2group or a –COOH group and so it will not react with amino acids. Therefore, option D is the correct response to this negative question.
9 A Separation by chromatography occurs because the different species have different Rfvalues. Some have different pH values that would result in different Rfvalues, but many have the same pH but different Rf
values, so option B is incorrect. The reaction with ninhydrin (C) is how the separated amino acids are located, not how they are separated. Their molar mass (D) is irrelevant.
10 A There are five different hydrogen environments in the molecule — that in NH2, in CH, in CH2, in the alcoholic OH group and in the acidic COOH group.
Chapter 13 Organic nitrogen compounds
1 The first step is to hydrolyse the halogenoalkane by warming it with aqueous sodium hydroxide and a few drops of ethanol for several minutes. The C–Cl and the C–I groups are hydrolysed to Cl–and I–ions.
The solution is cooled and excess nitric acid is added, followed by aqueous silver nitrate solution. Precipitates of AgCl and AgI will be obtained. The colour will be somewhere between white and pale yellow, depending on the relative amounts of chloride and iodide, so the colour does not prove which halogens are present.
Dilute aqueous ammonia is then added. Some of the precipitate will dissolve. The solution is filtered.
The filtrate is acidified with dilute nitric acid. If a white precipitate appears, the original organic compound contained chlorine.
If the residue is pale yellow and insoluble in concentrated aqueous ammonia, the organic compound also contained iodine.
eAfter hydrolysis the solution must be made acid, otherwise the sodium hydroxide still present would form a black precipitate with the silver nitrate.
After addition of silver nitrate the mixture must be filtered, because it is difficult to tell whether any of the precipitate dissolved in the dilute ammonia. Acidification of the filtrate reprecipitates the silver chloride, thus proving that the organic compound contained chlorine.
2 The presence of the C=C groups can be detected using bromine water. The brown bromine water will turn colourless.
The presence of the –OH group can be detected by adding phosphorus pentachloride to the dry substance.
Clouds of misty fumes (of hydrogen chloride) prove the presence of an –OH group. This result could be obtained with an acid or an alcohol, so the absence of a –COOH group has to be shown by adding a solution of sodium carbonate to the unknown. An acid would produce bubbles (of carbon dioxide), but an alcohol, such as linalool, would show no reaction.
eThe colour before (brown) and afterwards (colourless) must be stated when giving the results of the bromine water test.
The alcohol group could be detected by warming linalool with some ethanoic acid and a few drops of concentrated sulfuric acid, then pouring the mixture into a solution of sodium carbonate (to remove excess acids). An alcohol gives rise to the characteristic smell of an ester after this reaction.
3 a The decolorisation of bromine water shows that carvone must contain a C=C group.
b The failure to change the colour of acidified potassium dichromate(VI) solution shows that carvone cannot be oxidised easily. Therefore, it is not a primary or a secondary alcohol, or an aldehyde.
c A yellow precipitate with Brady’s reagent shows that carvone is a ketone. (The test in part b has already shown that it cannot be an aldehyde.)
eIn this example, the positive Brady’s test does not show that the unknown is either an aldehyde or ketone, since an aldehyde has been excluded by an earlier test.
4 a The production of steamy fumes means that compound Y contains an –OH group and so is either an acid, an alcohol or both.
b The fact that carbon dioxide is evolved on the addition of sodium carbonate means that Y is an acid. It could also contain an alcohol group since it has three oxygen atoms.