Objetivo VIII – Inserción Inteligente al Mundo Iniciativa 95 – Inserción Política
$) Iniciativa 24 - Fomento de la
4.6. El Plan Belgrano
6H(g)
6C(s) 3H2(g)
∆H1 ∆H2
∆H3
∆Hf ‘ ’
Unit 5 Transition metals, arenes and organic nitrogen chemistry
Chapter 12 Arenes and their
derivatives
2 The double bond between the carbon and the oxygen is delocalised between the two oxygen atoms in the ethanoate ion, thus stabilising it:
eIt is the stabilisation energy due to the delocalisation that makes ethanoic acid an acid, as the anion is stabilised relative to the undissociated acid. The hydrogensulfate ion, HSO4–, has three resonance structures, stabilising it considerably and making sulfuric acid a strong acid.
3 The aromatic isomers of C6H4ClNO2are:
eThe 1,6-structure is identical to the 1,2-isomer, as is the 1,5-structure to the 1,3-isomer.
4 The aromatic isomers of C6H3Br2OH are:
Chapter 12 Arenes and their derivatives
ePut one bromine atom next to the OH (in the 2-position) and work out how many different places the other bromine can go (three more). Then put one bromine in the 3-position and see where the other bromine can go (one more).
Putting a bromine in the 4-position would give one of the isomers that you have already drawn, so there are only six isomers.
Note that 3,6-dibromophenol is the same as 2,5-dibromophenol.
5 a This is an addition reaction and so free-radical conditions are required. Mix liquid bromine with the nitrobenzene and heat under reflux in the presence of ultraviolet light.
eThe reaction is slow and requires heat, so the reagents must be heated under reflux otherwise they would boil off.
b This is an electrophilic substitution reaction, so a catalyst of anhydrous aluminium chloride, iron(III) chloride or iron(III) bromide must be used under dry conditions. The mixture is heated under reflux.
eThe iron(III) bromide is usually made in situ, by adding iron filings to the reaction mixture, which contains excess bromine. The iron reacts with some of the bromine:
Fe + 1 1–2Br2→ FeBr3
6 a The equation for the formation of the NO2+electrophile is:
HNO3+ H2SO4→ NO2++ H2O + HSO4–
eThe first step is the protonation of the HNO3molecule by the stronger acid, sulfuric acid. The water produced causes the ionisation of another sulfuric acid molecule.
b The equation for the production of the Br+electrophile is:
FeBr3+ Br2→ Br++ FeBr4–
eIf AlCl3is used as the catalyst, the equation is:
AlCl3+ Br2→ Br++ AlCl3Br–
c The equation for the production of an alkyl electrophile, such as C2H5+, is:
AlCl3+ C2H5Cl → C2H5++ AlCl4–
7
eThe stability of the benzene ring is regained in the second step. This is the driving force of the reaction.
CH3C+
=
OChapter 12 Arenes and their derivatives
8 a
Conditions: warm under reflux with anhydrous aluminium chloride as catalyst
eThe electrophile will attach at either the 2- or the 4-position in this reaction and the subsequent reactions in this question.
b
Conditions: warm at 50°C under anhydrous conditions with some iron filings or with a catalyst of iron(III) chloride
eA mixture of 2-bromomethylbenzene and 4-bromomethylbenzene is produced.
c
Conditions: warm with a mixture of concentrated nitric and sulfuric acids
eThe temperature of the mixture must not be allowed to go above 55°C or a second NO2group will substitute into the ring.
9 a The electrons on the O–of the phenate ion are partially drawn into the ring and become delocalised with the π-electrons of the benzene ring. This stabilises the anion relative to the unionised phenol molecule, making phenol acidic. This delocalisation is not possible in the C2H5O–ion formed from ethanol. The ethoxide anion is not stabilised and the molecule is barely acidic.
b Both ethanol and phenol can form intermolecular hydrogen bonds. This is because they each have a δ+ hydrogen atom and a δ–oxygen atom. However, the dispersion (instantaneous induced dipole–induced dipole) forces are stronger between phenol molecules, because phenol has more electrons than ethanol.
Thus, more energy is required to separate phenol molecules than ethanol molecules and phenol has the higher boiling temperature.
eThe strength of covalent bonds is irrelevant to the boiling temperatures of these substances because no covalent bonds are broken on boiling. The strength of the dispersion forces (sometimes called van der Waals forces) depends mostly on the number of electrons and not on the masses of the molecules.
c Both substances can form hydrogen bonds with water molecules, which makes them soluble. However, the large, non-polar benzene ring is hydrophobic and reduces the solubility of phenol relative to that of the smaller ethanol molecule.
NO2
+ HNO3 + H2O
CH3 CH3
Br
+ Br2 + HBr
CH3 CH3
C
+ CH3COCl + HCl
CH3 O
CH3 CH3
Chapter 12 Arenes and their derivatives
eSolubility in water is due mostly to hydrogen bonds between solute molecules and water molecules. Do not use statements such as ‘like dissolves like’ — they are meaningless.
10 a
eDo not draw a covalent bond between the O and the K — the product, potassium phenate, is an ionic compound.
b
eA white precipitate would be observed. The reaction is of the same type as that between phenol and bromine in aqueous solution.
c
eThe product is the ester phenyl benzoate.
11 Phenol is a weak acid and dissociates according to the equation:
C6H5OH H++ C6H5O–
acid dissociation constant Ka= = 1.3 × 10–10mol dm–3
Ka= [H+]2/[C6H5OH]
[H+]2= Ka× [C6H5OH]
[H+] = Ka× [C6H5OH] = 1.3 × 10−10× 0.20 = 2.6 × 10−11
= 5.10 × 10–6mol dm–3
pH = –log [H+] = –log (5.10 × 10–6) = 5.29
eThe assumptions made in this calculation are that [H+] = [C6H5O–], that [C6H5OH]eq≈ [C6H5OH]initialand that the temperature of the solution is 25°C.
[H+][C6H5O–] [C6H5OH]
O C O
Cl Cl
OH
Cl O– K+
Chapter 12 Arenes and their derivatives
12 a The equation for the reaction of phenol with bromine is:
The bromine reacts with phenol in a 3:1 ratio.
molar mass of phenol = (6 × 12.0) + (6 × 1.0) + 16.0 = 94.0 g mol–1 amount of phenol = mass/molar mass = 1.23 g/94.0 g mol–1= 0.0131 mol amount of bromine = 3 × 0.0131 = 0.0393 mol
mass of bromine, Br2= mol × molar mass = 0.0393 × (2 × 79.9) = 6.28 g
eif you keep all the figures on the calculator during the calculation (which is a better method), the answer is 6.27 g.
b theoretical amount (moles) of 2,4,6-tribromophenol produced = 0.0131 mol theoretical mass of 2,4,6-tribromophenol, C6H2Br3OH, produced = mol × molar mass
= 0.0131 × [(6 × 12.0) + (3 × 1.0) + (3 × 79.9) + 16.0] = 0.0131 × 330.7 = 4.332 g percentage yield = mass produced × 100/theoretical mass
= (4.25 × 100)/4.332 = 98.1%
eAnother method is to calculate the moles of product formed (4.25/330.7 = 0.01285) and then to calculate the percentage yield as (moles produced × 100)/theoretical yield in moles.
A common error is to think that the percentage yield is (mass of product × 100)/mass of reactant.
13 A nucleophile has a lone pair of electrons, which it uses to form a new covalent bond. In the reaction of phenol with an acid chloride, the lone pair on the oxygen of phenol attacks the δ+carbon atom in the acid chloride, forming an ester and hydrogen chloride. An example is the reaction:
C6H5OH + CH3COCl → CH3COOC6H5+ HCl 14
eThe hydrogen bond is due to the attraction between the lone pair of electrons on the δ–oxygen atom of the C=O group and the δ+hydrogen of the –OH group in another benzoic acid molecule. This makes the OHO bond angle 180° (two pairs of electrons repelling) and the COH angle about 120° (one σ-pair, one lone pair and one pair in the hydrogen bond repelling).
15 If the temperature goes above about 65°C, a significant amount of 1,3-dinitrobenzene is formed. If the temperature falls much below this value, the nitration of benzene is too slow to be noticeable.
eOnce some nitrobenzene is formed, it competes with unreacted benzene in the nitration reaction. As the benzene ring is slightly deactivated by the NO2group, a higher temperature is needed to nitrate nitrobenzene than benzene, which is why a temperature below 65°C prevents formation of a significant amount of dinitrobenzene.
H
H
C C
O
O
O
O
C6H5 C6H5
180°
180°
120°
Br Br
+ 3Br2 + 3HBr
OH OH
Br
Chapter 12 Arenes and their derivatives
16 Formation of electrophile
17 The deductions are:
I As it burns with a smoky flame, compound X is probably an arene.
I The precipitate with Brady’s reagent means that it is either an aldehyde or a ketone.
I The lack of a silver mirror with Tollens’ reagent means that it is a ketone and not an aldehyde.
I The precipitate of iodoform indicates that it has a CH3C=O group.
I The simplest compound that X could be is:
18 eGoogle ‘Lister’ and follow leads. Also Google ‘phenol’ and go to www.3dchem.com.
O
C CH3
NO2+ CH3
H
NO2 and +
H +
NO2 Step 1
Step 2
+ H2SO4 HSO4–
HNO3 + H2SO4 H2NO3+ + HSO4– H2NO3+ H2O + NO2+
NO2
CH3 CH3
NO2 and
+ H2SO4
HSO4–
CH3 CH3
CH3
H NO2
H NO2
+ CH3
Chapter 12 Arenes and their derivatives
Summary worksheet (www.hodderplus.co.uk/philipallan)
1 C amount (moles) of nitrobenzene, C6H5NO2= mass/molar mass = 12.3 g/123.0 g mol−1= 0.100 mol amount (moles) of 1,3-dinitrobenzene actually produced = 0.100 × 44.4/100 = 0.0444 mol mass produced (yield) = 0.0444 mol × molar mass of C6H4(NO2)2= 0.0444 × 168.0 = 7.46 g
In options B and D C6H5(NO2)2 has been used as the formula of 1,3-dinitrobenzene, rather than C6H4(NO2)2. In options A and B it has been assumed that 0.1 mol of reactant would produce 0.1 mol of product, therefore ignoring the fact that the yield is only 44.4%.
2 C An addition reaction does occur between benzene and hydrogen under these conditions, so option A is incorrect. Option B is not true because benzene is stabilised by delocalisation of the π-electrons. Benzene is at a lower energy level than the theoretical cyclohexa-1,3,5-triene and so less, not more, energy is given out on hydrogenation. Therefore option C, and not D, is correct.
3 B The first step is addition of Br+. This is followed by loss of H+resulting in overall substitution. The interme-diate loses H+so that it regains the resonance stabilisation energy. The reaction occurs at room temperature, so the activation energy must be reasonably low. Therefore option A is incorrect. Bromine is an electrophile under these conditions because, after reaction with the catalyst, it becomes Br+. Therefore, options C and D are incorrect.
4 D Step 1 is acylation, so B is not the correct response. Step 2 is nitration, so C is not the correct response.
Step 3 is addition of hydrogen (also can be called reduction) and so A is not the correct response. None of the steps involves oxidation, so D is the correct response to this negative question.
5 B The first step in both acylation (Friedel-Crafts) and nitration is attack by an electrophile, so C and D are wrong. The overall reaction is substitution not addition, so A is also wrong.
6 B Lithium aluminium hydride is the reagent for reduction of polar π bonds and so is not the correct response.
Aluminium oxide is never a catalyst in arene chemistry, so C is wrong. Dry aluminium chloride is the catalyst for Friedel-Crafts or halogenation and so D is also wrong.
7 B Any compound with a low hydrogen-to-carbon ratio is liable to burn with a smoky flame unless oxygen is in excess. Option A is wrong as arenes with one or two oxygen atoms, for example phenol, will burn with a smoky flame. Species other than arenes can form resonance hybrids. An example is the anion of a carboxylic acid, so option C is incorrect. The statement in D is irrelevant.
8 A The reason that phenol is so reactive towards electrophilic substitution is that the lone pzpair of electrons on the oxygen atom becomes incorporated into the delocalised ring. This makes the ring more susceptible to attack by electrophiles. Options B and C are true statements, but are irrelevant to this question. Option D is not true. The Br+ion is a much stronger electrophile, but is not needed in the bromination of phenol because the ring is activated by the lone pair of electrons in the oxygen of the –OH group.
Chapter 12 Arenes and their derivatives
1 Step 1: the ethene has to be converted to a halogenoalkane Mix ethene with hydrogen chloride gas at room temperature:
CH2=CH2+ HCl → CH3CH2Cl
Step 2: react the chloroethane with excess concentrated ammonia solution in a mixture of water and ethanol.
Allow the mixture to stand at room temperature or heat it in a sealed tube:
CH3CH2Cl + 2NH3→ CH3CH2NH2+ NH4Cl
eA good way to work out the answer to this type of question is to start at the end. Ask yourself how a primary amine can be prepared — there are three ways. Reduction of a nitrile would not work as the –CN group would add an extra carbon atom. Amides can be reduced to amines, but ethene cannot be converted in one step to ethanamide. The only method that would work is the reaction of a halogenoalkane with ammonia. Step 1 is then obvious.
2 a The primary amine with a branched chain is:
b The symmetrical secondary amine with an unbranched chain is:
c The secondary amine with a branched chain is:
d The tertiary amine is:
eA primary amine has the general formula RNH2, a secondary amine the formula RR′NH and a tertiary amine RR′R′′N, where R, R′ and R′′ are alkyl groups.
3 The nitrogen atom in methylamine has three bond pairs and one lone pair, causing the molecule to be pyramidal. The four electron pairs repel each other, but the repulsion between the lone pair and the bond pairs is stronger than that between the bond pairs. This means that the H–N–H bond angle is about 107°, which is smaller than the tetrahedral angle of 109.5°.
eDo not make the common error of stating that the bonds or the atoms repel each other. The theory of shapes is called the valence-shell electron-pair repulsion theory (VSEPR). Another error is failing to realise that there is a lone pair of
N