2017 H2 Chemistry Paper 3 Suggested Solutions

2017 H2 Chemistry Paper 3 Section Section A Answer all the questions in this section.

1

(a)

Describe the thermal decomposition decompositi on of the hydrogen halides HCl, HBr and HI and explain any variation in their thermal stabilities. [3] Hydrogen halides undergo thermal decomposition to form hydrogen gas and their corresponding halogens halogens (as reddish-brown fumes in the case of Br 2 and dark purple fumes for I 2) as the only products: 2HBr → H2 + Br 2 2HI → H2 + I2 HCl is thermally stable and does not decompose on heating. The thermal stabilities of hydrogen halides decrease in the following order: HCl (most thermally stable), HBr and HI (least thermally stable). This is due to the increase in bond length. As a result, the effectiveness of orbital overlap and hence bond energy decreases from H – H –Cl Cl to H –Br –Br to H –I. –I. As such, the energy required to break the H – H –X X bond decreases from HCl to HBr to HI (hence thermal stabilities decrease from HCl to HBr to HI).

(b)

Table 1.1 gives the melting points, in °C, of the fluorides and chlorides of the two elements in Period 3. Table 1.1 magnesium

silicon

fluoride

1261

−90

chloride

714

−70

Explain, in terms of structure and bonding, the differences in melting point between (i) (i)

MgCl2 and SiCl 4,

[2]

MgCl2 is an ionic compound. It consists of Mg 2+ and Cl− held by strong ionic bonds between oppositely charged ions in a giant ionic lattice structure, which require a lot of energy to overcome during melting. Hence, MgCl 2 has a high melting point. SiCl 4 is a covalent compound. It exists as simple covalent molecules of SiCl4 held together by weak van der Waals’ forces of attraction between the discrete molecules, which require little energy to overcome during melting. Hence, SiCl 4 has a low melting point.

Copyright © 2017 by Lee Jun Hui

2017 H2 Chemistry Paper 3

(ii)

SiF4 and SiCl 4,

[1]

Both SiF4 and SiCl4 are covalent compounds existing as simple covalent molecules held together by weak van der Waals’ forces of attraction. As SiCl 4 has a larger electron cloud size than SiF 4, the electron cloud of SiCl 4 is more polarisable than that of SiF 4, resulting in stronger van der Waals’ forces in SiCl 4 than in SiF4. More energy is required to overcome the stronger van der Waals’ forces in SiCl 4 than in SiF 4, hence SiCl 4 has a higher melting point than SiF 4.

(iii)

MgF2 and MgCl 2.

[1]

Both MgF2 and MgCl 2 are ionic compounds compounds consisting of Mg 2+ and X− held by strong ionic bonds between oppositely oppos itely charged ions in a giant ionic lattice. The lattice energy (strength of ionic bonds) is directly proportional to the product of the cation and anion charges and inversely proportional proportional to the sum of the cationic and anionic radii. As − Cl is larger than F − (one more quantum shell), the sum of the cationic and anionic radii is larger for MgCl 2 than for MgF 2, hence the ionic bonds are stronger in MgF 2 than in MgCl 2. More energy is required to overcome the stronger ionic bonds in MgF 2 than in MgCl 2, hence MgF2 has a higher melting point than MgCl 2.

The halogens form many interhalogen compounds such as ClF 3 and BrCl. (c)

The central atom in ClF3 is surrounded by five pairs of electrons arranged in a trigonal bipyramidal shape. A trigonal bipyramidal arrangement is shown in Fig. 1.1.

Fig. 1.1 Three different molecular arrangements of ClF 3 are possible. (i)

Draw clear diagrams of these three molecular arrangements, arrangements , each showing the five pairs of electrons. State which arrangement, if any, would result in a molecule with no dipole moment. [2] Arrangement Arrangement 1: 2 lone pairs occupying occupying axial axial positions Arrangement Arrangement 2: 2 lone pairs occupying occupying equatorial equatorial positions positions Arrangement Arrangement 3: 1 lone pair pair occupying equatorial position; 1 lone lone pair occupying axial position



(Credit: http://www.chemguide.co.uk/atoms http://www.chemguide.co.uk/atoms/bonding/shap /bonding/shapesclf3.gif) esclf3.gif) Arrangement Arrangement 1 would result in a molecule with no dipole moment (lone pairs occupying axial positions are linearly arranged with respect to each other, hence cancelling dipole moment; halogen atoms occupying equatorial positions are arranged in a trigonal planar manner with respect to each other, hence cancelling dipole moment).

(ii)

Apply the principles of the VSEPR theory to discuss the relative stabilities of these three possible arrangements. arrangements. [2] VSEPR theory states that lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which in turn is greater than bond pairbond pair repulsion. The more stable arrangement should be such that lone pair-lone pair/lone pair-bond pair/bond pair-bond pair repulsions are minimised (i.e. oriented as far as possible). If there are more than 4 electron pairs arranged around the central atom, repulsions at angles greater than 90° can be neglected. In arrangement 3, two lone pairs are at 90° to each other, whereas in the other two arrangements, they are at more than 90° to each other and can be neglected (arrangement 1: 180°, arrangement 2: 120°). Hence, arrangement arrangement 3 is the least stable with the greatest amount of repulsion. In arrangement 1, each lone pair is at 90° to 3 bond pairs. Given two lone pairs, there are a total of six lone pair-bond pair repulsions. (Bond pair-bond pair repulsions can be neglected as they are at 120° to each other.) In arrangement 2, each lone pair is at 90° to 2 bond pairs. Given two lone pairs, there are a total of four lone pair-bond pair repulsions. (Each lone pair is 120° to one equatorial fluorine atom and to each other, which can be neglected.) The equatorial bond pair is at 90° to each axial bond pair above and below the plane, hence there are two bond pair-bond pair repulsions. Since lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, arrangement arrangem ent 2 has the least repulsions and is the most stable.

(d)

When benzene is heated with BrCl and AlCl 3, a monohalogenobenzene, C 6H5X, is formed.


2017 H2 Chemistry Paper 3 (i)

Suggest whether the product C 6H5X is bromobenzene. bromobenzene. Give a reason for your answer.

chlorobenzene

or [1]

Bromobenzene. Bromobenzene. As chlorine is more electronegative than bromine, Br bears a partial positive charge in BrCl, hence acting as the electrophile involved in the electrophilic electrophilic substitution reaction.

(ii)

Draw the mechanism of the reaction, showing any intermediates. intermediates.

[2]

Electrophilic Electrophilic substitution Cl

Cl δ+ Br

δ

+

−

Br

Cl

+

Cl

–

Al

Cl

Al Cl

Cl

Cl H +

Br

slow

Br

+

:Cl

H Br

Cl

–

Al

Br Cl

fast

+

+

HCl + AlCl 3

Cl

(e)

Silver chloride is sparingly soluble in water AgCl (s) ⇌ Ag+(aq) + Cl −(aq) (i)

Write the expression for the solubility product, Ksp, of silver chloride, stating its units. [1] Ksp = Ag [Ag+(aq)][Cl−(aq)]

(ii)

units: mol2 dm−6

An excess of solid silver chloride was stirred with 0.50 mol dm −3 AgNO3 until equilibrium was established. Calculate [Cl [ Cl−(aq)] in the resulting solution, given the value of K of Ksp is 2.0 −10 × 10 . [1] [ Ag Ag+(aq)] = [ Ag Ag+(aq)] from AgNO 3 + [ Ag Ag+(aq)] from AgCl ≈ Ag [Ag+(aq)] from AgNO 3 (AgCl is sparingly soluble) −3 = 0.50 mol dm Hence, [Cl [Cl−(aq)] = Ksp / Ag [Ag+(aq)] ≈ 2.0 × 10−10 / 0.50 = 4.0 × 10−10 mol dm−3



(iii)

Describe and explain how the solubility of AgCl is affected by • •

adding NH3(aq), adding NaCl(aq).

[2]

When NH3(aq) is added, a ligand exchange reaction takes place between Ag +(aq) and NH 3(aq) to form the complex Ag(NH 3)2+(aq). This decreases [ Ag Ag+(aq)] and causes the position of equilibrium to shift to the right, hence increasing solubility of AgCl. When NaCl(aq) is added, [Cl [ Cl−(aq)] increases, causing the position of equilibrium to shift to the left, hence decreasing solubility of AgCl due to common ion effect.

[Total: 18]


2017 H2 Chemistry Paper 3 2

There are over 20 different binary compounds containing only the elements nitrogen and hydrogen. The most common of these is ammonia, but others include hydrazine, diazene, and hydrogen azide. Their formulae are shown in Table 2.1. Table 2.1 compound

(a)

(i)

molecular formula

ammonia

NH 3

hydrazine

N 2H4

diazene

N 2H2

hydrogen azide

HN 3

Hydrazine has a boiling point of 114°C and ammonia has a boiling point of −33°C. Suggest a reason for the high boiling point of hydrazine compared to ammonia. [1] Hydrazine has a more extensive intermolecular hydrogen bonding than ammonia. On average, each hydrazine molecule can form two hydrogen bonds with neighbouring hydrazine molecules; whereas each ammonia molecule can only form one hydrogen bond with a neighbouring neighbouring ammonia molecule. Hence, more energy is required to overcome the more extensive hydrogen bonding in hydrazine, resulting in higher boiling point compared to ammonia.

(ii)

Write an equation for the reaction between hydrazine and sulfuric acid, identifying the conjugate acid-base pairs in the reaction. [2] N2H4 + H2SO4 → N2H5+ base acid conjugate acid

(b)

+

HSO4− conjugate base

There are two isomers of diazene which can be isolated at low temperatures. Suggest a structure for diazene, using the usual valencies of hydrogen and nitrogen. Predict its bond angles and explain how the two isomers arise. [3] H –N=N –N=N –H –H Each nitrogen atom has two bond pairs and one lone pair, i.e. V-shaped, bond angle <120° Diazene can exhibit geometric (cis-trans) isomerism, isomerism, due to the presence of a double bond between the two nitrogen atoms (restricted rotation), and two different groups bonded to each nitrogen atom (a hydrogen atom and a lone pair).


2017 H2 Chemistry Paper 3 H

..

N

..

H

H

trans-diazene

(c)

N

N ..

..

N H

cis-diazene

When ammonia is bubbled through a sample of hydrogen azide dissolved in an organic solvent, white crystals of compound A compound A,, N4H4, are precipitated. Heating solid A solid A to to 133°C in a test-tube causes it t o sublime. Moist red litmus paper held at the mouth of the tube turns blue. A solution of A of A in in water is a good conductor of electricity. Suggest the formulae of the particles in the gaseous mixture formed when solid A sublimes A sublimes and in a solution of A in water. [3] Sublimation: Sublimation: N4H4(s) →

NH3(g) + HN3(g) turns moist red litmus paper blue

Dissolution: N4H4(aq) → NH4+(aq) + N3−(aq)

(d)

Sodium azide, NaN 3, is used in some airbags. When NaN 3 is ignited, it undergoes rapid decomposition into nitrogen gas and sodium metal. All the sodium is then removed by reaction with an excess of sodium nitrate, NaNO 3, in the airbag. This produces sodium oxide, Na 2O, and more nitrogen gas. (i)

Construct a balanced equation for the overall process forming Na 2O and N2 from NaN3 and NaNO 3 in the airbag. [1] 5NaN3 + NaNO3 → 3 3Na Na2O + 8N2

(ii)

When 400 g of NaN 3 is ignited, it inflates an empty airbag to a volume of 100 dm3. Calculate the final pressure in the airbag, at a temperature of 298 K. [3] No. of moles of NaN 3 = 400 / (23 + 3 × 14) = 6.1539 mol No. of moles of N 2 = 8/5 × 6.1539 = 9.8462 mol pV = pV = nRT p = nRT/ nRT/V = (9.8462)(8.314)(298)/(100 (9.8462)(8.314)(298)/(100 ×10 −3) = 2.44 ×105 Pa

(e)

Hydrogen azide is used as a reagent in the Schmidt reaction, which inserts an –NH –NH – group – group into the C – C –C C bond adjacent to a carbonyl group. R –CO –CO –R’ + HN3 → R – –NH NH – –CO CO –R’ + N2


2017 H2 Chemistry Paper 3 (i)

Name the functional group that is formed during this reaction.

[1]

Secondary amide

The 2-stage synthesis of compound D from compound B , shown in Fig. 2.1, makes use of this reaction. O NH

step 2

step 1

O

C

+HN3 +

B

– N O O

D

Fig. 2.1 (ii)

Suggest a structure for intermediate C, and reagents and conditions for step 1. [2] O

+

– N O

O

C

Step 1: concentrated nitric acid, concentrated sulfuric acid, heat

Compound B can be made from benzene by the synthesis shown in Fig. 2.2. Cl

step 3

step 4

O

step 5

E

step 6

heat with K2Cr 2O7 +

B

dil. H2SO4

Fig. 2.2 (iii)

Suggest reagents and conditions for step 3 and for step 4. Step 3: CH3CH2Cl, AlCl3, heat Step 4: limited Cl 2(g), ultraviolet light


[2]

2017 H2 Chemistry Paper 3 (iv)

Suggest reagents and conditions for step 5, and the structure of intermediate E. [2] Step 5: NaOH(aq), heat OH

E

Compound D can be converted into compound F. NH

+

+

– N O

NH2

O

– N

O

O

O

F

D

Fig. 2.3 (v)

Suggest reagents and conditions for this reaction.

[1]

NaOH(aq), heat

(vi)

Describe and explain how the basicity of D would compare to that of F. [2] D is neutral whereas F is basic. This is because the phenylamine functional group in F has a lone pair available for donation in an acidbase reaction. However, the lone pair on the nitrogen atom in the amide group of D is resonance-stabilised in the amide bond, making it less available for donation in an acid-base reaction.

[Total: 23]



(a) (a)

State tw o chemical or physical properties of copper, apart from the colour of the metal, in which it differs from calcium. Explain the reasons for those differences.

[2]

Copper can exhibit multiple oxidation states, 0, +1 and +2, whereas calcium can only exhibit two oxidation states, 0 and +2. This is because the 4s orbital and 3d orbital in copper have very similar energies and there is not a huge jump in the amount amount of energy in removing a second second electron compared compared with 2+ + the first in forming Cu from Cu . However, for calcium, there is a significant jump in the amount amount of energy energy in removing a third electron compared compared with with the 3+ 2+ second in forming Ca from Ca , hence this is not energetically feasible. feasible. Copper has higher melting and boiling points than calcium because of stronger metallic bonding in copper as both 4s and 3d electrons are delocalised into the sea of electrons and involved in the metallic bonding due to similar energies of 4s orbitals and 3d orbitals; whereas only the 4s electrons are involved in metallic bonding in the case of calcium.

(b)

Aqueous copper(II) ions, Cu 2+(aq), are blue, whereas aqueous zinc(II) ions, Zn2+(aq), are colourless. (i)

Explain why Cu 2+(aq) are coloured, whereas Zn 2+(aq) ions are not. [4] When ligands bond with a transition metal ion, repulsion between the electrons in the ligands and electrons in the d orbitals of the transition metal ion raises the energy of the d orbitals, causing the d orbitals to split into two groups, one group with higher energy and the other group with lower energy. When light passes through a solution of the transition metal ion, some of the light is used to promote an electron from the lower energy set of orbitals to the higher energy set of orbitals. This is known as d-d transition. The particular wavelength which corresponds to the energy gap is absorbed, and the complementary complementary colour is observed. Cu2+(aq) has partially filled 3d orbitals, hence d-d transition is possible, resulting in a coloured solution. However, Zn 2+(aq) has completely filled 3d orbitals, making d-d transition impossible, impossible, hence resulting in a colourless solution.

(ii)

Explain the meaning of the terms ligand and ligand and complex, complex, using the reaction 2+ between Cu (aq) ions and ammonia to illustrate your answer. [3] In the reaction between Cu 2+(aq) and ammonia, ammonia acts as a ligand (i.e. a species with at least one lone pair of electrons to form a coordinate bond with the central metal ion) and undergoes a ligandexchange reaction with Cu 2+(aq) as follows: [Cu(H2O)6]2+(aq) + 4NH 3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)


2017 H2 Chemistry Paper 3 The resultant [Cu(NH 3)4(H2O)2]2+(aq) is a complex, i.e. a metal ion at its centre with a number of other molecules/ions (ligands) surrounding it.

(c)

An acidified solution, G, of a vanadium compound was divided into two portions. A 25.0 cm3 sample of one portion required 16.4 cm 3 of 2.00 × 10 −3 mol dm −3 KMnO4 solution for complete oxidation of the compound to the dioxovanadium(V) dioxovanadium(V) ion, VO 2+(aq). An excess of powdered zinc, E° (Zn2+/Zn) = −0.76 V, was added to the other portion of G. The mixture was stirred until the reaction was complete, and then filtered. The filtrate was solution H. A 25.0 cm3 sample of solution H required 24.6 cm 3 of 2.00 × 10−3 mol dm−3 KMnO4 for complete oxidation to the dioxovanadium(V) dioxovanadium(V) ion, VO 2+(aq). Deduce the oxidation states of vanadium in solution G and in solution H, and hence calculate the concentration of vanadium ions in solution G. [3] 1st portion (oxidation from higher oxidation state to +5): No. of moles of MnO 4− required = 16.4/1000 × 2.00 × 10 −3 = 3.28 × 10−5 mol No. of moles of e − involved in redox = 5 × 3.28 × 10 −5 = 1.64 × 10 −4 mol 2nd portion (oxidation from lower oxidation state to +5): No. of moles of MnO 4− required = 24.6/1000 × 2.00 × 10 −3 = 4.92 × 10−5 mol No. of moles of e − involved in redox = 5 × 4.92 × 10 −5 = 2.46 × 10 −4 mol Ratio of no. of moles of e − in 1st portion to 2 nd portion = 2 : 3 Hence, it can be deduced that oxidation state in G is +3 and oxidation state in H is +2. No. of moles of vanadium ions in 25.0 cm 3 of solution G = 1.64 × 10−4 / 2 = 0.82 × 10 −4 mol Concentration of vanadium ions in solution G = 0.82 × 10−4 / 0.0250 = 3.28 × 10 −3 mol dm−3

(d) (d)

Compound I, C6H8O2, decolourises bromine water, but does not react with sodium metal or alkaline aqueous iodine. A solution of I is optically active. When heated with H 2SO4(aq) it forms J , C6H10O3, which on heating with acidified KMnO 4 produces K , C3H4O3, as the only only product. K reacts with sodium metal and also with alkaline aqueous iodine. Suggest structures for I, J and K and explain the reactions described.

[7]

Compound I Decolourises bromine water  electrophilic addition of alkene or electrophilic electrophilic substitution of phenol Does not react with sodium metal or alkaline aqueous iodine  not alcohol, phenol, acid, methyl alcohol, methyl ketone


2017 H2 Chemistry Paper 3 Heated with H 2SO4(aq)  molecular formula increases by +H 2O, i.e. hydrolysis of ester, since only one hydrolysis product, I is a cyclic ester Optically active  presence of chiral carbon (marked with *) Compound J J is a hydrolysis product, product, i.e. secondary alcohol alcohol and carboxylic acid Reacts with KMnO 4 to produce K  oxidative cleavage of alkene; oxidation of secondary alcohol to ketone Compound K Reacts with sodium metal  acid-metal (redox) reaction of carboxylic acid or tertiary alcohol (cannot be primary or secondary alcohol due to oxidation with KMnO4) Reacts with alkaline aqueous iodine  triiodomethane reaction of methyl ketone

*

OH O

O

OH

OH

O

I

O

J

O

K

[Total: 19]


2017 H2 Chemistry Paper 3 Section Section B Answer one question from this section.

4

(a)

World supplies of methane (natural gas) are diminishing. Methods are being developed to produce methanol from the anaerobic fermentation of waste organic matter. The fermenting bacteria gain energy by metabolising organic compounds through a series of disproportionation reactions, which form methane, CH 4, and carbon dioxide, CO 2, only. (i)

Calculate the average oxidation number of carbon in butanoic acid, CH3CH2CH2CO2H. [1] oxidation number of H = 1 oxidation number number of O = −2 average oxidation number of C = x oxidation number of butanoic acid = sum of oxidation numbers 4x + 8(1) + 2(−2) = 0 4x = −4 x = −1

(ii)

Construct a balanced equation for the disproportionation disproportionation reaction of aqueous aqueous butanoic acid, CH 3CH2CH2CO2H, to produce CH 4 and CO 2 only, by using oxidation numbers or otherwise. [1] [R] CH3CH2CH2CO2H + 12H + + 12e → 4CH4 + 2H2O [O] CH3CH2CH2CO2H + 6H2O → 4CO2 + 20H+ + 20e 20e Overall: CH3CH2CH2CO2H + H2O → 2.5CH4 + 1.5CO2

(iii)

×2.5 ×1.5

Suggest a method for removing the CO 2 from the gaseous product mixture in (a)(ii). (a)(ii). [1] Bubble the gaseous product through sodium hydroxide solution.

(iv)

A sample of the gaseous product mixture from the complete reaction in (a)(ii), (a)(ii), at an initial pressure of 1.5 × 10 5 Pa, had all of its CO 2 removed. Calculate the pressure of the remaining methane, assuming the temperature and volume remain constant. [1] At constant temperature temperature and volume, volume, p is directly proportional to n pressure of remaining methane methane = 5/8 × 1.5 × 10 5 = 9.38 × 104 Pa


2017 H2 Chemistry Paper 3 (b)

ΔH Hf °, are given Some relevant standard enthalpy change of formation values, Δ in Table 4.1.

Table 4.1

(i)

compound

ΔH Δ Hf °/kJ mol−1

CH3CH2CH2CO2H

−534

H2O

−286

CH4

−75

CO2

−394

ΔH Hr °, for the reaction represented by Calculate the enthalpy change, Δ (a)(ii), using the data in Table 4.1. your equation in (a)(ii), [2]

− (− −286) = +41.5 kJ mol −1 ΔH Δ Hr ° = 2.5(−75) + 1.5(−394) − (−534) − (

(ii)

The value of Δ ΔG Gr ° at 298 K for the same reaction is −207 kJ mol −1. ΔS Sr ° for the reaction, and explain its sign with reference to Calculate Δ your equation in (a)(ii). (a)(ii). [2]

ΔG ΔG = ΔH ΔH – T ΔS ΔS −207 = +41.5 − (298)ΔS (298)ΔS −1 ΔS ΔS = 834 J K mol−1 The sign of the entropy change of reaction is positive due to increase in the number of moles of gaseous products (from 0 to 4 mol) and hence an increase in disorder and the number of ways energy can be distributed.

(c)

Light of a longer wavelength is lower in energy than light of a shorter wavelength. Chlorine Chlorine and bromine react with alkanes in the presence of light. (i)

Outline the mechanism of the reaction between methane and bromine to form bromomethane. bromomethane. [4] Free radical substitution Initiation: Br 2 → 2Br ∙ Propagation: CH4 + Br ∙ → CH3∙ + HBr CH3∙ + Br 2 → CH3Br + Br∙ Termination: CH3∙ + Br ∙ → CH3Br 2CH3∙ → CH3CH3 2Br∙ → Br 2


2017 H2 Chemistry Paper 3 (ii)

Use your mechanism to explain why the bromination of methane can be carried out with light of a longer wavelength wavelengt h (yellow light is adequate) than that needed for the chlorination of methane (blue or ultraviolet light is required). [1] In free radical substitution, the first step (initiation) involves the –X bond. Since Br –Br –Br is longer than Cl –Cl, –Cl, homolytic fission of the X –X the orbital overlap is less effective and hence bond energy is weaker. Thus, light of a longer wavelength wavelength (with lower energy) is sufficient to break the bond to form the free radicals.

There are three possible monochloroalkanes that can be formed from 3ethylpentane.

+

+ Cl

Cl

3-ethylpentane

Cl L

M

N

Fig. 4.1 (iii)

Predict the relative proportions of L , M and N that are likely to be produced from 3-ethylpentane. 3-ethylpentane. Explain your answer. [2] Only one hydrogen atom (the one bonded to the central carbon) can be substituted to produce L . Six hydrogen atoms (the ones bonded to the penultimate carbon in each ethyl group – group –C CH2CH3) can be substituted to produce M. Nine hydrogen atoms (the ones bonded to the terminal carbon in each ethyl group – group –CH CH2CH3) can be substituted to produce N. Therefore, based on probability, the relative proportions of L , M and N are 1 : 6 : 9.

(iv)

When the chlorination is carried out and the products are analysed, it is found that the mole ratio of L : N formed is about 2 : 1. Suggest an explanation for the difference between this ratio and the one you gave in (c)(iii). (c)(iii) . [2] The radical intermediate produced in the formation of L is a tertiary radical, whereas the radical produced in the formation of N is a primary radical.


2017 H2 Chemistry Paper 3 A tertiary radical (an electron-poor species) is more stable than a primary radical due to more electron-donating electron-donating alkyl groups present (3 alkyl groups in tertiary radical versus 2 in primary radical). Hence, the formation of the tertiary radical to produce L is more thermodynamically thermodynamically favoured than the formation of the primary radical carbocation to produce N.

(d)

2-chlorobutanoyl 2-chlorobutanoyl chloride, P, can be obtained from butanoic acid by heating it with Cl2 and PCl5. CH3CH2CH2CO2H + Cl2 + PCl5 → CH3CH2CHClCOCl + HCl + POCl 3 P P can then be used to produce S, C6H15NO, via compounds Q and R, as shown in Fig. 4.2.

P

NaOH(aq)

Q

CH3CH2NH2

R

LiAlH4

room temperature

S

C6H15NO Fig. 4.2

Suggest structures for compounds Q, R and S in Fig. 4.2.

[3]

Q: CH3CH2CHClCO 2H R: CH3CH2CH(NHCH 2CH3)CO2H S: CH3CH2CH(NHCH 2CH3)CH2OH

[Total: 20]



(a)

Describe reactions that illustrate the variation in acid-base behaviour of the oxides of elements in Period 3 (sodium to sulfur), using at least three oxides as examples. Write equations for all reactions you describe.

[6]

The variation in acid-base behaviour of the oxides of elements in Period 3 changes from basic oxide in sodium/magnesium to amphoteric oxide in aluminium and to acidic oxide in silicon/phosphorus/sulfur. silicon/phosphorus/sulfur. Basic oxide (e.g. Na 2O): Na2O(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) Amphoteric oxide oxide (e.g. Al2O3): Al2O3(s) + HCl → AlCl3(aq) + H2O(l) Al2O3(s) + 2NaOH + 3H 2O → 2NaAl(OH)4(aq) Acidic oxide (e.g. (e.g. SO2): SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l)

(b)

Chlorine dioxide, ClO 2, is a yellow gas. Although it is unstable and explosive when pure, it is an important commercial chemical. Over a million tonnes are produced annually for use in the bleaching of wood pulp for paper-making and the sterilisation of water supplies. It can be made in the laboratory by the action of concentrated sulfuric acid on potassium chlorate(V), KClO 3. In addition to ClO 2, the products of the reaction include KClO 4, KHSO4 and H2O. (i)

State the changes in oxidation numbers that occur during this reaction and use them them to construct the balanced balanced equation for the reaction. [2] The oxidation number of Cl decreases from +5 in KClO 3 to +4 in ClO 2. The oxidation number of Cl increases from +5 in KClO 3 to +7 in KClO4. [R] ClO3− + 2H+ + e → ClO2 + H2O [O] ClO3− + H2O → ClO4− + 2H+ + 2e 3KClO3 + 2H + 2H2SO4 → 2ClO2 + KClO4 + 2KHSO4 + H2O

Chlorine dioxide readily decomposes decomposes to its elements. 2 O=Cl=O(g) → Cl2(g) + 2O2(g) (ii)

ΔH Δ H° = −204 kJ mol−1

Construct a suitable energy cycle using this equation and use data from the Data Booklet and Booklet and your cycle to calculate the Cl=O bond energy in ClO2. [2] BE(Cl –Cl) –Cl) = 244 kJ mol −1 BE(O=O) = 496 kJ mol −1


2017 H2 Chemistry Paper 3 ΔH° = 4BE(Cl=O) − 244 − 2(496) = ΔH 2(496) = −204 kJ mol−1 BE(Cl=O) = +258 kJ mol −1

(iii)

ΔG G° would compare to the value and Explain how the value and sign of Δ sign of Δ [2] ΔH H° for the decomposition of ClO 2.

The sign of the entropy change for the decomposition of ClO 2 is positive due to an increase in the number of moles of gaseous products in the reaction from 2 moles to 3 moles, leading to an increase in disorder and the number of ways energy can be distributed. Since ΔG ΔG = ΔH – T ΔS ΔS, and ΔH ΔH is negative, ΔG ΔG is negative for all temperatures (assuming ΔH Δ H is negative for all temperatures). Hence, the absolute value (magnitude) of ΔG Δ G is larger than that of ΔH ΔH.

(c)

The following scheme shows a synthesis of the beta-blocker drug, metoprolol. metoprolol . Study Fig. 5.1 carefully and answer the questions that follow. OH

OH

OH

OH

ClCOCH3Cl

CH3ONa

H2 + Pd

step 1

step 2

step 3 O

O

Cl

O

O

NaOH

O OH

N H

O

+U

Br OH

step 7 O

–

+

O Na

O

Br 2(aq)

+T

step 6

step 5

O

step 4

O

O

Fig. 5.1 (i)

State the types of reaction that reaction that occur during each of the steps 3 and 4. [2] Step 3: reduction/hydrogenation reduction/hydrogenation Step 4: acid-base reaction

(ii)

Name the mechanism of the reaction in each of the steps 1, 2 and 6. [3]


2017 H2 Chemistry Paper 3 Step 1: electrophilic electrophilic substitution Step 2: nucleophilic substitution, S N2 Step 6: electrophilic electrophilic addition

(iii)

Suggest the identity of the reagent T in step 5.

[1]

CH2ClCH=CH 2

(iv)

Suggest the identity of the reagent U in step 7.

[1]

NH2CH(CH3)2

(v)

Explain why step 6 forms the isomer shown in Fig. 5.1 and not no t the isomer shown in Fig. 5.2. [1] O

OH Br

O

Fig. 5.2 The carbocation intermediate produced in the formation of the isomer shown in Fig. 5.1 is secondary whereas that produced in the formation of the isomer shown in Fig. 5.2 is primary. A secondary carbocation is more stable due to the presence of one more alkyl group which is electron-donating and disperses the positive charge. Hence, the formation of the isomer shown in Fig. 5.1 is more thermodynamically thermodynamically favoured.

[Total: 20]


2017 H2 Chemistry Paper 3 Suggested Solutions

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