section b (30 marks) - a level tuitiona-leveltuition.com/wp-content/uploads/201… · web...

1 (a) Samples of 2-bromo-2-methylpropane were dissolved in dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with sodium hydroxide solution. Several experiments were carried out at constant temperature. The initial rate of reaction was determined in each case.

Calculate a value for the rate constant. Hence write the rate equation for the reaction. [3]

Expt [(CH3)3CBr]/ mol dm-3

[OH¯]/ mol dm-3

Rate/ mol dm-3 s-1

1 0.020 0.010 20.22 0.020 0.020 20.23 0.040 0.030 40.4

Compare Expt 1 and 2,Keeping [(CH3)3CBr] constant, but double [OH¯], rate was constant.Order of reaction wrt to [OH¯] is ZERO. [With Working: 1m]

Compare Expt 1 and 3,Triple [OH¯] will not affect rate. But double [(CH3)3CBr] and rate was double.Order of reaction wrt to [(CH3)3CBr] is ONE. [With Working: 1m]

Rate = k [(CH3)3CBr]From Expt 1 data, 20.2 = k (0.020) k = 1010 s-1 [1m]

Rate = 1010 [(CH3)3CBr]

(b) The hydrolysis of 2-bromopropane is now investigated. Samples were dissolved in dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with sodium hydroxide solution. The rate equation is found to be as follows:

Rate = 0.24 x 10-5 [CH(CH3)2Br] + 4.7 x 10-5 [CH(CH3)2Br] [OH¯]

(i) The rate equation obtained indicates that the hydrolysis of 2-bromopropane exhibits a mixture of both first order and second order kinetics. Suggest why this may be so. [2]

2-bromopropane is a 2º alkyl halide. It poses some steric hindrance for the nucleophile to approach bromoethane compared with a 1o RX, if the reaction were to proceed by SN2. [1m]

On the other hand, if it were to proceed by SN1, 2 o carbocation is not as stable compared with a 3º alkyl halide. [1m]

As a result, 2-bromopropane undergoes both SN1 and SN2 reaction pathways, simultaneously.

ACJC 2010 9647/03/Aug/2010 [Turn over

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(ii) By deriving an expression in terms of [OH¯], show that the % rate due to SN2

is [1]

= [1m] It is alright if 100% is omitted.

(iii) Using the expression derived, calculate the % rate due to SN2 for various [OH¯] of 0.01M, 0.1M and 1.0M. [1]

At [OH¯] = 0.01M, % rate = 16.4%At [OH¯] = 0.1M, % rate = 66.2% [1m]At [OH¯] = 1M, % rate = 95.1%

(iv) In light of your answer to (iii), state how the % rate due to SN2 depends on the concentration of [OH¯] and explain why it varies in that manner. [2]

SN2 is favoured by strong nucleophile in high concentration, whereas SN1 is independent of the type or concentration of the nucleophile. [1m] Hence as [OH¯] increases, SN2 is favoured and % rate due to SN2 increases. [1m]

(v) By comparing the magnitude of two rate constants in the rate equation given in (b), comment on the ease of hydrolysis by the two different pathways.

The Arrhenius Equation is given as follows: [2]

k = A e–Ea / RT

SN2 component is characterized by a larger rate constant than the SN1 component. [1m]

At the same temperature, a larger rate constant means a smaller Ea value. For this reaction, SN2 has a smaller Ea value than SN1. Hence reaction proceeds via SN2 more than SN1. [1m]

(vi) Draw a labelled Maxwell-Boltzmann Curve for this reaction to illustrate the

ACJC 2011 9647/03/Aug/11 Preliminary Examination

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distribution of energy. Based on your answer to (v) indicate the activation energies clearly on the diagram for the two different reaction pathways. [3]

Ea,SN2 < Ea,SN1 : [1m] Legend and Axis Labels: [1m] Shape of Curve: [1m]

(vii) On the same axes, draw the energy profile illustrating both the SN1 kinetics and the SN2 kinetics components observed for this reaction. Indicate the activation energies clearly on the diagram, for the two reaction pathways. [2]

Ea,SN2 < Ea,SN1 : [1m] Shape of Curve, Axis Labels : [1m]

Should be exothermic.ACJC 2010 9647/03/Aug/2010 [Turn over

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EaSN2CH(CH3)2Br + OH¯

CH(CH3)2OH + Br¯

(viii) Explain how the rate constant will change if CH(CH3)2Cl is used instead of CH(CH3)2Br. [2]

As C–Cl bond is stronger than C–Br, it is harder for the C–Cl bond to break. Since reaction involves breaking the C-Hal bond, the activation energies for forming the carbocation and transition state will be higher for both SN1 and SN2 pathways. Hence the rate constant will decrease. [2m]

No explanation: [0]

(ix) Draw the structures of two possible organic by-products that may be formed in this reaction, other than CH(CH3)2(OH). Briefly account for their formations.

[2]

Structures: [½m] each

Acid base reaction between ethanol and NaOH leads to the formation of a small amount of C2H5O ̄ which competes with OH¯ as a nucleophile . [½m]

OH¯ also brings about elimination as it also functions as a strong base. [½m]

[Total: 20 marks]

2 Oxalic acid is an organic compound with the formula H2C2O4. This colourless solid is a dicarboxylic acid. In terms of acid strength, it is about 3,000 times stronger than acetic acid. Its conjugate base, known as oxalate (C2O4

2−), is a reducing agent as well as a chelating agent for metal cations.

Oxalic acid dissociates in water according to the following equations

HOOC-COOH + H2O HOOC-COO- + H3O+ I Ka1= 5.6 x 10-2 mol dm-3

HOOC-COO- + H2O -OOC-COO- + H3O+ II Ka2 = 5.4 x 10-5 mol dm-3

(a) (i) Explain why the value of Ka1 is larger than Ka2. [1]

HOOC-COOH is a stronger acid than HOOC-COO- , it is more difficult to remove a H+ from HOOC-COO-, a negative ion. [1]

(ii) Write expressions for acid dissociation constants for equation I and II above. [2]

Ka1= Ka2 =


5

(b) (i) The neutralization between oxalic acid and sodium hydroxide corresponds to the two equations given

HOOC-COOH + NaOH → HOOC-COONa + H2O step 1

HOOC-COONa+ + NaOH→ NaOOC-COONa + H2O step 2

A 25 cm3 sample of oxalic acid of concentration 0.100 mol dm-3 was titrated with sodium hydroxide of concentration 0.100 mol dm-3.The titration curve for the above titration was given below.

At the first equivalent point, X, the species formed is HOOC-COO-(aq), which is both an acid and a base where the relevant equilibriums are: HOOC-COO-(aq) + H2O(l) H3O+(aq) + (COO)2

2-(aq) HOOC-COO-(aq) + H2O(l) OH-(aq) + (COOH)2 (aq) It can be shown that in such an instance that the [H+] at the first equivalent point, X, can be given by expression,

[H3O+] =

Using the expression, determine the pH value at point X. [1]

[H3O+] = = 1.73 x 10-3

pH =2.75

(ii) Calculate the pH value at the second equivalent point, Y, given that the [OH-] can be assumed to be entirely due to the hydrolysis: -OOC-COO- + H2O HOOC-COO- + OH- [3]


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25 50 Vol of NaOH/cm3

pH

X

Y

Kb2 = , = =

[1]

[-OOC-COO-] = 25/1000 x 0.1 (25+50)/1000 = 0.0333 mol dm-3 [1]

1.85 x 10-10 = , [OH-] = 2.48 x 10-6 mol dm-3 , pOH =5.61

pH =8.39 [1]

(c) The concentration of oxalic acid in a solution can be determined by an acid-base titration. State another way by which it’s concentration can be determined volumetrically. [1]

Another method will to titrate an acidified solution of the oxalic acid with potassium manganate (VII) through redox titration.

(d) Oxalate, the salt from oxalic acid, is able to combine chemically with certain metals commonly found in the human body, it is also able to bond chemically, behaving as bidentate chelating agents, to transition elements, such as iron to form complex ion. A chelating agent is a ligand that is attached to a central metal ion by bonds from two or more donor atoms.

[Fe(H2O)6]3+ + 3 C2O42- [Fe(C2O4)3]3- + 6H2O Kstab = 5.00 x 104

(i) What do you understand by the term “bidentate” ligand. [1]

bidentate ligand is ligand which can form two dative bonds with the central atom or ion [1]

(ii) Suggest a reason why the stability constant of the above is greater than 1. [1]

Bidentate or polydentate ligands, bind more strongly with multiple dative bonds and hence form more stable complexes than Fe(H2O)6

3+. [1]

(iii)

Draw the structure of the complex ion, [Fe(C2O4)3] 3-

State the type of bonds formed between the ligands and the metal ion.

Hence suggest the shape for this complex [3]


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[1]

Bond between ligand and metal: dative bonds [1]

Shape: octahedral [1]

(e)

(i)

Oxalic acid was one of the products formed when an aromatic organic compound, A, with molecular formula C10H10O2 undergoes oxidation with acidified manganate(VII) to form another organic product, B, with the molecular formula C8H8O2. No other organic compound was formed in the oxidation. Compound B reacts readily with 2 mole of Br2(aq) to form compound E, C8H6O2Br2. Compounds A and B are both soluble in NaOH and both A and B reacts with 2,4 - DNPH. Compound A reacts with acidified dichromate to give an acid, C, C10H10O3. Compound C reacts with SOCl2 to form a sweet-smelling compound D, C10H8O2.

Deduce the structures of A,B,C,D and E. Explain your deductions [7]

Compound A: Compound C :

:


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Compound B: Compound D:

Compound E

[1] for structure A and C, [1] for structure B [1] for structure D and [1] for structure E

Deductions: - A and B are soluble in NaOH A and B consists of carboxylic acids

or phenol/B reacts with 1 mole of Br2(aq) B contains phenol. [1]- A and B reacts with 2,4-DNPH A and B contains the carbonyl

functional group. [1]- A reacts with acidified dichromate to form an acid, A contains the

aldehyde. [1]

3 The lac repressor is a DNA-binding protein which inhibits formation of proteins involved in the metabolism of lactose in bacteria. It is active in the absence of lactose, ensuring that the bacterium only invests energy metabolism of lactose when lactose is present. When lactose becomes available, it is converted into allolactose, which inhibits the lac repressor's DNA binding ability.

Structurally, the lac repressor is a homotetramer made up of four identical polypeptides, each consisting of 347 amino acid residues.

(a) (i) State the highest level of protein structure in the lac repressor. Quarternary structure. [1]

(ii) One of the most common features in the secondary structure of proteins is the α-helix. Draw a diagram of the α-helix, showing the bonding which maintains the structure. [2]


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[1] for showing hydrogen bonding between amide groups in adjacent coils of the alpha helix

(b) (i) An 18 residue section from the lac repressor was digested using two types of enzymes. The enzymatic hydrolysis gave rise to these fragments.

Enzyme 1 Enzyme 2

Tyr-Trp-Leu-Val-ArgGlu-Glu-Met-Lys

Val-AspAsp-Asp-Phe

Gly-Trp-Leu-Ala

Gly-Trp-Leu-Ala-GluLys-Tyr-Trp-Leu

Val-Asp-AspGlu-MetVal-ArgAsp-Phe

Suggest the primary sequence of this 18 residue polypeptide.

Gly-Trp-Leu-Ala-Glu-Glu-Met-Lys-Tyr-Trp-Leu-Val-Arg-Val-Asp-Asp-Asp-Phe [1] Answer + [1] working

(ii) Another segment of tripeptide, Thr-Glu-Lys, was subjected briefly to acidic hydrolysis which produced individual amino acids as well as various peptides due to partial hydrolysis. The resulting mixture buffered at pH 8 was separated in an electric field using electrophoresis.


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Either or

Amino acid

Mr = 131 Mr = 147 Mr = 146Abbreviation Thr Glu Lys

pI 5.60 3.22 9.74

Match each of the three amino acids to the spots A, B and C.

A – Lys; B – Thr; C – Glu.([2] If all three correct; OR [1] If two are correct)

(iii)

Given that longer peptide strands show greater resistance when migrating through the electrophoresis plate, suggest a structure of the substance at spot D when the electrophoresis was buffered at pH 8.

[1](iv)

Threonine (Thr) can be used as a buffer at pH 7. Show with the aid of equations how it serves as a buffer at this pH. [7]When a small amount of strong acid is added,


Cathode

Anode

A

BC

D Start

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pKa = 2.09 pKa = 2.19 pKa = 2.18

pKa = 4.25

[1]

When a small amount of strong base is added,

[1](c) lac repressor can be purified using affinity chromatography. In this technique, the

natural substrate of the lac repressor, a specific DNA sequence, is covalently attached to a chromatographic support.

When a mixture of proteins is poured into the chromatographic column, lac repressor proteins bind specifically to its natural substrate, while other proteins are washed through. This step is pH sensitive and a suitable buffer is used to maintain optimal binding pH.(i) Suggest two types of bonding that could take place between the following

amino acid side chains of the lac repressor and DNA nucleotides.

DNA sequence Section of peptide sequence from lac repressor


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[1] for each of the following bonding types; Max [2])

Type of Bonds Fnc group on DNA nucleotide Fnc group on aa residues

Ionic bonding Phosphate group Amino side chain (of lysine)Van der Waals

forces Heterocyclic arene Benzene side chain (of phenylalanine)

Hydrogen bonding

O atom on phosphate, ribose grp or heterocyclic

areneHydroxyl H atom (of serine)

H atom on ribose grp or heterocyclic arene Hydroxyl O atom (of serine)

After washing away impurities, the target molecule is eluted (i.e. released from the chromatographic column) using one of a variety of methods, such as changing the pH within the chromatographic column. This weakens the protein binding and causes pure lac repressor protein to be washed through.

(ii) For a protein that binds fully to the covalently bonded substrate, determine the pH for 70% of the protein to be eluted given that

Protein-Substrate complex (aq) + H+ (aq) Protein-H+ (aq) + Substrate (s) Kb = 8.81 x 105 mol-1 dm3

Kb =

For = 0.7,

Kb = 0.7 / [H+][H+] = 7.94 x 10-7 mol dm-3 [1]pH = -lg (7.94 x 10-7) = 6.10 [1]

(iii)

When pH in the chromatographic column is adjusted to elute the purified protein, care must be taken to maintain very mild pH conditions.

Suggest how the protein structure is affected when pH for elution is too high (e.g. pH 12) and explain your answer using lysine (Lys) and glutamic acid (Glu).

Denaturation of the protein occurs, affecting ionic bonding in tertiary and quaternary structure [1].Side chain of lysine becomes uncharged at pH 12, hence salt bridges responsible for ter. and quat. structure are disrupted [1].

(iv)

Uncoiling of the protein structure is favoured when temperature is higher than physiological conditions. With the aid of the equation,

ΔG = ΔH – TΔSsuggest why this process is favoured at higher temperature such as 70 °C.

[8]Although ΔH is positive [1] when Van der Waals forces and hydrogen bonds are overcomed at 70 °C, this is offset by the increase in entropy (ΔS) [1] of the protein when it uncoils.


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(d)

Sulfinic acids are reactive species with structure similar to carboxylic acid. Unlike carboxylic acids, the analogous sulfinic acids are chiral.

Deduce the shape of the molecule using and draw the pair of optical isomers. [3]In sulphinic acid, S has 3 bond pairs and 1 lone pair. [1]S has a trigonal pyramidal shape. [1]

[1][Total: 20m]

4 A chemist design an ion-specific probe for measuring [Ag +] in a NaCl solution saturated with AgCl using the following set- up.

(a) (i) Given the following standard half reactions.

Hg2Cl2(s) +2e 2Hg(s) + 2Cl-(aq) Eøred = +0.24 V

Ag+(aq) + e Ag(s) Eøred = +0.80 V

Obtain an overall balanced equation, including state symbols for the cell above and state the direction of the flow of electron in the cell.

[2]

2Ag+(aq) + 2Cl-(aq) + 2 Hg(s) → Hg2Cl2(s) + 2Ag(s) [1] From Pt to Ag [1]

(ii) An engineer wish to use the probe to analyse an ore sample. After pretreating the ACJC 2010 9647/03/Aug/2010 [Turn over

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Salt BridgePaste of Hg2Cl2 in Hg

Saturated KCl solution

Pt wire

NaCl solution saturated with AgCl

Ag wire

voltmeter

sample, the chemist measured the cell voltage, Ecell as 0.53V.

The Nernst equation can be used to measure the concentration of silver ions using the probe,

Ecell = Eøcell – log10 K

Where n is the no of moles of electrons transferred in the overall reaction and K is the equilibrium constant for the overall equation in a(i).

Assuming the concentration of Cl- is so high that it is essentially constant, use the Nernst equation to calculate the concentration of silver ions in the ore sample [3]

K = 1/ [Ag+]2 since [Cl-] is high and effectively constant) [1] n = 2 [1]

0.53 = 0.56 – log10 1/ [Ag+]2 [Cl-]2

[Ag+] = 0.311 mol dm-3 [1]

(iii) Given that, AgCl(s) + e Ag(s) + Cl-(aq) Eø

red = +0.22 V

Using the half equation above and any other relevant data from the Data Booklet, derive the Eø

cell for the overall reaction: Ag+(aq) + Cl-(aq) AgCl(s) and use the equation given in a(ii), calculate a value of the Ksp for silver chloride at equilibrium, if Ecell of any process at equilibrium is 0V. [3]

Ag+(aq) + Cl-(aq) AgCl(s) Eøcell = +0.80 – 0.22 = 0.58V [1]

Given that for a process at equilibrium, the Ecell = 0V K = 1 / Ksp(AgCl) and n = 1 [1]

0 = 0.58 - log10 1/Ksp

1/Ksp = 6.27 x 109

Ksp = 1.59 x10-10 mol2dm-6 [1]

(b) (i) To prepare the saturated solution of AgCl and NaCl. AgCl is added to a 200 cm3 solution of NaCl of concentration of 1 x 10-3 mol dm-3. Using the Ksp value obtained in a(iii), predict the maximum mass of AgCl to be added to obtain a saturated solution. If you had not been able to obtain a value for the Ksp in a(iii), assume a value of 1.59 x 10-10 mol2 dm-6 [2]Ksp = [Ag+][Cl–] = 1.59 x 10–10

Let s be the solubility of AgCl in the given NaCl solution.(s )(s+ 1 x 10-3) = 1.59 x 10–10 Assume s << 1.0 x 10–3,


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s = 1.59 x 10–7 mol dm–3 [1]Maximum mass of AgCl that could dissolve = 1.59 x 10–7 x 0.200 x 143.5 = 4.56 x 10–6 g [1]

(ii) The concentration of chloride ions from silver chloride in (b)(i) is less than the square root of the Ksp value of silver chloride. Explain why this is so. [1]

AgCl(s) ⇌ Ag+(aq) + Cl–(aq)Cl-(aq) from NaClcaused the above equilibrium to shift left by Le Chatelier’s Principle. Thus, the solubility of AgCl is suppressed due to common ion effect. [1]

(iii) Suggest why when aqueous sodium thiosulfate was added to the resultingsolution in (b)(i), more silver chloride was able to dissolve. [3] AgCl(s) ⇌ Ag+(aq) + Cl–(aq) S2O3

2– forms a complex with Ag+. [1]This decreases [Ag+]. [1]Hence, the above equilibrium shifts right which causes more AgCl todissolve. [1]

(c) In 1999, researchers in Israel reported a new type of alkaline battery, called the “superiron” battery. This battery used the same anode reaction as an ordinary alkaline battery.

The overall equation for the cell was found to be:

2 FeO42-(aq) + 8H2O(l) + 3 Zn(s) → 3Zn(OH)2(s) + 2 Fe(OH)3(s) + 4 OH-(aq)

(i)Construct Construct the half-equations, with state symbols, for each electrode reaction in alkaline conditions. [2]

Anode: Zn(s)+2OH(aq) →Zn(OH)2(s) + 2e

Cathode: FeO42(aq) + 4H2O (l) + 3e → Fe(OH)3 (s) + 5OH(aq)

(ii) A “super-iron” battery should last longer than an ordinary alkaline battery of the same size and weight.

Calculate the quantity of charge released by the reduction of 10.0 g of K2FeO4 to Fe(OH)3.

[3]

Amount of FeO42 = = 0.0505 mol

Amount of e = 3(0.0505) molQ = 3(0.0505) 96500 = C (missing units 1)

(iii) Suggest a reason why the “super-iron” battery can last longer, given that for anormal alkaline battery, the reaction at the cathode is:


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2MnO2(s) + 2NH4+(aq) + 2e → 2NH3(aq) + 2MnO(OH) (s) [1]

Storage capacity of alkaline batteries are cathode limited and hencethe cathode in the “super iron” battery accepts 3 moles of electrons - more electrons than the normal alkaline battery. 1 FeO4

2 Ξ 3e vs 1 MnO2 Ξ 1e

5 (a) Boron is used in the manufacture of boron steel and boron carbide is used for shielding in nuclear reactors and as control rods for nuclear reactors. Boron reacts with hydrogen to form a series of hydrides. Upon thermal decomposition, the simplest of the hydrides when analysed, is found to comprise 78.3% B and 21.7% H.Find the empirical formula of this hydride.

[1]

Element B H% 73.8 21.7Ar 10.8 1.0

%/Ar 6.83 21.7Ratio 1 3

The empirical formula of this hydride is BH3.[1]

(b) This hydride actually exists in the form of a gaseous dimer (diborane). This dimer is ‘unusual’ in the way some of the hydrogen atoms are involved in bonding. The structure of diborane is similar to Al2Cl6 as shown below.

[8](i) Unlike Al2Cl6, what do you think is ‘unusual’ about this dimer?

H has only one electron but can form 2 covalent bonds. (3 center-2 electrons)[1]

(ii) Using the above data and any relevant data where applicable from the Data Booklet, construct an energy cycle to calculate the bond energy for the bridging Hb – B bond.

ΔHf B2H6(g) = +36.4 kJ mol-1ΔHf BH3(g) = +89.2 kJ mol-1

Bond energy of B – Ha = +340 kJ mol-1 ΔH atomisation B(s) = +563 kJ mol-1

2B(s) + 3H2(g) B2H6(g)


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2B(g) + 6H(g)

36.4 = 2(536) + 3(436) - 4x – 4(340)

Bond energy for (Hb – B) = 259.4 kJ mol -1 [1] for energy cycle[1] for working; [1] for answer

[3](iii)

In the light of your answer to (b)(ii) do you then expect the B – Ha bonds to be of the same length as that of B – Hb bond? Explain any difference that is expected.No. One electron on Hb is shared with 2 B whereas in Ha is shared with 1 B.

[1]

(iv)

Despite the unusual characteristics, BH3 tends to exist as a dimer instead of a monomer. Given the following data to calculate the ΔGo for the dimerisation, show that B2H6 is energetically more favourable than BH3.

So BH3(g) = 188.2 J mol-1 K-1

So B2H6(g) = 232.1 J mol-1 K-1

2 BH3 B2H6

ΔHor = -142 kJ mol-1

ΔSor = -144.3 J K-1

ΔGor = -142 – 298(-144.3/1000) = -99.0 kJ mol -1

[2]

(v) Would you expect AlH3 to be able to exist in a form similar to diborane? Explain.No. Even though aluminium is in the same group as boron, size of aluminium atom is bigger than boron. This makes the formation of 3 center-2 electrons not viable.ORYes. Aluminium is in the same group as boron. Since they are in the same group, they are likely to react in the same way.

[1]

(c) When diborane is reacted with ammonia, a nucleophilic substitution reaction occurs resulting in a compound, B2H12N2 which is found to have electrical conductivity in the molten state. Given that one mol of diborane reacted with 2 moles of ammonia, Suggest a possible structural formula for this compound.

[1] (d) Diborane is a powerful electrophilic reducing agent and is used in the reduction of certain

functional groups. It attacks sites with a high electron density such as N atom in cyanides and nitriles and O in carbonyl compounds. However it is a highly reactive gas which catches fire spontaneously in air. Thus its direct use is restricted. Hence many organic syntheses (involving reduction) make use of the safer solid hydrides as reducing reagents. A typical example is sodium borohydride, NaBH4, which is commonly used to reduce carbonyl compounds. However, it cannot be used in reduction of alkenes to alkanes. Suggest a reason for this.


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[1]NaBH4 provide H- which behaves as a nucleophile. Therefore, NaBH4 is not suitable for the reduction of alkenes to alkanes.

[1]

(e) In practice, R-X (where X = Cl or Br) can react with ammonia to form an alkylammonium salt where the amine formed in the reaction mixture will react with HX. An alkali has to be added to the mixture to liberate the free amine. The reaction scheme for the formation of free amine is as follows:

Consider the three organic halogen compounds, A, B and C shown belowCH3CH2Cl CH2=CHCl CH2=CHCH2Cl A B C

(i) State the type of hybridization of the carbon atom of the C-X bond for all three halogen compounds.A – sp3

B – sp2

C – sp3

[1](ii) Separate samples of A, B and C were warmed initially with aqueous ethanolic sliver

nitrate and left to stand. A white precipitate was expected to appear in one or more of the samples.

The difference in reactivity between the different organic halogen compounds and the nucleophile can be owed to the different strength of the C-X bond. With reference to your answer to (i), which sample will yield a precipitate and why?

Samples A and C will result in ppt [1]

More energy is needed to break the C-Cl bond in B where carbon is sp2

hybridised. This is due to the double bond character (overlapping of p orbital of Cl with π electron cloud) which causes C-Cl bond to be strengthened.

[1]

(iii)

A can be converted to a primary amine by reacting it with ammonia is in a sealed tube. However for a good yield of the amine, A must be used in limiting amount. However when ethanoyl chloride is reacted with ammonia to form the primary amide, a good yield is obtained even if the ethanoyl chloride is in excess. Explain.A must be used in limiting amount and ammonia must be used in excess for the formation of primary amine to prevent multiple alkylation and that will ensure good yield of amine.

[1]Excess ethanoyl chloride is not necessary for the formation of amide with ammonia as amide is not nucleophilic due to electron withdrawal by C=O group. [1]

[4]


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(f) Explain the following observations:

(i) Nitration of methylbenzene gives approximately equal proportions of the 1,2- and 1,4- isomers, but nitration under the same conditions of the compound C6H5C(CH3)3 gives 90% of the 1,4-isomer.

Although attack of the 2-position is twice as likely as the 4-position.Steric hindrance due to the methyl group result in a 1:1.

[1]The very bulky C(CH3)3 group in C6H5C(CH3)3 poses a greater steric hindrance to substitution at 2-position much further.

[1]

(ii) HCl (g) will add to the C=C bond in CH2=CH2 but not to the C=O bond in CH3COCH3.

Adding of HCl (g) to C=C bond is by electrophilic addition where H in HCl which has a partial positive charge is the electrophile.

[1]H-Cl is not nucleophilic (or a weak nucleophile) and hence will not have reaction with C=O bond in CH3COCH3.

[1]

[4]


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