OCR€¦ · Web viewEnergy Core organic chemistry This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook

WORKBOOK ANSWERSOCR AS/A-level Chemistry AEnergy Core organic chemistry

This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook. They are not exhaustive and other answers may be acceptable, but they are intended as a guide to give teachers and students feedback.

Module 3 Periodic table and energyPhysical chemistry

Enthalpy changesΔH of reaction, formation, combustion and neutralisation1

a

For correct shape showing enthalpy of products as lower than the reactants. Labelling axes correctly. Labelling ΔH and Ea with appropriate arrow heads.

b i ΔH = – 42 kJ mol–1

© John Older and Mike Smith 2015 Philip Allan for Hodder Education 1

b ii The activation energy is the minimum energy needed for a reaction to take place.

b iii ΔH = + 80 kJ mol–1

It is best to include the ‘+’ sign but it is not essential.

2

a i Enthalpy change of combustion is the enthalpy change when 1 mol of a substance is burnt completely, in an excess of oxygen.

If standard enthalpy change of combustion is asked then ‘under standard conditions’ must be added.

a ii

a iii Standard conditions are 298 K and 100/101kPa.

OCR expects the pressure to be given as 100 kPa.

b i Standard enthalpy change of formation is the enthalpy change when 1 mol of a substance is formed from its elements, in their natural state, under standard conditions of 298 K and 101/100kPa .

b ii C(s) + 1½H2(g) + ½Cl2(g) → CH3Cl(g) + for state symbols

The equation must be given such that 1 mol of CH3Cl is formed.

3 Heat produced (Q) = 50 × 4.18 × 2.40 = 501.6 J or = 0.5016 kJ

Amount in mol of CaCO3 (n = m/Mr) = 1.2/100.1 = 0.012 mol

H = –Q/n = 0.5016/0.012 = –41.8 kJ mol–1 + for the minus sign.

4

a The standard enthalpy change of neutralisation is the enthalpy change when 1 mol of water is formed in a reaction between an acid and a base under standard conditions of 298 K and 101kPa .

b NaOH(aq) + ½ H2SO4(aq) → ½ Na2SO4(aq) + H2O(l)

c i The aqueous sodium hydroxide contains 0.080 mol.

The hydrochloric acid contains 0.075 mol.

Therefore, the NaOH is in excess.

c ii Temperature rise is 5.5oC.

Heat produced (Q) is 150 × 4.18 × 5.5 = 3448.5J = 3.4485 kJ

150 g is the combined mass of the two solutions.

The temperature rise is caused by the 0.075 mol of HCl

H = –Q/n = 3.4485/0.075 = –45.98 (46) kJ mol–1 + for the minus sign

c iii The percentage error = (1/100) × 100 = 1%.


c iv Two readings of the thermometer are taken each with a maximum error of 0.1oC.

The percentage error is ((2 × 0.1)/5.5) × 100 = 3.6%.

c v Any two from:

Stirring the reactants

Insulating the beaker

Providing a lid to the beaker

Exam-style question1

a The standard enthalpy change of combustion is the enthalpy change when 1 mol of a substance is burned completely in an excess of oxygen, under standard conditions.

For 2 marks it may not be necessary but it would nevertheless be sensible to include ‘standard conditions of 298K and 100kPa’.

b i The mass of ethanol burnt = 230.71 – 230.34 = 0.37g

Mr of ethanol = 46 g mol–1

Amount in mol of ethanol burnt = 0.37/46 = 0.00804 mol

The number of decimal places has not been specified so a judgement is called for here. It would probably not particularly matter if you included more.

b ii The volume of water used is 150 g and the temperature rise is 5.9oC

The heat generated therefore is 150 × 4.18 × 5.9 = 3699.3 J

This heat is generated by burning 0.00804 mol of ethanol.

Therefore H = – 3699.3/0.00804 = – 459 913 J or – 460 kJ mol–1

for the sign and for numerical answer to 3 significant figures.

The answer for H in joules has been obtained by keeping all the figures for the mol of ethanol in the calculator which should always be done throughout a calculation in several steps.

b iii % error in thermometer = (0.4/5.9) × 100 = 6.8%


Bond enthalpies1

a Bond enthalpy is the enthalpy change required to break and separate 1 mol of bonds in the molecules of a gas so that the resulting gaseous (neutral) particles exert no forces upon each other.

b The listed figure is an average figure taken for many compounds that contain a C–Cl bond . The figure for CCl4 is specific to that compound although this will also be an average for the successive breaking of each C–Cl bond .

2 The enthalpy required to break the bonds in the reactants is 347 + 6(413) + 3½(498) = 4568 kJ

The enthalpy gained when the products are formed = 4(805) + 6(464) = 6004 kJ

H = 4568 – 6004 = –1436 kJ mol–1

The sign must be correct.


a Bond enthalpy is the enthalpy change required to break and separate 1 mol of bonds in the molecules of a gas so that the resulting gaseous (neutral) particles exert no forces upon each other.

The average bond enthalpy is the average for many bonds taken from a wide range of compounds containing the bond.

There will be many different acceptable ways of wording the answer to this question.

b i In the reaction one C–H bond and one C–Cl bond are broken. One C–Cl bond is then formed and also one H–Cl bond.

The enthalpy required to break the necessary bonds of the reactants = 413 + 243 = 656 kJ

The enthalpy gained when the products are formed is (C–Cl) + 432

The enthalpy change for the reaction is –122 kJ mol–1 so:

656 – [(C–Cl) + 432] = –122

Therefore, the C–Cl bond enthalpy = 656 – 432 + 122 = 346 kJ mol–1

Remember that bond enthalpies are always positive.


b ii In the reaction assume that the C=C bond is broken and then reformed as a C–C bond.

The enthalpy required to break the necessary bonds of the reactants = 612 + 432 + 4(413) = 2696 kJ

The enthalpy gained when the products are formed is (C–Cl) + 347 + 5(413) = 2412 + (C–Cl) kJ

The enthalpy change for the reaction is –69 kJ mol–1 so:

2696 – 2412 – (C–Cl) = –69

Therefore, the C–Cl bond enthalpy = 284 + 69 = 353 kJ mol–1

c The figures that are used are average bond enthalpies and these will differ from the actual enthalpies involved.

Hess’ law and enthalpy cycles1

a

1 mark for the correct cycle.

ΔfH = –2219 = 3(–394) + 4(–286)

ΔfH = –107 kJ mol–1

b i The chain length of the hydrocarbon increases and this means more C and H atoms are burned and release enthalpy.

b ii Allow anything between 4110 and 4175 kJ mol–1

As the carbon chain lengthens each hydrocarbon contains an extra ‘CH2’, so the same extra enthalpy is released.

Any wording that expresses the reason is acceptable.


2

a

1 mark for the correct cycle.

ΔrH + ΔfH(C2H4) = ΔfH(C2H4Cl2(l))

ΔrH + 52.2 = –165.2

ΔrH = –217.4 kJ

b Two points should be made:

The figures for any reaction only represent the relative stabilities and not how readily a chemical will react. This depends on the activation enthalpy.

Neither chemical reacts by being converted into its elements (i.e. carbon and hydrogen) so the enthalpies of formation do not relate to their reactions.

3

a i The heat released is 108.11 × 4.18 × 7.5 = 3389J = 3.389 kJ

It is also acceptable to use 100 g instead of 108.11 g.

a ii Molar mass of magnesium sulfate is 24.3 + 32.1 + 4(16) = 120.4 g mol–1

Therefore, enthalpy per mole = –(3.389 × 120.4)/8.11 = –50.3 kJ mol–1

b i The heat absorbed is 108.23 × 4.18 x (–1) = 452.4J = – 0.4524 kJ

It is also acceptable to use 100 g instead of 108.23 g.

b ii Molar mass of magnesium sulfate is 120.4 + 7(18) = 246.4 g mol–1

Therefore, enthalpy per mole = (0.4524 × 246.4)/8.23 = +13.54 kJ mol–1

c

ΔrH + ΔH2 = ΔH1

ΔrH + 13.54 = –50.3

ΔrH = –63.84 kJ mol–1



a i Bubbling (effervescence)

Magnesium ribbon disappears to produce a colourless solution

Remember you have been asked to write down what you would see.

a ii Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

a iii Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

b Marks are awarded for:

insulated container (could be a polystyrene cup)

thermometer

measure mass of magnesium + volume (or mass) of hydrochloric acid

add acid to the container and take initial temperature of hydrochloric acid

add the magnesium ribbon and stir the contents of the container

record the highest temperature reached

precautions to minimise heat loss such as surrounding container by an insulated jacket or including a lid to the container.

7 marks overall

c i Heat produced in the reaction = 100 × 4.18 × 4.2 = 1755.6 = 1.7556 kJ

ram of Mg = 24.3 so enthalpy change per mole = –1.7556 × (24.3/0.10) = –426.6 kJ mol–1

c ii Heat produced in the reaction = 100 × 4.18 × 2.7 = 1128.6 = 1.1286 kJ

rmm of MgO = 24.3 + 16 = 40.3

so enthalpy change per mole = –1.1286 × (40.3/1.00) = –45.48 kJ mol–1

c iii

ΔfH + ΔH2 = ΔH1

ΔfH – 45.48 = –426.6

ΔfH = –426.6 + 45.48 = –381 kJ mol–1


Heat losses in experiments of this type are considerable so that the figure here is much less exothermic than the accepted figure.

Reaction rates1

a The two marking points are:

a reaction occurs by collision of the reacting particles

at higher concentrations collisions between the reactants will be more frequent

b The two marking points are:

an increase in the pressure forces gaseous reacting particles closer together

collisions are therefore more frequent

2

a Initially the curve is steep indicating the reaction is proceeding quickly.

The reactants are then used up and the reaction slows down.

until after 4½ minutes (approx.) the reaction ceases because nothing remains of one or both components of the reaction

b A tangent is drawn on the curve at time = 2 minutes

The tangent is then calculated and can be marked according to the particular tangent drawn but it should be in the region of:

(mass loss)/time = 2 g/2 min = 1 g min–1

Accurate measurement form graph drawn

Gradient calculated and units given

Don’t forget the units.

This question is for practice only and cannot be answered to great precision.



a

Marks for part (a) are for labelled axes and appropriate use of graph paper

Appropriate use of graph paper means more than 1/2 the grid should be used in both directions.

Points plotted correctly

Appropriate curve plotted going through all the points

b The steeper curve at the start shows the reaction is proceeding faster but as the gradient of the curve gets less the reaction is slowing.

The reaction ceases when the hydrochloric acid has been used up.

Hydrochloric acid must be mentioned as the question specifies that the marble is in excess.

c A tangent must be drawn at time = 0

Correct measurements of x- and y-axes made to allow the gradient to be determined

Gradient calculated with correct units

Gradient = (loss of mass)/time

Answer should be close to 4.5 g min–1 but allow 4.1–4.6 g min–1

d The curve drawn on the graph paper should become less steep and become horizontal at 153.1 g

Showing that the mass loss is only half the 5.4 g lost in the original experiment as there is now only half the amount in mol of HCl.

e The mass of carbon dioxide produced = initial mass of experiment – final mass of experiment = 155.8 – 150.4 = 5.4 g

1 mol of carbon dioxide molecules = 44 g

Therefore, 5.4/44 mol are produced.


CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

2 mol of HCl produces 1 mol of CO2

So there are 2 × (5.4/44) mol of HCl is the 50 cm3 used.

The concentration of the HCl is therefore 2 × (5.4/44) × (1000/50) = 4.9 mol dm–3

Catalysts1

a A catalyst speeds up the rate of a reaction by lowering the activation energy.

The mass of the catalyst is unchanged as a result of the reaction.

b A heterogeneous catalyst is one that is in a different phase to the reactants.

A homogeneous catalyst is one that is in the same phase (gas, liquid or solid) as the reactants.

c i N2(g) + 3H2(g) → 2NH3(g)

There are other possible reactions.

Iron

c ii CH3OH(l) + CH3COOH(l) → CH3COOCH3(l) + H2O(l)

Other alcohols and organic acids can be used and there are other acceptable answers.

Concentrated sulfuric acid

d Two from:

Increased rate lowers costs

Lower energy requirements

Lower carbon dioxide emissions from fuels used to supply energy

2

1 mark for general shape, 1 mark for warmer sample having lower maximum value, both lines should be asymptotic to the x-axis



a i Cl + O3 → ClO + O2

ClO + O → Cl + O2

a ii The chlorine radical is regenerated by the second step.

a iii O3 + O → 2O2

b

1 mark for the shape of the graph, 1 mark for showing the catalyst gives a lower energy requirement, 1 mark for showing or explaining that the catalyst leads to more particles being able to react.

Chemical equilibriumDynamic equilibrium and le Chatelier’s principle1

a An equilibrium is a reaction in which reactants and products are constantly interchanging so it is dynamic , but overall there is no change to the concentrations of the reactants and products .

b i The hydrogen bromide initially breaks down at a fast rate to form hydrogen and bromine , the hydrogen and bromine start reacting to reform hydrogen bromide again .

The rate of the decomposition of the hydrogen bromide decreases as its concentration decreases and the rate of reaction of hydrogen and bromine increases as their concentration increases until these two rates become equal .

Other wording expressing these ideas is acceptable.

b ii An increase in the pressure pushes the reagents closer together resulting in more successful collisions therefore the reaction rate will increase and equilibrium will be achieved more quickly .

The volume of gas on both sides of the equation is the same , therefore le Chatelier’s principle indicates that there will be no change in the composition of the equilibrium mixture .


2

a CH4(g) + 2H2O(g) ⇌ CO2(g) + 4H2(g)

b The forward reaction is endothermic so le Chatelier’s principle indicates that the equilibrium will respond to try and minimise the effect of the increase in the temperature. This will mean more CO2(g) and H2(g) will be formed.

Exam-style questions1

a The enthalpy change of reaction is (2 × –110.5) – (–393.5) = +172.5 kJ mol–1

for numerical answer, for units.

b The reaction to form carbon monoxide is endothermic so increasing the temperature would give a higher concentration of carbon monoxide .

c There is an increase in volume as the reaction occurs from 1 volume of carbon dioxide to 2 volumes of carbon monoxide.

Carbon is a solid so this does not affect the volume.

Therefore, an increase in pressure favours the formation of more carbon dioxide.

2

a To speed up the formation of the sulfur trioxide

b i By le Chatelier’s principle more oxygen will move equilibrium to the right to reduce the amount of sulfur dioxide left unreacted and encourage the formation of the sulfur trioxide .

b ii Since the reaction to form sulfur trioxide is exothermic , le Chatelier’s principle indicates a low temperature would favour the formation of sulfur trioxide .

b iii 425°C is a compromise between a high yield at equilibrium and a fast rate of formation of the equilibrium .

b iv Since the volume falls as sulfur trioxide is formed , le Chatelier’s principle suggests a high pressure should be used .

b v 1½ times standard atmospheric pressure is chosen because this gives a sufficient yield and avoids the extra cost of using a higher pressure .

The equilibrium constant, Kc

1

a

b Kc = [CO2(g)] )

Remember that solids are not included in the equilibrium constant expression.



a H2(g) + I2(g) ⇌ 2HI(g)

b

Therefore, Kc = (0.89)2/(0.50 × 0.010) = 158

c If the equilibrium constant decreases as the temperature is raised more of the HI must convert to hydrogen and iodine.

Therefore, the extra iodine would cause the colour of the contents of the flask would become more purple.


Module 4 Core organic chemistryBasic concepts and hydrocarbons

Basic concepts of organic chemistry1

a CnH2n

b CnH2n+1Cl

c CnH2n+1OH

2

a

Compound Hexane 2-methylpentane Butane-1,4-diol

Displayed formula

Structural formula

CH3CH2CH2CH2CH2CH3

or

CH3(CH2)4CH3

CH3CH2CH2CH2CH(CH3)CH3

or

CH3(CH2)2CH(CH3)2

HOCH2CH2CH2CH2OH

or

HO(CH2)4OH

Skeletal formula

OHOH

b Hexane and 2-methylpentane have the same molecular formula but a different structure and are structural isomers of each other.


c i C4H8

c ii C4H9Cl

3

a homologous series — a group of organic chemicals that have the same functional group, the same general formula and each member differs from the next by CH2.

b functional group — an atom or a group of atoms that are responsible for the characteristic reactions of the compound.

c alkyl group — formed when a H has been lost/replaced to give CnH2n+1 such as methyl (CH3-), ethyl (C2H5-) and propyl (C3H7-).

d aliphatic compound — a compound containing C and H joined together in straight chains, branched chains or non-aromatic rings.

e aromatic compound — a compound that contains benzene ring.


f alicyclic compound — an aliphatic compound containing non-aromatic rings.

g saturated compound — a compound containing C–C single bonds only.

h unsaturated compound — a compound containing either C=C double bonds, CC triple bonds or an aromatic ring.

Alkanes4

for bonds — must include wedged and dashed bonds

correct bond angle, covalent () bonding

5

a

Boiling points: C7H16 98oC , C9H20 151oC and C11H24 196oC (allow +/– 5oC)

b As the C–C chain length increases there are more induced dipole–dipole interactions between the molecules hence more energy is needed to break the intermolecular forces .

c Unbranched alkanes have more induced dipole–induced dipole interactions than branched alkanes because unbranched alkanes can pack close together .


d i

d ii Hexane has highest boiling point.

d iii Either 2,2- or 2,3-dimethylbutane has the lowest boiling points.

6 The C–C and C–H bond enthalpies are high (so difficult to break). The bonds have very low polarity (so do not attract electrophiles or nucleophiles).

7

a C8H18 + 12½O2 8CO2 + 9H2O

b C8H18 + 8½O2 8CO + 9H2O

c C8H18 + 4½O2 8C + 9H2O

8 Potential formation of CO, which is poisonous/prevents haemoglobin from transporting oxygen around the body.

9

a i A radical an atom, group of atoms or molecule that has a single unpaired electron.

a ii Homolytic fission breaking of a covalent bond in which each atom in the bond gets one of the shared pair of electrons:

X―Y X∙ + Y∙

b i Initiation — the formation of a radical:

e.g. X―Y X∙ + Y∙

b ii Propagation — maintaining the radical concentration:

X∙ + CH4 HX + ∙CH3

b iii Termination — loss of radical concentration:

X∙ + ∙CH3 CH3X


10

a C2H6 + Br2 C2H5Br + HBr

Initiation Br2 Br∙ + Br∙

Propagation Br∙ + C2H6 ∙C2H5 + HBr

∙C2H5 + Br2 C2H5Br + Br∙

Termination — any one from:

C2H5∙ + ∙C2H5 C4H10 or

Br∙ + ∙C2H5 C2H5Br or

Br∙ + ∙Br Br2

b i Any of:

Any two correctly named for 4 marks

b ii Radicals are very reactive and multiple substitution occurs (as well as substitution at different positions on the C chain).


11

CH3 CH2 CH2 CH CH3

Cl

CH3 CH2 CH CH2 CH3

ClCH3 CH2 CH2 CH2 CH2 Cl

CH3 CH2 CH CH2 Cl

CH3

CH3 CH2 C CH3

CH3

ClCH3 CH CH CH3

CH3Cl

CH2 CH2 CH CH3

CH3

Cl

CH3 C CH2Cl

CH3

CH3

(a) pentane

(b) methylbutane 4 different compounds

(c) dimethylpropane only one possible compound

3 different compounds


a i

Boiling point/oC 50 60 69

Compound A C B

a ii Induced dipole–induced dipole (which used to be referred to as van der Waals forces)

b i radical — an atom/group of atoms/molecule that contains a single unpaired electron.

Just writing ‘a single unpaired electron’ will not get the mark.

substitution — when 1 atom (or group) is replaced by a different atom (or group).

Try to avoid using the word ‘substitute’ in your answer.

b ii a bromine radical Br2 2Br∙

C6H13Br Br∙ + C6H14 HBr + ∙C6H13

Make sure you form the alkyl radical and the ∙ is on the C and not on the H. C6H13∙ might be marked wrong.

∙C6H13 + Br2 C6H13Br + Br∙

dodecane, C12H26 ∙C6H13 + ∙C6H13 C12H26

c i compound A 2

These are difficult to work out and it is easier to try and work out the number of different H environments in each isomer.


c ii compound B 3

c iii compound C 5

2

a i saturated an organic compound that has only single C–C bonds

hydrocarbon a compound that contains carbon and hydrogen only

a ii Empirical formula:

C = 85.7/12 = 7.14 H = (100-85.7) = 14.3 simplest ratio is 1:2

Hence CH2

Mass of CH2 = 14,

Molar mass = 70 g mol-1 therefore need 70/14 = 5 empirical units hence molecular formula is C5H10.

An alternative method:

Mass of C is (85.7/100) x 70 = 60

Hence 5 C atoms therefore must be 10 H in the molecule

a iii

b i

1 mark for the displayed formula and 1 mark for the bond angle

This sketch makes cyclohexane look planar when it is in fact not. Each C has 4 bonded pairs of electrons around so the shape around each C is tetrahedral. Ask your teacher for a Molymod set and try building cyclohexene.

b ii The mechanism is radical substitution,

Conditions: ultraviolet light is required

The fission is homolytic fission


Mechanism

Initiation Cl2 Cl∙ + Cl∙

Propagation 1 Cl∙ + C6H12 HCl + ∙C6H11

Propagation 2 ∙C6H11 + Cl2 C6H11Cl + Cl∙

Termination any 2 free radicals reacting together such as Cl∙ + ∙C6H11 C6H11Cl

b iii There are four possible isomers. They are:

3

a i unsaturated: B

a ii alicyclic: A and C

a iii has the highest molar mass: E

a iv are isomers of each other: D and F

b i mol of CO2 = 1920/24000 = 0.08

Hence hydrocarbon must contain 8 carbons

Therefore, either C or E

mol of H2O = 1.62/18 = 0.09

Hence hydrocarbon must contain 18 hydrogens

Therefore, E

b ii C8H18 + 12½O2 8CO2 + 9H2O


Alkenes1 An organic compound that contains C=C double bonds.

2

3

a b c

Bond angle 109.5o 120o 120o

Shape around bond angle

Angular Trigonal planar Trigonal planar

4 Compounds that have the same molecular formula, the same structural formula but a different spatial arrangement. (3D arrangement)

5

a

but-1-ene E-but-2-ene Z-but-2-ene methylpropene cyclobutane methylcyclopropane

b A E

b B Z

b C Z


6

7

a In an unsymmetrical alkene the addent other than hydrogen will always go to the C (in the C=C) with the least number of Hs attached.

b i

b ii Butan-2-ol would be the major product because it is formed via a secondary carbocation which is more stable than the primary carbocation that forms butan-1-ol

8

a

Propene


b

C CCH3

HCH3

CH3

c

9

Two repeat units of the polymer monomer

Name: pent-2-ene

Name: cyclohexene


a Structural isomers are molecules with the same molecular formula but different structural formula.

b

CH3 C C CH3

CH3 Hany two from

c i Isomer 1

c ii Same molecular and structural formulae but different spatial arrangement

c iii


isomer 1 (Z) (E)

d 2-methylbutan-1-ol and 2-methylbutan-2-ol and 2-methylbutan-2-ol would be the major product because it is formed via a tertiary carbonium ion which is more stable than the primary carbonium ion that forms 2-methylbutanol-1-ol

The last 2 marks can also be gained by quoting Markownikoff’s rule and explaining which C in the C=C bond has most H attached

2

a Add Br2 which would be decolourised .

b i An electrophile is an electron pair acceptor which forms a dative covalent bond.

b ii They are addition reactions and only form one product, hence no waste.

b iii

b iv The intermediate carbonium ion is unstable and quickly reverts back to the alkene. A secondary carbonium ion is more stable than a primary and therefore more likely to exist long enough to react with the bromide ion.


Alcohols, haloalkanes and analysis

Alcohols1 A saturated hydrocarbon chain with general formula CnH2n+1.

2

Isomer CH3 CH CH2 CH2 CH3

OH CH3 C CH2

CH3

CH3

OH CH3 CH CH CH3

CH3

OH

CH3 CH2 CH2 CH2 CH2 OH

Name Pentan-2-ol 2,2-dimethylpropan-1-ol

3-methylbutan-2-ol Pentan-1-ol

Classification Secondary Primary Secondary Primary

IsomerCH3 CH2 CH CH2 CH3

OH

CH3 CH2 C CH3

OH

CH3

CH3 CH2 CH CH2 OH

CH3

CH3 CH CH2 CH2 OH

CH3

Name Pentan-3-ol 2-methylbutan-2-ol 2-methylbutan-1-ol 3-methylbutan-1-ol

Classification Secondary Tertiary Primary Primary

3

a i Electronegativity measures the attraction for the electrons in a covalent bond of each atom in that bond.

a ii A dipole is made of two small opposite charges as a result of differences in electronegativity.

b O–―H+

4 Propane — the C–H bonds are polar but the symmetry of the molecule means the molecule is non-polar (or has very low polarity) so the only intermolecular forces are induced dipole–dipole forces .

Ethanol — is polar and has both hydrogen bonds and induced dipole–dipole forces which have to be broken when ethanol is boiled so the boiling point is higher .


5

6 Ethanol can form H bonds with water.

7


8

a CH3CH(OH)CH3 + 4½O2 3CO2 + 4H2O

The most common error is to balance the oxygen but to forget about the O in the alcohol which results in the incorrect equation: CH3CH(OH)CH3 + 5O2 3CO2 + 4H2O).

b CH3CH2CH2OH + 2[O] CH3CH2COOH + H2O

Often incorrectly written as CH3CH2CH2OH + [O] CH3CH2COOH + H2 . It is worth remembering that H2 is so reactive it is very rarely formed as a product.

Observations colour change from orange to dark green

c Equation

9

1 mark for each curly arrow and 1 mark for each lone pair shown below.


a i The OH group is bonded to a C that is bonded to 2 other C atoms/ the OH group is bonded to a CH.

a ii

2 marks for the diagram:

The dipole on the O–H bond allows the formation of H bonds between neighbouring molecules and therefore restricts the ease that it can boil/vaporise/energy needed to break H bonds

b i Concentrated sulfuric acid [1]

b ii C4H10O C4H8 + H2O

Usually alcohols are not written using their molecular formula. The equation C4H9OH C4H8 + H2O would not be accepted in this case.


b iii Stereoisomers have the same molecular and structural formulae but have a different spatial/3D arrangement.

Structural isomers have the same molecular formula but a different structural formula.

b iv

A B C

2

a i C

a ii D and F

a iii C6H12O

a iv 3-ethylcycloheptanol

a v F

b i

b ii Colour change from orange to (dark) green

b iii

OHOH

+ H2O or + H2O

or C9H18O C9H16 + H2O


Haloalkanes1

a A nucleophile is an electron pair donor which forms a covalent/dative bond.

b

2

a One by one a precipitate would be formed in each

Test tubes 1 = white ppt, 2 = cream ppt 3 = yellow ppt

Test 3 ppt 1st and test 1 ppt last

b Ag+(aq) + Br–(aq) AgBr(s)

c C–I bond is the weakest bond, therefore easiest to break and hence the fastest .

3 CF3Cl Cl∙ + ∙CF3

∙Cl + O3 ∙ClO + O2

∙ClO + O ∙Cl + O2


a V is a bromoalkane. Molar mass = 136.9, Br = 79.9, therefore the rest of the molecule has a mass = 57

Hence contains 4 C (=48) and 9H, therefore C4H9Br

W is an alcohol

Doesn’t react with acidified dichromate therefore must be a tertiary alcohol

W must be 2-methylpropan-2-ol


b i

CH3 C

CH3

CH3

OHCH3 C

CH3

CH2

CH3 C

CH3

CH2

CH3 C

CH3

CH2

BrH

CH3 C

CH3

CH3

Br

+ H2O 1 mark for equation

W X

+ HBr

X

Z

Y

1 mark for X

1 mark for equation

1 mark for Y

1 mark for Z

b ii Z is 2-bromo-2-methylpropane because it would be formed via the more stable tertiary carbonium ion/carbocation .

Organic synthesis1 Reflux involves continuous evaporation /condensation such that volatile compounds

cannot escape whereas distillation is evaporation followed by condensation that enables the most volatile components to escape/distil out

2 Use a separating funnel to separate the organic layer from aqueous layer dry the organic layer by anhydrous salt such as CaCl2 , filter to remove the CaCl2(s) .

3 Add drops of bromine water to a sample of each. The one that is rapidly decolourised is the cyclohexene. To the remaining three warm a sample of each with AgNO3 dissolved in ethanol. The one that gives the yellow ppt is iodopropane. Reflux samples of the remaining two with acidified dichromate, the one that turns green is propan-1ol and the remaining one is hexane . The order in which the tests are carried out can be varied.

4 Heat propan-1-ol with sulfuric acid to produce propene. React propene with steam at high temp/pressure and an acid catalyst to obtain propan-2-ol (Markownikoff addition)

Step 1 CH3CH2CH2OH CH3CH=CH2 + H2O

Step 2 CH3CH=CH2 + H2O CH3CH(OH)CH3

Analytical techniques1 The earth absorbs UV energy from the sun and re-emits infrared radiation back away

from the earth’s surface. Gases such as CO2, H2O vapour and CH4 absorb the infrared causing their bonds to vibrate and then re-emit the infrared back to the earth’s surface which is thought to be linked to global warming .

2 The first one is propanone because it has a C=O absorption at about 1700 cm–1. The second one is propan-1-ol gas, as it has an O–H absorption at about 3000 cm–1.


The third one is propanoic acid as it has absorptions for the C=O at about 1700 cm–1 as well as broad O–H at about 3300 cm–1.

3 The molar masses from the three spectra from top to bottom are 60, 60 and 58 and therefore the bottom spectrum on page 46 is propanone, as the other two both have masses of 60. Propan-1-ol is a primary alcohol so you would expect a fragment ion at 31 due to CH2OH+ (the top spectrum on page 46) and propan-2-ol is a secondary alcohol so you would expect CH3CH(OH)+ at m/z = 45 (the middle spectrum) .


a Empirical formula of compound M

C, 64.9/12 = 5.41 H, 13.5% O, 21.6/16 = 1.35

5.41/1.35 = 4 13.5/1.35 = 10 1

Empirical formula = C4H10O

Molecular ion peak is m/z = 74 so molecular formula is also C4H10O

Compound N is a ketone peak at approx. 1700 cm–1. It cannot be an aldehyde as the alcohol was refluxed and would have been oxidised to a carboxylic acid or the mass spec has a peak at m/z = 45. Therefore, compound M must be a secondary alcohol.

Compound N must be butanone and compound M must be butan-2-ol .

b C4H10O C4H8 + H2O

You must use the molecular formula.

Compound M is butan-2-ol so can be dehydrated to form but-1-ene or but-2-ene .

But-2-ene has E and Z isomers.


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OCR€¦ · Web viewEnergy Core organic chemistry This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook