suggested soln to foundation organic

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Practice Paper 7 Advo Education Group Pte Ltd

Suggested Solution to Foundation Organic ChemistryAlkanes

1. (a) A molecule can exhibit optical isomerism when it contains a chiral carbon atom which is bonded to four different groups of atoms, giving rise to non-superimposable mirror images. This pair of optical isomers/enantiomers rotate plane polarised light in opposite directions.

(b)

2. (a) (i)

(ii) (singly-branched and chiral)

CH3CH2CCH2CH2CH2CH2CH2CH3 OR CH3CH2C CH2CH2CH2CH3

A is Easy with ADVO!

Tel: (65) 6251 3359 / 8233 2753 www.advoedu.com 1 Goldhill Plaza #02-43 Singapore 308899

CH3C — C — CH3

CH3 CH3

CH3 CH3

2. (a) (iii) Equation for complete combustion of octane:

C8H18 + O2 8CO2 + 9H2O [1]

Amount of octane burnt = 100/114.0 = 0.877 mol

Amount of oxygen reacted = 25/2 0.877 = 11.0 mol [1]

Volume of oxygen reacted at r.t.p. = 11.0 24.0 = 263 dm 3 [1]

Since air contains 20% O2, 263 dm3 is only 20% of the vol. of air required for the combustion.

Volume of air at r.t.p. = 100/20 263 = 1.32 10 3 dm 3 [1]

(iv) Any ONE of the following:

Carbon dioxide formed from combustion contributes to e nhanced greenhouse effect resulting in global warming.

Carbon monoxide formed from incomplete combustion of octane combines with haemoglobin and prevents transportation of oxygen to all parts of the body.

Unburnt hydrocarbons formed from incomplete combustion of octane form photochemical smog /cause lung damage .

[1] for stating gas and source of gas

[1] for one environmental / health problem

(b) (i) Alkanes do not cause ozone depletion unlike CFCs. [1]

(ii) Alkanes are highly flammable. [1]

An asterisk is used to identify a chiral carbon.

[1]

*

H

CH3

[1]

H

CH2CH2CH3

*

Note:

CH3 —C — C — CH3

CH3 CH3

CH3 CH3

Not accepted

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3. (a) Free radical substitution [1] ;

Excess butane or limited Br2, ultra-violet (UV) light [1]

(b) X:

X exhibits optical isomerism as it has a chiral carbon atom

Y:

(c) Initiation step: (1)*

Br Br 2Br (2)

Propagation step: (1)*

Br + CH3CH2CH2CH3 CH3CHCH2CH3 + HBr (4)

CH3CHCH2CH3 + Br2 CH3CHCH2CH3 + Br (5)

Br

Termination step: (1)*

Br + Br Br2

CH3CHCH2CH3 + Br CH3CHCH2CH3

Br



*

H Br

H H

H H

H H [1]

H – C – C – C – C – H

Br H

H H

H H

H H [1]

UV

(6) (7)

Give any 2 possible equations

(3)

H – C – C – C – C – H

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CH3CHCH2CH3 + CH3CHCH2CH3 CH3CHCH2CH3

CH3CHCH2CH3

3. (d) Type of Stereoisomerism: Optical isomerism [1]

Diagrams:



7 () [3]5 – 6 () [2]2 – 3 () [1]1 () [0]

(1)*- labelling of all steps correctly

(2) - balanced eqn + half arrows + labelling of UV

(3) - correct radical formed with “” on correct C

(4) - correct balanced eqn

(5) - correct balanced eqn

(6) for each of any 2 possible (7) eqns, showing clearly the radicals involved.

C

CH2CH3

Br

H CH3

C

CH2CH3

BrH

CH3

() Non-superimposable mirror images

3 () [3]

() - correct structure and bond linkages

() - correct mirror image and bond linkages

Position of bonds must be drawn accurately. The followings are NOT acceptable: CH2CH3 CH3

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*H – C – C – C – C – H

H Br

H H

H H

H H

C

H – C – C – C – C – H

Br H

H H

H H

H H

X

Y


(e) X : Y = 2:3 [1]

Explanation:

Assuming that substitution is purely random, there are 4 possible H atoms (on 2 –CH2–) to be substituted to form X while there are 6 possible H atoms (on 2 –CH3 groups) to be substituted to form Y.

Ratio of X : Y

= 4 : 6

= 2 : 3

Note:

3. (f) Although molecule X has a chiral carbon atom, the reaction mixture contains an equal amount / equimolar / 50:50 mixture of both optical isomers (enantiomers) , a racemic mixture is formed. The rotating power to rotate the plane-polarised light of one isomer cancels that of the other. Thus, no optical activity is observed. [1]

Multiple-Choice Questions:

1. (Answer: D)

A Cl2 2Cl (represents the initiation step)

B CH3 + Cl CH3Cl (represents a termination step)

C CH3 + HCl CH3Cl + H (H radicals are not formed in the propagation step)

D CH3 + Cl2 CH3Cl + Cl (represents a propagation step between CH4 and Cl2)

2. (Answer: B)

Oxides of nitrogen are formed via the reaction of nitrogen with oxygen at high temperatures of the car engine. The high temperature is to provide the activation energy needed to break the strong NN in N2 and the strong O=O in O2.



Ha –C – C – C – C – Ha

Ha Hb

Ha Hb

Hb Ha

Hb Ha

Substitute a Hb

atom by a Br atom

Substitute a Ha

atom by a Br atom

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3. (Answer: D)

When methane reacts with chlorine in the presence of sunlight, free-radical substitution takes place whereby radicals such as Cl and CH3 are produced as intermediates.

A Radicals have high energy and are energetically unstable.

B Radicals have no charge.

C H radicals are not formed in the propagation step. Hence HCl cannot be produced in this way. HCl is formed from the reaction of CH4 + Cl CH3 + HCl and CH4 is the reacting molecule (reactant) not the intermediate.

D Radicals have unpaired electron they contain an odd number of electrons.

4. (Answer: D)

Free radical substitution is not selective (or Substitution process is a random). Both 2-chloropropane as well as 1-chloropropane would be formed from the substitution of respective hydrogen atoms by the chlorine atom. In addition, there is the possibility of multi-substitution to form further chlorine-polysubstituted products.

Due to a mixture of products will be formed, the yield of 2-chloropropane is low.

5. (Answer: C)

The more exothermic the ΔHr, the more stable the product and the more likely a reaction will occur.

Using ΔHr = +∑BE(broken) ─ ∑BE(formed)

Option Bonds formed Bonds broken ΔHr/ kJmol−−1

A C—H C—H 0

B C—C, H—Cl C—H, C—Cl −31

C C—C C—Cl −10

D C—C C—H +60

Thus, options A and D are eliminated. Single-step reaction (option C) is more favourable than multiple-steps reaction (option B). Hence, option C.

6. (Answer: D)



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Both have same Mr since they have the same molecular formula.

1. Both have simple molecular structure. Smaller amount of energy is required to overcome the weaker van der Waals’ forces between 2,2-dimethylpropane molecules than those between pentane molecules due to branching which results in a smaller surface area of contact between 2,2-dimethylpropane molecules.

Thus 2,2-dimethylpropane has a lower boiling point.

2. Covalent bonds in simple molecules are not broken during boiling.

3. Both are structural isomers (i.e same molecular formula but differ in the arrangement of carbon chain). Thus, both contain the same number of electrons per molecule.

7. (Answer: A)

Let R1 = CH3CH2 — and R2 = (CH3)2CH—

Possible alkane formed are R1— R1 ; R1— R2 ; R2— R2

1. Equiv to R2— R2



Alkenes

1. (a) (i) Reduction(ii) H2, Ni catalyst, 200 o C

(iii)

(b) (i) Electrophilic addition(ii) C l 2 in CC l 4/inert organic solvent, absence of light

(iii)

(c) (i) Electrophilic addition(ii) H I (g), room temperature



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(iii)

(d) (i) Electrophilic Addition(ii) Steam/H2O(g), H3PO4 catalyst, 300 o C , 65 atm

(iii)

(e) (i) (Mild) Oxidation(ii) COLD, alkaline/acidified KMnO4(aq)

(iii)

(i) (Strong) Oxidation(ii) acidified KMnO4(aq), HEAT under reflux

(iii)

(f) (i) Elimination(ii) Ethanolic NaOH/KOH, heat under reflux

(iii)



Apply Markovnikov's Rule for major product:+H of +HI (the atom bearing the partial positive charge) is added to the C of C=C with greater no. of H atoms directly attached to it. That will result in formation of a more stable carbocation.

Apply Markovnikov's Rule for major product.

Apply Zaitsev’s Rule for major product:

The major product is the most stable alkene, i.e. the alkene with the most number of substituents (e.g. alkyl group) attached to the carbon-carbon double bond.

R2C=CR2 > R2C=CHR > R2C=CH2 > RCH=CHR > RCH=CH2 > CH2=CH2

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2. (a) Reddish brown Br2 decolourises.

(b)

(c) Electrophilic addition

(d) (i) Geometric isomerism

2. (d) (ii) Optical isomerism

Non-superimposable mirror images

3. (a) 1-methylcyclohexene is an alkene while 1-methylcyclohexane is an alkane.

(Preferred test – simplest to carry out.)Test: Add Br2(aq) to each compound separately (in the absence of light).Obsn: For 1-methylcyclohexene, red-brown Br2(aq) decolourises but not for

1-methylcyclohexane.ORTest: Add acidified KMnO 4(aq) to each compound separately and heat.



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Obsn: Purple KMnO4 decolourises for 1-methylcyclohexene but not for 1-methylcyclohexane.

(b) Although propene and pent-2-ene are both alkenes, propene is a terminal alkene.

Test: Add acidified KMnO4 to each compound separately and heat. Pass any gas evolved through limewater.

Obsn: For both, purple KMnO4 decolourises.For propene, CO2 gas evolved forms a white ppt with limewater but not for pent-2-ene (no gas is evolved).

4.

Multiple-Choice Questions:



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5.

Ethene undergoes electrophilic addition with HBr. The organic intermediate is the CH3CH2

+

carbocation, which is an electrophile. Ans : C

6. Type of reaction: Elimination

Ans: D



Favorable as C = C formed are alternate with C – C (hyperconjugation) which helps to stabilise the compound

-+

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7.

Ans : D

8.

Ans: C



The carbocation (electrophile) formed is unstable and will be attacked by ANY nucleophile present in the mixture. In aqueous sodium nitrate, both H2O (majority) and NO3

– are nucleophiles, and the Br

–formed from the slow step also acts as a nucleophile. Thus all three nucleophiles (H2O, NO3

– and Br

–) attack the carbocation to form three different products.

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Arenes

1. (a) conc. HNO3, conc. H2SO4, 55 °C(b) Electrophilic substitution(c) HNO3 + H2SO4 NO2

+ + HSO4 + H2O

(1) – balanced equation for the generation of NO2+ electrophile

(2) – full arrow from –electron cloud of benzene ring to N of NO2+ electrophile

(3) – correct arenium ion with delocalisation of +ve charge over 5 sp2 C (not at the sp3 C)(4) – full arrow from C–H bond to the +ve charge of arenium ion(5) – correct product formed with balanced eqn (+ HSO4

on arrow) and regeneration of H2SO4 catalyst.

2. (a) (i) I: limited C l 2, UV lightII: Fe/FeC l 3/anhydrous A l C l 3, heat

(ii) I: Free radical substitutionII: Electrophilic substitution



(1)

(2)

(3)

(4)

(5)

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2. (b)

(c) If Fe is used as catalyst, Fe + Cl2 FeCl4 + Cl

+

If FeC l 3 is used as catalyst, FeCl3 + Cl2 FeCl4 + Cl

+

If A l C l 3 is used as catalyst, AlCl3 + Cl2 AlCl4 + Cl

+

(d)



Check!Draw displayed formula as requested by question: Show all bonds, trigonal planar shape wrt carboxyl C and bent shape wrt to O of OH.

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3. (a) AThis is because electron-donating CH3 group is 2,4-directing (i.e. direct the NO2

+ electrophile to the 2 or 4-position) and thus the 1,4-isomer will be one of the major products. For Route B, electron-withdrawing CO2H group (unsaturated) is 3,5-directing (i.e. direct the NO2

+

electrophile to the 3-position) and thus the 1,4 isomer will be the minor product.

(b) (i)

(ii)

Halogen Derivatives

1. (a) Reaction Reagents and Conditions Type of reactionI HBr(g) electrophilic addition

II limited Br2, UV light free radical substitutionIII PBr3, heat OR HBr, heat under reflux substitution

(b) Reaction Reagents and Conditions Type of reactionA ethanolic NaOH/KOH, heat under reflux eliminationB NaOH(aq)/KOH(aq), heat nucleophilic substitutionC NaCN/KCN, ethanol, heat under reflux nucleophilic substitutionD HC l (aq)/H 2SO4(aq), heat under reflux acid hydrolysis

(c) CH3CH2NH2



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1. (d) A primary alkyl halide usually undergoes SN2 mechanism unless otherwise stated (Both RX and nucleophile are involved in the rate determining step so this is a SINGLE STEP mechanism).

Reaction C

Check!(1) – on C and on Br of C–Br bond(2) – full arrow from lone pair on C atom of :CN– to C (to show “back–side attack”)(3) – full arrow from C–Br bond to Br

(4) – correct transition state showing the following: dotted lines to show partial bond formation of C–C bond and partial bond

breaking of C–Br bond square brackets with a net single NEGATIVE CHARGE

(5) – correct product (with inversion of stereochemistry) and balanced equation

Reaction E

Check!(1) – on C and on Br of C–Br bond(2) – full arrow from lone pair on N atom of :NH3

to C (to show “back–side attack”)(3) – full arrow from C–Br bond to Br

(4) – correct transition state showing the following: dotted lines to show partial bond formation of C–N bond and partial bond

breaking of C–Br bond square brackets with NO CHARGE

(5) – correct product (with inversion of stereochemistry) and balanced equation

2. (a) Since 3–bromo–3–methylhexane is a tertiary RX, hence it follows the SN1 mechanism (Only RX is involved in the rate determining step so the mechanism involves more than 1 step).

Reaction I



(1)(2)

(3)

(4) (5)

(1)(2)

(3)

(4) (5)

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Check!(1) – on C and on Br of C–Br bond(2) – full arrow from C–Br bond to Br

(3) – correct trigonal planar

of the carbocation + balanced equation

(4) – full arrow from lone pair on C atom of :OH– to C+;(5) – correct product (stereochemistry according to direction of OH– attack on C+)

and balanced equation.

Reaction II

Check!(1) – on C and on Br of C–Br bond(2) – full arrow from C–Br bond to Br



(3)

(5)

(4)

(2)

(2)

(1)

(5)(4)

(3)

(1)

(4)

(5)

(4)

(5)

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(3) – correct trigonal planar

of the carbocation + balanced equation

(4) – full arrow from lone pair on N atom of :NH3 to C+;(5) – correct product (stereochemistry according to direction of NH3 attack on C

and balanced equation (Note: Br on arrow and HBr as by-product).

2. (b) 3–bromo–3–methylhexane is a chiral alkyl halide. There is an equal probability for the

nucleophile :CN– or :NH3 to attack EITHER side of the trigonal planar of the carbocation intermediate to form a racemic mixture (OR equal amounts of optical isomers/enantiomers). The optical activity of the enantiomers cancel out one another and thus, the resulting mixture exhibits no net optical activity.

3. (a) Chloroethane is more readily hydrolysed than chlorobenzene.This is because p-p orbital overlap which results in the delocalisation of the lone pair of electrons on C l of chlorobenzene into the benzene ring , giving rise to partial double bond character in C-C l bond , thus strengthening the C-C l bond (i.e. difficult to break). Hence, chlorobenzene is resistant to nucleophilic attack.

(b) Reactivity towards hydrolysis: chloroethane < bromoethane < iodoethane

From Data Booklet:

E (C C l ) = +340 kJ mol 1 ; E (C Br) = +280 kJ mol 1 ; E (C I ) = +240 kJ mol 1

Since E (C C l ) > E (C Br) > E (C I ) , the C X bond strength decreases from C C l bond to C Br bond to C I bond . Hence the ease of breaking the C X bond increases from chloroethane to bromoethane to iodoethane.

OR

As atomic radius of X increases from C l to Br to I atom , C X bond length increases and thus the C X bond strength decreases from C C l bond to C Br bond to C I bond .Hence the ease of breaking the C X bond increases from chloroethane to bromoethane to iodoethane.

4. 1. Add NaOH(aq) to the compounds separately and heat.2. Add excess HNO3(aq) to remove excess NaOH.3. Add AgNO3(aq) to resulting mixture

For chlorobenzene, no ppt is formed.For chloroethane, white ppt of AgCl is formed.For iodoethane, yellow ppt of AgI is formed.



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5.

From molecular formula of M (C3H6O), M is CH3CH2CHO, an aldehyde () andhence L is CH3CH2CH2OH, a primary alcohol ().Therefore, K is CH3CH2CH2C l .

Explain your reasoning: Check you have all the ()s! State clearly the type of reaction a compound has undergone and what functional group or characteristic structure you can deduce a particular compound possess or do not possess (if no reaction).

Hydroxy Compounds

1. (a)

(b)

(c)

(d)

(e)

(f)

(g)



substitution

substitution

redox reaction

redox reaction

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1. (h)

(i)

(j)

(k)

(l)

(m)

(n)

(o)

(p)

(q)



acid-base reaction

oxidation

oxidation

oxidation

electrophilic substitution

electrophilic substitutionWhy phenol undergoes electrophilic substitution under milder conditions as compared to benzene?

electrophilic substitution

condensation

elimination

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2. (a) (i)

(ii)

(iii)

(iv)

(v)

(vi)

(a) (vii)

(b) Rxn with reagent (iii): white fumes of HC l evolved.Rxn with reagent (vi): orange K2Cr2O7 turns green.Rxn with reagent (vii): red-brown Br2(aq) decolourises and white ppt. is formed.



redox reaction

acid-base reaction

substitution

condensation

condensation

oxidation

electrophilic substitution of phenoland electrophilic addition of alkene

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3. (a)

(b) conc. H2SO4, heat under reflux

(c)

4. (a)

Test: Add neutral FeC l 3(aq) to each compound separately.Obsn: For phenol, violet colouration is observed but not for phenylmethanol.ORTest: Add aqueous Br2 to each compound separately.Obsn: For phenol, red-brown Br2(aq) decolourises but not for phenylmethanol.ORTest: Add PC l 5 to each compound separately.Obsn: For phenylmethanol, white fumes of HC l is formed but not for phenol.ORTest: Add acidified K2Cr 2O7(aq) to each compound separately and heat.Obsn: For phenylmethanol, orange K2Cr 2O7 turns green but not for phenol.The test of acidified KMnO4 with heating is not suitable as phenol may be oxidised by KMnO4

(oxidation of phenol is not in syllabus).(b)

Test: Add acidified KMnO4(aq) (OR acidified K2Cr2O7(aq)) to each compound separately and heat.

Obsn: For butan-1-ol, purple KMnO4 decolourises (or orange K2Cr2O7 turns green) but not for 2-methylpropan-2-ol.

(c)

Test: Add alkaline aqueous I 2 (or aqueous I 2 and aqueous NaOH) to each compound separately, and heat.

Obsn: For butan-2-ol, a pale yellow ppt. of CHI3 is formed but not for butan-1-ol.



Butan-2-ol has the characteristics structure of –CH(OH)CH3 which can be identified using iodoform test.

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5.

6. The p-p orbital overlap results in the d elocalisation of the lone pair of electrons on O atom into

the benzene ring, which disperses the negative charge and stabilises .

Hence, phenol is a stronger acid than ethanol.

Electronegative C l group is electron-withdrawing and thus disperses the negative charge and

stabilises .

Hence, 4-chlorophenol is a stronger acid than phenol.

Check! When explaining relative acidity,(1) – state the reason clearly (e.g. electron-donating group/ electron-withdrawing group/ p-p orbital

overlap)(2) – disperses or intensifies negative charge(3) – stabilises or destabilises <structure of anion>

Carbonyl Compounds



Check!Explain your reasoning. Make sure you have all the ().

(1)

(2) (3)

(1) (2)

(3)

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1. (a)

(b)

(c)

(d)

(e)

(f)

(e)

(f)



oxidation

reduction

nucleophilic addition

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2.

3. (a)



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3. (b)

(c)

4. Generation of nucleophile: HCN H+ + CN– (1)

Check!(1) – generation of nucleophile(2) – for both + on C and – on O (3) – full arrow from lone pair on C of CN– to C+ of C=O (check correct nucleophile)(4) – full arrow from C=O to O–

(5) – correct intermediate with negative charge on O(6) – balanced equation with HCN for fast step and CN– generated.

There is equal probability for the :CN – nucleophile to attack EITHER side (top or bottom) of the trigonal planar >C= of butanone, producing a racemic mixture/equal amount of both optical isomers.Note: Reject “…trigonal planar butanone…”. Only the shape wrt carbonyl C is trigonol planar but the shape wrt the each of the other carbon atoms is tetrahedral so the whole molecule is NOT planar.

5. (a) Pentan–3–one is a ketone and pentane is an alkane.



(6)(5)

(3)

(2) both partial charges

(4)

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Test: Add 2,4-DNPH to each compound separately.Obsn: For pentan–3–one, an orange ppt is formed but not for pentane.

(b) Pentan–2–one is a ketone with –COCH3 structural unit and pentan–3–one does not have the structural unit.Test: Add alkaline I 2(aq) (or aqueous I 2, aqueous NaOH) to each compound separately

and heat.Obsn: For pentan–2–one, a pale yellow ppt of CH I 3 is formed upon cooling but not for

pentan–3–one.

(c) Pentan-3-one is a ketone and pentanal is an aliphatic aldehyde.Test: Add Tollens’ reagent to each compound separately and heat.Obsn: For pentanal, a silver mirror is formed but not for pentan–3–one.ORTest: Add Fehling’s solution to each compound separately and heat.Obsn: For pentanal, a red-brown ppt is formed but not for pentan–3–one.ORTest: Add acidified KMnO4(aq) (OR acidified K2Cr2O7(aq)) to each compound separately

and heat.Obsn: For pentanal, purple KMnO4 decolourises (or orange K2Cr2O7 turns green) but not

for pentan-3-one.(d) Pentan-3-one is a ketone and pentan-3-ol is a secondary alcohol.

Test: Add 2,4-DNPH to each compound separately.Obsn: For pentan–3–one, an orange ppt is formed but not for pentan-3-ol.ORTest: Add PC l 5 to each compound separately.Obsn: For pentan–3–ol, white fumes of HCl is formed but not for pentan-3-one.

ORTest: Add acidified KMnO4(aq) (OR acidified K2Cr2O7(aq)) to each compound separately

and heat.Obsn: For pentan-3-ol, purple KMnO4 decolourises (or orange K2Cr2O7 turns green) but

not for pentan-3-one.



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6.

Carboxylic Acid & Derivatives



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1. (a) (i) I: acidified KMnO4(aq)/K2Cr2O7(aq), heat under refluxII: acidified KMnO4(aq)/K2Cr2O7(aq), heat under refluxIII: HC l (aq)/H 2SO4(aq), heat under reflux

(ii) I: oxidation II: oxidation III: acid hydrolysis(b) (i) LiA l H 4 in dry ether

(ii) A: CH3COC l B: CH3COOCH2CH3 C: CH3COO - Na + (iii)

IV: reductionV: substitutionVI: condensationVII: acid-carbonate reaction

2. (a) CH3COCl + H2O CH3COOH + HCl R

CH3COCl + NH3 CH3CONH2 + HCl

S

CH3COCl + CH3OH CH3COOCH3 + HCl

T(b)

(c) PC l 5, room temp. OR PC l 3, heat OR SOC l 2, heat(d) Ease of hydrolysis in decreasing order (easiest hardest): CH3COC l , CH 3CH2C l , C 6H5C l

C6H5Cl does not undergo hydrolysis because p-p orbital overlap results in delocalisation of lone pair of electrons on C l atom into the benzene ring , causes partial double bond character in C-C l bond and thus strengthening the C-C l bond , making it difficult to break.

CH3COCl is more readily hydrolysed than CH3CH2Cl because the c arbon in –COC l group is highly electron deficient as it is bonded to 2 electronegative atoms, O and C l as compared to CH3CH2Cl which has only one Cl bonded to its reactive carbon. Hence, the carbon in –COCl group is more susceptible to nucleophilic attack.

3. (a) secondary alcohol, carboxylic acid

(b) (i)

substitution

(ii)

substitution



Check!Question asks for displayed formula.

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3. (b) (iii)

redox reaction

(iv) acid-base reaction

(v)

oxidation

(vi) condensation

(vii)

condensation

4. (a)



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4. (b)

(c)

5. (a) (i) CH3CH2CO2H CH3CH2CO2─ + H+

CH3CH2CH2OH CH3CH2CH2O─ + H+

The p–p orbital overlap results in the delocalisation of lone pair of electrons on O

over the two O atoms in CH3CH2CO2. This disperses the negative charge and

stabilises CH3CH2CO2.

Whereas, the electron-donating ─CH2CH2CH3 group in propanol intensifies the negative charge and destabilises CH3CH2CH2O.

(ii) CH3CH2CO2H CH3CH2CO2─ + H+

CH3CH(CH3)CO2H CH3CH(CH3)CO2─ + H+

Electron-donating –CH 3/alkyl group intensifies the negative charge and destabilises CH3CH(CH3)COO − .Hence CH3CH(CH3)COOH is a weaker acid than CH3CH2COOH.



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5. (a) (iii) CH3CH2CO2H CH3CH2CO2─ + H+

CH3CH(OH)CO2H CH3CH(OH)CO2─ + H+

Electronegative O of –OH group is electron-withdrawing and thus disperses the negative charge and stabilises CH3CH(OH)CO2

.

Hence CH3CH(OH)COOH is a stronger acid than CH3CH2COOH.

(b) CH3CHClCO2H CH3CHClCO2─ + H+

CH2ClCH2CO2H CH2ClCH2CO2─ + H+

E lectronegative C l atom is further away from –COO in CH2ClCH2CO2 than in CH3CHClCO2

and thus electron-withdrawing effect of ─CH2CH2C l is weaker than ─CHC l CH 3. The negative charge on CH2C l CH 2CO2

─ is less dispersed and hence CH 2C l CH 2CO2─ is less

stable than the CH3CHClCO2─ anion.

Hence, 3-chloropropanoic acid is a weaker acid than 2-chloropropanoic acid.

6. (a) Test: Add Na2CO3(aq) or NaHCO3(aq) to each compound separately and pass any gas produced through limewater.

Obsn: For ethanoic acid, CO2 gas evolved form white ppt with limewater but not for phenol.

ORTest: Add PC l 5 to each compound separately.Obsn: For ethanoic acid, white fumes of HCl are given off but not for phenol.

ORTest: Add neutral FeC l 3(aq) to each compound separately.Obsn: For phenol, a violet colouration is observed but not for ethanoic acid.

ORTest: Add Br2(aq) to each compound separately.Obsn: For phenol, red-brown Br2(aq) decolourises but not for ethanoic acid .

(b) Test: 1. Add NaOH(aq) to each compound separately and heat.2. To the resulting mixture, add alkaline I 2 (aq) (OR I2(aq), NaOH(aq) ) and heat.

Obs: For CH3COOCH(CH3)CH2CH3, pale yellow ppt of CHI3 is formed upon cooling but not for CH3COOC(CH3)3.



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6. (c) Test: Add aqueous AgNO3 to each compound separately.

Obsn: For , white ppt. of AgCl is formed, but not for .

ORTest: Add 2,4-DNPH to each compound separately.

Obsn: For , orange ppt is formed, but not for .

ORTest: Add Tollens’ reagent, to each compound separately and heat.

Obsn: For , a silver mirror is formed, but not for .

7. (a) C H O % mass 35.8 4.5 59.7Amount/mol 35.812.0 = 2.98 4.51.0 = 4.5 59.716.0 = 3.73Mole ratio 1 1.5 1.25

4 6 5Empirical formula of R (diacid) is C4H6O5.

(b) R: diacid T H2(g) C4H6O5 C8H14O5

Orange ppt. V No oxidation

R:

V:

Organic Nitrogen Compounds



ethanol, conc.H2SO4,heat under reflux(condensation)

Na(s)(redox reaction) T: –OH group ()

acidified K2Cr2O7

(oxidation)

T is a primary or secondary alcohol

2,4-DNPH(condensation) ()

V is an aldehyde or ketone ()

Fehling’s orTollens’ reagent

V: not an aldehyde () Is a ketone () Since V is a ketone,

T is a secondary alcohol ()

()

()

() T is a diester or contains 2 ester groups ()

T:

C C C C

H OH

CH3CH2O H H OCH2CH3

OO

C C C C

H O

CH3CH2O H OCH2CH3

OO

C C C C

O H OH O

HO H H OH

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1. (a) (i) I: excess NH3, ethanol, heat in sealed tubeII: LiA l H 4 in dry ether OR H2, Ni, heat

(ii) I: Nucleophilic substitution II: Reduction(b) (i) CH3CH2NH3

+Cl

(ii) CH3COC l (iii)

III: Acid-base reactionIV: Condensation

2. (a)

I: conc HNO3, conc H2SO4, 55 °CII: 1. Sn, conc HC l , heat

2. NaOH(aq)(b) (i)

OR

Note: The following equation is not accepted as it only shows that phenylamine is a weak base. It does not show that phenylamine is acting as a base (i.e. react with an acid).

(ii) Relative basic strength in increasing order (weakest strongest): C , A , B

Compare B (i.e. aliphatic amine) versus A & C (i.e. aromatic amines):The p–p orbital overlap results in the delocalisation of the lone pair of electrons on N atom into the benzene ring, making the lone pair less available for protonation.Hence, A and C are weaker bases than B.

Compare A versus C:Electron–donating –CH3/alkyl group in A makes the lone pair of electrons on N atom more available for protonation. Hence, A is a stronger base than C.

(c)



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3. (a)

(b)

(c)

Proteins

4. (a) Thought Process involved :1) Assign pKa value to the respective –NH2/ –COOH group in glutamic acid.

-NH2 is a basic group, hence the pKa of its conjugate acid, -NH3+ is 9.5.

–COOH is stronger acid than side chain –COOH because of the electron–withdrawing N atom of –NH2 group is closer to the -COO– and hence, dispersing the negative charge of -COO– more, making -COO– more stable than side chain -COO–.

2)To determine the major species present at the respective pH values, check: pH > pKa predominantly exists as its conjugate base form; pH < pKa predominantly exists as its acidic form.

pH 1 pH 3pH < pKa of all 3 groups

predominantly acidic form

pH > pKa of –COOHpredominantly conjugate base form

pH 7 pH 11pH > pKa of –COOH AND side chain –COOH

predominantly conjugate base form

pH > pKa of all 3 groups predominantly conjugate base form

(b) Thought Process involved :Isoelectric point is the pH at which the amino acid exists only as zwitterion



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and thus will not migrate under the influence of an electric field.At pH > 2.1, the zwitterion is the major species present. At pH > 4.1, the net singly negative charged species is the major species present. Hence, the zwitterion is solely present in solution when pH of solution is between 2.1 and 4.1.

pI of glutamic acid is any value between 2.1 and 4.1 (e.g. 3.1).

5. Working:

arg–val–tyr

asp–arg–val

val–tyr–ile

ile–his–pro

pro–phe

asp–arg–val–tyr–ile–his–pro–phe

6. (a) peptide bond/linkage(b) HC l (aq)/H 2SO4(aq) of moderate concentration, prolonged heating under reflux

ORNaOH(aq)/KOH(aq) of moderate concentration, prolonged heating under refluxNote: Moderate concentration and prolonged heating are needed to break the numerous covalent bonds.

(c) During the reaction (i.e. hydrolysis), the amide bond is cleaved to form acidic –COOH and basic –NH2.If acid hydrolysis is employed, the three amino acids will exist as the following form in acidic medium:

H3N C

H

(CH2)4

C

O

OHH3N C

H

CH2

C

O

OH

OH

H3N C

H

CH2

C

O

OH

CO2HNH3

If alkaline hydrolysis is employed, the three amino acids will exist as the following form in alkaline medium:

6. (d) When a small amount of acid is added, H+ is removed as follows:



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When a small amount of base is added, OH is removed as follows:

7. (a) Primary structure of a protein refers to the sequence of the amino acid residues in a polypeptide chain.

(b)

Check! Draw “a section of a protein” NOT tripeptide, as requested in question.(c) –helix is a regular spiral configuration of the polypeptide chain, which is stabilised by

hydrogen bonds between the BACKBONE C=O group of one amino acid residue and the BACKBONE N–H group of the 4 th amino acid residue further down the chain .



Check!No H attached to N

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7. (c)Check! correct structure of –helix, showing:

C=O and N–H group of one peptide bond involved in the formation of hydrogen bonds

N and C must be on the polypeptide backbone

polypeptide chain must be labeled. at least 1 hydrogen bond correctly

drawn: on N and on H of N–H bond on O and on C of C=O dotted line between lone pair on O of C=O

and H of N–H bond clear labeling of hydrogen bonds

(reject “H–bond”)

(d) –pleated sheet consists of adjacent polypeptide strands stabilised by hydrogen bonds between the BACKBONE C=O group of one strand and the BACKBONE N–H group of the adjacent strand.

Check! correct structure of –pleated sheet,

showing: C=O and N–H group of one peptide bond

involved in the formation of hydrogen bonds

N and C must be on the polypeptide backbone

Strands of polypeptide chain must be labeled.

at least 1 hydrogen bond correctly drawn: on N and on H of N–H bond on O and on C of C=O dotted line between lone pair on O of C=O

and H of N–H bond clear labeling of hydrogen bonds

(reject “H–bond”)

8. (a) (i) Formed by oxidation of the SH groups in the cysteine residues:



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Check! Disulfide bridge is a covalent bond (solid line).(ii) Type of R

group interaction

Diagram illustrating R group interaction

ionic interaction

Note: When acidic –COOH comes close to basic –NH2, acid-base reaction will take place. –COOH will become –COO and –NH2 will become –NH3

+. These oppositely charged groups will form ionic interaction with one another (NOT hydrogen bond!).

hydrogen bond(between polar

R groups except between –NH2

and COOH)

Any one of the following:

van der Waals’ forces

(between non-polar alkyl R

groups)(iii) Lysine, serine, aspartic acid and asparagine. [1m]

Their side chains are hydrophilic and thus able to form hydrogen bond or ion–dipole interaction with water molecules surrounding the water–soluble globular protein.

8. (a) (iv) They can be found in the interior of the water–soluble globular protein. van der Waals’ forces occur between the residues with hydrophobic/non–polar R



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groups in the polypeptide chain.(b) (i) Denaturation results from the disruption of the interactions that maintain the shape

of a protein, leading to possible destruction of the secondary, tertiary and quaternary structures. This disruption alters the conformation/shape of the protein and causes the protein to lose its ability to perform its specific function.

(ii) Heavy metal ions, such as Ag+, form ionic interaction with –COO – which brings about the formation of insoluble protein salt (i.e. precipitation of protein). This disrupts the ionic interaction in the tertiary and quaternary structure s , resulting in the denaturation of the protein.

Heavy metal ions have a high affinity for sulfur and will bind tightly to – SH group of cysteine residue. This disrupts the disulfide bridges in tertiary and quaternary structure s , resulting in the denaturation of the protein.

At low pH, – COO – becomes – COOH while at high pH, – NH 3+ becomes – NH 2. This

disrupts the ionic interactions in the tertiary and quaternary structure s , resulting in the denaturation of the protein.

At high pH, – COOH becomes – COO – while at low pH, –NH2 becomes –NH3+. This

disrupts the hydrogen bonds in the tertiary and quaternary structures, resulting in the denaturation of the protein.

Note: pH changes WILL NOT disrupt hydrogen bonds in secondary structures as these hydrogen bonds are formed between C=O and N-H group of NEUTRAL amide linkage/ peptide bond.

(c) Extreme heat will disrupt the weak interactions such as the van der Waals’ forces in the tertiary and quaternary structures and the hydrogen bonds in the secondary, tertiary and quaternary structures of the protein.This alters the conformation/shape of the active site of the enzyme. As a result, substrate molecule cannot bind to the active site and hence the enzyme loses its catalytic activity, i.e. the enzyme is denatured.



Discuss the disruption of

each R group interaction separately.

suggested soln to foundation organic

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