Organic Chemistry Revision Notes and Questions 2014

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Catholic Junior College H2 Chemistry 9647

Contents Page

1. Synthesis/Conversion questions 2

2. Structural Elucidation questions 11

3. Distinguishing tests 17

4. Explanation based questions 20

5. Data based questions 22

6. Answers 29

Name: ______

Class: 2T____

1. Synthesis/Conversion type Questions

A. To synthesize a compound with a change in functional group.

Method 1: Change the functional group by substitution. If the functional group cannot be substituted to give the desired compound in a single step, substitute the functional group into another functional group that can undergo substitution readily (e.g. halogenoalkane)

Examples:

1) CH3CH2Br → CH3CH2OH

2) CH3CH2OH → CH3CH2NH2

Method 2: Convert the functional group by oxidation/reduction if the starting compound cannot undergo substitution.

Example:

1) CH3COCH3 → CH3CH(Cl)CH3

B. To synthesize a compound with a change in position of functional group (that may or may not be the same as the starting reagent).

Step 1: Convert the starting reagent into an alkene by elimination of the functional group. Step 2: Add the new functional group (same or different) back to the alkene.

Examples:

1) CH3CH2CH2Cl → CH3CHClCH3

2) CH3 CH3

CH3CHCH2Cl CH3CCH3

C. To synthesize a compound with an increase in carbon chain length.

Method 1: Substitute the nitrile group into the compound. (i.e. The compound needs to have a halogen group already) The nitrile group can be further converted into amine or carboxylic acid group. [This method does not increase the number of functional groups in the compound.]

Examples:

1) CH3CH2Br → CH3CH2CO2H

2) CH3CH2Br → CH3CH2CH2NH2

3) CH3CH2OH → CH3CH2CH2NH2

Method 2: Add the nitrile group into the compound using HCN. (i.e. The compound needs to have a carbonyl group already) The product formed will contain a hydroxyl group adjacent to the nitrile group. The nitrile group can be further converted into amine or carboxylic acid functional group. [This method increases the number of functional groups in the compound by 1.]

Examples:

1) CH3CHO → CH3CH(OH)CN

2) CH3CHO → CH3CH(OH)CO2H

3) O OH

CH3 C CH3 CH3CCH3

CH2NH2

Method 3: Employ other reactions that will increase in number of carbon atoms by more than one. (e.g. Reaction of alcohol with carboxylic acid, phenol with acyl chloride, amine with acyl chloride)

Example:

1) CH3CHO → CH3CH(OH)CN

D. To synthesize a compound with a decrease in carbon chain length.

Method 1: Oxidise the alkene to rupture the C=C double bond.

Example:

1) CH3CH=CHCH3 → CH3CH2OH

Method 2: Oxidise the compound that contain CH3CH(OH)- group or the CH3CO- group using alkaline aqueous iodine. This reduces the number of carbon atoms by one.

Example:

1) CH3CH(OH)CH2CH3 → CH3COCl

Method 3: Hydrolysis of esters and amides.

Exercises (Synthesis) Below are some two step syntheses. In each case, identify the reagents and conditions for each step and also identify the intermediate compound. (a)

CH2CH2Cl A CHCH3

(b) CH3 CH3

C O B CH C Br 3 CH3 H

Cl HO OH

(d) CH CH3 CH3 3

C C D CH3 C OH

CH CH 3 3 H

(e) O H

NO2 E CH3 C N

(f) CH3 CH3 O

C C F CH3 C

CH CH OH 3 3

PRACTICE QUESTIONS- Conversion/synthesis based questions

1. a) Suggest a 2-step synthesis of cyclohexanol, starting from cyclohexane, writing equations for all reactions. OH X Step 1 Step 2

cyclohexanol cyclohexane b) Suggest reagents and conditions for the dehydration of cyclohexanol to cyclohexene. c) Why is there only one isomer of cyclohexene, whereas there are two isomers of hex-3-ene, CH3CH2CH=CHCH2CH3? d) What reagents and conditions are needed to convert 1-bromopropane into the following? (i) CH3CH2CH2NH2; (ii) CH3CH2CH2CN; (iii) CH3CH=CH2 “A” Level Specimen paper/III/1(c) & (d)

2. Coumarin is a naturally occurring organic compound with a grassy odour. A series of reactions is shown below. CH=CHCOCl Q HCl(g) OH Step 1 O O heat coumarin

NaOH(aq), heat D2O(l)

P R S T H2SO4 (aq) KMnO4(aq) heat a) Name the type of reaction in Step 1. [1] b) Draw the structures of organic compounds P - T. [5] NJC Prelim 06/III/6a,b

*3. Propanone reacts with hydrogen cyanide to form a compound with molecular formula C4H7ON. a) Draw a displayed formula for the compound obtained. [1] b) The compound in a) can be used in the synthesis of methyl-2- CH3

methylpropanoate shown on the right. H2C C Outline a reaction scheme, stating reagents and conditions for the C O conversion. [3] NYJC Prelim 09/III/3c H3C O

*4. Below is a reaction scheme. Step 1 Step 2 Step 3 (CH3)2C=CH2 K L CH3CH=CH2 J M

Give the reagents and conditions required for each of the steps as well as the structures of the intermediate K and L. [5] IJC Prelims 09/II/3b

5. The following represents part of a synthetic pathway to obtain the amino acid, phenylalanine, and one of its other derivatives, D.

Suggest the reagents and conditions required to carry out Steps 1-5 as well as the structure of intermediate A formed. [6] MJC Prelim 06/II/5 (modified)

6. One type of hydroxyacid E, can be synthesized from compound D in three steps.

Give the reagents, conditions and intermediates for this synthesis. IJC Prelim 09/III/4c

7. 4-bromobenzoyl chloride can be synthesized from methylbenzene via multi-step transformation. CH3 COCl

Suggest reagents and conditions for each step and draw the structural formulae of the intermediates formed. [5] NYJC Prelim 08/III/4d

*8. Salbutamol, commonly known as Ventolin, is a drug that is used to treat acute asthma and to relieve symptoms associated with other conditions pertaining to reversible airway obstructions. In the laboratory, salbutamol can be made from the starting compound, 2-hydroxybenzoic acid via the following route:

a) State the reagents and conditions for Step 1 and Step 2. [2] b) i) Suggest a possible 3-step synthesis for the conversion of compound J to salbutamol. Your answer should include all essential reagents and conditions, as well as the structures of any intermediates formed. [3] ii) A lab technician suggested the use of chlorine gas under UV light to convert compound H to J in Step 3. Suggest why his choice of condition may not be appropriate for the conversion stated. [1] MJC Prelim 06/III/6a,b

*9. 5-nitro-2-propoxyphenylamine, T, is an artificial sweetening agent which is 4000 times as sweet as sucrose. It can be made from propoxybenzene, R.

Suggest a two-stage synthesis of T starting from propoxybenzene, R, identifying the intermediate compound S. The reactivity of propoxybenzene is similar to that of phenol. [3] “A” Levels/N08/III/5d

*10. Compound A can be used to synthesize compounds B and C via a series of steps. CH CH3 3 H N CONH(CH ) COOH O2N CH2Cl H N CH2Cl 2 2

A B C

a) Describe the reagents and conditions required to synthesize compounds B and C respectively. b) Compounds B and C can further react under suitable conditions to give compound D.

CH3 CH3

H N CH2 N CONH(CH2)2COOH

Give the structural formulae of the organic product(s) for each of the reaction of compound D with

(i) cold dilute hydrochloric acid; (ii) hot aq. KOH; (iii) aq. Br2

11. A cyclic amide can be synthesised from compound J as shown below.

a) Draw the structures of K and L. [2] b) A student suggests the use of hot, acidified potassium manganate(VII) as the oxidising agent in Step II. Explain why this choice of reagent is inappropriate and suggest a suitable alternative. [2] RI Prelim 13/III/4d

*12. A phenyl ester can be converted to a hydroxyl phenyl ketone via a rearrangement reaction known as the Fries rearrangement.

It involves migration of an acyl group of phenyl ester to the benzene ring. The reaction is ortho, para–selective and the position of acylation can be regulated by the choice of temperature. A

Lewis acid catalyst like AlCl3 is required.

R O OH O ortho–product

AlCl3 R O > 100 C

R O AlCl3 rt para–product

OH Using the Fries rearrangement in step II of the reaction scheme below, give the reagents and conditions for steps I – IV and deduce the structures for intermediates A – C.

[7] DHS Prelim 13/III/4c

2. Structural Elucidation Questions

Observations/Data from Qn Deduction From the molecular formulae - C: H ratio of about 1:1 Benzene ring is present. - C: H ratio of about 1:2 Unsaturation is present. (e.g. C=C, C=O) If compound only has C and H, but does not decolourise

Br2, cycloalkane is present. - C: H ratio > 1:2 Saturated compound. (No double bond) - Neutral compound with N,O Amide may be present. - Neutral compound with O Alcohols or ester may be present. Rotate plane polarized light/Optically A chiral carbon present in the compound. (Carbon with active compound four different groups attached to it) Can show geometrical isomerism 2 different substituents joined to each carbon in C=C present, and C=C is not in a rring Compound decolourises bromine. Alkene OR phenol OR phenylamine may be present. - If alkene is present, no. of moles of bromine reacted is equal to the number of moles of double bond. The molecular formula will also show an increase in no. of bromine atoms. - If phenol OR phenylamine is present, the molecular formula will show an increase in the number of bromine atoms with a decrease in no. of hydrogen atoms as substitution occurs.

Reaction with acidified KMnO4 leads to decolourisation of purple KMnO4. -If unknown compound contains - If there is no increase in no. of oxygen atoms, 2o oxygen alcohol is oxidised to ketone. - If there is an increase in one oxygen atom, 1o alcohol or aldehyde is oxidised to carboxylic acid.

-If CO2, ketone or carboxylic acid is C=C present. Oxidative cleavage has occurred. formed and compound contains - CO2 evolved indicates presence of terminal alkene double bond (H2C=) or alkene with the structure =CHCH=. - If carboxylic acid formed, RCH= group is present. - If ketone is formed, RR’C= group is present.

-If benzene is likely to be present in Side chain oxidation has occurred to give benzoic acid. compound. - The no. of –CO2H on the benzene ring indicates no. of carbon side chain on the ring.

- CO2 evolved indicates presence of ethyl side chain that contains more than one carbon atom. 0 Reaction with acidified K2Cr2O7 leads - If there is no increase in no. of oxygen atoms, 2 to colour change from orange to green alcohol is oxidised to ketone. (If unknown compound contains - If there is an increase in one oxygen atom, 10 alcohol oxygen) or aldehyde is oxidised to carboxylic acid. Compound has oxygen atom but does Absence of primary, secondary alcohols and aldehydes. not turn orange acidified K2Cr2O7 green

Reaction with Na Presence of a –OH group. (Depending on the molecular formula, compound may be alcohol, phenol or carboxylic acid) Reaction with NaOH(aq) at room Carboxylic acid, acyl chloride or phenol or salt of amine temperature (reactions of acid present. (Combine with reaction with sodium carbonate functional groups) or use molecular formula to determine compound present) Heat with NaOH(aq) under reflux Depending on molecular formula of reactants and products, halogenoalkane, esters, amides present

Followed by excess HNO3(aq), No precipitate implies halogenoarene present or there is

AgNO3(aq) no halogenoalkane White precipitate of AgCl implies chloroalkane present Cream precipitate of AgBr implies bromoalkane present Yellow precipitate of AgI implies iodoalkane present

Reaction of AgNO3(aq) and gives Acyl chloride present. white precipitate readily

Reaction with HCl(aq) or H2SO4(aq) at Amine or salt of phenol or salt of carboxylic acid present. room temperature (reactions of basic Depending on molecular formula of reactants and functional groups) products.

Heat with HCl(aq) or H2SO4(aq) Nitrile or acyl chloride or ester or amide present. under reflux Depending on molecular formula of reactants and products. Reaction of HCl(g) or HBr(g) Alkene present Effervescence with (aqueous) sodium Carboxylic acid or acyl chloride present. carbonate

Reaction with PCl5 or SOCl2 gives Alcohol or carboxylic acid present. steamy fumes of HCl Unknown gives orange ppt with 2,4- Aldehyde or ketone present. dinitrophenylhydrazine (2,4-DNPH) Violet colouration with neutral iron(III) Phenol present. chloride

Unknown compound gives yellow ppt Presence of CH3CH(OH)- or CH3CO- group . with alkaline aqueous iodine Combine with reaction with 2,4-DNPH to determine if functional group is an alcohol or carbonyl.

Silver mirror with Tollen’s reagent Aldehyde present. (R-CHO) (silver diammine solution)

Red precipitate of Cu2O with Fehling’s Aliphatic aldehyde present. (Cannot be benzaldehyde) solution (alkaline solution of Cu(II) complex ion)

Alkaline gas is evolved after heating NH3 is evolved. with NaOH(aq) Alkaline hydrolysis of nitriles (–CN) or amides (–

CONH2) has occurred.

Exercises (Structure Elucidation) 1. a) Reaction of an alkene A with hot concentrated potassium manganate (VII) produced the two

compounds, CH3COCH3 & CH3CH2CO2H in equimolar amounts. Suggest the structural formula of A.

b) B, C8H10, on boiling under reflux with aqueous acidified potassium manganate (VII) gives a

compound C, C8H6O4. C reacts with methanol and hot concentrated sulfuric acid to give D,

C10H10O4, which has a sweet smell. Suggest the structural formulae of B, C and D.

c) E, C2H5NO, is neutral. With cold aqueous alkali, there is no reaction, but on heating under reflux, ammonia is evolved.

d) F, C8H9NO, is neutral. Hydrolysis with aqueous NaOH gives a salt C2H3O2Na and a

compound, C6H7N, which will form a white precipitate with aqueous bromine.

e) G, C6H12, does not react with bromine in tetrachloromethane at room temperature.

PRACTICE QUESTIONS- Structural elucidation questions

1. An organic compound, A, has the following composition by mass: C, 40.0%; H, 6.65%; O, 53.3% i) What is the empirical formula of A? ii) A shows the following properties or reactions. State what can be deduced from each statement. a. It is optically active b. It gives a brisk effervescence with aqueous sodium carbonate c. It gives a yellow precipitate when warmed with alkaline aqueous iodine Suggest an identity of A and draw its displayed formula iii) Give the displayed formula of the organic product formed when the compound of A is warmed with acidified potassium dichromate (VI). iv) When A is warmed with concentrated sulfuric acid, a cyclic compound of molecular formula

C6H8O4 can be isolated. Suggest a displayed formula for C6H8O4 and state what functional group is present in the molecule. v) Suggest simple chemical test to distinguish ethanamide from ammonium ethanoate. “A” Levels/J98/II/5

*2. An organic compound A, C10H11O2Br, does not react with sodium hydrogen carbonate solution but reacts slowly on heating with sodium hydroxide to form a water soluble substance B,

C3H5O3Na, and an insoluble oil C, C7H8O. C on oxidation gives benzoic acid. B with sodium

produces hydrogen gas and D, C3H4O3Na2. Acidification of B gives an acid which is found to be optically active. Explain the chemistry involved and deduce the structures of all the compounds involved in the reactions.

3. This question is about compound F, C8H8O2 which is formed when phenyl ethanoate,

CH3CO2C6H5, is warmed with aluminium chloride. - Compound F is insoluble in water, but dissolves in NaOH(aq) - It reacts with 2,4-dinitrophenylhydrazine but not with Fehling’s solution - With bromine water, F gives G, C8H6O2Br2 - With alkaline aqueous iodine followed by acidification, F gives H, C7H6O3

Compound H dissolves in both NaOH(aq) and Na2CO3(aq). Compound H can also be obtained

by reacting oil of wintergreen, J, with hot NaOH(aq), followed by H2SO4(aq). OH

J CO2CH3

Deduce the structures of compounds F, G and H. Explain the chemistry of the reactions described, writing equations where appropriate. [There is no need to comment on the chemistry of the formation of F from phenyl ethanoate.] “A” Levels /N02/III/8 either

*4. Compound A, C5H10Cl2, is a chiral compound which reacts with warm aqueous potassium hydroxide to form compound B. Oxidation of B produced a compound that does not react with 2,4-dinitrophenylhydrazine. B does not give a yellow precipitate on heating with alkaline aqueous o iodine. When B is added to excess concentrated sulfuric acid at 170 C, compound C, C5H8 is

formed. On reacting C with acidified potassium manganate(VII), compound D, C3H4O3 is formed. Draw the structural formulae of compounds A to D. [4] NYJC Prelim 09/III/4e (modified)

5. The aromatic compounds H, J, K and L are isomers with the molecular formula C9H12O. All four reacted with sodium metal. Compound K also reacts with aqueous sodium hydroxide, and with aqueous bromine.

On warming with acidified K2Cr2O7 the reagent does not change colour with isomers K and L, but

turns green with H and J producing compounds M (C9H10O2) and N (C9H10O) respectively. Only isomer J reacts with alkaline aqueous iodine.

When heated with alkaline KMnO4, isomers H, J and L give the acids P, Q and R respectively. CO H CO2H CO2H 2

HO2C CO2H

CO2H P Q R Suggest a structure for each lettered compound, and explain the reactions involved. “A” Levels/N03/II/7

6. A di-substituted product C with molecular formula C6H12Cl2 was obtained in the reaction of chlorine with 2-methylpentane.

C does not exhibit optical activity. On warming with alcoholic KOH, a major product D is formed. When D is treated with hot acidified potassium manganate(VII) solution, effervescence was observed and the only organic product remaining gives a yellow precipitate with alkaline aqueous iodine.

Use this information to deduce the structure of C and D, explaining your reasoning. [5] NYJC Prelims 13/III/1c

*7. An alkane H with the formula C6H14 reacts with chlorine to yield three compounds with the

formula C6H13Cl: J, K and L. Of these only K and L can react with ethanolic potassium hydroxide. Moreover, K and L yield the same product. Deduce the structure of H, J, K and L. IJC Prelims 09/III/1f

8. Reagents like potassium manganate(VII) and sodium borohydride, NaBH4, are of great help in elucidating the structure of organic compounds.

Compound C of molecular formula, C6H10O2, produces a yellow precipitate when treated with an alkaline solution of iodine. C does not react with an ammoniacal solution of silver nitrate. Treatment of C with excess hot acidified concentrated potassium manganate(VII) gives

compound D, C4H6O3, as the only organic product. Compound C produces compound E,

C6H12O2, when reacted with NaBH4.

Suggest possible structural formulae of C, D and E, explaining your reasoning. NJC Prelim 13/III/2e

9. Compound F has a molecular formula of C9H10O2. Compound F is insoluble in water, but dissolves in NaOH(aq). It reacts with 2,4-dinitrophenylhydrazine as well as Fehling’s solution.

F reacts with bromine water to give G, C9H9O2Br.

F reacts with hot acidified KMnO4 to gives H, C8H6O5.

Compound H reacts with Na2CO3 in a 1:1 mole ratio.

Deduce the structures of compounds F, G and H and explain the chemistry of the reactions described. Write separate balanced equations for the reactions of F with Fehling’s solution, and with hot

acidified KMnO4. [10] TPJC Prelim 09/III/5c

10. Compound P has a molecular formula of C5H9O2Cl. Compound P is sparingly soluble in water but readily soluble in aqueous sodium hydroxide.

Heating compound P in aqueous sodium hydroxide, followed by adding excess acid to the

product yields compound Q, C5H10O3, which reacts with warm alkaline aqueous iodine to give a yellow precipitate.

On treatment with hot concentrated sulfuric acid, compound Q forms a mixture of three isomeric

compounds R, S and T, with the formula C5H8O2, two of which are geometrical isomers of each other.

Compound R is a sweet smelling liquid. Compound S reacts with 1 mole of aqueous bromine to give a number of products. Both S and T undergo reaction with acidified potassium

manganate(VII) to give U, C3H4O4, and ethanoic acid. 1 mole of U liberates 1 mole of CO2 on reaction with excess sodium carbonate. i) Identify the five compounds Q – U, explaining the chemistry of the reactions described. ii)Write a balanced equation for the reaction of compound Q with warm alkaline aqueous iodine.[11] AJC Prelim 13/III/3d(modified)

*11. Compound A, C14H19NO2, is the active substance in Cayenne pepper and it is basic. On heating with aqueous sodium hydroxide, compound A produces 2 compounds. One of them is - + compound B, C7H8NO Na . Compound B reacts with excess CH3Cl to give compound D, + - C11H18NO Cl .

The other compound from A gives compound C, C7H12O2, upon acidification. Compound C liberates carbon dioxide from aqueous sodium hydrogencarbonate, and catalytic hydrogenation

of C gives G, C7H14O2. When C is heated with acidified potassium manganate(VII),

HO2CCH2CO2H and H, C4H8O2, are obtained. H can be synthesised from 2-bromopropane in a two-step reaction.

Compound B can be synthesised via the following sequence of reactions.

Suggest a structure for each lettered compound, and explain the reactions involved. [8] SAJC Prelim 10/III/3c

3. Distinguishing test Questions

PRACTICE QUESTIONS- Distinguishing tests questions

REMEMBER: 1. When using a distinguishing test, remember to describe its reagents and conditions. 2. Do not use “heat under reflux” as a condition! Remember the test is to be done using test tubes. Just write “heat” if applicable. 3. Write observations clearly for each compound tested. Observations include effervescence of gas (identify the gas), formation of precipitates (state colour and identity), colour change of reagents added (state original colour change to new colour). Compounds that do not give the observation, do not write “no observation/no visible reaction/no change”. Write “no gas evolved/no colour change observed/no precipitate formed” (whichever is applicable). 4. Distinguishing test is a chemical test! Do not use the sense of smell, or an acid-base reaction or use litmus paper (not conclusive) to distinguish! 5. If there are more than 2 compounds to be distinguished, more than one distinguishing test is needed. 6. If the compounds involved esters and amides, hydrolyse the compound first before doing distinguishing tests. This can usually be achieved through the acidic/alkaline medium of some reagents together with heat.

*1. Phenol, phenylmethanol, cyclohexanol and benzoic acid all contain a hydroxyl group. None of these compounds is particularly soluble in water. Using aqueous sodium hydroxide, it is possible to divide them into two pairs. The two compounds of one pair can be further distinguished. OH CH2OH OH COOH

Phenol Phenylmethanol cyclohexanol benzoic acid

a) Divide the four compounds into two pairs based on their reactions with aqueous sodium hydroxide. b) State the reagent that can further distinguish the pair of compounds. Describe how each compound behaves with appropriate equation (s) and suggest a reason for the difference.

2. Phenylethanone, C6H5COCH3, is used to create fragrances that resemble almond, cherry, honeysuckle, jasmine and strawberry. It is used in chewing gum. Upon reduction of phenylethanone, a new organic compound A will be formed. a) Suggest a simple chemical test where both phenylethanone and compound A will give the same positive results. b) Suggest a simple chemical test to distinguish phenylethanone and compound A.

NJC Prelim 06/II/5a (modified)

3. Describe a simple chemical test which would enable you to distinguish the following pairs of compounds. For each test, state the reagents and conditions, and state what would be observed with each compound. Note: The distinguishing of the pair may rely on preliminary breaking up of the compounds and subsequent testing of the reaction products. a) b) O O CH2CHO CONH2 O O

CH3 H3C CH H3C 3 A & B X & Y

NYJC Prelim 09/III/2b

4. Describe a chemical test by which you could distinguish the following pairs of organic compounds. You should state the reagents and conditions, observations and write balanced equations. a) H3CH2C CHO H3C CH2CHO

*b)

*c)

*d)

*5. Compound C is used to to give a raspberry flavour while compound D is used to impart apple flavours to food. The structures of compounds C and D are shown below. compound B compound C compound D

O O O

O O

OH Suggest methods by which the compounds C and D could be distinguished from compound B by chemical tests, giving reagents, conditions and observations for each compound. The distinguishing test(s) may rely on the preliminary breaking-up of the compounds and the subsequent testing of the reaction products. AJC Prelim 09/III/2c

*6. Suggest a simple chemical test to distinguish between compounds A and B, A and C and B and C. The same test cannot be used more than once.

RJC Prelim 06/III/6 (modified)

7. Describe a simple chemical test which would enable you to distinguish the following pairs of compounds. For each test, state the reagents, and conditions, and state what would be observed with each compound.

i) CCl3CH2COCH3 CCl3COCH2CH3

CH3 CH2Cl Cl

ii) NYJC Prelim 08/III/4c

8. Suggest a simple chemical test to distinguish between each pair of compounds. For each test, give reagents and conditions, and state what would be seen with each compound.

CH CH (i) C H 3 2 3 (ii) NH2 CONH2

and and

CH3 (iii) CH2CH2Cl CH2CH2I

and

SAJC Prelim 10/III/4d

9. For each of the three compounds given below, suggest a simple chemical test that will identify it from the other two via a positive result. For each test, give reagents and conditions and state the observations for each of A, B and C. Your test(s) may rely on a preliminary breaking-up of the compound, and subsequent testing of the reaction products.

HCI Prelim 13/III/2c

4. Explanation based Questions

PRACTICE QUESTIONS - Explanation based questions

1. Compare and explain the difference in the relative rates of hydrolysis of compound R, S and T. State the reagents and conditions for the relevant hydrolysis reactions.

IJC Prelim 09/III/2c (modified)

2. Compare the basicity of N1 and N2 atom of oseltamivir. Explain which O O CH3 nitrogen atom has a lower pKb value. DHS Prelim 09/II/6 CH3

C1 CH3 H2N O 1 2 O NH CH3 oseltamivir

3. Trimecaine, one of the first clinical local anesthetics drug administered for surgery has the following structure. One molecule of trimecaine contains two nitrogen atoms. Circle the nitrogen atom(s) which you think will cause trimecaine to be basic in nature. Explain your reasoning. CH 3 H N CH2N(C2H5)2 C O H3C CH3

Trimecaine TJC Prelim 10/II/6 4. Explain

a) the difference in the reactivities of CH3CH2Cl and C6H5Cl towards NH3.

b) the difference in physical states of CH3CH2Cl and C6H5Cl at room temperature. NJC Prelim 09/III/5b

5. Explain why phenylamine, C6H5NH2 is a weaker base than benzylamine, C6H5CH2NH2 but reacts much more readily with aqueous bromine than benzylamine. MJC Prelim 08/III/5b

6. Phenyalmine, C6H5NH2, and N-methylpropylamine, CH3CH2CH2NHCH3, are also Bronsted-Lowry bases. i) Explain what is meant by Bronsted-Lowry base? ii) Describe and explain why the basicity of propylamine differs from those of phenylamine and N-methylpropylamine. HCI Prelim 08/III/5b

7. An optically pure sample of 2-chlorobutane reacts with ethanolic sodium cyanide to yield an optically pure product, X. Explain how the rate of reaction changes when 2-chlorobutane is replaced by 2-iodobutane. [2] RJC Prelim 06/III/8b

8. A B C

OH CH2OH OH

Cl CH3 Cl i) Arrange compounds A, B and C in order of increasing pKa. Explain your reasoning. ii) Explain why compound C is insoluble in water but soluble in aqueous sodium hydroxide. [5] TJC Prelim 08/III/1c

9. i) Explain the differences in Ka values between the 3 compounds shown below. [4]

Acids Ka -5 C6H5CO2H 6.46 x 10 -5 CH3CH2CO2H 1.34 x 10 -16 CH3CH2OH 1.0 x 10

ii) The pKa of the acid, is 4.8 where Y is a substituent.

Predict the product(s) formed when Cl2 is added to in the presence of FeCl3 in the dark. Explain your answer. [2] IJC Prelim 07/III/4c

10. The following compounds differ in their reactivities with aqueous sodium hydroxide: O

Br Br and C , Br

Describe and explain the differences. [4] RI Prelim 09/III/4c

11. The halogenoethanes C2H5Cl, C2H5Br and C2H5I differ in their physical properties and reactivities. i) Suggest and explain how the boiling points of these three compounds differ. ii) Suggest and explain how the C-X bond polarities in these three compounds differ. iii) Describe and explain how the reactivities of the three compounds differ towards nucleophilic reagents. [5] “A” Levels/N08/III/1d

12. State and explain the relative solubility of 2-aminobenzoic acid and 4-aminobenzoic acid in water.[3]

NH2

COOH H2N COOH

2-aminobenzoic acid 4-aminobenzoic acid SRJC Prelim 07/III/4

5. Data-based Questions

PRACTICE QUESTIONS- Data based questions

Some suggestions: 1. Read question carefully to understand story, take note of important key phrases, and data and clues. 2. Link to concepts learnt and figure out what is being tested and to be applied. 3. Read questions then look at data or story at beginning of question will help you focus on what information is important and to be used. 4. The subparts, (i), (ii), (iii) etc often are linked and depend on each other. Very often answers in subsequent subparts depend on whether you see the links to previous subparts. Ponder on why the subparts are arranged in a particular order and what all the subparts are testing on in common. 5. The parts, (a), (b), (c) etc may or may not be linked to each other. Do not give up even if you do not know how to do the earlier parts. You need not do parts in the order given, if it is possible to skip the more difficult parts and do the others first, do it.

1. Human serum albumin, HSA, is an important protein that transports long chain fatty acids and other hydrophobic molecules in the bloodstream. The HSA molecule is a single chain of 570 amino acids, 67% of which are incorporated into an α-helix. It is a globular protein and takes up a roughly spherical shape in water.

Eight of the most common amino acids in the HSA molecule are listed below.

(i) What is understood by the term primary structure of a protein?

(ii) Use three of the above amino acids to construct the displayed formula of a possible section of the protein chain of HSA.

(iii) Describe how a polypeptide chain is held in the shape of an α-helix.

(iv) The cysteine residues in the HSA molecule can form disulphide bridges. Illustrate this process by means of a chemical equation. Assuming all disulphide bridges are intramolecular, how many bridges could be made within each HSA molecule?

(v) The roughly spherical shape of the HSA molecule is due to the attractions between the various side chains. Apart from the disulphide interaction mentioned in part (iv), describe two other types of side-chain interaction, illustrating your answer with suitable pairs of amino acids from the table above.

(vi) Long chain fatty acids such as stearic acid (n = 17 in the general formula in H-(CH2)n-CO2H) are carried through the bloodstream in the inside of the HSA molecule. Suggest two amino acids in the above table which would interact with the long chain on stearic acid.

(vii)The amino acids whose side chains are on the outside of the HSA molecule are likely to interact strongly with water molecules. Suggest three amino acids in the above table which are likely to do this. [14] “A” Levels/N07/III/5(b)

2. In 1849 the German chemist Adolf Kolbe reported his investigations into the electrolysis of aqueous solutions of the sodium salts of some carboxylic acids, using inert platinum electrodes. Alkanes are formed at the anode by the following reaction. - - 2 R-CO2  R-R + 2 CO2 + 2e (R = CH3, C2H5 etc.) When a mixture of the sodium salts of two different mono-carboxylic acids is electrolysed, a mixture of alkanes is produced. - - R’CO2 + RCO2  R-R + R-R’ + R’-R’ Electrolysing a mixture of the sodium salts of the two mono-carboxylic acids A and B produced three alkanes C, D and E, which could be separated by fractional distillation.

(a)On titration, a solution containing 0.100g of the acid A required a volume of 11.4cm3 of 0.100 -3 mol dm NaOH to reach the end-point. Calculate the Mr of A and hence suggest a possible structure. [2] (b) A gaseous sample of 0.20g of C took up a volume of 87cm3 at a temperature of 380K and a 5 pressure of 1.01 x 10 Pa. Use these data to calculate the Mr of C and hence suggest its molecular formula. [2] (c) A 1.00g sample of D was burned in an excess of oxygen, and the gases that were produced were first passed through a U-tube containing phosphorus pentoxide (to absorb the water vapour) and then bubbled through concentrated NaOH(aq) (to absorb carbon dioxide). The phosphorus pentoxide U-tube increased in mass by 1.55g, and the NaOH(aq) bottle increased in mass by 3.03 g. (i) Write equations for the reaction of water vapour with phosphorus pentoxide and the reaction of carbon dioxide with sodium hydroxide. (ii) Calculate the number of moles of water, and of carbon dioxide, produced. (iii) Use these data to calculate the H:C ratio in alkane D, and hence suggest its molecular formula. [6] (d) Use your results in (a), (b) and (c) to deduce possible structures for the alkane E and the acid B. [2] “A” Levels/N08/III/2(c) to 2(f)

3. a) Oxytocin and vasopressin are two rather small polypeptides with strikingly similar structures. In spite of the similarity of their amino acid sequences, these two polypeptides have quite different physiological effects. Oxytocin occurs only in the female of a species and stimulates uterine contractions during childbirth. Vasopressin occurs in males and females; it causes contraction of peripheral blood vessels and an increase in blood pressure. Its major function, however, is as an antidiuretic; physiologists often refer to vasopressin as an antidiuretic hormone.

leu asn NH2 O O H O O N H N H2N N N NH NH2 O H cys gly O gln pro S O O S HN O O cys H oxytocin N H N N 2 H O Ile

OH arg tyr HN NH2 NH

NH2 O asn O H O O vasopressin N H N H2N N N NH NH2 gly O H cys O gln pro S O O S HN O O cys H N H N N 2 H O phe

tyr OH

i) Apart from the peptide bond in oxytocin, state two other functional groups in oxytocin. ii) State the name of the linkage that is highlighted in the two structures above which is common in both polypeptides. iii) Oxytocin and vasopressin has different primary structures, where oxytocin has leucine, vasopressin has arginine, and where oxytocin has isoleucine, vasopressin has phenylalanine. State the type of interactions arginine and phenylalanine are able to form in maintaining the tertiary structure of proteins. iv)The melting points of leucine, along with two other compounds, are given in the table below.

Compound Mr Melting Point/oC

Leucine 131 293 1,7-Heptanediamine 130 25 Heptanoic acid 130 -7

Explain the difference in melting points between I Leucine and 1,7-Heptanediamine and, II 1,7-Heptanediamine and Heptanoic acid [7]

b) Denaturation is a process in which proteins lose the quaternary structure, tertiary structure and secondary structure which is present in their native state, by application of some external stress or compound such as a strong acid or base.

A classic example of denaturing in proteins comes from egg whites, which are largely egg albumins in water. Fresh from the eggs, egg whites are transparent and liquid. Cooking the thermally unstable whites turns them opaque, forming an interconnected solid mass.

Considering the types of R group interactions that are affected during the denaturation of proteins, explain the following phenomena, highlighting the changes in the bonds that lead to denaturation.

I Casein is the predominant protein found in milk. The conversion of milk to cheese involves the addition of the Lactobacillus bacterium which produces lactic acid.

II Alcohol (ethanol) solutions are used as disinfectants on the skin as it can penetrate the bacterial cell wall and denature the proteins inside the cell. The part of the protein molecule which is affected by the ethanol added is as follows:

ACJC Prelim 13/III/5(a,b)

4. 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.

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]

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 Gly-Trp-Leu-Ala-Glu Tyr-Trp-Leu-Val-Arg Lys-Tyr-Trp-Leu Glu-Glu-Met-Lys Val-Asp-Asp Val-Asp Glu-Met Asp-Asp-Phe Val-Arg Gly-Trp-Leu-Ala Asp-Phe Suggest the primary sequence of this 18 residue polypeptide.

25 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.

pKa = 2.09 pKa = 2.19 pKa = 2.18 O O O

H2N CH C OH H N CH C OH H2N CH C OH 2

CH CH OH 2 CH2

CH CH Amino acid CH3 2 2

CH C O 2

CH2 OH pKa = 4.25 NH2

Mr = 131 Mr = 147 Mr = 146 Abbreviation Thr Glu Lys pI 5.60 3.22 9.74

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

D Start

B C Anode 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. 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]

26 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.

NH2

N N

N O O O HO N O H H H H H2N CH C N CH C N CH C OH H H O H O CH2 CH2 CH2 O P O- OH CH2 NH O CH2 N O O CH2 H H NH2 H H O H

O P O-

OH DNA sequence Section of peptide sequence from lac repressor

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)

5 -1 3 Kb = 8.81 x 10 mol dm

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).

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

ΔG = ΔH – TΔS

suggest why this process is favoured at higher temperature such as 70 °C. [8] ACJC Prelim10/III/3(a) to 3(c)

5. Ozone can be “good” or “bad” for people’s health and for the environment, depending on its location in the atmosphere. The stratosphere or “good” ozone layer extends upward from about 8 to 50 km and has been gradually depleted by man-made chemicals like chlorofluorocarbons (CFCs). Once trichlorofluoromethane (CFC-11) is in the stratosphere, there is an initial homolytic fission of a C–Cl bond in CFC-11 to give Cl• radical. Ozone depletion then proceeds via a four-step mechanism.

Step 1 The initial reaction is between the Cl• radical and ozone to give ClO• radical and O2 molecule. Step 2 Two ClO• radicals react to give a ClOOCl molecule. Step 3 The ClOOCl molecule dissociates to give ClOO• and Cl• Step 4 The ClOO• radical dissociates to give a Cl• radical and an oxygen molecule

i) Suggest the energy source for the homolytic fission of the C–Cl bond in CFC-11. ii) Use the information given above to outline the full mechanism for the reaction between CFC-11 and ozone. You are advised to draw full structural formulae for all species, so that it is clear which bonds are broken and which are formed. Indicate any unpaired electrons by a dot (•). iii) A single Cl• radical would keep on destroying ozone for up to two years. What is the role of the Cl• radical in the reaction between CFCs and ozone? VJC Prelim13/II/4(c)

Answers

1. Synthesis/Conversion Questions

Exercises (Synthesis) Below are some two step syntheses. In each case, identify the reagents and conditions for each step and also identify the intermediate compound. (a)

CH2CH2Cl A CHCH3

(b) CH3 CH3

C O B CH3 C Br

CH3 H

Cl HO OH

(d) CH CH3 CH3 3 C C D CH3 C OH CH3 CH3 H

(e) O H

NO2 E CH3 C N

(f) * CH3 CH3 O

C C CH C F 3 CH CH OH 3 3 CH CH3 CH3 + 3 O KMnO4/H 1. I2/NaOH(aq), heat C C C O CH C heat under 3 CH CH 2. HCl 3 3 reflux CH3 OH 29

PRACTICE QUESTIONS 1 a) b) Cl OH Step 1 Step 2

Cl2, UV light NaOH (aq) heat cyclohexanol cyclohexane c) Cyclohexene has only 1 isomer as the molecule only exhibits cis isomerism. The trans form would distort the ring with great strain and hence is not stable and non-existent. Hex-3-ene has geometrical cis-trans isomerism due to the restriction of rotation caused by the C=C double bond.

H H H CH2CH3 C C C C H3CH2C CH2CH3 H3CH2C H

Cis isomer Trans isomer d) i) concentrated NH3, heat in sealed tube ii) Ethanolic KCN, heat iii) Alcoholic NaOH, heat

2 (a) Condensation/Esterification/Addition-Elimination (b) Cl

P CH=CHCO2D R Q

- + CO2 Na OH O-Na+ O O or Cl

S T CO2H

O O CO2H OH OH

3 (a) H H C H H H C O C C N H H

CH CH (b) 3 3 O CH3 dilute HCl excess conc. H2SO4 O

HO C CN HO C C H2C C C heat under 170oC OH OH CH3 reflux CH3 CH3OH, trace amt. of conc.

H2SO4 heat under reflux

CH3 O H2C C C O CH3 30

+ 4 Step 1: KMnO4/H , heat

Step 2: LiAlH4 in dry ether, (followed by water) OR H2, Ni catalyst

Step 3: conc sulfuric acid, 170 °C OR Al2O3, 350 °C

K: L: C H 3

H C OH

CH 3 OH 5 Step 1: HCN with trace amount of NaOH or NaCN, 10-20oC CH2CHCN Step 2: PBr5, heat

Step 3: NH3, heat in a sealed tube

Step 4: dilute H2SO4, heat under reflux

Step 5: limited CH3CH2Cl, heat [to ensure only secondary amine is formed] Compound A 6

COCl CH3 CH3 COOH

1 2 3

Br Br Br

1: Br2, anhydrous AlBr3

2: KMnO4, dilute H2SO4/ heat under reflux

3: PCl5, rt

8 a) Step 1: LiAlH4 in dry ether Step 2: CH3COCl, room temp b)i) Compound J to Salbutamol

ii) Substitution may occur with other C-H bonds present in the molecule.

*Note: 1.

The mixture of Sn and conc HCl gives the reducing agent that reduces the nitrobenzene group to a phenylamine. 2. Using limited amounts of the reducing agent will ensure that only one of the nitro groups is reduced.

10 a) To synthesize compound B via 2 steps: Step 1: Sn, conc. HCl, heat followed by excess NaOH

Step 2: CH3Cl (alcoholic), heat

To synthesize compound C via 3 steps: + Step 1: KMnO4/H , heat

Step 2: PCl5, r.t.p.

Step 3: NH2CH2CH2CO2H, r.t.p.

b) i) ii)

CH3 CH3 - + - + H N CH2 N COO K + NH2CH2CH2COO K

iii) CH3 Br CH3 Br H N CH2 N CONHCH2CH2COOH

Br Br Br Br

11 a)

b) If hot, acidified potassium manganate(VII) is used, side chain oxidation occurs, resulting in formation of benzene-1,2-dicarboxylic acid and the loss of the N atom. Hence the cyclic amide cannot be formed.

Suitable alternative: I2(aq) and NaOH(aq), heat

12 I: CH3COCl (OR NaOH followed by CH3COCl), r.t.

II (Fries rearrangement): AlCl3, heat at > 100C

III: dilute HNO3, r.t. OR conc. HNO3 & conc. H2SO4, heat under reflux

IV: excess CH3I, heat

2. Structural Elucidation Exercises 1. CH3 CH2CH3

CH3 H 2. B C

CH3 CH3 HOOC COOH

H CO C CO CH 3 2 2 3 3. neutral compound with N and O  amide

Ammonia evolved  -CONH2 present

4. O H N CCH3

5. does not react with Br2  absence of C=C

PRACTICE QUESTIONS 1. i) Atom C H O Mole ratio 40.0/12 6.65/1 53.3/16 1 2 1

Empirical formula = CH2O H ii) a. A contains chiral carbon present A: b. A is acidic, contains carboxylic acid H C H

C O HO C c. A contains group H O H O O H iii) H C C C O H O

H H C O C H iv) Ester functional group present H3C C O C CH3

v) Ethanamide, CH3CONH2(s) and ammonium ethanoate, CH3CO2NH4(s) To separate samples of each compound, add NaOH(aq) at room temperature.

CH3CO2NH4 would react with NaOH(aq) to liberate NH3 which turns moist red litmus paper blue.

No such observation is seen with CH3CONH2.

2. A has a C:H ratio of about 1:1, showing the presence of an aromatic ring. A does not have a carboxylic acid group as it does not react with sodium hydrogen carbonate solution. A undergoes ester hydrolysis and nucleophilic substitution of bromine to give B and C. C is monosubstituted as oxidation of C gives benzoic acid. B reacts with 1 mole of Na indicating possible presence of one –OH group. Acidification of B gives a chiral compound.

3. F is insoluble in water, but dissolves in NaOH(aq)  F contains a large non-polar group, likely to be benzene. F is acidic since it can react with NaOH(aq). F reacts with 2,4-dinitrophenylhydrazine but not with Fehling’s solution  F contains ketone or benzaldehyde

With bromine water, F gives G, C8H6O2Br2 (disubstitution)  F contains phenol and has one of the positions 2 or 4 occupied.

F reacts positively with iodoform test to give H, C7H6O3  F has methyl cabonyl group, CH3CO-

H dissolves in both NaOH(aq) and Na2CO3(aq)  H contains carboxylic acid

4. A B C D CH2CH2Cl CH2CH2OH CH=CH2 COOH

CH3 C H CH3 C H CH3 C CH3 C O

CH2Cl CH2OH CH2

5. H, J, K and L has molecular formula C9H12O and reacts with Na  All four contain OH gp but none of them is a carboxylic acid as they only have 1 O atom K reacts with NaOH: K is a phenol K reacts with Br2(aq): K contains a phenol. o o o K & L have no reaction with acidified K2Cr2O7: K & L are neither 1 nor 2 alcohola, could be 3 alcohol o H reacts with acidified K2Cr2O7 to give M (C9H10O2): 1 alcohol oxidised to acid (since there is an increase in O atoms) o J reacts with acidified K2Cr2O7 to give N (C9H10O): 2 alcohol oxidised to ketone - Reaction of J & I2/OH : J has CH3CH(OH)- group

H, J, & L have side-chain alkyl group oxidised by hot alkaline KMnO4

CH2OH CO2H

H3C CH3 HO2C CO2H H P

6. C does not exhibit optical activity as it doesn’t contain any chiral centre (no carbon with 4 different groups). Alkyl halide in C undergoes elimination with alcoholic KOH to form alkene in D. Alkene in D undergoes oxidative cleavage to form the following products:

undergoes oxidation to form yellow precipitate with alkaline aq iodine. The terminal alkene is oxidised to CO2. Ethanedioic acid formed is also oxidised spontaneously to CO2.

C: D:

C CH3 8. C gives yellow ppt with alkaline aq I2  C contains either methyl carbonyl group or OH

C CH3

H structure. C has no reaction with Tollen’s reagent  C is not an aldehyde. C reacts with excess KMnO4 to give D, C4H6O3, as the only organic product  C is oxidized (vigorous oxidation/ oxidative cleavage of C=C bond). D contains a carboxylic acid group (increase in no. of O atoms)

C loses 2 carbon atoms as CO2 gas. i.e. oxidative cleavage of C=C bond resulted in ethanedioic acid which

is further oxidized to CO2 /C=C bond is between C2 and C3.

C produces compound E, C6H12O2, when reacted with NaBH4  C is reduced. OH

O C CH3

C contains an alcohol group and a methyl carbonyl group C CH3 which is reduced to H in E. H H H H O O H O H H H H OH H C C C C C CH H C C C C C CH 3 HO C C C CH3 3 OH H H OH H H

C D E

9. Compound F has high C:H ratio/C:H almost 1, hence likely to be benzene derivatives. Compound F is insoluble in water but dissolves in NaOH → F is acidic. (F undergoes CONDENSATION reaction with 2,4-DNPH → F is carbonyl compound.) F reacts with Fehling’s soln → F is a aliphatic aldehyde. F undergoes electrophillic substitution readily with bromine water to form G, which (only has one H of F substituted with Br ) → F is a disubstituted phenol + F undergoes side chain alkyl gp oxidation with hot H /KMnO4 to give an acid, H. (H, C8H6O5 has one carbon less than F → F has an ethyl side chain) H reacts with Na2CO3 in 1:1 ratio → H is dibasic acid

10. i) Observation Type of reaction Deduction P is sparingly soluble in water (P is not a salt, but may contain hydroxyl or

CO2H group)

P is readily soluble in NaOH acid–base reaction P contains a CO2H group

P reacts with hot aq NaOH to give nucleophilic Q contains an alcohol functional group.

Q, C5H10O3 substitution

Q reacts with alkaline I2(aq) to give oxidation Q contains CH3CH(OH)– or yellow ppt. (iodoform reaction) CH3 C O structure

Q reacts with concentrated H2SO4 self–esterification R is an ester. to give R, which is a stable sweet– (R must be a 5 or 6 membered cyclic ester smelling liquid as it is stable, hence minimal ring strain)

Q contains both the –OH and –CO2H group

Q reacts with concentrated H2SO4 elimination S and T contains C=C to give S and T, with the formula

C5H8O2 S and T exhibits geometrical isomerism

S reacts with 1 mol of Br2(aq) electrophilic addition S contains only 1 C=C (a bromohydrin product is formed as the major product)

S and T reacts with KMnO4 to give (strong) oxidation The C=C in S/T is cleaved, there is no

U, C3H4O4 and ethanoic acid (oxidative cleavage) terminal C=C as no CO2 is produced.

1 mole of U reacts with excess acid–base reaction OH Na CO to give 1 mole of CO . 2 3 2 C O U contains 2 U is a dicarboxylic acid. OH O H3C O CH3CCH2CH2COOH

H Q R

H H CH3 H C C C C HOOCCH COOH 2 CH3 CH2CO2H H CH2CO2H

S T U

ii) OH O O + 7NaOH + 4I2 CH3CCH2CH2COOH +- - + warm Na OCCH2CH2CO Na + CHI3 H Q + 5NaI + 6H O 2

11.

A is basic A contains amine group. A reacts with NaOH to produce two cpds. A is an ester undergoes hydrolysis, A is formed from a phenol and a carboxylic acid.

B reacts with excess CH3Cl to form D B is a primary amine, which undergoes nucleophilic substitution, to form a quarternary salt, D.

C liberates CO2 with NaHCO3 C contains carboxylic acid.

C, on hydrogenation, gives G, C7H14O2, C has contains 1 C=C bond. C forms two different compounds with C undergoes vigorous oxidation.

KMnO4.

Synthesis of H

3. Distinguishing tests

1. a) Phenol and benzoic acid can both react with NaOH (aq) while phenylmethanol and cyclohexanol does not. - + OH O Na + NaOH + H2O

COOH COO-Na+ + NaOH + H2O

b) A: phenol and B: benzoic acid

Reagent Solid sodium carbonate / aqueous bromine Condition Room temperature Observations Effervescence is observed in B which will form a white ppt in limewater when the gas evolved is bubbled through. A will not form effervescene and no white ppt formed in limewater. / Brown bromine decolourises and white ppt formed in A while brown colour remains in B. Br OH OH + 3Br2(aq) + 3HBr Br Br white ppt COOH COO-Na+ + Na2CO3 2 + H2O + CO2

CO2 + Ca(OH)2 CaCO3 (s) + H2O white ppt

A: phenylmethanol and B: cyclohexanol

Reagent K2Cr2O7 or KMnO4 followed by 2,4-DNPH Condition Acidified and Heat heat Observations Both A and B will turn orange dichromate to green. However, the product of oxidation of B will give an orange ppt with 2,4-DNPH but not the product of oxidation of A.

2. A: C6H5CH(OH)CH3 and B: phenylethanone

Reagent I2 / NaOH (aq) Condition Heat

Observations Both A and B will form yellow ppt of CHI3. (Iodoform test)

A: C6H5CH(OH)CH3 and B: phenylethanone Reagent 2,4-dinitrophenylhydrazine Condition Heat Observations Orange ppt formed with phenylethanone while no orange ppt is observed with A.

3. a) reagents and conditions: Add Fehling’s reagent to A and B seperately, and warm. Observations: For A, reddish brown precipitate formed. For B, no precipitate formed. Other possible answers: (Write their observations) Tollen’s reagent, warm

K2Cr2O7 with dil H2SO4 and heat 2,4-DNPH, warm

b) reagents and conditions:

Heat both X and Y separately with K2Cr2O7 / dil H2SO4.

Observations: For X, orange K2Cr2O7 turns green. For Y, orange K2Cr2O7 remains.

Other possible answers: KMnO4 with dil H2SO4 and heat [The acid present will hydrolyse the ester then immediately followed by oxidation of the alcohol]

4. a) Add Fehling’s solution to both compounds and warm. st For the 1 compound, a brick red ppt of Cu2O is obtained. For the 2nd compound, no brick red ppt is obtained.

b) Add I2(aq), NaOH(aq) to each compound separately and heat. st For the 1 compound, yellow ppt of CHI3 is obtained. (Hydrolysis of ester takes place first to form ethanol which undergoes triiodomethane test) For the 2nd compound, no yellow ppt of is obtained. c) Add Tollens’ reagent to both compounds separately with warming. For the 1st compound (aldehyde), there would be a silver mirror formed. For the 2nd compound, there would be no silver mirror formed.

d) First, add dil H2SO4 (aq) to both compounds separately, and heat. Then, cool and add aqueous bromine to both samples. For the 1st compound, the reddish-brown bromine remains, with the absence of the formation of any ppt. For the 2nd compound, there would be a decolourisation of reddish-brown bromine, with the formation of a white ppt. Or

Add K2Cr2O7 (aq), dil H2SO4, heat For the 1st compound, orange solution turns green. For the 2nd compound, orange solution does not turn green.

5. C and B

To separate samples of C and B, add neutral FeCl3. C will give a purple colouration whereas no such colouration is observed for B.

B and D

To separate samples of B and D, heat with KMnO4 in dilute sulfuric acid.

Purple colour of KMnO4 decolourises for B, whereas no such decolourisation is observed for D.

6. A and B Reagent PCl5 Condition Room Temperature Observations White fumes are evolved in B while A will have not form white fumes.

B and C Reagent solid sodium carbonate Condition Room Temperature Observations Effervescence is observed in B which will form a white ppt in limewater when the gas evolved is bubbled through. No effervescence is observed in B.

A and C Reagent NaOH (aq) , followed by HNO3 (aq) and AgNO3 (aq) Condition Heat [with NaOH] Observations White ppt observed in A while C has no white ppt formed.

7. i) Use aq I2, aq NaOH, warm. CCl3CH2COCH3: Yellow ppt formed. CCl3COCH2CH3: No yellow ppt formed.

ii) Use aqueous AgNO3, warm.

CH3 CH2Cl Cl

No ppt formed. : White ppt (of AgCl) formed.

+ 8. i) Reagents and conditions: Add KMnO4/H , heat Observation: Ethylbenzene - Purple manganate(VII) decolorised, colourless gas evolved formed white ppt with aqueous calcium hydroxide. 1,4-dimethylbenzene - Purple manganate(VII) decolourised but no gas evolved. ii) Reagents and conditions: aqueous bromine at room temperature Observation: Phenylamine - orange-yellow solution decolourised, white ppt formed. Benzylamide - orange-yellow solution remained and no white ppt formed.

iii) Reagents and conditions: Heat with aqueous sodium hydroxide, followed by HNO3(aq) and finally aqueous silver nitrate. Observation: Chloroethylbenzene – White ppt formed. Iodoethylbenzene – Yellow ppt formed.

9. Test 1: Reagents and Conditions: 1. dilute H2SO4, heat 2. neutral FeCl3 (aq) Observations: A: violet coloration B & C: No violet coloration

Test 2: Reagents and Conditions: NaOH(aq), heat

Observations: B: pungent NH3 gas evolved that turned moist red litmus blue A & C: No pungent NH3 evolved

Test 3: Reagents and Conditions: Br2(aq) C: orange-yellow Br2(aq) decolourizes, (white ppt) A & B: No decolourization of orange-yellow Br2(aq)

4. Explanation based questions

1. Compound S undergoes hydrolysis more easily / faster reaction compared to R. T does not undergo hydrolysis at all. Reagent & condition: Compound R undergoes hydrolysis when heated with aqueous sodium hydroxide, whereas compound S undergoes hydrolysis with water. Explanation: In compound S, the carbon atom is highly/more electron deficient due to the presence of two electronegative atoms, O & Cl. Hence, more susceptible to attack by OH- nucleopile and is the most reactive of the three compounds. In compound T, the carbon chlorine bond is strengthened by the overlap between the p-orbitals of Cl and the pi electron system of the benzene ring. This gives the C-Cl bond a partial double bond character, and hence makes it harder to break.

2. N1 is more basic than N2. The lone pair on N1 is more readily available for donation than N2 due to the

presence of 2 alkyl (electron donating groups) on N1. The availability of the lone pair on N2 atom is reduced due to the electron withdrawing C=O group.

3. The tertiary amine N will cause trimecaine to be basic in nature. Alkyl groups are electron donating, hence lone pair on nitrogen is more available for donation to acid/ dative bonding to a proton, thus rendering it to be basic. Electron withdrawing character of the carbonyl group reduces the electron density on the nitrogen atom, making the lone pair of electrons less available for dative bonding to a proton.

4. a) – partial double bond character present in the C-Cl bond present due to the overlapping of the p-orbital on Cl atom with those of the C atoms of the ring resulting in a stronger C-Cl bond. - C-Cl bond in chloroethane is a single bond hence weaker and easier to break. b) - Chlorobenzene has a larger electron cloud size than chloroethane resulting in its stronger temporary dipole-induced dipole interactions. - The energy at room temperature is insufficient to break all the forces between the chlorobenzene molecules but sufficient to break all the forces between the chloroethane molecules, causing chloroethane to exist in vapour form.

5. Phenylamine is a weaker base than benzylamine due to the delocalisation of the lone pair of electron on the nitrogen atom into the benzene ring. Hence, the lone pair on N atom is less available to accept a proton.

The electron donating alkyl group (C6H5CH2-) increases the electron density on the lone pair of N atom. Hence, the lone pair on N atom is more available to accept a proton. Phenylamine is more reactive than benzylamine towards electrophilic substitution. For phenylamine, there is delocalisation of the lone pair of electron on the nitrogen atom into the benzene ring which increases the electron density in the ring which makes the benzene ring more susceptible to electrophilic substitution.

6. The strength of the base depends on the availability of the lone pair on the nitrogen atom (the more available the lone pair, the more the above equilibrium lies to the right, the stronger the base) In phenylamine, the lone pair is delocalised into the benzene ring, making it much less available to accept a proton. ∴ phenylamine is a weaker base than propylamine In N-methylpropylamine, there are extra electron-donating alkyl groups on the N-atom, this enhances the availability of the lone pair. ∴ N-methylpropylamine is a stronger base than propylamine.

7. Rate will increase when 2-iodobutane is used. C-I bond is weaker than the C-Cl bond due to less effective overlap in the former. Rate of reaction is dependent on the ease of breaking the carbon-halogen bond.

8. i) Increasing pKa: C < A < B Alcohol B is least acidic as the negative charge of the alkoxide ion is localized on the O atom while for phenol A and C, the negative charge on O atom can be delocalized onto the benzene ring, giving a more stable phenoxide anion. C is more acidic than A as C has an electron-withdrawing Cl atom which can further disperse the negative

charge on O atom of the phenoxide ion, thus stabilizing the anion. For A, the –CH3 group is electron releasing and this will intensify the negative charge on the O atom, so destabilizing the phenoxide ion.

ii) Compound C is insoluble in water due to the hydrophobic benzene ring which forms Van der Waals forces with water, releasing insufficient energy to overcome intermolecular hydrogen bonds in water and in compound C. Compound C is acidic and forms a soluble salt sodium phenoxide with aqueous NaOH. Sodium phenoxide is ionic and forms ion-dipole attractions with water molecules, releasing sufficient energy to overcome intermolecular hydrogen bonds in water and in compound C.

9. i) Higher Ka suggests higher acidity of the aqueous compound.Carboxylic acids are stronger acids than phenol.

Carboxylic acids (propanoic acid and benzoic acid) are stronger acids than phenol because the carboxylate ion, RCO2- is resonance stabilized due to the delocalization of the negative charge over the O-C-O bond as compared to the phenoxide ion where it is localized on O. Hence, phenol has the lowest Ka value.

Comparing 2 acids: Benzoic acid in turn is a stronger acid than propanoic acid.

Benzoic acid is more acidic as its carboxylate ion is more stabilised by the delocalisation of the negative charge on O atom into the benzene ring. Hence benzoic acid dissociates to a greater extent as the benzoate ion is more stable and thus the position of equilibrium lies to the right. Hence is the most acidic with the largest Ka value. OR Propanoic acid is a weaker acid than ethanoic acid as the propyl group is electron donating, hence intensifying the negative charge on the anion, making the carboxylate ion less stable. Hence propanoic acid dissociates to a smaller extent than benzoic acid as the position of equilibrium lies to the left.

ii) Y is an electron donating group as the presence of Y decreases the acidity of benzoic acid. Hence, Y is 2, 4 directing.

O 10. C reacts with aq NaOH at room temperature, B r reacts with Br aq NaOH only on warming.

This is because C6H5COBr undergoes nucleophilic substitution more readily than C6H11Br which requires

heating for the reaction with aq NaOH to form C6H11OH. C6H5Br does not react with aq NaOH. Their differences in reaction with aq NaOH is due to differences in structures which result in different ease of substitution.

Due to the presence of electronegative O attached to the carbonyl C, the carbonyl C in C6H5COBr is more

partially positive compared to the C to which Br is attached in C6H11Br.

2 The carbonyl carbon in C6H5COBr is also sp hybridised so that it is less sterically hindered than the 3 sp hybridised carbon to which the Br atom is attached in C6H11Br. Due to the overlap of the orbital containing lonepair of electrons on Br with electron cloud of benzene ring, the CBr has partial double bond character so that it is not easily broken and does not react with aq NaOH even on heating.

11. i) Boiling point increases in the order: C2H5Cl < C2H5Br < C2H5I Down Grp VII, the number of electrons increases, therefore the strength of van de Waals forces increases between the molecules. Hence more energy is required to overcome these forces down the group. ii) C-X bond polarities decrease down the group, due to the decreasing electronegativity difference between C and X as X decreases in electronegativity from Cl to Br to I. iii) Reactivity increases down the Group as the C-X bond strength decreases from C-Cl to C-Br to C-I, thus requiring less energy to break.

12. 4-aminobenzoic acid is more soluble in water. Both 2-aminobenzoic acid and 4-aminobenzoic acid have simple molecular structures There is more extensive intermolecular hydrogen bonding with water molecules in 4-aminobenzoic acid. As intramolecular hydrogen bonding in 2-aminobenzoic acid renders the formation of intermolecular hydrogen bonding with water molecules less feasible.

5. Data based questions

1. i) The primary structure of a protein refers to the order (or sequence) of amino acid residues in a protein (or polypeptide chain). ii)

The question asks for displayed formula, so the molecule should have all bonds drawn, especially the peptide bonds. iii) The chain is held in the shape of α-helix by hydrogen bonds formed between C=O of one peptide linkage and N-H group of another peptide linkage in another part of the chain. It is good to give a sketch of the α-helix to show clearly where the hydrogen bonds are and how they are formed.

iv) -CH2-SH + HS-CH2- +[O]  -CH2-S-S-CH2- + H2O OR -CH2-SH + HS-CH2-  -CH2-S-S-CH2- + 2[H] or H2

Since there are 36 cysteine residues per molecule of HSA, there can be 18 disulfide bridges The equation written should be balanced. The –CH2 chain is required since is part of the R-group of cysteine.

v) Any two (Illustrate with diagrams): Ionic bonding between charged groups – glutamic acid, lysine

Hydrogen bonding between polar groups – threonine, serine

van de Waals’ forces between non-polar group – leucine, valine, phenylalanine

The pair of glutamic acid and lysine is NOT accepted as an example of hydrogen bonding since they exist as ions at physiological pH.

Either glu OR lys with any of the amino acid listed here would be accepted as suitable examples of hydrogen bonding as well. E.g. lys-thr, glu-thr

vi) Since the long carbon chain is non-polar, the amino acids that would interact are:- leucine, valine, phenylalanine (any two).

vii) Any polar amino acid (charged or uncharged) can interact with water. Thus the possible amino acids are:- glutamic acid, lysine, threonine, serine (any three).

2. a) No. of mol of NaOH = x 0.100 = 1.14 x 10-3 mol

Since NaOH ≡ RCO2H -3 No. of mol of RCO2H = 1.14 x 10 mol Mr of A = = 87.7

General formula of carboxylic acid: Cn-1H2n-1COOH 12n + 2n + 16(2) = 87.7 n= 4 Possible structures: CH3CH2CH2COOH or (CH3)2CHCO2H

b) pV = RT

Mr = = = 71.9

C is an alkane, CnH2n+2 12n + 2n + 2 = 71.9 n = 5 Molecular formula of C: C5H12

c) i) P4O10 + 6H2O → 4H3PO4 CO2 + 2NaOH → Na2CO3 + H2O or CO2 + NaOH → NaHCO3

Many thought that the formula for phosphorus pentoxide was PO5 rather than P2O5 or P4O10. Another

common mistake was failure to give balanced chemical equations.

ii) nH2O = = 0.0861 mol

nCO2= = 0.0689 mol By the conservation of mass, the total mass of reactants must be equal to the total mass of products. Thus, the increases in masses result from the absorption of H2O and CO2 respectively. Common mistakes include incorporating the Mr values of H3PO4 and Na2CO3 in calculations. iii) H: C ratio = 2 x 0.0861 : 0.0689 = 1 : 2.5 = 2 : 5 Empirical formula is C2H5. The molecular formula must be C4H10. To obtain the moles of hydrogen, it is necessary to multiply the moles of water by 2. Do not round off the ratio to 1:3. Since D is an alkane, it should have the general formula of CnH2n+2. Other molecular formula will not fit into the general formula for alkanes.

d) If A: CH3CH2CH2CO2H, C: CH3CH2CH2CH2CH3, D: CH3CH2CH2CH3, then B must be CH3CH2CO2H and E must be CH3CH2CH2CH2CH2CH3.

OR If A: (CH3)2CHCO2H, C: (CH3)2CHCH2CH3, D: CH3CH2CH2CH3, then B must be CH3CH2CO2H and

E must be Having discovered that acid A contained a C3 hydrocarbon chain and that alkane C was a C5 alkane, the structures of acid B and alkane E can be deduced EVEN IF a mistake was made in part (e). Alkane E must be symmetrical and has the formula C6H14. Acid B has to be C2H5CO2H.

3. a) i) Primary amine, tertiary amine, phenol, primary amide ii) Disulfide iii) Arginine forms hydrogen bonds or ionic bonds Phenylalanine forms van der waals forces of attraction iv) I. Leucine has a much higher melting point than 1,7-Heptanediamine because the electrostatic attractions (ionic bond) in the crystal lattice of the zwitterionic form is stronger than the intermolecular hydrogen bonding between 1,7-Heptanediamine molecules. II. 1,7-Heptanediamine has a higher melting point than Heptanoic acid. Both have hydrogen bonds between their molecules and similar Mr (similar strength of VDW). However, 1,7-Heptanediamine can form more extensive hydrogen bonds than Heptanoic acid

b) I: Lactic acid dissociates to give H ions, thus lowering the pH, The H+ ions protonate the negatively charged –COO- R groups, thus disrupts the ionic bonds in the protein

II: Ethanol forms hydrogen bonds with the –OH groups in the R groups, hence disrupts the intermolecular hydrogen bonds.

4. a) i) Quaternary structure ii)

Either or

b) i) Gly-Trp-Leu-Ala-Glu-Glu-Met-Lys-Tyr-Trp-Leu-Val-Arg-Val-Asp-Asp-Asp-Phe ii) A – Lys; B – Thr; C – Glu iii) O O O

H H - H2N CH C N CH C N CH C O

CH OH CH2 CH2

CH3 CH2 CH2

C O CH2

- O CH2

NH2 iv) When a small amount of strong base is added, O O

+ - - H3N CH C O H2N CH C O + OH- + H2O HC OH HC OH

CH 3 CH3

When a small amount of strong acid is added, O O

- + + + - H2N CH C O H H3N CH C O

HC OH HC OH

CH CH 3 3

c) i) 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 Benzene side chain (of Heterocyclic arene forces phenylalanine) O atom on phosphate, ribose grp Hydroxyl H atom (of serine) Hydrogen or heterocyclic arene bonding H atom on ribose grp or Hydroxyl O atom (of serine) heterocyclic arene

[Protein H ] ii) Kb = [P  S complex] [H ]

 For [Protein  H ] = 0.7, [P  S complex] + Kb = 0.7 / [H ] [H+] = 7.94 x 10-7 mol dm-3 pH = -lg (7.94 x 10-7) = 6.10

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

iv) Although ΔH is positive when Van der Waals forces and hydrogen bonds are overcomed at 70 °C, this is offset by increase in entropy (ΔS) of the protein when it uncoils. This results in l TΔS l > l ΔH l so that ΔG is negative.

5 i) Ultraviolet (UV) rays from the sun ii) Step 1

Step 2

Step 3

Step 4

iii) Cl radical acts as a homogenous catalyst because it is not consumed in the reaction.