Question 80: [Organic > Carboxylic Acids and Derivatives]

Mevalonic acid is an intermediate in the biosynthesis of cholesterol, and is shown below.

Which properties does mevalonic acid have?

1 It has only 1 chiral carbon atom.

2 It can be esterified both by ethanoic acid and by ethanol, in the presence of H⁺ ions.

3 It contains both primary and secondary alcohol groups.

Reference: Past Exam Paper –June 2003 Paper 1 Q40

Solution 80:

Answer: B.

Fact 1 is correct since the central carbon atom is bonded to 4 different groups, whereas the remaining carbon atoms have bonds to 2 or more H atoms, or have a double bond to an oxygen atom. Fact 2 is also correct, since mevalonic acid has a carboxylic acid group on the left-hand side and an alcohol group on the right-hand side. This allows esterification by both ethanolic acid and ethanol. Fact 3 is incorrect since mevalonic acid has a primary alcohol (2 or more H atoms) group on the right-hand side, and a tertiary alcohol (No H atoms) in the centre. Thus option B is the correct answer.

Question 79: [Organic > Hydroxyl Compounds]

When the apparatus below was used with compound Z, a brick-red precipitate formed in the right-hand tube.

Which compound could be Z?

1 CH₃CH(OH)CH₃

2 CH₃CH₂CH₂OH

3 CH₃OH

Reference: Past Exam Paper –June 2003 Paper 1 Q39

Solution 119:

Answer: C.

1, 2 and 3 are all alcohols. When an alcohol is presented to heated acidified potassium dichromate they will oxidise. In 1, the alcohol has 1 H atom bonded to the alcohol-bearing C atom and is secondary which will form a ketone. In 2 and 3 the alcohols have 2 or more H atoms bonded to the alcohol-bearing C atom and are primary which will form aldehydes. Aldehydes turn into a brick red precipitate when presented to Fehling’s reagent. Thus 2 and 3 are the only possible compounds that Z can be. Thus the correct answer is option C

Question 70: [Organic > Carbonyl Compounds]

The product of the reaction between propanone and hydrogen cyanide is hydrolysed under acidic conditions.

What is the formula of the final product?

A CH₃CH(OH)CO₂H

B CH₃CH₂CH₂CO₂H

C (CH₃)₂CHCONH₂

D (CH₃)₂C(OH)CO₂H

Reference: Past Exam Paper –June 2003 Paper 1 Q30

Solution 70:

Answer: D.

First observe the formula for propanone: (CH₃)₂C=O. The reactants undergo nucleophilic addition. The CN from HCN bonds with the central carbon atom, breaking the double bond with oxygen into a single bond. The oxygen has a polar charge attracting the remaining H atom from HCN, to form an alcohol group. Under acidic conditions the CN group is oxidised into a carboxylic acid. This is best shown by the product in option D.

Question 69: [Organic > Hydroxyl Compounds]

In a preparation of ethene, ethanol is added a drop at a time to a heated reagent Y. To purify the ethane it is bubbled through a solution Z and then collected.

What could reagent reagent Y and solution Z be?

Reference: Past Exam Paper –June 2003 Paper 1 Q29

Solution 69:

Answer: C.

Alcohol undergoes dehydration to form an alkene. Concentrated H₂SO₄ is among the possible conditions for this reaction. This eliminates options A and D as possible answers. After dehydration occurs there may be remaining H₂SO₄ mixed with the alkene. This remainder must be neutralised. It is acidic and in option B, the addition of more H₂SO₄ will contaminate the alkene more, thus is an incorrect answer. A basic solution will neutralise the acid, such as the dilute NaOH in option C, which is the correct answer.

Question 68: [Organic > Carbonyl Compounds]

In its reaction with sodium, 1 mol of a compound X gives 1 mol of H₂(g).

Which compound might X be?

A CH₃CH₂CH₂CH₂OH

B (CH₃)₃COH

C CH₃CH₂CH₂CO₂H

D CH₃CH(OH)CO₂H

Reference: Past Exam Paper –June 2003 Paper 1 Q28

Solution 68:

Answer: D.

1 alcohol group reacts with 1 Na atom to produce 1 hydrogen atom. The same occurs for 1 carboxylic acid group. 2 form 1 mole of H₂ the compound would need to produce 2 H atoms. This can only be done if the compound has 2 alcohol groups, or 2 carboxylic acid groups, or 1 of each. In option A, there is only 1 alcohol group. In option B, there is also only 1 alcohol group. In option C, there is 1 carboxylic acid group. In option D, there is 1 carboxylic acid group and 1 alcohol group. Thus option D is the correct answer.

Question 67: [Organic > Halogen Derivatives]

Chlorofluoroalkanes, commonly known as CFCs, undergo homolytic fission by ultraviolet irradiation in the stratosphere.

Which radical could result from the irradiation of CHFClCF₂Cl?

A CHFClĊFCl

B ĊHClCF₂Cl

C ĊHFCF₂Cl

D ĊFClCF₂Cl

Reference: Past Exam Paper –June 2003 Paper 1 Q27

Solution 67:

Answer: C.

In the given CFC above, there are C-H bonds, C-F bonds, C-Cl bonds and a C-C bond. In all the options the C-C bond remains intact. The order of bond strength of the remaining 3 bonds is: C-H>C-F>C-Cl. Thus a C-Cl bond will be broken since it requires the least energy. This only occurs in option C. In option A, the CFC loses a F atom, suggesting that a C-F bond is broken. In option B, The CFC loses a different F atom, yet still suggesting that a C-F bond is broken. In option D, the CFC loses a H atom, suggesting that a C-H bond is broken. According to the order of bond strengths these options are incorrect. Thus option C is the correct answer.

Question 66: [Organic > Halogen Derivatives]

The reaction scheme outlines the production of one of the monomers of nylon 66 from compound X.

Which compound could be X?

A BrCH₂CH₂CH₂CH₂Br

B CH₂=CHCH=CH₂

C HOCH₂CH₂CH₂CH₂OH

D HO₂CCH₂CH₂CH₂CH₂CO₂H

Reference: Past Exam Paper –June 2003 Paper 1 Q26

Solution 66:

Answer: A.

In option A, the halogenoalkane has Br atoms which are replaced with CN groups by nucleophilic substitution. In option B, the alkene does not react with KCN. In option C, the alcohol does not react with KCN. In option D, the carboxylic acid does not react with KCN. Thus option A is the correct answer.

Question 65: [Organic > Carbonyl Compounds]

Which reaction is not an electrophilic addition?

A CH₂=CH₂ + HI à CH₃CH₂I

B CH₃CH=CH₂ + Br₂ à CH₃CHBrCH₂Br

C CH₃CH=CH₂ + H₂O à CH₃CH(OH)CH₃

D CH₃CHO + HCN à CH₃CH(OH)CN

Note: in option C there is a catalyst - conc H₂SO₄

Reference: Past Exam Paper –June 2003 Paper 1 Q25

Solution 65:

Answer: D.

In organic reactions electrophilic addition occurs wherever a double bond is broken. In option A, a double bond is replaced with H and I atoms. In option B, the double bond is replaced with bromine atoms. In option C, the double bonds are replaced with an H atom and an OH group. In option D, there is no double bond and no electrophilic addition, since the reaction between aldehyde and HCN is nucleophilic addition. Thus option D is the correct answer.

Question 64: [Organic > Halogen Derivatives]

Cyclohexa-1,4-diene is treated with a solution of bromine in tetrachloromethane.

Reference: Past Exam Paper –June 2003 Paper 1 Q24

Solution 64:

Answer: B.

Double bond present on the right and left sides of cyclohexa-1,4diene. They undergo electrophilic addition due to presence of bromine molecules. 4 carbon atoms on the sides are unsaturated, where the bromine atoms will be added. This is depicted best by option B. In option A, the 2 bromine atoms on the sides are correctly placed, however the bromine atoms on the top and bottom are misplaced since the carbon atoms they are bonded to were already saturated. In option D, all the bromine atoms are misplaced for the same reason as the top and bottom bromine atoms in option A. In option C, the double bonds never broke which is completely illogical as they must break in presence of bromine solution. Thus options A, C and D are all incorrect.

Question 103: [Organic > Hydrocarbons]

The complete combustion of alkanes to produce carbon dioxide and water is an important exothermic reaction.

Which line on the graph shows the relationship between the number of carbon atoms in the alkane and the number of moles of oxygen gas needed for complete combustion of the alkane?

Reference: Past Exam Paper –June 2003 Paper 1 Q23

Solution 63:

Answer: B.

1 mole of oxygen gas has 2 oxygen atoms. In the combustion of 1 carbon atom, 1 mole of CO₂ is formed. This requires 1 carbon atom and 2 oxygen atoms. Thus 2 carbon atoms requires 2 moles of oxygen gas (which actually has 4 atoms of oxygen), to form 2 moles of CO₂. Hence the number of moles of oxygen gas and number of carbon atoms are directly proportional, in a 1:1 ratio. This is only depicted by the line in option B. Option A and C are curved thus do not show a constant proportionality, thus are eliminated. Option D suggests the amount of oxygen gas required for the combustion of any number of carbon atoms is the same, which is illogical, thus is eliminated.