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 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 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 62: [Organic > Carbonyl Compounds]

The compound hex-3-en-1-ol, P, has a strong ‘leafy’ smell of newly cut grass and is used in perfumery.

P is:        CH₃CH₂CH=CHCH₂CH₂OH

What is produced when P is treated with an excess of hot concentrated acidic KMnO₄?

A CH₃CH₂CH(OH)CH(OH)CH₂CH₂OH

B CH₃CH₂CH=CHCH₂CH₂CO₂H

C CH₃CH₂CHO and OCHCH₂CH₂OH

D CH₃CH₂CO₂H and HO₂CCH₂CO₂H

Reference: Past Exam Paper –June 2003 Paper 1 Q22

Solution 62:

Answer: D.

Double bond breaks. Final carbon in CH₃CH₂CH is bonded to H and C atoms thus, under the said conditions, will oxidise into an aldehyde: CH₃CH₂CHO. Aldehyde will then oxidise into a carboxylic acid: CH₃CH₂CO₂H. Going back to the other product of the initial double bond break, first carbon is bonded to H and C atoms thus, under said conditions, will also oxidise into an aldehyde. Furthermore, the primary alcohol group will also oxidise into an aldehyde, thus producing: HO₂CCH₂CO₂H. These 2 products are only shown in option D, thus making it the correct option.

Question 40: [Organic > Carbonyl Compounds]:

In the reaction between an aldehyde and HCN catalysed by NaCN, which statements about the reaction mechanism are true?

1 A new carbon-carbon bond is formed.

2 In the intermediate, the oxygen carried a negative charge.

3 The last stage involves the formation of a hydrogen-oxygen bond.

Reference: Past Exam Paper – November 2002 Paper 1 Q40

Solution 40:

Answer: A.

The CN^- ions attack the organic compound creating a new C-C bond, thus 1 is correct. In the intermediate the oxygen will carry a negative charge in order to attract the remaining H⁺ ion, thus 2 is also correct. Due to the formation of the O-H bond 3 must also be correct thus A is the valid option. 

For the sake of argument, to ensure that 3 is completely correct the O-H bond must be the last stage. This is proved by the fact that the mechanism of this reaction: nucleophilic addition. This means that first the nucleophile (CN^-) attacks then the remaining ions (H⁺) react last, hence completely proving 3.

Question 28: [Organic > Carbonyl Compounds]:

Ethanal may be converted into a three-carbon acid in a two-step process.

Which compound is the intermediate?

A CH₃CO₂H

B CH₃CN

C CH₃CH₂CN

D CH₃CH(OH)CN

Reference: Past Exam Paper – November 2002 Paper 1 Q28

Solution 28:

Answer: D.

As the number of carbons is increasing from 2 (ethanal) to 3, the addition of a cyanide ion is necessary as that is the only method to increase the carbon chain length. Therefore A is eliminated.

B is also eliminated as there are now only two carbons and it is impossible to get that compound from ethanal.

C has 3 carbons however does not have an alcohol group which can be oxidized to form an acid.

D is the only option that satisfies all conditions.

Question 27: [Organic > Carbonyl Compounds]:

Burnt sugar has a characteristic smell caused partly by the following compound. It has two functional groups indicated by Q and R.

	2,4-dinitrophenylhydrazine	Fehling’s reagent
A	Q and R	Q and R
B	R only	Q and R
C	Q and R	R only
D	Q only	R only