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 61: [Organic > Introductory Organic]
The isomers, citric acid and isocitric acid, are
intermediates in the Krebs cycle of the oxidation of glucose in living cells.
How many chiral centres does each acid possess?
|
|
citric acid
|
isocitric acid
|
|
A
|
0
|
1
|
|
B
|
0
|
2
|
|
C
|
1
|
1
|
|
D
|
1
|
2
|
Reference: Past Exam Paper –June 2003 Paper 1 Q21
Solution 61:
Answer: B.
All carbon atoms in CO2H group are not chiral.
All carbon atoms in CH2 groups are not chiral. In citric acid the
only remaining left-middle carbon atom is also not chiral since the bottom
group exactly matches the top, cancelling out options C and D. In isocitric
acid the left-middle carbon atom is chiral since the groups attached to it are:
H, CO2H, CH2CO2H and C(OH)CO2H,
which are all different. The bottom-left carbon atom is also chiral since the
groups attached to it are: H, OH, CO2H and the remaining top, which
are also all different. Thus the correct answer is option B.
Question 78: [Inorganic > Group 7]
Chloroethane can be formed from bromoethane in two steps.
Which statements about these steps are correct?
1 Step X involves a nucleophilic substitution.
2 Hot aqueous sodium hydroxide is the reagent in step X.
3 Hot aqueous sodium chloride is the reagent in step Y.
Reference: Past Exam Paper - June 2003 Paper 1 Q38
Solution 78:
Answer: B.
Statement 2 is true because a chiral centre is defined as a carbon with four different substituents and a chiral centre will always display optical isomerism therefore Statement 1 is also true.
Statement 3 could be true however no condition that will prove it happening every time so cannot be true.
Question 77: [Organic > Introductory Organic]
What will always be a characteristic of a compound containing a single carbon atom with four
different groups bonded to it?
1 It will have an optical isomer.
2 It will have a chiral centre.
3 It will have a structural isomer.
Reference: Past Exam Paper -June 2003 Paper 1 Q37
Solution 77:
Answer: B.
Statement 2 is true because a chiral centre is defined as a carbon with four different substituents and a chiral centre will always display optical isomerism therefore Statement 1 is also true.
Statement 3 could be true however no condition that will prove it happening every time so cannot be true.
Question 76: [Inorganic > Group 7]
Why is the addition of concentrated sulphuric acid to solid potassium iodide unsuitable for the
preparation of hydrogen iodide?
1 Hydrogen iodide is not displaced by sulphuric acid.
2 Iodide ions are oxidised to iodine.
3 The product is contaminated by sulphur compounds.
Reference: Past Exam Paper -June 2003 Paper 1 Q36
Solution 76:
Answer: C.
Statement 1 is false because sulphuric acid is a stronger oxidizing agent than hydrogen iodide so iodine can be displaced by sulphuric acid
Statement 2 is true as H2SO4 displaces HI to H2S and I2 and H2O
Statement 3 is true as there are sulphur compounds present, H2S and SO2
Question 75: [Inorganic > Nitrogen & Sulphur]
Which statements are reasons why sulphur dioxide is used as a food preservative?
1 It is a reducing agent and therefore an anti-oxidant.
2 It prevents alcohols forming sour-tasting acids.
3 It does not smell and therefore can be used in more than trace quantities.
Reference: Past Exam Paper -June 2003 Paper 1 Q35
Solution 75:
Answer: B.
Statement 1 is true; a reducing agent is oxidised itself and therefore prevents other substance from being oxidises therefore an anti-oxidant.
Statement 2 is true because it prevents alcohols from being oxidised to acid as it is an anti-oxidant
Statement 3 is false because it does have a pungent smell
Question 74: [Physical > Chemical Bonding]
Which of the following statements are correct for the sequence of compounds below considered
from left to right?
NaF MgO AlN SiC
1 The electronegativity difference between the elements in each compound increases.
2 The formula-units of these compounds are isoelectronic (have the same number of
electrons).
3 The bonding becomes increasingly covalent
Reference: Past Exam Paper -June 2003 Paper 1 Q34
Solution 74:
Answer: C.
Statement 1 is false because NaF has the greatest difference in electronegativity; one is from group 1 while the other is in group 7.
Statement 2 is true, they all have 18 electrons
Statement 3 is true because initially the bonding is between a non-metal and a metal but by the end it is between two non-metals.
Question 73: [Physical > Chemical Bonding]
The conversion of graphite into diamond is an endothermic reaction (ΔH = +3 kJ mol–1).
C(graphite) → C(diamond)
Which statements are correct?
1 The enthalpy change of atomisation of diamond is smaller than that of graphite.
2 The bond energy of the C–C bonds in graphite is greater than that in diamond.
3 The enthalpy change of combustion of diamond is greater than that of graphite.
Reference: Past Exam Paper -June 2003 Paper 1 Q33
Solution 73:
Answer: A.
Statement 1 is true because diamond is at a higher level of energy than graphite so it would be easier for it to reach the gaseous state.
If Statement 1 is true then 2 is too because if diamond has a smaller enthalpy change of atomisation then it means the C-C bonds are weaker in diamond than in graphite
Statement 3 is true because the intermolecular forces are higher in graphite so if we were to apply the same heat as to something like diamond, the diamond would break apart easier, and release much more heat, while the same heat applied to graphite will not release as much.
Question 72: [Physical > Chemical Bonding]
Long-chain alkanes are converted on an industrial scale into alkylsulphates for use as detergents,
e.g. sodium lauryl sulphate.
What deductions about the properties of this substance can be made from this structure?
1 Part of the structure is polar and is water–attracting.
2 The alkyl chain is soluble in oil droplets.
3 All the C-C-C bond angles are tetrahedral.
Reference: Past Exam Paper -June 2003 Paper 1 Q32
Solution 72:
Answer: A.
Option 1 is true that sodium lauryl suplhate is polar because it is made up of more than one type of atom and it does not have a symmetrical structure therefore has permanent dipoles.
Option 3 is also true because each C has four other single bonds which creates a tetrahedral structure.
Question 71: [Physical > Chemical Bonding]
Silicon tetrachloride, SiCl4, is a liquid of low boiling point. In the presence of water it decomposes
to form silicon(IV) oxide and hydrogen chloride.
What types of bonding occur in SiCl4(I)?
1 co–ordinate bonding
2 covalent bonding
3 van der Waals forces
Reference: Past Exam Paper -June 2003 Paper 1 Q31
Solution 71:
Answer: C.
Silicon tetrachloride is a molecule with covalent bonds therefore option 2 is true.
There are no coordiante bonds within silicon tetrachloride however as it has a simple molecular structure, VDW forces are present between molecules therefore option 3 is true.
.png)
