Chemistry 2013 Jamb Examination

The structure consists of a five-carbon chain with a double bond at carbon 1 and a methyl substituent at carbon 4. Numbering starts from the end nearest to the double bond to give the lowest possible number to the double bond. The methyl group is then attached to the fourth carbon atom, giving 4-methylpent-1-ene as the name. The suffix 'ene' indicates the presence of a double bond, and 'pent' denotes a five-carbon chain. This naming follows IUPAC rules for alkenes where the double bond has priority in numbering.

Dehydration of an ammonium salt of an alkanoic acid involves heating the salt formed from a carboxylic acid and ammonia. The process removes a molecule of water to produce an amide. The general reaction is \(RCOONH_4 \rightarrow RCONH_2 + H_2O\). Hence, the compound formed has the general formula \(RCONH_2\), which corresponds to an amide. Amides are produced when the -OH group of the carboxylic acid and a hydrogen atom from the ammonium ion are eliminated as water during heating.

Assume 100 g of the compound.
Carbon = 60 g → \(\dfrac{60}{12} = 5\) mol
Hydrogen = 13.3 g → \(\dfrac{13.3}{1} = 13.3\) mol
Oxygen = 26.7 g → \(\dfrac{26.7}{16} = 1.67\) mol

To get the simplest ratio, divide through by the smallest value (1.67):
C : H : O = \(\dfrac{5}{1.67} : \dfrac{13.3}{1.67} : \dfrac{1.67}{1.67} = 3 : 8 : 1\)

Therefore, the empirical formula is C₃H₈O. This represents a molecule containing three carbon atoms, eight hydrogen atoms, and one oxygen atom in the simplest whole-number ratio.

In 2-methylbutan-2-ol, the hydroxyl group (-OH) is attached to a carbon atom that is connected to three other carbon atoms, making it a tertiary alcohol. The naming shows that the -OH group is on the second carbon of the chain and the methyl group is also attached to the same carbon. Tertiary alcohols are those where the hydroxyl-bearing carbon has no hydrogen atoms directly attached to it, confirming its classification as a tertiary alkanol.

Ammonia reacts with an ester such as ethyl ethanoate in a nucleophilic substitution reaction. The ammonia attacks the carbonyl carbon, leading to the replacement of the ethoxy group (-OC₂H₅) with an amino group (-NH₂). The products are ethanol and ethanamide. The overall reaction is:
CH₃COOC₂H₅ + NH₃ → CH₃CONH₂ + C₂H₅OH
Hence, the organic amide formed is ethanamide and the alcohol is ethanol.

Decarboxylation is the removal of a carbon dioxide molecule from a carboxylic acid. When ethanoic acid (CH₃COOH) undergoes decarboxylation, it loses the -COOH group, producing methane (CH₄) and carbon dioxide (CO₂). The reaction is represented as:
CH₃COOH → CH₄ + CO₂
This occurs because the remaining carbon atom is bonded to hydrogen atoms only, forming the simplest alkane, methane.

Primary amides are derived from carboxylic acids where the hydroxyl group is replaced by an amino group (-NH₂). Hence, their general formula is RCONH₂. In this structure, R represents an alkyl or aryl group, and the amide nitrogen is attached to two hydrogen atoms. The carbonyl carbon is directly bonded to the nitrogen atom, forming the amide linkage (CONH₂).

Cracking is used to convert long-chain hydrocarbons into shorter, more useful ones such as petrol and alkenes. Diesel oil, which contains long-chain hydrocarbons, is suitable for this process. It is heated to high temperatures and subjected to catalysts or steam to break the large molecules into smaller ones. This makes diesel oils an ideal raw material for cracking.

When sulphur (IV) oxide (SO₂) dissolves in water, it forms sulphurous acid (H₂SO₃), which ionises to give trioxosulphate (IV) ions. These ions act as reducing agents by readily donating electrons to other substances. The reaction proceeds as:
SO₂ + H₂O → H₂SO₃
H₂SO₃ ⇌ 2H⁺ + SO₃²⁻
The trioxosulphate (IV) ion, SO₃²⁻, is therefore responsible for the reducing properties of sulphur (IV) oxide in aqueous solution.

Potassium forms soluble trioxosulphate (IV) salts such as potassium trioxosulphate (IV) (K₂SO₃). The solubility of trioxosulphate (IV) salts depends on the metal ion; salts of Group I elements like potassium and sodium are generally soluble in water, while those of Group II metals like barium are insoluble. This solubility arises from the high lattice energy of potassium salts being overcome by hydration energy in aqueous solution.