Give equations for the reaction of iodine with aqueous sodium hydroxide, the preparation of HCl from NaCl and H2SO4, why HI cannot be prepared similarly, and a suitable method for preparing HI.

SECTION A: INORGANIC CHEMISTRY

A) i) Give an equation for the reaction of iodine with aqueous sodium hydroxide.
When iodine reacts with cold, dilute aqueous sodium hydroxide, it undergoes disproportionation.

$$\text{I}_2(\text{aq}) + 2\text{NaOH}(\text{aq}) \longrightarrow \text{NaI}(\text{aq}) + \text{NaIO}(\text{aq}) + \text{H}_2\text{O}(\text{l})$$

A) ii) Write an equation to show how HCl can be prepared from NaCl and concentrated sulphuric acid.
Hydrogen chloride gas can be prepared by reacting solid sodium chloride with concentrated sulfuric acid.

$$\text{NaCl}(\text{s}) + \text{H}_2\text{SO}_4(\text{conc}) \longrightarrow \text{NaHSO}_4(\text{s}) + \text{HCl}(\text{g})$$

A) iii) Explain why HI cannot be prepared in a similar way.
Concentrated sulfuric acid is a strong oxidizing agent. Hydrogen iodide (HI) is a strong reducing agent. If concentrated sulfuric acid were used to react with an iodide salt, the sulfuric acid would oxidize the iodide ions (and thus the HI formed) to iodine ($I_2$), instead of simply forming HI.

$$2\text{HI}(\text{g}) + \text{H}_2\text{SO}_4(\text{conc}) \longrightarrow \text{I}_2(\text{s}) + \text{SO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$

C) Give a suitable method for preparing HI.
Hydrogen iodide can be prepared by reacting an iodide salt (e.g., sodium iodide) with a non-oxidizing acid, such as concentrated phosphoric acid.

$$\text{NaI}(\text{s}) + \text{H}_3\text{PO}_4(\text{conc}) \longrightarrow \text{NaH}_2\text{PO}_4(\text{s}) + \text{HI}(\text{g})$$

SECTION C: ORGANIC CHEMISTRY

a) Give the systematic name of the following compounds:

• i) $\text{CH}_3\text{CHFCCl}_2\text{CH}_2\text{OH}$

  • The longest carbon chain containing the hydroxyl group has 4 carbons (butanol). The hydroxyl group is at C1.
  • Substituents: two chloro groups at C2, one fluoro group at C3.
  • Name: $\boxed{\text{2,2-dichloro-3-fluorobutan-1-ol}}$

• ii) $\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CHO}$

  • The longest carbon chain containing the aldehyde group has 4 carbons (butanal). The aldehyde carbon is C1.
  • Substituent: methyl group at C2.
  • Name: $\boxed{\text{2-methylbutanal}}$

• iii) $\text{CH}_3\text{CH}_2\text{CONH}_2$

  • This is an amide with a 3-carbon chain.
  • Name: $\boxed{\text{Propanamide}}$

• iv) $\text{CH}_3\text{CH(NH}_2)\text{COOH}$

  • This is an amino acid with a 2-carbon chain. The carboxylic acid carbon is C1.
  • Substituent: amino group at C2.
  • Name: $\boxed{\text{2-aminoethanoic acid}}$

b) Draw the structural formula of each of the following compounds:

• i) 2-methylbutanoic acid

  $$\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOH}$$

  or

  $$\begin{matrix} & & \text{O} \\ & & \parallel \\ \text{CH}_3 & - & \text{CH}_2 & - & \text{CH} & - & \text{C} & - & \text{OH} \\ & & & & | \\ & & & & \text{CH}_3 \end{matrix}$$

• ii) Hexan-2-one (2-Hexanone)

  $$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$$

  or

  $$\begin{matrix} \text{CH}_3 & - & \text{C} & - & \text{CH}_2 & - & \text{CH}_2 & - & \text{CH}_2 & - & \text{CH}_3 \\ & & \parallel \\ & & \text{O} \end{matrix}$$

• iii) 2,4,6-tribromophenol

  $$\begin{matrix} & & \text{OH} \\ & & | \\ & \text{Br} & - & \text{C} & - & \text{C} & - & \text{Br} \\ & & \parallel & & \parallel \\ & \text{C} & & & & \text{C} \\ & & \backslash & & / \\ & & \text{C} & - & \text{C} \\ & & | \\ & & \text{Br} \end{matrix}$$

  (A benzene ring with -OH at C1, and Br at C2, C4, C6)

c) Carbonyls are a class of organic compounds consisting of aldehydes and ketones.

• i) What is the common feature in this class of organic compounds? The common feature is the carbonyl group ($\text{C=O}$).

• ii) What general method may be used to prepare both aldehydes and ketones? Both aldehydes and ketones can be prepared by the oxidation of alcohols. Primary alcohols are oxidized to aldehydes, and secondary alcohols are oxidized to ketones.

• iii) How would you distinguish between aldehydes and ketones? Aldehydes can be distinguished from ketones using Tollens' reagent or Fehling's solution.

  • Tollens' reagent: Aldehydes react with Tollens' reagent (ammoniacal silver nitrate) to form a silver mirror on the inside of the test tube, while ketones do not react.
  • Fehling's solution: Aldehydes reduce the blue copper(II) ions in Fehling's solution to a brick-red precipitate of copper(I) oxide, while ketones do not react.

• iv) Why do the carbonyl compounds have higher boiling points than corresponding alkanes? Carbonyl compounds (aldehydes and ketones) are polar molecules due to the significant electronegativity difference between carbon and oxygen in the $\text{C=O}$ bond. This polarity results in dipole-dipole interactions between molecules. Alkanes, being non-polar, only exhibit weaker London dispersion forces. The stronger dipole-dipole interactions in carbonyl compounds require more energy to overcome, leading to higher boiling points compared to alkanes of similar molecular mass.