Describe the reactions in the scheme, identifying A and D.

Here are the answers to the questions based on the provided reaction scheme:

a)

i) Step I: Propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$) is converted to propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) using $\text{MnO}_4^-/\text{H}^+$. This is an oxidation reaction.

ii) Step II: Propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$) is converted to propene ($\text{CH}_3\text{CH=CH}_2$) by heating with concentrated $\text{H}_2\text{SO}_4$. This is a dehydration reaction.

b)
Step III: Sodium propanoate ($\text{CH}_3\text{CH}_2\text{COONa}$) is heated to produce ethane ($\text{CH}_3\text{CH}_3$). This is a decarboxylation reaction, specifically Kolbe electrolysis or soda lime decarboxylation. Assuming soda lime decarboxylation:

$$\text{CH}_3\text{CH}_2\text{COONa} + \text{NaOH} \xrightarrow{\text{heat}} \text{CH}_3\text{CH}_3 + \text{Na}_2\text{CO}_3$$

c)

• A: Propan-1-ol reacts with potassium metal to form solution A and $\text{H}_2$. Solution A is potassium propoxide.

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{K} \longrightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OK} + \frac{1}{2}\text{H}_2$$

So, A is potassium propoxide ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OK}$).

• D: Ethane ($\text{CH}_3\text{CH}_3$) reacts with 1 mole of $\text{Cl}_2$ to form compound D and $\text{HCl}$. This is a free radical substitution reaction.

$$\text{CH}_3\text{CH}_3 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{CH}_2\text{Cl} + \text{HCl}$$

So, D is chloroethane ($\text{CH}_3\text{CH}_2\text{Cl}$).

• R: In the cleaning agent X, R is an alkyl chain. So, R is an alkyl group.

d)
Product C is formed from the reaction of propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) with Butan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$) in the presence of concentrated $\text{H}_2\text{SO}_4$ and heat (Step IV). This is an esterification reaction.

$$\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4/\text{heat}} \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$$

The structural formula of product C (butyl propanoate) is:

\begin{center} \begin{tikzpicture}[scale=0.8] \node (C1) at (0,0) {C}; \node (C2) at (1,0) {C}; \node (C3) at (2,0) {C}; \node (O1) at (2.5,0.5) {O}; \node (O2) at (3,-0.5) {O}; \node (C4) at (3.5,-0.5) {C}; \node (C5) at (4.5,-0.5) {C}; \node (C6) at (5.5,-0.5) {C}; \node (C7) at (6.5,-0.5) {C}; \draw (C1) -- (C2) -- (C3); \draw (C3) -- (O1); \draw (C3) -- (O2); \draw (O2) -- (C4) -- (C5) -- (C6) -- (C7); \node at (0,-0.5) {H$_3$}; \node at (1,-0.5) {H$_2$}; \node at (2,0.5) {O}; \node at (3.5,-1) {H$_2$}; \node at (4.5,-1) {H$_2$}; \node at (5.5,-1) {H$_2$}; \node at (6.5,-1) {H$_3$}; \end{tikzpicture} \end{center}

Or written linearly: $\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$.

e)
Compound E is formed from propene ($\text{CH}_3\text{CH=CH}_2$) by reaction with $\text{HCl}$. This is an addition reaction following Markovnikov's rule.

$$\text{CH}_3\text{CH=CH}_2 + \text{HCl} \longrightarrow \text{CH}_3\text{CHClCH}_3$$

So, E is 2-chloropropane.

f)
Compound B is formed from propene ($\text{CH}_3\text{CH=CH}_2$) through a polymerization reaction. The structure shown for Compound B is a repeating unit of propene.
The type of reaction involved in its formation is addition polymerization.

g)
Compound B is polypropene, with the repeating unit $\text{[-CH}_2\text{-CH(CH}_3)\text{-]}_n$.
The relative molecular mass of B is 35700.
The molecular mass of one repeating unit ($\text{CH}_2\text{CH(CH}_3)$) is:

$$(2 \times 12.01) + (3 \times 1.008) + (12.01) + (3 \times 1.008) = 24.02 + 3.024 + 12.01 + 3.024 = 42.078 \text{ g/mol}$$

Using approximate atomic masses: C=12, H=1.
Mass of one repeating unit = $(3 \times 12) + (6 \times 1) = 36 + 6 = 42$.
The relative molecular mass of B is $n \times (\text{mass of one repeating unit})$.

$$35700 = n \times 42$$ $$n = \frac{35700}{42}$$ $$n = 850$$

The value of n is $\boxed{\text{850}}$.

h)

i) Identify the cleaning agent suitable to be used with water containing magnesium chloride.
Magnesium chloride ($\text{MgCl}_2$) causes hard water.
Cleaning agent X is a soap ($\text{R-COONa}$). Soaps form scum with hard water ions like $\text{Mg}^{2+}$.
Cleaning agent Y is a synthetic detergent ($\text{R-O-SO}_3^-\text{Na}^+$). Synthetic detergents do not form scum with hard water ions.
Therefore, cleaning agent Y is suitable.

ii) State one advantage of using cleaning agent Y.
One advantage of using cleaning agent Y (a synthetic detergent) is that it does not form scum with hard water, making it effective in both hard and soft water.