1. First reaction: ethanol with $P + I_2$

   Red phosphorus and iodine generate $PI_3$ in situ, which converts alcohols to alkyl iodides.

   $3CH_3CH_2OH + PI_3 \rightarrow 3CH_3CH_2I + H_3PO_3$

   So, $A = CH_3CH_2I \quad (\text{ethyl iodide})$

2. Second reaction: $A$ with Mg in dry ether

   Alkyl iodide reacts with magnesium in dry ether to form the Grignard reagent.

   $CH_3CH_2I + Mg \xrightarrow[\text{dry ether}]{} CH_3CH_2MgI$

   Hence, $B = C_2H_5MgI \quad (\text{ethyl magnesium iodide})$

3. Third reaction: Grignard reagent with formaldehyde

   Grignard reagents react with formaldehyde to give, after hydrolysis, a primary alcohol with one extra carbon.

   First addition step: $CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI$

   So, $C = CH_3CH_2CH_2OMgI$

4. Fourth reaction: hydrolysis

   On hydrolysis: $CH_3CH_2CH_2OMgI + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)I$

   Therefore, $D = CH_3CH_2CH_2OH$

   This is n-propyl alcohol.

5. Check options

   • A: butanal $\to$ incorrect
   • B: n-butyl alcohol $\to$ incorrect
   • C: n-propyl alcohol $\to$ correct
   • D: propanal $\to$ incorrect

Thus, the final compound $D$ is n-propyl alcohol.