Question

An organic compound 'A', with molecular formula $C_2H_6O$ reacts with active metals such as sodium to give compound 'B' and hydrogen gas. 'A' on treatment with iodine and sodium hydroxide gives 'C' and in presence of $H_2SO_4$ at $413~K$ gives 'D' ($C_4H_{10}O$). 'D' on reaction with excess of HI gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write all the reactions involved.

Hint:

Temperature matters: $H_2SO_4$ at $413~K$ gives ethers, but at $443~K$ gives alkenes.

Answer 1

Step 1: Concept
Reactions of alcohols and ethers.

Step 2: Meaning
Reactivity with Na indicates an alcohol. Iodoform test confirms a $CH_3CH(OH)-$ group.

Step 3: Analysis
1. Formula $C_2H_6O$ releasing $H_2$ with Na is Ethanol ('A').
Reaction: $2CH_3CH_2OH \\ \text{ (A)} + 2Na \rightarrow 2CH_3CH_2ONa \\ \text{ (B)} + H_2$. So 'B' is Sodium ethoxide.
2. Ethanol reacts with $I_2/NaOH$ (iodoform reaction) to give a yellow precipitate of Iodoform ('C').
Reaction: $CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3 \\ \text{ (C)} + HCOONa + 5NaI + 5H_2O$.
3. Ethanol heated with $H_2SO_4$ at $413~K$ undergoes intermolecular dehydration to form an ether ('D').
Reaction: $2CH_3CH_2OH \xrightarrow{H_2SO_4, 413~K} C_2H_5-O-C_2H_5 \\ \text{ (D)} + H_2O$. 'D' is Diethyl ether (Ethoxyethane).
4. Cleavage of ether 'D' with excess HI yields alkyl iodides.
Reaction: $C_2H_5-O-C_2H_5 + 2HI \rightarrow 2C_2H_5I \\ \text{ (E)} + H_2O$. 'E' is Ethyl iodide (Iodoethane).

Step 4: Conclusion
The sequential transformations map directly to the identified structures.

Final Answer:
A: Ethanol ($CH_3CH_2OH$)
B: Sodium ethoxide ($CH_3CH_2ONa$)
C: Iodoform ($CHI_3$)
D: Diethyl ether or Ethoxyethane ($C_2H_5-O-C_2H_5$)
E: Ethyl iodide or Iodoethane ($C_2H_5I$)
Reactions are detailed in Step 3.