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: Identify compound A.
Molecular formula $C_2H_6O$, releases $H_2$ with sodium metal — this confirms an alcohol. The compound is ethanol ($CH_3CH_2OH$, A). Compound B (sodium ethoxide): \[ 2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa~(B) + H_2 \]
Step 2: Identify compound C (iodoform reaction).
Ethanol reacts with $I_2$ and $NaOH$ because it contains the $-CH_3-C(OH)-$ group. A yellow precipitate of iodoform ($CHI_3$, C) is formed: \[ CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3~(C) + HCOONa + 5NaI + 5H_2O \]
Step 3: Identify compound D (intermolecular dehydration).
Ethanol heated with concentrated $H_2SO_4$ at 413 K undergoes intermolecular dehydration to form an ether: \[ 2CH_3CH_2OH \xrightarrow{H_2SO_4,~413~K} C_2H_5-O-C_2H_5~(D) + H_2O \] D is diethyl ether (ethoxyethane, $C_4H_{10}O$).
Step 4: Identify compound E (cleavage of ether).
Diethyl ether reacts with excess HI to give ethyl iodide: \[ C_2H_5-O-C_2H_5 + 2HI \rightarrow 2C_2H_5I~(E) + H_2O \] A = Ethanol; B = Sodium ethoxide ($CH_3CH_2ONa$); C = Iodoform ($CHI_3$); D = Diethyl ether ($C_2H_5OC_2H_5$); E = Ethyl iodide ($C_2H_5I$).