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:

Remember the temperature dependence: Ethanol with $H_2SO_4$ at $413$ K gives an ether, but at $443$ K it gives an alkene (ethene).

Answer 1

Concept:

• Alcohols react with active metals to release $H_2$ gas.

• Compounds with $CH_3CH(OH)-$ group undergo the iodoform test.

• Intermolecular dehydration of alcohols at $413$ K yields ethers.

• Ethers react with concentrated HI to form alkyl iodides.
Step 1: Identifying A, B, and C.
Molecular formula $C_2H_6O$ and reaction with Na suggests 'A' is Ethanol ($CH_3CH_2OH$). \[ 2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa (B) + H_2 \uparrow \] Reaction with $I_2/NaOH$ (Iodoform test) yields a yellow precipitate of Iodoform (C): \[ CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3 (C) + HCOONa + 5NaI + 5H_2O \]

Step 2: Identifying D and E.
Heating Ethanol with $H_2SO_4$ at $413$ K leads to dehydration to form Diethyl ether (D): \[ 2CH_3CH_2OH \xrightarrow[413 K]{H_2SO_4} CH_3CH_2OCH_2CH_3 (D) + H_2O \] Reaction of ether with excess HI yields Ethyl iodide (E): \[ C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I (E) + H_2O \]