Alcohols, Phenols and Ethers Part 2

Concept:Mixed ether cleavage by hot concentrated HI gives two different alkyl iodides.These alkyl iodides on aqueous NaOH hydrolysis yield the corresponding alcohols Q and R.A yellow precipitate with NaOI (iodoform test) indicates the presence of a $\mathrm{CH_3CH(OH)-}$ group in the alcohol.Explanation:Ether P reacts with excess hot conc. $\mathrm{HI}$: $\mathrm{R-O-R' + 2HI \rightarrow R-I + R'-I + H_2O}$.Then $\mathrm{R-I + NaOH(aq) \rightarrow R-OH + NaI}$; similarly for $\mathrm{R'-I}$ gives $\mathrm{R'-OH}$.Both Q and R give iodoform test, so each alcohol must contain a $\mathrm{CH_3CH(OH)-}$ group.Hence the alkyl groups R and R' must be such that their corresponding alcohols are ethanol ($\mathrm{CH_3CH_2OH}$) or a secondary alcohol with a methyl group at the $\alpha$‑carbon (e.g., isopropanol $\mathrm{CH_3CH(OH)CH_3}$).Therefore, both alkyl groups must be either ethyl ($\mathrm{C_2H_5}$) or isopropyl ($\mathrm{CH_3CH\!-\!CH_3}$).Since P is a mixed ether, one alkyl group is ethyl and the other is isopropyl.Thus P is ethyl isopropyl ether: $\mathrm{CH_3CH_2-O-CH(CH_3)_2}$.Answer:The correct option is the one whose structure represents ethyl isopropyl ether (Option B).