A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

When an ether is treated with excess hot conc. $$HI$$, both the $$C-O$$ bonds finally cleave to give the two corresponding alkyl iodides:

$$R-O-R' \;+\;2\,HI \;\xrightarrow[\text{conc.}]{\text{hot}}\; R-I \;+\; R'-I \;+\;H_2O$$

The two alkyl iodides formed here are then converted into the corresponding alcohols on treatment with aqueous $$NaOH$$:

$$R-I \;+\;NaOH_{(aq)} \;\longrightarrow\; R-OH$$

According to the statement, both alcohols $$Q$$ and $$R$$ give a yellow precipitate with $$NaOI$$ (iodoform test).
The iodoform test is given only by:

• Ethanol, $$CH_3CH_2OH$$ (via the $$CH_3CHO$$ group on oxidation)
• Any 2° alcohol of the type $$CH_3CH(OH)R$$ (because oxidation gives a methyl ketone).

Hence each of the two alcohols produced must possess the $$CH_3-C(OH)$$ group. The simplest pair satisfying this condition is:

$$Q : CH_3CH_2OH \quad(\text{ethanol})$$
$$R : (CH_3)_2CHOH \quad(\text{2-propanol})$$

Their corresponding iodides (obtained from ether cleavage) must therefore be:

$$CH_3CH_2I \quad\text{and}\quad (CH_3)_2CHI$$

An ether that yields exactly these two iodides on cleavage is the mixed ether formed by joining the two alkyl groups through oxygen:

$$CH_3CH_2-O-CH(CH_3)_2$$

This is ethyl isopropyl ether.

Among the given choices, Option A represents $$CH_3CH_2OCH(CH_3)_2$$, i.e. ethyl isopropyl ether.

Therefore, the mixed ether $$P$$ is Option A which is: $$CH_3CH_2OCH(CH_3)_2$$.