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) :

Solution

Chemistry ma’am · 3 steps

Break both C–O bonds, turn each piece into an alcohol, then ask if each has a CH₃CH(OH)⁻ group.

1. Excess hot HI. With excess hot concentrated HI, both C–O bonds of the ether break, so each alkyl group leaves as an alkyl iodide. Aqueous NaOH then turns each alkyl iodide into the matching alcohol, so Q and R are the two alcohols that make up the ether.
2. Iodoform test. NaOI gives a yellow precipitate of CHI₃ only with an alcohol that has a CH₃CH(OH)⁻ unit, such as ethanol or a methyl secondary alcohol. Tertiary alcohols cannot be oxidised, and benzyl alcohol has no CH₃ on the carbinol carbon, so neither gives the test.

   Option	Two alcohols formed	Iodoform from both?
   (a) ethyl 1-phenylpropyl ether	CH₃CH₂OH and C₆H₅CH(OH)CH₂CH₃	No: the second has an ethyl group, not a methyl, next to the CH(OH)
   (b) ethyl sec-butyl ether	CH₃CH₂OH and CH₃CH(OH)CH₂CH₃	Yes, both
   (c) benzyl sec-butyl ether	C₆H₅CH₂OH and CH₃CH(OH)CH₂CH₃	No: benzyl alcohol fails
   (d) ethyl tert-pentyl ether	CH₃CH₂OH and (CH₃)₂C(OH)CH₂CH₃	No: the tertiary alcohol fails

3. So. Only (b), ethyl sec-butyl ether, gives ethanol and butan-2-ol, and both of them give yellow iodoform with NaOI.

The idea to keep. For iodoform, look for CH₃CO⁻ or CH₃CH(OH)⁻. Ethanol passes, methyl secondary alcohols pass. Tertiary alcohols and benzyl alcohol fail.

Answer from the NTA final answer key. Solution by PrepWiser.