COMEDK 2023 Shift 1 Paper

Let's analyze the given information and determine the starting compound C6H14O(X) through a step-by-step approach:
Step 1: Analyzing the reaction with HI
Reaction of an ether with HI results in the cleavage of the ether, yielding a haloalkane (A) and an alcohol (B). Depending on the structure of the ether, we can deduce probable structures for compounds (A) and (B).
Step 2: Consideration of subsequent reactions
Compound (A), a haloalkane, reacts with aqueous NaOH to give an alcohol (C). This implies hydrolysis of the haloalkane involves substitution by an OH group.
Step 3: Reactions of alcohols (B) and (C) with CrO3
Both alcohols react with CrO3 in an anhydrous medium to yield butanone and ethanal, respectively. This implies:
Alcohol (B) must have 4 carbon atoms to be oxidized to butanone, thus it should be butanol.
Alcohol (C) must have 2 carbon atoms to be oxidized to ethanal, suggesting it is ethanol.
Step 4: Considering the structure of compounds
The ether must be structured such that its cleavage with HI could plausibly form butanol and a chlorine derivative capable of transforming into ethanol. This implies a 4-carbon unit connected to a 2-carbon unit (possibly through an oxygen atom in the case of an ether).
Among the options provided, the most plausible is 2-Ethoxybutane (Option B). This ether would cleave to form 1-butanol (butanol, 4 carbons) and ethyl chloride ( 2 carbons, which is then hydrolyzed to ethanol).
On cleaving 2-ethoxybutane with HI
C4H9OCH2CH3⟶C4H9‌OH+CH3CH2I
1-butanol (B) and ethyl iodide (A).
Ethyl iodide (A) reacts with NaOH to form ethanol (C).
1-butanol (B) with CrO3 forms butanone, and ethanol (C) with CrO3 forms ethanal.
Therefore, the correct answer is Option B - 2-Ethoxybutane.