Identify the starting compound from the following data: C6H14O(X) on reaction with HI yields a haloalkane (A) and an alcohol (B). Compound (A) on reaction with aqueous NaOH gives an alcohol (C). Compounds (B) and (C) on reaction with CrO3 in anhydrous medium yields Butanone and Ethanal respectively.
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.