An ether (A), $$C_5H_{12}O$$, when heated with excess of hot concentrated HI produced two alkyl halides which when treated with NaOH yielded compounds (B) and (C). Oxidation of (B) and (C) gave a propanone and an ethanoic acid respectively. The IUPAC name of the ether (A) is :

The ether (A) has the molecular formula $$C_5H_{12}O$$. When heated with excess hot concentrated HI, it undergoes cleavage to form two alkyl iodides. The reaction of an ether with HI is:

$$ \text{R-O-R'} + 2\text{HI} \rightarrow \text{R-I} + \text{R'-I} + \text{H}_2\text{O} $$

These alkyl iodides are then treated with NaOH, which hydrolyzes them to alcohols:

$$ \text{R-I} + \text{NaOH} \rightarrow \text{R-OH} + \text{NaI} $$

$$ \text{R'-I} + \text{NaOH} \rightarrow \text{R'-OH} + \text{NaI} $$

So, the compounds (B) and (C) are alcohols. Oxidation of (B) gives propanone ($$CH_3COCH_3$$), which is a ketone. Ketones are formed by the oxidation of secondary alcohols. The only secondary alcohol that oxidizes to propanone is propan-2-ol ($$CH_3CH(OH)CH_3$$):

$$ \text{CH}_3\text{CH(OH)CH}_3 \rightarrow \text{CH}_3\text{COCH}_3 $$

Oxidation of (C) gives ethanoic acid ($$CH_3COOH$$), which is a carboxylic acid. Carboxylic acids are formed by the oxidation of primary alcohols. The primary alcohol that oxidizes to ethanoic acid is ethanol ($$CH_3CH_2OH$$), as ethanol oxidizes first to ethanal and then to ethanoic acid:

$$ \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{COOH} $$

Therefore, (B) is propan-2-ol and (C) is ethanol. The alkyl groups in the ether (A) correspond to these alcohols. Propan-2-ol comes from the isopropyl group ($$(CH_3)_2CH-$$), and ethanol comes from the ethyl group ($$CH_3CH_2-$$). Thus, the ether (A) is isopropyl ethyl ether, with the structure $$(CH_3)_2CH-O-CH_2CH_3$$.

To determine the IUPAC name, identify the longest carbon chain. The isopropyl group has a three-carbon chain (propane), and the ethyl group has two carbons. The parent chain is propane, with the ethoxy group attached to carbon 2. Therefore, the IUPAC name is 2-ethoxypropane.

Verifying with the options:

• Option A: 2-ethoxypropane matches the structure $$(CH_3)_2CH-O-CH_2CH_3$$.
• Option B: ethoxypropane is ambiguous but typically implies 1-ethoxypropane ($$CH_3CH_2CH_2-O-CH_2CH_3$$), which would give propan-1-ol and ethanol upon reaction; oxidation of propan-1-ol gives propanoic acid, not propanone.
• Option C: methoxybutane (e.g., $$CH_3O-CH_2CH_2CH_2CH_3$$) would give methanol and butan-1-ol; oxidation gives formaldehyde/formic acid and butanoic acid, not propanone or ethanoic acid.
• Option D: 2-methoxybutane ($$CH_3O-CH(CH_3)CH_2CH_3$$) would give methanol and butan-2-ol; oxidation gives formaldehyde/formic acid and butanone, not propanone or ethanoic acid.

Hence, only 2-ethoxypropane produces the correct products.

So, the answer is Option A.