Question

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 :

• A. 2-ethoxypropane
• B. ethoxypropane
• C. methoxybutane
• D. 2-methoxybutane

Answer 1

The Correct Option is A

To solve this problem, let's break down the steps based on the chemical reactions and the information provided:

1. The given ether is C_5H_{12}O, indicated as compound (A). When treated with excess hot concentrated HI, it produces two alkyl halides.
2. The alkyl halides produced are a result of the cleavage of the ether linkage. We need to determine the likely structure of these alkyl halides, given the formula C_5H_{12}O.
3. After the cleavage of the ether with HI, the two possible alkyl halides indicate the following:
   • One alkyl group combines with I to form an alkyl iodide.
   • The reaction of these alkyl iodides with NaOH results in the formation of alcohols (B) and (C).
4. Now, consider the oxidation results:
   • Oxidation of compound (B) gives propanone (commonly known as acetone), which is CH_3COCH_3.
   • Oxidation of compound (C) gives ethanoic acid, which is CH_3COOH.
5. These oxidation results suggest that compound (B) was originally a secondary alcohol—most likely isopropanol (CH_3CHOHCH_3)—and compound (C) was ethanol (CH_3CH_2OH).
6. The ether that produces these alcohols upon hydrolysis must be 2-ethoxypropane because:
   • 2-Ethoxypropane has the formula (CH_3)CHOCH_2CH_3, which fits the given molecular formula C_5H_{12}O.
   • Upon treatment with HI, it breaks into isopropyl iodide and ethyl iodide.
   • On treatment with NaOH, these iodides are converted into isopropanol and ethanol, respectively, leading to the given oxidization products.
7. Therefore, the IUPAC name of the ether is 2-ethoxypropane.

Therefore, the correct answer is 2-ethoxypropane.