Question

An ether [A], $C_5{H}_{12}O,$ When heated with excess of hot concentrated $HI$ produced two alkyl halides which when reacting with aqueous $NaOH$ yield 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. Methoxybutane
• C. Ethoxypropane
• D. 2- Methoxybutane

Answer 1

The correct option is A 2-Ethoxypropane

Oxidation of primary alcohol will give carbocylic acid and oxidation of secondary alcohols will ketone.
Since, the product of oxidations are ethanoic acid and propanone, the reactant of oxidation reaction will be ethanol and 2-Propanol.
Ether (A) on reacting with excess of hot conc. HI will give alkyl iodides. These iodide are on reacting with $NaOH$, the $O{H}^{-}$ group will replace the $I$ group by $S_{N}2$ reaction.
So, the compound (B) and (C) are Ethyl iodide and 2-Iodopropane.
Thus, the alkyl chain of (B) and (C) forms the ether compound (A).

Hence, the IUPAC name of compound (A) is 2-Ethoxypropane