question_answer

An ether [A], \[{{C}_{5}}{{H}_{12}}O\], 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

Correct Answer: A

Solution :

[a] hence the IUPAC name of compound is \[\underset{2-\,ethoxy\,propane}{\mathop{C{{H}_{3}}-C{{H}_{2}}-O-\underset{3}{\mathop{\overset{2}{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,}}\,H-\overset{1}{\mathop{C}}\,{{H}_{3}}}}\,\]