question_answer

\[{{C}_{3}}{{H}_{6}}C{{l}_{2}}\] on reaction with \[NaOH\] forms \[{{C}_{3}}{{H}_{6}}O\] which gives yellow precipitate on heating with \[NaOH\] and \[{{I}_{2}}\]. Thus, \[{{C}_{3}}{{H}_{6}}C{{l}_{2}}\] is

A) 1, 1-dichloropropane  

B) 1, 2-dichloropropane

C) 2, 2-dichloropropane  

D) 1, 3-dichloropropane

Correct Answer: C

Solution :

\[{{C}_{3}}{{H}_{6}}O\](carbonyl compound) gives iodoform test (yellow ppt. with \[{{I}_{2}}/NaOH\]) This suggest that, it contains \[-\underset{O}{\mathop{\underset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}\]group is due to hydrolysis of Then \[{{C}_{3}}{{H}_{6}}C{{l}_{2}}\] will be2, 2 dichloropropane