A) \[CHC{{l}_{2}}N{{O}_{2}}\]
B) \[CHC{{l}_{2}}N{{O}_{3}}\]
C) \[CC{{l}_{3}}N{{O}_{2}}\]
D) \[CHC{{l}_{2}}.HN{{O}_{3}}\]
Correct Answer: C
Solution :
\[\underset{chloroform}{\mathop{CHC{{l}_{3}}}}\,+HN{{O}_{3}}\xrightarrow{{}}\underset{chloropicrin}{\mathop{CC{{l}_{3}}N{{O}_{2}}}}\,\] Thus chloroform reacts with con. \[HN{{O}_{3}}\] to produce \[CC{{l}_{3}}.N{{O}_{2}}\].
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