A) Tetrafluoro ethane
B) Tetrafluro propene
C) Difluorodichloro ethane
D) Difluoro ethane
E) Trifluoro ethane
Correct Answer: A
Solution :
\[\underset{\text{Tetrafluoro ethane}}{\mathop{nC{{F}_{2}}=C{{F}_{2}}\,}}\,\,\xrightarrow{{}}\,\,\,\,\underset{\text{Teflon}}{\mathop{{{[-C{{F}_{2}}-C{{F}_{2}}-]}_{n}}}}\,\]You need to login to perform this action.
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