A) Glycerol
B) Glycol
C) 1, 2-dibromoethane
D) Benzene
Correct Answer: C
Solution :
\[Pb{{({{C}_{2}}{{H}_{5}})}_{4}}\xrightarrow{\text{heat}}Pb+\underset{\text{Ethyl}\,\text{radical}}{\mathop{4C{{H}_{3}}CH_{2}^{{}}}}\,\] \[\underset{\underset{Br\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,-\underset{\underset{Br\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,}}\,}{\mathop{C{{H}_{2}}}}\,+Pb\xrightarrow{{}}\underset{\text{Ethene}}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\,+\,\underset{\text{Lead bromide}}{\mathop{PbB{{r}_{2}}}}\,\] As leaded gasoline burns, lead metal gets deposited in the engine which is removed by adding ethylene dibromide. The lead bromide is volatile and is carried off with the exhaust gases from the engine
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