A) Tetraethyl lead
B) Tetraethyl bromide
C) Both A and B
D) None of the above
Correct Answer: A
Solution :
\[4{{C}_{2}}{{H}_{5}}Br+4Pb/Na\to \underset{\begin{smallmatrix} \text{Tetra}\,\,\text{Ethyl}\,\,\text{lead}\, \\ \,(\text{TEL}) \end{smallmatrix}}{\mathop{{{({{C}_{2}}{{H}_{5}})}_{4}}Pb}}\,+4NaBr\]
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