A) \[LiBr\]
B) \[E{{t}_{4}}Pb\]
C) \[{{({{C}_{4}}{{H}_{9}})}_{4}}Sn\]
D) \[{{({{C}_{2}}{{H}_{5}})}_{4}}Pb\]
Correct Answer: C
Solution :
\[4C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}-Li+SnC{{l}_{2}}\to {{({{C}_{4}}{{H}_{9}})}_{4}}Sn\].
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