Question

Ethyl bromide reacts with sodium lead alloy to form
A. Tetraethyl lead
B. Ethyl sodium
C. Ethane
D. Ethene

Answer 1

Hint: When ethyl bromide is reacted with sodium lead alloy, it forms tetraethyl lead as major products with the elimination of sodium bromide as a byproduct. Lead has covalency 4 thus it acquires its tetravalency with ethyl molecules.

Complete step by step answer: $TEL$ or tetraethyl Lead is produced by reacting chloro-ethane with a sodium–lead alloy
$4NaPb$ + $4 C{H_3}C{H_2}Br$ $\rightarrow$ $(C{H_3}C{H_2})_4Pb$ + $4NaBr$ + $3Pb$
sodium and lead bond breaks and there is the formation of positive sodium and negative lead in the solution.
So sodium attack with bromine and lead reacts with ethyl ion
The product is recovered by steam distillation, leaving a sludge of lead and sodium chloride
$TEL$ is a viscous colorless liquid. Because $TEL$ is charge neutral and contains an exterior of alkyl groups, it is highly lipophilic and soluble in petrol (gasoline).
 no reactions were found to improve upon this rather difficult process that involves metallic sodium and converts only $25%$ of the lead to $TEL.$
 tetramethyl lead was commercially produced by a different electrolytic reaction. A process with lithium was developed but never put into practice.
Hence, the correct option is (A) Tetraethyl lead.

Note: sodium and lead bond breaks and there is the formation of positive sodium and negative lead in the solution. Positively charged sodium attack on negatively charged bromine and negatively charged lead attack on positively charged ethyl ion.
Thus there is formation tetraethyl lead