Reaction:
Lithium di(3-pentyl)cuprate reacts with ethyl bromide to produce an alkane C7H16.

Explanation:
The reaction involves the substitution of the bromine atom in ethyl bromide with the alkyl group attached to the lithium atom in lithium di(3-pentyl)cuprate. The cuprate reagent acts as a nucleophile and attacks the electrophilic carbon in the ethyl bromide, resulting in the formation of a new carbon-carbon bond.

Mechanism:
The reaction proceeds through a nucleophilic substitution reaction. Here is the step-by-step mechanism:

1. Formation of the cuprate reagent:
Lithium di(3-pentyl)cuprate is prepared by treating 3-pentyl bromide with lithium metal in the presence of anhydrous ether. The reaction occurs through the formation of an organolithium intermediate, which then reacts with copper(I) iodide to form the cuprate reagent.

2. Nucleophilic attack:
The cuprate reagent, which has a nucleophilic carbanion, attacks the electrophilic carbon in the ethyl bromide. The nucleophilic attack leads to the formation of a new carbon-carbon bond.

3. Elimination of lithium bromide:
After the nucleophilic attack, lithium bromide is eliminated, and the alkyl group from the cuprate reagent is transferred to the carbon atom previously bonded to the bromine atom.

4. Formation of the product:
The final product is an alkane with seven carbon atoms, C7H16. The structure of the product is determined by the alkyl group transferred from the cuprate reagent. In this case, the transferred alkyl group is an ethyl group, resulting in the formation of 3-ethylpentane.

Conclusion:
The structural formula of the product formed by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide is 3-ethylpentane (option C). This is because the ethyl group from the cuprate reagent is transferred to the carbon atom previously bonded to the bromine atom in ethyl bromide.