Compounds X and Y are both C5H11Cl products formed in the radical chlorination of pentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C5H10 alkene (ignoring double bond stereochemistry). Both X and Y react in SN2 fashion with sodium iodide in acetone; Y reacts faster than X. What is the structure of Y?

C5H11Cl with the Cl group attached to the chiral carbon

Explanation:

radical chlorination is given by:

C5H12 + Cl2 → C5H11Cl + HCl

Such reactions are unselective, the product  could either 1-chloropentanes, 2-chloropentanes, or 3-chloropentanes, all with the same molecular formula C5H11Cl but different structure(depending on on the carbon the chlorine is attached)

1-chloropentanes has its Cl group attached to a primary carbon

2-chloropentanes has its Cl group attached to a second carbon

3-chloropentanes  has its Cl group attached to a third  carbon

SN2 reactions occurs fastest on the primary carbon(chiral) which. Therefore Y is 1-chloropentane which has its Cl group attached to a primary carbon with no beta-branching will proceed the fastest.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.