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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?

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ProfLincoln

Answer:

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.

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