Answer:
a) 1-Chlorohexane
b) 1-Bromopentane
c) 2-Chloropentane
d) 2-bromo-5-methylhexane
e) 1-bromodecane
Explanation:
One important fact that we must have behind our minds as we approach the problem is the fact that sodium iodide in acetone will lead to an SN2 reaction because the nucleophile (iodide ion) is naked and not solvated. Also, the iodide ion is a good nucleophile and the solvent is a polar aprotic solvent. All these factors together make the reactions to tilt towards SN2 mechanism.
In a) 1-Chlorohexane offers less steric hinderance and more ease of SN2 reaction hence the answer.
In b) 1-Bromopentane offers less steric hinderance being a primary alkyl halide hence the answer.
c) In c) 2-Chloropentane contains chlorine which is more polarizable and a better leaving group than fluorine.
In d) 2-bromo-5-methylhexane is a secondary alkyl halide and undergoes SN2 reaction with less steric hindrance. 2-Bromo-2-methylhexane is a tertiary alkyl halide and does not undergo SN2 reaction due to steric hindrance.
e) 1-bromodecane is a primary alkyl halide and offers less steric resistance to the approaching nucleophile than 2-Bromopropane.
