Question

Select the member of each pair that shows the greater rate of \(S_{N} 2\) reaction with KI in acetone. (a) CCCCCl (b) CCCCCl CCCCBr (c) CC(C)CCCl (d) CCCC(C)Br

Short answer

a) Compound A: CCCCCl or Compound B: CCCCCl? b) Compound A: CCCCCl or Compound B: CCCCBr? c) Compound A: CCCCl or Compound B: CC(C)CCCl? d) Compound A: CC(C)CCCl or Compound B: CCCC(C)Br? Answer: a) Both have the same rate b) Compound B: CCCCBr c) Compound A: CCCCl d) Compound B: CCCC(C)Br

Step-by-step solution

(a) Comparison of the two compounds

Both compounds are the same, and hence have the same rate of \(S_{N}2\) reaction with KI in acetone.

(b) Comparison between Cl and Br leaving group

Between the two compounds: CCCCCl CCCCBr The one with the bromine atom (CCCCBr) will have a greater rate of \(S_{N}2\) reaction because bromide ion is a better leaving group than chloride ion. This is due to the larger size and lower charge density of bromide ion, which allows it to better stabilize the negative charge formed when it leaves the molecule.

(c) Effect of methyl branching on the \(S_{N}2\) rate

For the compound CC(C)CCCl, the reactive site (the carbon bearing the chloride group) is situated next to a branching methyl group. This steric environment around the reactive site will hinder the approach of the nucleophile (KI) and therefore decrease the rate of the \(S_{N}2\) reaction compared to the non-branched compound CCCCl.

(d) Effect of branching position on \(S_{N}2\) rate

In the compound CCCC(C)Br, the branching methyl group is one carbon unit away from the reactive site (the carbon bearing the bromine atom). This will have less steric hindrance on the reactive site compared to the branched compound in part (c), resulting in a faster \(S_{N}2\) reaction compared to CC(C)CCCl. To summarize, the compounds with greater rate of \(S_{N}2\) reaction with KI in acetone are: (a) Both have the same rate (b) CCCCBr (c) CCCCl (d) CCCC(C)Br