Question

You want to convert 2-chloropentane(an alky halide) into 1-pentene (an alkene). Which of the following reagents would you choose? (A)NaOH/water (B)KOH/methanol

   (C)CH₃ONa/CH₃OH (D) (CH₃)₃COK/(CH₃)₃COH

1. Consider the S_N2 reaction of butyl bromide with hydroxide ion , Assuming no other changes, what effect on the reaction rate would result from simultaneously doubling the concentrations of both butyl bromide and hydroxyl ion  A. No effect (B)It would double the rate (C)it would increase the rate six times

(D) It would increase the rate four times.

5. Select the rate law for a second order reaction: (A) rate = k[RX] (B) rate = k[RX]²[OH^-]

(C) rate = k[RX][OH^-] (D) rate k[RX]²[OH^-]²

Answer 1

Answer 1):-

Option (D) : "(CH​​​​​​₃)₃COK / (CH₃)₃COH " is correct answer.

(CH₃)₃COK is a bulky base. Thus, it abstracts the less-hindered proton in the elimination of HCl molecule from 2-chloropentane. Thus, a less-substituted alkene (1-pentene) is formed. While, other bases given in other options give more-subsituted alkene (2-pentene).

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Answer 2):-

SN2 reaction follows second order kinetics, that is, rate depends on both the concentration of alkyl halide (butyl bromide) and nucleophile (hydroxide ion).

Rate = [butyl bromide] [OH^-] ...........,eq.(1)

Now, if concentrations of both butyl bromide and hydroxide ion are doubled , then

New Rate = [2 butyl bromide] [2 OH^-] .............eq.(2)

Dividing eq.(2) from eq.(1), we get,

New Rate = 4 × Rate

Thus, on doubling the concentrations of both butyl bromide and hydroxide ion, the rate increases four times.

Option (D): "It would increase the rate four times" is correct one.

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Answer 5):-

For Second order reaction,

Rate = k [RX] [OH^-]

Option (C): is correct answer.

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