Which of these would be the rate law for an S_N 1 reaction? a) rate = k [RX] b)

Question

Which of these would be the rate law for an S_N 1 reaction? a) rate = k [RX] b) rate = k [RX] [Nucleophile] c) rate = k [RX]^2 d) rate = k [Nucleophile]^2 For Problems 13 and 14, consider this reaction: CH_3 CH_2 CH_2 Br + NaSH rightarrow CH_3 CH_2 CH_2 SH + NaBr What effect does doubling the concentration of NaSH have? a) doubles the rate b) halves the rate c) quadruples the rate d) has no effect on the rate What effect does doubling the concentration CH_3 CH_2 CH_2 Br have? a) doubles the rate b) halves the rate c) quadruples the rate d) has no effect on the rate In the term "E_2". what does the E stand for and what does the 2 stand for? Why does this reaction not give the indicated product? Looking at the mass spectrum, how could you tell whether you had CH_3 Br or C_2 H_10? What would you see in the spectrum for CH_3 Br, and what would you see in the spectrum of C_7 H_10?

Explanation / Answer

The given reaction is over all second order reaction. It is first order in NaSH and first order in CH3CH2CH2Br. So, upon doubling the concentration of NaSH reaction rate will be doubled. (option a). It is first order in CH3CH2CH2Br. So, upon doubling the concentration of CH3CH2CH2Br reaction rate will be doubled. (option a). E = Elimination, 2 = bimolecular This reaction does not occur because Br attached to an alkene is not good leaving group. In the mass spectrum of CH3Br there will be M and M+2 peaks with 1:1 intensity ratio because of two isotopes of Br (79 and 81) where as M+2 peak will not be seen in the mass spectrum of C7H10.

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