Hydrolysis of t-butyl chloride (1) gives the alcohol (2) as the main product. Th

Question

Hydrolysis of t-butyl chloride (1) gives the alcohol (2) as the main product. The same reaction, run in the presence of 1 equivalent of sodium formate (3), gives the formate ester (4) as the main product. The rate of consumption of 1 is the same for both reactions. Which statement(s) about the reactions is/are true? H20 , (1) OH 2 Br H2O 4 Select one: O a. The alcohol forms first in both reactions, and the formate (4) forms from the alcohol (2) in a subsequent Sy2 step. O b. The alcohol (2) forms in an SN1 reaction, whereas the formate (4) forms in an S 2 reaction Oc. The alcohol (2) and the formate (4) form from the same intermediate. O d. The formate ion is a better nucleophile than water and it reacts faster with 1

Explanation / Answer

starting reactant is the tartiary butyl bromide which has 3 0  carbon attached to the leaving group bromine. It is well known that 3 0  carbon group is not good for SN2 reaction due to steric hindrance of bulkier group during back side attack for nucleophile. But  3 0  carbon group forms most stable carbocation for sn1 reaction. Here, both the products will formed through sn1 reaction via same carbocation intermediate and formation of carbocation intermediate is the rate determinate step (rate does not depends on nucleophile). therefore, statement "c" is right.

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