Dehydration of alcohol by conc. `H_(2)SO_(4)` takes place according to following steps:
`CH_(3)-overset(CH_(3))overset("| ")underset("H ")underset("| ")"C "-CH_(2)overset(* *)underset(* *)OHoverset(H^(+))underset("step - 1")rarrCH_(3)-overset(CH_(3))overset("| ")underset("H ")underset("| ")"C "-CH_(2)overset(+)OH_(2)overset(-H_(2)O)underset("step - 2")rarr CH_(3)-overset(CH_(3))overset("| ")underset("H ")underset("| ")"C "-overset(+)CH_(2)`
`overset(1,2-H^(-)" shift")underset("step - 3")Rarr CH_(3)-overset(CH_(3))overset("| ")underset("+ ")C-CH_(3)overset(-H^(+))underset("step - 4")rarr CH_(3)=overset(CH_(3))overset("| ")"C "-CH_(3)`
The lowest and fastest steps in the above reaction are

Options

A. step 1 is lowest, by 3 is fastest
B. step 2 is lowest while 3 is fastest
C. step 2 is lowest, while 4 is fastest
D. all steps proceed at equal rate

Answer

C

Explanation

Step 2 involves the formation of carbonium ion by the loss of weakly basic `H_(2)O` molecule. It is slowest step. Step 4 involves the conversion of an unstable (or intermediate) into a quite stable product, hence it is fastest step.