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

Dehydration of alcohol by conc. H_(2)SO_(4) takes place according to f

1.	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
Answer» step 1 is lowest, by 3 is fastest step 2 is lowest while 3 is fastest step 2 is lowest, while 4 is fastest all steps PROCEED at equal rate Solution :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.