
\(3^{\circ}>2^{\circ}>1^{\circ}\)
The order of leaving group in \(E 2\) elimination reaction is shown below :
\(R - I > R - Br > R - Cl\)
Therefore, the correct order for reaction with alcoholic \(KOH\) is as follows :
\(a > c > b > d\)
$(1)\,\,(CH_3)_3C-Br$ $(2)\,\,(C_6H_5)_2CH-Br$
$(3)\,\,(C_6H_5)_2C(CH_3)Br$ $(4)\,\,(CH_3)_2CH-Br$
$(5)\,\,C_2H_5Br$


