Question

Place in order of increasing E1 reactivity:

Bromomethane, 2-chloropropane, 1-bromobutane, 2-bromopropane, 2-bromo-2-methylpropane

Answer 1

mechanism of El elimination: H H H H Step 1 - ċ-x C + xo R-c- H H H 4 substrate Leaving group goes first forming carbocation. This is the rate deferemining step. Reactivity of substrate will depend on stability of carbocation and nature of leaving group. Step 28 H H REH=CH2 RC J stability of carbocation a no of d-Hydrogens CH₂ Bre - CH₂=CH-CH3 CH₃ CH₂ CH2 CH 2 - Bre Bromomethane 2-chloropropane - 1-bromobutane Bre bre CH3-CH-CH3 1 CH2 - C - CH3 2 Bromopropane CH3 0 2 B 2-Bromo-2 methylpropane CH3 CH3 - Bre Pro CH₂ Bre < CH3 CH2 CH₂ CH 2 Br < CH₂=CH-CH 3 C CH3-CH-CH3 <CHz-(-ut 3 - CH3 CH3 Reason: CH3 C X Bro CHA - C - Bre-> Che - - CH3 9x-H highly stable CH3 - CH3 H3-CH - Bre CH₂ - CHO - 66-H CH3 ② CH3-CH-u - CH3-CH @ tuo - CH3 1 66-H But Bromide is a good leaving group than chloride ion. So, 2-bromopropane has high reactivity.

6 CH ₂ CH 2 CH ₂ CH ₂ -Br → CH 3 CH 2 CH 2 CH 2 + Bp 24-H ② CH₂ - Br - Choo + Bro no at present least stable = least reactive.