1
Question
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markownikoff's addition to alkenes because:
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markownikoff's addition to alkenes because:
Open in App
Solution
The correct option is C One of the steps is endothermic in both the cases
I. Mechanism for the radical addition (HCl) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−Cl⟶R−OH+∙Cl ; ΔH=+ve
Chain propagation
RCH=CH2+∙Cl⟶R∙CH−CH2X ; ΔH=−veR∙CH−CH2Cl+HCl⟶RCH2−CH2Cl+∙Cl ; ΔH=+ve
I(b) is highly endothermic as H−Cl bond is stronger and may not be cleaved by a free radical.
II. Mechanism for the radical addition (HBr) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−Br⟶R−OH+∙Br ; ΔH=−ve
Chain propagation
RCH=CH2+∙Br⟶R∙CH−CH2Br ; ΔH=−veR∙CH−CH2Br+HBr⟶RCH2−CH2Br+∙Br ; ΔH=−ve
The addition of HBr is thermodynamically favourable as all the steps are exothermic.
III. Mechanism for the radical addition (HI) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−I⟶R−OH+∙I ; ΔH=−ve
Chain propagation
RCH=CH2+∙I⟶R∙CH−CH2I ; ΔH=+veR∙CH−CH2I+HI⟶RCH2−CH2I+∙I ; ΔH=−ve
H−I bond is weaker and iodine radical is easily produced. As the first propagation step is endothermic, iodine free radicals combine to form iodine molecules instead of addition to the double bond.
Therefore, peroxide effect is observed for HBr only.
I. Mechanism for the radical addition (HCl) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−Cl⟶R−OH+∙Cl ; ΔH=+ve
Chain propagation
RCH=CH2+∙Cl⟶R∙CH−CH2X ; ΔH=−veR∙CH−CH2Cl+HCl⟶RCH2−CH2Cl+∙Cl ; ΔH=+ve
I(b) is highly endothermic as H−Cl bond is stronger and may not be cleaved by a free radical.
II. Mechanism for the radical addition (HBr) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−Br⟶R−OH+∙Br ; ΔH=−ve
Chain propagation
RCH=CH2+∙Br⟶R∙CH−CH2Br ; ΔH=−veR∙CH−CH2Br+HBr⟶RCH2−CH2Br+∙Br ; ΔH=−ve
The addition of HBr is thermodynamically favourable as all the steps are exothermic.
III. Mechanism for the radical addition (HI) to an alkene:
Chain initiation
(a) R−O−O−RΔ⟶2R−∙O(b) R−∙O+H−I⟶R−OH+∙I ; ΔH=−ve
Chain propagation
RCH=CH2+∙I⟶R∙CH−CH2I ; ΔH=+veR∙CH−CH2I+HI⟶RCH2−CH2I+∙I ; ΔH=−ve
H−I bond is weaker and iodine radical is easily produced. As the first propagation step is endothermic, iodine free radicals combine to form iodine molecules instead of addition to the double bond.
Therefore, peroxide effect is observed for HBr only.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
