JEE Advanced levelSingle correct optionDifficulty 3 of 3o2093
Use the bond dissociation energies CHX3−H=435, H−Cl=431 and H−Br=366kJmol−1. The enthalpy changes (kJ/mol) for the hydrogen abstraction steps CHX4+X⋅CHX3⋅+HX for X = Cl and X = Br respectively are
A+4 and -69
B+4 and +69
C+69 and +4
D-4 and -69
Answer: B
ΔH=D(C−H)−D(H−X), bond broken minus bond formed. For Cl: 435−431=+4. For Br: 435−366=+69. The much more endothermic bromine step has a late transition state and hence is more selective.