Bond enthalpies of $$A_2, B_2$$ and $$A B$$ are in the ratio $$2: 1: 2$$. If bond enthalpy of formation of $$A B$$ is $$-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$$. The bond enthalpy of $$B_2$$ is
The order of reactivity of the compounds $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{Br}$$ and $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right)\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{Br}$$ in $$\mathrm{S}_{\mathrm{N}} 2$$ reaction is
The major product of the following reaction is
$$\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{OH} \xrightarrow[\text { Excess }]{\mathrm{HBr}} \text { Product }$$
The product '$$A$$' gives white precipitate when treated with bromine water. The product '$$B$$' is treated with barium hydroxide to give the product $$C$$. The compound $$C$$ is heated strongly to form product $$D$$. The product $$D$$ is