MCQ
The product of reaction between propene and $HBr$ in the presence of a peroxide is
- ✓$C{H_3} - C{H_2} - C{H_2}Br$
- B$C{H_3} - CHBr - C{H_3}$
- C$C{H_3} - C{H_2}Br$
- D$C{H_3} - CH = CHBr$
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| Reaction | Energy Change (in $kJ$ ) |
| $Li(s) \to Li(g)$ | $161$ |
| $Li(g) \to Li^+(g)$ | $520$ |
| $\frac {1}{2}F_2(g)\,\to F(g)$ | $77$ |
| $F(g) + e^- \to F^-(g)$ | (Electron gain enthalpy) |
| $Li^+ (g) + F^-(g) \to LiF(s)$ | $-1047$ |
| $Li (s) + \frac {1}{2}F_2(g)\to LiF(s)$ | $-617$ |
Based on data provided, the value of electron gain enthalpy of fluorine would be.....$kJ\,mol^{-1}$