Explanation:
1. Assertion (A): Correct. - In \( C_6H_5CH_2Br \), the \( CH_2-Br \) bond is connected to a benzyl group. The phenyl ring allows for stabilization of the transition state via resonance, facilitating the \( S_N2 \) reaction. This makes the reaction proceed more readily compared to \( CH_3CH_2Br \), where no such stabilization exists.
2. Reason (R): Correct. - The unhybridized \( p \)-orbital formed during the trigonal bipyramidal transition state interacts with the conjugated system of the phenyl ring, providing extra stabilization.
3. Conclusion: Both (A) and (R) are correct, and (R) is the correct explanation for (A).
Final Answer is option (3).
Consider the gas phase reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] At equilibrium for a particular temperature, the partial pressures of \( CO \), \( O_2 \), and \( CO_2 \) are found to be \( 10^{-6} \, {atm} \), \( 10^{-6} \, {atm} \), and \( 16 \, {atm} \), respectively. The equilibrium constant for the reaction is ......... \( \times 10^{10} \) (rounded off to one decimal place).
Molten steel at 1900 K having dissolved hydrogen needs to be vacuum degassed. The equilibrium partial pressure of hydrogen to be maintained to achieve 1 ppm (mass basis) of dissolved hydrogen is ......... Torr (rounded off to two decimal places). Given: For the hydrogen dissolution reaction in molten steel \( \left( \frac{1}{2} {H}_2(g) = [{H}] \right) \), the equilibrium constant (expressed in terms of ppm of dissolved H) is: \[ \log_{10} K_{eq} = \frac{1900}{T} + 2.4 \] 1 atm = 760 Torr.
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]