An isomer of C₆H₁₂ reacts with Br₂ in the presence of light. The reaction type is free radical halogenation, typically prioritizing hydrogen abstraction from the most substituted carbon to form the most stable radical. Since the reaction yields only one isomer C₆H₁₁Br, the isomer must have symmetry that limits bromine addition to a single position capable of forming a single product.
Analyzing potential structures of C₆H₁₂, cyclohexane is the most likely candidate, given its symmetric nature and lack of methyl or ethyl groups which would lead to multiple substitution locations. Thus, C₆H₁₂ is cyclohexane.
Upon reaction of cyclohexane with Br₂/light, bromination at one of the equivalent methylene groups occurs, forming bromocyclohexane.
The compound C₆H₁₁Br (bromocyclohexane) reacts with AgNO₃. Due to the weak C-Br bond in bromides and the precipitating action of Ag⁺ ions forming AgBr, the reaction is facilitated. Bromine is displaced, resulting in the formation of cyclohexanol as the alcohol group replaces Br in a nucleophilic substitution reaction.
Therefore, Y, the major product formed from reacting C₆H₁₁Br with AgNO₃, is cyclohexanol.
Observe the following reactions:
\( AB(g) + 25 H_2O(l) \rightarrow AB(H_2S{O_4}) \quad \Delta H = x \, {kJ/mol}^{-1} \)
\( AB(g) + 50 H_2O(l) \rightarrow AB(H_2SO_4) \quad \Delta H = y \, {kJ/mol}^{-1} \)
The enthalpy of dilution, \( \Delta H_{dil} \) in kJ/mol\(^{-1}\), is:
Kc for the reaction \[ A(g) \rightleftharpoons T(K) + B(g) \] is 39.0. In a closed one-litre flask, one mole of \( A(g) \) was heated to \( T(K) \). What are the concentrations of \( A(g) \) and \( B(g) \) (in mol L\(^{-1}\)) respectively at equilibrium?