




The reaction proceeds via an \( S_N1 \) mechanism where the chlorine group leaves, forming a carbocation.
After the formation of the carbocation, a 1,2-hydride shift occurs to stabilize the carbocation, resulting in a more stable tertiary carbocation.
Subsequent attack by hydroxide ion (\( \text{OH}^- \)) leads to the formation of the major product:
The major product is a tertiary alcohol.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: H2Te is more acidic than H2S.
Reason R: Bond dissociation enthalpy of H2Te is lower than H2S.
In light of the above statements, choose the most appropriate from the options given below:


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)