




Step 1: Analyzing the given reaction.
The given reaction involves the reduction of a molecule containing a carbonyl group (C=O) and an amide group (-NH\(_2\)) using lithium aluminum hydride (LiAlH\(_4\)), followed by hydrolysis with H\(_3\)O\(^+\).
1. Step 1: Reduction with LiAlH\(_4\)
Lithium aluminum hydride (LiAlH\(_4\)) is a strong reducing agent. When it reacts with a carbonyl compound (such as a ketone or aldehyde), it reduces the carbonyl group to a primary or secondary alcohol. \item In this case, the carbonyl group (C=O) of the given compound will be reduced to a hydroxyl group (-OH), resulting in an intermediate amide being reduced to a primary amine group (-NH\(_2\)). The structure of the intermediate product will be: \[ \text{H}_2\text{NC} - \text{CH}_2 - \text{CH}_2\text{OH} \] where the ketone group is reduced to an alcohol group.
2. Step 2: Hydrolysis with H\(_3\)O\(^+\)
After reduction, the product is treated with an acidic solution (H\(_3\)O\(^+\)), which will hydrolyze the intermediate and result in a final product where the amide group has been converted to an amine group (-NH\(_2\)) attached to a hydroxyl group (-OH). This confirms the product as: \[ \text{H}_2\text{N} - \text{CH}_2\text{OH} \] This is the final product, and it is a primary amine with a hydroxyl group attached to the adjacent carbon. Thus, the product 'X' is \( \text{H}_2\text{N} - \text{CH}_2\text{OH} \).
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 : \(A_2 \rightleftharpoons 2A\)