Consider the given reaction, Identify 'X' and 'Y' : 

\[ \begin{array}{c} \text{OH} \\ | \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CH}_2\text{NH}_2 \\ | \\ \text{CH}_3 \end{array} \]
This product is formed by the reduction of a nitrile group (-CN) to an amino group (-CH\(_2\)NH\(_2\)). Reversing this step, the intermediate cyanohydrin must have been:\[ \begin{array}{c} \text{OH} \\ | \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CN} \\ | \\ \text{CH}_3 \end{array} \]
This cyanohydrin is formed by the addition of HCN to a ketone. Reversing the cyanohydrin formation, the original carbonyl compound must have been Butan-2-one (or methyl ethyl ketone):\[ \begin{array}{c} \text{O} \\ || \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CH}_3 \end{array} \]
So, the reaction starts with Butan-2-one, even if the diagram in the question is ambiguous.\[ \text{Butan-2-one} + \text{HCN} \xrightarrow{\text{NaOH (X)}} \text{2-hydroxy-2-methylbutanenitrile} \]
Reaction 2: Reduction of the nitrile group.\[ \text{2-hydroxy-2-methylbutanenitrile} \xrightarrow{\text{1. LiAlH}_4 \text{ 2. H}_3\text{O}^+} \text{1-amino-2-methylbutan-2-ol (Y)} \]
The reducing agent LiAlH\(_4\) reduces the -CN group to a -CH\(_2\)NH\(_2\) group.Convert Ethanal to But-2-enal
Write structure of the products of the following reactions: 


