\(\xrightarrow{NH_2-NH_2/KOH}\;\; Product\)The Wolff–Kishner reduction is a chemical reaction used to reduce ketones and aldehydes to alkanes. In this reaction, the carbonyl group (\(C=O\)) of a ketone or aldehyde is converted into a methylene group (\(CH_2\)). This is achieved using hydrazine (\(NH_2-NH_2\)) and a strong base such as \(KOH\).
The given reaction can be represented as follows:
| + | \(NH_2-NH_2\) | \(\xrightarrow{KOH}\) | Alkane | + | \(N_2\) |
Conclusion: The Wolff–Kishner reduction converts the ketone in the given compound to an alkane. Therefore, the correct answer is Alkane.
Given below are the four isomeric compounds \(P, Q, R, S\): 
\(P\): Aromatic compound containing an \(-\mathrm{OH}\) group
\(Q\): Aromatic compound containing an \(-\mathrm{CHO}\) group (aldehyde)
\(R\): Aromatic compound containing a ketone group
\(S\): Aromatic compound containing a ketone group Identify the correct statements from below:
[A.] \(Q, R\) and \(S\) will give precipitate with \(2,4\)-DNP.
[B.] \(P\) and \(Q\) will give positive Baeyer’s test.
[C.] \(Q\) and \(R\) will give sooty flame.
[D.] \(R\) and \(S\) will give yellow precipitate with \(I_2/\mathrm{NaOH}\).
[E.] \(Q\) alone will deposit silver with Tollens’ reagent. Choose the correct option.
Match the LIST-I with LIST-II 
Choose the correct answer from the options given below: