An organic compound with molecular formula C$_8$H$_9$NO when reacts with KOH/BBr$_2$ forms 'P' which on diazotisation forms 'Q' followed by its reaction with CuCN forms 'R' which on hydrolysis (acidic) formed 'S' (S can also be made by hydrolysis of original compound (X) C$_8$H$_9$NO). 'S' can react with KMnO$_4$/H$^+$ forms 'T' which has two types of hydrogen. X will be:
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In organic reactions, understanding the functional groups and their reactivity can help predict the products of each step. For example, amides can undergo hydrolysis to form carboxylic acids.
Concept:
The reaction is a nucleophilic substitution on Methyl Bromide. The solvent CH$_3$OH is polar protic. In polar protic solvents, nucleophilicity increases down the group (larger size, less solvation) and decreases with electronegativity/stability.
Step 1: Analyze Halides.
In methanol (protic), larger ions are less solvated and thus better nucleophiles.
Order: I$^-$>F$^-$.
Step 2: Analyze Oxygen Nucleophiles.
Ethoxide (C$_2$H$_5$O$^-$) vs Phenoxide (PhO$^-$).
Ethoxide has a localized negative charge, making it a stronger base and stronger nucleophile than Phenoxide, where the charge is delocalized via resonance.
However, I$^-$ is generally a better nucleophile than alkoxides in substitution reactions due to high polarizability.
Step 3: Combine Order.
I$^-$ (Highest polarizability)>C$_2$H$_5$O$^-$ (Strong base)>PhO$^-$ (Resonance stabilized)>F$^-$ (Highly solvated/High electronegativity).
Conclusion: Option (3).
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