Q.What happens when
i. 2-Nitropropane is boiled with HCl?
ii. Nitrobenzene undergoes electrolytic reduction in a strongly acidic medium?
iii. tert-Butylamine is oxidised with KMnO4?
iv. Acetone oxime is oxidised with trifluoroperoxyacetic acid?

Step 1. i. 2-Nitropropane is a SECONDARY nitroalkane; boiling it with HCl (or H2SO4) hydrolyses it by the secondary-nitroalkane rule -- giving a ketone, not a carboxylic acid: (CH3)2CH-NO2 + HCl/H2O, boil, gives acetone, (CH3)2C=O, + N2O + H2O.

Step 2. ii. Nitrobenzene's electrolytic reduction in a STRONGLY acidic medium first gives phenylhydroxylamine (C6H5-NHOH), but under those same strongly acidic conditions this intermediate undergoes an acid-catalysed rearrangement, migrating the -OH to the para ring position -- giving p-aminophenol as the final product (contrast with a WEAKLY acidic medium, which gives aniline directly instead).

Step 3. iii. tert-Butylamine, (CH3)3C-NH2, oxidised with aqueous KMnO4 is the unit's own stated route to a TERTIARY nitroalkane: it gives 2-methyl-2-nitropropane, (CH3)3C-NO2, + H2O -- since there is no other general way to install a tertiary nitro group, this oxidation is specifically how the unit builds one.

Step 4. iv. Acetone oxime, (CH3)2C=N-OH, oxidised with trifluoroperoxyacetic acid, is the unit's stated oxime-oxidation route to a SECONDARY nitroalkane: it gives 2-nitropropane, (CH3)2CH-NO2.

Match each transformation directly to its own named rule from this unit: secondary-nitroalkane hydrolysis gives a ketone; strongly-acidic electrolytic reduction gives p-aminophenol via rearrangement; tertiary-amine oxidation gives a tertiary nitroalkane; oxime oxidation gives the corresponding nitroalkane.

• Answering acetic acid for part (i) by applying the PRIMARY-nitroalkane hydrolysis rule -- 2-nitropropane is SECONDARY, so its hydrolysis product is a ketone (acetone) + N2O, not a carboxylic acid, which is exclusively the primary-nitroalkane outcome.