Q.What happens when isobutylene is treated with acidified potassium permanganate?

Step 1. Isobutylene (2-methylpropene), (CH3)2C=CH2, has an UNSYMMETRICAL double bond: one alkene carbon carries two methyl groups (no H), and the other is a terminal =CH2 (two H's, no alkyl substituent).

Step 2. Hot, acidified KMnO4 is a strong oxidant that cleaves the C=C bond entirely, converting each alkene carbon into a carbonyl-type product depending on how many H's it originally carried.

Step 3. The fully-substituted carbon, (CH3)2C=, carries NO hydrogen, so it can only be oxidised as far as a KETONE -- giving propan-2-one (acetone), (CH3)2C=O -- since there is no C-H left for further oxidation to a carboxylic acid.

Step 4. The terminal =CH2 carbon carries TWO hydrogens, so under the strong, hot, acidic conditions it is oxidised all the way past the aldehyde and carboxylic-acid stages to CO2 and H2O (a terminal, disubstituted-with-H carbon is fully degraded rather than stopping at a stable organic product). …

Look at each alkene carbon's own substitution (how many H's it carries) separately before deciding its oxidation product: no H -> ketone (stops there); one H -> carboxylic acid (via an aldehyde); …

• Assuming BOTH alkene carbons give the same TYPE of product -- they generally do not, unless the alkene happens to be symmetric; here one side gives a stable ketone and the other is fully oxidised away. …