The four alcohols, ethanol, butan-1-ol, butan-2-ol and 2-methylpropan-2-ol, are added to test tubes and reacted with acidified potassium...

These alcohols can be classified according to the carbon they are attached to:

• ethanol and butan-1-ol are primary alcohols


• butan-2-ol is a secondary alcohol


• 2-methylpropan-2-ol is a tertiary alcohol

Given that we have primary, secondary and tertiary substituents, and because hydroxides are a terrible leaving group, we can assume that we're going to see a variety of reactions (probably not Sn2 though) and that the hydroxide has a good chance of being oxidized by the chromate ion, losing its hydrogen atom and then undergoing further stabilizations. Reduction of the chromate would be indicated by a color change from orange to green for primary and secondary alcohols. 

• The ethanol, butan-1-ol and butan-2-ol would be oxidized, forming a carbocation and negatively charged oxygen, which would then stabilize into a carbon-oxygen double bond. Due to having an extra hydrogen on the carbocation, the ethanol and butan-1-ol could be further oxidized into carboxylic acids. All three solutions would show the expected color change from orange to green.


• There would be no reaction for 2-methylpropan-2-ol. The oxidation from the chromate ion depends on having two hydrogens on lesser-substituted carbons, but none exist in this molecule. It's possible that the hydroxide would be protonated by the acid, but this wouldn't produce much of a response from the chromate, and there would be no color change even if a reaction did take place.