How can I know the colour change of this reaction? After you react rotundone ($\ce{C_{15}H_{22}O}$) with water it loses all of the double bonds (except the carboxylic bond, $\ce{C=O}$) and gets two of the functional group $\ce{-OH}$ (alcohol). Two compounds are formed, one (A) where both the alcohol groups are tertiary and another (B) where one is tertiary and the other is primary.

I'm asked to state and explain the colour change of both A and B when they react separately with acidified potassium dichromate(VI) solution. 

Now I can see that compound A would have no colour change because the $\ce{-OH}$ groups are tertiary so the potassium dichromate won't react or oxidise them.
I can also see that there will be a colour change on compound B because one of the $\ce{-OH}$ groups is primary, so the potassium dichromate will oxidise and react with it.

Here's my problem:

I'm expected to know that when compound B reacts, it changes from orange to green. How would I know that rotundone is orange in the first place? And what makes it go green?
I know that I'm expected to use intuition as we definitely aren't required to know the colour of rotundone.
Potassium dichromate (VI) is the species that turns from orange and to green when reduced. Most of the organic substances I came across for testing purposes are colourless, so I would assume that compound B would be as well.

It is just like you are expected to know the colour change of potassium manganate (VII) from purple to colourless when it is reduced.

Those two oxidants were used frequently in my labs, and I'm pretty sure you'll be using them often as well.

I hope this answers your question! :)