On reaction with acidified potassium dichromate(vii), two of the isomers are oxidized in two steps to produce different products. draw the structural formula of the two products formed from one of the isomers.

First, you don't specify which are the isomers to be oxidized. So I will show you some examples.

Explanation:

Oxidizing the different types of alcohols

The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, then the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is as follows:

Cr2O2−7 + 14H+ + 6e− → 2Cr3+ + 7H2O (1)

Primary alcohols

Primary alcohols can be oxidized to either aldehydes or carboxylic acids, depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde, which is then oxidized further to the acid.

An aldehyde is obtained if an excess amount of the alcohol is used, and the aldehyde is distilled off as soon as it forms. An excess of the alcohol means that there is not enough oxidizing agent present to carry out the second stage, and removing the aldehyde as soon as it is formed means that it is not present to be oxidized anyway!

If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal,  CH3CHO . The full equation for this reaction is fairly complicated, and you need to understand the electron-half-equations in order to work it out.

3CH3CH2OH + Cr2O2−7 + 8H+ → 3CH3CHO + 2Cr3+ + 7H2O

(See image 1 and 2)