The product of oxidation of ethanol with K₂Cr₂O₇ (potassium dichromate) depends on the reaction conditions, specifically the concentration of the oxidizing agent and the presence of a catalyst.

Here are the possible outcomes:

1. Primary Alcohol Oxidation:

* Mild conditions (dilute K₂Cr₂O₇, acidic medium): Ethanol is oxidized to ethanal (acetaldehyde).

CH₃CH₂OH + [O] → CH₃CHO + H₂O

* Strong conditions (concentrated K₂Cr₂O₇, acidic medium): Ethanol is oxidized further to ethanoic acid (acetic acid).

CH₃CHO + [O] → CH₃COOH

2. Secondary Alcohol Oxidation:

* If the reaction is carried out with a secondary alcohol, like propan-2-ol, the product will be a ketone (propanone/acetone) regardless of the concentration of the oxidizing agent.

It's important to remember that K₂Cr₂O₇ acts as an oxidizing agent in the presence of an acid, typically sulfuric acid (H₂SO₄).

Let me know if you have any more questions!