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When alcohols are treated with acidic potassium dichromate (K₂Cr₂O₇/H₂SO₄), they undergo oxidation reactions. However, the type of product formed depends on whether the alcohol is primary, secondary, or tertiary.
Primary alcohols are oxidized first to aldehydes and then further to carboxylic acids. For example, ethanol gives acetic acid on oxidation.
Secondary alcohols are oxidized to ketones. For example, isopropanol yields acetone.
Tertiary alcohols, however, behave differently. They resist normal oxidation because they do not have a hydrogen atom directly bonded to the carbon carrying the –OH group. Instead, under strong oxidizing and acidic conditions, they undergo elimination. The hydroxyl group is removed along with a hydrogen atom from the adjacent carbon atom, producing an alkene.
This is why the correct answer to the question is tertiary alcohols. For instance, tert-butanol (CH₃)₃COH when treated with concentrated sulfuric acid and an oxidizing agent like potassium dichromate, undergoes dehydration and elimination to form isobutene (2-methylpropene).
The mechanism involves the protonation of the hydroxyl group, which makes water a good leaving group. After losing water, a carbocation intermediate forms. Tertiary carbocations are highly stable, which explains why tertiary alcohols readily undergo this reaction. A β-hydrogen is then removed, and an alkene is formed.
This property of tertiary alcohols is very useful in synthetic organic chemistry, where controlled dehydration and elimination are used to prepare specific alkenes.
Therefore:
Primary alcohol → aldehyde → carboxylic acid
Secondary alcohol → ketone
Tertiary alcohol → alkene
Hence, the alcohol that forms an alkene with acidic potassium dichromate is the tertiary alcohol.
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