An alkanol can be prepared by the reaction of an alkene with

Analyze the question: We want to make an alkanol (an alcohol) from an alkene. This is a hydration-type reaction, so we need to find which reagent can add -OH across the C=C double bond.

Concept: Alkenes react with concentrated H₂SO₄ (tetraoxosulphate(VI) acid) by electrophilic addition. The acid adds across the double bond to give an alkyl hydrogen sulphate. When this is diluted with water and warmed (hydrolysed), it gives the alkanol.

Step-by-step solution:

1. Addition of the acid to the alkene: CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogen sulphate)

2. Hydrolysis with water: CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄

Overall, the alkene is hydrated to give ethanol, an alkanol.

Checking the other options:

• B. Bromine in tetrachloromethane: This is the test for unsaturation. It gives a dibromoalkane (e.g. 1,2-dibromoethane), not an alkanol.
• C. Aqueous KMnO₄: Cold, dilute, alkaline KMnO₄ oxidises alkenes to a diol (e.g. ethane-1,2-diol). This is a dihydric alcohol, not a simple alkanol. It is also an oxidation, not the standard preparation of an alkanol.
• D. NaOH solution: Aqueous NaOH converts haloalkanes into alkanols. It does not react with alkenes in this way.

Answer: The standard route from an alkene to an alkanol among these options is concentrated H₂SO₄ followed by hydrolysis, which is option A.

Tip: Remember "alkene + conc. H₂SO₄, then water → alcohol". Also remember:

• Br₂ gives dibromo compounds.
• KMnO₄ gives diols, and the purple colour disappears.
• NaOH(aq) with haloalkanes gives alcohols.