Question

Alkyl chloride reacts with aqueous KOH to form alcohol while in the presence of alcoholic KOH, alkene is obtained as major product. Explain. (2)

Hint:

Aqueous \( OH^- \) acts as nucleophile (substitution to alcohol); alcoholic KOH gives alkoxide, a strong base that removes a beta-hydrogen (elimination to alkene).

Answer 1

Step 1: The same reagent, KOH, plays two different chemical roles depending on the solvent, and that role decides the product.
Step 2: Dissolved in water, KOH supplies free hydroxide ions that are small and electron rich, i.e. strong nucleophiles. They walk in on the electrophilic carbon attached to chlorine, kick out chloride, and bond in its place. The result is a substitution product, the alcohol.
Step 3: Dissolved in alcohol, KOH reacts with the alcohol to make alkoxide ions. These are bulkier and act primarily as strong bases rather than nucleophiles. They pull off a proton from the carbon next to the C-Cl carbon (the beta carbon).
Step 4: Removing that beta proton and the chloride simultaneously creates a pi bond between the two carbons, so a molecule of HCl is effectively eliminated and an alkene forms as the main product. Thus aqueous medium favours substitution to alcohol, while alcoholic medium favours elimination to alkene.