Chemistry · Haloalkanes And Haloarenes · NEET
This is the most tested point in NEET. AQUEOUS KOH (KOH dissolved in water) supplies OH- ions that act as a nucleophile. OH- attacks the carbon and replaces the halogen, giving an ALCOHOL. This is substitution. ALCOHOLIC KOH (KOH dissolved in alcohol) supplies alkoxide/OH- that acts as a strong BASE. The base pulls off a beta-hydrogen, HX leaves, and a C=C double bond forms, giving an ALKENE. This is elimination (dehydrohalogenation). So: aqueous = alcohol (substitution), alcoholic = alkene (elimination).
OH- has two jobs: it can act as a nucleophile (attack carbon) or as a base (grab a hydrogen). In water, OH- is heavily surrounded by water molecules (hydration), so it acts more like a soft nucleophile and attacks the carbon = substitution. In alcohol, OH- is less hydrated and behaves as a stronger base, so it prefers to remove a beta-hydrogen = elimination. Same ion, different environment, different job. That is why the solvent decides the product.
For NEET, use simple rules. Substrate: primary halides favour SN2; tertiary halides favour SN1 or E1; a strong bulky base on any substrate favours E2. Reagent: aqueous KOH / weak nucleophile = substitution; alcoholic KOH / strong base = elimination. Heat: heating pushes the reaction toward ELIMINATION (alkene). So if you see 'alc. KOH, heat (delta)', think elimination almost every time.
Tertiary (3 degree) alkyl halides give the most elimination. They have the most beta-hydrogens and form the most stable (most substituted) alkene. They also cannot easily do SN2 because the crowded carbon blocks the nucleophile. Order of elimination rate: tertiary > secondary > primary. Primary halides mostly do substitution unless a strong bulky base and heat force elimination.
The carbon holding the halogen is the alpha (a) carbon. The carbons next to it are beta (b) carbons. A beta-hydrogen is a hydrogen on a beta-carbon. In elimination, the base removes a beta-H while X- leaves from the alpha-carbon, forming a C=C between the alpha and beta carbons. NO beta-hydrogen means NO normal elimination is possible, so only substitution can occur. Always check for beta-H first.
Use Saytzeff's (Zaitsev's) rule: the major alkene is the MORE substituted, MORE stable one (the double bond has more alkyl groups attached). For example, 2-bromopentane gives pent-2-ene (major) rather than pent-1-ene. The only exception is when a bulky base (like tert-butoxide) is used — then you get the LESS substituted alkene by Hofmann's rule.
For the following reactions: (a) CH3CH2CH2Br + KOH(alc) -> CH3CH=CH2 + KBr + H2O; (b) CH3CHBrCH2CH3 + KOH(aq) -> CH3CH(OH)CH2CH3 + KBr; (c) cyclohexene + Br2 -> trans-1,2-dibromocyclohexane. Which statement is correct?
The major product formed in the dehydrohalogenation of 2-bromopentane is pent-2-ene. This product formation is based on:
Major products A and B: 2-methylcyclohexan-1-ol ->[PBr3] A (major) ->[alc. KOH, heat] B (major) are:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Heating favours elimination. Higher temperature helps the base pull off a beta-hydrogen and form the more stable alkene. If a NEET question shows 'delta' (heat) with alcoholic KOH, expect the alkene (elimination) as the major product.
No. Normal (beta) elimination needs a hydrogen on the carbon next to the halogen. If there is no beta-hydrogen, the alkyl halide can only undergo substitution. Always check for a beta-H before choosing elimination.
They behave similarly. Both are strong bases that cause elimination (dehydrohalogenation) to form alkenes. Alcoholic KOH is the classic NEET reagent, but any strong base can drive elimination over substitution.
For tertiary halides with a strong base, elimination usually wins because the crowded carbon blocks the nucleophile from attacking (no SN2) and there are many beta-hydrogens. With a weak nucleophile and no heat, tertiary halides may still do SN1 substitution.
