Q.Predict the major product formed when sodium ethoxide reacts with tert.Butyl chloride.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Dehydrohalogenation
Dehydrohalogenation — First Look
Imagine you have a molecule that is "unstable" in a specific way — it carries a halogen atom (like Cl, Br, I) on one carbon and a hydrogen atom on the neighbouring carbon. If you treat it with a strong base, the base can pull off that hydrogen, and simultaneously the halogen leaves as a negative ion. The two carbons that lost these atoms now form a double bond between them. That's the core idea: dehydrohalogenation is the elimination of H and X (halogen) from adjacent carbons, producing an alkene.
The name itself tells you what happens: dehydro (removal of hydrogen) + halogenation (removal of halogen). So you are literally removing a hydrogen halide (HX) from the molecule.
The Precise Reaction
A haloalkane (alkyl halide) is treated with alcoholic KOH (potassium hydroxide dissolved in ethanol). The KOH acts as a strong base. The reaction follows this general pattern:
R−CH2−CH2−XKOHalcoholicR−CH=CH2+KX+H2O
For example, bromoethane gives ethene:
CH3−CH2−BrKOHalcoholicCH2=CH2+KBr+H2O
Aqueous KOH (KOH in water) does not cause elimination — it gives substitution (an alcohol). The alcoholic medium is essential because it keeps the base strong enough to pull off the hydrogen, and it does not favour the competing substitution reaction.
Why Alcoholic KOH and Not Aqueous?
In water, the hydroxide ion (OH−) is heavily solvated — surrounded by water molecules — which reduces its basic strength. In ethanol, the solvation is weaker, so OH− is a much stronger base. A strong base is needed to abstract the β-hydrogen (the hydrogen on the carbon next to the one bearing the halogen). The reaction proceeds via a one-step concerted mechanism (E2) where the base pulls the H, the halogen leaves, and the double bond forms — all at once.
Saytzeff's Rule — Which Alkene Forms?
When the haloalkane has more than one possible β-hydrogen (i.e., the carbon next to the halogen is attached to two different sets of hydrogens), more than one alkene can form. Saytzeff's rule tells you which one is the major product:
In dehydrohalogenation, the alkene with the more substituted double bond (the one with more alkyl groups attached to the double-bonded carbons) is the major product.
Why? More substituted alkenes are more stable (hyperconjugation and inductive effects). The reaction favours the pathway that leads to the more stable alkene.
NCERT's own example uses 2-bromopentane, and the preference is just as clear there:
Example: 2-bromobutane has two possible β-hydrogens: …
Sodium ethoxide is a strong, bulky base, and tert-butyl chloride is a tertiary halide, so elimination (E2) dominates over substitution. …
A strong bulky base (C2H5O−Na+) on a 3∘ halide gives β-elimination (E2), so the major product is 2-methylpropene.
Concept. The competition between substitution and elimination for haloalkanes — a core CBSE Class-12 haloalkanes-and-haloarenes idea.
…
- AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQQ.The number of α-hydrogens present in the major product (X) in the given reaction is [FIGURE] alc. KOHΔ X (A) 3 (B) 5 (C) 7 (D) 9
›Reveal solutionSolution
Alcoholic KOH with heat drives dehydrohalogenation (E2); the Saytzeff major product is 2-methylbut-2-ene, which carries 9 α-(allylic) hydrogens.
Reaction type. Alcoholic KOH under Δ removes H and the halogen from adjacent carbons (β-elimination, E2), giving the more substituted (Saytzeff) alkene as the major product X.
Major product and its α-hydrogens. The major alkene is 2-methylbut-2-ene:
(CH3)2C=CH−CH3.
The α-hydrogens are those on the carbons directly attached to the doubly-bonded carbons (the allylic positions):
- two methyls on the C-2 side: 3+3=6 H …
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.Among the following organic halides in the increasing order of their dehydrohalogenation reactions in the presence of alcoholic KOH: CH3CH2Br (A), CH3CH2CH2Br (B), CH3CH(Br)CH3 (C), (CH3)3CBr (D) (A) A<B<C<D (B) D<B<C<A (C) A<C<B<D (D) B<A<C<D
›Reveal solutionSolution
E2 dehydrohalogenation reactivity with alcoholic KOH rises with the degree of substitution of the halide: primary < primary(longer chain) < secondary < tertiary.
Concept and Intuition
In E2 elimination, the rate depends on how stabilized the developing alkene (and the transition state leading to it) is. More substituted alkenes are more stable (Zaitsev's rule, due to hyperconjugation/inductive donation from alkyl groups into the forming π system), and the alkyl halide's own substitution pattern similarly stabilizes the transition state via hyperconjugation. This makes tertiary halides eliminate fastest, followed by secondary, then primary. Among primary halides of different chain lengths, the longer/more substituted chain (n-propyl vs ethyl) offers slightly more hyperconjugative stabilization, making it marginally more reactive than the shorter ethyl halide.
Step-by-Step Solution
- Classify each halide: A = CH3CH2Br (ethyl, 1°), B = CH3CH2CH2Br (n-propyl, 1°, longer chain), C = CH3CHBrCH3 (isopropyl, 2°), D = (CH3)3CBr (tert-butyl, 3°).
- General reactivity order for E2 dehydrohalogenation: 3° > 2° > 1°, so D is most reactive, C next, and A/B (both 1°) are least reactive. …
- AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQQ.Identify the major product formed when 3- Bromo- 2- cyclohexene is treated with alcoholic KOH (A) Cyclohexane (a fully saturated six-membered ring, C6H12) (B) 1,3-Cyclohexadiene (a six-membered ring with two conjugated double bonds) (C) Cyclohex-2-en-1-ol (a six-membered ring with one double bond and an −OH group on the adjacent ring carbon) (D) Benzene (C6H6)
›Reveal solutionSolution
Dehydrohalogenation of the allylic bromide with alcoholic KOH gives the conjugated 1,3-cyclohexadiene as the major (more stable) product.
Concept and Intuition
Alcoholic KOH promotes E2 elimination (dehydrohalogenation) of alkyl/allylic halides to form alkenes. When the substrate is allylic (a C-Br bond adjacent to an existing C=C), elimination of HBr from the carbon bearing Br and an adjacent allylic hydrogen can generate a second, conjugated double bond. Conjugated dienes are more thermodynamically stable than non-conjugated ones (due to delocalisation of the pi electrons across the conjugated system), so the reaction preferentially forms the conjugated diene as the major product — an extension of Zaitsev's rule (more substituted/more stable alkene favoured).
Step-by-Step Solution
- Identify the substrate: a cyclohexene ring bearing Br at the allylic position (C3), with the existing double bond between C1-C2.
- Alcoholic KOH removes an H beta to the Br and eliminates Br−, forming a new C=C bond. …
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.The following reactions are examples for(i) and (ii).(i) H2C(Br)−CH2(Br)KOHH2C=CH2+KBr+H2O(ii) H2C=CH2Br2H2C(Br)−CH2(Br) (A)(i) Substitution reaction;(ii) Addition Reaction (B)(i) Elimination reaction;(ii) Addition Reaction (C)(i) Substitution reaction;(ii) Elimination Reaction (D)(i) Elimination reaction;(ii) Substitution Reaction
›Reveal solutionSolution
Converting a saturated dihalide to an alkene by removing atoms is elimination;
converting the alkene back to the dihalide by adding Br2 across the double
bond is addition — the reverse pair of a classic alkene synthesis/reaction set.
Concept and Intuition
Elimination reactions remove atoms/groups from adjacent carbons to CREATE a
multiple bond (unsaturation increases). Addition reactions do the opposite: atoms
add across an existing multiple bond, converting it to a single bond
(unsaturation decreases). This pair of reactions is the standard
alkene-dihalide interconversion used to teach the concept.
Step-by-Step Solution
- Reaction (i): H2C(Br)−CH2(Br)KOHH2C=CH2+KBr+H2O — starting from a saturated 1,2-dibromoethane, a double bond is formed while Br (and effectively H, consumed with KOH) are removed — this is a dehydrohalogenation, an ELIMINATION reaction. …
- AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQQ.The following reaction is a/an _________ CH3−CH2−CH2−BrKOHCH3−CH=CH2+KBr+H2O (A) Substitution reaction (B) Addition reaction (C) Electrophilic substitution reaction (D) Elimination reaction
›Reveal solutionSolution
Treating an alkyl halide with alcoholic KOH removes H and Br from adjacent carbons to form an alkene — this is a dehydrohalogenation, i.e., an elimination reaction.
Concept and Intuition
Alkyl halides can react with a nucleophile/base in two competing ways: substitution (the nucleophile replaces the halide) or elimination (the base removes a proton from a carbon adjacent to the halide, and the halide leaves, forming a π bond). Alcoholic KOH is a classic base used to favor elimination (E2) over substitution, because the alkoxide/ethanol solvent system and heat favor forming the more stable alkene product rather than an ether/alcohol.
Step-by-Step Solution
- Starting material: CH3−CH2−CH2−Br, an alkyl bromide with a β-hydrogen (on the middle carbon).
- Reagent: alcoholic KOH — this specific condition (alcoholic, not aqueous, KOH) is the standard signal for an elimination reaction in these problems.
- KOH (as base) abstracts a β-hydrogen while the C−Br bond breaks simultaneously, releasing Br− (which combines with K+ to give KBr) and forming a new C=C double bond.
- Product: CH3−CH=CH2 (propene), plus KBr and H2O (from KOH + HBr conceptually). …
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