Q.What happens when n-butyl chloride is treated with alcoholic KOH?

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The major product formed in the following reaction is $\text{H}_3\text{C}-\text{CH}_2-\underset{\underset{\text{Br}}{|}}{\text{CH}}-\text{CH}_3 \xrightarrow[\Delta]{\text{Alc.KOH}}$

   (a) 1-butene

   (b) 2-butanol

   (c) 1-bromobutane

   (d) 2-butene

   ›Reveal solutionSolution

   Alcoholic KOH removes HBr from 2-bromobutane; by Saytzeff's rule the more substituted alkene, but-2-ene, is the major product.

   The reactant $\text{H}_3\text{C}-\text{CH}_2-\underset{\text{Br}}{\text{CH}}-\text{CH}_3$ is 2-bromobutane. Alcoholic KOH is a strong base that promotes $\beta$-elimination (dehydrohalogenation), not substitution.

   A $\beta$-hydrogen can be removed from either the C-1 side or the C-3 side of the C-Br carbon:

   • Removal from C-3 (CH$_2$) gives $\text{CH}_3\text{CH}=\text{CHCH}_3$ (but-2-ene) — a disubstituted double bond.
   • Removal from C-1 (CH$_3$) gives $\text{CH}_2=\text{CHCH}_2\text{CH}_3$ (but-1-ene) — a monosubstituted double bond. …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set A1 markMCQ
   Q.Ethyl bromide on boiling with alcoholic caustic potash gives

   (a) Ethyl alcohol

   (b) Ethylene

   (c) Acetylene

   (d) Ethane

   ›Reveal solutionSolution

   Alcoholic KOH removes HBr from ethyl bromide (E2 elimination) to give ethylene.

   Alcoholic potassium hydroxide furnishes the strong base ethoxide/hydroxide in an alcoholic medium, which favours elimination over substitution. Ethyl bromide loses a molecule of HBr:

   CH3-CH2-Br --(alc. KOH, heat)--> CH2=CH2 + KBr + H2O

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion [A]: 1-chlorobutane on heating with alc. KOH gives mainly Bute-1-ene. Reason [R]: Bute-1-ene is more stable than Bute-2-ene.

   (a) Both [A] and [R] are true and [R] is the correct explanation of [A].

   (b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].

   (c) [A] is true, but [R] is false.

   (d) [A] is false, but [R] is true.

   ›Reveal solutionSolution

   1-Chlorobutane does give but-1-ene on heating with alcoholic KOH — but only because it is the ONLY elimination product possible (Cl is on the terminal carbon), not because but-1-ene is more stable than but-2-ene. In fact but-2-ene (especially the trans isomer) is MORE stable than but-1-ene, so the Reason is false.

   Assertion: 1-Chlorobutane is $CH_3CH_2CH_2CH_2Cl$ — the leaving group (Cl) is on C1, a terminal carbon. β-elimination (E2, Zaitsev/Hofmann considerations) can only remove a β-hydrogen from C2 (there is no carbon "before" C1 to eliminate towards). This necessarily gives:

   $CH_3CH_2CH_2CH_2Cl \xrightarrow{alc.\ KOH,\ \Delta} CH_3CH_2CH=CH_2\ (\text{but-1-ene}) + KCl + H_2O$

   So yes, but-1-ene IS the (essentially only) product — the Assertion is TRUE.

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 markMCQ
   Q.C2H5Cl on heating with alcoholic KOH will produce:

   (a) C2H5OH

   (b) C2H4

   (c) C2H2

   (d) C2H6

   ›Reveal solutionSolution

   Alcoholic KOH promotes elimination (dehydrohalogenation), so ethyl chloride gives ethene, not the substitution product ethanol.

   When an alkyl halide is heated with $KOH$ dissolved in alcohol, the hydroxide ion acts as a strong base rather than a nucleophile, abstracting a $\beta$-hydrogen and eliminating the halide to form an alkene (E2 elimination):

   $CH_3CH_2Cl \xrightarrow[\Delta]{alc.\ KOH} CH_2=CH_2 + KCl + H_2O$ …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set ANNUAL1 mark
   Q.Write the products of the following reaction: $CH_3-CH_2-CH_2-CH(Br)-CH_3 \xrightarrow{Alc.\ KOH} A + B$

   ›Reveal solutionSolution

   E2 dehydrohalogenation of the secondary bromide 2-bromopentane with alcoholic $KOH$ gives two alkenes; by Zaitsev's rule, the more-substituted one dominates.

   Numbering the given chain from the bromine end for IUPAC purposes: $CH_3(C1)-CHBr(C2)-CH_2(C3)-CH_2(C4)-CH_3(C5)$ — this is 2-bromopentane. Two sets of $\beta$-hydrogens are available for E2 elimination: on C1 (the adjacent methyl) and on C3.

   Elimination toward C3 (removing a $\beta$-H from C3) gives the internal, more substituted (disubstituted) alkene:

   $CH_3-CH=CH-CH_2-CH_3\quad(\text{pent-2-ene})$

   By Zaitsev's rule, the more substituted, more stable alkene (better stabilised by hyperconjugation/alkyl substitution across the double bond) is the major product (A).

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Alkyl halides react with alcoholic KOH to give:

   (a) Alkanes

   (b) Alkenes

   (c) Alcohols

   (d) Ethers

   ›Reveal solutionSolution

   Alcoholic KOH is a strong base used for elimination; it removes H and X from adjacent carbons of an alkyl halide to give an alkene.

   When an alkyl halide (R-CH2-CH2-X) is treated with KOH dissolved in alcohol, the hydroxide ion acts as a base rather than a nucleophile. It abstracts a β-hydrogen (a hydrogen on the carbon next to the one bearing the halogen), and the halide ion leaves simultaneously (E2 mechanism), forming a carbon-carbon double bond:

   CH3-CH2-X + KOH(alc.) → CH2=CH2 + KX + H2O

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2023Set 56/2/11 markMCQ
   Q.The conversion of an alkyl halide into an alkene by alcoholic KOH is classified as (A) a substitution reaction (B) an addition reaction (C) a dehydrohalogenation reaction (D) a dehydration reaction

   ›Reveal solutionSolution

   The reaction of an alkyl halide with alcoholic KOH is a dehydrohalogenation — it eliminates H and X from adjacent carbons to form a double bond, not a substitution or addition.

   The key here is to recognise what alcoholic KOH does. In organic chemistry, the solvent and reagent together decide the reaction path. Aqueous KOH gives you substitution (OH replaces the halogen). But alcoholic KOH is a strong base in a poorly nucleophilic medium — it favours elimination over substitution.

   Let’s break it down.

   1. What is the reactant? An alkyl halide ($R-X$). It has a carbon–halogen bond, polarised so that carbon is slightly positive ($\delta^+$) and halogen is slightly negative ($\delta^-$). This makes the carbon electrophilic.

   2. What is alcoholic KOH? KOH dissociates into $K^+$ and $OH^-$ ions. In alcohol (usually ethanol), the $OH^-$ ion is a strong base but a weaker nucleophile than in water (because alcohol solvates it differently). The high basicity of $OH^-$ in alcohol means it prefers to abstract a proton rather than attack the carbon.

   3. What actually happens? The $OH^-$ pulls a hydrogen from the $\beta$-carbon (the carbon next to the one bearing the halogen). The electrons from that C–H bond move to form a $\pi$ bond between $\alpha$ and $\beta$ carbons, and the halogen leaves as $X^-$. This is an E2 elimination — bimolecular, one step, no intermediate.

      The general equation:

   $R-CH_2-CH_2-X + KOH \ (\text{alc.}) \longrightarrow R-CH=CH_2 + KX + H_2O$

   4. Why not substitution? Substitution ($S_N2$) would require the $OH^-$ to attack the $\alpha$-carbon. But in alcoholic medium, the base is strong and the solvent is not very polar protic for stabilising the transition state of substitution — elimination is kinetically favoured, especially with secondary or tertiary halides.

   5. Why not addition? Addition reactions happen across multiple bonds (like alkenes or carbonyls). Here we start with a saturated alkyl halide — no multiple bond to add to. So addition is out. …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2023Set ANNUAL1 mark
   Q.How would you convert the following? But-1-ene to but-2-ene

   ›Reveal solutionSolution

   Markovnikov addition of HBr moves the halogen to the internal carbon of but-1-ene (2-bromobutane), and subsequent base-induced elimination follows Zaitsev's rule to give the more substituted, more stable but-2-ene as the major product — net effect: the double bond has migrated from the terminal to the internal position.

   Step 1 — Markovnikov hydrobromination: Addition of $HBr$ (no peroxide present) to but-1-ene proceeds by the ionic mechanism: protonation of the alkene occurs so as to generate the more stable (secondary) carbocation, i.e. $H^+$ adds to the terminal $CH_2=$ carbon, placing the positive charge on C-2; $Br^-$ then attacks this secondary carbocation:

   $CH_2=CH-CH_2CH_3 \xrightarrow{HBr} CH_3-CHBr-CH_2CH_3 \;\;(\text{2-bromobutane})$

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2022Set E1 markMCQ
   Q.Monohalogen derivative on reaction with alcoholic KOH gives

   (a) Alkane

   (b) Alkene

   (c) Alkyne

   (d) Alicyclic hydrocarbon

   ›Reveal solutionSolution

   Monohaloalkane + alcoholic KOH → dehydrohalogenation (elimination) → alkene.

   With alcoholic KOH, a monohalogen derivative (alkyl halide) loses a molecule of HX in a beta-elimination (dehydrohalogenation) reaction, forming a carbon-carbon double bond, i.e. an alkene.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2021Set A1 markMCQ
    Q.The product of the reaction C2H5Br --KOH/Alcoholic--> is

    (a) CH2 = CH2

    (b) CH3CH2OH

    (c) CH3CH2OCH2CH3

    (d) none of these

    ›Reveal solutionSolution

    Alcoholic KOH promotes beta-elimination (dehydrohalogenation) of an alkyl halide, giving an alkene.

    The reagent is critical here. Alcoholic KOH favours elimination (E2, dehydrohalogenation), removing HBr from bromoethane to form the alkene:

    CH3CH2Br + KOH(alcoholic) ---> CH2=CH2 + KBr + H2O

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2018Set ANNUAL1 mark
    Q.Name the reagent used in the dehydrohalogenation of haloalkanes.

    ›Reveal solutionSolution

    Alcoholic KOH removes HX from a haloalkane (dehydrohalogenation) to give an alkene.

    Concept. Dehydrohalogenation (a $\beta$-elimination) removes a hydrogen halide from a haloalkane to form an alkene. It requires a strong base in an alcoholic medium. Alcoholic potassium hydroxide (KOH in ethanol) is the standard reagent — the ethoxide/hydroxide acts as a base and abstracts the $\beta$-hydrogen.

    Example. …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2018Set 56/11 mark
    Q.Predict the major product formed when sodium ethoxide reacts with tert.Butyl chloride.

    ›Reveal solutionSolution

    A strong bulky base ($C_2H_5O^-Na^+$) on a $3^\circ$ halide gives $\beta$-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.

    …

    🔒 Start your 14-day free trial to see the full solution →