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Hydrocarbons question

2023 · 29 Jan · Shift 2 · Q12
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  5. /2023 · 29 Jan · Shift 2 · Q12

Hydrocarbons question

2023 · 29 Jan · Shift 2 · Q12

JEE MainChemistryHydrocarbonsMCQ+4 / −1
The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)
  1. A
    2-Bromo-3,3-dimethylpentane
  2. B
    2-Bromopropane
  3. C
    1-Bromo-2-methylbutane
  4. D
    2-Bromopentane
View written solutionFree

Correct answer: D

  1. Idea: In dehydrohalogenation of an alkyl halide, the halogen-bearing carbon is called the α-carbon and hydrogen is removed from an adjacent β-carbon.

    Each distinct β-carbon can give an alkene. We count the number of distinct isomeric alkenes formed, excluding rearrangement.

  1. Option A: 2-Bromo-3,3-dimethylpentane

    Structure: CH3−CH(Br)−C(CH3)2−CH2−CH3\text{CH}_3-\text{CH(Br)}-\text{C(CH}_3)_2-\text{CH}_2-\text{CH}_3CH3​−CH(Br)−C(CH3​)2​−CH2​−CH3​

    The α-carbon is C-2. Adjacent β-carbons are:

    • C-1: has hydrogens
    • C-3: this is C(CH3)2\text{C(CH}_3)_2C(CH3​)2​, so it has no hydrogen

    Hence elimination can occur only from C-1.

    Product: CH2=CH−C(CH3)2−CH2−CH3\text{CH}_2=\text{CH}-\text{C(CH}_3)_2-\text{CH}_2-\text{CH}_3CH2​=CH−C(CH3​)2​−CH2​−CH3​

    Number of alkenes = 1

  1. Option B: 2-Bromopropane

    Structure: CH3−CH(Br)−CH3\text{CH}_3-\text{CH(Br)}-\text{CH}_3CH3​−CH(Br)−CH3​

    α-carbon = middle carbon. Both adjacent methyl groups are equivalent.

    Elimination from either side gives the same alkene: CH2=CH−CH3\text{CH}_2=\text{CH}-\text{CH}_3CH2​=CH−CH3​

    Number of alkenes = 1

  1. Option C: 1-Bromo-2-methylbutane

    Structure: BrCH2−CH(CH3)−CH2−CH3\text{BrCH}_2-\text{CH(CH}_3)-\text{CH}_2-\text{CH}_3BrCH2​−CH(CH3​)−CH2​−CH3​

    α-carbon = C-1. Only one adjacent β-carbon exists, C-2.

    Elimination gives: CH2=C(CH3)−CH2−CH3\text{CH}_2=\text{C(CH}_3)-\text{CH}_2-\text{CH}_3CH2​=C(CH3​)−CH2​−CH3​

    Number of alkenes = 1

  1. Option D: 2-Bromopentane

    Structure: CH3−CH(Br)−CH2−CH2−CH3\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_3CH3​−CH(Br)−CH2​−CH2​−CH3​

    α-carbon = C-2. Adjacent β-carbons are:

    • C-1
    • C-3

    So two positional alkenes are possible:

    (i) Elimination from C-1: CH2=CH−CH2−CH2−CH3\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3CH2​=CH−CH2​−CH2​−CH3​ This is 1-pentene

    (ii) Elimination from C-3: CH3−CH=CH−CH2−CH3\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3CH3​−CH=CH−CH2​−CH3​ This is 2-pentene

    Now 2-pentene shows geometrical isomerism:

    • cis-2-pentene
    • trans-2-pentene

    Therefore total distinct alkenes from D: 1-pentene+cis-2-pentene+trans-2-pentene=31\text{-pentene} + \text{cis-2-pentene} + \text{trans-2-pentene} = 31-pentene+cis-2-pentene+trans-2-pentene=3

    Number of alkenes = 3

  1. Comparison of all options

    • A: 1 alkene
    • B: 1 alkene
    • C: 1 alkene
    • D: 3 alkenes

    Hence the compound giving the maximum number of isomeric alkenes is: D: 2-Bromopentane\boxed{\text{D: 2-Bromopentane}}D: 2-Bromopentane​

  1. Comparison with stored correct answer

    Stored correct answer = D

    Our derived answer = D

    So they agree.

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