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Basics of Organic Chemistry question

2024 · 31 Jan · Shift 2 · Q23
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Basics of Organic Chemistry question

2024 · 31 Jan · Shift 2 · Q23

JEE MainChemistryBasics of Organic ChemistryNumerical+4 / −1
Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7

  1. Write the structure of 2-methylbutane

    2-methylbutane has the structure: CH3−CH(CH3)−CH2−CH3\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}CH3​−CH(CH3​)−CH2​−CH3​

    Let us identify all nonequivalent hydrogen positions.

  2. Identify different types of hydrogens

    In monochlorination under sunlight, one hydrogen is replaced by chlorine via free-radical substitution.

    In 2-methylbutane, the hydrogens are present at:

    • Type A: The two equivalent methyl groups attached to carbon-2 CH3−CH(CH3)−\mathrm{CH_3-CH(CH_3)-}CH3​−CH(CH3​)− These two methyl groups are equivalent by symmetry.

    • Type B: The terminal methyl group at the other end −CH2−CH3\mathrm{-CH_2-CH_3}−CH2​−CH3​

    • Type C: The methylene group −CH2−\mathrm{-CH_2-}−CH2​−

    • Type D: The tertiary hydrogen on carbon-2 −CH(−)\mathrm{-CH(-)}−CH(−)

  3. Form products by replacing one H from each nonequivalent position

    (i) Substitution at Type A methyl groups

    Replacing one H from either of the two equivalent methyl groups gives the same constitutional product: CH2Cl−CH(CH3)−CH2−CH3\mathrm{CH_2Cl-CH(CH_3)-CH_2-CH_3}CH2​Cl−CH(CH3​)−CH2​−CH3​ This molecule has a chiral center at carbon-2, so it exists as two enantiomers. Hence, number of isomeric products from this substitution = 2.

    (ii) Substitution at Type B terminal methyl group

    Product: CH3−CH(CH3)−CH2−CH2Cl\mathrm{CH_3-CH(CH_3)-CH_2-CH_2Cl}CH3​−CH(CH3​)−CH2​−CH2​Cl This also creates a chiral center at carbon-2, so again there are two enantiomers. Hence, number of isomeric products = 2.

    (iii) Substitution at Type C methylene group

    Product: CH3−CH(CH3)−CH(Cl)−CH3\mathrm{CH_3-CH(CH_3)-CH(Cl)-CH_3}CH3​−CH(CH3​)−CH(Cl)−CH3​ Here carbon-3 becomes a chiral center, so this also exists as two enantiomers. Hence, number of isomeric products = 2.

    (iv) Substitution at Type D tertiary hydrogen

    Product: CH3−C(Cl)(CH3)−CH2−CH3\mathrm{CH_3-C(Cl)(CH_3)-CH_2-CH_3}CH3​−C(Cl)(CH3​)−CH2​−CH3​ The substituted carbon has two identical methyl groups, so it is not chiral. Hence, number of isomeric products = 1.

  4. Total number of isomeric monochloro products

    Therefore, 2+2+2+1=72+2+2+1=72+2+2+1=7

  5. Comparison with stored answer

    The stored correct answer is 6, but careful counting of all isomeric products (including optical isomers) gives 7.

    If the question intended only constitutional isomers, then the answer would be 4. But since it asks for isomeric products, stereoisomers should also be counted, giving 7.

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