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Aldehydes Ketones and Carboxylic Acids question

2021 · 1 Sep · Shift 2 · Q1
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Aldehydes Ketones and Carboxylic Acids question

2021 · 1 Sep · Shift 2 · Q1

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
Experimentally reducing a functional group cannot be done by which one of the following reagents?
  1. A
    Pt-C/H2H_2H2​
  2. B
    Na/H2H_2H2​
  3. C
    Pd-C/H2H_2H2​
  4. D
    Zn/H2OH_2OH2​O
View written solutionFree

Correct answer: B

  1. Interpret the question

    We need to identify which reagent cannot be used experimentally for reduction of a functional group.

    In organic chemistry, common reduction methods include:

    • catalytic hydrogenation: H2/Pt\mathrm{H_2/Pt}H2​/Pt or H2/Pd\mathrm{H_2/Pd}H2​/Pd
    • metal in water/alcohol/acid systems depending on the substrate
  2. Check each option

    Option A: $\mathrm{Pt!-

C/H_2}$ Platinum-catalyzed hydrogenation is a standard reducing system. It is used for reduction of many functional groups such as:

  • C=C\mathrm{C=C}C=C
  • C≡C\mathrm{C\equiv C}C≡C
  • in some cases other reducible groups

So this can be used experimentally.

Option C: $\mathrm{Pd!-

C/H_2}$ Palladium on carbon with hydrogen is also a standard catalytic hydrogenation reagent. It is widely used in laboratory reductions.

So this can be used experimentally.

Option D: Zn/H2O\mathrm{Zn/H_2O}Zn/H2​O

Zinc in appropriate medium can act as a reducing system in certain cases. Metal-mediated reductions are experimentally known. Hence this is not the impossible one among the given choices.

Option B: Na/H2\mathrm{Na/H_2}Na/H2​

Sodium metal with hydrogen gas is not a practical or standard reducing reagent for functional group reduction. Sodium is highly reactive, and this combination is not used as an experimental reduction method for organic functional groups.

Therefore, this reagent cannot be used experimentally.

  1. Conclusion

    The reagent that cannot be used experimentally for reduction is: Na/H2\boxed{\mathrm{Na/H_2}}Na/H2​​

  2. Compare with stored correct answer

    Stored correct answer: B\mathrm{B}B

    Our derived answer: B\mathrm{B}B

    They match.

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