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Hydrocarbons question

2024 · 4 Apr · Shift 2 · Q20
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  5. /2024 · 4 Apr · Shift 2 · Q20

Hydrocarbons question

2024 · 4 Apr · Shift 2 · Q20

JEE MainChemistryHydrocarbonsMCQ+4 / −1
JEE Main 2024 (Online) 4th April Evening Shift Chemistry - Hydrocarbons Question 26 English In the above chemical reaction sequence "A\mathrm{A}A" and "B\mathrm{B}B" respectively are
  1. A
    H2O,H+\mathrm{H}_2 \mathrm{O}, \mathrm{H}^{+}H2​O,H+ and NaOH(alc)/I2\mathrm{NaOH}_{(\mathrm{alc})} / \mathrm{I}_2NaOH(alc)​/I2​
  2. B
    O3,Zn/H2O\mathrm{O}_3, \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}O3​,Zn/H2​O and KMnO4\mathrm{KMnO}_4KMnO4​
  3. C
    H2O,H+\mathrm{H}_2 \mathrm{O}, \mathrm{H}^{+}H2​O,H+ and KMnO4\mathrm{KMnO}_4KMnO4​
  4. D
    O3,Zn/H2O\mathrm{O}_3, \mathrm{Zn} / \mathrm{H}_2 \mathrm{O}O3​,Zn/H2​O and NaOH(alc)/I2\mathrm{NaOH}_{(\mathrm{alc})} / \mathrm{I}_2NaOH(alc)​/I2​
View written solutionFree

Correct answer: D

The reaction scheme image/details are not visible in the prompt, so the exact structures in the sequence are missing. However, we can still infer the intended pair of reagents from the options.

1. Analyze the reagent pairs

The options give two reagents AAA and BBB.

Option A

  • A=H2O,H+A = \mathrm{H_2O, H^+}A=H2​O,H+
  • B=NaOH(alc)/I2B = \mathrm{NaOH_{(alc)}/I_2}B=NaOH(alc)​/I2​

Option B

  • A=O3,Zn/H2OA = \mathrm{O_3, Zn/H_2O}A=O3​,Zn/H2​O
  • B=KMnO4B = \mathrm{KMnO_4}B=KMnO4​

Option C

  • A=H2O,H+A = \mathrm{H_2O, H^+}A=H2​O,H+
  • B=KMnO4B = \mathrm{KMnO_4}B=KMnO4​

Option D

  • A=O3,Zn/H2OA = \mathrm{O_3, Zn/H_2O}A=O3​,Zn/H2​O
  • B=NaOH(alc)/I2B = \mathrm{NaOH_{(alc)}/I_2}B=NaOH(alc)​/I2​

2. Identify the characteristic roles of these reagents

Reagent AAA

  • H2O,H+\mathrm{H_2O, H^+}H2​O,H+ is used for acid-catalyzed hydration of alkenes.
  • O3,Zn/H2O\mathrm{O_3, Zn/H_2O}O3​,Zn/H2​O is reductive ozonolysis, used to cleave a double bond and form carbonyl compounds.

Reagent BBB

  • KMnO4\mathrm{KMnO_4}KMnO4​ is generally an oxidizing agent.
  • NaOH(alc)/I2\mathrm{NaOH_{(alc)}/I_2}NaOH(alc)​/I2​ represents the iodoform reaction conditions, used to detect/convert compounds containing the CH3CO−\mathrm{CH_3CO-}CH3​CO− group or alcohols of type CH3CH(OH)−\mathrm{CH_3CH(OH)-}CH3​CH(OH)−.

3. Likely logic of the sequence

In many hydrocarbon reaction-sequence questions, the first step converts an alkene into carbonyl compounds by ozonolysis, and the second step identifies a methyl ketone or acetaldehyde derivative via the iodoform reaction.

This makes the combination

A=O3,Zn/H2O,B=NaOH(alc)/I2A = \mathrm{O_3, Zn/H_2O}, \qquad B = \mathrm{NaOH_{(alc)}/I_2}A=O3​,Zn/H2​O,B=NaOH(alc)​/I2​

most chemically consistent.

That corresponds to Option D.


4. Check against stored answer

Stored correct answer: D

My derived answer: D

So the derived answer agrees with the stored answer.

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