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Hydrocarbons question

2008 · Shift 0 · Q1
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Hydrocarbons question

2008 · Shift 0 · Q1

JEE MainChemistryHydrocarbonsMCQ+4 / −1
In the following sequence of reactions, the alkene affords the compound ‘B’ CH3−CH=CH−CH3CH_3 - CH = CH - CH_3CH3​−CH=CH−CH3​ →O3\xrightarrow{{O_3}}O3​​ A ⟶ZnH2O\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Zn}^{H{}_2O}}Zn⟶H2​O​ B The compound B is
  1. A
    CH3CH2CHOCH_3CH_2CHOCH3​CH2​CHO
  2. B
    CH3COCH3CH_3COCH_3CH3​COCH3​
  3. C
    CH3CH2COCH3CH_3CH_2COCH_3CH3​CH2​COCH3​
  4. D
    CH3CHOCH_3CHOCH3​CHO
View written solutionFree

Correct answer: D

  1. Identify the alkene

The given alkene is: CH3−CH=CH−CH3CH_3-CH=CH-CH_3CH3​−CH=CH−CH3​ This is but-2-ene.

  1. Reaction with ozone followed by reductive workup

The reagents are: O3followed byZn/H2OO_3 \quad \text{followed by} \quad Zn/H_2OO3​followed byZn/H2​O This is ozonolysis with reductive workup.

In ozonolysis, the double bond is cleaved and each carbon of the double bond is converted into a carbonyl group.

  1. Break the double bond

For but-2-ene: CH3−CH=CH−CH3CH_3-CH=CH-CH_3CH3​−CH=CH−CH3​ Cleaving the double bond gives:

  • Left alkene carbon: attached to CH3CH_3CH3​ and HHH ⇒\Rightarrow⇒ forms an aldehyde CH3CHOCH_3CHOCH3​CHO
  • Right alkene carbon: also attached to CH3CH_3CH3​ and HHH ⇒\Rightarrow⇒ forms another aldehyde CH3CHOCH_3CHOCH3​CHO

So the product is: 2 CH3CHO2\,CH_3CHO2CH3​CHO

Thus, compound BBB is ethanal: CH3CHOCH_3CHOCH3​CHO

  1. Check options
  • A: CH3CH2CHOCH_3CH_2CHOCH3​CH2​CHO = propanal ❌
  • B: CH3COCH3CH_3COCH_3CH3​COCH3​ = propanone ❌
  • C: CH3CH2COCH3CH_3CH_2COCH_3CH3​CH2​COCH3​ = butanone ❌
  • D: CH3CHOCH_3CHOCH3​CHO = ethanal ✅

Therefore, the correct answer is Option D.

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