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Aldehydes Ketones and Carboxylic Acids question

2024 · 27 Jan · Shift 2 · Q20
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Aldehydes Ketones and Carboxylic Acids question

2024 · 27 Jan · Shift 2 · Q20

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
The molecular formula of second homologue in the homologous series of mono carboxylic acids is
  1. A
    C2H4O2\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2C2​H4​O2​
  2. B
    C2H2O2\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2C2​H2​O2​
  3. C
    CH2O\mathrm{CH}_2 \mathrm{O}CH2​O
  4. D
    C3H6O2\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}_2C3​H6​O2​
View written solutionFree

Correct answer: A

  1. Identify the homologous series

    Mono carboxylic acids have the functional group −COOH-\mathrm{COOH}−COOH and their general formula is: CnH2nO2\mathrm{C}_n\mathrm{H}_{2n}\mathrm{O}_2Cn​H2n​O2​

  2. Write the first few members of the series

    • First homologue: Methanoic acid =HCOOH=CH2O2= \mathrm{HCOOH} = \mathrm{CH}_2\mathrm{O}_2=HCOOH=CH2​O2​
    • Second homologue: Ethanoic acid =CH3COOH=C2H4O2= \mathrm{CH}_3\mathrm{COOH} = \mathrm{C}_2\mathrm{H}_4\mathrm{O}_2=CH3​COOH=C2​H4​O2​
    • Third homologue: Propanoic acid =C3H6O2= \mathrm{C}_3\mathrm{H}_6\mathrm{O}_2=C3​H6​O2​

  3. Find the second homologue

    The second homologue is ethanoic acid, whose molecular formula is: C2H4O2\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2C2​H4​O2​

  4. Check the options

    • A: C2H4O2\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2C2​H4​O2​ ✅
    • B: C2H2O2\mathrm{C}_2\mathrm{H}_2\mathrm{O}_2C2​H2​O2​ ❌
    • C: CH2O\mathrm{CH}_2\mathrm{O}CH2​O ❌
    • D: C3H6O2\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2C3​H6​O2​ ❌ (this is the third homologue)

Therefore, the correct option is A.

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