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NEET CHEMISTRYHydrocarbonsMedium

Question

What is the correct order of acidity among the following compounds?

A

CHCH>CH3CCH>CH2=CH2>CH3CH3CH \equiv CH > CH_3-C \equiv CH > CH_2=CH_2 > CH_3-CH_3

B

CHCH>CH2=CH2>CH3CCH>CH3CH3CH \equiv CH > CH_2=CH_2 > CH_3-C \equiv CH > CH_3-CH_3

C

CH3CH3>CH2=CH2>CH3CCH>CHCHCH_3-CH_3 > CH_2=CH_2 > CH_3-C \equiv CH > CH \equiv CH

D

CH2=CH2>CH3CH3>CH3CCH>CHCHCH_2=CH_2 > CH_3-CH_3 > CH_3-C \equiv CH > CH \equiv CH

Step-by-Step Solution

  1. Hybridisation and Electronegativity: The acidity of hydrogen attached to a carbon atom depends on the s-character of the carbon's hybrid orbitals. Higher s-character means the electrons are closer to the nucleus, making the carbon more electronegative and the attached hydrogen more acidic .
  • Ethyne (HCCHHC\equiv CH): sp hybridised (50% s-character).
  • Ethene (H2C=CH2H_2C=CH_2): sp² hybridised (33.3% s-character).
  • Ethane (H3CCH3H_3C-CH_3): sp³ hybridised (25% s-character).
  • Order based on hybridisation: Alkyne > Alkene > Alkane.
  1. Substituent Effect (+I Effect): Comparing ethyne (CHCHCH\equiv CH) and propyne (CH3CCHCH_3-C\equiv CH), both contain sp-hybridised carbons with acidic hydrogens. However, the methyl group (CH3-CH_3) in propyne exerts an electron-donating inductive effect (+I effect). This increases electron density on the carbon atom and destabilises the conjugate base (alkynide ion) formed after proton removal. Therefore, propyne is less acidic than ethyne .

  2. Final Order: CHCHCH\equiv CH (sp) > CH3CCHCH_3-C\equiv CH (sp, +I effect) > CH2=CH2CH_2=CH_2 (sp²) > CH3CH3CH_3-CH_3 (sp³).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonscorrectacidityfollowingcompounds

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