JEE Main · 2020 · Shift-ImediumGOC-499

The correct order for acid strength of compounds: CH≡CH, CH3-C≡CH, CH2=CH2, CH3-C≡CH, CH2=CH2 is as follows:

General Organic Chemistry (GOC) · Class 11 · JEE Main Previous Year Question

Question

The correct order for acid strength of compounds: \ceCH≡CH\ce{CH≡CH}, \ceCH3−C≡CH\ce{CH3-C≡CH}, \ceCH2=CH2\ce{CH2=CH2}, \ceCH3−C≡CH\ce{CH3-C≡CH}, \ceCH2=CH2\ce{CH2=CH2} is as follows:

Options
  1. a

    \ceCH≡CH>CH2=CH2>CH3−C≡CH>CH3−CH3\ce{CH≡CH > CH2=CH2 > CH3-C≡CH > CH3-CH3}

  2. b

    \ceCH3−C≡CH>CH≡CH>CH2=CH2>CH2=CH2\ce{CH3-C≡CH > CH≡CH > CH2=CH2 > CH2=CH2}

  3. c

    \ceCH3−C≡CH>CH≡CH>CH3−CH3>CH2=CH2\ce{CH3-C≡CH > CH≡CH > CH3-CH3 > CH2=CH2}

  4. d

    \ceHC≡CH>CH3−C≡CH>CH2=CH2>CH3−CH3\ce{HC≡CH > CH3-C≡CH > CH2=CH2 > CH3-CH3}

    ✓
Correct Answerd

\ceHC≡CH>CH3−C≡CH>CH2=CH2>CH3−CH3\ce{HC≡CH > CH3-C≡CH > CH2=CH2 > CH3-CH3}

Detailed Solution

Step 1: Acid Strength and Hybridization

Acid strength of C-H bonds depends on hybridization of the carbon:

  • sp carbon (alkyne): highest s-character (50%), electrons held closer to nucleus, more acidic
  • sp2sp^2 carbon (alkene): 33% s-character
  • sp3sp^3 carbon (alkane): 25% s-character, least acidic

Step 2: Order of pKa

| Compound | C-H type | pKa | Acid strength | |---|---|---|---| | \ceHC≡CH\ce{HC≡CH} (ethyne) | sp | ~25 | Strongest | | \ceCH3−C≡CH\ce{CH3-C≡CH} (propyne) | sp | ~25 | Slightly less (inductive effect of CH3) | | \ceCH2=CH2\ce{CH2=CH2} | sp2sp^2 | ~44 | Weaker | | \ceCH3−CH3\ce{CH3-CH3} | sp3sp^3 | ~50 | Weakest |

Order: \ceHC≡CH>CH3−C≡CH>CH2=CH2>CH3−CH3\ce{HC≡CH > CH3-C≡CH > CH2=CH2 > CH3-CH3} (ethane, not propane)

Step 3: Match to Options

Option (d) matches this order.

Key Points:

  • More s-character = electrons closer to nucleus = more stable conjugate base = more acidic
  • sp (50%) > sp2sp^2 (33%) > sp3sp^3 (25%)
  • Methyl group slightly reduces acidity of propyne vs ethyne (inductive electron donation)

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