1. Tardigrade
  2. Question
  3. Chemistry
  4. On the basis of the fact that NaNH2 can deprotonate HC≡ CH but not CH2=CH2 and the hybridization effect, order the following. four acids according to acidity undersetI mathopHC≡ CH undersetII mathopCH2=CH2 undersetIII mathopCH3-CH3 undersetIV mathopNH3

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Q. On the basis of the fact that $ NaN{{H}_{2}} $ can deprotonate $ HC\equiv CH $ but not $ C{{H}_{2}}=C{{H}_{2}} $ and the hybridization effect, order the following. four acids according to acidity $ \underset{I}{\mathop{HC\equiv CH}}\,\underset{II}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\underset{III}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\underset{IV}{\mathop{N{{H}_{3}}}}\, $

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Solution:

Since, $ NH_{2}^{-} $ can deprotonate $ HC\equiv CH,N{{H}_{3}} $ is weaker acid than $ HC\equiv CH. $ On the other hand, $ NH_{2}^{-} $ cannot deprotonate $ C{{H}_{2}}=C{{H}_{2}} $ therefore $ N{{H}_{3}} $ is stronger acid than ethane. Ethane is weakest acid due to hybridisation effect.