Chenpo asked in 科學化學 · 9 years ago




1 Answer

  • 9 years ago
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    一點題外話: 這道題, 還有你另外問的兩道題, 都給我很大的違和感,


    純粹好奇一下, 這些問題, 你是從香港大學或香港高考的來源取得的嗎?

    還有, 你會問這些程度頗高的有機化學, 轉過頭卻在問關於緩衝溶液的基本計算, 實在是很奇怪..

    問題是英文, 我也用回英文好了; 有困難再提出.


    your answer is correct.

    they're all n-butanol, so acidic proton is on O-atom.

    obviously main point is the nitro- group.

    -NO2 is strongly, negatively-inducting, i.e. electron-drawing.

    when alkoxide ion is formed after deprotonation, the ion is stabilized by dispersing (delocalizing) the -ve charge over the ion.

    as the dispersing is by induction (not resonance effect), distance between the groups plays important role.

    nearer the groups --> stronger electron-withdrawing effect --> stronger dispersing --> more stable ion --> acidity is higher


    answer is correct.

    alcohols are surely more acidic than alkyne.

    alkyne is (exceptionally) more acidic than other hydrocarbons, but is still too weak compared to alpha-hydroxyl ketones, alcohols and acids.

    again, inductive effect comes with important role.

    -CF3 is strongly electron-withdrawing, thus dispersing the negative charge on alkoxide ion and stabilizing it.

    alkyl groups are slightly electron-giving (via hyper-conjugation), which destabilize the alkaoxide ion a little.

    the more alkyls the less stable.

    ethanol: 1 alkyl on C-OH

    tert-butanol: 3 alkyls on C-OH


    acid is surely the most acidic; phenol comes next; benzyl alcohol (3rd from left) acts as simple alcohol, least acidic.

    for the two phenols, -CH3 can destabilize the alkoxide ion by e-giving effect; -CN is not only e-drawing, but can further disperse the (-)charge via resonance effect. try to draw the resonance structure of it!

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