A) \[C{{H}_{3}}-C\equiv CH>CH\equiv CH>C{{H}_{2}}=C{{H}_{2}}\]
B) \[C{{H}_{3}}-C\equiv CH>C{{H}_{2}}=C{{H}_{2}}>HC\equiv CH\]
C) \[HC\equiv CH>C{{H}_{3}}-C\equiv CH>C{{H}_{2}}=C{{H}_{2}}\]
D) \[CH\equiv CH>C{{H}_{2}}=C{{H}_{2}}>C{{H}_{3}}-C\equiv CH\]
Correct Answer: C
Solution :
As the s-character of \[CH\text{ }bond\] increases, acidity increases. Thus, acidity order \[\underset{\begin{smallmatrix} sP \\ (50%) \end{smallmatrix}}{\mathop{CH\equiv CH>}}\,\underset{\begin{smallmatrix} s{{P}^{2}} \\ (33.3%) \end{smallmatrix}}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\] +1 effect (\[-C{{H}_{3}}\]group) reduces acidity Thus, acidity order well be \[CH\equiv CH>C{{H}_{3}}-C\equiv CH>C{{H}_{2}}=C{{H}_{2}}\]
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