MCQ
Keto-enol tautomerism is found in
  • A
    ${{H}_{5}}{{C}_{6}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-H$
  • B
    ${{H}_{5}}{{C}_{6}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}$
  • C
    ${{H}_{5}}{{C}_{6}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{3}}$
  • $(b)$ and $(c)$ both

Answer

Correct option: D.
$(b)$ and $(c)$ both
d
(d) $\mathop {{C_6}{H_5} - \mathop {\mathop C\limits^{||} }\limits^O - C{H_3}}\limits_{{\rm{(Keto}}\,{\rm{form)}}} \,{\rm{and}}\,\mathop {{C_6}{H_5} - \mathop {\mathop {C = }\limits^{|\,\,\,\,\,\,} }\limits^{OH} }\limits_{{\rm{(enol form)}}} C{H_2}$

$\mathop {{C_6}{H_5} - \mathop {\mathop C\limits^{||} }\limits^O - C{H_2} - \mathop {\mathop C\limits^{||} }\limits^O - C{H_3}}\limits_{{\rm{(Keto form)}}} \,$and

$\mathop {{C_6}{H_5} - \mathop {\mathop C\limits_{||} }\limits_O - CH = \mathop {\mathop {C\,\,\,}\limits_{|\,\,\,\,} }\limits_{OH} - C{H_3}}\limits_{{\rm{(enol form)}}} $

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