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NEET · Chemistry · STD 12 -8.1. Aldehydes and ketones

નીચેનામાંથી ચલરૂપક સંયોજનોની સ્થિરતાનો કમ ....  \(\mathop {\begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_2} = C - C{H_2} - C - C{H_3}} 
\end{array}}\limits_{(I)}  \rightleftharpoons \) \(\mathop {\begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_3} - C - C{H_2} - C - C{H_3}} 
\end{array}}\limits_{(II)}  \rightleftharpoons \) \(\mathop {\begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_3} - C = CH - C - C{H_3}} 
\end{array}}\limits_{(III)} \)

  1. A \(II > I > III\)
  2. B \(II > III > I\)
  3. C \(I > II > III\)
  4. D \(III > II > I\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(III > II > I\)

Step-by-step Solution

Detailed explanation

The enols of \(\beta\) -dicarbonyl compounds are more stable because of conjugation and intramolecular \(H-\) bonding. Thus, the order of stability is \(\underset{\begin{smallmatrix} 
 (Stabilised\,\,by\,\,conjugation \\ 
 and\,\,H\,-\,bonding )\\ III 
\end{smallmatrix}}{\mathop{\begin{matrix}
   OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O  \\
   |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||  \\
   {{H}_{3}}C-C=CH-C-C{{H}_{3}}  \\
\end{matrix}}}\,\) \(>\) \(\mathop {\begin{array}{*{20}{c}}
  {O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_3} - C - C{H_2} - C - C{H_3}} 
\end{array}}\limits_{(II)}\)  \(>\) \(\mathop {\begin{array}{*{20}{c}}
  {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {C{H_2} = C - C{H_2} - C - C{H_3}} 
\end{array}}\limits_{(I)}\)
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