Among the following compounds, which is the correct order of % enol content?a)I > II > IIIb)III > II > Ic)II > III > Id)I > III > IICorrect answer is option 'B'. Can you explain this answer?

Question

Among the following compounds, which is the correct order of % enol content?a)I &#62; II &#62; IIIb)III &#62; II &#62; Ic)II &#62; III &#62; Id)I &#62; III &#62; IICorrect answer is option 'B'. Can you explain this answer?

Answer

In the 1st compound the double bond is at the bridgehead carbon , as a result this compound cannot exist in enol form because such conformation would give an impossibility strained ring .Such dis stabilizing effect is absent in 2 nd compound . Whereas the 3 rd compound exclusively exist in enol form due to the formation of intramolecular hydrogen bonding

According to britz's rule at bridge head position -ve charge is unstable so; 3>2>1

Answer

In the 1 St compound the double bond is at the bridgehead carbon , as a result this compound cannot exist in enol form because such conformation would give an impossibility strained ring .Such dis stabilizing effect is absent in 2 nd compound . Whereas the 3 rd compound exclusively exist in enol form due to the formation of intramolecular hydrogen bonding

Answer

Because in structure 1  the double bondis formed at bridge head carbon which can't be flip in planar structure either of the side.Therefore structure 1 can't form enol .B ut in structure 2 there is one side where it can form enol form While 3 can form two enol form in  both to the side which further supported by NO2 group

Answer

According to britz's rule at bridge head position -ve charge is unstable so; 3>2>1

Answer

Bycyclo compound 3degree carbon does not partice pated in tautomerism

Answer

In the 1st compound the double bond is at the bridgehead carbon , as a result this compound cannot exist in enol form because such conformation would give an impossibility strained ring .Such dis stabilizing effect is absent in 2 nd compound . Whereas the 3 rd compound exclusively exist in enol form due to the formation of intramolecular hydrogen bonding

According to britz's rule at bridge head position -ve charge is unstable so; 3>2>1

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