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O, , O, , (a), , (b), O, , O, , (c), , (d), O, , O, Ans.- (d), O, , Explanation- In order to show keto-enol tautomerism, the compound must, contain at least one α-hydrogen atom attached to sp3 hybridized carbon atom., Here,, (1) The compounds given in option (a) and (b) contain α-hydrogen atoms, attached to sp3 hybridized carbon atom., , O, , O, α, , α, , α, , Hence, they will show keto-enol tautomerism as under-, , H, O, , ⇌, , O─H, , Keto form, H, O, , Enol form, ⇌, , O─H, , Keto form, , Enol form, 2

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O, Ans.- (d), , Explanation- In order to show keto-enol tautomerism, the compound must, contain at least one α-hydrogen atom attached to sp3 hybridized carbon atom., Here,, (1) The compounds given in option (a) and (b) contain α-hydrogen atoms, attached to sp3 hybridized carbon atom., , O, , O, α, , α, , Bridge α-carbon, , Hence, they will show keto-enol tautomerism. Here, sp3 hybridized bridge, carbon of option (b) also contains α-hydrogen, but it cannot participate in, tautomerization, because sp2 hybridization (planarity) is not possible in bridge, carbon atom., , H, ⇌, , O, , O─H, , Keto form, H, , Enol form, ⇌, , O, , O─H, , Keto form, , Enol form, , (2) The compound given in option (c) does not contain α-hydrogen atom attached, to sp3 hybridized non-bridged carbon atom, but it involves conjugation., , O, α, β, , γ, 4

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Explanation- In order to show keto-enol tautomerism, the compound must, contain at least one α-hydrogen atom attached to sp3 hybridized non-bridged, carbon atom. Here,, (1) The compounds given in option (a) contains α-hydrogen atoms attached to sp3, hybridized carbon atoms, but these are bridged carbon atoms where, sp 2, hybridization is not possible. Hence, it will not show keto-enol tautomerism., , O, Bridge α-carbon, O, No α-hydrogen on sp3 hybridized non-bridged carbon atom, (Does not show keto-enol tautomerism), (2) The compound given in option (b) contain α-hydrogen atom attached to sp3, hybridized non-bridged carbon atom, hence, it undergoes tautomerization., , O, , (3) The compound given in option (c) does not contain sp 3 hybridized nonbridged α-carbon atom, but it contains hydrogen atoms attached with sp3, hybridized non-bridged carbon present in conjugation. Hence, it will show ketoenol tautomerism., , sp3 hybridized carbon, , O, , (4) The compound given in option (d) contains sp3 hybridized non-bridged αcarbon atom with acidic hydrogen atom. Hence, it will show keto-enol, tautomerism., , O, , sp3 hybridized carbon, , C, Q.5.- The number of keto-enol tautomers possible for 2-butanone is6

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(a) 1, , (b) 2, , (c) 3, , (d) Unpredictable, , Ans.- (c) 3, Explanation- 2-butanone contains two types of acidic hydrogen (α-hydrogen), that can participate in tautomerization., , O, CH3─ C─ CH2 ─ CH3, 1, 2, Here, participation of α-hydrogen atoms numbered 1 gives only one enol, tautomer while, participation of those numbered 2 gives two enol tautomers (E, and Z-forms)., , O─H, , O, +, , H, , ⇌, , H, O, , OH, , OH, , H+, , ⇌, , +, , H, , E-form, , Z-form, , Thus, total number of products formed will be 3. Here, But-2-en-2-ol is major, product of the reaction., , Q.6.- Which of the following contains the highest content of enol formO, O, (a), , (b), , O, O, , O, , O, , (c), , (d), , O, , O, 7

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Ans.- (c), Explanation- Here,, (1) Option (a) and (d) are monocarbonyl compounds (monoketones) which, almost exclusively exist in keto form (99.99%)., (2) Option (b) is acyclic diketone (1,2-dicarbonyl compound or α-dicarbonyl, compound) which also exist mainly in keto form. Thus, the percentage of enol, form is much lesser than keto form in equilibrium mixture., (3) Option (c) is also acyclic diketone but, it is 1,3-dicarbonyl compound or βdicarbonyl compound which contains active methylene group having more acidic, hydrogen. Here, enol form is stabilized by intramolecular hydrogen bonding and, resonance due to presence of conjugation. Thus, it involves more enolization and, therefore, it contains more percentage of enol form as compared to others in, equilibrium mixture. In fact, in this compound, enol form (76.4%) predominates, keto form in equilibrium mixture., , O, , O, , O, , O─H, , ⇌, Enol form, , H, Keto form, (23.6%), , or, H, O, , O, , Intramolecular H-bonding in Enol form, (76.4%), , 8

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Q.7. Theoretically following two enol forms are possible for phenyl, acetoneOH, , OH, , (A), Which of these two is more stable(a) (A), (c) Both are equally stable, , (B), , (b) (B), (d) Unpredictable, , Ans.- (a) (A), Explanation- Here, (A) involves extended conjugation and therefore, it is more, stable than (B) which does not involve extended conjugation., , OH, , OH, , (A), (Extended conjugation), , (B), (No extended conjugation), , Q.8. Consider following compoundsO, , O, , O, , O, H, , (A), O, , (B), O, , O, , (C), , O, , (D), , Enol form of which one of these is the most stable(a) (A), (b) (B), (c) (C), (d) (D), 9

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Ans.- (a) (A), Explanation- Here,, (A) involves extended conjugation on both the sides and therefore, its enol form, will be the most stable amongst all., (B) involves extended conjugation on one of the sides and therefore, its enol, form will be less stable than that of (A)., (C) involves extended conjugation on one of the sides, but it has methyl group on, other side, whose +I effect opposes resonance in enol form and decreases its, stability. Hence, its enol form will be less stable than that of (B)., (D) does not involve extended conjugation on any of the sides, but it has methyl, groups on both the sides, whose +I effect opposes resonance in enol form and, decreases its stability. Hence, its enol form will be the least stable amongst all., , Q.9. Consider following compoundsO, , O, , O, , O, H, , (A), O, H, , (B), O, , O, , O, , H, , (C), (D), The correct sequence of enol content in equilibrium mixture is(a) (A) < (B) < (C) < (D), (b) (A) > (C) > (B) > (D), (c) (A) < (B) < (D) < (C), (d) (A) > (B) > (D) > (C), Ans.- (b) (A) > (C) > (B) > (D), Explanation- Here,, (1) Extension of conjugation in enol form increases its stability and facilitates, enolization i.e. increases the enol content in the equilibrium mixture., (2) +I effect of alkyl group opposes resonance stabilization of enol form and, decreases its stability. Hence, it discourages enolization i.e. decreases the enol, content in the equilibrium mixture., Let us consider following compound10

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O, , O, , G1, Here, if(1) G1 =, (2) G1 =, (3) G1 =, (4) G1 =, (5) G1 =, (6) G1 =, , G2, , G2 = C6H5-,, C6H5-, G2 = H, C6H5-, G2 = CH3G2 = H,, CH3-, G2 = H,, G2 = CH3-,, , then enol content, then enol content, then enol content, then enol content, then enol content, then enol content, , ≈, =, =, =, =, =, , 100 %, 95 - 100 %, 90 - 95 %, 85 - 90 %, 80 - 85 %, 75 - 80 %, , Q.10. Consider following compoundsO, , O, , O, , O, OC2H5, , (A), O, H5C2O, , (B), O, , O, , O, , OC2H5, , (C), (D), The correct sequence of enol content in equilibrium mixture is(a) (A) < (B) < (C) < (D), (b) (A) > (D) > (B) > (C), (c) (A) < (B) < (D) < (C), (d) (A) > (B) > (D) > (C), Ans.- (b) (A) > (D) > (B) > (C), Explanation- Here,, (1) Extension of conjugation in enol form increases its stability and facilitates, enolization i.e. increases the enol content in the equilibrium mixture., (2) +I effect of alkyl group opposes resonance stabilization of enol form and, decreases its stability. Hence, it discourages enolization i.e. decreases the enol, content in the equilibrium mixture., (3) Cross conjugation of alkoxy group strongly opposes the resonance, stabilization of enol form and decreases its stability to a very large extent. Hence,, 11

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it strongly discourages enolization i.e. decreases the enol content in the, equilibrium mixture to a very large extent., Let us consider following compound-, , O, , O, OC2H5, , Here, cross conjugation of lone pair of ethoxy group strongly opposes the, resonance stabilization of enol form and decreases its stability to a very large, extent. Hence, it strongly discourages enolization i.e. decreases the enol content, in the equilibrium mixture to a very large extent (approx. 7.2 %)., , O, , O, , O, , O, , -, , H, Keto form, , + OH, .., O, .. ─ C2H5 - H2O, , .., , .., O, .. ─ C2H5, , Carbanion (Cross conjugation), O, , O, , .., O, .. ─ C2H5, Enolate ion, Similarly, compound (C) involves, double cross conjugation, so that the, hydrogen atom of active methylene group is the least acidic. Hence, it does not, prefer enolization at all and contains the least enol content (about 1%)., , O, , O, , H5C2O, , OC2H5, , Involves cross conjugation on both the ends, (Least enol content), 12

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The percentage of enol content in some most common carbonyl compounds are, listed here-, , Compound, O, , O, , Enol Content (%), ≈ 10-4, , H, O, , H5C2O, , ≈1, , OC2H5, O, , ≈ 1.2, O, O, , O, , O, , O, , ≈ 7.2, , H5C2O, ≈ 76, O, , O, ≈ 90, , O, , O, ≈ 100, OH, ≈ 100, , Phenol, 13