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Arrange in their increasing order of equilibrium constants for hydration:

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(a) A < B < C < D < E
(b) A < C < B < E < D
(c) A < C < E < B < D
(d) C < A < B < E < D

How to compare figures E and D? Formaldehyde (figure D) doesn't have steric repulsion. Figure E has steric repulsion along with an electron withdrawing group. Given answer is (b). Does this mean steric hindrance is dominant over inductive effect while comparing? Please explain!

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To my knowledge, comparing the steric hindrance of the carbonyl group is not the path you want to follow to answering this question, because the trigonal planar configuration of the carbonyl carbon provides plenty of room in order for the inductive effect to outweigh the steric hindrance. We have to examine the partial positive charge of the carbonyl carbon resulted from the C=O and the other substituents. The C=O bond has a great dipole moment resulting in a partially positive carbonyl carbon. This charge can be stabilized or increased by alkyl substituents (hyperconjugation) or electronegative atoms (inductive effect) respectively. The greater the positive charge on the carbonyl carbon the higher the equilibrium constant for hydration (due to the mechanism). Therefore, we can easily say that (A) < all, because of the two alkyl substituents. C > A, because the chlorine reduces the hyperconjugation effect. B > C (only 1 alkyl substituent). E > B (because of the same reason as the C > A scenario. And lastly D > E, because of the abscense of any alkyl groups (same reason as the B > C scenario). Now, why is B > C and D > E? Why is less alkyl substitution increasing the charge to a greater extent than 1 more alkyl group with an electronegative atom? If the chlorine was attached to the carbonyl carbon itself (attach the chlorine in E to the C=O), it would greatly icrease the charge and in fact this molecule would have the highest Constant of all. However, we have to keep in mind that the inductive effect greatly deminshes the further away you go from the chlorine atom. The carbonyl carbon is 2 σ bonds away from the chlorine (as in C and E), so yes it will feel the induction but the alkyl group will still donate elenctron density via hyperconjugation to a lower extent. In conclusion, the chlorine in C and E is reducing the hyperconjugation effect but it doesn't neutralize it completely. Therefore, B > C and D > E

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