I'm not sure how to approach this problem. Do I look at the EWG's and EDG's of the substiuents in the products, or find the pka value of each product? please explain so I can do it myself Predict the equilibrium of the following reactions. Which of the following reaction is Keq <1?a)Meb)MeMe MeF3C CF3HMeHOH+ H₂OH+ MeMgBr+ H₂O+ H₂O+PCCHOOHMe MeBrMg-O MeXMe MeHOOHF3C CF3HO OHxHHMe MeKetone formation is not favorable.

The given reactions are types of a nucleophilic addition reaction in which a nucleophile attacks the electron deficient atom and forms a sigma bond with that electron-deficient species.In carbonyl compounds, the carbon-oxygen bond is polar due to the difference in the electronegativities of carbon atom and oxygen atom . As the oxygen atom is highly electronegative, oxygen carries a partial negtaive charge and carbon carries the partial positive charge. And as the carbonyl carbon has a partial positive charge, it behaves as an electrophile.a) MeMgBr is a Grignard reagent. The reaction of ketone with MeMgBr is favorable as Me (-CH3) which carries a partial negative charge goes to C atom of carbonyl bond and MgBr which carries a partial positive charge goes to O atom of carbonyl atom. Thus, its Keq >1.b) F atoms are highly electronegative , thus -CF3 shows -I effect and thus makes the positive carbon more positively charged. Higher the electron deficiency on carbon, higher the rate of reaction towards the nucleophilic addition reaction. Thus, the formation of hydrate is favored. Now, the given compound has two CF3 group, thus, the equilibrium will shift in the forward reaction. Hence, Keq >1.c) Normally, hydrates cannot be isolated and thus the reaction is highly reversible but in case of formaldehyde, the C=O bond is comparatively weaker and the small size of hydrogen atoms favour the formation of product, Hence, the equilibrium shifts in forward direction. d) PCC is an oxidising agent and it oxidizes an alcohol to an aldehyde. Thus, the reaction will not go in forward direction. Therefore, equilibrium shifts to reactant side. Hence, Keq <1.e) Now, CH3 groups are electron donating groups, thus they make the carbon less electron deficient and the carbonyl compound less active for the addition of H2O. Hence, the reaction shifts to the reactant side. And Keq <1.