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5

For the compound given in the question, the priority order of Grignard reagent attack follows, $\ce{R-OH > R-CO-R' > R-COOR'}$ Reason: Acid-base neutralisation are very fast. Carbonyl carbon is more nucleophilic than carboxylic carbon, as the latter one is stabilized by resonance. Therefore, first alcohol is attacked and hence P is (a), then the ...


2

There are several points that I'd throw out: As a technicality, you probably can't buy protonated alcohols; you'd need to use the alcohol plus an acid like HCl in a non-aqueous solvent. Now, once the amine drops off, it will mop up your catalytic acid, so you had probably better use at least stoichiometric acid, and possibly quite a bit more. Amides are ...


1

As @user55119 says it is unlikely the di-iodide is formed under iodoform conditions. According to this paper1, acetoacetate gives only mono-iodination at the methylene centre under more powerful iodinating conditions than the standard iodoform $\ce{I2/OH-}$. The mono-iodo product in the presence of excess base will exist as the iodo-enolate effectively ...


0

Well, yes, the mesomeric effect is shown by the carbonyl group. It's an electron withdrawing group. What happens here is that since the oxygen is a highly electronegative atom, it will tend to pull the $\pi$ bond electrons towards itself. Thus, the carbon becomes electron deficient i.e. it has it's electrons drawn towards the oxygen. So, we can safely say ...


0

Iodoform test is used to detect the carbonyl carbon and since acid anhydride is not a carbonyl compound , it does not give this test . The lone pair of oxygen ( the middle one ) and the π bond of carbonyl group are in resonance . When you actually see the reaction mechanism , you observe that the methyl group attached to the carbonyl should have acidic ...


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