25

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic ...


24

The mechanism of the Cannizzaro reaction is illustrated below. The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate. Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon ...


22

The correct answer is the 1,2-addition product (i.e. the allylic alcohol 1). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone. This exact transformation was reported by Akai and coworkers recently (Org. Lett. 2010, 12, ...


21

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon: $\hspace{2.4cm}$ $\ce{NaBH4}$ is ...


20

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations ...


19

In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation: Due ...


19

So, shouldn’t the order be the other way round? Yes, it is the other way around. Ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25). The figure below shows the resonance structures for the enolate anion of a ketone and an ester. Resonance structures II and IV stabilize the enolate anion in the ketone and ester respectively. In the ester, there is ...


19

First, the methoxy- group can be demethylated by $\ce{AlCl3}$, beginning with $\ce{AlCl3}$ coordination to the ether oxygen, followed by an $\mathrm{S_N2}$ reaction: $\hspace{3.6cm}$ $\hspace{0cm}$ The oxygen activates naphthalene's position 1, and the acid chloride will undergo a Friedel-Crafts acylation (as you noted): $\hspace{.5cm}$ Being a strong ...


19

As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate. But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered. Hence, the triiodo intermediate is well anticipated. However, a quick glance at spatial ...


19

Does folic acid contain a benzyl or a phenyl group? This is the question asked in the title. At the first glance to the structure, one would say folic acid consists of phenyl function but not benzyl function because the question did not define what is the phenyl group. In reality, phenyl group $(\ce{Ph})$ is $\ce{C6H5}$. As Poutnik pointed out in the ...


18

It isn't that good a generalization: always look at the data first. Here is a table of most of the aldehydes and ketones with 6 or fewer carbons (the labels are used in the chart later): Now plot this on a chart: Branches is the number of branches in the carbon chain. Note that while for 3 and 4 carbons the ketones do have higher boiling points, it is ...


18

The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester. Look at the resonance structures drawn below. Both the ketone and ester have a resonance ...


18

1) Why is only triphenyl phosphine used here. Why not some stronger nucleophile like say trimethyl phosphine? Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems. Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine. It would form in high yield, but in the ...


18

If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide. The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However,...


18

The iodoform reaction proceeds by the mechanism shown below. (The final proton transfer need not occur between the two partners, any other hydroxide could abstract the acid’s proton and any water molecule can protonate iodoform.) This can fail for carboxylic acids or esters in one of two ways: The first step. The $\mathrm{p}K_\mathrm{a}$ value of the ...


18

You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ ...


18

Your analysis is slightly incorrect, actually. If you do a $\mathrm{S_N2}$ on the chloride, hydrolyse the nitrile, and decarboxylate, you end up with the wrong oxidation state: Instead, if you actually get to the cyanohydrin, you could (in theory) simply lose $\ce{CN-}$ to form the ketone: However, a $\mathrm{S_N2}$ reaction at that centre is going to be ...


18

This is based off my memory, as far as I can remember what my teacher taught me. Note, instead of $\ce{Pt}$, you can go for $\ce{Ni}$ (or Raney Nickel) or even $\ce{Pd}$ as a catalyst. Functional groups such as acid halide, cyanide, nitro, aldehyde, ketone, alkene, and alkyne are easily reduced by $\ce{H2/Ni}$. Some other groups can be reduced with $\ce{H2/...


17

Here are the two half reactions: $$\begin{align} \ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\ \ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-} \end{align}$$ which together yield the overall reaction $$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$ Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the ...


17

Here is my "old school" explanation. Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left. As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well ...


17

You have asked quite a number of questions in your post so let's tackle it one by one. Why does the oxidation take place at that particular nitrogen atom? To understand this, we first need to acknowledge that peroxyacids $\ce {RCO3H}$ are electrophilic oxidising agents. Specifically, the oxygen atom attached to the $\ce {H}$ is the electrophilic atom. ...


16

Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks. This is not ...


16

Your reasoning is along the right lines, but it is incomplete. enol of B is more stable because of more substituted alkene. The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest. However, the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is ...


16

Significant amount of geminal diol of benzaldehyde exists in an aqueous solution of benzaldehyde at 25 °C because $\mathrm{p}K_{\text{hyd}} = 2$ (Ref. 1) The $\mathrm{p}K_{\mathrm a}$ of benzyl alcohol is listed as 15.40 (Wikipedia). Thus, one can reasonably assume that the given value of $\mathrm{p}K_{\mathrm a}$ 14.9 represents a composite equilibrium ...


15

For this approach I am basically employing Frontier Molecular Orbital Theory (FMO) to predict the reactivity of carbonyl compounds towards nucleophiles. For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack ...


15

Regarding this question, there are two things that should be stated beforehand. Firstly, the mechanism of the Clemmensen reduction is not well-understood. There are two plausible pathways that have been proposed:1 so, the mechanism for this rearrangement is equally unclear. Secondly, this compound does not actually undergo rearrangement in the Clemmensen ...


15

Of your three "thoughts", you are correct that an aldol reaction is not an option. Not only is the product "crowded", but the reaction is reversible. [BTW, an aldol condensation occurs when water is eliminated from the initial aldol product. In the case of a benzoin aldol product, elimination of water is impossible.] Your second premise is a good one. ...


14

Tropone, or 2,4,6-cycloheptatrien-1-one, is an aromatic, non-benzenoid hydrocarbon. If you look at the resonance structures in the drawing below you'll see that structure B depicts a molecule with a continuous, planar loop of 7 p-orbitals that contain 6 pi-electrons. Any planar (or near-planar), cyclic system with a continuous loop of p-orbitals is ...


14

It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that ...


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