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It seems that NickT was looking for an experimental solution. My post deals, however, with a computational solution.
How would I quantify how significant the interaction is?
Determining the interaction energy between two defined monomers such as your aromatic triazole and amide is a rather straightforward process. This process is referred to as the ...
22
I am uncertain about the "Popoff's" rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.
The Baeyer-Villager oxidation is an oxidation of ketones to esters using a peracid in the ...
20
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 ...
20
The rate-limiting step of the Fehling’s test reaction with aldehydes is the formation of the corresponding enolate:
The subsequent reaction of the enolate with copper(II) proceeds through a single electron transfer mechanism.
Aldehydes that lack alpha hydrogens, such as benzaldehyde or pivalaldehyde (2,2-dimethylpropanal) cannot form an enolate and thus do ...
19
In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.
Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ...
18
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 ...
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 ...
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
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 ...
17
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 ...
17
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 ...
17
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 ...
16
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 ...
16
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 ...
15
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 ...
15
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, ...
15
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. ...
14
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 ...
14
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 ...
13
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 ...
13
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 ...
13
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 $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ ...
13
This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI:
10.1016/0040-4039(95)00502-4)
The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond:
The ...
13
Probably any DNP could work if you could get it. But, because of the directing character of substitutents in electrophilic substitution you inherently favor the 2,4 isomer:
the hydrazine group (or more accurately the chlorine atom which is converted to a hydrazine group) is ortho/para directing.
the nitro groups are deactivating and meta directing.
So ...
13
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 ...
13
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. ...
12
This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.
For example a Google search for "reduce ester to ether" brought up the following:
The reaction appears to be complete within 3 hours ...
12
As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3.
The enolate of 1 (sorry for the circle in the structure, i'm to lazy to fix it) and 2 undergo an aldol reaction yielding 3. With a bit of charge shifting, we get again a nicely stabilized enolate (4), which ...
answered Dec 28 '13 at 21:36
Klaus-Dieter Warzecha
40.7k7 gold badges80 silver badges145 bronze badges
12
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 ...
12
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 ...
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