11

Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very ...


9

Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic. Alternatively, I ...


8

For the enolates that can be formed, the others have already given comprehensive answers: All four enolates are possible and since we are under thermodynamic control (weak base), we expect o observe them all. Numbering the chain carbons according to IUPAC, and noting the non-carbonyl end of the enolate, I expect the enolate ratio to be $3>6>1\approx 8$....


8

1. The acetate aldol reaction Before looking at selective variants, its worth pointing out what an acetate aldol is. An acetate aldol (below), is the simplest kind of aldol in which the β-hydroxyketone product has no methyl group at the α-position to the carbonyl. This is in contrast to the (more usual?) propionate aldol reaction in which the α-position ...


8

Partially yes. The big problem with aldol reactions in general is that there are a lot of species that can react together to form stable compounds resulting in low yield. There are numerous ways to counter this: Specific enolate equivalants, like lithium enolates, silyl enol ethers, aza enolates, enamines, etc. These are among the most important. Lack of ...


8

I see what went wrong. Such questions are often confused with aldol or cannizarro as they have a carbonyl group and a base is given. Actually this is a completely different name reaction called Favorskii rearrangement. The key to identify it is that its applicable to α-halo ketones. In Favorskii rearrangement, the base takes α-hydrogen from that side of ...


7

It certainly could undergo the acid-base equilibrium you described, if there's enough energy in the system. Your problem is omitting an important piece of information, which is the temperature at which the reaction is taking place. Lithium enolates (the right compound) are usually formed at low temperatures, down to -78 C. The hope/assumption is that the ...


6

This problem has probably been on everyone's mechanistic chemistry exam at some time or another. Does the alkyl (allyl) just "jump" from one side of the carbonyl group to the other? In most examples, the allyl group is a saturated alkyl group. Under such circumstances I am in full agreement with @orthocresol as to the mechanism of the process (1 --> 9). The ...


6

Your claim is worded very poorly, and taken in the broadest sense, is not true. However, it's more complex than an outright yes or no. If we're interested in fundamental reactivity, it suffices to look at cyclohexanecarbaldehyde, as there is no reason that the dialdehyde should be any different (apart from the likelihood that it gives very messy reactions ...


6

Claisen and aldol condensations are thermodynamically controlled,* so it is not a question of which enolate forms, it is a question of which product forms. In your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two ...


6

If in doubt, draw: Deprotonation at C-6 under thermodynamic control results in the more stable enolate (1c) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction. As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene. However, if that is your target ...


6

Conceptually, we can think of this reaction as occurring in three stages. I've provided diagrams for the self-condensation of acetaldehyde, however you should be able to draw an analogous mechanism for the self-condensation of butanal (your aldehyde). Your mechanism is essentially correct, except for the final elimination step, which as you've drawn it gives ...


6

Intramolecular-Aldol reactions involves heat, as you have mentioned, and the base used (NaOH) isn't much affected by steric factors. Therefore, the preferred product should be thermodynamically controlled product. Moreover, as you commented about the formed carbanion being less stable in case of 2° carbon, actually the carbanion is converted to an enolate ...


5

Not every α-hydrogen automatically means that a ketone is enoliseable. It would need to be an α-hydrogen capable of aligning with the π system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the ...


5

There is no reason why the aldehyde cannot undergo an aldol reaction with itself. Enolates are not rendered unstable by the presence of additional alkyl substituents; in fact they are considered to be more stable. (See: formation of silyl enol ethers with $\ce{Et3N}$/$\ce{Me3SiCl}$.) However, there is no way it can eliminate $\ce{H2O}$ to form the α,β-...


5

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS. The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted ...


5

B.Anshuman has given a short answer, but I felt more explanation is needed to understand this situation. This reaction can be done in kineticlly controlled conditions (e.g., using $\ce{LDA/THF}$ as a base at $\pu{-78 ^\circ C}$) or in thermodynamically controlled conditions as in this case. Each case may give totally different major products. For example, ...


4

I believe your conception of the need for deprotonation under acidic conditions is misconceived. Whenever you are under acidic conditions, your mechanisms must necessarily start with protonation. The key concept here is the enol tautomerization. The enol is more favored than for propanal as compared to ethanal because it results in more substituted alkene....


4

The more stable enol would involve $\ce{C^6}$, with the aldol condensation you would get a six member ring with $\ce{-COCH3}$ attached to the ring. With the enol involving $\ce{C^2}$, the product would have a methyl and $\ce{-CHO}$ attached to the ring. I don't know the distribution between the two products; it could be 51:49 or as much as 99:1. You should ...


4

Your immediate concern seems to be which hydrogen is abstracted, 1 or 2, in structure 1 (see below). Examination of pKa’s is useful. pKa’s: EtOH, 16; alkyl ketone, ~20; aliphatic ester, ~25; and β-diketone, ~9; aliphatic carboxylic acid, ~5. Clearly, alkyl ketones are more acidic than aliphatic esters. The methyl ketone hydrogen 3 in structure 1 is ...


4

If you look carefully, the reagent for the first step is $\ce{O3/H2O}$, which is used for Oxidative ozonolysis. If the mechanism of ozonolysis is drawn, you will realise that a molecule of Hydrogen peroxide($\ce{H2O2}$) is released by the oxidation of the alkene to aldehyde and if that $\ce{H2O2}$ is not removed from the reaction mixture, it will oxidize the ...


3

It is clear that the first step is ozonolysis with a slightly reductive workup, yielding two ketones. You seem to have already established that. The second step features a base and an alcohol as a solvent. The hydroxide can react with the ketones in two ways: 1) nucleophilic attack at the electrophilic carbonyl carbon 2) abstraction of the mildly acidic $\...


3

I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here: H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons. Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they ...


3

For most aldol reactions, the relative 1,2-diastereoselectivity is a consequence of the enolate geometry.* Both tin and titanium enolates are easily formed, and have been shown to exist predominantly in the (Z) configuration. When reacting a cyclic (Zimmerman Traxler) TS, its commonly observed that (Z) enolates afford 1,2-syn aldol adducts, due to the ...


3

Step P is std base-driven alpha alkylation of ketone. Step Q Wacker details here takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol mechanism here gives the product shown. So the answer is (b).


3

First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction). Counter to your analysis, the following reaction will not occur since $\ce{H}$ bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to $\mathrm{p}K_\mathrm{a}$ values given by Mathew Mahindaratne. Under the given ...


3

Is the alpha hydrogen acidic enough to react with base? The answer is yes. As you said +I effect of the methyl group stabilizes the carbocation. It is weakly acidic but not weak enough to not be taken by any base. Will it proceed through aldol reaction or will Cannizzaro reaction occur? It will proceed through Cannizzaro mechanism even though it has an ...


3

I'm interested in the OP's curiosity. I'd agree with Waylander about disfavoring to make 4-membered rings. Yet, it is possible to have two six-membered ring products. The first $\alpha$-proton abstraction followed by Michael addition resulted the first intermediate as OP suggested: Now, this newly formed intermediate contain 4 types of eligible $\alpha$-...


2

We expect the aldehyde to be more reactive toward nucleophilic attack for its smaller steric hindrance (kinetics). The product ratio may vary with time, kinetics vs. thermodynamics at equilibrium. Attack followed by elimination helps lock in the product.


2

You can actually take either. The other two $\alpha$ hydrogens are more acidic thermodynamically, but if you used a hindered (kinetic) base at a low temperature, you can predominatly de-protonate at the less hindered methyl (as in your diagram). So, the answer is, it does sometimes, but it depends on the base and reaction conditions. This is why organic ...


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