New answers tagged

1

It appears that your teacher is wrong! The synthesis of pyridine from acetylene and hydrogen cyanide over red hot iron was originally reported by William Ramsay in 1876 wikipedia. Further examples have been reported JACS paper here, patent with Rhodium and Cobalt catalysis here The second reaction is the Pechmann Pyrazole synthesis article here. The ...


1

I assume you are looking for activated manganese dioxide (AMD). Apart from the Attenburrow process mentioned in @Buttonwood's answer ($\ce{MnSO4 + H2SO4 + NaOH}$), I found three other process that might suit your need (I noticed OP has loads of $\ce{MnSO4}$. So, the aforementioned process will work. The processes mentioned in this answer doesn't involve $\ce{...


3

Your question suggests you did not access Leo Paquette's Encyclopedia of Organic Reagents (14 volumes, some schools have access to the electronic version eEROS), or its one-volume siblings by topic, e.g., Oxidizing and Reducing Agents (this one by Burke and Dannheiser). These resources cover physical / chemical properties of frequently used reagents and ...


2

It seems like you figured out that filtration of the reaction mixute through a Celite$^®$ pad would work in this case since you have used nonpolar solvent such as toluene. However, if you have used tert-butoxide (e.g., $\ce{KOC(CH3)3}$) insteard as in the given mechanism, the byproduct is tert-butanol, which is soluble in toluene. In this case, you can do ...


-1

To compare, we have to use something called the second ortho effect. From wiki: The protonation of Substituted aniline is inhibited due to steric clashes. Upon protonation the hybridization of Nitrogen in amino group changes from sp2 to sp3 making the group non planar. This causes the steric hindrance between the ortho substituted group and H atom of amino ...


-2

The answer to this is by something known as the first ortho effect, from Wikipedia: When a group is present ortho to carboxylic acid group in substituted benzoic acid, the steric hindrance forces the carboxyl group to twist out of the plane of the benzene ring. This inhibits the resonance of carboxyl group with phenyl ring which increases the acidity of ...


1

I used toluene as solvent. Ionic compound Et3NHBr (although it's unclear that it is really formed) is insoluble in organic solvent such as toluene, so it can be filtrated through Celite pad.


6

From here: Reimer and Tiemann (Ber., 1876, 9, 824) state that, contrary to experience in the Kolbe-Schmidt reaction, in their synthesis of hydroxy-aldehydes the proportion of $ortho-$ and $para-$ derivatives is unaffected by changing from sodium hydroxide to potassium hydroxide The improvement in technique afforded by the determination of aldehydes by ...


4

No. Example 1 - neopentyl halide: slow SN2 as it has considerable steric hindrance; slow SN1 as it is a primary carbon, so it does not form stable carbocation intermediate. Example 2 - benzyl halide: fast SN2 being a primary carbon; fast SN1 due to a resonance-stabilized benzyl carbocation intermediate.


9

You have two choices for choosing the parent hydride here in this case. 1. Choosing the cyclohexene ring as the parent hydride If we consider the cyclohexene ring as the parent hydride, we get the cyclopentadiene ring to be a substituent leading the name to become 1-(cyclopenta-1,3-dien-1-yl)cyclohex-1-ene 2. Choosing the cyclopentadiene as the parent ...


10

If you use "cyclopenta-1,3-diene" in the end of the name (your first attempt), then this would be the stem system. In case like the one presented in the question -- two rings connected via a single bond -- it is recommended to recognize the cyclohexene as the stem (like parent), and the cyclopentadienyl moiety as a substituent (like child). ...


4

Short answer: no, it is not correct. You can have compounds that are thermodynamically/kinetically impeded for a variety of reasons and present both slow SN1 and SN2 reactivities. The exceptions would be carefully handcrafted “everything else being equal” lists.


0

TLC Visualization by the color formation when used p-Anisaldehyde or vanillin in sulfuric acid as the developing agent is known in Natural Products Chemistry for long time. OP's suggestion of color development is due to inclusion of p-methoxybenzyl group (PMB) or dimethoxybenzyl group (DMB) not correct. It seems like these staining compounds make colored ...


2

Converting the comments into an answer; Yes, DMS can be used in place of $\ce{Zn/H2O}$ during the reductive workup for ozonolysis. The mechanism involves DMS reducing a $\ce{O-O}$ fragment of the trioxolane intermediate produced from ozonolysis by one oxygen and itself oxidizing to DMSO (dimethyl sulfoxide). (Image source) I did some literature survey and ...


8

This is a rather unusual and interesting case of aromaticity, which has been given a special name: homoaromaticity. The Wikipedia page does a quite nice job of explaining what's going on. As you state, protonation of cyclooctatetraene generates a $\ce{C8H9^{+}}$ cation containing an $\mathrm{sp^3}$-hybridised carbon atom between all the other $\mathrm{sp^2}$-...


2

In addition to the answer by @Waylander, I would add something since his answer is about symmetrical benzil. In case the substituents on the two phenyl rings are different, it must be noted that the first step occurring in this reaction is attack of $\ce{OH^-}$ as a nucleophile on the carbonyl $\ce{C=O}$. Nucleophilic addition is preferred at that carbon ...


2

This is the Benzilic Acid rearrangement - diagram sourced from here The Wikipedia entry is as good a place to start as any Wikipedia


0

Let me start from the basics, Distillation is used to purify a mixture that has volatile liquid and non-volatile impurities. We use Steam distillation in which the oil has a significantly high boiling point than water atleast and it is also immiscible with water. 1)Steam does not increase its external pressure rather it increases the mixture's vapor pressure ...


4

Since primary amino acids like glycine, alanine can't be decarboxylated under oxidative conditions, what would be the best catalyst for this reaction? Decarboxylation of $\alpha$-amino acids is known enzymatic procedure in biological systems. Non-enzymatic decarboxylation of $\alpha$-amino acids is also a long-known reaction, which leads to amines with a ...


1

Thanks to @orthocresol for helping me via the comments. Here is a reasonable explanation for it - The final product comprises of two different diastereomers, of configurations RR and RS. Diastereomers, unlike enantiomers, have some inherently different properties, other than optical rotation. Consequently, the reaction that forms two different diastereomers ...


6

Given enough exposure time, the reaction will lead to formation of various products, even the fully chlorinated compound: dodecachlorocyclohexane. Increasing the concentration of chlorine also help in the formation of this product. Doing some literature survey confirms this: [...] During most of the course of this reaction (photochemical chlorination of ...


4

There may be experimental data for the specific cycloalkane in question, for example by a single-crystal diffraction analysis. Beside entries in the primary literature, there are databases with the relevant structure data including the atomic coordinates; the later ones may be used to calculate the angles in question. If experimental data are not (yet) ...


-1

Order of -I in Halogens is −F > −Cl > −Br > −I Order of +M in Halogens is −F > −Cl > −Br > −I So whenever halogens are in resonance (as in p-fluorophenol and p-chlorophenol), we take combined effect as: −F < −Cl < −Br < −I and consider halogens as an overall EWG. Hence, p-chlorophenol is more acidic than p-fluorophenol.


3

One paper1 suggested use of pyridoxal as catalyst: When α-amino acid are heated with pyridoxal in dilute aqueous solution in the absence of metal ions, two closely but independent reaction takes place: $$ \begin{align} \ce{RR'CNH2COOH ->[Pyridoxal] RR'CHNH2 + CO2} & \tag{R1}\\ \ce{RR'CNH2COOH ->[Pyridoxal] RR'C=O + CO2 + Pyridoxamine} & \tag{...


3

It is entirely possible you form aniline, but it gets consumed. During the reduction nitrosobenzene would also be formed, and Delmagro et al. [1] report that in basic solution the aniline can couple with the nitrosobenzene to give azobenzene. This coupling depends on the nitrogen in aniline acting as a nucleophile, so under acidic conditions as in a tin/...


5

According to this site here citing Carey & Sundberg and Morrison & Boyd "In case of 2-methyl cyclohexanone, this planarity would cause steric clashes between the methyl group and the pyrrolidine hydrogen atoms. Due to this, formation of the less stable 6-position carbanion is preferred and hence forces reactions to proceed at this position."...


5

First, there is a little mistake in your reasoning. You said that the carbanion is more stable on the unsubstituted side but actually, forming an enolate on the more substituted side will give a more stable substituted alkene. Remember, the negative charge is not concentrated on carbon. It is concentrated on the oxygen atom as the system is conjugated. That ...


6

The Von-Richter reaction's accepted mechanism is given by- As you can see, this is not a simple case of say, removal of $\ce{NO2-}$ ion, instead it involves removal of $\ce{N2}$ which provides thermodynamic stability. the tables you use have been calculated on the basis of $\ce{pK_a}$ values which doesn't give the right answers in many cases due to solvent ...


1

Which of these functional groups is soluble in aqueous HCl and/or NaOH? Functional groups don't have solubility. Their presence in a compound, however, can have an effect on solubility. My approach before was this : Esters do not dissolve in cold aqueous bases or acids, nor do amides or alcohols. Amines dissolve in aqueous acids. Phenols and Carboxylic ...


1

Assuming that your end products are correct I think this reaction would proceed in a similar manner to the way it will proceed if you use hot $\ce{KMnO4}$ as a reagent. The reason I think that is because $\ce{HClO4}$ is practically similar to $\ce{KMnO4}$ structure and is an even stronger oxidizing agent ($\ce{Cl^{+7}}$ is much more unstable (and hence much ...


-2

For a benzyne mechanism formation of benzyne intermediate is always the slow step or rate determining step but unlike normal nucleophilic addition elimination reactions formation of final products in benzyne mechanism is very different. Here the nucleophile simply doesn't go and attack the leaving group but instead the nucleophile of the base obstructs the ...


-3

The answer to your question is no! Ei is an elimination reaction which doesn't doesn't undergo any carbocation rearrangements and also it does not depend on carbocation stability, the carbocation rearrangement generally occurs in acid catalysed dehydration. pyrolytic elimination reactions proceed through a cyclic transition state, this is a beta elimination ...


4

Methanol is a polar solvent; heptane is very non-polar. If you start with methanol as solvent, you can dissolve a little heptane into it; if you keep adding heptane, you reach the solubility limit of the solute heptane in methanol, and the excess heptane floats. If you start with heptane as solvent, you can dissolve a little bit of solute methanol into it; ...


0

The disappearance of boundary is an optical illusion. Whenever we see two boundaries in a liquid, it means that their refractive index is different. If the composition of two phases is such that their refractive indices are very close or match, the phase boundary disappears. It does not mean that separate phases do not exist. Heptane and methanol are ...


2

This might be the mechanism based on the answer in your book. The isopropyl group might shift because of the resonance stabilization due to phenyl group and inductive effect of the methyl group in the resulting carbocation. This carbocation might be the most stable one compared to the other carbocations' possible.


0

Organic nitro compounds can tautomerise to give its aci-nitro form. As we know tautomers exist in a dynamic equilibrium, the enol form of nitro compound is called its aci-nitro form, which is present in a small amount in the equilibrium mixture. Aci-nitro form is the acidic form of a respective nitro compound, which is able to liberate an acidic hydrogen ...


2

The OP's comment: I'm just using the chloroform to get rid of other compounds, so once I separate it out, I expect my molecule to be in the aqueous phase, and then to vacuum distill it out. This indicate OP's in the process of isolating a water soluble compound from a (bio)metrix. Probably a natural product. In old days, there was a procedure we have ...


1

Alcohols contain a hydroxyl functional group. This is different from the hydroxide ion, $\ce{OH-}$. By way of example, the formula for the alcohol present in alcoholic drinks, ethanol or ethyl alcohol, is $\ce{C2H5OH}$. If you want to emphasize the hydroxyl functional group, you could also write $\ce{C2H5-OH}$. To symbolize any alcohol, you could use $\ce{R-...


-1

Here is the phase diagram, good luck! Addition of water will eventually give two phases but your extract will most likely distribute between them. Depending on what you were extracting the advice to carefully vacuum distill at very low temperature the solution is the best idea; then I would carefully progressively extract the residue with solvents of ...


0

Bromine will react with methane to give a kinetic or thermodynamic mix of products basically because the HBr bond is moderately weaker than HCl and the reaction with the bromine radical is more selective. This article seems to discuss this in detail: https://www.mdpi.com/2227-9717/8/4/443/htm. I did not find the relative reaction rates between CH4 and ...


1

The OP asks why doesn't AlCl$_3$ react with benzene - without Cl$_2$! So, what does aluminum chloride do in benzene? It is slightly soluble in benzene, forming an oily compound (Ref 1, 2). So if you mix AlCl$_3$ with benzene, the "reaction" just goes to a complex and awaits further reactant. Perhaps the Cl atom on an alkyl chloride or on a Cl$_2$ ...


6

In the given compound the $\ce{-OH}$ group and the $\ce{-NH2}$ and later $\ce{N2^+}$ group are not in anti position with respect to each other. Equatorial positions in cyclohexane are not anti with respect to each other. Below I have drawn newmann projection of the reactant which should help you visualise that a carbon (encircled) is instead in anti-relation ...


6

I don't think there is a "primary source" for types of isomerism. The types of isomerism are human-defined categories that different authors find conceptually or pedagogically useful in different ways. It isn't like isomerism classification emerges from some unsupervised analysis of all possible structures. If it were my opinion, I'd follow the ...


4

The reaction can also be done if catalyst is applied: [...] we developed a unique cross-coupling reaction of alkyl halides with organomagnesium or organozinc reagents catalyzed by using a 1,3-butadiene as the additive. This reaction follows a new catalytic pathway: the Ni or Pd catalyst reacts first with R−MgX to form an anionic complex, which then reacts ...


3

This reaction does not work for Grignards due to competing deprotonation and reduction reactions, however it is possible if you first modify the Grignard with copper to make the cuprate (Gilman reagent). There is a detailed explanation here


3

From Section P 14-3.5 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, IUPAC Blue book, we have, P-14.3.5 Lowest set of locants The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of ...


4

The reaction wouldn't happen because you forget that $\ce{Cl2}$ is also present in the solution (because $\ce{AlCl3}$ is a strong lewis acid). An acid-base reaction is way faster than an aromatic electrophilic substitution. Once the acid base has happened I doubt that $\ce{AlCl4-}$ would still behave as an electrophile. Footnote: $\ce{AlCl3}$ is a strong ...


1

It is not easy to bind an organic molecule to silicium oxide. It is necessary to synthesize the Si-O bond. It cannot be done starting from $\ce{SiO2}$. A possibility is to react $\ce{CH3Cl}$ over elementary silicium at high temperature to carry out the reaction : $\ce{2 CH3Cl + Si > Si(CH3)2Cl2}$. Another approach is to make methyl sodium $\ce{NaCH3}$ ...


8

The general idea in the answer above is correct, but some of the details need to be looked at more closely. The first step is as drawn in the original post. Deprotonation of the ketone, followed by nucleophilic addition to the alkyne gives an allene-type intermediate. Note that the anionic intermediate only has one geometry, it doesn't have cis and trans ...


5

Your attempted mechanism is correct. There is an active $\ce{\alpha-H}$ in the your final intermediate ($2$ $\ce{-COOEt}$s are present on the carbon (well one of them is attached indirectly, but it still shows it's $-M$ character so it is more or less the same) The full version of the reaction should look like this (I couldn't find a primary source (like it ...


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