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The abnormal Beckmann rearrangement differs from the normal one that there is a nitrile as product of reaction. As compiled by William Reusch (formerly Michigan State University) in his freely accessible virtual textbook organic chemistry, the assumed mechanism of the abnormal Beckmann rearrangement is e.g., (image credit) For comparison, the normal ...


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The reaction schemes of both Beckmann and abnormal Beckmann rearrangement has been already provided by @Buttonwood. I am going to explain the mechanism of the reaction and provide some literature to look at. Ketoximes undergo rearrangement reactions involving cleavage of the carbon–carbon bond adjacent to the imino carbon. The normal rearrangement is called ...


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Have you checked Wikipedia? Typically the reaction to form Grignard reagents involves the use of magnesium ribbon. All magnesium is coated with a passivating layer of magnesium oxide, which inhibits reactions with the organic halide. Many methods have been developed to weaken this passivating layer, thereby exposing highly reactive magnesium to the organic ...


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Step 1 - benzylic bromination with NBS, Benzoyl peroxide Step 2 - displace bromine with cyanide using KCN/DMSO Step 3 - ester hydrolysis to acid with LiOH/THF/water Step 4 - acid reduction to alcohol with BH3 Step 5 - nitrile hydrolysis with KOH Step 6 - lactone formation with cat. PTSA/toluene reflux.


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An alternative approach to the proposal of @Waylander for the synthesis of lactone 7 from ester 1 depends on the efficacy of the Barton nitrite photolysis, 2$\rightarrow$3. Examples of this reaction are typically conducted with secondary nitrite esters in rigid, steroidal frameworks with few degrees of freedom, an advantage that the transformation 2$\...


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Basically, you are on the right track and answered the question. First, NaNH$_2$ is used as a base to deprotonate the terminal alkyne (see this Reagent Friday). Second, the generated nucleophile opens the epoxide in an S$_\mathrm{N}$2 fashion, usually adding to the lesser substituted carbon which in your case yields a quaternary alcohol after acidic workup (...


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Finally I find a primary reference by Oae et al.[1] They react benzenesulfonic acid with fused potassium hydroxide using isotopically labelled chemicals. The isotopic labelling results favor a direct nucleophilic substitution of hydroxide for sulfite, so that the only ring position involved in the reaction is that originally holding the sulfonate function. ...


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The Beckmann rearrangement is an acid-mediated isomerization of an oxime (or N-haloimine or nitrone) functional group to corresponding substituted amides (Cyclic oximes or cyclic N-haloimine yield lactams). The rearrangement was discovered by the German chemist, Ernst Otto Beckmann (1853–1923) in 1886 (Ref.1). The rearrangement, which bears its discoverer’s ...


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There is no solvated molecule of either of 4 ionic compounds. All interactions ( or lack of ) happen on hydrated ionic level, including eventual precipitation or formation of ionic pairs. All four salts are soluble, including magnesium chromate with the solubility $\pu{137 g/100 mL}$ at $\pu{20^{\circ}C}$ (solubility table). When dissolved, salts form ...


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I have used the achiral (Z)-vinylsilane 1 as an example with one of the p-orbitals of the double bond colored red. When the electrophile reacts on the "top" of the black p-orbital the carbon rehybridizes from sp2 to sp3 producing the chiral carbocation conformation 2. The same operation on the "bottom" lobe of the black p-orbital affords ...


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This question shows up a couple times in a google search, e.g. Source: https://faculty.uscupstate.edu/cbender/Web%20page%20folder%20enmass/CHM111L/NIE%20answers%20worksheet%20update%20Su2013.pdf or Source: https://www.fusd1.org/site/handlers/filedownload.ashx?moduleinstanceid=10419&dataid=21159&FileName=Net%20Ionic%20Equations%20worksheet.pdf ...


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It starts with the formation of brominium ion, as in direct bromination of alkenes. If you do not know the mechanism, you can refer to it here, at MasterOrganicChemistry. But here, the $\ce{MeOH}$ can itself act as a potential nucleophile. It is similar to if $\ce{H2O}$ is used as a solvent - refer the formation of halohydrin here. Since $\ce{MeOH}$ is ...


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$\ce{Mg3N2}$ is magnesium nitride. It reacts with $\ce{MeOH}$ to generate magnesium methoxide and ammonia according to online report by Buske [1]. The $\ce{NH3}$ thus generated is acting as nucleophile to react with the methyl ester to give the primary amide. It also removes two trifluoracetamide groups (or the $\ce{MeO-}$ does) to reveal the amino groups. ...


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According to Bordwell $\mathrm{p}K_\mathrm{a}$ Table, $\mathrm{p}K_\mathrm{a}$ of $\ce{PhCH2COSPh}$ is $16.9$, which is compatible with alcoholic $\ce{OH}$. We all know that Grignard reagents and organolithium reagents are very susceptible to acidic hydrogens because they are strongly basic as well as nucleophilic. Between $\ce{t-BuLi}$ and $\ce{n-BuLi}$, $\...


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On OP's comment: I just don't see how the carbon can become part of the chain! shows confusion on the mechanism. However, the comments by user55119 and Ron correctly suggested the path of the mechanism. I thought it would be helpful to put the matter as a solution. Following scheme shows the possible mechanism to achieve the sought product: Note that, ...


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I think Wikipedia's equation is not a correct representation. In the Mineralreich (Wiki link is based on Scribd, whereas the same is available on Internet Archive). With a little bit of effort one can locate the relevant paragraph Verwendung. Strontianit ist das am leichtesten zu verarbeitende Rohmaterial für Strontiumpräparate und wird daher in Westfalen, ...


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The stereochemistry of this type of cyclization is well-understood. An extensive review may be found here. As you have realized the role of the Lewis acid stannic chloride is to generate the carbocation 1. The vacant p-orbital is aligned in the sigma sense with the π-bond of the double bond to permit cyclization, in the Markovnikov sense, to carbocation 2. ...


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$\ce{SO2}$ does react with $\ce{H2O}$ according to your second reaction (which is a half-equation) : $$\ce{SO2 + 2 H2O -> SO4^{2-} + 4 H+ + 2 e^- }$$ But $\ce{SO2}$ cannot react according to the first equation, which produces two non-combined (nascent) $\ce{H}$ atoms. Usually the electrons produced in this half-equation do not react with $\ce{H+}$ to ...


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Catalysts very much do affect end products because they may act differently on competing reactions. For instance, given ethylene and oxygen a suitable catalyst may promote formation of ethylene oxide and not as strongly promote oxidizing the ethylene to carbon dioxide and water. (In this particular case, a silver catalyst with carefully controlled ...


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The reaction would theoretically occur through an SN2 mechanism because the hydroxyl (OH) group is bound to a primary carbon, and SN1 will usually only occur under tertiary (steric) conditions with a weak nucleophile. What you are trying to do is referred to as "activation of alcohols". The OH- is not a good leaving group because OH- is a stronger ...


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If a catalyst is not supposed to affect the reaction's final equilibrium position how do we explain the catalyst selectivity seen here? If you wait long enough so that all three reactions attain equilibrium, the presence or absences of catalysts have no effect on the product mixtures. In the examples, however, the reactions without catalysts are all slow. ...


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Copper deposited on the zinc surface does not protect the remaining zinc to be dissolved. Copper does not form a real layer on the zinc surface. As long as some metallic zinc remains in contact of the copper sulfate solution, the reaction will proceed. It may be irregular and produce holes in the zinc plate. The reaction will stop when and only when the zinc ...


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Like so much organic chemistry the answer is it depends. As I said in the comments you can buy Allyl Grignard in ether solution, but you can also dimerise it by allowing it to reach room temperature Organic Syntheses prep here. Assuming the cyclohexenyl Grignard behaves similarly (and I think it will) then you can get the expected Grignard addition product ...


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