17

Sara Jane: Lest you miss the point of the stimulating discussion between Waylander and Zhe, I would like to sum up their thoughts with a diagram. Your first clue should have been the opening line: "Recall that epoxides rearrange with Lewis acid...". It is common that catalytic Lewis acids rearrange 2,3-dialkyl oxiranes [epoxides] to ketones. In the case of ...


10

I think your friend is thinking of the cyclobutyl carbocation which does ring contract to the cycylopropyl carbinyl carbocation (and also equilibrates with the methallyl carbocation). However, just as you thought, the cyclobutyl carbinyl carbocation does ring open to the cyclopentyl carbocation (ref_1, ref_2, ref_3). This rearrangement is driven by ...


10

The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete ...


9

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.)...


9

Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for the migrating group to be anti to the leaving group ($\ce{N2+}$), for orbital overlap reasons. (apologies for hand-drawing - doing this on my phone) You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. ...


8

I took the opportunity to read the McCasland paper1 to which the OP's link referred. The products were isolated as their 2,4-dinitrophenylhydrazones. In the case of the cis-isomer 4, the reaction mixture was "apparently richer" in the aldehyde 3. As @PCK has argued for the trans-isomer 1 the highlighted bonds are aligned properly for a ring contraction. ...


8

I have spent a lot of hours to find the evidence to support OP's idea of formamide may undergoes Hoffmann degradation reaction, but I can't find any. However, I find at least one publication, which may suggest formamide may decompose to formic acid and ammonia (ironically) in hot alkaline medium, before exchanging amide $\ce{N-H}$ to $\ce{N-Br}$ (please keep ...


7

Potassium carbonate is a perfectly good base for the alkylation of phenol ($\mathrm{p}K_\mathrm{a} = 10$) with a good electrophile, in this case 3,3-dimethylallylbromide. The reaction you are looking for is Claisen rearrangement which proceeds by a 3,3-sigmatropic rearrangement mechanism. Image from ref 1


7

You are correct if this epoxide is a alkyl peroxide. However, this is a 2-phenyloxirane (styrene oxide or phenylethylene oxide), which can stabilize a protonated intermediate by resonance, specifically with 2-hydroxyphenyl group. Thus, it can isomerize to phenylacetaldehyde through 1,2-hydride shift (similar to pinacol-pinacolone rearrangement). The ...


7

The formation of structure 6 (your D) occurs as follows. Photolysis of dienone 1 affords excited state 2 that forms cyclopropane 3. In turn, 3 is capable of breaking a different cyclopropane bond to form 5-membered ring 4. Structure 5 is a resonance structure of 4. Structure 5 closes to the [3.1.0] product 6 (your D). I don't know how you formed A (12 here) ...


7

The answer to your question, "can a six member ring expand to achieve octet completion to stabilize a carbocation?" is yes. The most specific and relevant example for that rearrangement is Büchner–Curtius–Schlotterbeck reaction (Wikipedia page). Your second question is "does the expansion lead to the major product?". The answer is yes, but it is based on ...


7

The reaction which we are talking about here is called Tiffeneau–Demjanov rearrangement. According to Wikipedia, the Tiffeneau–Demjanov rearrangement (TDR) is the chemical reaction of a 1-aminomethyl-cycloalkanol with nitrous acid to form an enlarged cycloketone. The Tiffeneau–Demjanov ring expansion, Tiffeneau–Demjanov rearrangement, or TDR, provides an ...


7

As your heading correctly states, this reaction take the path of Fritsch–Buttenberg–Wiechell rearrangement (Wikipedia). The reaction mechanism of Fritsch–Buttenberg–Wiechell rearrangement (FBW) is described as: The strong base deprotonates the vinylic hydrogen, which after alpha-elimination forms a vinyl carbene. A 1,2-aryl migration forms the 1,2-...


6

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does ...


6

Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction ...


5

I think Sameer Thakur was in right track when started to write the mechanism. But the path got lost at the end. I don't see reason to have a methide shift followed by a hydride shift and then proton abstraction. The 1,2-methide shift gives you a very stable $3^\circ$-carbocation, which is compatible with the initial $3^\circ$-carbocation given by elimination ...


5

Your questions: Q1) How the rearrangement from (G) to non-1-yne can happen (i.e., its mechanism)? For example, see the following mechanism suggested in Ref.1 as the Acetylene Zipper reaction: Q2) Why Sodium Amide in particular yields terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne? To answer your question, I ...


5

Let me expand a little on my earlier comment. If we only consider the inversion of the ammonia molecule (we ignore rotation and vibrations other than the inversion mode), the Schrödinger equation is given by $$ -\frac{\hbar^2}{2\mu_\text{red}}\frac{\text{d}^2\psi{z}}{\text{d}z^2}+V(z)\psi(z)=E\psi(z), $$ with $\mu_\text{red}$ the reduced mass and $z=d_\text{...


5

One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.


5

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve ...


5

There seems to be an issue with both mechanisms on the front that an SN1 reaction would not take place since the carbocation formed is a vinylic carbocation which is highly unstable. The actual reaction follows a termolecular mechanism where the rate of reaction is given to be: $$\text{Rate}=[\ce{HX}]^2[\text{alkyne}]$$ Now, according to Advanced Organic ...


4

The first step is nucleophilic substitution. It is possible for nucleophile to attack directly at the allylic position, displacing the leaving group in a single step, in a process referred to as $\mathrm{S_N2'}$ substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to ...


4

May be Baeyer Villiger oxidation is not for aldehydes in the past. But that has changed recently (Ref.2). The abstract of this reference states that: A conceptually distinct, modern strategy for Baeyer−Villiger oxidation (BVO) was developed. Our novel method involves initial hydration of water to carbonyl compounds, followed by ligand exchange of ...


3

As I mentioned in the comments, this is a simple homolytic bond fission to give a carbene and a stable aromatic compound (Phenanthrene). Here's the mechanism:


3

As orthocresol commented (vide supra), migratory aptitude in cationic reactions has been given as: $\ce{H}=\mathrm{Ph} \gt \ce{(CH3)3C \gt (CH3)2CH \gt CH3CH2 \gt CH3}$ (Wikipedia and Migratory aptitude in pinacol-pinacolone rearrangement). In your question, the secondary carbocation formed when protonated has three groups, which can migrate to give more ...


3

The Baeyer-Villiger oxidation is conducted under acidic conditions such as an alkyl peroxide in the presence of a mineral acid (e.g.; H2SO4) or a peracid (e.g.; peracetic acid or m-chloroperbenzoic acid). In the reactions described in the diagram, the reactions of aldehyde 1 or α-diketone 7 are initiated by proton of the carbonyl oxygen followed by ...


3

Czakó/Kürti$^{[1]}$ suggest a [1,2] shift, which in your case would mean a shift of the 5-6-bond to the 1-carbon; in concert with a free electron pair of the anionic oxygen neutralizing the 6-carbon. The carbonyl oxygen becomes anionic and accepts a proton in the next step. So instead of the 1-6-bond breaking, the 5-6-bond breaks. Responding to a comment ...


3

Simple answer is consider the stability of carbocations. You already got the tertiary carbocation in hand that is much stable than secondary carbocation, which would be resulted by ring expantion. Sure, a methide migration would give you another tertiary carbocation, but before you get that, the reaction progress has to climb another actvation barrier, which ...


3

Do you think that the mechanism of "Beckmann" and "dehydration" are different? Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments) The mechanism for Beckmann rearrangement is as follows, (Source: ...


2

We can envisage the first steps of the reaction from the dibromide form the internal alkyne, then this undergoes the zipper reaction to isomerize to the terminal alkyne due to $\mathrm{p}K_\mathrm{a}$ differences. See e.g. Alkyne zipper reaction. The fused KOH is not sufficently strong a base to effectively catalyze this isomerization.


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