16

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 ...


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

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 ...


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

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

Potassium carbonate is a perfectly good base for the alkylation of phenol (pKa 10) with a good electrophile, in this case 3,3-dimethylallylbromide. The reaction you are looking for is a Claisen rearrangement which proceeds by a 3,3-sigmatropic rearrangement mechanism. image from ref 1


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-...


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

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 ...


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 $\ce{PCl5}$.


4

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{...


4

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 ...


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 SN2' substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to those seen ...


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 ...


2

Forget the Beckmann, the oxime is still nucleophilic through N (imagine the N-hydroxy enamine tautomer if that helps). Have it attack the second carbonyl intramolecularly (and hence favoured) to give a dihydropyridine. Dihydropyridines are unstable with respect to aromatisation unless substituted with strongly electron withdrawing groups. The N-hydroxy-...


2

Your approach is feasible. Thermal cyclizations of appropriate dienes, enones, enynes, and other related unsaturated systems are known in literature. For example, there are extensive examples of intramolecular ene-reactions in Organic Synthesis and some of them are reviewed (Ref.1 and Ref.2). To justify your first step, I'd include the abstract of reference ...


2

It will not work the way the OP likely envisions. To make an aromatic ring you need to move the methyl group into the bottom ring, but all of these bottom positions are saturated. No place for the proposed rearranging group to land. There is, however, another possibility if the system from a nonclassical carbocation. To make the nonclassical carbocation, ...


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.


2

Your three photons represent the total energy added to the system: $$E_\mathrm{photon} = 3 \cdot \frac{hc}{\lambda} = \pu{13.98 eV}$$ That, less the sum of ionization energy and kinetic energy of the removed electron, leaves the change of the internal energy of the cation: $$E_\mathrm{int} = \pu{13.98 eV} - \pu{9.10 eV} - \pu{1.4 eV} = \pu{3.48 eV}$$ ...


1

You are correct acyl hydrazide can be converted into acyl azide by adding $\ce{HNO2}$ also you wrote that the $\ce{NH2}$ group of hydrazide will be diazotisated making a $\ce{N2+}$ group hence I am assuming that you know the mechanism, after diazotisation the adjacent $\ce{N}$ donate the lone pair to the $\ce{N2+}$ group making the final product as acyl ...


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