# Tag Info

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Enders SAMP/RAMP hydrazone chemistry will do this(1). The Wikipedia article on this technique is a good introduction: (1): Corey, E. J.; Enders, D. (1976). "Applications of N,N-dimethylhydrazones to synthesis. Use in efficient, positionally and stereochemically selective C-C bond formation; oxidative hydrolysis to carbonyl compounds". Tetrahedron Letters. ...

4

Due to the aromatic character of the cyclopentadienyl ligands, ferrocene can undergo electrophilic aromatic substitution reactions typical of aromatic compounds such as benzene. Ferrocene shows very high reactivity towards electrophiles, making it more closely comparable to phenol than to benzene, references and examples here and here.

1

Anthracene oil (CAS registry number: [90640-80-5]): Both European Chemical agency (ECHA) and US Toxicology Date Network (TOXNET) classify anthracene oil as: A complex combination of polycyclic aromatic hydrocarbons obtained from coal tar having an approximate distillation range of $\pu{300 ^{\circ}C}$ to $\pu{400 ^{\circ}C}$ ($\pu{572 ^{\circ}F}$ to $\pu{... 0 An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may ... 0 One other aspect is that the oxygen in the nitro group is not a particularly good electron donor for the hydrogen bond. True, it has a negative formal charge; but this is shared between two atoms and the polarity of the bonds in the nitro group is relatively low (compared with, for example, a carbonyl or hydroxyl oxygen). It is possible to eliminate meta ... 2 In the case of nitration of aniline intermediate stability determines products.The following figure depicts possible electrophillic attack on aniline. Structures 4 and 12 are additional stable resonance structures obtained by attack of nitronium electrophile at ortho and para positions. Structure 8 is relatively unstable due to negative effect of Nitrogen. ... 1 In case of nitration in aniline, the Nitric Acid protonates the aniline to form the Anilinium ion. Now since the nitrogen atom has no lone pair to conjugate, it has no mesomeric effect on the ring, the but as nitrogen is now protonated, it has a high negative inductive effect. Which simply leads to the fact that the position farthest from the nitrogen ... 6 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 You add dissicant until it suspends easily in the solvent when you swirl the flask. Some people say: "Until it snows". When you add a bit of magnesium sulfate or other fine powdered dissicant, it forms a lump and goes to the bottom. Many times it just sticks to the bottom when you swirl the flask. You add more solid and swirl until it suspends and then ... 1 On dilution, ionisation increases, although pH increases due to the fact that, volume increases dominantly over dissociation. Hence concentration of$\ce{H+}$ions decreases instead of the extent of ionisation increases. 1 only 4 products are are formed. 4 Cyclopent-2,4-dien-1-one is not aromatic as you suggested by the resonance structure (see A on the following diagram). That resonance is forbidden and does not exist, ever (I marked it with a red cross). Also, Note that the conjugated$\mathrm{sp^2}$carbons of the original structure are not cyclic. Nonetheless, its allowed resonance is given in B, and ... 4 With molecule (2), you have two possible cations from addition of Br+; one is tertiary, one is secondary. Tertiary is much more stable so you will get very little product resulting from water reacting the secondary cation = high selectivity observed. With molecule (1), Br+ addition gives 2 possible secondary cations; nothing much to choose between them = ... 0 " What is Carbon tri chloride purpose?" is not answered in the link provided but the answer starts with a carbene attack on furan . Trichloromethane in presence of a base undergoes 1,1-elimination (or an alpha elimination) because both groups are removed from the same carbon.The electron-withdrawing effect of the halogens makes the hydrogen acidic enough to ... 0 The answers already given explain it quite well. I would lend another perspective of saying the exact same point, i.e, of transition state energies. So if we look at the reaction co-ordinate diagram of a generic chemical reaction, this pops up: Pardon my laziness, but in the image you can just replace the "No Catalyst" pathway with that of primary ... 0 To acknowledge OP's request, I am putting this as my answer: What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that: A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ... 4 Albeit carboxylic acids and their derivatives lose carbon dioxide under a variety of experimental conditions, the literature on the decarboxylation of$\alpha,\beta$-unsaturated carboxylic acids is not common. The reason for this is all$\alpha,\beta$-unsaturated carboxylic acids seemingly do not undergo decaboxylation under mild conditions, except for those ... 1 The decarboxylation of cinnamic acid 1 and related structures has been studied by Johnson and Heinz. The reasonable mechanism below was suggested. W. S. Johnson and W. E. Heinz, J. Am. Chem. Soc., 1949, 71, 2913. 0 The Keto form given in the solution would be the one which is the most stable. The hydrogen of the Hydroxyl group form H bond with the O of the carbonyl group , making this type of structure highly stable. Of course other types of structures are possible and will surely exist , but this is the most likely and most stable form of it 0 Benzene ring is resonance stabilized and is highly stable . According to Huckel's Rule it has$(4n+2)e^-$s (where$n = 1$), which makes this ring aromatic. This is the reason why$\ce{KMnO4}$did not cleave the benzene ring. 0 The question in its current form ONE KETO FOMRM implicitly implies a single form , while three keto forms are possible 4 , 3 and c. Structure 4 can enolize to 3 since enol form 4 can tautomerise to keto form 3.Remember in keto enol tautomerism , equilibrium can be achieved from keto end and also enol end. Energetics would determine the extent of one form ... -1 I would rather not solve this by comparing potential energy values. Instead there is another theoretical way of doing so. When SN₁ happens, either the attacking reagent( In this case water) can attack from above or below the planar carbocation. In the first case, Br⁻ leaves from above, and now water can easily attack from below. In the second case, Br⁻ ... 0 It's not certain properties, but a very careful observation if and with what products glucose reacts to other chemicals. Took many years. And even then Fischer did not know the absolute configuration, because at his time nobody knew the absolute configuration of anything. Only when the first absolute configurations were known from X-ray diffraction ... 0 As you can see from above mechanism the dibromo needs to be in an axial positions for elimination 1 The only difference between substrates A and C is disulfide vs sulfone nature at 9-position. Actually, that is good enough to display this major stereochemical outcome. As explained in Ref.1, the disulfide nature at 9-position of A has an advantage of having two lone pair on sulfur, one of which is proximate enough to attack backwardly on$\ce{C-Cl}$bond ($\...

1

2-chloro-1,3-dinitrobenzene is shown below: I believe your question is referring to the lack of conjugation between the nitro groups and the aromatic ring. this is best shown in an X-ray crystal structure such as this one: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=722415 Due to the sterics between the nitro groups and the chloro substituent, the ...

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Because of the absence of a lone pair, C undergoes usual $\mathrm{S_N2}$ transformation under the conditions. However, the availability of a lone pair in A undergoes the nucleophilic substitution of cyanide ion via neighboring-group participation (see above comments by Waylander and J. Deans) by the sulfur atom under the conditions (Ref.1). A very reliable ...

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If they are off the plane due to steric effect, then resonance wouldn't occur, hence the (-)R effect of - COOH group vanishes and hence "generally" acidity of the respective acid increases.

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It should probably be option C as in one of its resonating structures, there is an alpha hydrogen(as it is saturated) wrt the carbocation formed. Hence, it can show hyper conjugation

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Magnesium oxide can be an effective catalyst in transesterification reaction of vegetable oil samples mostly containing high carbon esters: $\ce{MgO}$ is used as catalyst to increase the rate of transesterification reaction of sunflower oil in supercritical methanol. The reaction in case of soybean oil has also been reported. This method was developed ...

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I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as ...

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A keto or an aldo group with phenyl ring is less reactive towards nucleophillic addition then their aliphatic counterparts(acetaldehyde or acetone).With phenyl group the positive charge on carbonyl carbon is delocalised via resonance (structures 2 , 3 , 4 , 5 ) . Resonance with phenyl ring is due to conjugation. such conjugation is absent in aliphatic ...

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Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution. If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic. If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic. ...

4

It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but: The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions. One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing. ...

5

Cyclobutyne cannot exist. There is no way to enforce 90° bends at both ends of a triple bond, but the carbanion formed by the Grignard reagent synthesis finds an easy way around this problem. Let's say the magnesium reacts at position 1. Then the negative charge would couple with the conjugated triple bond to spread to position 3: $\ce{Br\overset{+}{Mg}\... 0 As I said in the comments, it entirely depends on the conditions. It is easy to select mild conditions to cleave the dioxolane. Elimination would require very strong forcing conditions (concentrated acid and heating). The first site of protonation is the carbonyl, so you would form the hemi-acetal of the product long before elimination or substitution ... 0 Acoording to the IUPAC rules for nomenclature of the halo compounds the$\ce{Cl}$here is treated like$\ce{CH3}$. And also according to rule, the double bond have preference over the groups other than functional groups. 6 Yes, that is used as a pathway for synthesis of amines from amides, albeit rarely. I quote from the abstract of a communication that reports the same: (emphasis mine) The reduction of amides to their corresponding amines has been used extensively in the synthesis of many drugs and their intermediates. This reaction has usually been carried out by means ... 4 Your questions: Q1) How the rearrangement from (G) to non-1-yne can happen (i.e., its mechanism)? For example, see the following mechanism suggested in Ref.1 as the Acetylene Zipper reaction: Q2) Why Sodium Amide in particular yields terminal alkyne, unlike the fused Potassium hydroxides why almost always yield internal alkyne? To answer your question, I ... 3 The −C≡C− is normally drawn as linear about each C atom as that is what is found in nature for most alkyne x-ray crystal structures. We can use hybridization theory to explain this as the C atoms would be expected to adopt an sp hybridization, which gives two hybrid orbitals along the axis of the p orbital hybridized with the s to make the sp hybrids: https:... 5 Triple bond in carbon means that the carbon atom is sp hybridized. sp hybridized carbon has the two bonded groups at 180 to each other. Therefore it is displayed linearly which is closest to the actual structure. Single and double bonds are at an angle of nearly 109.5 and 120 degree therefore are not drawn linearly. 2 We can envisage the first steps of the reaction from the dibromide form the internal alkyne, then this undergoes the zipper reaction to isomerize to the terminal alkyne due to$\mathrm{p}K_\mathrm{a}$differences. See e.g. Alkyne zipper reaction. The fused KOH is not sufficently strong a base to effectively catalyze this isomerization. 0 Simply check the order of acidic strength and reverse it to find basic strength as a weaker base will be stronger acid and vice versa. Among them trifluoro-o-cresol is more acidic because after the removal of acidic hydrogen the negative charge thus generated is stabilized by the −I effect of three fluorine atoms. Whereas in the former the$\ce{CH3}$... 3 First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction). Counter to your analysis, the following reaction will not occur since$\ce{H}$bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to$\mathrm{p}K_\mathrm{a}$values given by Mathew Mahindaratne. Under the given ... 3 I think you have posed an interesting question. Upon looking through some literature, I have chanced upon the answer to your question in Fleming (2010), p. 30-31. The diagram below was also taken from the discussion in the book. What you have proposed are two ways of constructing the$\pi$molecular orbitals of butadiene from the$\pi$molecular orbitals ... 4 No, this is not practicable because aldehydes are higher in free energy than either alcohols or acids. Thus, the equilibrium constant of the Cannizaro reaction lies strongly against the aldehydes. The case of formaldehyde may be even more extreme than the example reaction you gave, but it is what I could find data for. The eQuilibrator database lists the ... 2 If you are a using reversed-phase (RP) column, it is okay to directly inject your crude sample in the running HPLC solvent system. But, make sure to filter the injecting sample through a membrane filter to remove minute particles (from glass beats, etc.) from the original sample after dissolving it in suitable chosen solvent. Please also note that your ... 2 The Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of two molecules of a non-enolizable aldehyde to give a primary alcohol and a carboxylic acid (Wikipedia). The haloform reaction requires a methyl ketone as the substrate. Two compounds (P and R, respectively) in choice [A] give you 4-methylbenzaldehyde,$\ce{...

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The parallel process to deposition of limescale ( that is inhibited by acidic jam) is deposition of products of oxidation of epigallocatechine gallate(EGCG) and similar compounds, that are getting adsorbed on limescale particles and cup walls. Epigallocatechin gallate (EGCG), also known as epigallocatechin-3-gallate, is the ester of epigallocatechin and ...

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As @andselisk suggested the problem lied in hard water. And calling that thingy "layer" instead of "membrane" helped to google the problem properly: cooking.stackexchange.com >, www.teamuse.com > "The scum on the top of the tea is due to hard water (ie calcium carbonate) deposits combining with the tea and reacting with oxygen." To dissolve that layer we ...

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