17

The reaction of benzene over $\ce{V2O5/ PtAu}$ catalyst at lower temperatures, can convert benzene to phenol with some success. See this book here Direct hydroxylation of benzene [The original question before editing was: What is the Ratta Maar Reaction?] Someone has played a prank with you with the named reaction. "Ratta maar" is a slang for "rote ...


11

The best way to download bulk data from PubChem is actually FTP, as documented in their documentation. For example, if you want the unfiltered SMILES of every CID in PubChem, the URL is ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/Extras/CID-SMILES.gz You can also download subsets using the PubChem Structure Download service And as mentioned above, there ...


11

It turns out that the opposite of what happens in polar solvents takes place when a non-polar solvent is used. At the same temperature, o-nitrophenol is more soluble in benzene than it's m and p isomers. Sidgwick et al.1 did a study of this and obtained the following results. (Note that the solvent they used was toluene and not benzene, but they are similar ...


10

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated. The mechanism (on paper) seems to be as follows Dehydration $1$ For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows: Dehydration $2$ Similar to the ...


9

Biphenyl 2 is the only optically active compound here. These stereoisomers are due to the hindered rotation about the 1,1'-single bond of the compound (Ref.1). Biphenyl 3 is not optically active, because partially allowed rotation about the 1,1'-single bond of the compound (rotation is only partially restricted). To illustrate this phenomenon, I depicted the ...


9

Initially one would assume that cumene (isopropyl benzene) would be the major product. However, kinetic control does have a say in this reaction, according to Gilman and Means.1 At - 2°, n-propyl bromide with benzene and aluminum chloride gave n-propylbenzene, identified as its sulfonamide. Genvresse (6) obtained both n-propylbenzene and isopropylbenzene by ...


8

(3) is not chiral, so maybe it is (2).


8

Nothing against Safdar, but this is technically not a two-step reaction. When lithium amalgam is added to 2-bromofluorobenzene in presence of furan, the reaction goes directly to the product, a Deals-Alder adduct: However, in the absence of furan, the reaction proceeds to give biphenylene and triphenylene (Ref.1&2). The tentative reaction mechanism can ...


7

Nitrosyl cation, $\ce{NO^+}$, is a better electrophile than molecular $\ce{HNO3}$ (you don't get $\ce{NO_2^+}$ in this system without sulfuric acid or other strong auxiliary acid) and so you get nitrosobenzene in the substitution reaction. But then the nitrosyl group with its nonbonding electron pair is an attractive target for oxidation by the nitric acid ...


7

I don't think this is an either this or that situation. As with most of the equilibria we know it depends very much on the circumstances we look at it. Many examples of keto-enol-tautomerism are heavily influenced by the solvent. Due to time and resource restrictions I won't be able to go in as much detail as I did in explaining whether enolates get ...


7

Biphynyls are a class of compounds with two phenyl rings attached by $\ce{C-C'}$ carbons in each phenyl ring (See Figure $\bf{\text{A}}$): The phenyl rings can rotate around this $\ce{C-C'}$ bond (acts as an axis). Keep in mind that any molecule capable of rotating one part of it around fixed axis (a bond) is never planner. As a rule, if the rotational ...


7

Benzocyclobutadiene has an aromatic benzene ring as well as an anti-aromatic cyclobutadiene ring. This gives it the characteristics of both aromatic and anti-aromatic compounds. Now, according to Cyclobutadiene and its related compounds, [1,p 180]: Benzocyclobutadiene, the monobenzo derivative of cyclobutadiene, is intermediate in structure between the ...


7

The nitrophenols have completely different physical behavior based on the position of nitro group: $$ \begin{array}{c|ccc} \hline \text{Compound} & \text{Melting point} & \text{Boiling point} & \text{Water solubility at } \pu{25 ^\circ C}\\ \hline \text{2-Nitrophenol} & \pu{43-45 ^\circ C} & \pu{215 ^\circ C} & \pu{2 g/L} \\ \text{...


7

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4. There are two factors to ...


7

This is an another way to answer the question: On the way, I also want to show OP that why central ring bearing $\ce{CF3}$ group has the highest electron deficiency at ortho- and para-positions (meaning, the highest electron density is at meta-position in this ring). Let's look at the nitration of trifluoromethylbenzene: According to this University of ...


7

This is a question based on two reactions. The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]: A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]: Reference ...


7

I think the third option as greatest number of resonance structures can be alluded to it. The following may suffice: In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it. In the second one the lone pair on oxygen, the ...


6

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.) In order for the resonance to stabilise a negative ...


6

Chiavarino, et al. [1] report that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen. Nucleophiles such as methanol prefer boron. We can explain that result in terms of both molecular orbitals and the Wheland intermediate. In the molecular orbital explanation, recall the familiar ...


5

TL;DR: It's 2 (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl) and maybe even 3 (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl) at higher temperatures. Introduction According to the IUPAC gold book chirality is defined the as the geometric property of a rigid object of being non-superposable on its mirror image. Orr in other words, such an object has ...


5

The answer is as simple as it will be unhelpful to your original question: There is no tautomer possible. Coumarin does not have a keto group, and there are no enolisable protons either. Coumarin consists of a benzene ring, connected to a heterocycle with the functional group lactone. The molecule is flat and the conjugation extends throughout it. What you ...


5

1.Why are they carcinogenic: From my point of view as a medicinal chemist, polycyclic aromatic hydrocarbons compounds tend to be metabolized by specific cytochrome P450 (CYP 450) isofroms to form highly electrophilic species, which are highly reactive and can react with the DNA of the cell and damage it. Specifically, they can react with the nucleophilic ...


5

As to your first reaction that does not utilize copper, a reference to the source would be useful. Nonetheless, the Meerwein Arylation Reaction has been reviewed in Organic Reactions. The reaction is conducted in the presence of catalytic cuprous ion. Meerwein proposed an aryl cation but a radical mechanism appears plausible. Reduction of the aryldiazonium ...


5

This is simply the summery of all comments involved: $$\ce{Cl-C6H4-NO2 ->[Sn/HCl] Cl-C6H4-NH3+Cl-} \tag1$$ $$\ce{Cl-C6H4-NH3+Cl- ->[\text{a. } NaNO2/HCl \ (\pu{0-5 ^\circ C})][\text{b. } CuCN] Cl-C6H4-CN} \\ \tag2$$ $$\ce{Cl-C6H4-CN ->[H+/H2O][\Delta] Cl-C6H4-COOH} \\ \tag3$$ Note: The reaction $(1)$ is a typical reduction reaction, which reduce ...


5

Aniruddha Deb gave an excellent answer to your question. Yet, there is another important point of differences on two Friedel-Craft processes: Alkylation vs acylation. I admit that rearrangement is the major drawback on alkylation process (using alkyl halides with a catalyst). However, another fact is you cannot stop the reaction after monoalkylation. The ...


5

In this question, we need to find the most electron-rich carbon out of positions $1,2,3,4,5,6$ in the compound given below: We proceed using the process of elimination. Note: Not sure how scientific this is, however an attempt I made in finding the effect of the inductive effect was to run a DFT B3LYP/3-21 G on trifluoromethylbenzene. The result showed the ...


4

Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. ...


4

This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA ...


4

Here's the 3D structure of the substrate: As you can see, the $\ce{-N(Me)2}$ group is clearly larger than the $\ce{-CONH2}$ group. Also notice that the $\ce{-N(Me)2}$ group is planar with the benzene ring. This is because this structure is the most favourable for $\mathrm{p-\pi}$ conjugation 1. Your guess that A would not form is correct. The large size of ...


3

It's a reversible reaction, so at higher temperature the thermodynamic para product is formed from the kinetic ortho product. Further reading here 1


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