# Tag Info

## New answers tagged aromatic-compounds

0

In this case, the dominance form of the compound depends on which phase it is in. Ref.1 states that: In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to ...

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 ...

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 ...

3

The mechanism should provide for the successive processes of carbocation formation, rearrangement and proton loss. Observations(Points to consider): The tendency to attain stablity by aromaticity. The high ring strain in the cyclobutane ring. The first step step should involve protonation of each hydroxy group attached. Protonation produces hydronium ion ...

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 ...

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 ...

0

$\ce{CF3}$ loves to pull electrons, hence it is easier to polarize the molecule in this way (which is inaccurate because you cannot polarize it completely, but considering partial effects). This polarized structure is made considering the requirement of negative charge at the carbon positioned adjacent to CF3. position 3 and 5 maybe a dispute but, as a known ...

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

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 ...

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 ...

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

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{...

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 ...

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 ...

2

I'm not sure if G is an activating or deactivating compound. You may want to look at the relative order of activation/deactivation of some substituents, (source: www.chem.ucalgary.ca) As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{... 3 The rate of reaction depends on stability of intermediate. The given reaction follows$\ce{S_N2Ar}\$ mechanism through "Meisenheimer intermediate". So, the stability of "Meisenheimer" primarily drives the reaction. ortho and para isomer are almost same except their position, which is the main reason why is the former reacts much slowly ...

1

the electron pair that is creating negative charge on ring is sp2 hybridized and its orbital direction is away from the pi orbital electronic cloud of benzene ring so not able to resonate along with pi electrons. OMe inductively stabilize this electron pair as it is electron with-drawing inductively. thats why NH2 group attaches at meta position rather than ...

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