11

Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol ...


9

The traditional explanation for the reactivity of indole at C–3 is that attack at C–3 does not disrupt the aromaticity of the benzene ring in the cationic intermediate (see any organic chemistry book; for heterocycles a good book is Joule and Mills as has been suggested in the comments): You have drawn many of the resonance structures for the intermediate, ...


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 ...


6

You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where $\ce{HCl}$ comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by $\ce{AlCl3}$, which generates in situ the "acylium" ion ($\ce{...


6

The simplest stable acyl chloride is ethanoyl chloride or acetyl chloride; methanoyl chloride (formyl chloride) is not stable at room temperature, although it can be prepared at –60 °C or below. (Wikipedia) The instability of $\ce{HCOCl}$ is caused by ease of elimination of HCl from its molecules. Cl is decent leaving group and after it's gone, remaining ...


5

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

The reaction given in options (a) and (b) are called formation of nitrate esters. The following nitrate ester is called "ethyl nitrate", Firstly, $\ce{HNO3}$ and $\ce{H2SO4}$ reacts to form the $\ce{E+}$. As we know, the former is relatively weaker acid than the latter, therefore $\ce{HNO3}$ donates $\ce{OH-}$ and neutralisation takes place, $$\ce{...


4

The reaction of benzene and γ-butyrolactone with AlCl3 was reported by Truce and Olson1 in 1952. The authors found that the ratio of γ-phenylbutyric acid to α-tetralone was dependent on the AlCl3/lactone ratio. The larger the latter ratio; the more α-tetralone. (See Table 1 from the paper). The ratio employed in the Organic Syntheses ...


4

The paper by Order and Lindwall [1] agrees with you that the R-T reaction gives indole-3-carboxaldehyde. Since its isolation in 1903 by Hopkins and Cole (1), 3-indole aldehyde has been investigated very little. It was first prepared by Ellinger (2) from indole through the use of the Reimer-Tiemann reaction. This method was improved on later by Boyd and ...


4

The key in this reaction is ketone in propiophenone is acting as Lewis base to $\ce{AlCl3}$ Lewis acid. If $\ce{AlCl3}$ is in catalytic amount, it activate $\alpha$-$\ce{H}$ in side chain sufficiently enough (similar to acid would do to a carbonyl compound) and progress to produce $\alpha$-substituted product. If $\ce{AlCl3}$ is in excess, it seemingly ...


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

tBu-benzene is prepared exactly as you describe, see this preparation here. Follow this by nitration and reduction. There is no obviously comparable strategy of acylation that can compete with this. Making a tBu group from acetophenone is not straightforward and is illogical when there is a perfectly good direct route.


3

What you have gone wrong in this question is you completely disregard the fact that mentioned stronger acids than water are all in aqueous medium, in other words, dissolved in water. Thus, fast acid-base reaction ($\ce{H2O}$ acts as a base here) happens to give $\ce{H3O+}$ as the only acid in these mixtures (leveling effect of water). On the other hand, ...


3

To start, it is important that you understand that the methyl and hydroxy groups are activating only in the ortho and para positions. For example, if I were to substitute plain phenol, which contains an activating hydroxy group, which position would my electrophile (in this a bromine electrophile) add to? You can see that the intermediary sigma complex ...


3

Of course, there must be some boundary as to whether or not nitrobenzene undergoes a particular electrophilic aromatic substitution (EAS) since nitro-group is very deactivating. I leave it to you to decide that fact on the yield obtained in following reactions, because there are some data you may fail to find but, nonetheless, exist in literature. ...


3

Electrophile can substitute 2-naphthol ring on 3 possible positions: 1-position through intermediate I, 6-position through intermediate II, and 3-position through intermediate III (See following diagram). Among these three intermediates, intermediate I is the most stable (You must visit Mechanism of formation of 2-naphthol red dye to get excellent ...


2

In ipso substitution first step is usually attack of the nucleophilic species at the ipso carbon of the aromatic ring (the carbon bearing the leaving group in this case). Scheme 1 The first step is followed by elimination of the leaving group and regeneration of the aromatic ring. Scheme 2 This mechanism greatly resembles the tetrahedral mechanism . The ...


2

Chlorination of aniline with chlorine is rapid and gives 2,4,6-trichloroaniline according to the references cited in this paper here. The earliest reference cited is 1845 by AW Hofmann so it has been around a long time. More modern approaches use N-Chlorosuccinimide as the reaction is more controllable


2

The answer to the question is (1) Aniline. This occurs because under standard strongly acidic nitrating conditions (typically cH2SO4/HNO3) the aniline NH2 is protonated removing the ability of nitrogen lone pair to donate any electrons. Consequently the electron-withdrawing polar effect makes it m-directing. Some p-nitroaniline is formed because there is a ...


2

The following is the abstract of K.M. Zhang "Selective Nitration Of Benzyl Chloride And Its Application In The Synthesis Of Nitrobenzaldehyde"(link). The full pdf is behind paywall. The selective nitration of benzyl chloride and its application in the synthesis of 4-nitrobenzaldehyde were studied. Nitration of benzyl chloride with $\ce{HNO3}$ in the ...


2

Electron giving groups like $-OCH_3$ (by resonance) and $-CH_3$ (by hyperconjugation) are Ortho-Para directing. But, in cases where there is no increased stability by H-bonding or other factors at ortho position, the para substituted product will give major yield. This can simply be explained by stating that there is less steric hindernace on the para ...


2

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present ...


2

Thank you for the question. This is a very basic physical chemistry question but the way it has been asked is interesting. In order to understand the probability of displacement of a H from any X-H bond (X is any atom), you need to know its pKa. The pKa values are as follows (ref.): pKa of water = 14 pKa of OH (phenol) = 9.95 pKa of CH (benzene) = 43 If ...


2

On Friedel-Crafts-acylation reactions, two types of Friedel-Crafts-acylation mechanisms, namely an ion pair mechanism and a dipolar ion mechanism, have been proposed (Ref.1). Normal acylations are presumed to proceed via the ion Pair mechanism, which seems to be more important in sterically hindered reactions: The product complexes with aluminum chloride ...


2

The carbonyl group attached to the benzene ring is an electron-withdrawing group and deactivates the ring towards electrophilic substitution. Thus, $\ce{AlCl3}$ in excess is required to brominate the ring in this case. If a catalytic amount is used, then the $\alpha-\ce{H}$ atom is substituted by a standard Ketone halogenation mechanism.


2

Based on literature, I'd say the unsuccessful Friedel-Crafts reaction between benzoyl chloride and mesitylene is not due to steric reasoning. My argument is based on at least one report of Friedel-Crafts acylation on following polymethylated aromatic compounds (Ref.1): The mono- and di-acylation have been taken place between 1,3-substitutions in each ...


1

Let me put it this way: Every compound exist in a 3N dimensional configuration space (may be simple XYZ coordinates of the system, may be internal coordinates, based on your choice), and based on the configuration, its potential energy is defined. The potential energy surface (PES) can be calculated quantum mechanically or using some force field classically. ...


1

A is correct because the question specifically says an electrophilic substitution reaction (D will also react, but it's an electrophilic addition reaction). However, you also need a Lewis acid catalyst like FeCl$_3$. This is because, while the delocalised system of electrons in benzene is indeed electron rich, the fact that it's aromatic (there is a system ...


1

Sodium Phenoxide is much more activating than phenol itself. Because here directly one negative charge comes and bond with the C below and the pi bond shifts to the ortho position as negative charge. And it also answers your doubt that why ortho product is major. You can refer to this mechanism. After reacting with OH- then the tautomerization takes place ...


1

First we have to understand what affects basic nature of a compound. Tendency to donate lone pair Electronic effects— Inductive, resonance etc Factors stabilizing conjugate acid (or acidity of conjugate acid Nature of solvent Tendency to donate lone pair is affected by the acidity of conjugate acid and electronic effects. Grab a pen and paper and draw ...


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