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16

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile. In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. ...

15

During a electrophilic aromatic substitution, it is always possible to have multiple substitutions during one reaction. However, your example is not the ideal one for a discussion. As noted, Oscar Lanzi has questioned even aromatic ring of nitrobenzene is active enough to give even one $\ce{Br}$ substitution. To clear that, I have found a reliable reference: ...

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 ...

11

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, ...

9

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

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 ...

7

Page 513 of Organic Reaction Mechanisms by Bansal references the work of Melander in Arkiv för Kemi, 2, 211 (1950). This states that the rates of nitration for benzene vs benzene-d6 are about the same, implying that the addition of the nitronium ion is rate controlling and not the weakening of the C-H / C-D bond. However in case of sulphonation H containing ...

7

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

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer ...

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 ($\ce{... 6 This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations. First of all, it is not true that the para-product is always favoured with the$\ce{NH3+}$ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining ... 5 The reaction given in options (a) and (b) are called formation of nitrate esters. The following nitrate ester is called "ethyl nitrate", Firstly,$\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{... 5 In electrophilic aromatic substitution the activation energy is very high ( as it should be, since the aryl carbocation transition state is very unstable) The role of$\ce{FeBr_3}$or any lewis acid in general is not exactly promoting the electrophilic aromatic substitution as much as making a good electrophile.$\ce{AlCl_3}$plays the same role much better* ... 5 Friedel-Crafts reactions of anisole is known and the relevant acylation is performed in most academic institutions during undergraduate sophomore organic chemistry laboratories. For instance, Aluminum Chloride Catalyzed Friedel-Crafts reaction of anisole with Epoxide has been published (Ref.1). The authors have performed the reaction in various solvents ... 4 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. ... 4 This is an example of an inverse kinetic isotope effect. When a transition state increases in coordination then differences gained from vibrational bending energy for the$\ce{C-D}$vs$\ce{C-H}$bond favor$\ce{C-D}$, lowering the transition state energy relative to starting materials, and increasing reaction rate. In this case, the intermediate is the ... 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 Technically, the teacher's claim might be slightly modified. Deactivated benzenoid aromatics do fail to undergo Friedel-Crafts reactions, but according to a summary description from the University of Calgary: Deactivated benzenes are not reactive to Friedel-Crafts conditions, the benzene needs to be as or more reactive than a mono-halobenzene. So, a ... 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 ... 4 First of all, the linked paper clearly points out the following: Hint: In second step (A$\rightarrow$B), combination of lithium bromide and cerium(IV) ammonium nitrate (CAN) is used as a brominating reagent From a problem solving perspective, this step is trivial for someone who has read the problem well (not me, unfortunately :) Ceric Ammonium Nitrate ... 4 The reaction mechanism shown on the Wikipedia page for the Perkin Reaction is perfectly clear. The first step is deprotonation of acetic anhydride by acetate anion. The second step is nucleophilic attack by the anion of acetic anhydride on the aromatic aldehyde; thus the aldehyde is the electrophile. 4 According to this patent here, N-ethyl 1,8-naphtholactam nitrates in the 4 position i.e. structure a. It seems entirely likely the N-unsubstituted material will do likewise. 4 The reaction wouldn't happen because you forget that$\ce{Cl2}$is also present in the solution (because$\ce{AlCl3}$is a strong lewis acid). An acid-base reaction is way faster than an aromatic electrophilic substitution. Once the acid base has happened I doubt that$\ce{AlCl4-}$would still behave as an electrophile. Footnote:$\ce{AlCl3}$is a strong ... 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 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 ... 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 OP quote: I know that Friedel Crafts alkylation reaction are not possible for aniline and phenol since they form complexes with Lewis Acids. This is true partially for Lewis acids (Note: I said partially because Lewis acids work with anisole and other phenol ethers; Ref.1). However, Brønsted Acid has been used in Friedel-Crafts alkylation of phenol and ... 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, ...

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