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In the first step, due to the selectivity of bromine in a free radical reaction, the tertiary carbon which has a $\ce{-CH3}$ group attached to it forms the radical. Now, since the radical is a $\ce{^{.}CR3}$ radical, the carbon radical in the intermediate becomes $\ce{sp^2}$. This means that $\ce{^{.}Br}$ can attack from both the top and the bottom since ...

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 ...

7

Given enough exposure time, the reaction will lead to formation of various products, even the fully chlorinated compound: dodecachlorocyclohexane. Increasing the concentration of chlorine also help in the formation of this product. Doing some literature survey confirms this: [...] During most of the course of this reaction (photochemical chlorination of ...

6

The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$). In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base ...

5

Chlorination and bromination of alkenes are very general reactions, and mechanistic study of these reactions provides additional insight into electrophilic addition reactions of alkenes. Although much less detail is known about fluorination and iodination of alkenes, it is believed that iodination follows the similar mechanistic steps of much studied ...

5

This is not a radical process. $\ce{BrCl}$ is a source of $\ce{Br+},$ so this is the addition of $\ce{Br+}$ to the alkene followed by capture of the cation (bromonium ion) by $\ce{Cl-}$ [1]. The product is that of $\ce{Cl-}$ addition to the most stable cation centre i.e. the tertiary centre, so 1-methyl-1-chloro-2-bromocyclohexane is the product. Reference ...

4

Check if you may skip using elemental bromine as starting material altogether. An alternative approach may be running the bromination in a heterogenic, biphasic setup with the couple $\ce{NaBrO3/NaHSO3}$ in water (e.g., 1998JOrgChem6023, 2000OrgProcRes Dev.30), or $\ce{NaBr}$ / sodium perborate (e.g., 1998SynthCommun925), or pyridinium tribomide (e.g.,...

4

As a follow-up to the excellent comments provided thus far, the following thoughts are added. Electrostatic Potential Map of azulene (1a) shows that there is high electron density in the 5-membered ring (red color signifies high electron density). The 5-membered ring is reminiscent of the aromaticity of cyclopentadienide anion and the 7-membered ring bears a ...

3

While there is no formal carbocation intermediate, the cyclic bromonium intermediate has a few different resonance forms (reference): As you can see on the right, in the structure representing the resonance hybrid, partial positive charges build up on the electrophilic carbons connected to the bromine. The carbon that is more substituted with electron ...

3

According to this Wikipedia article: "In electrophilic halogenation the addition of elemental bromine or chlorine to alkenes yields vicinal dibromo- and dichloroalkanes (1,2-dihalides or ethylene dihalides), respectively. The decoloration of a solution of bromine in water is an analytical test for the presence of alkenes." Alkanes tend to be ...

3

The reason you use the respective halides for halogenation is that the halogen and Lewis acid can inter-react to form a chemical equilibrium if you don't. $$\ce{FeBr3 + Cl2 <=> Cl+ + [FeBr3Cl]- <=> Br+ + [FeBr2Cl2]- }\\ \ce{AlCl3 + Br2 <=> Br+ + [AlBrCl3]- <=> Cl+ + [AlBr2Cl2]- }$$ Given these equilibrium reactions you can see that ...

3

My prof says that I am missing my R/S. I can clearly see my chiral centres, but how do I determine the orientation of the H? Red is the major product, blue is the minor product, and magenta is the very minor product. You don't determine the orientation of the H you just acknowledge the formation of both the R and S stereo isomers in your products in 50/50 ...

3

An electrophilic aromatic substitution reaction is one in which the aromatic ring reacts (as a nucleophile) with an electrophilic reagent. In all of the reactions that follow this naming convention, the adjectives "nucleophilic" and "electrophilic" describe the reagents, not the substrate. For example, the following reaction is a ...

2

I'd like to point out that the ketone would actually disfavor an attack at its alpha position: Placing two positive charges on adjacent carbon atoms is much less favorable than not, especially when there is an oxygen perfectly poised to help stabilize this positive charge, as user55119 pointed out.

2

It does seem odd as a blanket statement. However, the bond dissociation energies (BDEs) for Cl2 and Br2 are quite low, 243 and 193 kJ/mol, respectively, while BDEs for the C-H bond in alkanes are much higher (eg. 435 kJ/mol for methane). So, it's not necessarily the case that events involving only bond breaking will be the rate-determining step, particularly ...

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 expected products of mono-chlorination of 2,4-dimethylpentane are depicted in following scheme: The substrate has 12 primary hydrogens (on 1-,1'-, 5-, and 5'-methyl groups), 2 secondary hydrogens (on 3-methylene group), and 2 tertiary hydrogens (on 2-, and 4-methine groups), as numbered in the scheme. Altogether, total of 3 different hydrogens are ...

2

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the ...

2

In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy. Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, ...

2

There is plenty of information about all these compounds if you look at "Interhalogen compounds" through Google. All these compounds are known : AX- type : ClF, BrF, BrCl, ICl, IBr AX3-type: ClF3, BrF3, (ICl3)2, AX5-type: ClF5, BrF5, IF5, AX7-type: IF7. Their synthesis, structures and properties are nicely described, much better than what I could ...

2

This is to do with the relative stabilities of the free radicals produced. Here is a schematic diagram showing the two possibilities for propene (similar mechanism, also with two free radical possibilities like propane): As you can see, a more stable and less stable free radical is produced. The stability depends on how many carbons are attached to the ...

2

There multiple means to generate the radicals you would need for a reaction of $$\ce{\mbox{2-Bromopentane} + Br2 -> \mbox{2,3-Dibromopentane} + HBr}$$ An alternative to radiation is the combination of heat and some of radical initiator which is typically deployed at gentle $70-90\,\pu{^\circ{}C}$ like AIBN or dibenzoyl peroxide. To moderate the radical ...

2

TLDR: Sulfuric acid act as catalyst in the isomerization reaction of parafin hydrocarbons and sometimes it produce undesired sideproducts like sulfonate salts, napthenic acid, bisulfite ions etc. from incomplete oxidation. It follows "hydrogen exchange mechanism. Long answer: @andselisk cited some papers that shows that sulfuric acid can act as a catalyst ...

2

Here are the six structures that your source is probably referencing.

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It is impossible to compare the stability of methane and carbon tetrachloride directly as the do not have a common frame of reference. At the same time you cannot claim that the dichlorine molecule is less stable than the hydrogen chloride molecule. The only true observation that you can make is that the reaction of methane and chlorine to chloromethane and ...

2

It starts with the formation of brominium ion, as in direct bromination of alkenes. If you do not know the mechanism, you can refer to it here, at MasterOrganicChemistry. But here, the $\ce{MeOH}$ can itself act as a potential nucleophile. It is similar to if $\ce{H2O}$ is used as a solvent - refer the formation of halohydrin here. Since $\ce{MeOH}$ is ...

1

In free radical substitution, a free radical is formed that is an electron deficient species and is stabilised by electron donating species bonded to atom on whose orbital it is present. So, electron donating groups that is +I ,+M increase chances of formation of free radical on it and presence of electron withdrawing species (-I,-M, Backbonding,by Cl) ...

1

I think a very interesting way to perform this reaction, absence light or elevated temperature, especially since Br2 is a liquid at room temperature, is via sonolysis of either bromine or an alkyl bromide: $\ce{Br2 + Energy (sonolysis) -> {*}[Br2] -> .Br + .Br}$ Here is a description of the mechanics from comments in ScienceDirect on sonolysis (...

1

The mechanism of the gas phase halogenation reactions is discussed in some detail in a paper by Benson and Bus[1]. In the following, I submarised section II ( mechanism and rate laws). The mechanism of the reaction thermal reaction is $\ce{X2 + M -> 2X^. + M \;\; k_1 \quad(1) }$ $\ce{2X^. + M -> X2 + M \;\; k_{-1} \quad(-1) }$ \$\ce{X^. + RH -> ...

1

The key feature in the product that shows what is going on is the trans configuration of the two bromines. This is characteristic of the addition of elemental bromine via a brominium ion mechanism here. So where does the Br2 come from as we started with HBr? This is where the hydrogen peroxide comes in. It is well documented example here that hydrogen ...

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