# Tag Info

4

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 ...

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

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 ...

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

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.

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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 ...

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 ...

1

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 ...

1

https://en.wikipedia.org/wiki/Iodolactonization I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic ...

1

Acid Catalysed Halogenation of ketone In the halogenation of ketones, both in acidic and basic medium, enols are reactive intermediates. After addition of first halogen to double bond, deprotanation occurs immediately (Scheme 1). The second halogen does not add$\ce{^1}$. Scheme 1 The introduction of second Bromine is slower then the first ,since ...

1

I'm not entirely sure but I'll have a go. As a strong acid, HCl will be dissociated into Chloride and protons in water. I'd suggest that the mercury will first associate with the acetylene forming an Hg- Pi complex which will weaken the triple bond. Then the chloride ion will attack one side of the acetylene and the electron density one of the triple bonds ...

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I like the ringo's point, but literature evidence does not support the classic cyclic bromonium ion mechanism. Effects of a carbonyl group on the ring opening of a neighboring bromonium ion in some bromination of reactions of $\alpha,\beta$-unsaturated ketones have been studied. The authors have used methyl acrylate and methyl crotonates (cis- and trans-...

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First off, we can see that both reactions are exothermic If we do the thermodynamic analysis. Given that your starting materials have the same heat of formation: $$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$ \ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}...

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