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2

Clear distinction has to be made between Arrhenius-equation and transition state theory. The formulas for TST can be derived, as opposed to Arrhenius equation, which by itself, is not based on a solid physical derivation. Basically any process can be modeled by an Arrhenius equation that has the formula: $$r \propto \exp{(-E_a/RT)}$$ Any molecular ...

3

It does not make sense for the constant to be zero. The constant is usually expressed as follows: $$K_M = \frac{k_{\mathrm{reverse}} + k_{\mathrm{catalytic}}}{k_{\mathrm{forward}}}$$ Since the $k$ values are strictly non-negative, the constant would only be zero if $k_{\mathrm{reverse}} = k_{\mathrm{catalytic}} = 0$. In theory, $k_{\mathrm{reverse}}$ ...

0

Solid bleaching powder is essentially a mixture of calcium hydroxide and calcium hypochlorite. The discussion in the link you provided is quite incorrect Ask IITians. I think it originates from historical concept which is perhaps still taught in India as "active chlorine" or so called "nascent chlorine". There is no free chlorine in bleach but it is rather ...

4

It isn't unstable at all. 1,2-diiodoethane is characterized with a melting point of 80–82 °C, and a mass spectrum with five major peaks. It even serves as the iodine source in some rare earth iodide syntheses. Read about it on Wikipedia.

4

I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case. There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred ...

0

See, we know that nucleophiles which are in polar protic solvents are solvated and therefore nucleophilicity is directly proportional to the size of nucleophile 'cause they will be less solvated. And in polar aprotic solvent the anions are not solvated so they will show their usual nucleophilicity behaviour and is inversely proportional to the stability of ...

1

This elimination is unfavorable because C--C bonds are generally quite stable. Decarboxylation would involve the formation of an unstable carbanion before proceeding to the elimination as you have outlined. (Note, this is more favorable under different functional group conditions, namely when a neighboring carbonyl can stabilize the carbanion through the ...

0

Here the issue in question is about rate. Of course, in a real life scenario, both reactions probably take place, the question is which reaction is faster? Well, looking at the reactants here, you are having a strong acid. The hydroxy group is way more basic/nucleophilic than the double bond simply due to electronic effects. Hence, the Sn1 substitution is ...

0

My suspicion is that the resonance structures accessible to the allylic carbocation provide a lower energy state than the formation of the C--H single bond and the isolated carbocation that your expected pathway would require. In order of reactivities, you are right to think that the alcohol is the controlling factor because of its pKa. Under acidic ...

0

According to some people: If you look carefully, you will find both the aldehyde radical and the carboxyl radical in formic acid: The red line shows the aldehyde functional group character and the blue line indicates the carboxyl functional group character. As a result, it has mainly characteristics of carboxyl compound, but sometimes shows carbonyl ...

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 ...

3

$\ce{H2A}$ and $\ce{A^2-}$ forms of a biprotic acid seldom coexist as major components, expecially if the acid is weak, unless $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ are similar. Instead, coexistence of major components $\ce{H2A / HA-}$ or $\ce{HA- / A^2-}$ occurs. $\ce{SO3^2-}$ starts to form in about neutral or alkalic solutions, while $\... 1 It should first be noted that the circle drawn derives from experimental results. If you were to mark specific oxygens, e.g. by introducing radioactive$\ce{^18O}$in place of normal$\ce{^16O}$, you can predict whether and where the ester product will contain$\ce{^18O}$or not and whether the water will contain$\ce{^18O}$or not.$\begin{align}\ce{R-CO-... -1 Esterification involves ANDN mechanism which is basically attack of one nucleophile and the departure of another from the same molecule.here OR- group attack the carboxylic acid at the carbonyl carbon causing OH- group to leave. OH- will combine with the H+ ions present in the system to form water. 0 The reaction between the bis salt of proline and phenylsuccinic acid will give free phenyl succinic acid and proline hydrochloride as HCl is a much stronger acid than phenylsuccinic acid. All you are really doing is exchanging one acid for another so it is hardly an acid/base reaction. 1 Your reaction goes way back to 1930s when Ingold and coworkers have done pioneering work on nitration of aromatic compounds using nitric acid in organic solvents in place of sulfuric acid (vide infra). Let's see how this has been developed: Historical perspectives: The earliest recorded experiments on the kinetics of aromatic nitration has been performed ... 0 at first i thought your mechanism was very comprehensive but upon closer inspection i think it has some flaws because in this reaction conc sulphuric acid is used as a catalyst and it must be regenerated at the end and in the above mechanism only ethanol is generated This must be the mechanism shown in your book. 3 This is not a magical reaction. Yet, it is a two step reaction as depicted in diagram below: As @Waylander's comment elsewhere, benzyl chloride might have converted to benzyl alcohol before get oxidized to benzoic acid in acidic\ce{KMnO4}$solution (the first step). This oxidation is well known (see here). The second step is heating (may be$\gt \pu{200 ^\...

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