78

Your marble chips react on the surface. In the case of hydrochloric acid, the resulting salt, calcium chloride, is highly soluble in the acid, dissolves and provides further attack to the (new) surface. With sulfuric acid, the highly insoluble calcium sulfate is formed on the surface of the marble chip. With other words: Calcium sulfate acts like a ...


61

Great question! When I was teaching, Anslyn and Dougherty was a decent text for this. Here are some general comments: First, please note that you cannot be sure about a mechanism. That's the real killer. You can devise experiments that are consistent with the mechanism but because you cannot devise and run all possible experiments, you can never be sure ...


53

Intriguing question. First, the best yield would be achieved by selectively producing one enantiomer instead of the other. In this case, White wants D-methamphetamine (powerful psychoactive drug), not L-methamphetamine (Vicks Vapor Inhaler). Reaction processes designed to do this are known as "asymmetric synthesis" reactions, because they favor production ...


40

Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the ...


38

From explainthatstuff.com: Why doesn't glue stick to the tube? Adhesives are designed to work when they leave the tube—and not before. Different adhesives achieve this in different ways. Some are dissolved in chemicals called solvents that keep them stable and non-sticky in the tube. When you squeeze them out, the solvents quickly ...


38

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products. (source) As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and ...


38

It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$. $$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$ They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$...


37

This doesn’t exactly concern the actual mechanism you asked for, but as part of my PhD thesis, I performed an amide alkyne coupling the mechanism of which had been researched by Arndt et al.[1] Analysing how they established an accepted mechanism may help understanding how these are accepted. The authors note five proposed mechanisms at the beginning of ...


36

Thermodynamic and kinetic control This is a classic example of the concept of thermodynamic versus kinetic control of a reaction. Take a look at this energy profile diagram.1 The horizontal axis is a reaction coordinate, and the vertical axis represents Gibbs free energy. In this scenario, the starting material $\ce{A}$ can react to form either $\ce{B}$ or ...


35

In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the ...


33

tl;dr: I don't think there is any mechanism that is 100% correct, and, in cases like this especially, I think it would completely depend upon what set of carbonyl/ylid/base/solvent etc. was used. But, of course, we like being able to generalise, and to my knowledge theres a lot more evidence to support a concerted type mechanism. General Background The ...


33

The symbol is called "double dagger" (sometimes also "double cross") and is used to denote transition state (a maximum in an energy diagram; also often denoted with "*" or "TS") or a related physical property. Note, however, that a transition state and an intermediate are two different terms. The symbol has the peculiar origins: as written by H. Eyring, it'...


32

In contrast to many other electrophilic aromatic substitution reactions, aromatic sulfonation is reversible, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate. This makes sulfonation a nice method to ...


30

You were on right path; tert-butyl carbocation is quite stable$\ldots$ so stable that the acylium cation, which normally reacts as an electrophile itself, instead decomposes via decarbonylation (loss of stable carbon monoxide molecule). The t-butyl carbocation isn't as stable as the acylium cation (which is stabilised by resonance), but the difference is ...


29

Ammonium nitrate ($\ce{NH4NO3}$) is widely used in the fertilizer industry and is one of the most concentrated forms of nitrogen fertilizer (35% of $\ce{N}$). At the same time, it has also been widely used as an explosive material for detonation in mines. Because of its explosiveness, $\ce{NH4NO3}$ is associated with various hazards including fire and ...


25

First part It won't decide the issue but the Organic Chemistry text by Clayden, Greeves, Warren and Wothers also mentions that the matter might not be as clear-cut as the majority of your textbooks make it seem. This might strengthen the position of the textbook you're using a bit. But again, there are no references given. Here is the relevant passage (...


25

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic ...


24

An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products. Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones) An intermediate differs from a transition state in that the intermediate ...


24

The mechanism of the Cannizzaro reaction is illustrated below. The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate. Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon ...


24

Just heat it up: Isonitriles can be thermally rearranged to nitriles. $$\ce{R-NC ->[\Delta] R-CN}$$ Please have a look at some references, such as: Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc., 1972, 94, 2704-2706 DOI Michael Meier, Barbara ...


23

According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $...


23

Let's first look at the acid-catalyzed hydration of an alkyne. Just like in the analogous reaction of alkenes, the new proton goes to the least substituted position, to give the most substituted carbocation. Water attacks the carbocation, and after deprotonation, an enol is produced. Under acidic conditions, the enol tautomerizes the the corresponding ketone....


23

A large pile of grey magnesium powder, when lit in air, produces a smouldering pile which cools down to reveal a crusty white solid of magnesium oxide. However, if you break apart the mound, you can find something quite strange in the middle - a clearly brownish powder that wasn't there before. Seeing is believing! The author of the video also has a clever ...


22

Meth doesn't have to be optically pure to be "pure". A mixture of d,l-methamphetamine is still pure, but I get where you're going with this. He has a few options: Chiral resolution - he could make the racemic meth and they resolve it by selectively crystallizing out the desired enantiomer. Chiral acids like tartaric acid can be used to do this. He could ...


22

The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions. These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much ...


21

The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst. Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent ...


21

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon: $\hspace{2.4cm}$ $\ce{NaBH4}$ is ...


21

It is well known that SN1 reactions often give incomplete racemisation: Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 ...


20

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations ...


19

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen. So in "first" step: $\ce{CH_4 + O_2 -> CO_2 + 2H_2}$ but then the hydrogen will be also ...


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