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2

It is interesting to consider that either outcome is possible, depending on the order of addition. If a solution containing 0.5 moles of HCl is added to a solution of 1 mole of K2CO3, slowly, with mild stirring, cold, the products will be 0.5 KCl + 0.5 KHCO3. Nothing is lost to the atmosphere. However, if a solution of 1 mole of K2CO3 is added to a ...


2

The reaction of epoxyketone 1 affords diketone 2 under photochemical and thermal conditions ostensibly through the diradical.1 Boron trifluoride provides the diketone 3 through acyl bond migration and not via alkyl group migration as in your intermediate 3a. I have not located a proton-catalyzed reaction. Unfortunately, diketone 3 was not one of your choices....


2

Energy usually comes from reactions not one component of the reactions The problem with the way the description of oxygen as "energy rich" is that it doesn't make sense outside of the context of reactions oxygen can participate in. And, in that context, it doesn't make sense to partition the "energy* between the things that are reacting. Chemical reactions ...


3

Boiling off some water is a decent plan. Perhaps it should be done at reduced pressure so as to keep the temperature low. Magnesium and calcium bicarbonates decompose at temperatures above 60C in water to yield insoluble carbonates, and several data sheets for sodium bicarbonate suggest the same. One said decomposition was complete at 100C. I suppose you ...


6

[OP:] It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions. An enzyme is a catalyst, so it does not change the enthalpy of a reaction. That part is correct. "Oxygen only enables the release of energy" is incorrect. Oxygen is one of the reactants, and the oxygen atoms also ...


2

I present another (almost) two-step synthesis, involving the famous Wittig reaction. The first step details the preparation of a phosphonium ylide, which subsequently is allowed to react with the carbonyl compound (benzophenone in our case) to yield the final product, 1,1-diphenyl-1-butene. For mechanistic details, visit NotEvans.'s answer to Which is the ...


6

Two steps: Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be down in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group. Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq ...


1

Karsten Theis' answer is a perfect one for your question. He also did his best to explain why is he used the bond-making bond-breaking sign conversion but you are still in mind set of product-reactant explanation. So I decided to explain it in different direction. Remember the law of conservation of energy, which states that the total energy of an isolated ...


3

Here is the tally of bonds: One $\ce{H-H}$ bond broken One half $\ce{O=O}$ double bond broken Two $\ce{O-H}$ bonds made Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $\ce{O-H}$ bond depends on what else ...


0

This is because when number of electron donating alkyl group on OH-bonded carbon atom increases, polarity of carbon oxygen bond also increases, which further facilitates the cleavage of carbon oxygen bond. Therefore reactivity increases.


2

In solvolysis of simple primary cyclopropylmethyl systems the rate is enhanced because of participation by the $\sigma $ bonds of the ring. The ion that forms initially is an unrearranged cyclopropylmethyl cation that is symmetrically stabilized, that is, both the 2,3 and 2,4 $\sigma $ bonds help stabilize the positive charge(page 464 ,JerryMarch ,...


3

The first step is nucleophilic substitution . It is possible for nucleophile to attack directly at the allylic position, displacing the leaving group in a single step, in a process referred to as SN2' substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to those seen ...


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The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete ...


6

Potassium carbonate is a perfectly good base for the alkylation of phenol (pKa 10) with a good electrophile, in this case 3,3-dimethylallylbromide. The reaction you are looking for is a Claisen rearrangement which proceeds by a 3,3-sigmatropic rearrangement mechanism. image from ref 1


2

First passage problems This is a first passage problem, asking when a system reaches a certain final state for the first time . If you were to use a simulation to explore this, you would erase parts of trajectories right after reaching the final state before summing up results. The Backwards Master Equation is especially suited for first passage problems (...


-1

Aldehydes and ketones prefer nucleophilic addition reactions Instead of electrophilic addition reactions this is because in case of nucleophilic addition reaction a stable intermediate is formed as compared to the electrophilic addition reactions where the intermediate formed is carbocation which is highly unstable.


3

Example 1 Ethylene glycol is often used to make cyclic acetals; its acetals are called ethylene acetals (or ethylene ketals).This interconversion makes acetals attractive as protecting groups to prevent ketones and aldehydes from reacting with strong bases and nucleophiles Example 2 Reference Page 861, Organic Chemistry , Eight edition , L.G.Wade


3

The hint is to form the dioxolane (cyclic acetal) of the ketone with ethylene glycol then reduce the ester (LiAlH4 will do this, the dioxolane will not react) and finally deprotect the ketone. The reduction can also be accomplished by hydrolysing the ester and reducing the resulting acid with borane.


0

I tried to work out a mechanism from the information I could gather about the reaction. If someone notices something odd/wrong please comment and I'll fix it. The steps are either showing the protonation or skipping directly to the protonated forms of the reactants, them being in acidic medium. Formation of isobutylene Not very complicated, in acidic ...


0

$\ce{CaCO3}$ is easily decarboxylated to $\ce{CaO}$ in the heat of the fire. By dumping the ash into water you get a solution of $\ce{Ca(OH)2}$ and $\ce{K2CO3}$, and since $\ce{CaCO3}$ is less soluble than $\ce{Ca(OH)2}$ and $\ce{K2CO3}$ you end up with $\ce{KOH + Ca(OH)2}$ solution in one step and $\ce{CO2}$ dissolving into this solution will deplete ...


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