# Tag Info

## New answers tagged organic-chemistry

3

Yes, such a reaction exists. In fact, it can not only confirm the presence of ring forms in solution but also how prevalent they are. In 1962, Walker Jr, Gee and McCready published Complete Methylation of Reducing Carbohydrates, in which they subject solutions of sugars in dry DMF to silver(I) oxide and methyl iodide. The reaction proceeds according to ...

0

I tell a true story about myself. In my senior year of college, I took as an elective, a course in Pre-engineering Physics. At that point in my studies, I had already taken a lot of math (Advanced Calculus, Linear Algebra, Statistics, Theory of Differential Equations,..., and even graduate-level credit in Numerical Analysis, Numerical Methods,...). To my ...

0

More simply, starting with Cl2 + NaOH implies: Cl2 + NaOH = NaCl + NaOCl with the above reaction being the basis for the commercial production of aqueous chlorine bleach. So, equivalently, we should be able to start with commercial bleach (which is actually not pure NaOCl, but sodium hypochlorite plus NaCl). Interestingly, if there is an electrochemical ...

0

Now, a reported reaction, to quote: "CCl4 + 6 NaOH → Na2CO3 + 4 NaCl + 3 H2O Carbon tetrachloride reacts with sodium hydroxide to produce sodium carbonate, sodium chloride and water. Sodium hydroxide - concentrated solution. The reaction takes place in a boiling solution." Source: https://chemiday.com/en/reaction/3-1-0-3670 As such, I would similarly ...

5

In their 2013 Annual Report titled Evolving to a Specialty Care Biopharma Company, a well-known pharmaceutical company, Bristol-Myers Squibb has introduced the following statement under the title, Developing Innovative Drug Platforms: As R&D evolves its focus, it is also investing in technology platforms that concentrate on new ways to affect disease ...

-2

In this reaction if peroxide is not present then reaction follows markovnikov addition in which carbonation is formed as intermediate. By stability due to resonance and 2° carbonation 5-methylhexa-1-ene cation with C+ at 3 is formed. Hence 3-bromo-5-methylhexa-1-ene is formed as product.

7

You are correct if this epoxide is a alkyl peroxide. However, this is a 2-phenyloxirane (styrene oxide or phenylethylene oxide), which can stabilize a protonated intermediate by resonance, specifically with 2-hydroxyphenyl group. Thus, it can isomerize to phenylacetaldehyde through 1,2-hydride shift (similar to pinacol-pinacolone rearrangement). The ...

4

I disagree with the idea to forget about Lewis formulae because of their age. If you browse across the table of contents of scientific journals, however, you see Lewis formualae are widely used. From the first December issue of Organic Letters, for example is taken the following screen photo: If Lewis formulae were so off and wrong, chemists would use ...

4

Wikipedia reports that calculations show orthocarbonic acid spontaneously decomposes to the ordinary (meta)carbonic acid $\ce{H2CO3}$ plus water. However, orthocarbonate esters $\ce{C(OR)4}$ do exist. The orthoacid may, however, become stable at high pressure [1, with a nontechnical summary given in 2] and possibly exist within Uranus and Neptune. It is ...

0

You could try reacting the mixture with excess calcium hydroxide to precipitate insoluble calcium citrate/tartarate. You could then filter off this mixture to obtain calcium citrate/tartarate. Reacting these with just enough sulfuric acid till the original pH is reached would give you citric acid/tartaric acid (and calcium sulfate which can be filtered ...

1

I am afraid the measurement of the pH at half titration does not give a useful information on the pKa. There has been a detailed series of measurements of the pH at half titration for different concentrations of phosphoric acid. See Cecile Canlet, BUP 938, 1, p. 1129, November 2011. For the first H atom, the theoretical pKa should be 2.12. But the pH ...

3

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off. If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving ...

4

From the article by van Asselt and van Krevelen [1]: Cyclohexanol is oxidized to cyclohexanone. The cyclohexanone reacts with the nitrous acid to yield isonitrosocyclohexanone producing 2-nitro-2-nitrosocyclohexanone by reaction with nitric acid. With attendant ring opening and uptake of water the latter compound is transformed into 6-hydroxyimino-6-nitro ...

0

I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.

-1

The IUPAC name is 3-carbamoylcyclopentane-1-carboxylic acid. Its CAS registry numbers are 19042-33-2. This compound doesn't have a trade name. You can find physical properties of the molecule in Pubchem and Chemspider by this information.

7

Your first and third structures are good but the arrow pushing in the second structure doesn't get the job done. As you have noted, diol 1 forms carbocation 2a the more stable of the two options. Now the lefthand cyclopropane bond with its pair of electrons migrates to the cationic center to form resonance-stabilized species 2b and on to ketone 3. This type ...

3

@Jan has provided a well-reasoned explanation for the OP's question regarding the Cannizzaro reaction. However, the transition state provided in Figure 1 is not in accord with the experimental results. In 1956 Hauser, et al.1 demonstrated that benzaldehyde-d1 (1-d1) in the presence of aqueous KOH efficiently affords benzyl alcohol-d2 (3-d2). On the other ...

4

In the Org. Syn. link you provided phthalic anhydride 1 and p-chlorophenol 2 in 95% H2SO4 and boric acid at 200oC undergo condensation and cyclization. Phthalic anhydride 1 under acid catalysis reacts with p-chlorophenol 2 at the ortho position as shown below to form ketoacid 3. Chlorine is a weak o,p-director and boric acid, I believe, is complexing with ...

2

Bond strength (bond dissociation energy, BDE) refers to homolytic cleavage of a bond (C-H in this case). Heterolytic cleavage is measured by pKa. The two processes are the numerical inverse of one another. In the case of acetylene with sp-hybridization (50% s-character) of carbon, the electron density of the C-H bond lies closer to carbon when compared to ...

2

Your benzyl acetal groups should be stable to all varieties of hydride reducing agents. For the enoate reduction I suggest Dibal-H or LiBH4/Et2O. I don't have references for this as I no longer have Scifinder access, and Google is not finding me any references that are not behind paywalls.

7

The first thing to point out is that the products of the Cannizzaro reaction do not necessarily correspond to the most stable products, i.e. the reaction is not necessarily thermodynamically controlled. Instead, one should consider the transition state—a six-membered ring as taught generally in organic chemistry—and note that after the reaction the compound ...

-1

The solution given by user16347 seems logical but is not really the correct solution. According to the answer the anti form is more stable since it has the least energy, this is actually false. Experimentally, the gauche form with 3 gauche interactions ( that is structure with energy 2.7 ) is more stable than the anti form. The reason being that in the anti ...

4

You have to evaluate shortcomings as compared to something. You can make a list and say "but InChI doesn't represent X" where X is ranging around mixtures Markush groups polymers from repeat units reactions toasters The point is: so what? All representations have some limitations. But limitations aren't shortcomings unless you have a different ...

2

The group -COO- is called "carboxylate" if it is included in a molecule with some organic radicals attached on its left and on its right-hand side. It is a definition, due to the fact that there is a carbon atom bound to an oxygen atom. Joined together, it makes carboxy. This terminology is used when no simpler way of naming this structure exists. Examples. ...

3

If you look at complete mechanism, this wasn't hard to understand why only oxygen get protonated instead of readily available nitrogen, which is supposedly more nucleophilic than oxygen. OP has given N-phenylnitrosamine as the direct product from reaction between aniline and nitrosonium ion, avoiding the first intermediate I (see following reaction mechanism ...

4

Iodide ion is recycled in the system. $\ce{KI}$ is quite soluble in acetone $(\pu{1.31 g}/\pu{100 ml}),$ $\ce{KCl}$ is essentially insoluble. You are quite correct that the first reaction is $\ce{CN-}$ displacing iodide giving $\ce{Cl-CH2-CH2-CH2-CN}$ and $\ce{KI}.$ The soluble $\ce{I-}$ will then do a nucleophilic attack on the $\ce{CH2-Cl}$ to give $\ce{... 7 The vapour pressure can be estimated by equating the rate of leaving the liquid surface with that returning to it from the vapour. This leads to an expression of the form$\displaystyle p\sim n_{l}RTe^{-\Delta E/RT}$where$\Delta E$is the energy needed to remove a molecule from the liquid to the vapour phase and$n_{l}$is the number density (number / unit ... 0 According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids 1 Don't count bonds. Count electrons. Copper has$11$valence electrons as a neutral atom, as the$d$orbitals are able to participate in reactions. In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets. But then the ... 1 A reading of the Corey-Schmidt1 paper on the oxidation of non-conjugated primary alcohols and aldehydes to the corresponding carboxylic acids with dipyridinium dichromate (PDC) in dimethylformamide (DMF) makes no explicit mention of drying DMF but flame-dried glassware is employed. Perhaps this implies that the solvent was dried. @Ringo has speculated (Ringo'... 1 The dye has probably migrated a bit into your ball`s surface, so whatever (solvent) you try to get it out with again, it might take a while. Remember the classic school (chromatography) experiment: You draw a line with a felt pen on a piece of paper, 2 cm from the edge, and stick the paper into a glass filled only 1cm high with acetone. You can try to ... 0 The following three products wil be formed: Your book may be giving the answer to the number of monochloro products formed, as there are only two distinct monochloro products formed on chlorination of isobutane. 1 The initial chemiluminescent (CL) species transfers its excited state energy to the dye. This depends on the overlap of the CL emission spectrum with the absorption spectrum of the dye and their spatial separation, i.e. better spectral overlap and higher concentration more transfer and more dye emission. (see Forster energy transfer for mechanism). The dyes ... 7 My answer is intended to support Waylander's suggestion elsewhere. I thought it'd be better to give OP some insight of this reduction mentioned (Ref.1). To my knowledge, after doing thorough literature search, the aromatic nitro group reduction by$\ce{SnCl2.2H2O}$is the best method so far in the presence of other sensitive groups such as nitrile group on ... 5$\ce{SnCl2.2H2O}$in refluxing$\ce{EtOAc}$as reported by Bellamy & Ou here will do this very cleanly. Work up into ice then add concentrated$\ce{NaOH}$until you get phase separation. I have done this transformation. 5 Ludwig Gattermann's The Practical Methods of Organic Chemistry (translated into English, 1896) has two references to reduction of aromatics with zinc dust: The first one refers to Liebigs Annalen, issue 140, page 295 (1866): Ueber die Reduction aromatischer Verbindungen mittelst Zinkstaub, Adolf Baeyer This roughly translates to "On the reduction of ... 3 This is a variant of the Mannich reaction. The methoxymethyl groups on the amine are a form of captured Mannich intermediates or masked formaldehyde equivalents. The Lewis acid unmasks them by binding to the oxygen of the methoxy group, creating the functional equivalent of H2C=N+. These are reactive electrophiles which react with the active methylenes of ... 1 The acid will protonate the O of the carbonyl carbon, so a positive charge is generated. The oxygen of the alcohol with the free pair of electrons will act as nucleophile. So the alcohol is added to the cycle and a pair of electrons of the double bond of the carbonyl will compensate the positive charge generated. As the medium is acid, in the second step the ... 3 Actually one s orbital and three p "mix" together, hybridize to form four sp3 orbitals. Maybe this could help: -1 The 2s and 2p orbitals are close to one another in energy. In the case of a single hydrogen atom, these states are in fact the same in energy, although you would not see a hydrogen atom with an electron in the 2p orbital unless it had been excited by light or a collison. Because these orbitals are close in energy, the hybridization concept promotes the ... 0 Honestly, leaving groups usually do not determine whether it is syn or ante elimination. Hence, although the leaving group is an amine, it does not change how you view the problem. What determines syn or ante elimination is the type of reaction occurring, whether it is E1, E2 or E1cB. E2 would always go through an anti elimination due to the requirement of ... 0 Starch makes a glue or a paste with water. Humidity from the air may be sufficient to transform your starch into a glue. Water produces strong hydrogen bridges between starch molecules. This transforms a pile of independent starch molecules into a strong block of interconnected starch molecules. 3 Most people would call it vinyl acetate. If you want the systematic name, then it is ethenyl ethanoate. It is formally derived from "vinyl alcohol" and acetic acid, from which the name follows easily. See here (https://en.wikipedia.org/wiki/Ester#IUPAC_nomenclature), for example. The compound is https://en.wikipedia.org/wiki/Vinyl_acetate. 0 The reason behind this is that in the 3rd step the electron of the C -R bond is breaking and electron is going to each of them but if the R is tertiary then due to the positive inductive effect the electron density gets high in the R and as a result of this it becomes difficult for the new electron to hold by R. In case of primary there is no such a type ... 4 Chiral amino alcohols such as L-valinol are generally prepared by the reduction of corresponding α-amino acids and suitable reducing reagent. Most commonly used reducing agent is$\ce{LiAlH4}$since most common reducing reagent,$\ce{NaBH4}$does not reduce carboxylic acid to corresponding alcohols. However, use of$\ce{LiAlH4}$has notable disadvantages ... 0 Notwithstanding the convention that you should offer you ideas before receiving an answer here is a viable route. Step 1 - hydroborate the alkene then oxidise the resulting alkylborane with N-Methyl Morphiline-N-Oxide/Tetrapropyl Ammonium Perruthenate to give the aldehyde as described here Step 2 - Wittig reaction with Methyltriphenylphosphonium Iodide/LDA ... 4 In this answer a hydrocarbon anion called Kuhn's anion has the formula$\ce{C_{67}H_{39}^-}$has reported$pK_b=8.1$which would correspond to the neutral hydrocarbon having$pK_a=5.9\$. The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem....

3

This doesn't qualify as racemisation, it is epimerization. Racemization would require both stereocentres to invert to form the enantiomer and - as you rightly suppose - the stereocentre bearing the methyl group is untouched by this reaction. It is also unlikely to lead to a 1:1 mixture of products since the two diastereomers (and the transition states ...

1

It results in racemisation if only one chiral carbon atom is present in the compound. If there are many, it merely leads to formation of compounds with both R and S configurations of the chiral carbon on which substitution takes place.

4

As a practical matter, vitamins generally can't be provided to consumers in their pure forms. The problem is that we need so little of them. For instance, the RDA for d-alpha tocopherol is 15 mg. Consumers simply don't have the training or equipment necessary to precisely measure out 15 mg (or the time or patience to do this daily!). Most likely they would ...

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