# Tag Info

6

It isn't just both bromine atoms that are lost, although that might be expected in mass spec given the relative weakness of the carbon-bromine bond. You also lose a hydrogen atom. And that is key. We might suppose that a single bromide ion (or bromine atom plus electron) comes off first, forming a $\ce{C4H8Br^+}$ ion. That is the peaks at 135 and 137, ...

3

What is the endo-selectivity of Diels-Alder reactions? If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly ...

2

May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes. image and further information here

0

I have been working on a similar project (VirtualChemist), which is an organic reaction software focusing on regiochemistry and stereochemistry. Using reactant Lewis structure as the only user input, it can (1) describe detailed reaction mechanisms at the level of electron flows in elementary reaction steps, ensuring that all reaction steps are not only atom ...

1

In fact, yes. It can be explained by a HOMO-LUMO interaction. I am currently running NBO (natural bond orbital) computations for a theoretical study of hydrogen bonds in some models. And NBO interprets hydrogen bond as the donation of electrons from the lone pair (non-bonding electrons) of the hydrogen acceptor to the (LUMO) antibonding orbital of the ...

0

First of all, a molecule only has one HOMO and LUMO. Don't confuse MOs with localized orbitals from VB theory. You can basically approximate this mechanism with four MOs: LUMO+1: Mainly $\pi^*_{C=C}$ with contributions from $\sigma^*_{C-H}$ LUMO: Mainly $\pi^*_{C=O}$ HOMO: Mainly $\pi_{C=C}$ with contributions from $\sigma_{C-H}$ HOMO-1: Mainly $p_O$ with ...

0

Strictly same retention time means exact same signal in GC (i.e."apparently" same compound). This means you'd have one peak for the mixture as if it was a single compound. On an unknown sample, you might miss the fact that there are two compounds coeluting. To avoid the issue, you would always inject the sample on (at least) two columns with ...

-1

According to a 1951 article in J. Am. Chem. Soc., water is soluble in diethyl ether to the extent of 1.468%, although other investigators had found between 1.19 and 2.66%. https://pubs.acs.org/doi/abs/10.1021/ja01150a531 And the solubility of ether in water is about the same. Water and glycerol are completely miscible - no insolubility at all. According to ...

4

Adderall Adderall is a drug(trade name) that contains a combination of four salts of amphetamine (equal parts racemic amphetamine and dextroamphetamine, which produces a (3:1) ratio between dextroamphetamine and levoamphetamine, the two enantiomers of amphetamine). Racemic amphetamine: levoamphetamine and dextroamphetamine respectively. Notice the dash-wedge ...

1

At this point, I would just go and finish burning that char in air. You need air to complete the combustion, don't do it under vacuum. 300-400 C should do the trick. Although, if you are very concerned about the temperature, just do a TGA to the crystal bond under air and determine the temperature where it starts to combust; then apply that temperature to ...

-1

The difference is that it is not the same molecule (group $\ce {NH_2}$ for one, group $\ce {NH-CH_3}$ for the other) so not the same properties.

0

This is straightforward. 4-Toluenesulfonyl Chloride is readily attacked by nucleophiles and allylamine is a good nucleophile. Dissolve 2.05 equivalents of allylamine (or 1.05 eq allylamine and 1 eq of an tertiary organic base such as triethylamine) in dichloromethane, cool in an ice/water bath (reaction is exothermic) then slowly add 1 eq of 4-...

4

Have a look at this one from reference (1) [my highlight]: In the simple case, the use of retention data for identification purposes involves merely a comparison between the retention time of an unknown and that of a series of authentic compounds. Since the essential ability of chromatography is to separate substances, a negative conclusion - these two ...

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] Complete overlap (= identical retention) is a regular feature of experimental chromatography with very different compounds and complex mixtures. It just happens when the selectivity happens to be equal to 1.00. You can have one, two, or three different compounds hidden under a single peak and this usually happens in early eluting peaks in all modes of ...

1

Crystalbond™ 555-HMP is an adhesive. The SDS lists only hazardous ingredients, none of which should decompose at 150$^o$C to char. The char is probably from the non-hazardous ingredients, probably something like a hot-melt adhesive. The adhesive char now bonded to the SiO$_2$ layer probably involves carboxyl oxygens, and if the char is not removed by the ...

1

There isn't a good overall formula that I know. You can look at C/H ratio and easily tell a bit here. There must be at least one of the following, but not two: carbon ring carbon double Bond (1) 1-Butene (2) 2-Butene which has two stereoisomers cis-2-Butene trans-2-Butene (3) Isobutylene (4) cyclobutane (5) methylcyclopropane

1

Just a quick answer without looking too much into it. You have been using a lot of polar solvents without any luck. The reason is that the molecule no longer has the groups that make it water-soluble because of the thermal decomposition. Only hydrocarbons chain remain. You need to use an organic solvent (toluene, benzene, hexane). Since you probably don't ...

0

The simple rule is that if there are two chains of equal sizes, then the chain with more number of side chains is to be selected as part of the parent chain... Here, the two chain are $\ce{-CH2CH(CH2CH3)CH2CH3}$ and $\ce{-CH2C(CH3)2CH2CH3}$ . The second one has two side chains ($\ce{-H3C}$ and $\ce{-CH3}$) while the first one has only one side chain ($\ce{-... 1 Some hints are provided by Wikipedia:Lemon liqueur: To produce the Lemon liqueur requires sugar, water, lemon zest, mixing in organic salutes liquor, and time to mature. Lemon zest is soaked in high proof neutral spirits to extract from it the lemon oil (an essential oil). The extraction is then diluted with simple syrup. In other words, lemon zest is ... 4 Chlorotrimethylsilane, like other organosilicon compounds, is easily attacked by nucleophiles and readily hydrolyzes in air yielding hydrogen chloride: $$\ce{2 Me3SiCl(l) + H2O(l) -> Me3Si-O-SiMe3(l) + 2 HCl(g)}$$ Hydrogen chloride reacting with traces of ammonia$\ce{NH3}$in the air builds up white solid ammonium chloride$\ce{NH4Cl}$depositing on the ... 0 A canonical definition seems hard to find. If you look in the Wikipedia, polar molecules are defined as those with net dipole moments: A polar molecule has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from polar bonds arranged asymmetrically. The closest IUPAC comes to a definition of "... 0 It depends on the reaction, some are more temperature sensitive than others. You might get a slightly reduced yield or, at the other end of the spectrum, a runaway reaction leading to a redecorated fumehood. For the Jacobsen asymmetric epoxidation this quote from the Wikipedia article here seems relevant: The degree of enantioselectivity depends on numerous ... 2 Aminoxyl (trivial name: a nitroxyl radical or a nitroxide) denotes a radical functional group with general structure$\ce{R2N–O^•}$. It is well known that sterically unhindered aminoxyls baring$\alpha$‐hydrogens are unstable and undergo rapid disproportionation to nitrones and hydroxylamines (Ref.1: A new mechanism for the decomposition of$\alpha$‐hydrogen ... 0 it cant possibly cause wood is not a compound. It has many different compounds. A substance burns because it releases vapors which burn, for instance if we can remove those vapor . Therefore burning would not take place. I think then we can melt wood in absence of oxygen then there is a possibility that it can melt otherwise we have to change the composition ... 2 Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets? They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel,... 3 Probably not Part of the problem with the question is geometry. Any two points always form a plane. What I suspect you mean is "is the allyl double bond in the same plane as the benzene ring?" in which case the answer is probably not. To make the situation clearer here is a different numbering of the atoms in the allyl benzene molecule: The ... 3 This reaction is very similar to Reimer-Tiemann reaction which also consists of the reagent$CHCl_{3}$+$NaOH$. The mechanism of that reaction is believed to have dichlorocarbene intermediate too. A good point to remember is that the dichlorocarbene intermediate is often used to form cyclopropane or carbonyl groups. This is the mechanism for the reimer ... 4 The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compare to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the ... 5 Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced$\pi$-electron bonding difference. Less than meets the eye Suppose you were to use ... 2 Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy. I would argue the phenolate ion has five mesomeric structures, two with the charge assigned to oxygen (analogous to the two resonance structures of phenol), and three with the ... 0 The basic idea of geometrical isomerism is the difference between the spatial arrangement of atoms/groups between two compounds which have same molecular formulae and connectivity. One way of judging GI is analysing the difference between the distances of groups. Cumulenes having an odd number of double bonds are capable of showing GI (granted that the two ... 1 The reaction mechanism shown on the Wikipedia page for the Perkin Reaction is perfectly clear. mechanism here The first step is deprotonation of acetic anhydride by acetate anion. The second step is nucleophilic attack by the anion of acetic anhydride on the aromatic aldehyde; thus the aldehyde is the electrophile. 0 I would like to give a very rough answer as to why equivalent resonance structures have a stabilizing effect. To begin, we assume that resonance can be represented as a discrete n-state system. Feynman has used this simple approximation to explain many of the essential features of chemical bonding and resonance in the Feynman Lectures on Physics Vol III, ... 0 Benzylamine is$\ce{C6H5CH2NH2}.$Therefore,$\ce{NH2}$group is bonded to$\ce{C6H5-CH2}$unit which has greater +I effect. Therefore, it increases electron density on$\ce{N}$and makes benzylamine more basic. -1 2-(6-nitrocyclohex-3-en-1-yl)ethanoyl chloride should be the correct name. According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a lower locant should be assigned to the substituent only written as a prefix (i.e. the nitro group) than a double bond '-ene' ending, since the unsaturation is considered to ... 4 The Hofmann-Loeffler-Freytag reaction as suggested by @user55119 in the comments above is the way to go. First you need to reduce your nitro group. The reduction of alkyl nitro compounds is less simple than aromatic nitro groups and lower yielding in general. I would suggest the use of Al/Hg amalgam. Acetylate the resulting amine. N-chlorinate the amide - ... 3 I'd like first to answer your question: My question is, when writing the formula for$K_\mathrm{D}$, is the organic phase always in the numerator and the aqueous phase in the denominator? The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook): Partition Ratio ($K_\mathrm{D}$): ... 1 As of Loong's suggestion, I'd put the answer removing it from comment section: The name should be 4,4-dibromo-3,5-dichloro-2,5-difluoro-3,5-diiodo-pentan-2-ol as Maurice pointed out. However, keep in mind that there are 3 chiral centers within the molecule. But, the indicated name is okay since stereochemistry of the molecule is not shown. If$\ce{OH}$group ... 3 The accepted answer is incorrect There are some subtleties in the determination of stereochemistry in this question which need to be carefully explained with reference to the official rules, as laid out in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book). The gist of it is similar to that provided in other sources,... -3 It's wrong. Vinyl loses at position 2 of distance 2 against both isopropyl and sec-butyl and therefore gets priority 3. sec-butyl beats isopropyl at the first position of distance 3 and therefore gets priority 1 thus isopropyl gets priority 2. Phantom atoms are less worth than real atoms but more worth than real atoms with lower atomic number. 3 Let's compare the two compounds. Solubility: Both compounds are miscible with water, so there is no difference. With the straight chain, 1-propanol may be slightly more lipophilic, and thus better able to act as a surfactant. However, 1-propanol is a really poor surfactant. Volatility: Isopropanol has a lower boiling point (82.5 degrees C) than 1-propanol (... 3 You are correct that the nitro group can be displaced by methoxide, this has been reported examples here. My instinct is that the less hindered nitro group is the one that will be displaced. 0 The formation of a distinct carbocation followed by migration of a neighboring group could follow your scenario. Thinking about it tho the nearest Lewis base to the incipient C+ ion is the adjacent group and it seems likely a concerted shift ionization would be a lower energy transition state and your scenario is best described as a$\pu{S_N}2$reaction and ... 1 Overall I agree with Alchimista's answer, though I think it might be quite complex to grasp, seeing the comments that followed it. If it helps, here's my down-to-earth version, which in fact covers chirality in general, not just cyclohexanes. The 'executive summary' of my answer would be: not only it is correct to use an 'unnatural' or 'high energy' ... 2 What you have shown in red is the second propagation step in the benzylic free radical bromination of ethylbenzene. The bond dissociation energy (BDE) of bromine is +46 kcal/mol while the overall heat of reaction of the step is -12 kcal/mol. Thus, the formation of the C-Br bond must be -58 kcal/mol. The BDE for the C-Br bond in (1-bromoethyl)benzene is +58 ... 2 Are amides more acidic than alcohols? This is very broad question. The very short and general answer would be both have similar acidities (as Mithoron pointed out in his comment elsewhere). To support this statement, there is some evident in the Table provided by UMass website that mentioned by Waylander in his comment. For example, the$\mathrm{p}K_\...

3

I am not sure the 3D conformation stills as reported on PubChem are all that accurate. Although crystal structures are sometimes referenced for specific molecules, some of the displays are in fact models, some of which do not make sense to me. Here are some observations regarding the three molecules you noted. 2-Nitrobenzoic acid The PubChem model shows both ...

-4

I personally don't think the other two previous reasons given are correct. I've many a times come across an organic compound for which the nomenclature seems to be defying the longest chain rule. I can't get an example right off my head but I'll edit my answer soon to include an example proving the same. So, the more convincing reason I think is: In this ...

3

Operatively: A plane of symmetry in a 2-D projection of a cyclohexane is a sufficient condition. Alternatively, and lengthy, you conduct an analysis involving conformers but not limited to the flipping of the ring, that means you must let the chloromethyl rotate, in your example. In more details: The answer to your title question is YES providing that the ...

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It's so because while doing IUPAC naming we have to always choose longest branch where the number of functional group (here Amine) should be at least number. That's why we here call it 2-methylpropan-2-amine and not 2,2-dimethylethanamine. You can clearly see in the actual name it's propane which mean 3 carbon atom while as the name you want to give it has ...

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