18

Does folic acid contain a benzyl or a phenyl group? This is the question asked in the title. At the first glance to the structure, one would say folic acid consists of phenyl function but not benzyl function because the question did not define what is the phenyl group. In reality, phenyl group $(\ce{Ph})$ is $\ce{C6H5}$. As Poutnik pointed out in the ...


16

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile. In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. ...


15

During a electrophilic aromatic substitution, it is always possible to have multiple substitutions during one reaction. However, your example is not the ideal one for a discussion. As noted, Oscar Lanzi has questioned even aromatic ring of nitrobenzene is active enough to give even one $\ce{Br}$ substitution. To clear that, I have found a reliable reference: ...


14

You wanted to measure the distance "across tropolone" but you didn't specify from where to where. The distance from a remote hydrogen to each of the oxygens and to the hydroxyl hydrogen are shown below. You can do the measurements you want yourself here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ESPotential/ESPmisc2.html


12

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated. The mechanism (on paper) seems to be as follows Dehydration $1$ For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows: Dehydration $2$ Similar to the ...


12

According to ChemDoodle's ruler tool, this molecule is probably about 5.35 Angstroms across. I'm a little annoyed that the font doesn't seem to have a proper Angstrom symbol. The symbol is part of the symbol selection widget, but doesn't get used by the ruler tool. Oh well, still cheaper than ChemDraw. EDIT: The angstrom symbol issue is font related. If I ...


11

Initially one would assume that cumene (isopropyl benzene) would be the major product. However, kinetic control does have a say in this reaction, according to Gilman and Means.1 At - 2°, n-propyl bromide with benzene and aluminum chloride gave n-propylbenzene, identified as its sulfonamide. Genvresse (6) obtained both n-propylbenzene and isopropylbenzene by ...


11

It turns out that the opposite of what happens in polar solvents takes place when a non-polar solvent is used. At the same temperature, o-nitrophenol is more soluble in benzene than it's m and p isomers. Sidgwick et al.1 did a study of this and obtained the following results. (Note that the solvent they used was toluene and not benzene, but they are similar ...


11

It is important to remember that the purine scaffold is one of the most fundamental organic structures in all lifeforms. From an evolutionary standpoint, this means that there must have been a simple, effective way for this scaffold to be prepared in the pre-biotic earth. After all, the first organisms to contain purine scaffolds did not have "access&...


9

The correct order is in fact $(\bf{d}) \gt (\bf{b}) \gt (\bf{c}) \gt (\bf{a})$, the reason for this is as follows. $(\bf{a})$ is definitely last in this order of basicity since its lone pair is delocalised by the phenyl ring. Now we need to compare the other three. We can split this into two parts, a comparison between $(\bf{d})$ and $(\bf{b})$ $(\bf{b})$ ...


9

It is complicated, as this paper from Olah et al. here shows, but this passage offers a possible explanation: With the reactive nitronium salts, the isomer distribution of the nitration of anisole shows the highest ortho/para ratio (2.7-2.4), reflecting the "early" (i.e., starting aromatic-like) nature of the transition state of highest energy. ...


8

Benzocyclobutadiene has an aromatic benzene ring as well as an anti-aromatic cyclobutadiene ring. This gives it the characteristics of both aromatic and anti-aromatic compounds. Now, according to Cyclobutadiene and its related compounds, [1,p 180]: Benzocyclobutadiene, the monobenzo derivative of cyclobutadiene, is intermediate in structure between the ...


8

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4. There are two factors to ...


8

This is a question based on two reactions. The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]: A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]: Reference ...


8

Nothing against Safdar, but this is technically not a two-step reaction. When lithium amalgam is added to 2-bromofluorobenzene in presence of furan, the reaction goes directly to the product, a Deals-Alder adduct: However, in the absence of furan, the reaction proceeds to give biphenylene and triphenylene (Ref.1&2). The tentative reaction mechanism can ...


8

A quick way to do this would be to enter the molecule in name in the search form on ChemSpider (maintained by the UK Royal Society of Chemistry). Select the 3D representation by hitting the 3D button. In the JSMol window, you can select an atom by double clicking it. To get the largest atom-to-atom distance, I've selected the hydroxyl hydrogen. By moving the ...


8

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 compared 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 ...


8

Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861: On heating, terephthalic acid sublimes without previously fusing(melting). The sublimate, which is indistinctly crystalline, has the same composition and properties as the original acid, and therefore, unlike other bibasic acids, ...


7

I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some ...


7

The nitrophenols have completely different physical behavior based on the position of nitro group: $$ \begin{array}{c|ccc} \hline \text{Compound} & \text{Melting point} & \text{Boiling point} & \text{Water solubility at } \pu{25 ^\circ C}\\ \hline \text{2-Nitrophenol} & \pu{43-45 ^\circ C} & \pu{215 ^\circ C} & \pu{2 g/L} \\ \text{...


7

This is an another way to answer the question: On the way, I also want to show OP that why central ring bearing $\ce{CF3}$ group has the highest electron deficiency at ortho- and para-positions (meaning, the highest electron density is at meta-position in this ring). Let's look at the nitration of trifluoromethylbenzene: According to this University of ...


7

I think the third option as greatest number of resonance structures can be alluded to it. The following may suffice: In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it. In the second one the lone pair on oxygen, the ...


7

Chiavarino, et al. [1] report that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen. Nucleophiles such as methanol prefer boron. We can explain that result in terms of both molecular orbitals and the Wheland intermediate. In the molecular orbital explanation, recall the familiar ...


7

Since we're comparing: molden has a zmat builder and a force field (amber) implemented. You can start with the obvious choices of getting a structure from a database, or generating a starting structure with Open Babel, or simply click it together. The result will come to something like this (I admit this is even less prettier than VMD, but you can use some ...


7

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions. The reaction conditions in the ...


7

There are three common classes of methods: Separate the components of the mixture, then detect the amounts of the substances. Use a fancy method that can identify multiple substances in a mixture (such as training a dog to smell different substances, or nuclear magnetic resonance) Use a method that detects only your substance of interest, even when in a ...


7

This can be explained by comparing the stability of the carbocations formed in Friedel-Crafts acylation and Friedel-Crafts alkylation using chloroacetylchloride. This is because the rate determining step in both reactions is the formation of the carbocation. As you can see, the acyl carbocation is resonance stabilized, whereas the carbocation formed in the ...


6

Aniruddha Deb gave an excellent answer to your question. Yet, there is another important point of differences on two Friedel-Craft processes: Alkylation vs acylation. I admit that rearrangement is the major drawback on alkylation process (using alkyl halides with a catalyst). However, another fact is you cannot stop the reaction after monoalkylation. The ...


6

Usually the designation of a solvent as aprotic is based on the unmodified solvent molecule, which in the case of pyridine is aprotic. Pyridine is not unique in terms of this question. We would generally expect polar aprotic solvents such as DMSO to accept a proton at their negative charge centers (such as the oxygen in DMSO) given a suitable acid. Such a ...


6

Another nice free (but limited in function) tool for this job is NIH PubChem's viewer: The computed distance (not clearly visible in the figure) is $\pu{6.10 Å}$ The structure is also from PubChem: https://pubchem.ncbi.nlm.nih.gov/compound/Tropolone#section=3D-Conformer You can alternately process the coordinates with a tool such as Matlab/Octave: data=[-2....


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