# Tag Info

2

The egg yolk contains a lot of lecithin, which is a phospholipid with 1 fatty acyl replaced with phosphatidylcholine. It acts like emulsifier, what is used in industrial food production.

6

Yes there are, many in fact. One of the examples is having a complex consisting of a fluorene molecule and of cp altogether with iron, described by Decken in Acta Cryst. E (doi: 10.1107/S1600536804028077): (source) You are not limited to complexation with iron, though. Already by the title «Synthesis and characterisation of Ti, Cr, Mo and W bis(fluorene) ...

0

In free radical substitution, a free radical is formed that is an electron deficient species and is stabilised by electron donating species bonded to atom on whose orbital it is present. So, electron donating groups that is +I ,+M increase chances of formation of free radical on it and presence of electron withdrawing species (-I,-M, Backbonding,by Cl) ...

2

This is to do with the relative stabilities of the free radicals produced. Here is a schematic diagram showing the two possibilities for propene (similar mechanism, also with two free radical possibilities like propane): As you can see, a more stable and less stable free radical is produced. The stability depends on how many carbons are attached to the ...

0

B is the correct answer, but not for the reasons you have given. 4-Me-benzaldehyde cannot self-condense, and is a far more reactive electrophile than acetophenone. Acetophenone has a pKa of around 16 so will readily undergo base-catalysed aldol reaction with the aldehyde. This is the only reaction that will occur. All the other sets of reactants shown are ...

0

For comparing the rate of decarboxylation just compare the stability of the carbanion formed. It's an easy jee concept.

0

Your hypothesis concerning a "bouncing" contribution to the reaction kinetics would be incredibly difficult to observe or quantify even with the best supercomputers. The likelihood of a molecule to react in a manner like the SN2 reaction is actually an incredibly rare event at the molecular time scale (femtoseconds 10^-15 s). This may sound contradictory ...

0

I believe you are misreading the question. Rather than asking for the number of stereoisomers the product can have (this is where 2^n comes in), it is asking for the number of stereoisomeric products formed. Note that that the two products are diastereomers, which is a type of stereoisomeric relationship. It's not looking for any fancy calculation; just for ...

1

Let's focus for now on the two circled crosspeaks. In a COSY sequence, the frequency in the indirect dimension (the vertical axis) corresponds to the spin which the magnetisation is on during the $t_1$ period. The frequency in the direct dimension (horizontal axis) corresponds to the spin which you measure the magnetisation of during $t_2$. It turns out ...

-2

grignard reagent is formed by the reaction of organic halide with magnesium. The reaction is as follows. RX + Mg = RMgX since x is halide so R can not be an elctronegative group. That means the existance of XCN (eg: CNI) is not known

1

The reaction shown here is a Finkelstein reaction and SN2 reaction mechanism is followed.is an organic reaction that uses an alkyl halide exchange into another alkyl halide through a reaction wherein the metal halide salt is used. This reaction takes place at an equilibrium process by taking the advantage of poor acetone solubility in metal halide salt that ...

-1

According to the IUPAC nomenclature of hydrocarbon: Number the carbons of the parent chain from the end that gives the substituents the lowest numbers. When comparing a series of numbers, the series that is the "lowest" is the one that contains the lowest number at the occasion of the first difference. If two or more side chains are in equivalent positions, ...

0

Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers. We know, that 3° carbocation is a more stable intermediate than an allyl ...

0

If you have a basic knowledge of IUPAC nomenclature of hydrocarbons then you should know that we start naming a hydrocarbon by first choosing the longest carbon chain (if the hydrocarbon is a saturated one otherwise we take the chain containing double or single bond) and then by assigning the lowest possible number (which stands for its position ) to the ...

1

The most questions about keto-enol tautomerism can be answered by their NMR spectral data (e.g., evidence of existing enol-form of acetylacetone at room temperature). According to its $\ce{^1H}$-NMR spectral data, it was declared that phloroglucinol, in its neutral form, exists nearly exclusively in its aromatic form (Ref.1). However, their dianion form (e.g....

3

As small as six-membered rings with allenes can exist, albeit fleetingly. Cyclohexa-1,2-diene has been known since a 1966 report by Georg Wittig (Angew. Chem. Int. Ed. 1966, 5 (9), 846), in which it is prepared by direct elimination of HBr from a vinyl bromide using t-BuOK: For a more modern take on the matter, see e.g. Nat. Chem. 2018, 10 (9), 953–960 ...

0

On the subject of 4-pyridone, Wikipedia says this: 4-Pyridone is an organic compound with the formula C 5H 4NH(O). It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol. So as far as I know, the aromatic enol form of this guy is lesser in equilibrium content than that of phloroglucinol. The reason for this ...

0

In case of pholoroglucinol, its ketoform has 3 $\ce{C=O}$ bond which make its keto form more stable as compared to the keto form of 4-pyridone which has to offer only one $\ce{C=O}$.

1

Alkyl halides with bad-leaving groups (like $\ce{F-}$) having EWG at $\ce{\beta}$-position undergoes these reactions, also known as $\ce{E_{1CB}}$, viz. elimination unimolecular conjugate base. Example: aldol condensation photo-induced decarboxylation Reagent: bulky or strong bases or both (like $\ce{K+(^tBu)-, LDA, NaOH, KNH2}$, etc) along with some ...

3

The $\ce{LiAlH4}$ will reduce the cyclic anhydride to the diol. $\ce{KMnO4}$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would ...

4

Consider the resonance structures of the pyranone. MeMgCl attacks the structure shown second. The perchloric acid workup hydrolyses the enol ether to give the diketone shown bottom right. NaOH promotes the internal aldol cyclisation to give 3,5-dimethylphenol.

3

An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin. source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://...

0

1.The selective reduction of the carboxylic group is explained in this article. Link:https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114. It could reduce esters but is not favourable at all and difficult.So if u only have an ester grp and its treated with bh3, then you can reduce it. However if both carboxylic and ester grps are present, carboxylic acid gets ...

0

If hetro atom with lone pair is neighboring to carbocation, then that lone pair can be donated to empty orbital on carbocation (see figure below). These orbitals are said to be in conjugation. Conjugation leads to delocalization of electrons resulting in resonance structures. Quoting from "Organic Chemistry" by T.W. GRAHAM SOLOMONS , CRAIG B. FRYHLE .SCOTT ...

0

I'd try to give my explanation with the principles simple enough so that it would familiar to your level of education: Rahul Verma has given simple enough explanation to answer most part of your question. I said here 'simple enough' because of your subsequent comments. However, I have to point out that mesomeric effect ($M$) almost always predominate ...

3

For preferred IUPAC names, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), explicitly stipulates when locants are omitted. Most of these rules apply to the omission of the locant ‘1’. In particular: P-14.3.4.2 The locant ‘1’ is omitted: (…) (c) in monosubstituted homogeneous ...

1

Forget the Beckmann, the oxime is still nucleophilic through N (imagine the N-hydroxy enamine tautomer if that helps). Have it attack the second carbonyl intramolecularly (and hence favoured) to give a dihydropyridine. Dihydropyridines are unstable with respect to aromatisation unless substituted with strongly electron withdrawing groups. The N-hydroxy-...

1

It is because of the fact that a double bond is not usually stable at the bridge head position in smaller fused rings. Therefore the hydrogen cannot be considered as acidic.

2

How to decide whether +M effect or -I effect will operate in this case? The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$. In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ ...

10

You can find a nice discussion of diethynyl ether and some of its derivatives here: Ooi, I. H., & Smithers, R. H. (1989). Bis [(trimethylsilyl) ethynyl] ether: a moderately stable C4H2O derivative. The Journal of Organic Chemistry, 54(6), 1479-1480. https://pubs.acs.org/doi/pdf/10.1021/jo00267a053 They write that "little or nothing is known about ...

0

"Aromatic" is a qualitative term. "Aromatic" is also ambiguous. Mitheron's reference to the question about "Is 1,2-dihydronaphthalene aromatic?" lists some of the rules that are applied, but the final judge is your teacher: what would the teacher say? Long ago, my professor said that aromaticity was the measure of surprise that an experimenter had when he ...

7

At Last! Success! I found this research paper, which describes a "Self-assembling nanocapsule", which is held together by plenty of hydrogen bonding, and can assemble around a linear molecule of all-anti tetradecane, $\ce{(C14H20)}$ using the energy afforded by the hydrogen bonding to squeeze/twist the alkane into a helical all-gauche conformation. Have a ...

1

The first step is called Michael addition. Have a look at this wikipedia page. The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated ...

-1

Simple method is measure 100 ml Isopropyl alcohol water mixture heat at 80 degrees centigrade and after distilling left of liquid,cool and measure the volume. You get water content and the balance quantity is IPA.///

0

The OH group of phenol is so activating that the nitration mixture rapidly overheats even with dilute nitric acid. The result is oxidation with the evolution of copious amounts of nitrogen oxides. Instead, you heat the phenol with sulfuric acid to get mostly phenol-2,4-disulfonic acid.The sulfonic acid groups are so electron-withdrawing that nitration can ...

5

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate ...

4

Your question is how can the CH2 be split into a dd when the -OCHO shows no triplett, right? Well, the second splitting at the $\ce{CH2}$ is very small, and the $\ce{CHO}$ looks like it is very nearly split. That´s not just a broad peak, it has a rather clear shape imo. If you do just a bit of zero filling and apodisation, the triplett might well (should) ...

0

The apparent answer is yes, to quote from a source (composed from extracts of old chemistry journal): Calcium chloride forms addition compounds with the alcohols. On evaporating a solution in ethyl alcohol at a low temperature rectangular plates of $\ce{2CaCl2.7C2H5OH}$ are deposited. The compounds $\ce{CaCl2.3C2H5OH}$ and $\ce{CaCl2.CH3OH}$ have also ...

1

It is well known fact that branched-chain alkanes are more stable than the straight-chain alkanes with the same number of carbon atoms. That means branched-chain alkanes have higher values of enthalphy of formation ($\Delta H_f^\circ$) than that of straight-chain alkanes with the same number of carbon atoms. During the complete combustion reaction with ...

1

You may wish to give this suggestion a try, courtesy of the literature on constructing nano-particles. First, dissolve the protein samples in say ethanol or other flammable solvents. Next, place it into an alcohol lamp with a wick composed preferably from glass fiber, but one may be able to employ woven cotton or a cotton string. Lastly, light the lamp ...

1

This question has already had some pretty good answers by user55119, James Gaidis and so on. However, since the bounty setter was looking for a definitive answer in pertinence with some tests, and I am quite well-versed with the particular exam he is interested in(i.e. the JEE), I would like to provide my final insights along with a summary of the concerned ...

6

There is no evidence (e.g. aromaticity) to support your claim that the compound is planar. In fact, crystalographic studies [1] revealed existence of both cis- and trans-3,6-dimethylpiperazine-2,5-dione. Corresponding CCDC numbers are LCDMPP01 and TRDMPP01, respectively — feel free to open the links and play around with the 3D structures in JSmol to ...

0

You have to delve into the concept of SIR. Steric Inhibition to Resonance takes place because the groups are so huge that they can not remain in the same plane as the rest of the compound, which implies that they must be in a plane which is at an angle to the molecular plane(the plane where the major portion of the structure is). Hence, while calculating ...

1

HCL is a stronger acid so it will displace the bitartarate forming choline hydrochloride and tartaric acid

3

Naturally occuring amino acid with basic sidechain - L-Arginine: Oral supplementation with L-arginine at doses up to 15 grams daily are generally well tolerated. source here

-1

Some suggestions: First, collect the fingernail samples by applying a nail file to a subject's nail. This creates a fine powder. Second, precisely weigh the sample and place the nail powder in distilled water.[EDIT] Here is a possible dissolution of nail path per a reference that should be verified in its accuracy, which claims: COCO COLA IS CAPABLE OF ...

-2

These analyses are difficult to carry out at a high school level. Nails are not soluble in any solvent. They must be burned and the gases coming out are analyzed and weighed. To burn them, they have to be mixed with a great excess of copper oxide, put in a long horizontal glass tube, heated from below. A current of oxygen gas should be sent through the glass ...

2

The Lemieux-Johnson oxidation (details here) using catalytic $\ce{OsO4}$ or $\ce{RuO4}$ turned over with $\ce{NaIO4}$ will do the transformation you want in one step. These conditions should be selective for the exo double bond.

1

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect. But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation . In ...

0

To confirm, you want to make (R)-4-acetyl-1-methylcyclohex-1-ene? Do you have to start from (S)-limonene? If not, consider the Diels-Alder reaction of isoprene and methyl vinyl ketone (e.g. https://doi.org/10.1039/B701333G , though this is not an enantioselective synthesis I imagine you can add chiral auxiliaries somewhere to increase that selectivity)

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