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20

This question raises more important issues than just the technical "why methyl ester," so I'll address those too. The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. This report says it is approximately $10\times$ less sweet (see Table VI on page 2689 and the entry '$\ce{Asp-Phe-OEt}$' with "$++$" ...


16

Dissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex. Why many esters smell fruity while most thiols smell rather unpleasant ...


15

Overview The only interesting thing about lactones is that if the ring size is relatively small, they necessarily adopt the (E) conformation. Therefore, the question essentially boils down to: for a generic ester, why is the (E)-conformer 1 more acidic than the (Z)-conformer 2? This is a question that is easier to investigate, since we remove all other ...


14

Aspartame’s sweetening properties were discovered by accident. the G.D. Searle company was trying to develop drugs to treat ulcers in the 1960s when one of the scientists accidentally contaminated his finger and licked it. Knowing how many versions of molecules get tested in drug development, my guess is that an ethyl ester version was made, and probably ...


12

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are usually formed by placing the alcoholic component in front of the name as a separate word. The particular subsections concerning esters from a single alcoholic component and multiple acid components read ...


11

The most likely reason to use methanol instead of ethanol is cost. [1] Ethanol is roughly twice as expensive as methanol. Although both aspartame and the ethanol analog can be produced with reasonably high yield[2], the ethanol analog has not yet been produced commercially, so it would be even more expensive to develop industrial scale production of the ...


11

Yes, it is possible. One reagent to do so is acidic bromate salts. This paper discusses a number of optimizations and characterizations of the bromate-powered ether oxidation reaction. $$\ce{R-CH2-O-CH2-R + BrO3- ->[H+][H2O; {r.t.}] R-CO-O-CH2-R + Br^-}$$ Hydrolysis of the formed ester leads to an undesired byproduct: the free acid $\ce{RCOOH}$. In ...


11

Surely the negatively charged oxygen would protonate before the other oxygen would. Most of the time, but not all of the time. The amine group is more basic than the "alcohol group" that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)? Most of the time, but not all of the time. Your ...


10

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$) $$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} & \text{Bond} & \text{Enthalpy} \\ \hline \ce{C-C} & 350 & \ce{Si-Si} & 226\\ \ce{N-N} & 163 & \ce{P-P} & 201\\ \ce{O-O} & 146 & \ce{S-S} & 226\\ \ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\ \...


8

The term aromaticity originated with the discovery of unusually stable hydrocarbons that also happened to have strong smells. Many hydrocarbons smell, but not all are aromatic. Nowadays, a compound being classified as "aromatic" has little to nothing to do with its smell and everything to do with its electron configuration. There are also non-aromatic ...


8

Double alkylations of the $\ce{CH2}$ centre of malonic esters do work! Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23 EDIT 1 The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to ...


8

Understanding how this reaction is actually performed in the lab is key to answering your question. Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place. (image source) Also, keep in mind that both ...


8

The first four steps (i.e. the main reaction) are conducted under basic conditions due to the presence of alkoxide. When that reaction is complete, the result is the deprotonated ketoester--the formation of that stable species is what drives the reaction to completion. Note that the first 3 steps are shown as equilibria, whereas the fourth step is ...


7

The IUPAC recommends s-cis and s-trans for the rotamers of conjugated dienes. The nomenclature derives from having a "cis-like" or "trans-like" geometry about a sigma bond. I feel like something similar should be appropriate, however that same IUPAC link recommends E/Z or sp/ap (syn-periplanar/anti-periplanar) for the N-alkyl amides (5,6). Amide 5 ...


7

A methyl cation definitely won’t leave by itself. However, you can consider bromide ions floating around in solution. These bromide, being nucleophilic can attack the methyl group in an $\mathrm{S_N2}$ manner, because any positively charged oxygen is a good leaving group. Therefore, the side product would be bromomethane. Other than that, your mechanism is ...


7

Prediction software always has its limitations, and there is always a degree of error in the calculation. For the ChemDraw predictions, you will see that for the 3 aromatic environments, it has done 3 independent calculations, and has happened to arrive at the same chemical shift. This simply means that these shifts are coincident, not equivalent. Remember, ...


7

To answer this,Think about how an ester is formed. In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$ $\ce{RCOOH + R'OH -> RCOOR' + H2O}$ Now what we have found by replacing the oxygen with an isotope of oxygen is that $\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$ What this reveals is that the acid ...


7

The esterification is not the only thing catalysed by sulphuric acid. Basically, it’s catalytic activity is protonation of whatever feels happy enough to be protonated. In your desired esterification mechanism, we want to protonate the acid: $$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=\overset{+}{O}H)-OH + HSO4-}\tag{1}$$ This will ease the nucleophilic attack ...


7

Amide analogue of lactons are lactams, their tautomeric forms are lactims. Citing from the IUPAC Nomenclature of Organic Chemistry (Preferred names 2013): P-66.1.5.1 Lactams and lactims Intramolecular amides of amino carboxylic acids, $\ce{-CO-NH\bond{-}}$, are called ‘lactams’ and their tautomers, $\ce{-C(OH)=N\bond{-}}$, are ‘lactims’. Lactams ...


6

Option 2 generally gives higher yields. This is because option 1 (Fischer esterification) usually results in an equilibrium mixture which contains significant amounts of products and reactants, limiting the yield. This is mitigated by using a stoichiometric amount of a concentrated acid catalyst, such as sulfuric acid, which is also a strong dehydrating ...


6

You are getting too specific as though these are biological processes. They do not have to be... ANY ester made is an esterification reaction. So any ester can be transesterified into a different alkyl alkanoate and any carboxylic acid or derivative (even nitriles) can be esterified. See the diagram:


6

It's unnecessary. In the reaction you are suggesting, the enolate would have to attack the carbonyl double bond from above. The optimum angle of attack is roughly $107^\circ$ but since both the attacking enolate and the diester substrate are reasonably bulky, there I suspect there will be a considerable amount of steric hindrance here, creating a large ...


6

I doubt your book is asking you to convert this molecule into a lactone. This would be very hard to do. It could, however, be easily converted into a lactam. 5 and 6 membered lactams are formed by simple dehydration. For your γ-lactam, this can be achieved by heating the amino carboxylic acid, forming the thermodynamically favourable, stable 6-membered ...


5

The biodiesel process should work well, you will need just slightly more aggressive conditions (i.e. slightly more catalyst, slightly higher EtOH excess, and/or slightly longer time) to make ethyl esters instead of methyl esters, as ethanol is slightly less reactive than methanol. For skin care, you will need higher purity than biodiesel processes usually ...


5

Traditional malonic ester synthesis post-processing includes hot acidic hydrolysis, which should lead to substituted malonic acid. However, such acids are prone to decarboxylation, leading to tertiary carbon. In case you deviate from the tradition, there is no problem with getting quaternary carbon (though general bulkiness still applies and may prevent ...


5

Try to understand it in a layman's way, greater the extent of overlap between the atoms, the closer their nucleus will be. The stability of the molecule will thus increase as stability of a molecule is inversely proportional to the bond length of the atoms in the molecule. However this is applicable only upto some extent like your situation but if the ...


5

Both methods are used in practice (and a a lot more exist). Diazomethane is a last resort method when everything else fails. For example you might have a molecule that is acid sensitive or heat sensitive and then $\ce {MeOH/H^+}$ is out of the question. Although diazomethane is an excellent reagent to do the job, as it works selectively with almost every ...


5

Reaction of a Grignard reagent with an ester is a standard method for producing tertiary alcohols where at least two of the substituents (the "$\small\ce{R_2}$" group attached to the Grignard) are the same. Judicious choice of the starting ester allows for the preparation of a tertiary alcohol where all 3 substituents are the same. Alternately, you could ...


5

The individual parts of the proposed name are correct; however, their order of citation in the name is wrong. The name of the parent hydride is ‘bicyclo[2.2.1]hept-2-ene’. Note that the ending ‘ene’ receives the lowest locant possible (here: ‘2’ since the bicyclic ring system is numbered starting with one of the bridgeheads). The principal characteristic ...


5

The Oxford dictionary is correct. A nitro compound is defined by the $\ce{-NO2}$ functional group, but only if that group is attached to a carbon atom. Once an additional oxygen is added, the molecule is basically inorganic nitrate and thus belongs to a wildly different compound class. You think that is confusing? Well, an alcohol is defined by a hydroxy ...


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