# Tag Info

5

{OP:] It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions. An enzyme is a catalyst, so it does not change the enthalpy of a reaction. That part is correct. "Oxygen only enables the release of energy" is incorrect. Oxygen is one of the reactants, and the oxygen atoms also ...

1

This isn't best solution to this problem, as its usage would be quite limited, but nonetheless it is still a solution: using enzymes specific to the enantiomer you wish to remove. As an example, using enzymes specific for D-amino-acids would leave you with its L-counterpart untouched. They can then be separated as the "product" form of the D-amino-acid is ...

3

The radical monobromination of bicyclo[2.2.1] heptane could give $\ce{1}$ ,$\ce{2}$ and $\ce{3}$. However , $\ce{2}$ and $\ce{3}$ , are unstable .On observing $\ce{2}$ , planar radical at bridgehead is unstable due to angular strain introduced by planarity of radical at bridgehead . Similarly in $\ce{3}$ , due to angular strain introduced in to the ...

0

Moffat & Pfitzner have accidentally found their famous oxidation while working on mono-nucleotides (acually 5'-phosphates). Breifly, when pyromidium thymidine-5'-O-phosphate was reacted with dicyclohexylcaebodiimide in anhydrous DMSO at room temperature resulted a violent reaction to give 90% of 5'-aldehyde product (Ref.1). They further pursued this ...

2

I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl ...

1

The following schematic (not meant to represent the actual changes in $H$) illustrates the problem: The first step (from leftmost reagents to middle products) represents the formation reaction (not balanced), the second step (from middle to oxidation products on the right) the combustion reaction. It pays to be careful using words such as "greater" or "...

3

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too? Many molecules exist in both forms in nature. One fun example are the enantiomeric terpenoids R-(–)-carvone and S-(+)-carvone. The R-form smells like spearmint while the S-form smells like caraway. The difference in smell shows that properties other than the ...

3

I'd say the suggested reaction of benzene-1,4-diol (hydroquinone) with $\ce{PCl3}$ to get 1,4-dichlorobenzene would not work (see Waylander's comment). However, you can buy 1,4-dichlorobenzene very cheap (e.g., USD 21.50 for $\pu{100 g}$ bottle from Sigma-Aldrich) so why you need this trouble? The suggested reaction of 1,4-dichlorobenzene with $\ce{NH3}$ ...

0

Enzymes are very specific for the reaction to enantiomers. For certain products you have only one enantiomer in nature. It may be the case that there is just one enantiomers which can react with an enzyme because of steric hindrance. But it is also possible that a racemic mixture can occur in fermentation products. So there is no general answer. The ...

-1

(Z) is the same like cis (E) is the same like trans BUT cis and trans is usually only for double bonds with 2 different substituents. Confusion can arise when, for instance, you have 2-halobut-2-enes, because some take the highest priority when determining if this is cis or trans whereas others may take the identical groups. You can use (Z) and (E) for ...

-1

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show geometrical isomerism.

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From Chemgapedia: Most of the aminoacids in nature have different spatial structures. These properties are due to a chiral center (on the C-alpha atom), which has four different substituents. One differentiates a D-form (from Latin dextro = right) in which the amino group of the C-alpha atom in the Fischer projection to the right, and an L-shape (from levo =...

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Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$. $\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure). $\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically active compounds formed after ozonolysis is $\ce{= 1}$.

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The given answer is correct. The product scheme you drew is correct as well. However, as I marked in your scheme (see below), products A and B are essentially the same enanthiomer (both have (2S)-configuration). The compounds you have drown in right-hand side are also an identical compound, which is not optically active. Therefore, ozonolysis has given only ...

2

For the customs of ChemSE, the question is rather broad; perhaps even too broad to provide you specific compositions, times and temperatures to improve your process of producing extrudates. On the other hand, you already collected a number of plausible parameters influencing the target property, which you named as «pill strength». As general picture, you ...

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You do not really need aqueous acid to get the Grignard reagent to react. Rather you need it to convert the product of the Grigalbard reaction to an isolatable form. Say you are reacting methylmagnesium iodide with acetone to get tert-butyl alcohol. The anionic moiety combines with the carbonyl group so you have the required carbon-carbon bond structure. ...

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If the extraction procedure does not involve HBr, then I would doubt it is the hydrobromide because: hydrobromide salts don't dissolve well in acetone; hydrobromide salts are rare in nature. According to the work by Bracci et al. [1], the major component of MeOH extraction of Brugmansia is atropine. Śramska et al. [2] used ethyl acetate extraction and ...

1

https://en.wikipedia.org/wiki/Iodolactonization I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic ...

1

Mesomerism and resonance are identical, completely fungible, used interchangeably. Hückel-Lewis-Projection Method: A “Weights Watcher” for Mesomeric Structures: Y. Carissan, D. Hagebaum-Reignier, N. Goudard, S. Humbel J. Phys. Chem. A. 2008, 112, 13256–13262

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I present another (almost) two-step synthesis, involving the famous Wittig reaction. The first step details the preparation of a phosphonium ylide, which subsequently is allowed to react with the carbonyl compound (benzophenone in our case) to yield the final product, 1,1-diphenyl-1-butene. For mechanistic details, visit NotEvans.'s answer to Which is the ...

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You are forming the urea because there is always an excess of phenylpiperazine present to react with the carbamoyl chloride you are forming. You need to add the phenylpiperazine to the triphosgene. The paper by Varjosaari et al. [1] has a procedure (described on Organic Chemistry Portal) that should work. @Mathew Mahindaratne's suggestion will also work: ...

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Two steps: Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be down in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group. Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq ...

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In nature fat is stored as triglycerides (TGs): Three fatty acid (FA) chains bound together by a glycerol backbone. Olive oil is mainly composed of TGs, though it also contains small amounts of free fatty acids (FFAs). An example of Triglyceride (from Wikipedia) Soap is commonly made via the reaction of triglycerides with a strong base: usually Sodium ...

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The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize. However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, ...

2

$\ce{AlCl3}$ is a Lewis acid and will bind to the oxygen of the ethylene oxide giving a species that functions as $\ce{[R2O-AlCl2]+}.$ This cationic species will be attacked at carbon by the electron-rich aromatic ring in much the same way as a standard Friedel-Crafts reaction intermediate generated from an alkyl halide to give $\ce{[Ph-CH2CH2-OAlCl2]+}.$ ...

3

I see two possible answers. Not (A) and (B) but rather (A) and (D). Chakravarthy Kalyan is responsible for reminding us that (D) can be an answer too (refer to comments). For option (D): $\ce{HC ≡ CH + 2AgNO3 + 2NH4OH→ AgC ≡ CAg↓ + 2NH4NO3 + 2H2O}$ The greyish precipitate of silver acetylide will readily form when ethyne is passed through freshly ...

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I'm not going to address all of the reagents, but I want to highlight a specific approach in this partial answer. Occasionally, knowing some history of chemistry is quite useful when thinking about how to solve problems. In this specific case, I direct you to the work of Emil Fisher (chemistry Nobel 1902), who determined the structure of all hexose sugars. ...

2

You are correct that there are few differences between glucose and xylose simply based on functional groups. The only real difference is that one has six carbons and the other five; they are members of what we normally call hexose and pentose sugars. This is not my area of expertise, but a quick Google search suggests that these can be differentiated using ...

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The steps for conversion of Benzene to 1,3-dichloro-5-(propan-2-yl)benzene are shown below (Scheme).

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In photochemical isomerization absorbed light is responsible for conversion of cis to trans where as in thermal isomerization molecule uses its own thermal energy where molecules donot need to absorb photos .In addition no excitation takes place.

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You have to protect the amino group because the amino group is so activating that it is hard to stop multiple substitutions from occurring. By reacting it with the anhydride, the lone pair on the amino group gets delocalized due to conjugation and is therefore less available to the ring, making it easier for one addition. However, it seems that you want the ...

0

This reaction is an example of nucleophilic acyl substitution, whose mechanism is addition-elimination (as mentioned by Soumik Das), which is neither SN1 nor SN2 (which one might call nucleophilic alkyl substitutions).

2

For chair-like transition states the angle at which you draw your chair can make a significant difference to how easy it is to visualise stereoelectronic interactions. Sadly, there is no real way to know which is the best way to draw it, except by either trial and error, or prior experience (or leveraging on other people's experience, i.e. copying it from ...

0

Bond order is directly proportional to stability of a compound. In aromatic compounds bond order is maximum because all the electrons goes in bonding molecular orbitals and all electrons are paired . Thus according to the formula of bond order bond order = 1/2(number of electrons in bonding molecular orbitals - number of electrons in anti bonding molecular ...

2

The expected products of mono-chlorination of 2,4-dimethylpentane are depicted in following scheme: The substrate has 12 primary hydrogens (on 1-,1'-, 5-, and 5'-methyl groups), 2 secondary hydrogens (on 3-methylene group), and 2 tertiary hydrogens (on 2-, and 4-methine groups), as numbered in the scheme. Altogether, total of 3 different hydrogens are ...

0

The video you linked to mentioned a particular research paper which you should read if interested (either just ask your institution for a copy or use you-know-what-website). The conclusion writes: “...backside SN2-b barriers increase along the nucleophiles F-, Cl-, Br-, and I- and decrease along the substrates CH3F, CH3Cl, CH3Br, and CH3I. Frontside SN2-f ...

2

$\ce{H2O2}$ is Selective The reason why hydrogen peroxide can be used for such diverse applications is the different ways in which its power can be directed -- termed selectivity. By simply adjusting the conditions of the reaction (e.g., $\mathrm{pH}$, temperature, dose, reaction time, and/or catalyst addition), $\ce{H2O2}$ can often be made to oxidize one ...

1

As one already pointed out, your sesamin molecule has extra $\ce{O}$ atom. Assuming it will be corrected, the answer for your question is yes, all hydrogens should have different chemical shifts since all of them are either chiral or diestereptropic (aliphatic hydrogens) or in different environment (aromatic protons) except for two methylene groups on the ...

3

Horseradish peroxidase is not hydrogen peroxide, $\ce{H2O2}$, but rather an enzyme that breaks $\ce{H2O2}$ into $\ce{H2O}$ and $\ce{O2}$. In general, enzymes ending in -ase are lytic enzymes, catalyzing the breakdown of a similar-sounding substance. So, if you add ground horseradish to hydrogen peroxide, bubbles of oxygen are released. BTW, horseradish roots ...

2

For sesamolin, I think the pairs of protons at A and N would have the same chemical shift. In the spectrum this would appear as a singlet with an area corresponding to 4 protons due to the accidental equivalence of the A and N site. It looks like you did this in assignment of sesamin, which looks good to me. Assigning shifts to geminal protons can be tricky....

1

Potassium cyanide ($\ce{KCN}$) is a predominantly ionic compound ($\ce{K+}$ is Hard Acid while $\ce{CN-}$ is Soft base) and would dissociate completely to give cyanide ions in solution. Cyanide ion ($\ce{CN-}$) is considered an ambient nucleophile, which means that the negative charge can associate with either carbon or nitrogen. However, the negative charge ...

0

This is because when number of electron donating alkyl group on OH-bonded carbon atom increases, polarity of carbon oxygen bond also increases, which further facilitates the cleavage of carbon oxygen bond. Therefore reactivity increases.

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In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups, etc.) on either side of functional group (i.e., $\ce{-O-,-S-, -NH-, -C(=O) -,}$ etc.) are called metamers and the phenomenon is called metamerism (Wikipedia). Thus, you agree that all three compounds are metamers. Positional isomers are ...

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There is no IUPAC definition for metamers or metamerism in the gold book (not even one that makes it obsolete), and Wikipedia doesn't even have a real article about it, see Metamerism. In chemistry, the chemical property of having the same proportion of atomic components in different arrangements (obsolete, replaced with isomer). In organic chemistry, ...

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gaseous hydrochloric acid or anhydrous, we have an alkene where each side is attached to the same number of hydrogens —> both “equally substituted”. Products in this case, 3-chloropentane and 2-chloropentane.

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The safest solvent might be supercritical carbon dioxide, $\ce{CO2}$. Supercritical carbon dioxide is also used in "green" dry-cleaning. Use for dissolving polymers requires a pressure vessel for ~5 to 50 MPa, able to work at ~40 to 100°C. There are (paywalled) articles at ACS and NCBI, if you want to investigate further. According to Agilent, 1,2,4-...

2

In solvolysis of simple primary cyclopropylmethyl systems the rate is enhanced because of participation by the $\sigma$ bonds of the ring. The ion that forms initially is an unrearranged cyclopropylmethyl cation that is symmetrically stabilized, that is, both the 2,3 and 2,4 $\sigma$ bonds help stabilize the positive charge(page 464 ,JerryMarch ,...

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Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite,which gives various stereochemical products as shown here. Extending this to the current question , alkyl chloro sulfite 2 is formed.${Cl^-}$ attacks in a ${S_N^2}$ mode to give 3. Base deprotonates 3 to give 4. Alkoxide and chloride in 4 are in correct trans configuration for ...

0

The wikipedia does a good job of explaining the concept of biodegradation: In practice, almost all chemical compounds and materials are subject to biodegradation processes. The significance, however, is in the relative rates of such processes, such as days, weeks, years or centuries. Saying that something is "biodegradable" is not very informative unless ...

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The first step is nucleophilic substitution . It is possible for nucleophile to attack directly at the allylic position, displacing the leaving group in a single step, in a process referred to as SN2' substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to those seen ...

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