# Tag Info

1

According to the IUPAC nomenclature, preference for placing prefixes is given in accordance with alphabetical order (just like in a standard English dictionary). It has nothing to do with the number of letters in a prefix. Since "P" in heptyl comes before "X" in hexyl, heptyl will be placed before hexyl in IUPAC nomenclature. Hence, the ...

3

The scientist who coined the term polymer, Jöns Jacob Berzelius, used the word to refer to different substances which had the same empirical formula. In this sense, benzene is a polymer of ethyne (acetylene), because benzene $\ce{C6H6}$ and ethyne $\ce{C2H2}$ have the same empirical formula, namely $\ce{CH}$. In other words, all substances with molecular ...

2

Short answer: Yes and No, depending if we consider polymer(1) or polymer(2). Long answer: The first step in decision "Is X A or not ?" must be clarification what we mean by "A". Benzene is a cyclic trimer of ethyne. Trimers are a special case of polymers(1) as repeated monomer structures, being a superset of oligomers and polymers(2). ...

2

Is this a 18-electron aromatic system? No, because as the OP says in the question, the rings are not in a single plane. Is this compound considered aromatic? Yes, because it has three aromatic rings, as the OP stated. They all have 6 delocalized electrons, so they conform to the $4n + 2$ rule. "What if" all carbon atoms were in a plane? The C14 and ...

0

Solubility order is (I) > (II) for the reasons stated above in the question.

0

Yours is a primary alcohol and it has $\beta$ hydrogen atoms also. Thus, it will undergo $\pu E2$ mechanism. In this mechanism removal of protonated hydroxyl group and $\beta ~\pu H$ happen at once (in a single step). Therefore, there will be no scope of hydride shifting. You may compare this with the dehydration of neo-pentyl alcohol.

-5

This compound can be named 1 ethyl 3,4 dimethyl cyclohexane and 4 ethyl 1,2 diethyl cyclohexane. There are no problems in any name.

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Note: The anhydride would be hydrolysed first and would form acid which does not give the iodoform test. Even if we consider that they were not hydrolysed , what would happen ? $\ce{CH3COO-}$ would be removed and hence $\ce{CHI3}$ will not be formed. Hence anhydrides will not respond to this test.

0

A basic principle in OChem says that stability of intermediate is the key factor in deciding the major product, in general. Another factor is stability of final product. Will a kinetically favoured (minor) product (c) form? It will form, as far as it's intermediate is known to exist (and it exists). However, that will be present in very minor amount, as ...

2

Formally, the structural formulas of n-butanol, n=1,2,3,4 are possible, similarly as n-ethanol, n=1,2. But due the symmetry, there is redundance, as 4-butanol is identical to 1-butanol, 3-butanol to 2-butanol and 1-ethanol to 2-ethanol. Therefore IUPAC rules use just 1-butanol, 2-butanol and ethanol. Using eventually 3-butanol for tertiary butanol ( or t-...

0

According to me, while comparing acidic strength, we should remove $\ce{H+}$ from both the compounds, hence we get $\ce{CF3-O-}$ and $\ce{CCl3-O-}$. Now, we will compare stability using priority order: Aromaticity > Resonance > Hyper-conjugation > Inductive effect Now, we can see that, in $\ce{CF3O-}$ only inductive effect is applicable whereas in $\... 1 I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here: H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons. Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they ... 3 The reactivity of the SN2 reaction is determined by the stability of its transition state. More stable the transition state(hereby referred to as TS), lower will the activation energy be for the reaction, and therefore, the reactivity increases. Let's first have a look at the transition state of an SN2 reaction.$^{[1]}$What you see inside the square ... -1 I think the question wants you to consider the SN1 mechanism as well in your rankings. As Q and S each have a more electron rich ring than R, they generate a more stable carbocation on loss of the chloride ion than R. Therefore, the rate of SN1 has increased considerably and so a smaller proportion of the molecules go through the SN2 route. The carbocation ... 2 This is a really good question. The reason you got it wrong is because you only identified one feature that leads to faster rate of bimolecular substitution, and that is secondary electrophiles react more slowly than primary electrophiles. That's good because you are trying to identify patterns of reactivity. The reason that the secondary electrophile reacts ... 1 It is impossible to be definitive with such a simple picture. But, if we assume the colour code is black for carbon and that hydrogens have been omitted, the simplest answer is (to use a non-systematic name) tri-isopropyl methane. See this (very basic) description for systematic names. 2 The prefix neo- is only used with pentane. It has a historical origin. In the$19$th century, the first$\ce{C_5H_{12}}$that has been discovered was our n-pentane. At that time it was simply called pentane. After some time, a second$\ce{C_5H_{12}}$was discovered. Today we know that it is methylbutane. As it was an isomer of pentane, the chemists decided ... 0 With respect to the question "if a chemical reaction is possible or not", here is some background material on oxygen species. In particular, Singlet oxygen apparently reacts with oxygen (also referred to as dioxygen) creating a very strong (but transient) oxidizing atomic oxygen. Examples of possible reactions of interest, including atomic oxygen, ... 1 It is known that terminal alkynes reacts$1$with$\ce{NaNH2}$and produces salt, much like an acid-base reaction, hence it is a fast reaction. Furthermore, the aldehyde group is an EWG, which makes it easy to lose a proton, hence drives the reaction! Whereas, cannizzaro reaction happens through multiple reversible steps as seen in the following image, ... 1 Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on ... 0 But why does this happen? Well, if it was a plain old carbon in the middle, this migration wouldn't happen. It would sit there forever. Boron isn't carbon, though: it has fewer protons, is less electronegative, and doesn't like holding onto electron density all that much. That electron density gets pushed onto the adjacent carbons, which make them rather ... 6 The first pair is not stereoisomers. You have correctly explain why. When you are dealing with stereoisomers, that's what you have to do first: Identify stereo centers. Accordingly, each compound of second pair has two stereo centers. Then, mark (R,S) configurations of all stereo centers. As I marked, the first compound has (R,S) configuration, while second ... 6 Your first set of compounds are identical. In the Fischer projection of a chiral centre, you can rotate three of the connected groups in a clockwise/anti-clockwise direction without changing the configuration of the chiral centre. Upon performing this operation (keeping the top fixed, rotating bottom three clockwise) on the first compound in the first set, ... 2 According to Wikipedia, solubility of aspirin in water is$\pu{3 g/L}$or$\pu{3 mg/mL}$at$\pu{25 ^\circ C}$. Thus, it is not hard to make a$0.3\% (w/v)$aquious aspirin solution (maximum concentration, which equals to$\pu{300 mg}/\pu{100 mL}$at$\pu{25 ^\circ C}$). Therefore, your target,$\pu{10 mg}/\pu{100 mL}$at$\pu{25 ^\circ C}$can be easily ... 9 This reaction was conducted by Ho and Liao1. Nitro compound 1 is reduced to the nitroso compound 2 which undergoes a [3+3] electrocyclization to form 3 (4). Tautomerization of 4 to 5 is followed by a retro-"Diels-Alder" reaction to open to 6. Conjugate addition of the nitrogen to the unsaturated aldehyde moiety of 6 to afford dihydro pyridine 7. ... 2 I'd say your mechanism is acceptable. However, the final product, pentafulvene (PIN: 5-Methylidenecyclopenta-1,3-diene) is said to be not stable at room temperature. Generally, fulvenes are thermally unstable, sensitive to oxygen and photosensitive (Ref.1). They are also prone to acid- and cation-catalysed polymerisations. For example, there have been ... 3 Brønsted Acid has been used in Friedel-Crafts alkylation before (for example, Ref.1). According to this reference, when toluene use as the solvent (), the F-C alkylation reaction of toluene is fruitful for tertiary and secondary alkyl bromides, tosylates, and alkenes (which give both carbocations upon protonation). However, the only primary bromide, which ... 0 Your solution manual correctly predict that phenol has a higher enol content due to aromaticity. For example, you won't see strong carbonyl peak around$\pu{1700 cm-1}$in FT-IR, which proves keto-form does not exists. It is 100% enol-form as phenol. On the other hand, 4-pyridone is a different story. The dominance form, whether it is 4-pyridone or 4-... 0 In this case, the dominance form of the compound depends on which phase it is in. Ref.1 states that: In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to ... 2 You have misinterpreted what Wikipedia has said. I'd put whole quote here: Cyclobutadiene is an organic compound with the formula$\ce{C4H4}$. It is very reactive owing to its tendency to dimerize. Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable. Since the ... 2 I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of$\pi$electrons when it is present ... 0 In real-life, Cyclobutadiene dimerizes at temperatures above 35 K by a Diels-Alder reaction (source: Wikipedia). However, this seems more like a theoretical reaction rather than a practical one, so treat it as such and stick to the task of finding the dehydration product of this cyclobutadiene derivative. Yes, this step is correct (although the arrow-pushing ... 1 When the ozonide is reduced to Ketone/Aldehyde, We're concerned with what is called the "reductive workup" . We can summarize it like this: If we look a little further into the mechanism (of a different alkene): Step (e) is what we're concerned with. Water is not having an impact on this reaction in any manner. The main compound here that's ... 1 (S)-(-)-$\alpha,\alpha$-Diphenyl-2-pyrrolidinemethanol is a chiral ligand, which is commercially available (e.g., for$\pu{53.00 USD}$/$\pu{5 g}$from Spectrum; Manufacturer: TCI). That is much cheaper price compared to other vendors and the trouble preparing it (assuming this is not the project you are assigned for). To be honest, I didn't have experience ... 4 The perspective about ethyl acetate should be revisited. EtOAc is slightly soluble in water; not infinite (as e.g. ethanol), nor this little as hexane(s). It is greener than halogenated solvents such as dichloromethane, and the boiling point higher than the for dichloromethane or diethyl ether*) is an advantage during summer. Thus, EtOAc is an often used ... 2 Steric effects play an important role in determining the planarity, aromaticity of a molecule. Cyclobutadiene is sterically unstable(90 degrees angle between bonds), hence having high energy, and thus anti-aromatic. A lot of compounds change their conformations for reducing their overall energy. Check for the conformations of glucose, for instance. A ... 3 These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question: The order (B > A > C > D) was said to be correct for the conc.$\ce{H2SO4}$/$\Delta$case since there would be no effect of$\ce{CH-}$acidity since the protonation of the$\ce{-OH}$group takes place and so the ... 6 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 ... 3 Learning the unknown sample is called analysis and is a subject of analytical chemistry. But your question is probably not about this. You mean "can i guess something without knowing chemistry at all?" And answer is yes! Look at the food salt. Do you know it without any knowledge of chemistry? Yes, you do! But how much? You know what it looks like, ... 4 Stay with the carbonate. In the course of the reaction$\ce{HCl}$is formed. The aim is to remove this from the reaction mixture both as soon as well as complete as reasonably possible. Compare The addition of sodium acetate will increase the complexity of your reaction mixture.$\ce{HCl}$is soluble in methanol (reference), and upon dissociation ($\ce{...

2

Yes, theoretically speaking, you CAN use it (not that you should), but only with suitable organic solvent and it also depends on the grade of Calcium chloride you have. I'll try to keep my answer as short and precise as possible. Two things to consider, as I said before, are the grade of Calcium chloride and the organic solvent used. • Calcium chloride can ...

3

Similar to @MathewMahindaratne's comment: if you know the drying agent suits your needs well enough, keep it. In particular here, sodium sulfate is a neuter one (as is calcium sulfate, too) is a safe option because while drying the organic layer, the agent will neither protonate, nor deprotonate your product. Thus, if the protocol states $\ce{Na2SO4}$, ...

1

The key, as emphasized in Zhe'a comment to Mathew's answer, is that the cyclic intermediate is achiral. This ensures a perfect racemic product mixture ("erasing" the stereochemical information, i.e. you could have started with the other enantiomer as a reactant and obtained the same product mixture). Here is a 3D rendering of the stereochemistry ...

1

An acid catalysed dehydration reaction of alcohols will always proceed with the removal of -OH as H2O molecule and therefore, will be dependent on the formed carbocation's stability because the rate deternining step for this process is formation of carbocation. Carbocation Carbocation can be stabilized by inductive effect, field effect, mesomeric effect (or) ...

2

Your basic assumptions are correct. It can be observed as such: In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest. Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also ...

5

The full name of the compound is 3,4-methyl​enedioxy​methamphetamine, which is typically abbreviated as "MDMA". It is commonly used recreationally, and is casually referred to as "ecstasy", "molly", "E" and many other names.

2

Background Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated. The key ...

1

I highlighted the some of your attempts: $\ce{H_d}$ has $\mathrm pK_\mathrm{a} \approx 9$, so it will not be the most acidic. This assumption is correct. For aspartic acid, $\mathrm pK_\mathrm{a}(\ce{H3N+}) = 9.6$. $\ce{NH3+}$ has very high -I effect but no mesomeric effect, so $\ce{H_c}$ appears to be quite acidic. But at the same time it is at the ...

3

A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion. A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point. Let's ...

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