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50

I'd like to throw a tentative explanation for the ortho effect into the ring: In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively, can be assumed. In the first case, the electron density on the N atom is (slightly) reduced and ...


36

I would like to back up Klaus' answer with some Quantum Theory of Atoms in Molecules (QTAIM) results, based on a DF-BP86/def2-SVP calculation. Note that these are results, obtained without the consideration of solvation or condensed phases. I believe they still prove a valid point in the case of electronic structure theory. I revisited this question in ...


35

There are 2 cases, both related to the acid-base reactions. Both are also partial reasons why so many fish recipes use lemon juice. Fish, especially sea fish, naturally contain trimethylamine-N-oxide $\ce{(CH3)3N-O}$ that, after death, gets enzymatically reduced to trimethylamine $\ce{(CH3)3N}$, the source of ammonia-like fish odour. Trimethylamine N-...


25

It can be explained another way : o-toluidene is less basic than aniline cause of a different reason. See, what happens is : when aniline acting as a base becomes $\ce{NH3+}$ (on top of a benzene ring), it is usually stabilised by solvation. but if there is a substituent on the ortho position, it inhibits solvation. Thus the tendency to act like a base is ...


23

In the conversion of primary alcohols to primary amines $$\ce{R-CH2OH -> R-CH2NH2}$$ direct alkylation of ammonia normally is the last thing you want to do in the lab. Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent ...


21

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity. Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It ...


21

how is anilinium ion meta directing for electrophiles? Actually, anilinium is not meta directing (I know it is often taught that way), but rather it inductively deactivates the entire aromatic ring. To explain the electrophilic substitution pattern observed with any aromatic system we must consider both resonance and inductive effects. Resonance Effects: ...


19

In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from. Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ...


15

Background Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons. The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$)...


15

If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide. The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. ...


15

This is a really interesting question and the answer is that the reaction of benzenediazonium chloride with aniline is a bit different to most of the reactions of benzenediazonium salts in that the initial product is compound 1, diazoaminobenzene. It is possible to run the reaction to isolate diazoaminobenzene prep here. These diazoaminobenzene compounds ...


14

(Will do more research into ammines come February.) Premise Ammonia is spelled with two 'm's. The more natural derivative of the word in a linguistic sense would also have the same number 'm's. Thus whichever predates the other, ammine complexes or amines, would claim the throne. In a structural sense, when awknowledgement of the first ammine complex came ...


13

You may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below. The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side. The same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups ...


13

As you point out, the tert-butyl group wants to remain in an equatorial position, else both the tert-butyl group and the trimethylammonium group end up being axial which is disfavoured for two bulky groups. The obvious role of tert-butoxide: The tert-butoxide would obviously rather act as a base than a nucleophile (in-fact under ordinary circumstances we ...


13

In general, for anilines reaction at nitrogen is kinetically faster than reaction at carbon. However, the C-substituted product is usually more stable than the N-substituted product, so can prevail under thermodynamic conditions. As noted by Mayr et al. (the discussion below is taken from the same article):[1] Similar regioselectivities are found in azo ...


12

An R group is used in structural formulae as a placeholder for a range of possible subtituents, e.g. hydrogen, alkyl chains or aryl groups etc. It can also be used as an abbreviation when R is a predefined rest. An amide is a derivative of an organic acid where an $\ce{OH}$ group is substituted by $\ce{NR2}$ (with R = H or any organic rest, see here, for ...


12

Hydrogenolysis typically involves a metal surface ($\ce{Pd, Pt}$, etc.) and hydrogen gas. In the reaction a weak bond, that is, a bond that is either strained or a bond that can generate a stabilized radical is broken. In your examples there is no strain, but a resonance stabilized benzylic radical can be generated in one case. In the benzylic amine ...


10

Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride: However, in addition to substitution reactions by loss of $\ce{N2}$, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are diazo couplings, where the nucleophile is an ...


10

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

Firstly a note about terminology. The word "terminus" is reserved for the N- or C-termini of a polypeptide chain. For a free amino acid, you should refer to the carboxyl and amino groups as the $\alpha$-$\ce{COOH}$ and $\alpha$-$\ce{NH2}$ groups respectively. Anyway, the -$\ce{COOH}$ group is acidic; above a certain pH, typically around 2, it can be ...


10

In the second row of the periodic table, elements have relatively small differences between the size their s- and p-orbitals. Therefore, the orbitals of $\ce{NR3}$ can go from $sp^3$ to $sp^2$ with relatively little energy increase, so an amine can become planar and then reorient with the inverted stereochemistry. The same occurs with carbanions. I'm not ...


10

$\ce{Fc}$ stands for ferrocenyl. So the first compound is an amide of ferrocenecarboxylic acid and tetramethylene diamine, and the second compound is p-ferrocenyl aniline.


10

The proposed strucutre of these compound are, correct me if i am wrong.


10

Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required: e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]...


10

The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise, though there are some interesting techniques using carbonyl azoles ...


9

I'm not aware of a $\ce{NaBH4}$-based reduction of nitroalkenes to saturated amines, such as in: It is however possible to reduce nitroalkenes to saturated hydroxylamines using $\ce{BH3*THF}$ in the presence of catalytic amounts of $\ce{NaBH4}$. (DOI) If there's a chance to perform the desired transformation in one step via catalytic hydrogenation at ...


9

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the name ‘disilazane’ is no longer recommended for $\ce{H3Si-NH-SiH3}$. Such structures are now named as amines. (Also note that the name ‘hexamethyldisilazane’ is ambiguous and therefore ...


8

Between 1 and 2, 2 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 2. I agree that more methyl groups would result in a greater +I effect. But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option ...


8

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained names “aniline” and “phenol” are preferred IUPAC names (PINs). Parent structures having retained names are called functional parent compounds.      For both retained names, substitution is permitted at any ...


8

Both sources are correct. However, they are referring to two different $\mathrm{p}K_\mathrm{a}$ values. $$\ce{H3C-NH3+ <=>[$K_\mathrm{a1}$] H3C-NH2 + H+ <=>[$K_\mathrm{a2}$] H3C-NH- + 2 H+}$$ The second value is the one that your marking scheme refers to when it asks for the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3C-NH2}$. However, it is of little ...


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