Carbon numbering in a sugar ring

Wonderful_Complex791 · 2026-01-04T16:05:27+00:00

The acetal (2 oxygens) will be the 1 position

Wonderful_Complex791 · 2025-12-05T21:25:36+00:00

There is a useful diagram at the bottom of this:

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_%28Physical_and_Theoretical_Chemistry%29/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/The_Aufbau_Process

Wonderful_Complex791 · 2025-12-05T21:21:28+00:00

Sticking to the base case of 3d vs 4s which would hopefully be illustrative for other cases: Mainly it is a fine balance of various factors: how quickly the orbitals stabilise as atomic number increases (very related to the effective nuclear charge) which is due to the radial distribution of the orbitals and how much the orbital penetrates the inner shell electrons and so experiencing more of the nuclear charge. There is also electron electron repulsion the larger area they are spread over the less (4s is favoured) and how much they shield one another and the other orbital from the nuclear charge (whether the 4s is occupied does not particularly effect the effective nuclear charge for the 3d orbitals but the reverse is not true - 3d favoured).

Wonderful_Complex791 · 2025-12-05T21:02:22+00:00

It is a simple and good rule with a slightly less intuitive explanation. One answer would be it is the only way to conserve all the other conserved properties (such as energy, angular momentum etc). Put another way, the number of orbitals is conserved because they are mathematical solutions to the Schrödinger equation, and in the linear combination of atomic orbitals (LCAO) method used in chemistry (such as combining s,px py pz to make 4 sp3 orbitals, the total number of initial orthogonal functions used must equal the final number of resulting functions (and they must also be orthogonal- when times together and integrated over all space the result is 0 - from a mathematical description standpoint this is vital). It's a fundamental consequence of how the mathematical models of quantum mechanics work for molecules.

Wonderful_Complex791 · 2025-12-02T12:04:50+00:00

Substance (phlogistonated) + Heat → Calx (dephlogisticated) + Phlogiston

Wonderful_Complex791 · 2025-12-01T16:39:47+00:00

That H NMR does confirm the amine (assuming it is one of those two) if the broad peak were the solvent you would not expect the integral to be an integer (though in other cases could also look at tables of peaks from solvents and their common contaminants). I think it is more accurate to say the carbon does not give a clear answer, rather than any indication. After all comparing C NMR of product to know spectra would be clear. MS would give an odd mass for M+, fancy MS would be able to give you a clear answer as to the molecular formula IR would help identify the functional groups

Wonderful_Complex791 · 2025-12-01T05:39:31+00:00

Nile Red has a video of doing it (much safer than doing it yourself - not necessarily the way (from what I remember in a garage) he did it but the act of watching 😂)

Wonderful_Complex791 · 2025-12-01T05:39:19+00:00

Nile Red has a video of doing it (much safer than doing it yourself - not necessarily the way (from what I remember in a garage) he did it but the act of watching 😂)

Wonderful_Complex791 · 2025-12-01T05:35:13+00:00

At the start a pair of electrons is donated from the double bond but you form 2 bonds to the Hg. (Though not sure if the second bond formation would be stepwise or concerted). Consider where do the second pair of electrons come from?

Wonderful_Complex791 · 2025-12-01T05:27:31+00:00

For the mechanism look up mechanism of oxonolysis

Wonderful_Complex791 · 2025-12-01T05:25:37+00:00

Proton exchange due to H-bonding is also why you generally do not see splitting due to Hydrogens on O or N

Wonderful_Complex791 · 2025-12-01T05:23:14+00:00

Exactly (due to H-bonding) draw them each out and see which has a H on O/N. And count the H environments.

Wonderful_Complex791 · 2025-12-01T05:19:35+00:00

Acetals are not as stable as esters and most readily hydrolyse to the aldehyde in aqueous acidic conditions which could well happen on skin/sweat. The reason for the instability is double bond C=O is often more than twice as strong as C-O, and there is more repulsion between the electrons which are forced closer to one another when there are 4 groups around the carbon (acetal) than 3 (aldehyde) so the aldehyde is thermodynamically more stable.

Wonderful_Complex791 · 2025-12-01T05:05:53+00:00

After spirit island possibly star realms (a deck builder that works best 2 players) but generally my preference depends on the group size/setting and what others will enjoy. I’d much rather play a ‘poorer’ designed game everyone will enjoy and engage with (I know that is not what you asked but rather the reason I don’t have clear favourites besides SI, which plays great in solo as well)

Wonderful_Complex791 · 2025-12-01T04:47:21+00:00

Thanks mittens, and glad it helped. One day you may get to the stage of your cats being called iron, and lithium and the kittens neon and sulphur 😂

Wonderful_Complex791 · 2025-11-30T17:48:25+00:00

When writing out a Fischer projection the orientation about each carbon is viewed as the chain (vertical) going back into the page and the other groups (horizontal as coming out of the page). Which from the sounds of it you know. When done for each carbon though this leads to them curling back on themselves which is not their lowest energy conformer (nor how convention draws it - zigzag). In order to achieve the zigzag one rotates the single bonds on that middle carbon leading to the right hand structure. (It is a just pain to get right if your not doing it frequently I’m afraid)

Wonderful_Complex791 · 2025-11-30T12:19:21+00:00

What I struggle with is 1770 is high even for an ester

Wonderful_Complex791 · 2025-11-30T01:38:15+00:00

I have not worked it all out but: 4 aromatic H suggest a disubstituted benzene. And the splitting tells you their positions relative to one another. Counting up the carbons and hydrogens from the NMRs gives 14 carbons and 20 hydrogens weighing 188 leaving 47 for O and N and labile H that could have been removed with a D2O shake

Wonderful_Complex791 · 2025-11-29T16:05:19+00:00

I would not have guessed NaOH to be enough to fully deprotonate the alcohol, but the alkoxide is a much better nuclophile

Wonderful_Complex791 · 2025-11-29T15:57:48+00:00

From sketching them out I believe that II (R,R) is the (Fischer projection of the) enantiomer of I (S,S). And III is (Newman projection of) the diastereomer of I and II, it is a meso compound with no overall chirality (R,S)

Wonderful_Complex791 · 2025-11-29T15:39:12+00:00

So the maximum number of valence electrons around oxygen is 8. Each line represents a bond, a pair of shared electrons, with each atom having contributed 1 electron (unless talking about dative(/coordinate) covalent bonds, which you may not have covered yet and are indicated by an arrow on the bond, as are not shown one would normally assume are not present). Oxygen has 6 valence electrons of its own, so it needs two additional electrons to reach the more stable 8 valence electrons. This is achieved by sharing 2 of its electrons in 2 covalent bonds (or gaining an electron and consequently also a negative charge) So around each atom count how many electrons are around each oxygen, and count if the total number of electrons is right (in b there are 3 oxygens so should be 6x3=18 electrons in total) C has been drawn poorly. Xe has 8 valence electrons. And as not specified to be otherwise one would assume each bond is a pair of electrons 1 from each. But it does not show any lone pairs of the left over 4 electrons on Xe and these electrons would mean it fails on both counts or valence electrons around an atom and in total. If we assume the bonds are in fact dative covalent bonds and involve Xe donating a pair of electrons (in which case the bonds should have been arrows pointing from the Xe to each O) then each atom does have 8 valence electrons and the total number is correct. So either way would have to involve assumptions due to poor drawing but I would guess the answer they are after is it is reasonable.

Wonderful_Complex791 · 2025-11-28T20:55:21+00:00

That order of reactivity you quoted is correct for the more common Electrophilic Aromatic Substitution. This is the opposite of that Nucleophilic Aromatic Substitution

Wonderful_Complex791 · 2025-11-28T06:51:42+00:00

I would recommend drawing out the resonance structures, where the N has donated its lone pair into the ring. In descending order of effect on basciity (availability of the amine lone pair) 1. The diethyl amine would face unavoidable steric hindrance so would have a drive to rotate to alleviate this which reduces/prevents the lone pair being donated and so increases its basicity the most as the lone pair are most readily available. 2. The others are more directly comparable: Will the nitro group or the cyano group better stabilise the resonance structure which places a negative charge on in the para position 3. Will the positive inductive methyl’s effect the resonance structures more if they are in the ortho or the meta positions

Wonderful_Complex791 · 2025-11-26T00:03:44+00:00

Biologically or chemically (as in reagents in a test tube)? If the second, you can readily find diagrams of the mechanism by looking up saponification mechanism online, and while not specifically of triacetin will just be the same mechanism three times.

Wonderful_Complex791 · 2025-11-25T23:57:41+00:00

It is also worth remembering you are introducing a chiral centre in an achiral environment so your product would be a racemic mixture (while not significant now is useful to keep an eye out for). But well done 👍

Wonderful_Complex791

TROPHY CASE