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Acid protonates one methoxy group of 2,2-Dimethoxypropane. methanol departure gives the oxocarbenium species $\mathrm{M}{\mathrm{e}}_2\mathrm{C}(\mathrm{OMe}{)}^{+}$, which is captured by one diol oxygen. After proton transfer, protonation and loss of the remaining methoxy group generate a second oxocarbenium center. Intramolecular attack by the other diol oxygen closes the five-membered ring; deprotonation regenerates acid. The net transacetalization exchanges two methoxy groups for the two diol oxygens, releasing two methanol molecules.

The two starred carbon centers retain their original configurations, and the two carbon side chains on the acetonide are trans. No C–O bond at either original stereocenter is broken; reaction takes place at the reagent-derived CMe${}_2$ acetal carbon. The apparent change in wedge/dash presentation comes from redrawing the backbone into the five-membered ring, not inversion at a starred center. In the planar pentagon used in the diagram, draw the alkynyl-bearing side chain behind and the $\mathrm{C}{\mathrm{H}}_2\mathrm{OPMB}$ side chain in front. The unstarred methyl-bearing center also retains its configuration. The acetal carbon itself is not stereogenic because its two methyl substituents are identical.

Product 5 is 1-methylcyclohexan-1-ol. The tertiary ring C–H bond of methylcyclohexane is replaced by C–OH. This is a single constitutional product; the two ring paths at that carbon are equivalent, so no new stereocenter is created there.

The manganese–oxo species abstracts the relatively weak tertiary hydrogen, then hydroxyl rebound converts the carbon radical into the tertiary alcohol. The unfunctionalized catalyst has no crown-ether receptor to preassociate the ammonium substrate. Its selectivity therefore mainly follows intrinsic C–H reactivity: the tertiary site of methylcyclohexane outcompetes the secondary C–H sites of the linear chain. The cationic ammonium group also electronically deactivates nearby sites. This explains the dominant product under the idealized selectivity described in the question.

The terminal alkene of D first undergoes Wacker oxidation, giving the side chain $\mathrm{Ar}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{COC}{\mathrm{H}}_3$. Alkene coordination to Pd(II), water attack, proton transfers and beta-hydride elimination give the ketone. Copper and oxygen regenerate the active palladium oxidation state.

Base removes a proton from the terminal methyl of that ketone. Its enolate attacks the neighboring aromatic aldehyde intramolecularly, closing a six-membered ring. Dehydration gives a conjugated cyclohexadienone-type intermediate, and keto-enol tautomerism furnishes aromatic naphthol E. The retained methoxy and MOM-protected oxygen groups follow the original aromatic-carbon positions.

F is the O-allyl ether of E. Carbonate forms the naphthoxide and allyl bromide undergoes an SN2 reaction at its terminal carbon. Heating F then causes a [3,3]-Claisen rearrangement: the O–allyl bond breaks, the terminal allyl carbon bonds to the adjacent naphthalene C1, and the phenolic carbon temporarily becomes a ketone. Rearomatization gives G, with a free phenolic OH and a C1 allyl group. The C1 position gives the favored alpha-substituted naphthol and avoids the alternative less favorable annulation pattern.

The final step is a dearomative Michael addition, not a Nef conversion of the nitro group. Hydrogen bonding from the thiourea activates nitroethylene. The nucleophilic C1 resonance form of G attacks the terminal carbon of $\mathrm{C}{\mathrm{H}}_2=\mathrm{CHN}{\mathrm{O}}_2$, and proton transfer converts the nitronate to a 2-nitroethyl substituent. The phenolic carbon becomes a ketone.

H is the corresponding 1-allyl-1-(2-nitroethyl)naphthalen-2-one, retaining the methoxy and MOM substituents of G. The second ring now has a C2=O group and a C3=C4 bond, while the other ring remains aromatic. The reported ${}^{13}$C signal near $202$ ppm is consistent with that newly formed ketone. Related dearomative additions establish this reaction class.^[ref]

C1 is now an all-carbon quaternary stereocenter bearing both allyl and 2-nitroethyl groups. Both the drawn thiourea and the starting materials are achiral, so H is racemic. A chiral catalyst would be needed to select one absolute configuration.

For a tetrahedral stereocenter, rank its four substituents using stereochemical sequence rules. With lowest-priority group pointing away, a clockwise $1\to 2\to 3$ sequence is $R$ and a counterclockwise sequence is $S$. Looking from the other side reverses the apparent sequence, so the viewing condition is essential. The geometric letters A/B/C/D in the tetrahedral mirror-image test carry no assigned priority. Neither $R$ nor $S$ determines positive or negative optical rotation.^[ref]

Ribose is a monosaccharide with molecular formula ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5$. Its biologically familiar form is D-ribose, an aldopentose whose open-chain Fischer projection has $\mathrm{CHO}$ at the top, ${\mathrm{CH}}_2\mathrm{OH}$ at the bottom, and hydroxy groups on the right at C2, C3, and C4. The opposite horizontal substituents are hydrogen. See McMurry, Organic Chemistry, 25.3: D-ribose and D-glucose configurations.

Read the drawing by counting groups: $\mathrm{CHO}$ contributes $\mathrm{CHO}$, three $\mathrm{CHOH}$ groups contribute ${\mathrm{C}}_3{\mathrm{H}}_6{\mathrm{O}}_3$, and ${\mathrm{CH}}_2\mathrm{OH}$ contributes $\mathrm{C}{\mathrm{H}}_3\mathrm{O}$. Their sum is ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5$. The count checks composition; the right/right/right pattern additionally specifies the stereochemistry of D-ribose.

In RNA, ribose occurs as a ring within each nucleotide, joining a nucleobase to the phosphate-linked backbone. Its C2 position retains a hydroxy group, unlike deoxyribose in DNA. In a nucleotide, the sugar positions carry primes, so this is C2′. See Alberts et al.: nucleotide sugars in RNA and DNA. Free ribose in water is also predominantly cyclic, with several ring forms in chemical equilibrium and very little open-chain form; the Fischer projection is valuable for comparing configurations, not for estimating their abundances. See Mason et al., The Conformation of a Ribose Derivative in Aqueous Solution: A Neutron Scattering and Molecular Dynamics Study (2013), introduction.

At a stereocenter, exchanging two attached groups produces a different spatial configuration. The open-chain ribose and glucose examples have carbon stereocenters bearing four different groups; C2 of deoxyribose has two identical hydrogen substituents and is not a stereocenter. See Moss, Basic Terminology of Stereochemistry, IUPAC Recommendations 1996: stereogenic units.

Stereochemistry studies the spatial arrangement of chemical atoms in molecules. A Fischer projection records this information at successive stereocenters, whereas a molecular formula only counts chemical atoms. See McMurry: representing carbohydrate stereochemistry.

Tartaric acid has two hydroxy groups and two carboxylic acid groups. Its meso compound form illustrates why two stereocenters need not make a chiral molecule.

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.