Disconnect the tertiary amide into cyclohex-3-ene-1-carboxylic acid and the secondary amine $\mathrm{EtNHCH}(\mathrm{Me})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Ph}$. The acid-derived ring is an immediate Diels–Alder disconnection: buta-1,3-diene plus the indicated methyl acrylate gives methyl cyclohex-3-ene-1-carboxylate. Heating, or suitable Lewis-acid activation of the dienophile, forms the two new sigma bonds and leaves the diene's internal double bond as the ring alkene.

Hydrolyse this ester to the acid. Convert it to its acid chloride with oxalyl chloride or thionyl chloride, and react with the secondary amine in the presence of a base to capture HCl. The tertiary amide now contains the complete cyclohexene acyl fragment; direct conjugate addition of the amine to methyl acrylate before ring construction would give the wrong connectivity.

Prepare the amine by reductive amination of 4-phenylbutan-2-one, $\mathrm{MeCOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Ph}$, with ethylamine, using sodium triacetoxyborohydride or another iminium-selective reducing system. This places nitrogen on $\mathrm{CH}(\mathrm{Me})$, not on a terminal methylene. If a route from simpler carbon compounds is wanted, crossed aldol condensation of acetone with benzaldehyde gives $\mathrm{MeCOCH}=\mathrm{CHPh}$, and selective C=C reduction gives the required ketone before reductive amination.

The acid branch retains the acrylic carbonyl as the amide carbonyl; the amine branch supplies both N-ethyl and N-(4-phenylbutan-2-yl) substituents. The unlabelled stereogenic centres may be racemic mixture. This sequence separates ring formation from amide formation and avoids invoking uncontrolled alkylation of a tertiary amide nitrogen.

Open the cyclic ester at its acyl–oxygen bond. The resulting delta-hydroxy acid isOxidation-state reversal at the benzylic alcohol gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{CH}(\mathrm{nPr})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{iPr})-\mathrm{C}(=\mathrm{O})\mathrm{Ph}$. This identifies the key Michael disconnection between the central $\mathrm{C}{\mathrm{H}}_2$ and $\mathrm{CH}(\mathrm{iPr})$ groups:The nucleophile is the enolate or enamine of the phenyl ketone. The electrophile's propyl substituent must be alpha to its ester carbonyl; using a beta-propyl acrylate would move the substituent to the wrong ring position.

Disconnect $\mathrm{PhCOC}{\mathrm{H}}_2\mathrm{iPr}$ further to the indicated acetophenone. Crossed aldol addition to acetone followed by dehydration gives $\mathrm{PhCOCH}=\mathrm{CM}{\mathrm{e}}_2$; selective conjugate reduction or controlled alkene hydrogenation then gives $\mathrm{PhCOC}{\mathrm{H}}_2\mathrm{CHM}{\mathrm{e}}_2$. This avoids requiring an efficient SN2 alkylation with a secondary isopropyl halide. The alpha-propyl acrylate can be obtained by alpha-hydroxymethylation/dehydration of ethyl pentanoate, $\mathrm{Et}{\mathrm{O}}_2\mathrm{CC}{\mathrm{H}}_2\mathrm{nPr}$, using formaldehyde.

In the forward direction, perform the Michael coupling, reduce the phenyl ketone selectively with sodium borohydride while retaining the ordinary ester, hydrolyse the ester, and cyclize the delta-hydroxy acid by controlled acid-catalysed lactonization or an activated-acid coupling. The six-membered ring then has the order carbonyl–$\mathrm{CH}(\mathrm{nPr})$–$\mathrm{C}{\mathrm{H}}_2$–$\mathrm{CH}(\mathrm{iPr})$–$\mathrm{CH}(\mathrm{Ph})$–O. This atom order checks all three substituent positions. No relative configuration is specified, so the scheme does not imply a single diastereomer from the Michael addition or reduction.

The target is 5-Aminohexanoic acid, $\mathrm{H}{\mathrm{O}}_2\mathrm{C}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CH}(\mathrm{N}{\mathrm{H}}_2)\mathrm{Me}$. Disconnect its terminal amine/carboxylic-acid relationship through hydrolysis of 6-methylpiperidin-2-one. This six-membered lactam follows from a regioselected Beckmann ring expansion of 2-methylcyclopentanone oxime.

Start with cyclopentanone. Form its enolate with lithium diisopropylamide in dry THF at low temperature, add methyl iodide, then quench to obtain 2-methylcyclopentanone. Control the methyl iodide/base stoichiometry to suppress a second alkylation. React the ketone with hydroxylamine to form its oxime and select the geometrical oxime isomer in which the bond to the methyl-substituted adjacent ring carbon is anti to the departing oxime oxygen group.

Activate the oxime, for example under appropriate acid-promoted Beckmann conditions. The anti methyl-bearing C–C bond migrates onto nitrogen as the N–O-derived leaving group departs, inserting nitrogen between the original carbonyl carbon and the methyl-substituted ring carbon. Hydration/tautomerization gives 6-methylpiperidin-2-one. The opposite oxime geometry would insert nitrogen on the other side and ultimately place methyl at a different position in the amino acid; oxime geometry is therefore an essential part of this proposal.

Hydrolyse the lactam with aqueous acid and heat, then adjust the work-up to the desired acid/ammonium form. Cleavage of its acyl–nitrogen bond opens the ring to $\mathrm{H}{\mathrm{O}}_2\mathrm{C}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CH}(\mathrm{N}{\mathrm{H}}_2)\mathrm{Me}$ after suitable neutralization. All five original cyclopentanone carbons are retained, and the one methyl carbon comes from methyl iodide. A racemic mixture product is appropriate to the stereochemically unspecified target. The reaction sequence is an organic ring-expansion synthesis and requires no biological transformation.

Let $R=\mathrm{CH}(\mathrm{Me})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The useful disconnections are benzoate formation, carbonyl reduction and thiol conjugate addition:The sulfur-bearing carbon is the enone beta carbon, so this is a Michael disconnection, not attack of thiophenol on the alcohol-derived central carbon. Reduce the saturated ketone with sodium borohydride after installing SPh, then benzoylate the alcohol with benzoyl chloride and pyridine or a tertiary amine.

Disconnect the enone to $\mathrm{RCOC}{\mathrm{H}}_3$ and formaldehyde. Kinetic deprotonation on the methyl side of the hindered unsymmetrical ketone, trapping with formaldehyde, and dehydration gives $\mathrm{RCOCH}=\mathrm{C}{\mathrm{H}}_2$. This order leaves the substituted $R$ side unchanged. An alpha-methylenation/Mannich-base-elimination variant is an equivalent way of installing the terminal enone.

Finally, $\mathrm{RCOC}{\mathrm{H}}_3=\mathrm{C}{\mathrm{H}}_3\mathrm{COCH}(\mathrm{Me})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$ disconnects by alpha alkylation to butan-2-one and 1-iodopropane. Use the more substituted, ethyl-side enolate of butan-2-one for this step, rather than its usual methyl-side kinetic enolate; an equilibrated/regioisomerically selected enolate or enol derivative is required, and the desired monoalkylated ketone should be isolated. Three carbons from propyl iodide and one from formaldehyde increase the original four-carbon ketone to the required eight-carbon alcohol/thioether skeleton. Choosing ethyl iodide here would leave the target short by one carbon.

This is a constitutional synthesis: the target drawing specifies no configuration at either newly formed stereogenic site, so no unproved stereoselective reduction or alkylation is needed. Regioisomeric enolate alkylation and competing overalkylation are the practical selectivity issues to control, rather than something that follows from writing “base”.

Solid $\mathrm{MoC}{\mathrm{l}}_2$ contains octahedral $\mathrm{M}{\mathrm{o}}_6$ clusters, while $\alpha$-$\mathrm{MoC}{\mathrm{l}}_3$ has paired molybdenum centres in a layered chloride lattice. The first material is better represented by $\mathrm{M}{\mathrm{o}}_6\mathrm{C}{\mathrm{l}}_{12}$ than by isolated bent or linear $\mathrm{MoC}{\mathrm{l}}_2$ molecules. Eight inner chlorides cap the triangular faces of an octahedral metal core, giving $[\mathrm{M}{\mathrm{o}}_6({\mu }_3-\mathrm{Cl}{)}_8{]}^{4+}$. Outer chloride coordination completes the solid: one useful connectivity notation is $\mathrm{M}{\mathrm{o}}_6\mathrm{C}{\mathrm{l}}_8\mathrm{C}{\mathrm{l}}_2\mathrm{C}{\mathrm{l}}_{4/2}$, with two terminal outer chlorides and four intercluster-shared outer chlorides per octahedron. The empirical total is $8+2+4/2=12$ chlorides for six metals.

Each Mo(II) is formally $d^4$, giving 24 metal-based electrons. These support extensive Mo–Mo bonding along the 12 octahedral edges; they are not localized as four free spins on each molybdenum. In $\alpha$-$\mathrm{MoC}{\mathrm{l}}_3$, Mo(III) is $d^3$ in edge-sharing $\mathrm{MoC}{\mathrm{l}}_6$ octahedral layers. The metal honeycomb distorts into short Mo–Mo pairs. A localized dimer picture uses the six d electrons to fill one sigma and two pi bonding combinations, producing a formal triple bond and pairing the spins. The bulk crystal remains an extended chloride lattice, rather than a collection of unconnected $\mathrm{M}{\mathrm{o}}_2\mathrm{C}{\mathrm{l}}_6$ molecules.

By comparison, $\mathrm{CrC}{\mathrm{l}}_2$ has predominantly localized high-spin Cr(II), $d^4$, in Jahn–Teller-distorted chloride octahedra; $\mathrm{CrC}{\mathrm{l}}_3$ has layered Cr(III), $d^3$, with three locally unpaired electrons. Their magnetic exchange is not evidence for the same strong short multiple bonds as in the molybdenum solids. The greater radial extent of Mo $4d$ orbitals enables larger metal–metal overlap than the more contracted Cr $3d$ orbitals. Real structure and magnetism still depend on ligand geometry and phase, so the orbital comparison is a trend rather than a prohibition of chromium–chromium bonds in other compounds. Diffraction and magnetic/spectroscopic studies directly distinguish the dimerized $\alpha$-$\mathrm{MoC}{\mathrm{l}}_3$ lattice from $\mathrm{CrC}{\mathrm{l}}_3$.^[ref]

${\mathrm{S}}_4^{2+}$ is approximately square planar; $\mathrm{S}{\mathrm{n}}_5^{2-}$ is a closo trigonal bipyramid. Both have 22 valence electrons, $4(6)-2=22$ and $5(4)+2=22$, but their different local orbital requirements prevent inferring identical shapes from this total alone.

For the sulfur ring, four S–S sigma bonds use eight electrons and four in-plane lone pairs use another eight, leaving six electrons in the perpendicular cyclic pi system. The resulting filled bonding and degenerate nonbonding pi levels give a six-pi-electron aromatic description of a four-membered ring, consistent with equivalent S–S links in the ideal square. The out-of-plane p orbitals are retained; this is not a tetrahedral four-vertex closo cage.

For tin, allocate one mainly s-rich lone pair to each atom. The remaining $22-5(2)=12$ electrons form six skeletal pairs, exactly $n+1$ for $n=5$. Wade's rules therefore predict a trigonal bipyramid, with three equatorial vertices and two axial vertices. The axial pair is not joined by a polyhedral edge; its bonding is part of the delocalized three-dimensional framework. The contrast is a planar aromatic sigma/pi ring versus a three-dimensional multicentre-bonded Zintl cluster, rather than cationic charge alone imposing planarity or anionic charge alone imposing a polyhedron.