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The cyclic trimer of formaldehyde, $(\mathrm{C}{\mathrm{H}}_2\mathrm{O}{)}_3$. Acid-catalysed acetal cleavage can regenerate formaldehyde, which is strongly hydrated in water.

A molecular point group with identity, one twofold axis, and two mirror planes containing that axis. With the twofold axis along $z$, its one-dimensional irreducible representations have Cartesian bases $z\sim A_1$, $x\sim B_1$, and $y\sim B_2$. For formaldehyde in the $xz$ plane, the vibrational representation is $3A_1+2B_1+B_2$.

In this C2v point group convention, $z\sim A_1$, $x\sim B_1$, and $y\sim B_2$. Therefore all six normal modes have symmetry-allowed infrared fundamentals: $\boxed{6\text{ IR features}}$, absent accidental coincidences or vanishing intensities.

For Raman spectroscopy, examine the polarizability tensor: $x^2,y^2,z^2\sim A_1$, $xz\sim B_1$, and $yz\sim B_2$; $xy\sim A_2$ is also Raman-allowed but no $A_2$ vibration occurs here. All six vibrations are therefore Raman-active too, giving $\boxed{6\text{ Raman features and }6\text{ IR/Raman coincidences}}$. The centrosymmetric mutual-exclusion rule does not apply to $C_{2v}$ formaldehyde.

With $z$ along C=O and the molecule in the $xz$ plane, formaldehyde has $E$, $C_2(z)$, ${\sigma }_{xz}$ and ${\sigma }_{yz}$. It belongs to the C2v point group, $\boxed{C_{2v}}$. Keeping this axis convention is essential when assigning the labels $B_1$ and $B_2$.

The C=O length is invariant under every operation, so its stretch is $A_1$. If $s_L,s_R$ are the two C–H bond-length changes, their sum and difference giveHence $\boxed{{\Gamma }_{\mathrm{stretch}}=2A_1+B_1}$. These are stretching symmetry coordinates; the two $A_1$ motions can mix in actual normal modes, and mass-weighted atomic motions need not have the schematic arrow lengths below.

Ring cleavage of 1,3,5-Trioxane regenerates formaldehyde, whose strong carbon–oxygen pi bond stabilizes the monomer; in water much of it is further stabilized as its hydrate. acetal hydrolysis, particularly under acid catalysis, provides a route between the cyclic trimer and these monomeric forms.

The corresponding formal depolymerization of a siloxane would create ${\mathrm{R}}_2\mathrm{Si}=\mathrm{O}$ silanone units. Such ordinary small-substituent silanones have much less favourable pi bonding and are very reactive toward forming new Si–O–Si linkages. The connected siloxane is consequently preferred over a persistent aqueous solution of silanone monomers. This explains the intended contrast, but aqueous stability is conditional: acid or base can catalyse siloxane cleavage to silanols, and those silanols can condense again. Neither neutral-water inertness of every siloxane nor instantaneous neutral-water dissociation of every trioxane should be inferred.

Write the reacting terminus as $R-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OC}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as ${\mathrm{O}}^{-}-{\mathrm{O}}^{+}=\mathrm{O}$, the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments $R-\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ and formaldehyde.

Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes $R-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.

The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.