= Solution
Write the reacting terminus as $R{-}\mathrm{CH_2{-}CH{=}CH_2}$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OCH_2C(=O)NMe_2}$ <substituent>. <ozone> undergoes 1,3-dipolar <cycloaddition> across the <terminal alkene> to give a <molozonide>. In a resonance description of <ozone> as $\mathrm{O^-{-}O^+{=}O}$, 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 $R{-}\mathrm{CH_2CHO}$ 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 \b[the retained chain becomes $R{-}\mathrm{CH_2CH_2OH}$]. The <terminal alkene> <carbon> becomes <formaldehyde> and can be reduced to <methanol>; it is no longer part of the main <product (chemistry)>. 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 (chemistry)>. The new terminal <alcohol> <carbon> is CH2 and therefore adds no <stereocenter>.
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