= Solution
The reactive fragment is the C6 amidine-like portion of the <purine> ring; the N9–<ribose> attachment is retained. Write it locally as $\mathrm{N1{=}C6(NH_2){-}}$, with the rest of the fused ring understood. One productive <acid catalysis> route protonates N1, making C6 more electrophilic. <water> then adds at C6 while the C6=N1 pi <electrons> move onto N1. Subsequent <proton transfers> generate a <carbinolamine> and then a protonated amino leaving group:
$$
\begin{aligned}
\mathrm{N1{=}C6(NH_2)}&\xrightarrow{\ +H^+\ }\mathrm{N1H^+{=}C6(NH_2)}\\
&\xrightarrow{\ +H_2O\ }\mathrm{N1H{-}C6(OH_2^+)(NH_2)}\\
&\xrightarrow{\ -H^+\ }\mathrm{N1H{-}C6(OH)(NH_2)}\\
&\xrightarrow{\ +H^+\ }\mathrm{N1H{-}C6(OH)(NH_3^+)}\\
&\xrightarrow{\ -NH_3\ }\mathrm{N1H{-}C6({=}OH^+)}\\
&\xrightarrow{\ -H^+\ }\mathrm{N1H{-}C6({=}O)}.
\end{aligned}
$$
These are local bonding fragments: C6 retains its bond to the adjoining ring <carbon> at every step. In the addition step, an <oxygen> <lone pair> forms C6–O and the C6=N1 <pi bond> opens. During collapse, an <oxygen> <lone pair> reforms C6=O and the C6–N6 bond breaks toward the departing <nitrogen>. Solvent mediates each <proton transfer>; no direct unsupported transfer across a long distance is required.
\b[<water> supplies the new C6 <oxygen>, and <ammonia> is expelled after <protonation> of the leaving group.] Final <deprotonation> gives the <lactam> form of <inosine>. In acidic bulk solution the released <ammonia> is largely trapped as <ammonium>. The catalytic <proton transfers> regenerate acid around the substitution cycle; <ammonium> trapping separately affects the net proton balance. This is <hydrolytic deamination>, not cleavage of the <glycosidic bond>.
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