Given below are two statements:

Statement-I: The condensation reaction between $CH_3-CH=O$ and under optimum pH will produce .

Statement-II: The molecule will generate $Ph-CH=O$ in the presence of dilute acid.

In the light of the above statements, choose the correct answer from the options given below:

In Statement-I, evaluating the structure of semicarbazide ($\text{H}_2\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2$), the lone pair of electrons on the $-\text{NH}_2$ group directly attached to the carbonyl carbon is delocalized due to resonance. The terminal $-\text{NH}_2$ group of the hydrazine part is not involved in resonance with the carbonyl group, making it significantly more nucleophilic.

Therefore, the condensation reaction of acetaldehyde with semicarbazide under optimum pH occurs at the terminal $-\text{NH}_2$ group to yield acetaldehyde semicarbazone:

$\text{CH}_3\text{CH}=\text{O} + \text{H}_2\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2 \rightarrow \text{CH}_3\text{CH}=\text{N}-\text{NH}-\text{C}(=\text{O})-\text{NH}_2 + \text{H}_2\text{O}$

The product provided in Statement-I indicates an attack by the less nucleophilic amide nitrogen to form $\text{CH}_3\text{CH}=\text{N}-\text{C}(=\text{O})-\text{NH}-\text{NH}_2$, which is incorrect. Thus, Statement-I is false.

In Statement-II, the given molecule $\text{Ph}-\text{CH}(\text{OH})(\text{OCH}_3)$ is a hemiacetal. Hemiacetals are unstable in the presence of dilute aqueous acid and undergo rapid hydrolysis to regenerate the corresponding carbonyl compound and alcohol.

$\text{Ph}-\text{CH}(\text{OH})(\text{OCH}_3) \xrightarrow{\text{dil. H}^+} \text{Ph}-\text{CH}=\text{O} + \text{CH}_3\text{OH}$

This hydrolysis reaction successfully generates benzaldehyde ($\text{Ph}-\text{CH}=\text{O}$). Thus, Statement-II is true.

We conclude that Statement-I is false but Statement-II is true.