The three most common doubts from this month's questions

Three doubts came up more than once this month. If you were about to ask one of these, you're not alone.

"Why does anisole react with HI to give phenol + methyl iodide, and not iodobenzene?"

It comes down to which bond is actually weaker. In anisole ($\text{C}_6\text{H}_5\text{-O-CH}_3$), the phenyl-oxygen bond has partial double-bond character from resonance with the ring, which makes it resistant to nucleophilic attack. The oxygen-methyl bond has no such stabilization, so once the ether oxygen is protonated, iodide attacks the methyl carbon instead:

$$\text{C}_6\text{H}_5\text{-O-CH}_3 + \text{HI} \longrightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I}$$

The rule "iodide attacks the less hindered carbon" that you learn for dialkyl ethers still applies here — it's just that with an aryl group, the ring bond was never a realistic option to break in the first place.

"If picric acid and phenol are both phenols, why is one so much more acidic?"

Because acidity here is entirely about how well the negative charge on the phenoxide ion gets stabilized after the O-H proton leaves. Picric acid (2,4,6-trinitrophenol) has three nitro groups, each pulling electron density away from the ring by both resonance and induction, spreading and stabilizing that negative charge far more effectively than phenol's ring alone can:

$$\text{p}K_a: \text{picric acid (0.38)} < 4\text{-nitrophenol (7.1)} < \text{phenol (10.0)}$$

More electron-withdrawing groups on the ring means a more stable conjugate base, which means a stronger acid — the same logic that ranks substituted phenols against each other in general.

"Why doesn't benzaldehyde give a positive iodoform test?"

The iodoform test only works on compounds with a $\text{CH}_3\text{-CO-}$ group (methyl ketones) or a $\text{CH}_3\text{-CH(OH)-}$ group (secondary methyl carbinols), because the mechanism specifically needs three alpha-hydrogens on a methyl group next to the carbonyl or hydroxyl. Benzaldehyde is $\text{C}_6\text{H}_5\text{-CHO}$ — the carbon attached to the carbonyl is the aromatic ring, not a methyl group, so there's no methyl ketone pattern for the reaction to act on. It's an easy trap: students see "aldehyde" and assume oxidation-related tests apply automatically, but the iodoform test is about a specific structural pattern, not the functional group class in general.