The increasing order of the following compounds towards HCN addition is:

The addition of cyanide ($$\text{CN}^-$$) from $$\text{HCN}$$ to a carbonyl group ($$-\text{CHO}$$) is a nucleophilic addition reaction. The rate of this reaction is driven by two main factors:

1. Electronic Factor: Electron-withdrawing groups (via $$-I$$ or $$-M$$ effects) decrease the electron density on the carbonyl carbon, making it more electrophilic (more positive) and highly reactive toward nucleophiles. Conversely, electron-donating groups (via $$+I$$ or $$+M$$ effects) make the carbon less positive and less reactive.
2. Steric Factor: Bulky groups positioned near the carbonyl carbon physically block the approach of the incoming nucleophile, slowing down the reaction.

Detailed Analysis of the Compounds:

• Compound (ii): 2-Nitrobenzaldehyde

  • The nitro group ($$-\text{NO}_2$$) is a powerful electron-withdrawing group via both inductive ($$-I$$) and resonance ($$-M$$) effects.
  • When placed at the ortho-position, its proximity maximizes the electron-withdrawing impact on the aldehyde carbon, making it exceptionally electrophilic and the most reactive compound in the set.



• Compound (iv): 3-Nitrobenzaldehyde

  • Here, the electron-withdrawing nitro group ($$-\text{NO}_2$$) is at the meta-position. At this position, it can only exert its weaker inductive ($$-I$$) effect, not its resonance ($$-M$$) effect.
  • It is more reactive than benzaldehyde due to the $$-I$$ effect, but significantly less reactive than the ortho-isomer.



• Compound (i): 4-Methoxybenzaldehyde (Anisaldehyde)

  • The methoxy group ($$-\text{OCH}_3$$) is at the para-position. While it is inductively electron-withdrawing ($$-I$$), its strong lone-pair delocalization creates a dominant electron-donating resonance effect ($$+M$$).
  • This increases electron density at the carbonyl carbon, reducing its electrophilic character and making it less reactive than unsubstituted benzaldehyde.



• Compound (iii): 2-Methoxybenzaldehyde

  • The methoxy group ($$-\text{OCH}_3$$) is at the ortho-position. In addition to donating electrons via resonance ($$+M$$), its close proximity creates significant steric hindrance directly next to the incoming nucleophile.
  • The combination of electronic donation and heavy steric crowding makes it the least reactive compound in the entire group.

Conclusion:

Combining these observations, the reactivity increases from the sterically hindered electron-rich system to the electron-deficient nitro system:

$$\text{(iii)} < \text{(i)} < \text{(iv)} < \text{(ii)}$$

Answer: Option C — (iii) < (i) < (iv) < (ii)