Identify the likely reaction sequence

The reagent set contains:

• $\mathrm{HNO_2}$ : used for diazotization of a primary aromatic amine
• $\mathrm{KI}$ : replaces diazonium group by iodine
• $\mathrm{Fe/H^+}$ : reduces a nitro group $(-\mathrm{NO_2})$ to amine $(-\mathrm{NH_2})$
• $\mathrm{H_2O/H^+}$ or warm water: hydrolysis of diazonium salt to phenol

So the sequence must involve:

• converting an $-\mathrm{NH_2}$ into diazonium,
• replacing it with $-\mathrm{I}$,
• reducing another $-\mathrm{NO_2}$ to $-\mathrm{NH_2}$,
• diazotizing again,
• then hydrolyzing to $-\mathrm{OH}$.

Construct the logical order

Suppose the starting compound contains both an amino group and a nitro group on benzene.

• First, the existing $-\mathrm{NH_2}$ is treated with $\mathrm{HNO_2}$ to form diazonium salt.
• Then $\mathrm{KI}$ substitutes the diazonium group by $-\mathrm{I}$.
• Next, the remaining $-\mathrm{NO_2}$ group is reduced by $\mathrm{Fe/H^+}$ to $-\mathrm{NH_2}$.
• Then this newly formed amine is diazotized again using $\mathrm{HNO_2}$.
• Finally, warming with water converts diazonium salt to phenol.

Thus the order is: $\mathrm{HNO_2} \;\to\; \mathrm{KI} \;\to\; \mathrm{Fe/H^+} \;\to\; \mathrm{HNO_2} \;\to\; \mathrm{H_2O/warm}$

Match with options

• A: $\mathrm{HNO_2}, \mathrm{Fe/H^+}, \mathrm{HNO_2}, \mathrm{KI}, \mathrm{H_2O/H^+}$
  Wrong, because reduction should not come before replacing the first diazonium by iodine.

• B: $\mathrm{HNO_2}, \mathrm{KI}, \mathrm{Fe/H^+}, \mathrm{HNO_2}, \mathrm{H_2O/warm}$
  Correct.

• C: $\mathrm{HNO_2}, \mathrm{KI}, \mathrm{HNO_2}, \mathrm{Fe/H^+}, \mathrm{H_2O/H^+}$
  Wrong, because diazotization cannot occur before reducing $-\mathrm{NO_2}$ to $-\mathrm{NH_2}$.

• D: $\mathrm{HNO_2}, \mathrm{Fe/H^+}, \mathrm{KI}, \mathrm{HNO_2}, \mathrm{H_2O/warm}$
  Wrong, for the same reason as A.

Final answer

The correct sequential order is: $\boxed{\mathrm{HNO_2},\ \mathrm{KI},\ \mathrm{Fe/H^+},\ \mathrm{HNO_2},\ \mathrm{H_2O/warm}}$

So, Option B is correct.