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Question 48

The correct sequential order of the reagents for the given reaction is

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This question asks about the correct sequential order of reagents for a given reaction. Since the question references a reaction diagram that is not available, we will reason based on the given answer.

The correct answer is Option B: $$HNO_2$$, KI, $$Fe/H^+$$, $$HNO_2$$, $$H_2O$$/warm

Since the starting amine group ($$-NH_2$$) is converted to a diazonium salt ($$-N_2^+$$), nitrous acid at 0-5 °C (from $$NaNO_2 + HCl$$) is used for diazotization:

$$Ar-NH_2 \xrightarrow{HNO_2} Ar-N_2^+Cl^-$$

After diazotization, the diazonium group is replaced by iodine using potassium iodide in a Sandmeyer-type reaction:

$$Ar-N_2^+Cl^- \xrightarrow{KI} Ar-I + N_2 + KCl$$

Next, a nitro group ($$-NO_2$$) present elsewhere on the ring is reduced to an amine group ($$-NH_2$$) with iron in acidic medium:

$$Ar-NO_2 \xrightarrow{Fe/H^+} Ar-NH_2$$

Since the newly formed $$-NH_2$$ group must be diazotized again, nitrous acid is once more employed:

$$Ar-NH_2 \xrightarrow{HNO_2} Ar-N_2^+Cl^-$$

Finally, hydrolysis by warming with water converts the diazonium salt into a phenol ($$-OH$$) group:

$$Ar-N_2^+Cl^- \xrightarrow{H_2O/\text{warm}} Ar-OH + N_2 + HCl$$

Therefore, the correct sequence is Option B: $$HNO_2$$, KI, $$Fe/H^+$$, $$HNO_2$$, $$H_2O$$/warm.

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