Starting with pure, unsubstituted benzene, a chemist performs the following flawless synthetic seque — Haloalkanes and Haloarenes Chemistry Question
Question
Starting with pure, unsubstituted benzene, a chemist performs the following flawless synthetic sequence: 1) Nitration ($HNO_3/H_2SO_4$) 2) Bromination ($Br_2/Fe$) 3) Reduction ($Sn/HCl$) 4) Diazotization ($NaNO_2/HCl, 0^\circ C$) 5) Balz-Schiemann reaction ($HBF_4$, heat). What is the mathematical sum of the numerical locants assigned strictly to the halogen substituents in the correct IUPAC name of the final isolated major product?
💡 Solution & Explanation
Step 1 yields nitrobenzene. Step 2 introduces a bromine atom. Because the $-NO_2$ group is strongly meta-directing, the product is $m$ -bromonitrobenzene. Step 3 reduces the nitro group to an amine, yielding $m$ -bromoaniline. Step 4 converts the amine to a diazonium salt, yielding $m$ -bromobenzenediazonium chloride. Step 5 replaces the diazonium group with fluorine. The final molecule contains a bromine and a fluorine atom in a meta relationship. Alphabetically, bromo comes before fluoro, so it is named 1-bromo-3-fluorobenzene. The sum of the halogen locants is $1 + 3 = 4$.