How do you name and draw 5 compounds that have the formula #C_7H_7Br#?

Well, you know that an alkane would have the general formula #C_nH_(2n+2)#, so this compound cannot be a straight-chained alkane.

That means you should try drawing alkenes and alkynes, and definitely try rings.

• I tried a five-membered ring with two sidechain carbons, and added conjugated double bonds to decrease the number of hydrogens. That's one isomer.
• I tried drawing a six-membered ring with one sidechain carbon. When I placed a bromine on that carbon, I achieved #C_7H_7Br#. I moved the bromine onto the ring for another isomer, and moved that bromine around the ring for two more isomers for a total of five.

Seems like there are at least 14 though. If you want, three of them are bromoene-ynes, which can be a challenge to name.

From left to right, top to bottom:

• 2-methyl-bromobenzene, or o-bromotoluene, or 1-bromo-2-methylbenzene
• 3-methyl-bromobenzene, or m-bromotoluene, or 1-bromo-3-methylbenzene
• 4-methyl-bromobenzene, or p-bromotoluene, or 1-bromo-4-methylbenzene
• I'm not sure how to name this one, so I looked it up and got:
  5-​(1-​bromoethylidene)​-1,​3-cyclopentadiene

This uses cyclopentadiene as the parent compound, and an ethylidene substituent (#"C"=stackrel("*")("C")-"CH"_3#, with bromine on the starred carbon), and the upper-left carbon on cyclopentadiene is carbon-1; we'd move counterclockwise.

• benzyl bromide, or bromomethylbenzene