How do you draw #(1)# 3-fluoro-4,5-dimethyl-1-heptyne and #(2)# 2-chloro-4-ethyl-1,3-cyclohexadiene?
Not sure why you have two questions.
- Since the compound is a heptyne, it is an alkyne with 7 carbons on the main chain; hept = 7, yne = alkyne. The "1" indicates the triple bond is on carbon-1, i.e. the terminal carbon.
- The 4,5-dimethyl indicates 1 methyl group on carbon 4 and 1 on carbon 5.
- The 3-fluoro indicates 1 fluorine on carbon 3.
Hence, we have:
The main chain (a seven-membered straight-chained hydrocarbon) is highlighted.
- Since the prefix is "cyclo" , this is a ring. Hence, choose one carbon as your carbon-1, and choose a direction to move (CW/CCW). Then carbon-3 is two away from carbon-1.
- Furthermore, "hexa" indicates a six-membered ring.
- Since the compound is a diene, it is an alkene that contains 2 double bonds. Based on "1,3- . . .", they are located on carbons 1 and 3.
- The 2-chloro indicates that a chlorine is on carbon-2.
- The 4-ethyl indicates that a two-carbon alkyl fragment, i.e.
#-"CH"_2"CH"_3#, is on carbon-4.
That's all you need to know to generate this:
The parent compound (a six-membered ring) is highlighted.