Question

How many configurational isomers are possible for the following compounds?

A. 4-chloro-3-hexen-2-ol

B. 2,4-hexadiene

C. 3-chloro-1,4-pentadiene

Answer 1

Answer: There are far too many to do all three for you, but the general approach is "be methodical". Draw the structural skeletal formula for the starting substance, then work at moving the important features (the double bonds) to other locations. Once all positions for the hexadienes have been located, remove an end carbon (a methyl) and begin placing it as an interior carbon substitution on the patterns already generated (where the double bonds are moved to different locations. These will all be methylpentadienes. After that, do the dimethylbutadienes.

For example, starting with the hexadienes you can write 1,2; 1,3; 1,4; and 1,5 hexadienes then 2,3; 2,4; 2,5 hexadienes followed by 3,4; and 3,5 hexadienes and finally 4,5 hexadiene - that gives us 10 isomers for hexadienes.

For the methylpentadienes we can write: 2-methyl-1, 3-pentadiene; 2-methyl-1, 4...; 3-methyl-1, 3...
3-methyl-1, 4...; 4-methyl-1, 3...;4-methyl-1, 2...;and 4-methyl-1, 4pentidiene thats 7 more isomers

Finally there is one dimethylbutadiene: 2,3-dimethyl-1,4-butadiene

The total is 18 isomers for the first one only.

If you'll sit down and write out each of these iosmers in the order presented you'll see what "be methodical" means. I use skeletal formulas when I'm planning like:
C=C=C-C-C-C and C=C-C=C-C-C and C=C-C-C=C-C and C=C-C-C-C=C
or.. C
.....|
C=C-C-C=C