Draw the formulas of the following compounds:

1) 2-Methylhexa-1,5-diene
2) 2,3,4-Trimethylocta-1,4,6-triene
3) 4-tert-Butyl-2-chlorohept-1-ene
4) 3-Ethyl-2,4-dimethylhept-3-ene
5) 3,4-Diisopropyl-2,5-dimethylhex-3-ene
6) Hept-1-ene
7) cis-Oct-3-ene
8) 3-ethylpent-2-ene
9) trans-1,4-Dibromobut-2-ene
10) 3-Chlorohex-2-ene
11) Buta-1,3-diene
12) Hexa-1,4-diene
13) 5-Methyl-3-propylocta-1,4,6-triene
14) 6-Methyl-6-propylnone-2,4,7-triene
15) 2,3,5-Trimethylocta-1,4-diene
16) 3-Propylhepta-1,5-diene.


SOLUTION:

1) 2-Methylhexa-1,5-diene

molecule 01
2) 2,3,4-Trimethylocta-1,4,6-triene

molecule 02
3) 4-tert-Butyl-2-chlorohept-1-ene

molecule 03
4) 3-Ethyl-2,4-dimethylhept-3-ene

molecule 04
5) 3,4-Diisopropyl-2,5-dimethylhex-3-ene

molecule 05
6) Hept-1-ene

molecule 06
7) cis-Oct-3-ene

molecule 07
8) 3-ethylpent-2-ene
molecule 08

I The cis notation indicates that the equal groups (hydrogens) of the 3,4 positions are oriented to the same side of the alkene.


9) trans-1,4-Dibromobut-2-ene

molecule 09
10) 3-Chlorohex-2-ene

molecule 10


IThe trans notation indicates that the equal groups of the double bond (hydrogens) are oriented to opposite sides.



11) Buta-1,3-diene

molecule 11
12) Hexa-1,4-diene

molecule 12
13) 5-Methyl-3-propylocta-1,4,6-triene

molecule 13
14) 6-Methyl-6-propylnone-2,4,7-triene

molecule 14
15) 2,3,5-Trimethylocta-1,4-diene

molecule 15
16) 3-Propylhepta-1,5-diene.
molecule 16