Answer
a) 2-pentene
b) 4-methyl-2-pentene
c) 3,3-dimethyl-1-butene
d) 3,4-dimethyl-1-hexene
Work Step by Step
a) No. of Carbon in straight chain = 5. So, main prefix will be 'Pent'. Numbering of Carbon atom is done so that Carbon associated with double bond gets the minimum number. Here, numbering is done from left side and the first Carbon associated with double bond gets number 2.
2 is added before main prefix and the ending name will be 'ene' for alkene.
b) No. of C in straight chain = 5. Hence, main prefix is ‘Pent’.
Numbering is done from right side so that first ‘C’ associated with double bond get minimum number. In this case, the first ‘C’ associated with double bond gets ‘2’ if numbering is done from right side. 2 and a hyphen will be added before ‘Pent’ i.e. ‘2-pent’.
Now, here a substitute methyl (CH3) is present at no. 4 Carbon. So, we need to add ‘4-methyl’ before main prefix.
Also, the ending name will be ‘ene’.
c) No. of C in straight chain = 4. Hence, main prefix is ‘but’.
Numbering will be done from right side. (first ‘C’ associated with double bond gets number 1) and the number 1 is added before ‘but’.
Two number of methyl group is present in no. 3 Carbon. So, ‘3,3-dimethyl’ is added before naming main prefix.
Ending name is ‘ene’.
d) No. of C in straight chain = 6. Hence, main prefix is ‘hex’.
Numbering will be done from right side. First ‘C’ associated with double bond gets 1. So, number 1 is added before ‘hex’.
Two number of methyl group is present in no. 3 and no. 4 Carbon respectively. So, ‘3,4-dimethyl’ is added before ‘1-hex’.
Ending name is ‘ene’.
