Question

3. Name each of the given compounds: b) Br

Answer 1

Longest carbon chain has 7 carbon atoms, hence rootword is 'hept' alkene is present at 3rd carbon, hence suffix is '3-ene' Same priority groups present on same side of alkene, hence alkene configuration is "Z" (Z)-3-ethyl-2,4,5-trimethylhept-3-ene Longest carbon chain has 8 carbon atoms, hence rootword is 'octa' alkenes are present at 3rd and 5th carbons, hence suffix is '3,5- diene Same priority groups present on same side of alkene, hence alkene configuration is "Z" (37,5%)-3-bromo-2,2,4,5,6-pentamethylocta-3,5-diene 31 5 Br Longest carbon chain has 7 carbon atoms, hence rootword is Thepta' 3 alkenes are present, hence suffix is '1,3,5-triene Same priority groups present on same side of alkene, hence alkene configuration is "Z" 3-ethyl-5-isopropyl-1-methylcyclohepta-1,3,5-triene six carbons present in cyclic form, hence rootword is 'cyclohex' The absolute configuration of chiral centers at 5 and 6 carbons is R 22 alkenes are present at 1st and 3rd carbons, hence suffix is '1,3- diene' 1 (5R,6R)-5-bromo-6-ethylcyclohexa-1,3-diene 12 26 Longest carbon chain has 10 carbon atoms, hence rootword is 'deca' 3 alkenes are present, hence suffix is '2,4,6-triene' 8 10 Same priority groups opposite side at 2 and 4th alkenes and hence alkene configuration is "E" (S2E,45,6Z)-8-chloro-4-ethyl-3,5,7,9- tetramethyldeca-2,4,6-triene 5 carbons present in cyclic form, hence rootword is 'cyclopent' The absolute configuration of chiral centers at 3 carbon is R alkenes are present at 1st and 3rd carbons, hence suffix is '1,3- diene (R)-3-ethyl-1,5- dimethylcyclopenta-1,3-diene

The IUPAC name of the any compound contains 3 parts prefix+ rootword + suffix Prefix = any substitutents rootword = longest chain containing most number of carbons if the chain contains 1 carbon = meth 2 carbon = eth 3 carbon = prop 7 carbon = hept 4 carbon = but 8 carbon = oct 5 carbon = pent 9 carbon = non 6 carbon = hex and so on Suffix = if the compound contains single bond (C-C) only then suffix 'ane double bonds (C=C) 'ene triple bonds 'yne For selecting prefix numbering, the substituents numbers has to be given in such way that the sum of the all substituents should be as less as possible. For selecting rootword chain, the chain with more number of carbons has to be selected In all the given compounds alkanes are present, hence the suffix is 'ane' in all the four compound For all the compounds, blue coloured part of the name is prefix. red coloured part of the name is rootword and pink coloured part of the name is suffix

To give the E and Z nomenclature to the double bond, if same priority number (molecular weight wise) on the same side then that alkene is Z, if same priority numbers are in opposite side then the configuration of alkene is E This side first atom is carbon on both side, isopropyl gets first priority than methyl first priority groups are on same side, hence olefin configuration is Z F gets 1 because it's atomic number is 9 (greater than C)

The configuration for the both estrogenic centers is R HO Estrogenic center or chiral center is the center that is attached to four different groups to it. In the given molecule, there are two such centers marked with asterisk (*). To know the absolute configuration at chiral carbons (carbons attached with four different groups), the priority of the attached groups has to given by Cahn-Ingold-Prelog (CIP) rules. According to CIP rules the group with highest molecular weight should get first priority (1) followed by the groups with next highest molecular weights. If the attached groups are same in the adjacent atoms, one has to move further to see different atoms. After giving priorties according to CIP rules, if the rotation of the priority groups (from 1st priority group to 3rd priority group) is clock wise then the configuration of that chiral center is R if the rotation of the priority groups (from 1st priority group to 3rd priority group) is anti clock wise then the configuration of that chiral center is S If the least priority group (40 is present above the plane then the configuration of the chiral center should be opposite. That means R becomes S and S becomes HO Clock wise orientation of the priority groups and hence configuration is R HO ZI anti clock wise orientation of the priority groups but the least priority group (H) present above the plane and hence configuration is R