Question

Draw the structures for the following compounds.
a) (3Z,5S)-4-bromo-5-chloro-2-methylhexa-1,3-diene
b) (5S,7S)-5-(bromomethyl)-7-chloro-oct-2-yne
c) (1S,5S,6S)-6-bromo-5-[(1R)-1-chloroethyl]cyclohex-3-enol
d) 2-bromo-4-[(1R)-1-hydroxyethyl]benzoic acid
e) (3S,4S,5S)-5-amino-3-hydroxy-4-methylhexanal
f) (Z,2R,4R)-6-bromo-4-chloro-N-ethyl-N-methylhept-5-en-2-amine
g) [(1S)-1-chloroethyl] (Z,3R)-5-chloro-3-methoxyhex-4-enoate
h) (Z,3R)-3-bromo-4-chloro-7-methyloct-4-enoic acid
i) (3S)-6-bromo-N-ethyl-N,3-dimethylhex-4-ynamide
j) (Z,3R,4S)-6-bromo-3-[(1S)-1-chloroethyl]-4-hydroxyhept-5-en-2-one

Answer 1

a) (32,5S)-4-bromo-5-chloro-2-methylhexa-1,3-diene TY b) (59,78)-5-(bromomethyl)-7-chloro-oct-2-yne 2 c) (15,59,6S)-6-bromo-5-[(1R)-1-chloroethyl]cyclohex-3-enol 106 d) Br CI 2-bromo-4-[(1R)-1-hydroxyethyl]benzoic acid A Br OH (3S,4S,5S)-5-amino-3-hydroxy-4-methylhexanal NH2 OH f) (7,2R, 4R)-6-bromo-4-chloro-N-ethyl-N-methylhept-5-en-2-amine Br

g) [(1S)-1-chloroethyl] (7,3R)-5-chloro-3-methoxyhex-4-enoate 4 2 2 (7,3R)-3-bromo-4-chloro-7-methyloct-4-enoic acid Br O 5 86 OH 2 (3S)-6-bromo-N-ethyl-N,3-dimethylhex-4-ynamide 3 V 1N Br24 21 2 (7,3R,4S)-6-bromo-3-[(1S)-1-chloroethyl]-4-hydroxyhept-5-en-2-one BrOH O 7 5

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 know the absolute configuration at chiral carbons (carbons attached with four different groups), the priority of the attached groups has to given by Chan-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 Sand S becomes R