The number of stereoisomers possible for the | vedclass

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(C) The given compound is $CH_3-CH=CH-CH(Br)-CH_2-CH_3$.This molecule contains one chiral carbon atom (indicated by the asterisk in the structure) and

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$4$

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(C) The given compound is $CH_3-CH=CH-CH(Br)-CH_2-CH_3$.This molecule contains one chiral carbon atom (indicated by the asterisk in the structure) and one carbon-carbon double bond.For a molecule with $n$ chiral centers and $m$ double bonds that can exhibit geometrical isomerism,where the molecule is unsymmetrical,the total number of stereoisomers is given by $2^n 	imes 2^m$ if the double bond is not part of the chiral center system.Here,$n=1$ (one chiral center) and $m=1$ (one double bond capable of $cis-trans$ isomerism).Total stereoisomers $= 2^1 	imes 2^1 = 2 	imes 2 = 4$.The four stereoisomers are: $(cis, R)$,$(cis, S)$,$(trans, R)$,and $(trans, S)$.

The number of stereoisomers possible for the following compound is $CH_3-CH=CH-CH(Br)-CH_2-CH_3$

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