we know DBE (double bond equivalent) The formula here; 7. DBE - one double bond will be present structures (according to
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we know DBE (double bond equivalent) The formula here; 7. DBE - one double bond will be present structures (according to
question double bonds only) frams isomer IN 14 He: He=e=ch 40/ of the write ½C=CH E 111 -CH₂ structures with CH₂ 50- c+l - - +4 CH₂ "c=no. of C-aton] H=no of H-atomy [x = no of halogenate N = No. of N-akm 441-1/2/20 5-4=1 (50, one double bond or a ring an each we should cis isomer
we know DBE (double bond equivalent) The formula here; 7. DBE - one double bond will be present structures (according to