I am trying to draw reactions like this
I tried it like this
\documentclass[11pt]{article}
\usepackage[version=4]{mhchem}
\usepackage{chemfig}
\begin{document}
\ce{ \chemfig{CH_3-CH (-[2]OH)-CH_3} + HI -> \chemfig{CH_3-CH(-[:90]I)-CH_3} + H2O}
\end{document}
the result looks very messy
.So how do i accomplish making such structures by showing only one bond between OH and Carbon at reactant and one bond at iodine and Carbon at product?
Best Answer
You want this result, right?
Then
CH3CHCH3
must be considered as one group of 6 atoms (C, H, C, H, C and H) where the bond is leaving from the third atom. You need to tell this tochemfig
using the bond's optional argument<departure>
:The code then is
\chemfig{CH_3CHCH_3-[2,,3]OH}