Drawing chemical reactions

Question

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?

Answer 1

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 to chemfig using the bond's optional argument <departure>:

<bond>[<angle>,<length factor>,<departure>,<arrival>,<tikz>]

The code then is \chemfig{CH_3CHCH_3-[2,,3]OH}

\documentclass{article}
\usepackage{chemfig}
\usepackage[version=4]{mhchem}

\begin{document}

\ce{ \chemfig{CH_3CHCH_3-[2,,3]OH} + HI -> \chemfig{CH_3CHCH_3-[2,,3]I} + H2O}

\end{document}

Answer 2

Remove the \ce command, that will fix the links. The arrow is done with the command \arrow.

Also remember to enclose it between \schemestart and \schemestop.

Output

Code

\documentclass[11pt]{article}
\usepackage[version=4]{mhchem}
\usepackage{chemfig}

\begin{document}
\schemestart
\chemfig{CH_3}\chemfig{CH(-[2]OH)}\chemfig{CH_3} + HI \arrow(.mid east--.mid west)\chemfig{CH_3(-[2]I)}\chemfig{CH CH_3} + \chemfig{H_2O}
\schemestop
\end{document}

Answer 3

I thing that your code work well with [version=3]{mhchem}:

\documentclass[11pt]{article}

\usepackage[version=3]{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}