a.

* Hướng dẫn giải

a. \(\begin{array}{l}
C{H_3}C{\rm{OOH}}\,\,\,\,\,{\rm{ + }}\,\,\,\,\,{{\rm{K}}^{\rm{ + }}}\,\,\,\, + \,\,\,\,\,HCO_3^ - \,\,\, \to \,\,C{H_3}C{\rm{O}}{{\rm{O}}^{\rm{ - }}}\,\,\, + \,\,\,{{\rm{K}}^{\rm{ + }}}\,\, + \,\,\,{H_2}O\,\,\,\, + \,\,\,\,C{O_2}\\
C{H_3}C{\rm{OOH}}\,\,\,\,\,{\rm{ + }}\,\,\,\,\,\,\,\,\,HCO_3^ - \,\,\, \to \,\,C{H_3}C{\rm{O}}{{\rm{O}}^{\rm{ - }}}\,\,\, + \,\,\,{H_2}O\,\,\,\, + \,\,\,\,C{O_2}
\end{array}\)

b. \(\begin{array}{l}
C{H_3} - CH = C{H_2}\,\,\,\, + \,\,\,\,\,HCl\,\,\,\,\,\, \to \,\,\,\,C{H_3} - CHCl - C{H_3}\,\,\,\,\\
C{H_3} - CH = C{H_2}\,\,\,\, + \,\,\,\,\,HCl\,\,\,\,\,\, \to \,\,\,\,C{H_3} - C{H_2} - C{H_{2\,}}Cl\,
\end{array}\)

c. \(\begin{array}{l}
C{H_2} = CH - CH = C{H_2}\,\,\,\, + \,\,\,\,\,B{r_2}\,\,\,\, \to \,\,\,\,C{H_2}Br - CHBr - CH = C{H_2}\,( - {80^o}C)\\
C{H_2} = CH - CH = C{H_2}\,\,\,\, + \,\,\,\,\,B{r_2}\,\,\,\, \to \,\,\,\,C{H_2}Br - CH = CH - C{H_2}Br\,({40^o}C)
\end{array}\)