Consider the following reactions:
$\left(\mathrm{a}\right)$ ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_3\mathrm{C}\mathrm{C}\mathrm{H}\left(\mathrm{O}\mathrm{H}\right)\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{c}.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }$
$\left(\mathrm{b}\right)$ ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{C}\mathrm{H}\mathrm{C}\mathrm{H}\left(\mathrm{B}\mathrm{r}\right)\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{a}\mathrm{l}\mathrm{c}.\mathrm{K}\mathrm{O}\mathrm{H}}{\to }$
$\left(\mathrm{c}\right)$ ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{C}\mathrm{H}\mathrm{C}\mathrm{H}\left(\mathrm{B}\mathrm{r}\right)\mathrm{C}{\mathrm{H}}_3\stackrel{{\left(\mathrm{C}{\mathrm{H}}_3\right)}_3{\mathrm{O}}^{\ominus }{\mathrm{K}}^{\oplus }}{\to }$
$\left(\mathrm{d}\right)$

Which of the reaction(s) will not produce Saytzeff product?

We have to know that a Saytzeff product is basically an alkene compound which has undergone many substitution reactions to form the alkene product. From Saytzeff rule, a stable alkene is formed when removal of hydrogen from beta carbon has a low number of hydrogen atoms.