Match list I with list II and select the correct answer using the codes given below the lists.

	List I (Reactions)		List II (Products)
$\left(\mathrm{p}\right)$	$\mathrm{C}{\mathrm{H}}_2=\mathrm{C}\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+{\mathrm{H}}_2/\mathrm{N}\mathrm{i}\to$	$1.$	trans- $\mathrm{P}\mathrm{h}\mathrm{C}\mathrm{H}=\mathrm{C}\mathrm{H}\mathrm{O}\mathrm{O}\mathrm{H}$
$\left(\mathrm{q}\right)$	$\mathrm{c}\mathrm{i}\mathrm{s}-\mathrm{C}{\mathrm{H}}_3\mathrm{C}\mathrm{H}=\mathrm{C}\mathrm{H}\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+\mathrm{B}{\mathrm{r}}_2/\mathrm{C}\mathrm{C}{\mathrm{l}}_4\to$	$2.$	racemic-2,3-dibromobutanoic acid
$\left(\mathrm{r}\right)$	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH(OH)CH}}_{\text{2}}\text{COOH}\text{+}{\text{H}}^{\text{+}}\stackrel{\text{Δ}}{\to }$	$3.$	$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}$
$\left(\mathrm{s}\right)$	$\mathrm{o}-\mathrm{H}\mathrm{O}\mathrm{O}\mathrm{C}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{O}\mathrm{H}+{\mathrm{H}}^{+}\stackrel{\hspace{1em}\mathrm{\Delta }\hspace{1em}}{\to }$	$4.$

• A

  (p)-3, (q)-4, (r)-2, (s)-1
• B

  (p)-2, (q)-3, (r)-1, (s)-4
• Correct answer
  C

  (p)-3, (q)-2, (r)-1, (s)-4
• D

  (p)-3, (q)-2, (r)-4, (s)-1

Solution

The reactions are as follows

Hence, the correct answer is option: 'C'.