Question

Consider the following reaction:

The correct set of reagents for these reactions

• A. $R_1$ - $H_2/Lindla{r}^{\prime }scatalyst$
  $R_2$ - $Na/liq.N{H}_3$
  $R_3$ - $\left(i\right)O_3\left(ii\right)H_{2}O$
  $R_4$ - $H_{2}O,H_{2}S{O}_4,HgS{O}_4$
• B. $R_1$ - $H_2/Lindla{r}^{\prime }scatalyst$
  $R_2$ - $Na/liq.N{H}_3$
  $R_3$ - $H_{2}O,H_{2}S{O}_4,HgS{O}_4$
  $R_4$ - $\left(i\right)O_3\left(ii\right)H_{2}O$
• C. $R_1$ - $\left(i\right)O_3\left(ii\right)H_{2}O$
  $R_2$ - $H_{2}O,H_{2}S{O}_4,HgS{O}_4$
  $R_3$ - $Na/liq.N{H}_3$
  $R_4$ - $H_2/Lindla{r}^{\prime }scatalyst$
• D. $R_1$ - $H_{2}O,H_{2}S{O}_4,HgS{O}_4$
  $R_2$ - $H_2/Lindla{r}^{\prime }scatalyst$
  $R_3$ - $\left(i\right)O_3\left(ii\right)H_{2}O$
  $R_4$ - $Na/liq.N{H}_3$

Answer 1

The correct option is B $R_1$ - $H_2/Lindla{r}^{\prime }scatalyst$

$R_2$ - $Na/liq.N{H}_3$
$R_3$ - $H_{2}O,H_{2}S{O}_4,HgS{O}_4$
$R_4$ - $\left(i\right)O_3\left(ii\right)H_{2}O$
$R_1$ - Lindlar's catalyst gives cis product
$R_2$ - $Na/liq.N{H}_3$ gives trans product
$R_3$ - undergoes oxymercuration-demercuration to give ketone
$R_4$ - Ozonolysis followed by hydrolysis give acid.