Carboxylic Acid Derivatives Question 2
Question 2 - 25 January - Shift 1
The correct sequence of reagents for the preparation of $Q$ and $R$ is :
(1) (i) $Cr_2 O_3, 770 K, 20 atm$;
(ii) $CrO_2 Cl_2, H_3 O^{+}$;
(iii) $NaOH$;
(iv) $H_3 O^{+}$
(2) (i) $CrO_2 Cl_2, H_3 O^{+}$; (ii) $Cr_2 O_3, 770 K, 20 atm$;
(iii) $NaOH$; (iv) $H_3 O^{+}$
(3) (i) $KMnO_4, OH^{-}$; (ii) $Mo_2 O_3, A$; (iii) $NaOH$; (iv) $H_3 O^{+}$
(4) (i) $Mo_2 O_3, \Delta$; (ii) $CrO_2 Cl_2, H_3 O^{+}$; (iii) $NaOH$; (iv) $H_3 O^{+}$
Show Answer
Answer: (1)
Solution:
Formula: Cannizzaro Reaction
$\widehat{(P)}$
$\xrightarrow{\text{ (i) } Cr_2 O_3, 770 K, 20 atm}$
$\overbrace0^{CH}$
$\xrightarrow[H_3 O]{\text{ (ii) } CrO_2 Cl_2}$
(iii) $NaOH \lvert, \begin{aligned} & \text{ Cannizaro } \\ & \text{ Reaction }\end{aligned}.$
0
(iv) $H_3 O$
$COOH \quad CH_2 OH$
自
(Q) (R)
