Question

Propene on reaction with each of the reagents given in column I gives products given in column II. Match the column and select the correct answer using the code given below:

Column I		Column II
(A)	$KMnO_4/H^+$	(1)	Propan-2-ol
(B)	$KMnO_4/OH^-$	(2)	Propan 1, 2,diol
(C)	$H_2O/H^+$	(3)	Acetic acid and $CO_{2}$
(D)	$O_3/Zn +H_2O$	(4)	Acetaldehyde + formaldehyde

• A. (A)-(4), (B)-(3), (C)-(2), (D)-(1)
• B. (A)-(3), (B)-(2), (C)-(1), (D)-(4)
• C. (A)-(3), (B)-(1), (C)-(2), (D)-(4)
• D. (A)-(3), (B)-(2), (C)-(4), (D)-(1)

Answer 1

Answer: (B)

(a) $\underset{\text{propene}}{CH_3CH} = CH_{2} CH_2 \xrightarrow[\text{(Oxidation)}]{KMnO_4/H^+}\underset{\text{Acetic acid}}{CH_3COOH} + CO_2$

(b) $\underset{\text{propene}}{CH_3CH} = CH_{2}CH_2 \xrightarrow[\text{(Hydroxylation)}]{KMnO_4/H^-}\underset{\text{propane- 1, 2-diol}}{CH_3COOH} - CH_2OH$

(c) $\underset{\text{propene}}{CH_3CH} = CH_{2}CH_2 \xrightarrow[\text{(Mark.addition)}]{H_O/H^+}\underset{\text{propan-2-ol}}{CH_3CHOH} - CH_3$

(d) $\underset{\text{Propene}}{CH_3CH = CH_2} \xrightarrow[(ii) Zn + H_2O \text{(Reductive ozonolysis)}]{(i)O_3}$

$\underset{\text{Acetaldehyde}}{CH_3 - CHO} + \underset{\text{Formaldehyde}}{O = CH_2}$