For the conversion shown, $n$-heptane $\xrightarrow{\text{(reagents)}}$ $Q$ + $R$, where $Q$ is benzoic acid ($\mathrm{C_6H_5COOH}$) and $R$ is benzyl alcohol ($\mathrm{C_6H_5CH_2OH}$), the correct sequence of reagents for the preparation of $Q$ and $R$ is:

(C) (i) $\mathrm{Cr_2O_3},\ 770\ \mathrm{K},\ 20\ \mathrm{atm}$; (ii) $\mathrm{CrO_2Cl_2},\ \mathrm{H_3O^+}$; (iii) $\mathrm{NaOH}$; (iv) $\mathrm{H_3O^+}$

(A) (i) $\mathrm{CrO_2Cl_2},\ \mathrm{H_3O^+}$; (ii) $\mathrm{Cr_2O_3},\ 770\ \mathrm{K},\ 20\ \mathrm{atm}$; (iii) $\mathrm{NaOH}$; (iv) $\mathrm{H_3O^+}$

(B) (i) $\mathrm{KMnO_4},\ \mathrm{OH^-}$; (ii) $\mathrm{Mo_2O_3},\ \Delta$; (iii) $\mathrm{NaOH}$; (iv) $\mathrm{H_3O^+}$

(D) (i) $\mathrm{Mo_2O_3},\ \Delta$; (ii) $\mathrm{CrO_2Cl_2},\ \mathrm{H_3O^+}$; (iii) $\mathrm{NaOH}$; (iv) $\mathrm{H_3O^+}$

Chapter_pending — JEE Main · 2023 · 25 Jan 2023 · Shift 1 Question with Solution

For the conversion shown,

n

-heptane

(reagents)

Q

R

, where

Q

is benzoic acid (

C_{6} H_{5} COOH

) and

R

is benzyl alcohol (

C_{6} H_{5} C H_{2} OH

), the correct sequence of reagents for the preparation of

Q

and

R

is:

(A)(i) $Cr O_{2} C l_{2}, H_{3} O^{+}$ ; (ii) $C r_{2} O_{3}, 770 K, 20 atm$ ; (iii) $NaOH$ ; (iv) $H_{3} O^{+}$
(B)(i) $KMn O_{4}, O H^{-}$ ; (ii) $M o_{2} O_{3}, Δ$ ; (iii) $NaOH$ ; (iv) $H_{3} O^{+}$
(C)(i) $C r_{2} O_{3}, 770 K, 20 atm$ ; (ii) $Cr O_{2} C l_{2}, H_{3} O^{+}$ ; (iii) $NaOH$ ; (iv) $H_{3} O^{+}$ ✓ Correct
(D)(i) $M o_{2} O_{3}, Δ$ ; (ii) $Cr O_{2} C l_{2}, H_{3} O^{+}$ ; (iii) $NaOH$ ; (iv) $H_{3} O^{+}$

Answer: (C)

Solution

n

-Heptane on heating over

C r_{2} O_{3}

770 K

and

20 atm

undergoes aromatisation (cyclisation with dehydrogenation) to give toluene. Toluene treated with chromyl chloride,

Cr O_{2} C l_{2}

, followed by hydrolysis is the Etard reaction, which oxidises the methyl group to the aldehyde, giving benzaldehyde. Benzaldehyde has no

α

-hydrogen, so with concentrated

NaOH

it undergoes the Cannizzaro reaction, disproportionating into sodium benzoate and benzyl alcohol; acidification of the sodium benzoate with

H_{3} O^{+}

liberates benzoic acid. Hence

Q

is benzoic acid and

R

is benzyl alcohol, obtained by treating

n

-heptane first with

C r_{2} O_{3}

770 K

and

20 atm

, then with

Cr O_{2} C l_{2} / H_{3} O^{+}

, then with

NaOH

, and finally with

H_{3} O^{+}

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Chapter_pending

Solution

More JEE Main questions on Chapter_pending