Which compound on oxidation with acidic KMnO_4 give Benzoic acid - Vidyadip

MCQ

Which compound on oxidation with acidic $KMnO_4$ give Benzoic acid

A
B
C
✓
All of these

✓

Answer

Correct option: D.

All of these

d

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$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their $\text{BDE}$ in Column $K$ is

Column $J$ Molecule	Column $K$ $\operatorname{BDE}( kcal mol -1)$
$(P)$ $H - C H \left( CH _3\right)_2$	${ (i) } 132$
$(Q)$ $H - CH _2 Ph$	${ (ii) } 110$
$(R)$ $H - C H = CH _2$	${ (iii) } 95$
$(S)$ $H - C \equiv CH$	${ (iv) } 88$

$(A) P - iii, Q - iv, R - ii, S - i$
$(B) P - i, Q - ii, R - iii, S - iv$
$(C) P - iii, Q - ii, R - i, S - iv$
$(D) P - ii, Q - i, R - iv, S - iii$
$(2)$ For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 \ kcal \ mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 \ kcal \ mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 \ kcal \ mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 \ kcal \ mol ^{-1}$.