- AAniline
- BMethylamine
- CEthyl benzoate
- ✓Phenol
So, Liebermann's test is answered by phenol.
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$(a)$ $NCl_3$ is formed when $Cl_2$ is present in excess.
$(b)$ $N_2$ is formed when $Cl_2$ is present in excess.
$(c)$ $N_2$ is formed when $NH_3$ is present in excess.
$(d)$ $NCl_3$ is formed when $NH_3$ is present in excess.
Correct option are
$I.$ ${Sn}-{HCl}$ $II.$ ${Sn}-{NH}_{4} {OH}$ $III.$ ${Fe}-{HCl}$ ${IV} . {Zn}-{HCl}$ $V.$ ${H}_{2}-{Pd}$ $VI.$ ${H}_{2}-$ રેની નિકલ
$k=A\,e-^{E/RT}$
In this equation, $E$ represents
Compound $X$ and $Y$ are respectively.
$(a)\,\,q + w$ $ (b)\,\,q$
$(c)\,\,w$ $ (d)\,\,H -TS$