In the above conversion of compound (X) to pr | vedclass

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(D) The conversion of $p$-nitrotoluene $(X)$ to $3,5$-dibromotoluene $(Y)$ involves the following steps: $1$. Reduction of the nitro group to an amino

Vedclass · Accepted Answer

$(I) Fe, H^{+}$ $(II) Br_2 (aq)$ $(III) HNO_2$ $(IV) H_3PO_2$

Vedclass · Answer

(D) The conversion of $p$-nitrotoluene $(X)$ to $3,5$-dibromotoluene $(Y)$ involves the following steps: $1$. Reduction of the nitro group to an amino group: $p$-nitrotoluene is reduced using $Fe/H^{+}$ to form $p$-toluidine. $2$. Bromination: $p$-toluidine is treated with $Br_2 (aq)$ to form $2,6$-dibromo-$4$-methylaniline due to the strong activating effect of the $-NH_2$ group. $3$. Diazotization: The amino group is converted to a diazonium salt using $HNO_2$ at $0-5 \ ^{\circ}C$. $4$. Deamination: The diazonium group is removed using $H_3PO_2$ and water to yield $3,5$-dibromotoluene $(Y)$.

In the above conversion of compound $(X)$ to product $(Y)$,the sequence of reagents to be used will be:

Similar Questions

Identify the products $y$ and $x$ in the following reaction sequence:
$C_6H_5CONH_2 \xrightarrow{NaOBr} y$
$C_6H_5CONH_2 \xrightarrow[(ii) H_3O^+]{(i) C_6H_5SO_2Cl / py, \Delta} x$

The amine $A$,when treated with nitrous acid,gives a yellow oily substance. The amine $A$ is

The product of this reaction is:

Order of basic strength: $(A) Ph-NH_2$,$(B) Ph-NH-CH_3$,$(C) Ph-N(CH_3)_2$,$(D) \text{o-toluidine (2-methylaniline)}$

The product$(s)$ of the following reaction sequence is(are):
Aniline $\xrightarrow[\text{ii) } KBrO_3/HBr]{\text{i) Acetic anhydride/pyridine}}$ $\xrightarrow[\text{iv) } NaNO_2/HCl, 273-278 \ K]{\text{iii) } H_3O^+, \text{heat}}$ $\xrightarrow{\text{v) } Cu/HBr} \text{Product}$

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