In the given reaction,identify the final product.

  • A
    $C_6H_5-NH-CH_2-CH_2-COCl$
  • B
    $C_6H_5-CH(NHCH_3)-CH_2-CH_3$
  • C
    $C_6H_5-CH_2-CH_2-CONHCH_3$
  • D
    $C_6H_5-CO-CH_2-CH_2-NHCH_3$

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Which one of the following methods is neither meant for the synthesis nor for separation of amines?

Which one of the following reaction sequences will give an azo dye?

Complete the following reactions:
$(i)$ $C_{6}H_{5}NH_{2} + CHCl_{3} +$ alc. $KOH \rightarrow$
$(ii)$ $C_{6}H_{5}N_{2}Cl + H_{3}PO_{2} + H_{2}O \rightarrow$
$(iii)$ $C_{6}H_{5}NH_{2} + H_{2}SO_{4}$ (conc.) $\rightarrow$
$(iv)$ $C_{6}H_{5}N_{2}Cl + C_{2}H_{5}OH \rightarrow$
$(v)$ $C_{6}H_{5}NH_{2} + Br_{2(aq)} \rightarrow$
$(vi)$ $C_{6}H_{5}NH_{2} + (CH_{3}CO)_{2}O \rightarrow$
$(vii)$ $C_{6}H_{5}N_{2}Cl \xrightarrow[(ii) NaNO_{2}/Cu, \Delta ]{(i) HBF_{4}}$

$A$ trinitro compound,$1,3,5$-tris-($4$-nitrophenyl)benzene,on complete reaction with an excess of $Sn/HCl$ gives a major product,which on treatment with an excess of $NaNO_2/HCl$ at $0^{\circ} C$ provides $P$ as the product. $P$,upon treatment with excess of $H_2O$ at room temperature,gives the product $Q$. Bromination of $Q$ in aqueous medium furnishes the product $R$. The compound $P$ upon treatment with an excess of phenol under basic conditions gives the product $S$.
The molar mass difference between compounds $Q$ and $R$ is $474 \ g \ mol^{-1}$ and between compounds $P$ and $S$ is $172.5 \ g \ mol^{-1}$.
$(1)$ The number of heteroatoms present in one molecule of $R$ is . . . . .
[Use: Molar mass (in $g \ mol^{-1}$): $H=1, C=12, N=14, O=16, Br=80, Cl=35.5$
Atoms other than $C$ and $H$ are considered as heteroatoms]
$(2)$ The total number of carbon atoms and heteroatoms present in one molecule of $S$ . . . . . .
[Use: Molar mass in $g \ mol^{-1}$]: $H=1, C=12, N=14, O=16, Br=80, Cl=35.5$
Atoms other than $C$ and $H$ are considered as heteroatoms
Give the answer for question $(1)$ and $(2)$.

How will you convert:
$(i)$ Ethanoic acid into methanamine
$(ii)$ Hexanenitrile into $1$-aminopentane
$(iii)$ Methanol to ethanoic acid
$(iv)$ Ethanamine into methanamine
$(v)$ Ethanoic acid into propanoic acid
$(vi)$ Methanamine into ethanamine
$(vii)$ Nitromethane into dimethylamine
$(viii)$ Propanoic acid into ethanoic acid?

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