A reaction of 0.1 mol of Benzylamine with br | vedclass

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Question

(B) The reaction of Benzylamine $(Ph-CH_2-NH_2)$ with bromomethane $(CH_3Br)$ is an exhaustive methylation reaction.Benzylamine reacts with $3$ moles

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The reaction of Benzylamine $(Ph-CH_2-NH_2)$ with bromomethane $(CH_3Br)$ is an exhaustive methylation reaction.Benzylamine reacts with $3$ moles of bromomethane to form Benzyl trimethyl ammonium bromide $(Ph-CH_2-N(CH_3)_3^+Br^-)$.The molar mass of Benzyl trimethyl ammonium bromide $(C_{10}H_{16}NBr)$ is calculated as:$M = (10 	imes 12) + (16 	imes 1) + (14 	imes 1) + (80 	imes 1) = 120 + 16 + 14 + 80 = 230 \ g/mol$.Given mass of product = $23 \ g$.Number of moles of product = $\frac{23 \ g}{230 \ g/mol} = 0.1 \ mol$.Since $1 \ mol$ of Benzylamine reacts with $3 \ mol$ of bromomethane to produce $1 \ mol$ of Benzyl trimethyl ammonium bromide,$0.1 \ mol$ of Benzylamine will react with $0.1 	imes 3 = 0.3 \ mol$ of bromomethane.Given that the number of moles of bromomethane consumed is $n 	imes 10^{-1} = 0.3$.Therefore,$n = 3$.

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