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Q.A saturated monoamine liberates nitrogen gas on reaction with nitrous acid in cold condition. On heating with methyl iodide it forms quarternary animonium iodide (mol. mass = 215). Deduce the formula of the amine. (Given at. mass of iodine = 127).

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2020Subjective· 2mImportance★★★★★
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The nitrogen-gas test identifies a primary amine, and matching the quaternary salt's molar mass to the given value pins it down to ethylamine.

Liberating N2N_2 gas with cold nitrous acid is characteristic of a primary amine (RNH2+HNO2→ROH+N2+H2ORNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O). On exhaustive methylation with excess methyl iodide, a primary amine converts to a quaternary ammonium iodide:

RNH2+3CH3I→RN+(CH3)3 I−+2HIRNH_2 + 3CH_3I \rightarrow R\overset{+}{N}(CH_3)_3\,I^- + 2HI

Given the quaternary salt's molar mass is 215215 (with I=127I=127):

M(R)+14(N)+3×15(CH3)+127(I)=215M(R) + 14(N) + 3\times15(CH_3) + 127(I) = 215

M(R)+14+45+127=215  ⟹  M(R)=215−186=29M(R) + 14 + 45 + 127 = 215 \implies M(R) = 215 - 186 = 29

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