Question Bank [2022] — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryQuestion Bank [2022]3 Marks
Question
Distinguish between methanamine, dimethanamine and trimethanamine using Hinsberg’s reagent.
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Answer
(i) Methanamine reacts with benzenesulphonyl chloride (Hinsberg’s reagent) to give N-methyl benzenesulphonamide which is soluble in alkali.
$\underset{\text { Methanamine }}{ CH _3 NH _2}+ C _6 H _5 SO _2 Cl \stackrel{ OH ^{-}}{\longrightarrow} \underset{\text { N-Methyl benzenesulphonamide (Soluble in alkali) }}{ C _6 H _5 SO _2 NHCH _3}$
(ii) Dimethanamine reacts with benzenesulphonyl chloride (Hinsberg’s reagent) to give a compound which does not dissolve in alkali.
$\underset{\text { Dimethanamine }}{\left( CH _3\right)_2 NH }+ C _6 H _5 SO _2 Cl \stackrel{ OH ^{-}}{\longrightarrow} \underset{ N , N \text {-Dimethylbenzene sulphonamide (Insoluble in alkali) }}{ C _6 H _5 SO _2 N \left( CH _3\right)_2}$
(iii) Trimethanamine does not react with benzenesulphonyl chloride (Hinsberg’s reagent) due to the absence of H-atom attached N-atom.
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