Question

Can someone please explain the mechanisms for the equation below

Part B: AMINES – The Hinsberg Test - Acidic Hydrogen O Η O Na RNH+ RNH2 + -š-cı OH S-CI -In O=o=o S-NR -S-NR + NaCl + 2H2O =O 1° Amine Water Soluble salt HCI -S-NR ID OH Insoluble

Can someone please explain the mechanisms for the equation below

Can someone please explain the mechanisms for the equation below

Answer 1

OH = l le S-N-R + R-NH2 lamine -Hol audic Na OH I O=CO Tz hydrogen To=y=0 s-a- r e S-N-R Water Soluble ! (Due to formation of charges) - HCL 0=úzo I->

i L-amine veact with þ-melhylbenzene sulfonyl chloxide to give a p-melħyl benzene sulfonamide. This has one auidic hydrogen which is abstracted by a base like Nach to give an ionic compound which is soluble in water. The addition of HCl protonates the nitrogen and it becomes insoluble in water and precipitates out from the . aqueous solution, a o= il 12 TR-NH + - yao NaOH. in Soluble R insoluble

Secondary amines © react with compound Ⓡ to gire compound ©. The compound. © loses a proton to gire compound (D. The compound ③ does not react with Nach due to absence of audic hydrogen and this remains insoluble in NaOH. Treating this solution will HCI do not give any reachen. | NR₃ + tertiary O=y=0 + NaOH 1 U osno Les amine @ -Nac! | NaOH H2O + NaCl + PRE + rs-o Na o=y=o insoluble Mc C. I O CI NR3 soluble

Tev mary amines antrow do not react with compound B But compound ® react with Nach to give componenst ☺ Tertiary amines remain insoluble in the Naol solution when HCl is added to this naoh Solution, it reacts I with ŇRz to give salt 14 R3 which is sclubie in water