Question

2. Provide a detailed explanation for the observation below. Use drawing and words in your explanation. No partial credit will be given for this question. "SOH

Answer 1

ig & region Answeri H.S04. is a very strong acid, Ht of this acid is highly reactive .:--> two lome pair Aniline with lone pair on nitrogen i NH2, electrons is a partially negative charge B region (8) So, Ht of highly reactive acid H2 504 which is positively charged gets grabbed by 8 MHz like below t >positively (Step 1 .ŏ SO2H region + HS0G As nitrogen donates its e lone pair to Ht, the nitrogen by loosing negatively charged es becomes positively, charged. So, as aniline / NH? I and H₂SO4 are mixed, first iNH3, charged H MHZ and HsOq ts are product formed Now the kit reacts with SO₃ In S03, as origen are more electronegative than relatively Sulphur : 69% sto s sulphur is a positively charged region because of di electro- - negativity difference in sulphur and oxygen

So So, with with tNH3 8t charge on sulphur when mixed , it will react on 5 region available on benzene ring The 3 types of reaction spots available o benzene ring are : for 503 NH3 ortho (o) meta (m) para (p) For aniline, the lone pair on nitrogen go in resonance with benzene ring making ortho & para position more e- dense that is 8 8-' for so; addition, like: However, due to H₂SO4 addition in the begining the lone pair of NH 2 is blocked by Ht forming Hz So, the ortho-para positions highly activated in case of simple addition of So3 are in fact very de actived due to introduction of H2SO4 in the begining, causing meta position as relatively most favourable position for Soz aszf forming end product as to below: touts HQ -> Oko Step 2 F ES=6 HSO (end product) (formed TOROSO Step 1)

the end exists in product is a type of zwitter in which two rearrangements : HH3 "SO₂H (final form)