Question

Qualitative analysis lab
Give structures for the following aromatic compounds.

e. p-methylphenol 2. Give structures for the following aromatic compounds: a. A (C H10) positive 2,4-DNP; positive iodoform. b. B (C,HO) insoluble in NAOH; positive 2,4-DNP; negative iodoform hot KMnO, oxidation yields p-methoxybenzoic acid. c. C (C,H0,) soluble in NaHCO oxidation with KMn0, yields phthalic acid; IR peak at 1690 cm d. D (C,H,0,) soluble in NaHCO oxidation with KMnO, yields phthalic acid; IR peak at 1710 cm e. E (CH120,) insoluble in NaHCO3; oxidation with KMnO, yields isophthalic acid; IR peak at 1740 cm f.F (C H,N) soluble in HCl; Hinsberg test precipitate insoluble both in excess base and acid (two possible structures). 10

Answer 1

2a) Co Hoo towards DBE = (C+1) - H/2 = (8+1) - 10/2 Compound shows positive response 2,4-DNO can be concluded that it aldehyde ketone. is einer a JAN.B ol If the compound in aromatic & contains groups total double bond equivalence 0 ) = GTI 4 PBE =s But from the formula We are go geting So question is not connect Positive iodofon & contairs – e-cit group. . Probable structure (3-may be in op as well) Coltoor DBE (oril - 10/2 I u 5 & must contain a benzene ring (DBE 4) & a group. (Indicated by the 0 2,4-DNP) Negative iodoform indicates absence of chur. Oxidation product indicates presence of p-ome group.

. Structure is Cocks 7 in It is insouble NaOH OCU3 (e) Coto2: DBE = 5 (as before) Soluble in Matteoz & most contain - cool group. It produces ptrau aad - Two on the substituent present in original compound Structure - COUH cood Kinney, court -elhcltz JR! 1690 cm! (GR effect of the ring decreases IR peal from ideal value of 1710 con Everything Ir peau not in is same as above except IR peau. appear at 1710 cont indicates – Coos is conjugation with prenyl ring. Structure Chcoon coot nmney -cz Cour ich croit IR 1710cml pthalic acid.