Question

explain

02. A compound with a molecular formula C&H SCIO, has the following 'H NMR spectrum. The IR spectrum shows strong absorption at 1800 cm!. Which of the following structures is consistent with this spectrum? PPM علی علی معلم A) I B) II C) III D) IV E) none of these

Answer 1

& Chu H12O2 Degree of unsaturation - 22-12 = 10 -5 2 It means there must be a benzene ring with a unsaturated functional group. Malysis of 1H NMR (qi) S: 0.9(+, 3H), 1.8 (@, 2H), 7:6(4, 2H) 7.4 (d, 2H), 8.1 (d, 2H) and 115,14) 0.914, 3H), 18(m, 2H) and 2.6(€, 24) indicates There are there the two signals groups are in para 7.9 8.1. substituted It in inclicates benzene. HCS There is no exchangeable peaks so no -OH gouf. Thus, the structure which fulfills all conditions is is 1.241) Heb) sk. Hou -Ha 1163, 14) the Colchestion option : J 7 -4 -811 ppm AJI 2:6 (4, 2H) bre, but

SAU 83 (SHUNO : Double bed equivalent = and equivalent 13-11 - = 1 means a double bond. IR : 3400 (m' =) -N- Stretching . . I Analysis of IH NMR : S: 0.9 ppm (4, 34), 14 ppm( dt, 2H), 2.4(7, 2H), 3.1 (8,3H) 8( s exchangeable, 11) 0.9 ppm (8,3H), 1,4 (dt, 2H) and 2.4(4, 2H) indicates i fragment AH H 3168,34) indicates 8 3 ,) indicates 2-I Combining all fragments 2.468,24). 0.961, 3H) -N (H2CH-CH3 HC 3.418,34) 1.4(, 24) 9 8( A, 2H) Correst oftian : (I)
Questions no is written.
83 Cg His COS : Degree of unsaturation - 18-15 -Ilus 18-16 means a double bond IR: 1800 cinta and Analysis of 'H NMR : l integration of all signals are given) 6, 8:H(4,6H), 19(quartet, 24), 2.764, 2H), 3.4(2, 4 H) 4.2 (1, 2H) NOW, 1(4,64), 3.4 (2, 4H) s) and 4.2(4, IH) means row fragment was HH 19 (?, 2H) and 2:7 (t, 2H) → v y fragment

from IR & Now combing all fragments obtained 1 H NMR, we get the structure G ied(1.9, 2, 2H). < o as () allows a 3.468,4H) ☺ (4.2 €, 1H) b (1.4 8,64) 2.7(,214) Correction oflion!