Question

A: C,H,O 4.2H pom Int. 1050.85 11.734 142 225.16 2.515 117 217.72 ) 2.431 421 210.21/ 2.348 505 202.60 2.253 163 110.91 1.239481 103.65 1.158 1000 95.87 1.071 362 S 1H 108 ppm

what is the structure and assignment of this molecule C3H6O2. Also calculate J values

Answer 1

The given Compound has molecular formula; C 3 H602 Double Bond equivalence = XH XH (DBE) =(3+)-(6/2) DBE = 1 1 double bund/unsaturation. • Looking at 11.7.ppm presence of — Cool group. (charactersti peaks o triplet of 3H (CH3 CH2 fragment o quartet at 2H CH₂ - CH2 fragment So, combining all links we reached to the conclusion that,

The compound is Propanoic acid 11.7 ppm 1.15pm 2.38 ppm Calculations of coupling constant * for – CH3 we get a triplet at 1.15ppm Splitting can be Shownas; * ! 103 95.87Hz Hotel He 1.239 | 21.079 1.158 Coupling constant TAX3 = 4239-154 JAB=(103.65-95.87) HZ JAB = 7.78 HZ * for – CH₂ , We get a quartet at 2.38 ppm. 217 210202 HA IHA 225 te Leak | | | | | |A 25/11 2.43 2.34 mont JBA = (217.72-210.21) Hz I BA = 7:51 Hz This values of Coupling constant have been calculated. mert 2.26