Question

1. Formula C,H4O 4000 3500 2H 3H 3H 2H 2H 2H 1.5 2.0 4.5 3.5 3.0 2.5 1.0 PPM 4.0 20es 0*250 T6 61 '8201 LSIT 60 99T 90 01L1 e'SPLT Or'8282

Answer 1

C8H4,0% Degree of Unsaturation ( U ) =(2n4+ng-ny+2구 Where nu Number of tetravalent atmsEC=8 Namber of toivaleat atoms EN =0 Number df monovalent atoms = 14 n3 1 J [2x8+0-14+2 2 7 -L16-14 +2 2- LJ 2 The Compound Contain too double bonds. U = 2

IR Spectra An absonption bands at 2995 m 2937 cml and 2878 cm due to c-H shretchin *An absorphon bands at 174s cml is due to c=O shetching in ester.. An absonphion band at aound 1716 cm may due to An shelching io kehooe absorption band at 1317 cmt due to C-o shetehingf ester. An absorphon band at 1647 cm may due to in enolate fom d Kehooe, CニC fe=e-007 H-NMR:It display dignals due to 6plitting of hydgeo The splitiox occurs according to is the number Protons where adiacent carbon Jf then dinglet (5) n 0 then doublet rd) then tripler than D=1 D 2 quartet (2) nE4 then quintet (2t) D=S then sextet 1st). n:3

The giveo HNm spechra display follouwing Signali do.9s pem t 3H due to-CH-CH 3 due to o-CH2- Ch od.4 ppm,t, 3H | d17 prm, 3,2, due to-cH2-CH2-CH 2-6ppm, t, 2M, dur to-cH-CH 3-5 ppm, S 2H due to cHа- d4.35 PpM, qugrtet 2, due to o-ch-CH From the above data the shruchre df Compound be. with m.F. C8HI403 was foungd to CHA-CH-O-CCHCCH2-CH-CH3