Question

Information: Each spectra below was obtained from a pure compound. Mass Spectrum parent peaks (M) are listed for all examples. IR peaks listed are strong (s) unless otherwise indicated for signals above 1500 cm 'H NMR Spectra, the integral is given in number of hydrogens (#H) or as a relative ratio. Important coupling constants (J-values) are listed next to the peaks for some examples. For some spectra, an inset (grey box) is also given showing a "zoom-in" on an important part of the spectrum. Mass Spectrometry (not shown): [M] = 147 (100%), (M+2] = 149 (100%) m/z Infrared Spectroscopy (not shown): 2969, 2935, 2249 cm 'H Nuclear Magnetic Resonance. quin 2H PPM 2H ™C Nuclear Magnetic Resonance. 12010080 .60 PPM 40 20

Draw the Skeletal Structure of the Unknown Compound.
Note - Formal skeletal structures do not include hydrogens on any carbons (including the aldehyde C).

Answer 1

M = 147 and 149 1.1 m/2 hence Br-atom must present. Base formula by rule of B=Mt = 147 13 13 Ci Hi5 bux Br-atom present hence add one Bretom and substret 6 - C-atom and 7 H-atoms Molecular formula becoms C5 H₂ B8 but, molecular weight is odd , hence one N-atom to be added. and one C-atom and two H-atom to be substicht molecular formula = Cu Ho Brn a Chant2 Double bond equivalence Cu H2x4+2 = Cu Hio Ho-H6 = 2 - indicans -CEN

probable structure may be CN-CH2-CH2-CH2- Br (c) (b) ca HNMR a = 8= 3. 5ppm, triplet, 2H. b = 8= 2.2ppm, quintet, at c =) S=1.8 ppm , triplet, 2H IS CNNIR a = 32 ppm b= 18 ppm с 20 рр d = 120 ppm