Question

Find the chemical structure using the wave spectrum shown

Problem I1.3 40 Exact M.S. (E)- 84.0575 UVAmax-209 (E-16,000) 328 (E-50) FT-IR (NEAT) 20 3000 2000 WAVENUMBERS (CM) 1800 1200 100 Mass Spec. El 70eV 40 20 50 40 44 55 70 m/z T 13C NMR 62.3 MHz COCla D D 192 11 190 CDCI Fwwwwluuluuluulnluulululul luulnluuluuluulun ad 0 ppm 0 HE 150 120 100 120 160 150 140 130 110 50 50 40 20 10 250 10 Ha/dw 250 MHz CDCla 4.2 Hz/mm 1H NMA 250 MHz CDCI 5.9 Hz/mm 3H 8.4 9.2 9.0 70 8.6 6.0 2H 60 74 50 4.0 30 2.0 1,0

Answer 1

¹³C NMR spectrum shows 5 peaks. Therefore, there are at least 5 carbon atoms in the compound.

The total integration value of ¹H NMR spectrum adds up to = (3 + 2 + 1 + 1 + 1) = 8. Therefore, there are 8 hydrogen atoms in the molecule.

The doublet at ~9.2 ppm in ¹H NMR spectrum and the peak at ~191 ppm in ¹³C NMR spectrum indicates the presence of an aldehyde group in the molecule.

Hence, the molecular formula of the compound = C₅H₈O

Exact mass, m/z = 84.06 (matches with the EI exact mass given)

Degree of unsaturation = (2C + 2 - H)/2

                                      = (2 x 5 + 2 - 8)/2

                                      = 2

IR spectrum peak assignment:

ww 40 C-H symmetric & anti-symmetric alkyl 20 FT-IR (NEAT) C-C stretch C-H stretches stretches C=O stretch 4000 1200 2000 M00 1O00 800 WAVENUMBER (CM-

¹³C NMR peak assignment:

D T B 13C NMR 62.3 MHz COCI CH3 E .C D B H2 192 191 190 CDCI 09 a 0 0 40 30 20 130 120 110 100 180 170 150 140 10 0 pem

¹H NMR peak assignment:

250 Jantaulalnd 10 Ha/dv. E 250 MHz CDCI 4.2 Hz/mm 1H NMR 250 MHz CDCI 5.9 Hz/mm 3H H Llulul 9.0 8.4 8.2 7.0 8.6 6.0 E C 2H B CH3 C H H B H2 7.0 5.0 4.0 3.0 2.0 10 0.0 CI A CHC HA

High intensity ($\epsilon$ = 16000) peak at
mar
= 209 nm indicates the
T T
transition in UV spectrum.

Low intensity ($\epsilon$ = 50) peak at
mar
= 328 nm indicates the
n T
transition in UV spectrum.

Hence, the chemical structure that satisfies the given spectral data: