ASR Interpretation of Spectra Tutorial

1 Anderson Serangoon Junior College 2025 H3 Chemistry Interpretation of Spectra 1 The spectra shown below were obtained from compound Q, which contains the element carbon, hydrogen, nitrogen and oxygen. Consider the NMR, mass and IR spectra in turn. Explain what information each spectrum gives about Q. Hence give a possible structure for the compound.

2 NMR  /ppm Multiplicity No of H Deduction 7-8 Multiplet 5 HH H H H Monosubstituted phenyl ring 2.1 Singlet 3 -CH3 with no H on adjacent C 3.1 Singlet 1 Since the peak disappear on adding D2O, it is a labile proton, attached to an electronegative atom. MS M : M+1 16.5 : 1.47 100 : 1.1 n No. of carbon atoms = (100/1.1)(1.47/16.5) = 8 Given Q contains C, H, O and N, Peak at m/e 135 could be molecular ion C8H9NO+. Structure of Q could be A or B. C O N CH3 H NH C CH3 O A B m/e Fragment 93 135 – 42 NH2 43 - C O CH3 Absence of m/e at 105 suggest absence of C6H5CO+. From the fragment ions, structure of Q is B. IR Presence of N-H at 3300 cm-1 Prsence of amide functional group.

3 2 Interpret the following spectra and use them to suggest a structure for the unknown compound.

4 MS peak at m/e 91 is [C6H5CH2]+ (benzyl or tropilium cation) peak at m/e 77 is [C6H5]+ peak at m/e 43 is [CH3CO] + UV/VIS Compound shows several absorptions within UV/VIS region, suggesting presence of a conjugated π-system IR C-H stretch at 3100cm-1 Strong absorption at 1750cm-1 due to C=O stretch. NMR δ / ppm No. of 1H Multiplicity Deductions 2.1 3 s -CH3 with no H on adj C Deshielded by electron -withdrawing ketone group 3.7 2 s -CH2- with no H on adj C Deshielded by ring current and electron- withdrawing ketone group 7.1 – 7.4 5 m Aromatic H on mono-substituted phenyl ring Structure:

5 3 Deduce the structure of compound A, giving your reasoning. [8]

6 IR absorption bond 1740 C=O 1200 C–O  Absence of peak at 3230 to 3500 cm–1 suggests absence of O–H stretch  No alcohol  The presence of a C–O (strong and broad) at 1200 cm–1 and C=O at 1740 cm–1 confirm the identity of an ester functional group. NMR MS m/e species 102 (C5H10O2)+ 57 (CH3CH2CO)+ 29 (CH3CH2)+ chemical shift /ppm splitting pattern No. of adjacent protons relative peak area group responsible 1.15 triplet 2 3 –CH3 (a) adj to a -CH2- 1.25 triplet 2 3 –CH3 (b) adj to a -CH2- 2.30 quartet 3 2 –CH2CO– (adjacent to a –CO– and –CH3) 4.15 quartet 3 2 –OCH2– (adjacent to a –O– and –CH3)

7 4 The 1H NMR signals of protons on carbon atoms adjacent to C=C bonds ca n be split by protons at the far end of the C=C bond. HC R R R HA R R HB For example, protons HA and HC with split each other with a small coupling constant JAC. Protons HB and HC will also split each other with coupling constant JBC. JAC and JBC are often not equal. Protons HA and HB may not be equivalent because of the restricted rotation around the C=C bond. (a) Propanone undergoes the following sequence of reactions. O C D Ba(OH)2 conc. H3PO4 warm The infra-red and 1H NMR spectra of compound C are shown in Fig. 2.1 and Fig. 2.2. The mass spectrum of C shows major peaks at m/z values of 15, 43, 58, 101 and a small M+ peak at m/z 116. (i) Use this information to deduce the structure of C. Explain your reasoning. [6]

8 IR: Wavenumber of absorbance / cm-1 Interpretation Strong absorption at 1700 cm-1 C=O stretch in ketone Strong absorption at 2800-3000 cm-1 Aliphatic C–H stretch Broad and strong peak at 3350 cm-1 O–H bond stretch in alcohol Presence of alcohol and ketone in C C has a M r of 116. Based on IR, it has two oxygen atoms (due to ketone and alcohol). Based on NMR, there are 12 hydrogen atoms. Hence, no of C is (116 ‒ 2x16 ‒ 12x1)/12 = 6 and molecular formula is C6H12O2. MS: m/z Interpretation 15 [CH3]+ 43 [CH3CO]+ 58 CH3 C OH CH2 101 116 ‒ 101 = 15 Loss of CH3 group. Hence m/z = 101 is C5H9O2 [CH3COCH2C(OH)CH3]+ or [COCH2C(OH)(CH3)2]+ 116 CH3 C CH3 OH CH2 C CH3 O NMR:  /ppm Multiplicity No. of adjacent protons No of H Deduction 1.5 Singlet 0 6 Two –CH3 adjacent to to C with no H 2.5 Singlet 0 3 –CH3 adjacent to C=O (no H on adjacent C) 2.7 Singlet 0 2 –CH2 adjacent to C=O (no H on adjacent C) 4.0 Singlet (broad) 0 1 Labile –O–H Structure of C: CH3 C CH3 OH CH2 C CH3 O + +

9 The infra-red and 1H NMR spectra of compound D are shown in Fig. 2.3 and Fig. 2.4. (ii) Use Fig. 2.3 and Fig. 2.4 to deduce the structure of D. Explain your reasoning. [4]

10 IR: Wavenumber of absorbance / cm-1 Interpretation Strong absorption at 1700 cm-1 C=O stretch in ketone Weak absorption at 3000 cm-1 Alkenes –C=H NMR:  /ppm Multiplicity No. of adjacent protons No of H Deduction 1.90 Doublet 1 3 –CH3 adjacent to C=C and split by the proton on the far end of C=C (HA or HB split by HC) 2.11 Doublet 1 3 –CH3 adjacent to C=C and split by the proton on the far end of C=C (HA or HB split by HC) 2.19 Singlet 0 3 –CH3 adjacent to C=O (no H on adjacent C) 6.0 Multiplet 6 1 =CH adjacent to (CH3)2C= and split by the protons on the two CH3 (HC split by HA and HB) Note: C C C C O CH3 C HA HA HA HB HB HB Hc Structure of D: CH3 C CH3 CH C O CH3