Question

Assistance needed with the NMR and fragmentation portions.

a. Formula: C12H1.0 b. IR (see information below) IR Peak Frequency 3062 2962 1759 1613 IR Peak Frequency 1509 1432 1202 1166 c TH-NMR (see information below) Integration Peak A B C D Chemical Shift 7.135 6939 2.391 2.306 2.224 1035 Splitting Singlet Singlet Doublet Singlet Multiplet Doublet 2 3 1 6 F d. MS Label Peak #1 (base) 2 Peak # Peak #3 Pal Peak #4 Peak #5 (M) M/Z value 47 8 5 107 135 Determine the correct structure for the compound, using the information above. Be sure you O Calculate the DOU of your compound and analyze what functional groups/structural motifs would be present based on that value (do this before you even look at the NMR or IR information) o Explain what functional groups are present using the IR information. Use the DOU information to confirm your choice(s). O For the NMR, you will need to do the following . Draw what you suspect to be the correct structure Determine how many sets of equivalent hydrogens are present (label them as 1, 2, 3, 4, 5, etc) Using the NMR information provided, match the hydrogens to the correct peaks (set 1 - peak s et 2-peak , etc) Be sure to explain your choices using the splitting information, chemical shift information, integration values, etc o Use the provided MS information to confirm your final structure by determining how thar structure will fragment to give the provided peaks. For each peak, draw the fragment that correlates to that m/z value, and explain why the molecule will fragment there to give that peak

Answer 1

Solution:

1. Degree of unsaturation =5

Structural formula = C12H16O2

Degree of unsaturation = 1/2 (2 + 2 x number of carbons - number of hydrogens)

= 1/2 (2 + 2 x 12 - 16)

= 1/2 (2 + 24 - 16)

= 1/2 (10)

= 5

From FTIR and NMR values analysis we can predict that on benzene ring and one ester group must be present

Degree of unsaturation = 5 = 4 for aromatic ring + 1 for ester functionality.

2. FTIR

FTIR shows characteristic band at 1759 cm-1 of ester functionality, 3062, 2962, 1613, 1509 cm-1 are representing the presence of aromatic ring. As DOU is 5 that is satisfied by benzene ring [DOU= 4 ] and ester [DOU = 1].

3. NMR

peak A= 7.135 ppp [s, 2H], ------------> 2H of aromatic ring

peak B= 6.939 [s, 2H], ---------------> 2H of aromatic ring

Note: as there are only two peaks appeared in the aromatic region, benzene ring must be para disubstituted.

peak C= 2.391 [d, 2H], -------------------> -CH2-CH [CH2 is attached to CH]

peak D= 2.306 [s, 3H], -------------------> -CH3 [CH3 must be attached to C which is not having any proton, i.e this CH3 is attached to aromatic ring]

peak E= 2.224 [m, 1H], --------------------> -CH- [-CH is attached to carbon/s conating multiple protons]

peak F = 1.035 [d, 6H] ----------------------> 2 CH3-CH [Two CH3 attached to CH]

Possible structure

2 H 5 3 1 6 5 1 6

Assignment of signals:

Peak B 6.939 Peak E 2.224 Peak CH 2.391 1.035 Peak F Peak A 7.135 1759 cm-1 [FTIR] 1.035 6.939 Peak B 2.306 Peak F 7.135 Peak A Peak D

4. Mass fragmentation

H I 0 + m/z = 192 m/z = 135 m/z = 57 H H + m/z = 85 m/z = 107 m/z = 192 HOH + m/z = 92 m/z = 135 m/z = 47 [+1]

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