NMR StructureProblems1 CompleteSolutions

7/27/2019 NMR StructureProblems1 CompleteSolutions

1/7

Additional NMR Problems 2012 Solutions

Problem A C4H10O2 DBE = 0 (EASY)

IR and D2O experiment tell us that this is an alcohol13C NMR: CH2O at 67

CH at 37 (NOT on O)

CH3 at 14 (normal methyl)

Total = C3H6O so missing CH4O but since the1H NMR shows 2H for the OH, this is a diol and we have

two CH2OH groups. There is only one way you can connect 2 x CH2OH, CH3 and a CH:

H3C

H

OH

OH

Does this match the 1H NMR? 4H m must be the CH2 and the shift is correct (if you want a detailed

explanation of the pattern observed, it is the same explanation as for the inequivalent protons of the CH 2

group in problem E); 1H octet is correct for the CH (7 H are within 3 bonds); 3H doublet for the methyl

group matches.

Problem B C9H11N DBE = 5 Aromatic + 1 (MODERATE)

IR: br 3350 cm-1 one peak and the fact there is a br singlet in the 1H NMR that exchanges with water

indicates we have a secondary amine, NH

13C NMR: 6 aromatic carbons with four CH matches the 1H NMR integrations for a C6H4 group

There are NO alkene carbons or carbonyls so the missing DBE MUST BE A RING: C6H4 + ring + NH

What else does the 13C show? 48.5 and 43.7 (-) must be CH2N (note amine is secondary) so the amine

is NOT directly on the ring.

Anything else? Yes, remaining C is at 28.9 (-) so we have anotherCH2 NOT on N.

So we have CH2-, -CH2NHCH2- and a C6H4 group. Connect the bifunctional groups first to give

-CH2CH2NHCH2- which is then connected to the C6H4 aromatic ring the connection MUST be ortho

on the C6H4 ring because it is simply not possible for this short of a chain to connect any other way:

NH

Note: 1H NMR supports this because there is a downfield isolated singlet due to the CH2 group on both N

and the aromatic ring.


2/7

Problem C C4H8O DBE = 1 (EASY)

13C NMR: =CH at 143 (up)

=CH2 at 113 (down) so this contains a mono-substituted alkene CH=CH2

Other C? OCH at 69 (downfield 1H pentet at 4.3 ppm indicates this is a CH not a CH3)

CH3 at 23 (up)

Total: C4H7O missing 1H but we have a br singlet at 2.5 strongly suggesting this is an alcohol

At this stage, we could simply note that the only multi-functional group is the CH, the rest are ALL

terminal: -CH=CH2, OH and CH3 so all we have to do is connect them to the CH to get the structure:

The other option is to look at the 1H NMR and consider the CH group. Since this H is far downfield it

must, at the very least, have an OH attached and it is a pentet indicating 4 near neighbour H. The alcohol

H will not couple (it is a singlet after all) so the near groups MUST be the CH3 and the alkene =CH.

Problem D C16H16O3 DBE = 9 (HARD)

The large number of DBE and aromatic resonances between 100 and 165 ppm, plus the absence of many

resonances below 60 ppm, strongly suggests we have two benzene rings.IR: 1680 cm-1 tells us we have a conjugated carbonyl so this is the other DBE; the fact that there is a 13C

resonance at 194 ppm that disappears in the DEPT indicates this is a conjugated ketone.

There are a few options as to how to proceed from here but I would try to decide two things:

1) What are the other two oxygens?

2) What is the substitution pattern like on the benzene rings?

Point 1 Note that the only peaks between 3.8 and 4.1 ppm are ALL singlets;

Also note that the remaining non-aromatic C are at 55.3 (CH3), 55.2 (CH3) and 44.0 (CH2)

This suggests we have two OCH3 groups to account for both the1H and 13C shifts and the 1H

multiplicity. The CH2 group is quite deshielded but this is not as far downfield as typically observed for a

CH2O.


3/7

Point 2 The benzene substitution pattern is actually very simple and consists of only four sets of doublets.

The key thing here is to recognize this as two (different) para-disubstituted benzene rings (there are

only 8 aromatic H by integration).

So we have: 2 x p-C6H4 + 2 x CH3O + C=O (on aromatic) + -CH2- for a total of C16H16O (correct)

We know that ketone is conjugated AND cannot be attached to the OCH 3 (or it would be an ester) so this

leaves 3 possibilities:

Can we tell which one? Yes. The structure in the box has two strongly shielded carbons ortho to the

OCH3 groups (the ortho C in the same ring are equivalent by rotation) but the other two only have one.

The 13C NMR shows two shielded C (114.1 and 113.8) so the structure in the box is correct.

Problem E C11H24O4 DBE = 0 (HARD)

First points to note: 1) highly symmetrical since there are only four unique C

2) all oxygens are ethers (no OH and no DBE)

13C NMR: 100 (up) no unsaturation so how can you get this far downfield? There MUST be two

oxygens on a CH carbon

62 (down) suggests a CH2O group

38 (down) more normal CH2

14 (up) typical CH3

1H NMR: 12H triplet at 1.2 suggests four equivalent CH3 next to CH2. There is only one group

of resonances that can be due to the CH2 protons and this is the 4H/4H pentets at around 3.6 ppm. The

chemical shift tells us that we must have four equivalent OCH2CH3 but the strange appearance of two

pentets seems to imply the CH2 protons are inequivalent (more on this below).


4/7

Four equivalent OCH2CH3 gives us C8H20O4 so we are missing C3H4 and there are only two types of C in

this group: 100 (up) which we assigned as an -O-CH-O- group and a normal CH2 at 38 (down).

Since there are four oxygens, we must have two -O-CH-O- groups and to maintain the symmetry the

CH2 must be connected to both giving the only possible structure:

The CH2 groups of the ethoxy fragments are inequivalent because there is no conformation that makes the

two H on the same CH2 equivalent: there is an inside and an outside type of H. Note there is a mirror

plane (plane of the page) and a rotation axis that makes the inside (or outside) H on all four arms the

same but it never makes the inside and outside H on the same C equivalent. One pentet is for the

inside H and the other for the outside H.

H

O

O

CH3

CH3

H

H

H

H

H

O

O

H3C

H3C

H

H

H

H

H H

H

H 'inside' H

'outside' H

Problem F C8H6O3 DBE = 6 Aromatic + 2 (MODERATE)

13C NMR 190 (d) indicates an aldehyde (confirmed by the resonance at 9.8 ppm in 1H NMR)

Aromatic ring is C6 so we only need one other C and it must be a triplet at 102 ppm. Again, this is

extremely far downfield for a CH2 but we do have two oxygens left and both would be needed to get to

this shift as a -O-CH2-O fragment.

You might be tempted to think we have an alkene and at first glance the presence of a singlet at 6.1 ppm

and the 13C resonance at 102 might seem consistent with a R2C=CH2 group (we do need one more DBE

after all). However, a check of the aromatic region of the 1H NMR rules this out because there are only 3

aromatic resonances (2 x doublet and a singlet) which means the ring is tri-substituted and it cannot be

if we invoke an alkene.


5/7

The singlet, doublet, doublet pattern in the aromatic region is consistent with 1,2,4-subsituted aromatic

ring and we also have a CHO and a -O-CH2-O- group. We still need one DBE though so there must be

another ring and the only way that can work and make chemical sense is if we have the structure shown

below (the O groups MUST be ortho because the OCH2O group is far too short to span meta or para

positions on the aromatic ring):

O

O

O

H

Problem G C4H4O2 DBE = 3 (EASY)

IR says it is not an acid or an alcohol but the carbonyl band at 1710 cm -1 might indicate a conjugated

(carbonyl side) ester OR a non-conjugated ketone. Which one? 13C carbonyl resonance at 173.2 is far

too low for a ketone so this is a conjugated ester.

This only leaves C3H4 and we need two more DBE! Not an alkyne according to13C but there are two

alkene C: 153.0 (d) =CH and 121.8 (d) =CH so we have a disubstituted alkene CH=CH-

But we still need another DBE and we only have a CH2 at 4.9 ppm in the1H and 72.1 ppm in the 13C so

there must be a ring. The ester is conjugated and must be connected to the alkene AND the CH2 is far

downfield and must be connected to an O with all of it tied into a ringonly choice is:

O

O

Problem H C11H14O3 DBE = 5 Aromatic + 1 (MODERATE)

IR: broad band between 3300 and 3400 cm-1 suggests an alcohol (an acid would likely extend well

below 3000 cm-1)

strong band at 1710 cm-1 band suggests a saturated ketone or a conjugated ester but the 13C rules

out a ketone or aldehyde so this is a conjugated (carbonyl side) ester

13C NMR: 168.0 ester C=O

162.1, 120.1 aromatic C-R131,7, 116.1 aromatic C-H

64.0 CH2O

31.1, 19.0 2 x CH2

14.1 CH3

Total C11 so all C accounted for


6/7

1H NMR: two aromatic doublets the goalposts = p-disubstituted aromatic ring C6H4

10.4 1H br s suggests a phenol or an acid (we think phenol because of the IR)

4.2 2H, t OCH2 next to a CH2

1.7 2H, pentet CH2 with CH2 on each side

1.5 2H, sextet CH2 with CH2 and CH3 on either side

0.9 3H, t CH3 next to a CH2

This is consistent with an OCH2CH2CH2CH3 chain + p-C6H4 + conj-CO2 + OH

Which leads to only one structural possibility:

HO

O

O CH2

H2C CH2

CH3

Problem I C6H11N DBE = 2 (EASY)

IR: Only notable feature is the br s at 3300 cm-1 suggesting a secondary amine; note there is NO nitrile

13C NMR: only 3 resonances = highly symmetrical

137 (up) =CH

116 (down) =CH2

51 (down) -CH2 (must be on N to account for the shift)

Total C3H5 but the CH2 is a 4H d in the1H NMR so this suggests two identical arms

with CH2-CH=CH2 on N (the CH2 is coupled to the alkene H but not the exchangeableNH); not this also accounts for the 2 DBE.

NH

This matches the 1H NMR well but if in doubt, you could have used the 1H NMR to establish the identity

of the allyl (C3H5) arms.


7/7

Problem J C11H13N DBE = 6 Aromatic + 2 (EASY)

IR: v. strong band at ca. 2130 cm-1 suggests a nitrile and accounts for the DBE

1H NMR: First thing to do is to note that there is no coupling at all and the aromatic substitution is

symmetrical with only 2H left on the aromatic ring = symmetrical and tetra-substituted

3.59 s CH22.35 s CH3 x 2

2.26 s CH3

Total = C6H2 + CH3 x 3 + CH2 + CN = C11H13N all accounted for

Note: CH2 must be connected to the CN since otherwise it would have to be connected to a CH3 and it

would show coupling. This means all 4 groups on the ring are identified: 3 x CH3 and one CH2CN

Since the structure is symmetrical, there are two possible options:

These can be distinguished with 2D experiments but not without. The 13C NMR data can be used to check

the structure but is not really necessary for solution.

END

Documents

NMR StructureProblems1 CompleteSolutions