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BT501: Problem set-III

1. NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given

in hertz downfield from the TMS standard. Convert the absorptions to δ units

a. 436 Hz

b. 956 Hz

c. 1504 Hz

Ans:

Chemical shift, δ(ppm) = , The values given above are . The reference

frequency, = 200 MHz. Therefore, the chemical shifts in δ units i.e. in ppm are:

a. 436 Hz = 2.18 ppm

b. 956 Hz = 4.78 ppm

c. 1504 Hz = 7.52 ppm

2. Is a nucleus that absorbs at 5.5 δ more shielded or less shielded than a nucleus that

absorbs at 3.2 δ? Does the nucleus that absorbs at 5.5 δ require a stronger applied field

or a weaker applied field to come into resonance than the nucleus that absorbs at 3.2 δ?

Ans:

The nucleus that absorbs 5.5 δ is less shielded than a nucleus that absorbs at 3.2 δ.

Consequently, the nucleus that absorbs at 5.5 δ requires a a weaker applied field to come

into resonance than the nucleus that absorbs at 3.2 δ.

3. What is the multiplicity of the hydrogen type shown in bold? (Hint: Tell if the peak

arising from these protons will be singlet, doublet, triplet, etc.)

Ans:

The protons shown in bold are equivalent; we therefore have only one proton type in

question. The adjacent carbon is having only one proton attached to it. Therefore, by n+1

rule, the multiplicity of the protons shown in bold is doublet.

4. What is the multiplicity of the hydrogen type shown in bold? Assume that the CH2 is

significantly different from the CH3.

a) Singlet

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b) Doublet

c) Quartet

d) 4 × 3 peaks

e) 7 × 3 peaks

5. What is the multiplicity of the hydrogen type shown in bold?

Ans: Doublet

6. What is the multiplicity of the hydrogen type shown in bold?

Ans: Singlet

7. Refer to the 1D proton NMR spectrum shown below for a compound with formula

C3H7Br. Answer the questions asked below.

a) How many types of hydrogen are present, assuming no serious overlap

between different types? Three

b) What is the chemical shift of the hydrogen type that is most upfield? Give

approximate values or range. ~ 1 Hz

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c) What best describes the most upfield hydrogen type?

i. Singlet

ii. Doublet

iii. Triplet

iv. Quartet

v. Doublet of doublets

d) What does this multiplicity indicate about the most upfield hydrogens?

i. There is one hydrogen present at this position

ii. There are two hydrogens present at this position

iii. There are three hydrogens present at this position

iv. There are two hydrogens on the neighbouring carbon

v. There are three hydrogens on the neighbouring carbon

e) What does the integration show about the most upfield hydrogens?

i. There is one hydrogen present at this position

ii. There are two hydrogens present at this position

iii. There are three hydrogens present at this position

iv. There are two hydrogens on the neighbouring carbon

v. There are three hydrogens on the neighbouring carbon

f) What can you conclude about the most upfield hydrogens, using the above

evidence?

i. They represent a CH2 group, attached to another CH2.

ii. They represent a CH2 group, attached to a CH3.

iii. They represent a CH3 group, attached to a CH2.

iv. They represent a CH3 group, attached to another CH3.

v. They represent a CH2 group, attached to CH.

g) What is the chemical shift of the hydrogen type that is most downfield? Give

approximate values or range. ~3.4 Hz

h) Knowing that the formula for this compound is C3H7Br, what does the

chemical shift of these downfield hydrogens suggest?

i. There is an aromatic ring attached at this position

ii. There is a bromine attached at this position

iii. The aromatic ring is at the other end of the molecule

iv. The bromine is at the other end of the molecule

i) How many hydrogens are present at this downfield position? Two

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j) How many hydrogens are present on the carbon adjacent to the downfield

position? Two

8. Resolution in atomic force microscopy is limited by probe geometry and diameter

while that of light microscopy is limited by wavelength of the light used .

9. The resolution of an atomic microscope is highest in z dimension (x, y, or,

z).

10. Which mode of an atomic force microscope can be used for obtaining the high

resolution frictional coefficient profile of a surface?

a. Contact mode

b. Intermittent contact mode

c. Non-contact mode

11. Infrared spectroscopy provides valuable information about

a. Molecular weight

b. Functional groups

c. Conjugation

d. Melting point

12. Mention any two advantages of AFM over EM.

Ans:

i. AFM can be used for imaging the samples in solution while EM cannot.

ii. AFM does not require any staining of the specimen while EM usually does require

staining with heavy metals

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13. Match the five C5H10O isomers below with the key infrared absorption bands listed:

(Note: the fingerprint bands and alkane C-H stretches are not reported.)

Refer to table 1 on Page 5 for typical vibrational wavenumbers of key functional groups

and the guidelines for identifying the functional groups.

Absorption band(s) Molecule

3400 cm-1

3400, 1660 cm-1

1715 cm

-1

1660, 1100 cm-1

1100 cm-1

14. One of the five compounds shown below was analyzed by infrared spectroscopy. Peaks

were observed at: 1639 cm-1

(weak) and 1714 cm-1

(strong) but were absent at 3200-

3400 cm-1

. Which compound was analyzed?

Ans:

i. A strong peak at 1714 cm-1

is suggestive of carbonyl group. Therefore compounds

A, B, and D are ruled out.

ii. Absence of any peak in the region 3200 – 3400 cm-1

suggests absence of hydroxyl

group. This further rules out compound A and D.

iii. Presence of a weak band at 1639 cm-1

could arise from an alkene. Therefore, the

compound analyzed was E.

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Table 1: Typical vibrational wavenumbers of key functional group

Wavenumber Vibration Functional groups Remarks

3500 – 3250 N–H Amine Medium

3500 – 3200 O–H Alcohols Strong, Broad

3100 – 3000 C–H Alkenes Medium

Aromatics Strong

3000 – 2850 C–H Alkanes Strong

1760 – 1665 C=O Ketone, Aldehyde, Ester, etc. Strong

1680 – 1640 C=C Alkene Medium/Weak

1000 – 1260 C–O Alcohol, Ether, Ester, Acid Strong

General guidelines for identifying the functional groups (Not to be taken as rules):

i. Begin by looking in the region from 4000-1300.

ii. Look at the C–H stretching bands around 3000:

a. Are any or all bands lesser than but close to 3000 Alkyl group

b. Are any or all bands higher than but close to 3000 C=C bond/aromatic group

iii. Look for a carbonyl in the region 1760-1665. If there is such a band:

a. Is an O–H band also present? Carboxylic acid

b. Is a C–O band also present? Ester

c. Is an aldehydic C–H band (2830–2695) also present? Aldehyde

d. Is an N–H band also present? Amide

e. Are none of the above present? Ketone

iv. Look for a broad O–H band in the region 3500-3200 cm-1

Alcohol/phenol

v. Look for a single or double sharp N–H band in the region 3500-3250 cm-1

. If there is

such a band:

a. Are there two bands? A primary amine

b. Is there only one band? A secondary amine