IR

Question 1

spectroscopy measures the bond
vibration frequencies in a molecule
and is used to determine the
functional group.

Answer 1

INFRARED (IR)

Question 2

fragments the molecule and
measures the masses.

Answer 2

MASS SPECTROMETRY (MS)

Question 3

spectroscopy detects signals from
hydrogen atoms and can be used to
distinguish isomers.

Answer 3

NUCLEAR MAGNETIC RESONANCE (NMR)

Question 4

spectroscopy uses electron transitions
to determine bonding patterns.

Answer 4

ULTRAVIOLET (UV)

Question 5

1. Analysis of gaseous, liquid, or solid samples
2. organic synthesis
3. biological research
4. environmental industry
5. food and pharmaceutical industry
6. forensic

Answer 5

IR APPLICATIONS

Question 6

IR is most useful in providing information
about the presence or absence of
specific functional groups.

Question 7

IR can provide a molecular fingerprint
that can be used when comparing
samples

Question 8

IR does not provide detailed information
or proof of molecular formula or
structure.

Question 9

IR Has very limited scope and must be
used in conjunction with other
techniques to provide a more complete
picture of the molecular structure.

Question 10

Relies upon interferences of
various frequencies of light to
produce a spectrum

Answer 10

FOURIER TRANSFORM INFRARED (FTIR) SPECTROPHOTOMETER

Question 11

A polar bond is usually IR-active.

Question 12

A nonpolar bond in a symmetrical molecule will absorb weakly or not at all

Question 13

Strongly polar bonds such as carbonyl groups (C=O) produce strong bands.

Question 14

Medium polarity bonds and asymmetric bonds produce medium bands.

Question 15

Weakly polar bond and symmetric bonds produce weak or non observable bands.

Question 16

ADVANTAGES OF USING ATR

Answer 16

Measurement using ATR is fast and simple

NO SAMPLING PREPARATION
REQUIRED

Very small amount of sample needed

Question 17

Carboxylic acid

Answer 17

O-H = 2500-3300 cm-1; very strong and broad
C=O = 1700 cm-1; very strong
C-H = 2900 cm-1

Question 18

Alcohol

Answer 18

O-H = 3200-3600 cm-1; strong absorption
C-O (SP3) = 1000-1100 cm-1; medium

Question 19

Aldehyde

Answer 19

C=O = 1700 cm-1
C-H = 2900 cm-1
-C-H = 2700 cm-1

Question 20

Ketone

Answer 20

C=O = 1700 cm-1

Question 21

Ester

Answer 21

C=O = 1700 cm-1
C-O (SP2) = 1200-1300 cm-1
C-O (SP3) = 1000-1150 cm-1; strong

Question 22

Ether

Answer 22

C-O (SP3) = 1000-1150 cm-1; medium

Question 23

Amine (primary)

Answer 23

double peak
3300-3500 cm-1; medium

Question 24

Amine (secondary)

Answer 24

single peak
3300-3500 cm-1; medium

Question 25

Amide

Answer 25

C=O = 1700 cm-1
3300-3500 cm-1; medium

Question 26

Alkane

Answer 26

-CH3 (SP3) = 2900 cm-1

-CH3 bend = 1365-1385
-CH, -CH2, -CH3 bend = 1400-1450

Question 27

Alkene

Answer 27

C=C = 1660 cm-1
=C-H (SP2) = 3000-3100 cm-1 ; weak to medium

Question 28

Alkyne

Answer 28

C TRIPLE BOND C = 2100-2201 cm-1
Triple bond-C-H = 3300 cm-1 (terminal alkyne); medium to weak

Question 29

Nitrile

Answer 29

C TRIPLE BOND N = 2200-2300 cm-1; medium

Question 30

Benzene

Answer 30

1450-1600 cm-1; strong to weak

Question 31

Conjugation lowers the frequency:
isolated C=C
1640-1680 cm-1

conjugated C=C 1620-1640 cm-1

aromatic C=C
approx. 1600 cm-1

Question 32

More polar bonds are generally stronger than less polar bonds

Question 33

The greater the H bonding, the lower is the absorption frequency.