第2章 官能基團，分子間的各種作用力及紅外光譜簡介
(Functional Groups, Intermolecular Forces and Infrared
Spectroscopy)
一) The Functional groups in organic chemistry
1) Hydrocarbons: Alkanes, alkenes, and alkynes aromatic compounds
Saturated compounds: the molecules contain only single bonds. They have the
maximum number of hydrogen atoms.
Unaturated compounds: (Alkenes and alkynes: CnH2n and CnH2n-2) the molecules
have fewer than maximum number of hydrogen atoms.
Aromatic compounds (benzene):
the carbon-carbon bonds of benzene are
all the same length (1.39 Å)
Six electrons associated with p orbitals are delocalized about all six
carbon atoms of the ring.
2) covalent
bonds and dipole moments
partially positive end
partially negative end
The more electronegative chlorine draws electron density away
from the hydrogen
dipole moment (μ; unit D; can be measured by experiment).
μ=e×d
Write δ+ and δ– by the appropriate atoms and a dipole moment vector for the
molecules: HF, IBr
Electrostatic potential maps
*
Explain
*
in detail
*
*
*
Having polar bonds, but no dipole moment
Unshared electron pair contributes a large moment directed away from the central
atom.
Explain SO2 (μ = 1.63D), CO2 (μ = 0)?
Explain CHCl3 has a larger dipole moment CFCl3?
Using three-dimensional formula, show the direction of the dipole moment of
CH3OH?
Write the structural formulas for C2H2Br2 and C2Br2Cl2, predict the dipole
moment of each one.
Cl
Cl
Cl
H
H
H
H
Cl
net
F
F
H
H
H
F
H
F
net
F
H
F
F
H
F
F
F
μ=0
μ=0
3) *functional group: a certain arrangement of atoms
A functional group is the site of most chemical reactivity of a
molecule
Alkanes do not have a functional groups
a) Alkyl group (R-): obtained by removing a hydrogen atom from an alkane
(CnH2n+2):
butyl , tert-butyl and sec-butyl?
b) Phenyl and benzyl groups:
c) Alkyl halides or haloalkanes: the hydrogen atom(s) in an alkane is (are)
replaced by halogen atom(s); Cl, F, Br, I.
Alkyl halides are classified as being primary (1o), secondary (2o) and tertiary (3o)
depending on the carbon atom to which the halogen is directly attached:
Write two constitutional primary isomers for C4H9Br: 1)a secondary alkyl bromide:
2) a tertiary alkyl bromide:
Propyl bromide, iospropyl flouride and phenyl bromide
d) Alcohols
CH3 OH
Methyl alcohol or methanol
Write two constitutional isomers for 1) two primary alcohols for C4H10O;
2) A secondary alcohol; 3) A tertiary alcohol
Write the structures for propyl alcohol and isopropyl alcohol.
e) Ethers
Write the structures for 1) Diethyl ether; 2) ethyl propyl ether; 3) propyl methyl ether;
4) diisopropyl ether; 5) methyl phenyl ether.
f) Amines
Amines can be considered as organic derivative of ammonia
The classification IS DIFFERENT from that of alcohols and alkyl halides:
Write the structures for 1)Isopropylamine; 2)
propylamine;3) trimethylamine; 4)
ethylisopropylamine;5) isopropylpropylamine;
6) tripropylamine; 7)mthylphenylamine; 8)
dimethylphenylamine.9) diethyl amine
g) Aldehydes and ketones: the compounds that contain the carbonyl group
g) Carboxylic acids:
h) Esters:
i) Amides
j) Nitriles:
• Summary (cont.)
二)分子間的各種作用力以及對化合物物理性質的影響
The strength of intermolecular forces (forces between molecules) determines the
physical properties (i.e. melting point, boiling point and solubility) of a compound.
1) Ion-Ion bond
2) Dipole-Dipole Forces
Hydrogen bonding:
Z: O; N; F
Draw the hydrogen boning interaction for 1) CH3OH, 2) CH3NH2, 3) CH3CO2H, 4)
HF 5) RCONHR’
Explain the boiling point of EtOH is higher than MeOMe.
Which compound in the following to have the higher boiling point? 1)
CH3CH2CH2CH2OH and CH3CH2OCH2CH3; 2) (CH3)3N and CH3CH2NHCH3;
3) CH3CH2CH2CH2OH and OHCH2CH2CH2OH
In addition to polarity and hydrogen bonding, A factor in melting points is that
symmetrical molecules tend to pack better in the crystalline lattice and have
higher melting points:
3) Van der Waals Forces:
* Van der Waals forces result when a temporary dipole in a molecule caused by a
momentary shifting of electrons induces an opposite and also temporary dipole in an
adjacent molecule
* These temporary opposite dipoles cause a weak attraction between the two
molecules
* Molecules which rely only on van der Waals forces generally have low melting
points and boiling points
Explain why CH4 becoming a solid at below -182.6oC?, what’s the force holding the
molecules together?
Explain in detail
Polarity( the ability of electrons to respond to a changing electron field): I > Br> Cl >
F
4) Solubility (water-soluble: minimum 3g/ 100 ml of H2O):
a) Ionic compounds:
These dipole-ion interactions are powerful enough to overcome lattice energy
and interionic interactions in the solid
b) Polar and non-polar compounds: “like-dissolve-like” rule
Decyl alcohol is only slightly soluble in water
三)紅外光譜簡介 （鑒定化合物官能基團的重要手段）
c = λν, ν = c / λ = c ν
C = 3 × 1010 cm/sec
0.75→ 1000 mm(400 cm-1-4000 cm-1) 紅外光區;分子振動能(stretching and
bending)級
T=
radiant power transmitted by a sample
=
radiant power incident on the sample
I
I0
Stretching bands:
1) Hydrocarbons:
The C-H OOPbending vibration
peaks located at 600-1000 cm-1 can
be used to determine the substitution
pattern of the double bond
2) Other functional groups:
a) Carbonyl compounds
Assign the structures
b) Alcohol, phenyl and amines
O-H stretching: 3590-3650 cm-1
Assign the structures
1o amines give two peaks and 2o amines give one peak, 3o
have no N-H bonds and do not absorb in this region
Exercise (Page 90):
2.20 Classified the following compounds:
OH
O
alkyne
ketone
C8H17
alcohol
O
OH
H
aldehyde
alcohol
C13H27
H
H
alkene
2.21 Identify the functional groups
O H
2
HO
C
carboxylic
acid
HO
hydroxyl
alkenyl
H2N
O
N
H H
amide
phenyl
Ph
O
CH3
O
ester
NH2
phenyl
HO
hydroxyl
amine
alkene
phenyl
CO2Et
Ph
ester
O
aldehyde
H
N
alkene
amine
O
i-Pr
O
O
O
alkene
i-Pr
ester
2.22 Write the structures with the formula C4H9Br, indicating whether it is
primary, secondary, or tertiary.
Br
Br
1o
Br
3o
Br
2o
1o
2.23 Write the seven isomeric compounds with the formula C4H10O
OH
OH
OH
OH
Alcohol
O
O
O
ether
2.24 Write the four structural formulas with the formula C3H6O, predict the IR
absorptions.
O
C=O stretching: ～1710 cm-1
OH
O
H
OH
O
C=O stretching: ～1740 cm-1; C-H stretching: ～2710 cm-1
C-O stretching: ～1200 cm-1; O-H stretching: 3590～3650 cm-1, C=C
stretching: ～1680 cm-1; =C-H OOP bending: 1000 cm-1; 900 cm-1
C-O stretching: ～1200 cm-1; C=C stretching: ～1680 cm-1; =C-H OOP
bending: 1000 cm-1; 900 cm-1
2.25 Classify the following alcohols as primary, secondary, or tertiary.
a)1o, b) 2o, c) 3o, d) 3o, e) 2o
2.26 Classify the following amines as primary, secondary, or tertiary.
a) 2o, b) 1o, c) 3o, d) 2o, e) 2o; f) 3o
2.27 Write structural formulas for each of the following:
g)
a)
O
b)
c)
O
OH
h)
OH
O
O
i)
OH
j)
O
O
O
H
N
l)
H3CO
CH3
Br
f)
EtO
NH2
NH2
O
H
Br
Br
m)
N
O
Br
n)
O
NH2
H
H
O
OH
k)
e)
O
H
H
d)
Br
Br
O
OH
Br
Br
N
H
2.28 Which compound would have the high boiling point:
a)
OH >
O
O
g)
b)
OH >
HO
OH >
OH
O
O
O
c)
OH >
d)
OH >
NH
e)
>
F
F
>
f)
O
>
h)
i)
N
O
>
F
F
2.29 Predict the IR absorption bands which would allow to distinguish the pair of
compounds in a), c), d), e), g), i):
a)
c)
d)
OH
O-H stretching
O
OH
O-H stretching
OH
O-H stretching
C=O
e)
g)
NH
i)
O
OH
O-H stretching
N-H stretching
O
C=O stretching
2.30
a) C3H7NO:
O
O
N
H
NH2
O
O
H
H
N
H
N
O
b)
H
would have the lowest m.p. because no hydrogen bonding can
N
be formed
O
O
c)
NH2
two N-H stretching
O
O
N
H
H
N
H
one N-H stretching
H
N
no N-H stretching
2.31 write the structures that would not be expected to exhibit absorption in
3200-3550 cm-1 and 1620-1780 cm-1
C4H6O:
O
H
H
H3C
O CH3
O
O
O
2.32
O
O
ester
Explain the lactone and lactam
2.33 Explain the difference between the boiling point of HF (19.34 oC) and
EtF (-37.7oC), two molecules almost having the same dipole moments.
(Hydrogen bonding)
2.34: cis-isomer has the higher dipole moments, the dipole-dipole
interaction is stronger than that of the trans-isomer.
2.35:
N (CH2)15CH3 Br
Is more soluble in H2O than EtOEt
2.36: NH3 is expected to dissolve the ionic compounds.
2.37:
H
F
H
F
H
H
H
O H
H
H
F
H
O
H
F
F
H
F
H
H
H
H
μ=0
H
Cl
F
F Be F
μ=0
F
H
H
H
F
F
F
μ=0
Cl
Cl
B
Cl
μ=0
O
H
2.38:
H
C≣C stretching, 2100 cm-1
symmetrical, no dipole change
2.39 Describe the hybridization, predict the geometry and dipole moment for each
of the following compounds:
O
H3C
N
CH3
sp3, having dipole; < 109o
H2C
H3C
CH3
CH3
CH3
sp3, having dipole; < 109o
H3C
B
CH3
sp2, having no dipole; ～120o
CH2
Be
sp, having no dipole; ～180o
2.40:
+
Ion-dipola interaction
Ag
2.41 For a molecule to be polar, the presence of polar bond is necessary, but it is
not a sufficient requirement _________ Right!
2.42 (Omitted)
2.43
H
O
O
O
H
O
2.44
Two peaks because of symmetrical
and unsymmetrical interation