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) ydrocarbons: 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: C n 2n and C n 2n-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: F, IBr
Electrostatic potential maps * Explain * in detail * * *
aving polar bonds, but no dipole moment Unshared electron pair contributes a large moment directed away from the central atom.
Explain S2 (μ = 1.63D), C2 (μ = 0)? Explain CCl3 has a larger dipole moment CFCl3? Using three-dimensional formula, show the direction of the dipole moment of C3? Write the structural formulas for C22Br2 and C2Br2Cl2, predict the dipole moment of each one.
Cl Cl Cl Cl net F F F F net F F F 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 (Cn2n+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 (1 o ), secondary (2 o ) and tertiary (3 o ) depending on the carbon atom to which the halogen is directly attached: Write two constitutional primary isomers for C49Br: 1)a secondary alkyl bromide: 2) a tertiary alkyl bromide: Propyl bromide, iospropyl flouride and phenyl bromide
d) Alcohols C 3 Methyl alcohol or methanol
Write two constitutional isomers for 1) two primary alcohols for C410; 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 ydrogen bonding: Z: ; N; F Draw the hydrogen boning interaction for 1) C3, 2) C3N2, 3) C3C2, 4) F 5) RCNR Explain the boiling point of Et is higher than MeMe. Which compound in the following to have the higher boiling point? 1) C3C2C2C2 and C3C2C2C3; 2) (C3)3N and C3C2NC3; 3) C3C2C2C2 and C2C2C2
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 C4 becoming a solid at below -182.6 o C?, 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 2): 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 = cν C = 3 10 10 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
radiant power transmitted by a sample T = radiant power incident on the sample = I I 0
Stretching bands:
1) ydrocarbons:
The C- Pbending vibration peaks located at 600-1000 cm-1 can be used to determine the substitution pattern of the double bond
2) ther functional groups: a) Carbonyl compounds
Assign the structures
b) Alcohol, phenyl and amines - stretching: 3590-3650 cm-1
Assign the structures
1 o amines give two peaks and 2 o amines give one peak, 3 o have no N- bonds and do not absorb in this region
Exercise (Page 90): 2.20 Classified the following compounds: ketone alkyne alcohol aldehyde alcohol C 8 17 C13 27 alkene 2.21 Identify the functional groups hydroxyl alkenyl carboxylic acid 2 C 2 N N amide phenyl Ph C3 ester
N 2 phenyl amine hydroxyl alkene phenyl Ph N C 2 Et ester alkene aldehyde amine i-pr i-pr ester alkene
2.22 Write the structures with the formula C49Br, indicating whether it is primary, secondary, or tertiary. Br Br Br Br 1 o 1 o 2 o 3 o 2.23 Write the seven isomeric compounds with the formula C410 Alcohol ether 2.24 Write the four structural formulas with the formula C36, predict the IR absorptions.
C= stretching: ~1710 cm-1 C= stretching: ~1740 cm-1; C- stretching: ~2710 cm-1 C- stretching: ~1200 cm-1; - stretching: 3590~3650 cm-1, C=C stretching: ~1680 cm-1; =C- P bending: 1000 cm-1; 900 cm-1 C- stretching: ~1200 cm-1; C=C stretching: ~1680 cm-1; =C- P bending: 1000 cm-1; 900 cm-1 2.25 Classify the following alcohols as primary, secondary, or tertiary. a)1 o,b) 2 o, c) 3 o, d) 3 o, e) 2 o 2.26 Classify the following amines as primary, secondary, or tertiary. a) 2 o, b) 1 o, c) 3 o, d) 2 o, e) 2 o ; f) 3 o
2.27 Write structural formulas for each of the following: a) b) c) d) g) h) i) j) Br Br Br Br k) N 2 N 2 e) 3 C C 3 Et l) N f) Br Br Br Br m) n) N N 2 N
2.28 Which compound would have the high boiling point: a) b) c) > > > g) > d) > h) > e) N > N i) > f) F F > 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) - stretching - stretching C= - stretching
e) N g) i) N- stretching 2.30 a) C37N: - stretching C= stretching b) N would have the lowest m.p. because no hydrogen bonding can be formed N 2 N N N c) N 2 N N N two N- stretching one N- stretching no N- stretching
2.31 write the structures that would not be expected to exhibit absorption in 3200-3550 cm-1 and 1620-1780 cm-1 C46: 3 C C 3 2.32 ester Explain the lactone and lactam 2.33 Explain the difference between the boiling point of F (19.34 o C) and EtF (-37.7 o C), two molecules almost having the same dipole moments. (ydrogen 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 (C 2 ) 15 C 3 Br Is more soluble in 2 than EtEt
2.36: N3 is expected to dissolve the ionic compounds. 2.37: F F F F F F F Cl F Be F μ = 0 F Cl F F B F Cl Cl μ = 0 μ = 0 2.38: μ = 0 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: 3 C C 3 sp3, having dipole; < 109 o 3 C N C 3 C 3 sp3, having dipole; < 109 o C 3 B 3 C C 3 sp2, having no dipole; ~120 o 2 C Be C 2 sp, having no dipole; ~180 o 2.40: + Ag Ion-dipola interaction 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 (mitted) 2.43
2.44 Two peaks because of symmetrical and unsymmetrical interation