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)

Transcription

1 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.

2 Aromatic compounds (benzene): the carbon-carbon bonds of benzene are all the same length (1.39 Å)

3 Six electrons associated with p orbitals are delocalized about all six carbon atoms of the ring.

4 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

5 Electrostatic potential maps * Explain * in detail * * *

6 aving polar bonds, but no dipole moment Unshared electron pair contributes a large moment directed away from the central atom.

7 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.

8 Cl Cl Cl Cl net F F F F net F F F F F F μ = 0 μ = 0

9 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?

10 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.

11 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

12 d) Alcohols C 3 Methyl alcohol or methanol

13 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.

14 e) Ethers Write the structures for 1) Diethyl ether; 2) ethyl propyl ether; 3) propyl methyl ether; 4) diisopropyl ether; 5) methyl phenyl ether.

15 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

16 g) Aldehydes and ketones: the compounds that contain the carbonyl group

17

18 g) Carboxylic acids:

19 h) Esters: i) Amides

20 j) Nitriles:

21

22 Summary (cont.)

23 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.

24 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

25 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

26 Explain why C4 becoming a solid at below 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

27 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

28 b) Polar and non-polar compounds: like-dissolve-like rule Decyl alcohol is only slightly soluble in water

29

30

31 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 = cm/sec mm(400 cm cm-1) 紅外光區 ; 分子振動能 (stretching and bending) 級

32 radiant power transmitted by a sample T = radiant power incident on the sample = I I 0

33 Stretching bands:

34

35 1) ydrocarbons:

36

37 The C- Pbending vibration peaks located at cm-1 can be used to determine the substitution pattern of the double bond

38

39 2) ther functional groups: a) Carbonyl compounds

40

41

42

43 Assign the structures

44 b) Alcohol, phenyl and amines - stretching: cm-1

45 Assign the structures

46

47 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

48 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

49 N 2 phenyl amine hydroxyl alkene phenyl Ph N C 2 Et ester alkene aldehyde amine i-pr i-pr ester alkene

50 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.

51 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 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

52 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

53 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

54 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

55 2.31 write the structures that would not be expected to exhibit absorption in cm-1 and cm-1 C46: 3 C C 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

56 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 = : μ = 0 C C stretching, 2100 cm-1 symmetrical, no dipole change

57 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!

58 2.42 (mitted) 2.43

59 2.44 Two peaks because of symmetrical and unsymmetrical interation