• Summary (cont.)

第2章 官能基團，分子間的各種作用力及紅外光譜簡介

(

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: C_nH_2n and C_nH_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

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

The more electronegative chlorine draws electron density away from the hydrogen

partially negative end partially positive end

Electrostatic potential maps

*

*

*

*

* Explain in detail

Unshared electron pair contributes a large moment directed away from the central atom.

Having polar bonds, but no dipole moment

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 H

Cl H

Cl

H Cl

H

F H

F H

net

H H

F F

net

F H

H 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 (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 (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 C4H9Br: 1)a secondary alkyl bromide:

2) a tertiary alkyl bromide:

Propyl bromide, iospropyl flouride and phenyl bromide

d) Alcohols

CH

OH

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:

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 Z: O; N; F

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 in detail

Polarity( the ability of electrons to respond to a changing electron field): I > Br> Cl >

F

Explain why CH4 becoming a solid at below -182.6^oC?, what’s the force holding the molecules together?

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 = cν 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

radiant power transmitted by a sample

T = 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

O-H stretching: 3590-3650 cm-1 b) Alcohol, phenyl and amines

Assign the structures

1^o amines give two peaks and 2^o amines give one peak, 3^o have no N-H bonds and do not absorb in this region

Exercise (Page 90):

2.20 Classified the following compounds:

O OH

H

O OH

H H

C₈H₁₇ C₁₃H₂₇

ketone alkyne alcohol aldehyde alcohol

alkene

2.21 Identify the functional groups

HO

hydroxyl alkenyl

H₂

C N

H

O CH₃ O

HO

H₂N H

O Ph

carboxylic O

acid amide ester

phenyl

NH₂ HO hydroxyl amine

phenyl alkene

N

Ph CO₂Et

amine phenyl

ester

H O

aldehyde

alkene

O O

O

O i-Pr

i-Pr ester

alkene

2.22 Write the structures with the formula C4H9Br, 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 C4H10O

2.24 Write the four structural formulas with the formula C3H6O, predict the IR absorptions.

OH OH OH OH

Alcohol

O O O

ether

O

H O

OH

O

C=O stretching: ～1710 cm-1

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)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

OH

2.27 Write structural formulas for each of the following:

a) b) c)

d)

e)

f)

g) h) i) j)

k)

l)

m)

n)

O O O

OH OH OH

OH

H₃CO CH₃ O

EtO H O

Br Br Br

Br

Br Br Br

Br

O H

O H

O H

O O O

NH₂ NH₂

HN

N

O

NH₂

O NH H

OH

2.28 Which compound would have the high boiling point:

OH O

> O O

>

O >

a) b) c) d)

e)

f)

g)

h) i)

OH > O OH >

OH > O

HO OH > OH

O

NH

> N

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)

O-H stretching OH

O-H stretching C=O OH

O

O-H stretching OH

e) g) _NH i)

N-H stretching O-H stretching OH O

C=O stretching O

2.30

a) C3H7NO:

b) would have the lowest m.p. because no hydrogen bonding can be formed

c)

NH₂ O

NH O

H N

H O

H N

O

H N

O

NH₂ O

NH O

H N

H O

H N

O

two N-H stretching one N-H stretching 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

O

H₃C O CH₃ H

O

H 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 (-37.7^oC), 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

H F

H

F

H F

F

H

H F

Cl F Be F μ = 0 H

H O

H H

F

F H

H

F

F F

F μ = 0

Cl B Cl Cl

O H

H H

H H H μ = 0

H O H 2.38:

C≣C stretching, 2100 cm-1 symmetrical, no dipole change H

μ = 0

2.39 Describe the hybridization, predict the geometry and dipole moment for each of the following compounds:

+Ag

Ion-dipola interaction 2.40:

2.41 For a molecule to be polar, the presence of polar bond is necessary, but it is not a sufficient requirement ^_________ Right!

O

H₃C CH₃ sp3, having dipole; < 109^o

H₃C N

CH₃ CH₃

sp3, having dipole; < 109^o

CH₃ H₃C B CH₃

sp2, having no dipole; ～120^o

H₂C Be CH₂

sp, having no dipole; ～180^o

2.42 (Omitted) 2.43

O H

O

O H O

2.44

Two peaks because of symmetrical and unsymmetrical interation