ORGANIC REACTION TYPES: - CHEM HOGchemrat.com/ChemHog2/Organic Chem_files/organicintro3.doc · Web viewAdd up the number of valence electrons in each atom to find the total number

SOME FUNCTIONAL GROUPS

ORGANIC CHEMISTRY INTRO 1

R RC

H

H

C

H

Halkane

CH

CH

alkene

C C

alkyne

H

HH

H

H Harene

RRRR

alkyl halide alcohol

O

phenol

R X O HR O

ether

R R C N

nitrile

RH

..

..

NR

HN

R

RN

H

H

1º amine 2º amine 3º amine nitro

NO

O

+-R R R R

aldehyde

C

O

H C

O

ketone

C

O

OH

carboxylic acid

C

O

O C

O

acid anhydride

R R R R R R

C

O

O

ester acid chloride

C

O

Cl

1º amide

NH

HC

O

2º amide

NR

HC

O

R R R R R

NR

RC

O

3º amide sulfide

S S

O

sulfoxide

S

O

O

sulfone

..R R R R R R R

HS

thiol

OHS

O

sulfinic acid

OHS

O

Osulfonic acid

O

O

SO OH

bisulfate

..

......

..

......R R R R

1º

::

: ::S

O

O

NH

HR

sulfonamide

R O O R......

..dialkylperoxide

R:_

MgX+

Grignard

R O O H......

..hydroperoxide


Ar = aryl groupe.g., phenyl

naphthyl

R = alkyl groupe.g., methyl CH3 ethyl CH3CH2

CH2 CH CH2

allyl group

CH2 CH

vinyl group

CH2

benzyl group

CHH3C

CH3

isopropyl group

CHH3C

CH3

CH2

isobutyl group

CH2CHH3C

CH3

CH2

isopentyl group

CHH3C CH2 CH3

sec-butyl group

C

CH3

H3C

CH3

CH2

neopentyl group

C

CH3

H3C

CH3

t-butyl group

H+ R'OH

R C

O

H

R'

R C

OH

H

O R'

R C

O

H

OR'

H+ R'OH+ ++ H2O

aldehyde hemiacetal acetal

+ N N

H

C

O

NH2C

semicarbazone

: :......N

H

HN

H

C

O

NH2

semicarazide

::......

C O

carbonyl

H2O_

Structures above this line must be memorized. Those below are for reference.

R N C O.. ..

..isocyanate

R N N R....

azo compound+

R N N....

_ .._R N N

..+

diazo compound

C

O

NN

urea

....::

C N..

imine

C N OH..

oxime

N OH....

..

hydroxylamine

....

_C

O

O :

carboxylate

O:_

alkoxide

C

O

acyl group

C

O

carbonyl group

: :R R R

WHAT IS ORGANIC CHEMISTRY?:Organic chemistry is the study of the chemistry of carbon (C) compounds. Exceptions include CO, CO2 , carbonates, bicarbonates, cyanates (CNO-), thiocyanates (CNS-), and elemental C compounds like diamond and graphite. Although these compounds contain carbon, their chemistry is typical of inorganic chemicals and they are generally classed as inorganic. Most organic compounds contain H. Many also contain N, O, P, S, Cl, Br and I.

Considering only C, H, O, and N, there are over 18 106 known C-containing compounds and this number increases by about 500,000 each year.

C is unique among the elements in its ability to bond with itself forming long chains of compounds from simple to immense, i.e., from methane (CH4 ) to DNA containing 10’s of billions of C atoms.

Synthetic organic compounds include medicines, dyes, paints, polymers, food additives, pesticides, fibers, etc.

Natural organic compounds include the matter contained in all living and once-living organisms, e.g., hair, skin, muscles, genes, food, etc. Aside from water, living organisms are made up primarily of organic compounds.

HISTORY: In 1807, J. J. Berzelius coined the name ‘organic’ chemistry for materials derived from

living organisms. (‘Inorganic’ refers to materials derived from minerals). Up until ~ 1800, the only source of organic chemicals was from living organisms (hence

the name ‘organic’). In 1826, Friedrich Wohler converted an inorganic C-containing chemical, ammonium cyanate into urea, a previously known ‘organic’ substance isolated from urine .

Wohler was intrigued that chemicals of the same elemental composition could be different and invented the term ‘isomerism’ to describe this.

Organic chemistry began with chemists synthesizing natural organic compounds. For example, aspirin, which was originally obtained as an extract from the bark of the willow tree, is now produced synthetically in millions of tons per year.

The availability of large, inexpensive sources of raw materials, i.e., petroleum, coal, and natural gas, has caused synthetic organic chemistry to flourish.

Unfortunately, < 10% of the fossil fuels consumed are used to make chemicals; > 90% is burned to supply energy.


NH4+ -OCN

heatC

O

NH2H2Nammonium cyanate

urea

Some Basics: Carbon atoms have 4 bonds or less, never 5 bonds!

The 4 bonds on carbon may exist in any of 4 arrangements.

Note that each line (---) in the structures represents a covalent bond, i.e., a pair of electrons that is shared between two atoms.

When carbon has 4 bonds it is neutral (as above), but when carbon has only three bonds it exists in one of 3 possible forms, a cation (‘carbocation’ or ‘carbonium ion’), an anion (‘carbanion’) or a neutral radical.

Electron Configuration of the Elements:In order to understand organic chemistry we must learn the electron configuration of the first 20 elements (H to Ca) plus Br and I.

Electrons are continuously buzzing around the nucleus at mind-boggling speeds (ca. 1/10 the speed of light). We don’t know the exact position of electrons from one moment to the next (Heizenburg uncertainty principle) but we do know that their movement is not entirely random.

Electrons fly within well-defined flight paths (orbitals) around the nucleus. Each orbital can hold a maximum of 2 electrons. Think about the heavier elements on the periodic table, with 100+ electrons flying around the nucleus in 50+ different flight paths. Inevitably, some of the orbitals overlap. Just imagine how busy their flight controllers must be while trying to prevent all those flying electrons from colliding.

The orbitals lowest to ground zero (the nucleus) are lowest in energy and are occupied by electrons before the outer, high-energy orbitals. The 50+ orbitals around all atoms are grouped into 7 different ‘energy levels’ (also called layers or ‘shells’) with n = 1 being the shell closest to the nucleus (lowest energy) and n = 7 being the farthest from the nucleus (highest energy). Study the order of orbitals and shells in the planetary model of the atom.

PLANETARY MODEL OF THE ATOM SHOWING ENERGY LEVELS n = 1 to 6


This model is not spatially correct. There is some overlap of orbitals in the 3rd shell and higher.

Within each shell, there exist subshells or types of orbitals. The types of orbitals are named s, p, d, f, g, h, etc.

The 1st shell has only an s orbital, named 1s. The 2nd shell has both s and p-type orbitals, named 2s and 2p. The 3rd shell has s, p and d-type orbitals, named 3s, 3p and 3d. The 4th shell has s, p, d and f-type orbitals, named 4s, 4p, 4d and 4f. etc.


Nucleus

K shell (n = 1)1s 1

L shell (n = 2)2s 1, 2p 3

M shell (n = 3)3s 1, 3p 3, 3d 5

N shell (n = 4)4s 1, 4p 3, 4d 5, 4f 7

O shell (n = 5)5s 1, 5p 3, 5d 5, 5f 7, 5g 9

P shell (n = 6)6s 1, 6p 3, 6d 5, 6f 7, 6g 9, 6h 11

ORGANIC CHEMISTRY INTRO

SHAPES OF ATOMIC ORBITALS

An s orbital

x

y

z

x

y

z

x

y

z

z

x

y

A px orbital A py orbital A pz orbital Three p orbitals

A dxy orbital A dxz orbital A dx2-y2 orbital A dz2 orbital

z

A dyz orbital

x

y

z

x

y

z

(between the x and y axes) (between the x and z axes) (between the y and z axes) (on the x and y axes) (on the z axis)

(on the z axis)(on the y axis)(on the x axis)

x y

x

y

z

y

x

z

(l = 2, ml = -2) (l = 2, ml = -1) (l = 2, ml = 0) (l = 2, ml = +1) (l = 2, ml = +2)

(l = 1, ml = -1) (l = 1, ml = 0) (l = 1, ml = +1)

(l = 0, ml = 0)

6

The number of each type of orbital, their shape and orientation are listed below. There is only one s-orbital in each shell and it is spherical. There are three p-orbitals in the 2nd and all higher shells. The three p-orbitals

are propeller shaped and are oriented along an x, y or z axis in space. They are named px, py and pz, respectively.

There are five d-orbitals in the 3rd and all higher shells. Four of the five d-orbitals look like four-leaf clovers each oriented differently around the nucleus. The fifth d-orbital looks like a propeller inside a donut. They are named dxy, dxz, dyz, dx2-y2 and dz2.

There are seven f-orbitals in the 4th and all higher shells. All but one have six lobes. Each one is oriented differently around the nucleus.

In writing the electron configuration of the elements we fill lowest energy orbitals first (Aufbau principle), with a maximum of 2 electrons per orbital –with opposite spins (Pauli Exclusion principle). Orbitals of the same energy level (‘degenerate orbitals’) are all singly filled (half-filled) before electrons pair up. This occurs, for example in the 2px, 2py, and 2pz orbitals.

The filling order (increasing energy level) of the various orbitals is shown in the following chart.

Reading the table left to right and top to bottom, the orbital filling order is as follows:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p


ns (n-2)f (n-1)d np

H 1s He

Li 2s 2p Ne

Na 3s 3p Ar

K 4s 3d 4p Kr

Rb 5s 4d 5p Xe

Cs 6s 4f 5d 6p Rn

Fr 7s 5f 6d 7p Uuo

Compare this table with a periodic table. The filling order is the same as the layout of the s-, p-, d- and f-blocks on the periodic table, i.e., this is the filling order. Note the patterns.

The orbital filling order is: ns, (n-2)f, (n-1)d, np.

The p-orbitals begin filling after the 2s orbital.

The d-orbitals begin filling after the 4s orbital.

The f-orbitals begin filling after the 6s orbital.

7

For the purpose of learning organic chemistry, we need only study the electron configuration of the first 20 elements (H to Ca) plus Br and I, i.e., in the following orbitals:

1s 2s 2p 3s 3p 4s

Br and I have electrons in the 4p and 5p orbitals respectively, but their electron arrangement is analogous to F and Cl.

The electron configuration of atoms is shown using a notation in which the number of electrons in each orbital is written as a superscript. The orbital is shown as a line, _ or as a circle, O. Each electron in the orbital is written as an arrow, . The direction of the arrow is either up, , (indicating clockwise rotation) or down, , (indicating counterclockwise rotation). Complete the following table.

Full Orbital Notation Simplified Notation

1s 2s 2px 2py 2pz

1H __ 1s1

2He __ 1s2

3Li __ __ 1s2 2s1

4Be __ __ 1s2 2s2

5B __ __ __ __ __ 1s2 2s2 2p1

1s 2s 2px 2py 2pz 3s6C __ __ __ __ __ 1s2 2s2 2p2

7N __ __ __ __ __ 1s2 2s2 2p3

8O __ __ __ __ __ 1s2 2s2 2p4

9F __ __ __ __ __ 1s2 2s2 2p5

10Ne __ __ __ __ __ 1s2 2s2 2p6

11Na __ __ __ __ __ __ 1s2 2s2 2p6

Problem: Write out the electron configuration for Mg through Ca in both 'Full Orbital Notation' and 'Simplified Orbital Notation'. Recall that the 4s orbital is filled before the 3d orbital.


1s 2s 2px 2pz2py

C126

outer, valence electronsbonding electrons

Write out the full electronconfiguration of carbon:

ENERGY

The 2px, 2py, & 2pzorbitals are equal inenergy ('degenerate').

8

Full Orbital Notation Simplified Orbital Notation3s 3px 3py 3pz 4s

12Mg

[Ne] [Ne] 3s2

13Al14Si15P16S

17Cl18Ar19K[Ca

The outermost occupied shell is referred to as the ‘valence’ shell. Orbitals of the valence shell are thus ‘valence orbitals’ and electrons in the valence orbitals are ‘valence electrons’.

The outer (valence) electrons are transferred or shared in chemical reactions. Chemistry is understood in terms of valence electron arrangement.

The number of valence electrons determines the ability of an atom to combine with other atoms

The number of covalent bonds an atom forms to become isoelectronic with its nearest noble gas is called its ‘covalence’. (Isoelectronic means ‘having the same valence electronic configuration’.)

The number of valence electrons in an atom is shown with a Lewis Symbol. One dot is drawn for each valence electron. The dots are placed into four positions (one for each of the one s plus three p orbitals) around the symbol of the element, i.e., north, south, east or west. Once all four positions are singly filled, electrons (dots) are paired up until a maximum of 8 valence electrons (dots) have been drawn.

Although Lewis symbols do not always show the lowest energy electron arrangement of an unbonded atom, they are a good depiction of the electron arrangement just prior to bonding.

When dots (electrons) are drawn, each orbital is first half-filled before electrons are paired up in orbitals.

The arrangement of elements in the periodic table is based on the number of valence electrons. For example, elements in Group IVA have 4 valence electrons. For all representative elements (A-group elements), the number of valence electrons equals the group number. Complete the following table. Note that He is an exception. Although it has 2

valence electrons like Be and Mg, it is unreactive (like other noble gases) and therefore has a valence of 0.


B.. .some Lewis symbols N

...

..C

.. .

.

9

Group # 1A 2A 3A 4A 5A 6A 7A 8A

# valence electrons

1 2 3 4 5 6 7 8

covalence (# bonds)

Period1

H He

Period2

Li Be B C N O F Ne

Period3

Na Mg Al Si P S Cl Ar

Period4

K Ca Br Kr

electronconfig.

s1 s2 s2 p1 s2 p2 s2 p3 s2 p4 Is2 p5

Xes2 p6

show lone pairs of

electrons after

bonding

H Be B C N O F

Note that covalence is the same as the group number for groups 1A to 4A, but covalence is equal to [8 - (group number)] for groups 5A to 8A.

It is very important to appreciate the relationship between the Lewis symbols and the number of covalent bonds formed by an atom (its covalence). Usually , in most stable organic compounds, the atoms form a covalent bond for each unpaired electron in the Lewis symbol of the atom.

Study the bonding arrangements of the neutral atoms shown below. Note that all single (unpaired) electrons in a Lewis structure will bond (as shown in the bonded structure). The non bonded electron pairs (‘lone pairs’) may either remain unbonded or form two additional bonds per electron pair. The two additional bonds may be two single bonds or one double bond

Octet Rule:Note however that 2nd period elements (B, C, N, O and F) will never have more than 4 bonds (8 electrons) around themselves as they can only use four orbitals - their 2s and 2p orbitals for bonding (‘octet rule’).

Hypervalent Atoms:


3rd period elements and higher (Si, P, S, Cl and Br) can form more than 4 bonds (more than 8 electrons) by using their d-orbitals. Examples include PCl5, SF6, ClF7, and BrF7. Such elements that exceed the octet rule are called ‘hypervalent’.

Group # 3A 4A 5A 6A 7A

LewisSymbol

BondedStructure

LewisSymbol

BondedStructure

Carbon and nitrogen are the only two elements that can form a triple bond.


ELECTRONEGATIVITY:One way to estimate the degree of ionic or covalent character in a chemical bond is to compare electronegativities of atoms involved. Electronegativity is a measure of the force of an atom’s attraction for electrons that it shares in a chemical bond with other atoms.

In the 1930’s, Linus Pauling assigned electronegativity values to all elements relative to F (the most electronegative element), which he gave a value of 4.0 .

Linus Pauling's Table of ElectronegativitiesH2.1Li1.0

Be1.5

B2.0

C2.5

N3.0

O3.5

F4.0

Na1.0

Mg1.2

Al1.5

Si1.8

P2.1

S2.5

Cl3.0

K0.9

Ca1.0

Sc1.3

Ti1.4

V1.5

Cr1.6

Mn1.6

Fe1.7

Co1.7

Ni1.8

Cu1.8

Zn1.6

Ga1.7

Ge1.9

As2.1

Se2.4

Br2.8

Rb0.9

Sr1.0

Y1.2

Zr1.3

Nb1.5

Mo1.6

Tc1.7

Ru1.8

Rh1.8

Pd1.8

Ag1.6

Cd1.6

In1.6

Sn1.8

Sb1.9

Te2.1

I2.5

Cs0.8

Ba1.0

La1.1

Hf1.3

Ta1.4

W1.5

Re1.7

Os1.9

Ir1.9

Pt1.8

Au1.9

Hg1.7

Tl1.6

Pb1.7

Bi1.8

Po1.9

At2.1

Fr0.8

Ra1.0

Ac1.1

Bear in mind that these values can vary slightly depending upon the chemical environment and so the values are average values.Note that EN increases across any period and decreases down any group (in most cases).

Pure covalent bonds involve equal sharing of the bonding electron pairs. Pure covalent bonds occur when both atoms involved have equal EN (i.e., EN = 0) ...For example, H2, N2, O2, F2, Cl2, Br2, I2, Cx, S8, PH3, and CS2 are all pure covalent.

Nonpolar covalent bonds are those in which EN 0.4. Examples include all the pure covalent compounds listed above as well as compounds, e.g., CH4 (EN = 0.4) and BH3

(calculate EN)

Polar covalent bonds are those in which the bonding pair of electrons is unequally shared (0.5 EN 1.7). Examples include HBr, HCl, etc. In polar covalent compounds, the more electronegative atom has a partial negative charge ( -) and the less electronegative atom has a partial positive charge ( +).

Ionic bonds are those in which EN 1.8 and are generally considered to have complete charge separation, i.e., considered to be made up of cations and anions. Examples include NaCl, Li3N, and CaO. Exceptions are HF and alkali metal iodides. EN in HF is 1.9 but


this compound behaves as a polar covalent compound. EN in LiI is only 1.5 but LiI behaves ionically. BF3 is also anomalous. It is covalent although EN = 2.0.

A scale of bond type versus EN follows...

In polar covalent bonds, the electron distribution is said to be polarized, i.e., not equally distributed but closer to the more electronegative atom. For example, in HCl....

EN = (3.0 – 2.1) = 0.9 i.e., polar covalentA separation of '+' and '-' charge is called a dipole. Dipoles are sometimes illustrated with an arrow pointing toward the more electronegative atom.

+ - The tail of the arrow is crossed to look like a + sign.The head of the arrow points in the direction of electron shift.

H Cl

The shifting of electron density through sigma bonds due to EN differences between atoms is called an 'inductive effect'. Electropositive elements (metals) such as Zn and Hg, inductively donate electrons through sigma (single) bonds with carbon. Electronegative elements (nonmetals) such as oxygen and chlorine inductively withdraw electrons from carbon through their sigma bonds with carbon.

Problem: Calculate EN values and show bond dipoles and dipole arrows.


covalentnon polar polar c ovale nt ionic

0 0.4

0.5 1.7

1.8 3.2EN

pure c ovale nt

13

BONDING (INTRAMOLECULAR FORCES):Atoms bond because the resulting compound is more stable (lower energy). For the representative elements, filled valence orbitals (isoelectronic with the nearest noble gas) is a stable arrangement.

Note the # of electrons in each PEL for the noble gases and the # of valence electrons..

period noble gas PEL = 1 PEL = 2 PEL = 3 PEL = 4 PEL = 5

1 He 2

2 Ne 2 8

3 Ar 2 8 8

4 Kr 2 8 18 8

5 Xe 2 8 18 18 8

For main group elements, bonds form in which the combining atoms obtain a noble gas electron configuration by either transferring electrons (ionic) or sharing electrons (covalent). A-Group element bond to become isoelectronic with their nearest noble gas.

Ionization energy (Ei) is the amount of energy added to remove an electron form an isolated atom (endothermic process). Metals have low Ei whereas metalloids and non metals have high Ei. Ei decreases down all groups (due to increased shielding and distance from the nucleus) and from right to left across all periods (due to increased distance from the nucleus).

Write the full electron configuration of the reactants and products in the reaction ...Na + Cl [Na+ + Cl-]

Na (1s2 2s2 2p6 3s1) + Cl (1s2 2s2 2p6 3s2 3p5) Na+ (1s2 2s2 2p6) + Cl- (1s2 2s2 2p6 3s2 3p6)

isoelectronic with Neon isoelectronic with Argon Ionic bonds result when metals lose electrons to nonmetals forming cations (+) and anions

(-). Electrostatic attraction holds the solid together, e.g. ...Na (g) Na+ (g) + 1e- Ei = + 119 kcal/molCl (g) + 1e- Cl- (g) Eea = - 83 kcal/molNa+ (g) + Cl- (g) NaCl (g) Elattice -121 kcal/mol

The lattice energy (Elattice) is the energy released due to coulombic (electrostatic) attraction as cations and anions combine in a crystallatice.

Ionic bonds are common in inorganic compounds, but are less common in organic compounds. The Ei of C is too high (+261 kcal/mol). C shows little tendency to act as a source of a cation or anion in an ionic bond so most of its compounds are characterized by covalent bonds.

Covalent bonding is readily shown using Lewis diagrams or Lewis structures.


-85 kcal overall

14

Using Lewis structures show the bonding in CH4, CH3OH, and H3O+ (hydronium ion)

Lewis structures show a bonding electron pair as a line ( ) and a nonbonded pair as two dots (.. or :).

Draw Lewis structures for H2, and N2 and include lone pair electrons.

Most of the 2nd and 3rd period representative elements we encounter in organic chemistry (C, N, O, F, Na, Mg, Al, Si, P, S, Cl, etc.) obey the ‘octet rule’, i.e., in forming compounds, atoms will gain, lose or share electrons to obtain 8 valence electrons. This produces a stable valence electron configuration (stable octet) like that of Ne and Ar.

H and Li also react to obtain a noble gas electron configuration, i.e., that of Helium with 2 valence electrons.

Reaction Mechanisms: We show the transfer of electrons with curved arrows. A one-electron transfer is shown with a half arrow head, while a two-electron transfer is shown with a full (double) arrowhead. Arrows point away from the donor atom and toward the acceptor atom. Be careful to draw covalent bonds as a line (a shared pair) but recall that an ionic bond is shown by a non bonded pair and electric charges on the anion and cation. .


3 H . C..

..+ + O..... . + H. C

H

H

H

O H..

..

O..... . + O

..

... . O....O

..

..

CH4

CH3OH

O2

H2O

4 H . C..

..+ C

H

H

H

H. . .. . ...C

H

H

H

H

Lewis structure

15

EXCEPTIONS TO THE OCTET RULE:The octet rule indicates that atoms bond to share enough electrons so that each atom has 8 valence electrons after bonding. Most main group elements (Group 1A to 7A) obey this but there are a few exceptions, divided into 2 kinds, i.e., molecules in which atoms contain fewer than 8 valence electrons and molecules in which atoms contain more than 8 valence electrons.

Molecules in which atoms contain less than 8 valence electrons include BeCl2, AlCl3 and BF3. Be has only 4 valence electrons around it while both Al and B have only 6 valence electrons around them. These compounds thus behave as Lewis acids or electrophiles (electron acceptors). Learn the Lewis structures for these compounds.

Molecules in which atoms contain more than 8 valence electrons.....Atoms of the 2nd period use 2s and 2p orbitals for bonding and these orbitals can contain only 8 valence electrons, hence, the octet rule.Atoms of the 3rd, 4th, and 5th period have ns, np, nd, etc. orbitals and can accommodate more than 8 electrons in their valence shells.Phosphorus and sulfur have 3s, 3p, and 3d orbitals and in some compounds have > 8 valence electrons. Study the Lewis structures for the following .....

PI3

PI5

SF6


PI

II

......:

:::

.... ..

P

I

I

I

I

:

: :

:

: :: :

..

..

..I

....

.. ..

16

DRAWING LEWIS STRUCTURES OF COMPOUNDS:1. The least electronegative atom is central, e.g., in CO2 O=C=Onot O=O=C

2. H is never central (max 1 bond), e.g., in H2O H-O-H not H-H-O

3. In oxyacids (‘ternary’ acids), e.g., HNO3, H2SO4, HClO3, etc., acidic H’s are bonded to O and O’s are bonded to the central atom, e.g., in HClO (hypochlorous acid) H-O-Cl not H-Cl=O

4. Join all atoms with single bonds. Add additional bonds and non-bonded electron pairs consistent with the Lewis symbols and normal covalence of atoms. Non-bonded e pairs may remain unbonded but single electrons always form bonds. Make each atom isoelectronic with its nearest noble gas but ‘3rd period plus’ atoms may be hypervalent. Check for normal bonding as shown on page 10.

5. Count (check) the number of valence electrons in the structure. Add 1 e- for each negative charge on an ion and subtract 1 e- for each positive charge on an ion.

Note: Oxygen does not normally bond to itself, except in peroxides, e.g., hydrogen peroxide, H-O-O-HAlternate method:1. Add up the number of valence electrons in each atom to find the total number of valence electrons

in the compound. Add 1 e- for each negative charge on an ion and subtract 1 e - for each positive charge on an ion.

2. Initially, connect all atoms by single bonds (2 electrons per bond) and then add electron pairs to each atom to complete their octet. Place any leftover electrons on the central atom. If there are not enough electrons to give the central atom an octet, try multiple bonds. Use one or more of the unshared pairs of electrons on the atoms bonded to the central atom to form double or triple bonds. Note that some atoms often have one or more lone pairs of electrons, i.e., group 5A (1 lone pair), groups 6A (2 lone pair) and group 7A (3 lone pairs).

3. Check that B, C, N, O, and F do not have more than 8 electrons around them after bonding.


C

OH

O HO

H2CO3

carbonic acid

(24 valence electrons)

sulfuric acidH2SO4

( 32 valence electrons)

phosphoric acidH3PO4


P

O

OO

OS

O

OO

O

:

....

: :

..

..H

H

H

: :

..

..

..

..

..

: :

....

H

H

..:....

sulfurous acidH2SO3


acetic acidCH3COOH


nitrous acidHNO2


OO

OH

S

..

....

:

:

....

C

H

HH

O

O H

: :....C N OOH

....

..

..

..H

17

Draw Lewis structures for thionyl chloride (SOCl2), phosphorus oxychloride (POCl3), and orthosilicic acid [Si(OH)4.

Thus far we have only drawn structures of neutral molecules. Charged polyatomic ions are also encountered in organic chemistry. Consider the following reaction.

We see that hydronium ion, a charged polyatomic cation, forms when water is ‘protonated’ (bonds with H+). In hydronium ion it is the oxygen atom (no longer the hydrogen ion) that has most of the positive charge. This can be seen by determining the ‘formal charge’ of each atom in the complex.

FORMAL CHARGE:Formal charge allows us to assign charges to all atoms in a polyatomic ion or molecule. A charged atom is usually a site of activity. The charge of an atom in a polyatomic ion or molecule is called its formal charge.

To derive the formal charge, follow these steps....

1. Write a correct Lewis structure for the molecule or ion.

2. Assign to each atom, all of its unshared (non bonding) electrons and ½ of its shared electrons (half of each bonding pair).

3. Compare this with the number of electrons in the neutral, unbonded atom. The difference is the formal charge.

4. The total charge on a molecule or ion is the sum of its formal charges.


thionyl chlorideSOCl2

..S

O

ClCl

: :

::..

....

..

phosphorus oxychoridePOCl3

P

O: :

Cl

Cl

Cl ::

: :

....

..

....

18

Formal Charge = the theoretical charge on a bonded atom assuming the atom owns all its non bonded electrons and half of its bonded electrons. Compare this to the Lewis symbol of the neutral atom.

Although oxygen has 8 valence electrons (1 non bonded pair + 3 shared pairs), it only ‘owns’ 5 valence electrons, 1 less than that seen in the Lewis symbol of neutral oxygen. Hence oxygen has a formal charge of +1 hydronium ion.

BH3 (borane) has an incomplete octet (an empty 2pz orbital). As a result it has a tendency to accept a pair of electrons from an electron donor, e.g., sodium hydride (NaH). The resulting product is a polyatomic anion in which boron has a formal charge of –1. Study the following Lewis structures.

Exercises:1. Draw Lewis structures (with a curved arrow) to show the reaction of lithium hydride

(LiH) with aluminum hydride (AlH3). Lithium aluminum hydride (LiAlH4) forms.

2. Draw the Lewis structure of sodium hydroxide (NaOH). Note that the Na to O bond is ionic, not covalent. Do not draw a line (covalent bond) showing a shared electron pair between Na and O. Rather the pair of electrons will appear as an extra non bonded pair on oxygen. Oxygen will have a formal charge of –1, while sodium will have a formal charge of +1. Hydrogen has a formal charge equal to 0.

3. Draw Lewis structures (with a curved arrow) to show the reaction of ammonia (NH3) with anhydrous hydrogen chloride gas (HCl). Ammonium chloride (NH4Cl) forms.

4. Draw the Lewis structure of sodium amide (NaNH2). As with NaOH, the Na to N bond is ionic, not covalent. The amide ion (NH2

-) is a polyatomic anion in which N has formal charge of –1.


Note that only valence electrons are considered when determining formal charge. Inner shell electrons are not involved in bonding and need not be considered. Even if inner shell electrons are counted, the result is the same so we use the simpler method of considering only outer valence electrons when calculating formal charge.

Bonding electron pairs are not necessarily shared equally so formal charges do not necessarily represent actual charges. However, in most cases, if the Lewis structure shows that an atom has a formal charge, it actually bears at least part of that charge. This partial transfer of '+' and '-' charge is called polarization.

SOME ACID BASE REACTION MECHANISMS:Neutralization of H2SO4 (a diprotic acid) with KOH (a monoprotic base) occurs in two steps. First potassium bisulfate (NaHSO4) forms and finally sodium sulfate (Na2SO4). The reaction mechanism can be shown by drawing Lewis structures and curved arrows.

Note that OH- donates a pair of non bonded electrons to remove H+ from the acid. H simultaneously leaves its covalently bonded electrons with oxygen in the acid (producing H+) and accepts the pair of electrons from OH- forming H2O. This leads us to two useful definitions of acids and bases:

A Lewis base, such as OH-, is an electron (pair) donor.

A Lewis acid, such as H2SO4 is an electron (pair) acceptor.

A Bronsted base, such as OH-, is a proton (H+) acceptor.

A Bronsted acid, such as H2SO4, is a proton (H+) donor.

K+ ion proceeds throughout the reaction unchanged and is thus called a ‘spectator ion’.

Study the mechanism of Na2CO3 (a diprotic base) reacting with aqueous HCl. Note that when HCl (and other very strong acids) dissolves in water it reacts completely forming hydronium chloride (H3O+ Cl-). No HCl remains in dilute aqueous solutions of the acid.


Recall from the rules for writing Lewis structures that acidic hydrogens are attached to the oxygen, not the central atom in oxyacids.

Exercises:1. Explain why Na2CO3 is a Lewis base.

2. Explain why H3O+ is a Lewis acid.

3. Explain why Na2CO3 is a Bronsted base.

4. Explain why H3O+ is a Bronsted acid.

5. Determine the formal charge on the carbon and oxygen atoms in HCO3- (Carbon is 0. One oxygen is –1, the other two have a formal charge of 0).

6. Draw Lewis structures and show the mechanism (with curved arrows) of the reaction of perchloric acid (HClO4) plus H2O. Perchloric acid is a stronger acid than HCl and it reacts completely with water producing perchlorate ion (ClO4

-) and hydronium ion (H3O+).

7. Sodium amide (NaNH2) is a stronger base than sodium hydroxide. Draw Lewis structures to show the mechanism of its reaction with acetic acid (CH3COOH) forming sodium acetate (Na+ CH3COO-) and ammonia (NH3).

Having become familiar with Lewis structures, now is good time to learn the Lewis structures of the generic functional groups shown on page 1 in this section. Note that ‘R’ stand for ‘the Rest of the molecule’ and is usually interpreted as some alkyl group such as methyl (CH3-), ethyl (CH3CH2-), etc.

Look closely at the Lewis structure of nitro compounds. Notice that the overall charge of a nitro compound is zero but N is +1 while one O is –1.

Exercise: Draw two Lewis structures of nitric acid (HNO3)..


The Lewis structure of nitric acid shows permanent charge separation within the molecule. Two resonance structures are possible (above). In resonance structures, the positions of all atoms are unchanged but electron arrangements differ.

We will most commonly see resonance when we study reaction mechanisms. Study the energy diagram for the combustion of methane.

Free radicals (unpaired electrons) are very reactive (high energy). If resonance could occur such that the free electrons could move between several atoms (instead of just one), then the free radical transition state would be less reactive and the activation energy would be lowered and the reaction would proceed more quickly.

RESONANCE:


As with free radicals, electric charges in an organic molecule indicates a high energy and high instability in the compound. Reaction intermediates (or transition states) are often charged and thus represent the highest energy level of a chemical reaction. The higher the energy of a transition state, the greater is the activation energy of a reaction and hence the slower the reaction proceeds.

Resonance is a means by which electric charge on an atom is spread (delocalized) over several atoms in a molecule. The effect of charge delocalization in a reaction's transition state is to decrease the activation energy (G*) and speed up a reaction. The chemical reactivity of many conjugated unsaturated compounds is enhanced by their ability to delocalized charge via resonance in a transition state. More resonance structures indicates greater charge delocalization and lower G*.

The student must become skilled at drawing resonance structures of conjugated organic molecules as this is often a way of understanding a reaction as well as a way of successfully predicting reaction products.

Rules for Resonance:

1. Electron(s) may move from an atom to an adjacent -bond(forming a -bond)

2. Electron(s) may move from a -bond to an adjacent atom(eliminating a bond)

3. A bond may flop from one side of an atom to anotherside, i.e., from atop one bond to atop an adjacent bond on the same atom.

Use a full arrowhead for showing movement of a pair of electrons and a half-arrowhead for movement of a single electron (a free radical):Benzyl Cation:

CC+

C C

Benzyl Anion:

C C CC-

Benzyl Free Radical:

C C CC


X+

X....

X

X+

X....

X

23

VALENCE BOND THEORY & VSEPR:Valence bond theory and Valence Shell Electron Pair Repulsion theory (VSEPR) are simple but effective models for understanding the bonding that occurs in organic compounds.

Covalent bonds are usually formed between two atoms by the overlap of 2 half-filled (singly-occupied) orbitals of the bonding atoms. For example, two hydrogen atoms form a covalent bond by the overlap of their half-filled 1s atomic orbitals.

Carbon is in group IVA of the periodic table and therefore has 4 valence (bonding) electrons, i.e., 4 electrons in its highest Principal Energy Level (PEL). Its ground state electron configuration is given as … 1s2 2s2 2px

1 2py1 2pz

0

Carbon normally forms 4 covalent bonds using its 4 valence shell electrons, however, its ground state electron configuration is not suitable for this because the 2s orbital is full and the 2pz orbital is empty. In a process called ‘hybridization’, one of the 2s electrons is promoted to the vacant 2pz orbital producing 4 half-filled orbitals. These orbitals alter their shape in order to achieve the best possible overlap with orbitals from other atoms and also to move as far away from each other as possible.

1s2 2s2 2px1 2py

1 2pz0 1s2 2s1 2px

1 2py1 2pz

1

Only the outer, valence electrons are involved in bonding and thus the inner shells of electrons will not be considered further.

The vast majority of organic compounds are formed from just a few elements, i.e., C, N, O, P, S and the halogens. We will look at their hybridization states in the following pages.


4 valence electrons

hybridization

H HH H+

1s11s1

H H

Valence Bond theory predicts thattwo, half-filled 1s atomic orbitalsof hydrogen overlap to form a single(sigma) bond.

This H2 molecule is stable (low in energy) because the electronsspend most of their time betweenH nuclei-drawing them together.

Each H atom inH2 is isoelectronicwith Helium

24

BONDING IN CARBON COMPOUNDS


The arrangement of the 4 atomicorbitals and the electronic configuration in the carbon atomare not suitable to form 4 bonds.

A change (hybridization) occursto the orbital shape and electronicconfiguration to facilitate bonding.The 3 hybridizations occurring incabon are shown below.

orbital shape

ground statevalence electronconfiguration

2s2 2px1 2py1 2pz0

C6

orbitalshape

hybridizedorbitals

electronconfigurationbonds

109.5º or

(sp3 orbitals x 4)

(forms 4 bonds) alkanese.g., ethane (C2H6)

H

HHCH

H

H

C

H

H

H

C H

H

H

CC

C

sp3 hybridized (tetrahedral)Carbon forms 4 single bonds Carbon forms a double bond

(sp2 orbitals x 3) + (p x 1)

(3 bonds) + (1 bond) alkenes and arenese.g., ethene and benzene (C 2H4) (C 6H6)

H

HH

H

HH

120º90º

HH

CH

H H

HH

H

H

H

C

C

C CH

H

H

H

sp2 hybridized (triangular planar)

180º

(sp orbitals x 2) + (p x 2)

(2 bonds) + (2 bonds) alkynese.g., acetylene (C2H2)

H

HC

C

C

Carbon forms 1 triple or 2 double bonds

sp hybridized (linear)

C C HH

also CO2C OO

Group 4A4 valence e’s. Covalence = 4

25

BONDING IN NITROGEN COMPOUNDS


The shape and orientation of the4 atomic orbitals in the nitrogen atom are not suitable for forming 3 (or 4) bonds.

A change (hybridization) occursto the orbital shape to facilitatebonding. The 3 hybridizationsoccurring in nitrogen are shownbelow.

orbital shape


2s2 2px1 2py1 2pz1

N7

.

....

.

(sp 2 orbitals x 3) + (p x 1)

(1 lone pair) + (2 bonds) + (1 bond) azo compoundse.g., (trans azobenzene)

120º

90º

N N....

NN

N:

..

:

:

sp2 hybridized

Nitrogen forms a double bond

..

.. ..

...

. . . .

.

NH3 NH4+Cl-

orbitalshape

hybridizedorbitalselectronconfigurationbonds

or

(sp3 orbitals x 4)

amines ammoniumammonia chloride

(4 bonds)or

107º

NN

H

+Cl -

1 lone pair +3 bonds

NHH

H

NHH

H

sp 3 hybridizedNitrogen forms 3 or 4 single bonds

.. ..

..

..


..

(sp orbitals x 2) + (p x 2)

(1 lone pair) + (1 bond) + (2 bonds) nitrilese.g., ethanenitrile

N C CH3:

CH3

180º

N

NC

..

..

Nitrogen forms 1 triple bond sp hybridized

..

26

BONDING IN OXYGEN COMPOUNDS


The shape and orientation of the4 atomic orbitals in the oxygen atom are not optimal for forming2 (or 3) bonds.

A change (hybridization) occursto the orbital shapes to facilitatebonding. The 2 hybridizationsoccurring in oxygen are shownbelow.

orbital shape


2s2 2px2 2py1 2pz1

O8

..

.. ..: :

(sp 2 orbitals x 3) + (p x 1)

(2 lone pair) + (1 bonds) + (1 bond)

e.g., acetone

H3C CH3

H3C CH3

(CH3CCH3)

O

CC

O

O

.. ..

: :

: :O 120º

90º

sp 2 hybridized Oxygen forms a double bond

..

. .

orbitalshape

hybridizedorbitalselectronconfigurationbonds

or

(sp3 orbitals x 4)

O105º

O

e.g., methanol (CH 3OH) hydronium ion (H3O+)

or2 lone pair +2 bonds

1 lone pair +3 bonds

....

.. ....

..

Oxygen forms 2 single bonds sp 3 hybridized

....

CH3

OH..

..O

H

+

H

..H


27

Note that a double bond is made of one and one bond.

Note that a triple bond is made of one and two bonds.

Halogens (groups VIIA elements) generally form only one bond in organic compounds. They do not reshape their orbitals (hybridize) when they bond.

Hydrogen , like the halogens, does not hybridize its 1s orbital when bonding. Silicon , like carbon, is a group 4A element with 4 valence electrons. As expected, silicon

forms sp3 hybridized tetrahedral compounds with 4 substituents. Simple examples include silicon tetrabromide (SiBr4) and tetramethylsilane [(CH3)4Si]. Silicon forms a few compounds in which it has double bonds, e.g., H2Si=CH2. However, silicon's large size makes p-orbital overlap for bonds less effective than in carbon compounds. Unlike 2nd period elements which cannot accommodate more than 8 electrons in their valence orbitals, Si, a 3rd period element can expand its valence shell to accommodate 10 electrons (sp3d hybridized – 5 substituents, e.g., SiF5

-) and even 12 electrons (sp3d2 hybridized – 6 substituents, e.g., fluorosilicic acid, H2SiF6). Phosphorus , like nitrogen, is a group 5A element with 5 valence electrons. As expected,

phosphorus forms sp3 hybridized compounds with 3 substituents. Simple examples include phosphorus tribromide (PBr3) and trimethylphosphine [(CH3)3P]. Phosphorus forms some compounds in which it has double bonds to oxygen, e.g., phosphoric acid (H3PO4). However, phosphorus’ large size makes p-orbital overlap for bonds less effective than in C or N compounds. Like other 3rd period elements, phosphorus can be bonded to 4, 5, and 6 atoms. e.g., phosphorus oxychloride (Cl3P=O), phosphorus dibromide trichloride (PBr2Cl3), and phosphorus hexafluoride anion (PF6

-). Sulfur , like oxygen, is a group 6A element with 6 valence electrons. As expected, sulfur forms

sp3 hybridized compounds with 2 substituents. Simple examples include dimethyl sulfide [(CH3)2S] and methyl mercaptan (methane thiol) (CH3SH). Sulfur can form bonds to three (H2SO3), four (H2SO4), five (SOF4) and six atoms (SF6).

A methyl cation has an sp2 hybridized carbocation with a vacant p orbital. A methyl radical has an sp2 hybridized carbon atom with a ½-filled 2p orbital. A methyl anion contains an sp3

hybridized carbanion with a lone pair in one of its sp3 orbitals. Draw them.


(3 lone pairs and 1 bond)

The shape and orientation of the4 atomic orbitals in the halogens are adequate for forming onesingle bond.

Hybridization does not occurwhen halogens form single bonds.

orbital shape


3s2 3px2 3py2 3pz1

Cl17

Cl CH

H

H

CH3Cl:....

methyl chloride(chloromethane)

27

In saturated compounds, all atoms have only bonds, whereas in unsaturated compounds, one or more bonds are present.

Conjugated unsaturation occurs when alternating and bonds are present. In such compounds, all p-orbitals in conjugated bonds overlap.

Isolated unsaturation occurs when bonds are separated by more than one bond. In such compounds, p-orbitals of one bond cannot overlap with p-orbitals of other bonds.

Cumulated unsaturation describes immediately adjacent unsaturation. Cumulated carbon-to-carbon compounds are not very stable and are rarely encountered.


CH2 CH CH CH2

C CC CH

H

H

H

H

H

1,3-butadiene

C CC CH

H

H

H

H

H

O

2-cylcopentenone

1,4-pentadiene

C CCH

H

H

H

CH

H

H

C

H

CH2 CH CH2 CH CH2

C CCH

H

H

HC

H

H

H

C

H

O

3-cylcopentenone

CH2C CH CH31,2-butadiene

28


Study the following table. In the last 3 columns Lewis structures are drawn as if the atoms were bonded. Learn these names and structures and identify their hybridization states.

Lewis

Symbol

#

valence

e- 's

#

bonds

#

unshared

e- 's + 1 F.C. neutral - 1 F.C.

B 3 3 0 none

C 4 4 0

N 5 3 2

O 6 2 4

F 7 1 6

Cl 7 1 6

Bromine and iodine are analogous to fluorine and chlorine. Draw the structures of bromonium and iodinium cations, bromide and iodide anions, and bromine and iodine.


B

CC +

carbonium ion

..

NN +

nitronium ion

C

..

-

carbide ion

B-

boride ion

..

:O

..

O+

oxonium ion

..

O

..:

an oxide ion

-

..:

F+

fluoronium ion

..:

+

chloronium ion

Cl

..

N:

nitride ion

-

F..

....

unhybridized

Cl..

....

unhybridized

F..

....

unhybridized

:

f luoride ion

-

Cl..

....

unhybridized

:

chloride ion

-

30

HOMOLYTIC AND HETEROLYTIC PROCESSES:The amount of energy required to dissociate a hydrogen molecule into 2 H atoms is called its ‘bond (dissociation) energy and is quite large (104 kcal/mol)

Bond breaking: There are 2 ways in which a covalent, 2-electron bond can break: in an electronically symmetrical way so that one electron remains with each product fragment or in an electronically unsymmetrical way so that both electrons remain in one product fragment, leaving the other fragment with an empty orbital.

Symmetrical cleavage is called a homolytic process and the unsymmetrical cleavage is called a heterolytic process.

A:B A + B homolytic cleavage (radical reaction)

A:B A+ + B:- heterolytic cleavage (polar reaction)

Bond making: Conversely, there are 2 ways in which a covalent 2-electron bond can form: in an electronically symmetrical (homogenic) way when 1 electron is donated to the new bond by each reactant, or in an electronically unsymmetrical (heterogenic) way when both bonding electrons are donated to the new bond by one reactant.

A + B A:B homogenic bond making (radical reaction)

A+ + B:- A:B heterogenic bond making (polar reaction)

Processes that involve symmetrical bond breaking and making are called radical reactions. A radical is an atom or group of atoms that contains one or more unpaired valence electrons and thus has one or more orbitals with only 1 electron...

Cl2 Cl + Cl (2 chlorine radicals/atoms)

O2 O + O (2 oxygen diradicals/atoms)

CH4 CH3 + H

Radicals are very reactive species and generally only exist for a short time.


HOMOLYTIC BOND DISSOCIATION ENERGY:Bond dissociation energy (D) is the amount of energy consumed when a bond is broken (cleaved). Homolytic cleavage produces free radicals: A B A . + B .Homolytic bond breaking is always endothermic but varies widely from I2 (36 kcal/mol) to HF (135 kcal/mol). Some values are listed in the following tables ....

Homolytic Bond dissociation Energies (D) of Various A B Bonds in kcal/mol

B in A BA in A B group H F Cl Br I OH NH2 CH3

H --- 104 135 103 87 71 119 107 104CH3 methyl 105 110 85 71 57 93 80 90

CH3CH2 1 98 107 81 69 53 92 77 89

CH3CH2CH2 1 98 107 81 69 53 91 78 85

(CH3)2CH 2 94.5 106 81 68 53 92 93 84

(CH3)3C 3 92 110 79 63 52 93 93 80

C6H5 CH2 benzyl 88 72 55 45 70 75

CH2 CHCH2

allyl 86 69 55 41 72 72

CH2 CH vinyl 108 84 92

C6H5 phenyl 111 116 96 81 65 103 91 101

Homolytic Bond Dissociation Energies (D) of Miscellaneous Bondscompound D (kcal/mol) compound D (kcal/mol)

CH3 OCH3 80 CH2 CH2 (both bonds) 165

CH3 OH 93 CH2 CH2 ( bond only) 62

CH3O H 104 C N (both bonds) 143

HO H 119 C C (all bonds) ~ 200

CH3O OCH3 36 C N (all bonds) ~ 213

N C H 130 C C H 132

C H (aldehydes) 87 C O (aldehydes & ketones) 176-179

Others: H2 =104, F2 = 38, Cl2 = 58, Br2 = 46, I2 = 36, RS H 80, Si Si 53 Bond dissociation energies are often measured in the vapor phase, which ignores other effects

such as salvation, by the solvent. Despite these limitations, bond dissociation energies provide a set of relative bond strengths that are helpful in understanding many reactions.

Note particularly the trend in the dissociation energies for the C H bond. There is a decrease with the progression from methyl to 1 to 2 to 3 C's. The same trend is seen in the dissociation of C X and C C bonds. This indicates an increasing stability of more substituted free radicals. In addition, conjugation stabilizes radicals due to resonance, i.e., allylic and benzylic radicals are unusually stable.


HETEROLYTIC BOND DISSOCIATION ENERGY:Heterolytic cleavage produces ions: A B A+ + B:-

Heterolytic bond breaking is more endothermic than homolytic. The separation of opposite electric charges requires additional energy. Some values are listed ....

Heterolytic Bond dissociation Energies of Various A B Bonds in kcal/mol

B in A BA in A B group H F Cl Br I OH

H --- 401 370 334 324 315 390

CH3 methyl 218 256 227 219 212 274

CH3CH2 1 182 191 184 176 242

CH3CH2CH2 1 185 178 171 235

(CH3)2CH 2 163 170 164 156 222

(CH3)3C 3 148 157 149 140 208

C6H5CH2 benzyl 166 157 149 215

CH2 CHCH2

allyl 173 165 159 223

CH2 CH vinyl 207 200 194

C6H5 phenyl 219 210 202 275

Note that the dissociation energies for C X bonds generally decrease with the progression from methyl to 1 to 2 to 3. The stability of carbocations is increased by alkyl substitution since alkyl groups are inductively electron donating.

In addition, carbocation stability is increased by resonance. Thus allylic and benzylic carbocations are unusually stable.

Vinyl and phenyl compounds dissociate less readily. No resonance occurs in vinyl and phenyl carbocations. Furthermore, the terminal carbon in vinyl and phenyl compounds are sp2 hybridized (more electronegative – holds the bonded electrons more strongly) compared to all others listed, in which the terminal carbon is sp3

hybridized (less electronegative – holds the bonded electrons less strongly). The relative stability of carbocations is as follows...


C

R

+R

R

C

R

+R

H

C+

H

H

H

C

R

+H

H

CH2+

H2C CH+

H2C CH CH2 +

+

3º

2º

1º methylphenylvinylallyl

benzyl

> > > > >

increasing C+ stability

33

In the case of heterolytic dissociation, removal of a halogen atom produces a carbocation (C+). The relative stabilities of some carbocations is shown.

Two reasons proposed for the increased stability of more highly substituted carbocations are inductive effects and hyperconjugation. Inductive effects refer to the shift of electron density through a sigma bond. Methyl

groups (and other alkyl groups) which are directly bonded to electron-deficient atoms (e.g., C+) shift electron density to that site and hence are found to be weak electron donors. Alkyl groups stabilize C + ’s by inductive effects .

Hyperconjugation refers to the shift of electron density by the partial overlap of C-H sigma orbitals with a vacant p orbital. A carbocation is sp2 hybridized. Its vacant p-orbital extends above and below the plane of the sigma bonds at right angles.

Note that the order of stability of free radicals is also 3º > 2º > 1º > methyl due to hyperconjugation with neighboring alkyl groups.


C

H

H

H +C

CH3

H

H +C

CH3

H

H3C +C

CH3

CH3

H3C +

tertiary 3º

secondary 2º

primary 1º

methyl

increasing carbocation stabilitymorestable

lessstable

> > >

C

CH3

CH3

H3C +

Inductive effects: Alkyl groups donate (shift) electrondensity through sigma bonds toelectron deficient atoms.This stabilizes the carbocation.

vacant p orbitalof a carbocation

sp2hybridizedcarbocation

Csp3-Hssigma bondorbital

overlap (hyperconjugation)

HYPERCONJUGATION

+C C

..H

H

H

34

ACIDS and BASES:Arrhenius Theory: In 1884, Svante Arrhenius defined acids and bases as follows...

An Arrhenius acid contains H and produces H+ (or H3O+) in water (e.g., HCl).

An Arrhenius base contains the OH group and produces OH- in water (e.g., NaOH).

Neutralization is the combination of H+ (or H3O+) and OH- ions to form HOH.

The Arrhenius theory explains reactions of protonic acids with metal hydroxides (hydroxy bases), e.g., HCl (aq) + NaOH HOH + NaCl

Bronsted-Lowry Theory: In 1923, J. N. Bronsted and T. M. Lowry defined acids and bases as

A Bronsted acid is a proton (H+) donor

A Bronsted base is a proton acceptor.

Neutralization was defined as the transfer of a proton from a proton donor (acid) to a proton acceptor (base).

The Bronsted Theory is broader than the Arrhenius and includes non aqueous reactions, e.g., HCl (g) + :NH3 (g) NH4Cl (s)e.g., H-CC-H + Na+NH2

- H-CC: - Na+ + :NH3

Conjugate acid / base pairs are defined as species that differ by a proton.

Na+ OH- + CH3COOH HOH + Na+ CH3COO-

base acid conj. acid conj. base

Strength of Acids and Bases:

A very strong acid (HCl) has a very weak conjugate base (Cl -), i.e., if the acid releases a proton readily, its conjugate base will not attract a proton strongly.

A very strong base (NaOH) has a very weak conjugate acid (H2O), i.e., if the base accepts a proton readily, its conjugate acid will not readily give up the proton.

Acid strength is quantified by Ka and pKa ....

HA + HOH H3O+ + A- and the degree of dissociation is given by...

K =[H O ] [A ]

[HOH]eq3

+ -

[ ]HAbut in dilute soln., [HOH] is a constant (55.5 M) and this is

combined with the equilibrium constant (Keq) giving another constant, Ka, which is a

measure of the strength of an acid... Ka = Keq [HOH] = Keq 55.5 = [ ]

[ ]H O

HA3

[A ]-

and note that: pKa = -log10 Ka

A strong acid (H3O+) has a very weak conjugate base (H2O)

Similarly, a strong base (OH-) has a very weak conjugate acid (H2O).

Similarly for bases: B + HOH BH+ + OH- and the degree of dissociation is ...


K = [BH ] [OH ] [HOH]eq

+ -

[ ]B but in dilute soln., [HOH] is a constant (55.5 M) and this is combined with

the equilibrium constant (Keq) giving another constant, Kb, which is a measure of the strength

of a base... Kb = Keq [HOH] = Keq 55.5 = [ ]

[ ]BH

B

[OH ]-

and pKb = -log10 Kb

The relationships that apply to conjugate acid base pairs are ...

Ka Kb = 10-14 or pKa + pKb = 14

For example, CH3COOH has pKa = 4.7. Its conjugate base, CH3COO- has pKb = (14 – 4.7) = 9.3

It is important that students of organic chemistry be familiar with pK values of acids and bases. Not only do they provide a quantitative measure of the strength of an acid or base but, as we will soon see, these values can also combined to predict the extent of reaction of acid/base reactions

In the following tables, the approximate strengths versus pK value is given.

pKa < 1 1 - 5 5 - 15 > 15acid strength strong moderate weak very weak

example H2SO4 H3PO4 H2CO3 HOH

pKb < 1 1 - 5 5 - 15 > 15base strength strong moderate weak very weak

example NaOH Na3PO4 NaHCO3 HOH

Consider Acids Stronger Than H3O+: H3O+ has a pKa = -1.74. Stronger acids such as HCl,

with pKa's lower than -1.74, protonate HOH producing H3O+ (hydronium ion).

HCl + HOH H3O+ + Cl- pKa = -7 pKb = 15.74 pKa = -1.74 pKb = 21

Thus H3O+ is the strongest acid which can exist in water. Any stronger acid will be 'leveled' (reduced) in strength to pKa = -1.74 by the solvent water. Water is said to be a leveling solvent.

Consider Bases Stronger Than OH-: OH- has a pKb = -1.74. Stronger bases such as

NaNH2, with pKb's lower than -1.74, are protonated by HOH producing OH-.NaNH2 + HOH NH3 + NaOH

pKb = -21 pKa = 15.74 pKa = 35 pKb = -1.74

Thus OH- is the strongest base that can exist in water. Any stronger base will be 'leveled' (reduced) in strength to pKb = -1.74 by the solvent water.

ACIDITY & BASICITY OF WATERWater is both weakly acidic and weakly basic, i.e., ‘amphoteric’ or ‘amphiprotic’.


Water acts as a base by accepting protons from acids...

H-O-H + H-Cl H3O+ + Cl-

Water acts as an acid by donating protons to bases ...

H-O-H + :NH3 NH4+ + OH-

Water undergoes self-ionization, or ‘autopyrolysis’ to form equal concentrations of H3O+ and OH- ions ...

H-O-H + H-O-H H3O+ + OH-

The extent of self-ionization of water is very small. At room temperature (25C) the concentrations of H3O+ and OH_ are 1.0 10-7 moles/L each, i.e., [H3O+]= [OH_] = 10-7

M

The equilibrium constant, Keq, for this reaction is given by ...

where 10-14 is ‘KW’, the ion product constant for water

As for all other acids and bases, Ka and Kb for water is calculated as follows ...

Ka = Kb = Keq [101.74] = 1010

1014

1 7415 74

.

. and thus pKa = pKb = 15.74 for water.

THREE CLASSES OF SOLVENTS

1. Amphiprotic solvents : Other ionizable solvents are also amphiprotic, e.g., methanol, ethanol, acetic acid, and ammonia. The autopyrolysis constants for some are listed.

Solvent - log Ks or (pKw)

water 14

acetic acid 14.5

ethylenediamine 15.3

methanol 16.7

ethanol 19.1

2. Non-polar Aprotic solvents : These have no protons and are not polar. They are nonionizable and inert, i.e., neither acidic nor basic, e.g., benzene, CCl4.

3. Polar Aprotic (Basic but not Acidic) : These are also nonionizable solvents but can accept a proton because of the presence of atoms such as O or N which have lone pairs of electrons, e.g., ether, dioxane, ketones, and pyridine. The are no known examples of solvents that are acidic but not basic.


K = [H O ] [OH ]HOH] HOH]

[10 ]55.5] [55.5] [10 ]eq

3+ - -7

1.74[ [[ ][ [ ].

10 1010

7 14

1 74

37

Predicting the extent of a reaction using the equation: (pKeq = pKa + pKb - 14)

Consider an acid in HOH: Consider a base in HOH:HA + HOH H3O+ + A- B + HOH BH+ + OH-

Ka = Keq [55.5] =[H O ] [A ]

[HA]3

+ -

Kb = Keq [55.5] =[BH ] [OH ]

[B]

+ -

Now consider an acid (HA) reacting with a base (B):

HA + B BH+ + A-

Keq =[BH ] [A ][HA] [B]

+ -

By manipulation of these expressions we derive an expression for predicting the extent of reaction ...

or Ka Kb = ( Keq ) ( Kw )

or Ka Kb = ( Keq ) ( 10-14 )

Taking the log of both sides and then multiplying the equation by (-1) yields ...

pKa + pKb = pKeq + 14

Finally rearranging this we have a simple expression for calculating the extent of an acid-base reaction ...

pKeq = pKa + pKb - 14

The equilibrium constant (Keq) can then be determined (if desired) by taking the negative antilog of pKeq calculated above, however it is easier to convert from pKeq to extent of reaction graphically (following). One can readily get a sense for extent of reaction from basic thermodynamics …

Keq = 1 (pKeq = 0) means that the extent of a reaction = 50%A reaction is said to go to completion when Keq 103 or pKeq -3

These calculations assume that reagents are 100% pure (not dilute) and yet they work quite well as estimates even in dilute aqueous solutions. For example, Pasto, et. al. state (p. 274) that "compounds containing acidic functional groups with pKa's of less than ca. 12 will dissolve in dilute aqueous (5%) NaOH".

HA + NaOH Na+ A- + HOHWe can understand this by calculating pKeq, i.e., [12 + (-1.74) - 14] = -3.74 = pKeq or Keq 5500 (quite favorable even in dilute aqueous solution)

On p.275, the same author states that "only acids with pKa's < 6 will dissolve in dilute (5%) NaHCO3" and we could also have predicted this because pKeq = [ 6 + 7.6 - 14] = -0.4 or Keq = 2.5 Using this equation, determine if 5% NaHCO3 will dissolve a) acetic acid, b) phenol. Note: in order to quantitatively titrate an acid base pair in aqueous media, pKeq must be -8.



Determine pKeq (the negative logarithm of the equilibrium constant, Keq) using the formula: pKeq = pKa + pKb –14.

Read the graph (pKeq vs. Extent of Reaction in %).Note: This only works for Acid/Base reactions.

39

More on Predicting Acid / Base Reactions: In addition to calculating pKeq we can also predict whether an acid/base reaction will occur by making use of a few simple principles. Let's begin with a familiar example which we know reacts essentially to completion ...

CH3COOH + NaOH HOH + Na+ CH3COO-

moderate strong weak weakacid base acid base4.7 -1.74 15.74 9.3 (pKa or pKb)

Note the pK values. In general, an acid and base will spontaneously react only if the reaction products are a weaker acid and a weaker base than the reagent acid and reagent base.

A stronger acid will donate H+ to a base whose conjugate acid is weaker (higher pKa) or

An acid with a lower pKa will donate H+ to a base whose conjugate acid has a higher pKa.

Assign pK values to all species in the following reactions and predict which of the following reactions will proceed as shown ...

CH3CH2O- + CH3COOH CH3CH2OH + CH3COO-

H2N- + CH3CH2OH NH3 + CH3CH2O-

H-CC-H + NaNH2 NH3 + H-CC:-Na+

C6H5OH + OH- C6H5O- + HOH

C5H5N + H3O+ C5H5NH+ + HOH


acid pKa conjugate base

pKb


weakestacids

strongestbases

41

CH4 ca. 55 :CH3- ca. -41C2H4 44 :C2H3- -30C6H6 43 C6H5- -29NH3 35 :NH2- -21H2 35 :H- -21

C2H2 25 :C2H- -11(CH3)2CO 20 CH3COCH2- -6

(CH3)3COH 18 (CH3)3CO- -4CH3CHO 17 CH2- CHO -3C2H5OH 16 C2H5O- -2

H2O 15.74 OH- -1.74HCO3- 10.3 CO3-2 3.7HPO4-2 12.3 PO4-3 1.7

CH3NH3+ 10.6 CH3NH2 3.4C6H5OH 9.9 C6H5O- 4.1

HCN 9.3 CN- 4.7NH4+ 9.2 NH3 4.8CH3SH 8 CH3S- 6H2PO4- 7.2 HPO4-2 6.8

H2S 7.1 HS- 6.9H2CO3 6.4 HCO3- 7.6C6H5SH 6 C6H5S- 8

C5H5NH+ 5.3 C5H5N 8.7HN3 4.7 N3- 9.3

CH3COOH 4.7 CH3COO- 9.3C6H5NH3+ 4.6 C6H5NH2 9.4C6H5COOH 4.0 C6H5COO- 10.0

HF 3.2 F- 10.8H3PO4 2.1 H2PO4- 11.9

CF3COOH 0.2 CF3COO- 13.8C2H5COH+NH2 -1 C2H5CONH2 15

HNO3 -1.4 NO3- 15.4H3O+ -1.74 HOH 15.74

CH3CH2OH2+ -3.6 CH3CH2OH 17.6H2SO4 -5 HSO4- 19

HCl -7 Cl- 21HBr -8 Br- 22HI -9 I- 23

C2H5CNH+ -10 C2H5CN: 24HClO4 -10 ClO4- 24HSbF6 -12 SbF6- 26


Cannot protonate water

Completelyprotonatedby water

Determinedin water.

Protonate water increasingly.

Determinedin water.

Protonated by water increasingly.

Protonatewatercompletely

Not protonatedby water

weakestbases

strongestacids

(pKa + pKb) = 14 for conjugate

acid-base pairs

42

Determine the extent of the following reactions(pKeq) Base

(pKb)Acid (pKa)

Conjugate Base (pKb)

Conjugate Acid (pKa)

(pKeq) reverse rxn

NaOH CH3COOH CH3COO- HOH

NaHCO3 HCl (aq) HOH H2CO3

CH3NH2 H3O+ HOH CH3NH3+

C5H5N H3O+ HOH C5H5NH+

C6H5NH2 H3O+ HOH C6H5NH3+

NaOH C2H5OH C2H5O- HOH

C2H5OH H3O+ HOH C2H5OH2+

NO2--NH2 H3O+ HOH NO2--NH3+

NaHCO3 HA(6.0)

A- H2CO3

NaOH HA(12.0)

A- HOH

NaHCO3 -OH -O- H2CO3

I- CH4 CH3- HI

B7

HA7

A- BH+

B8

H3O+ HOH BH+

NaOH HA8

A- HOH

NaOH -OH -O- Na+ HOH


Note: pKeq [forward rxn.] = -pKeq [reverse rxn.] and (% forward rxn. + % reverse rxn = 100%)

43

Typical pKa values of functional groups Typical pKb values of functional groups

pKa group example pKb group example

55 alkane CH4 24 nitrile CH3CN:45 alkene CH2=CH2 23 acid chloride ethanoyl

chloride35 ammonia NH3 22 aldehyde ethanal25 alkyne CH CH 21.2 ketone acetone20 ketone acetone 20.5 ester ethyl acetate17 amide methanamide 20 carboxylic acid acetic acid17 aldehyde ethanal 17.6 ether C2H5OC2H5

15 - 18 alcohol ethanol 16 - 18 alcohols CH3CH2OH15.74 water H2O 15.74 water H2O

10 phenol -OH 15 aliphatic amides ethanamide9.3 hydrocyanic acid HCN 9.3 acetate CH3COO-

8 aliphatic thiol CH3SH 9 aromatic amine -NH2

7 hydrogen sulfide H2S 7.6 bicarbonate NaHCO3

6 aromatic thiol -SH 6.0 aliphatic sulfide CH3S-

5 carboxylic acid CH3COOH 4.7 cyanide CN-

-1 sulfonic acid -SO3H 4.6 ammonia NH3

-1 bisulfate R-OSO3H 4 phenoxide -O-

-1.74 hydronium ion H3O+ 3 aliphatic amine CH3NH2

-2 protonated alcohol CH3CH2OH2+ -1.74 hydroxide OH-

-7 mineral acid HCl -2 alkoxide CH3CH2O-

borohydride NaBH4

aluminum hydride LiAlH4

Grignard CH3- +MgBr

-21 hydride Li+ :H-

-21 sodium amide Na+ :NH2-

-41 1 carbanion Li+ :CH3-

sodium Na0


LEWIS ACIDS AND BASES:A Lewis acid (electrophile, E+) is a substance that accepts an electron pair.A Lewis base (nucleophile, Nu: -) is a substance that donates an electron pair.As a result of this electron donation from a base to an acid, a bond is formed.Lewis Acids: Lewis acids must have vacant, low energy orbitals, or a polar bond to H, so H+ can be lost. Lewis acids include, but are much broader than Bronsted-Lowry and Arrhenius acids. For

example, metal cations, such as Al+3, are Lewis acids because they can accept a pair of electrons when they form a bond to a base.

In the same way, compounds of Group 3A elements, such as BF3 and AlCl3, are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases.

Similarly, many transition metals salts, such as TiCl4, FeCl3, ZnCl2, and SnCl4 are Lewis acids.

By means of Lewis structures, show the following acid-base reactions ...

1. HCl(g) + H2O

2. BF3 + (CH3)2O (dimethyl ether)

3. AlCl3 + (CH3)3N (trimethylamine)


:H Cl+O HH

.... O HH

..H

+ + Cl.... ..

..: :

_

Chemical reactions invovle the transfer of electrons from electron donors to electron acceptors Arrows show the electron transfer (from nucleophile to electrophile) (from Lewis base to Lewis acid).Arrows do not show movement of molecules, atoms or ions!!!!!!!

The large arrow shows that one of the lone pairs of electrons on oxygen (nucleophile) is used to form a covalent bond with the H atom in HCl (the electrophile). Notice that the oxygen atom in the hydronium ionproduct has one less pair of electrons than the oxygen in HOH and notice that the oxygen atom has a + charge.

The small arrow shows that the hydrogen to chlorine bond breaks as the shared (bonding) pair of electronsmoves to the chlorine atom which thus becomes a chloride anion. Notice that the chloride anion has one morelone pair of electrons than the chlorine atom in HCl.

Lewis base Lewis acid

45

Some Lewis Acids: (proton donors or electron pair acceptors): strong acids: H3O+, HCl, HBr, HNO3, H2SO4 weak acids CH3COOH, CH3CH2OH, C6H5OH (phenol), H2O

cations Cl+, Br+

compounds with vacant orbitals AlCl3, BF3, TiCl4, FeCl3, ZnCl2Lewis Bases:Lewis bases have nonbonding electron pairs that can be donated to Lewis acids. Most nitrogen-containing and oxygen-containing organic compounds have 1 or 2 lone pairs of electrons, respectively.

Examples of Bases: (proton acceptors or electron pair donors)

Note that some compounds can act as both Lewis acids or Lewis bases, depending upon the reaction conditions.

Alcohols and carboxylic acids act as acids by donating a proton but can also act as bases when their oxygen atoms donate an electron pair (to very strong acids), and accept a proton.

Draw Lewis structures showing the following acid-base reactions.....1. CH3OH + HBr

2. (CH3)2O + H2SO4

Note that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. For example, acetic acid can be protonated on the doubly-bonded or singly-bonded O atoms ...


CH3CH2 O H.... CH3CH2 O CH2CH3

..

.. CH3 C H

O

CH3 C CH3

O

alcohol ether aldehyde ketone

: : : :

: :

CH3 C Cl

O : :

CH3 C O

O

H....

: :

CH3 C O

O

CH3

....

acid chloride carboxylic acid ester amide

: :

CH3 C NH2

O..

CH3CH2 N CH2CH3

..

CH3

H OH..

..amine sulfide water

CH3 S CH3

..

46

Some Factors Affecting Acidity:HI > HBr > HCl > HF

decreasing acidity

Decreasing halogen size decreases acid strength, or conversely, increasing halogen size increases acid strength. In the same vertical group of the periodic table, the larger atoms (higher molecular weight), can disperse '-' charge over a larger region and thus add stability to the conjugate base. As the stability of the conjugate base increases (i.e., weaker conjugate base), the greater is the strength of the acid. I- is a weaker base than F- so HI is stronger than HF

Consider EN also: In the same row of the periodic table, the EN of the atom bonded to H increases from left to right across the table. Since more EN atoms can carry a negative charge more readily than a less EN atom, the acidity increases as shown....

(CH3)3C ---- H (CH3)2N ---- H CH3O ---- H F ---- H

increasing acidity

EN C < N < O < Fstability - CH3 < - NH2 < OH - < F -

acidity H ---- CH3 < H ---- NH2 < H ---- OH < H ---- Fbasicity - CH3 > - NH2 > OH - > F -

For organic acids, i.e., carboxylic acids, the proximity of an electronegative atom to an acidic H affects the acidity of the compound ...

(CH2Cl)CH2CH2COOH > CH3(CHCl)CH2COOH > CH3CH2(CHCl)COOH

increasing acidity

As previously mentioned in the section on resonance, delocalization of electrons in a conjugate base increases the stability of the anion, therefore increasing the acidity of its conjugate acid.....

CH3SOOOH > CH3COOH > CH3CH2OH

decreasing acidity


pKa’s of Binary AcidsGroup 4A Group 5A Group 6A Group 7A

CH4

55

NH3

35

H2O

15.74

HF

3.2

SiH4

35

PH3

27

H2S

7.1

HCl

-7

GeH4

25

AsH3

23

H2Se

3.8

HBr

-8

H2Te

2.6

HI

-9

Note that the arrows point in the direction of an increase. Increasing EN for an acid (H-B) increases its acidity. Increasing size of H-B also increases acidity. Acidity of binary acids increases left to right across each period as the EN increases. Acidity of binary acids increases down each group as the size of the heteroatom increases greatly, even

though EN is increasing up the group. Basicity trends of conjugate bases are opposite those of acidity trends of acids. See below.

pKb’s of Conjugate Bases of Binary AcidsGroup 4A Group 5A Group 6A Group 7A

CH3-

-41

NH2-

-21

OH-

-1.74

F-

10.8

SiH3-

-21

PH2-

-13

HS-

6.9

Cl-

21

GeH3-

-11

AsH2-

-9

HSe-

10.2

Br-

22

HTe-

11.4

I-

23

pKa’s of Ternary (Oxy-) AcidsTernary acids (oxyacids) contain hydrogen-oxygen-hereroatom links and conform to the general formula HmXOn. These compounds are acidic when X is a nonmetal or metalloid but are basic when X is metallic. With few exceptions (such as phosphorus oxyacids), H-atoms are bonded exclusively to O-atoms. Several factors affect the strength of an oxyacid.

1. If there is more than one ionizable hydrogen, Ka1 > Ka2 > Ka3. The successive acidity constants differ by ca. 5 powers of 10, i.e., 105, e.g., H3PO4, pKa1 = 2.1, pKa2 = 7.2, pKa3 = 12.4

2. Inductive and resonance effects are important. Acidity is greatest when the heteroatom is highly electronegative and is increased by the presence of electron-withdrawing groups (e.g., -F or –CF 3). Thus the acidity increases left to right in the following series:

H3PO4 < H2SO4 < HClO4 HIO3 < HBrO3 < HClO3

CH3CO2H < FCH2CO2H < F2CHCO2H < CF3CO2H

3. Acidity increases with increasing number of oxygens. This is also an inductive (and resonance) effect. The Ka’s increase successively in the following series by factors of ca.105, i.e., pka’s decrease by ca. 5:

HClO < HClO2 < HClO3 < HClO4


acidity, EN

aci

dity

, siz

e

basicity

bas

icity

EN

48

Electrophiles (E+) and Lewis acids are both electron pair acceptors (proton donors in the binary acids shown below). They have similar periodic trends as seen in the following table. Note that the arrows point in the direction of an increase.

pKa’s of Binary AcidsGroup 4A Group 5A Group 6A Group 7A

CH4

55

NH3

35

H2O

15.74

HF

3.2

SiH4

35

PH3

27

H2S

7.1

HCl

-7

GeH4

25

AsH3

23

H2Se

3.8

HBr

-8

H2Te

2.6

HI

-9

Nucleophiles (Nu:-) and Lewis bases are both electron pair donors, however their periodic trends are opposite in vertical columns (as seen in the following table). Note that the arrows point in the direction of an increase.

pKb’s of Conjugate Bases of Binary AcidsGroup 4A Group 5A Group 6A Group 7A

CH3-

-41

NH2-

-21

OH-

-1.74

F-

10.8

SiH3-

-21

PH2-

-13

HS-

6.9

Cl-

21

GeH3-

-11

AsH2-

-9

HSe-

10.2

Br-

22

HTe-

11.4

I-

23

The size (polarizability) of nucleophiles has the largest effect on nucleophilicity. Larger, more polarizable nucleophiles are more able to donate an electron pair (through distortion of their large valence electron clouds).


acidity, E+

aci

dity

, E+

basicity, Nu:-

Nu:

- ba

sici

ty

49

STRENGTH OF NUCLEOPHILES (NUCLEOPHILICITY)The relative rate at which a nucleophile (Nu:-) reacts to displace (substitute for) a leaving group is called ‘nucleophilicity’. Consider the following nucleophilic substitution reactions:

CH3OH + HI CH3I + HOHCH3OH + HCl CH3Cl + HOH

The first reaction is much faster than the second because I- is a much better Nu:- than Cl-. The leaving group (HOH) was the same in both cases. The nucleophilicity (relative reactivity) of various Nu:-’s is listed in the following table ...

Reactivity Nu:- Relative Reactivity

very weak HSO4-, H2PO4

-, RCOOH < 0.01weak ROH 1

HOH, NO3- 100

fair F- 500Cl-, RCOO- 20 103

NH3, CH3SCH3 ~ 300 103

good N3-, Br- ~ 600 103

OH-, CH3O- 2 106

very good CN-, HS-, RS-, R3P:, NH2- , I-, H-, R- > 100 106

Note that Nu:-’s are electron donors as are Lewis bases and reducing agents. Nu: - ’s are either uncharged (with nonbonded electrons) or they are anions, but they are never cations. Nu:-’s are basic, neutral, or sometimes weakly acidic, but not strongly acidic. Strong acids (HCl, H2SO4) and Lewis acids (AlCl3, SnCl2) are electrophiles (E+’s), i.e., electron acceptors as are oxidizing agents.1. Within any given row of the periodic table, nucleophilicity decreases from left to right

as basicity & polarizability decrease (because electronegativity of the central atom is increasing)

CH3- > NH2

- > OH- > F-

> NH3 > OH2 > HFPH2

- > SH- > Cl-

PH3 > SH2 > HCl

2. For nucleophiles with the same attacking atom, the anion is more basic and more nucleophilic than the neutral compound.

Cl- > HCl OH- > HOH RO- > ROH NH2- > NH3 CH3CO2

- > CH3CO2 H CN- > HCN3. Nucleophilicity increases down any column of the periodic table; as the polarizability

of atoms increases (despite decreasing basicity – Polarizability has greater influence).NH2

- OH- F-

H2P- HS- Cl-

H2As- HSe- Br-

H2Sb- HTe- I-


increasing nucleophilicityincreasing

50

Note the similarities and differences of nucleophiles and bases ... Nu:-’s and bases are both electron donors Basicity deals with equilibrium position (Keq). At equilibrium, a stronger base holds a

greater proportion of H+. Nucleophilicity deals with kinetics. A stronger Nu:- attacks faster than a weaker one.

Basicity deals with interaction with H+ while nucleophilicity is broader and also deals with interaction with other atoms, especially, but not only C atom.

Polarizability (‘squashiness’) of Nucleophiles: A polarizable nucleophile, e.g., I-, is large and soft (‘squashy’) because its valence (donor)

electrons are far from the nucleus (in the 5th period). The electron cloud is readily distorted during bond making and breaking which reduces the energy maximum in the transition state and thus speeds up reactions.

A non-polarizable nucleophile, e.g., F- is small and hard (rigid). Its outer valence electrons are close to the nucleus (in the 2nd period) and tightly held. F - forms strong bonds but its electron cloud is not easily distorted during bond formation and breaking so its transition states are high energy (slow reaction).

It is often true that good nucleophiles are also good leaving groups for the same reasons, i.e., they are polarizable and stabilize a negative charge (which leaving groups often have).

Efficiency of Leaving Groups: Groups which best stabilize a '-' charge are the best leaving groups, i.e., the weakest bases

are most stable as anions and are the best leaving groups. These are salts of strong acids (conjugate bases of strong acids), e.g., HI + H2O I- + H3O+

In the following table, the relative rate at which various groups will ‘leave’ in a substitution reaction are listed. Note that the weakest bases (which are least reactive –most stable) are the best leaving groups.

pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21

TosO- I- Br - Cl- F- HO- RO- H2N-

60,000 30,000 10,000 200 1 0 0 0

Note: F-, OH-, RO-, & NH2- are not easily displaced by nucleophiles, i.e., they are lousy

leaving groups.

Based on pKb values, where would the following leaving groups be placed in the preceding table? H2O ? ROH? carboxylate, RCOO- ?


Trends in Strength of Nucleophiles and Electrophiles

Nucleophiles (Nu:-) are electron donors. Nucleophiles are anions or neutral but never cations.Electrophiles (E+) are electron acceptors. Electrophiles are usually cations or neutral but occasionally are anions.Study the following tables and note that for nucleophiles with the same attacking atom (in the same column), the anion is a much

better nucleophile than the neutral atom for electrophiles of the same kind (e.g., H2O and H3O+), the cation is a much better electrophile

than the neutral species. some neutral molecules (like H2O) can be nucleophilic in the presence of a strong E+

(e.g., H2O + HBr) and can be electrophilic in the presence of a strong Nu:- (e.g., H2O + NH2-).

BH3 CH3+ NH4

+ H3O+ R-OH2+ HF

BH4- :CH3

- :NH3 H2O R-OH F-

NH2- OH- R-O-

O-2

PH4+ H2S R-SH HCl

:PH3 HS- R-S- Cl-

PH2- S-2

H3BO3 H2CO3 HNO3 H3O+ HBr

H2BO3- HCO3

- NO3- H2O Br-

CO3-2 OH-

O-2

HCN H3PO4 H2SO4 HClO3

CN- H2PO4- HSO4

- ClO3-

HPO4-2 SO4

-2

PO4-3

Nucleophiles donate loosely held electrons. These are electrons (e.g., CH2=CH2) or non bonded electrons (e.g., :NH3) but rarely are they electrons such as in Na-H. Other examples include metals such as Mg: and Na.

Some electrophiles have empty orbitals such as BH3 or AlCl3. Other electrophiles produce an empty orbital in order to accept electrons from the Nu:-.

Strong Bronsted acids (protic acids such as HCl H+) are good examples. Atoms with + charge are potential electrophiles, such as the C in a carbonyl group. It is +

and its weak bond to O can cleave to form an empty orbital (C+) . Similarly, a + C attached to a good leaving group (e.g., -Br) can cleave a bond to form an empty orbital (C+).


Nu:- strength

E+ strength

Nu:- strength

E+ strength

Nu:- strength

E+ strength

Nu:- strength

E+ strength

52

Study the following reactions. Write products where they are not shown and identify nucleophiles and electrophiles.


H Cl....

:

+ Na+ O H....:

-+H+Cl -

....: : + Na+Cl- + OH H

....

C

CH3

CH3

H3C O H....+ C

CH3

CH3

H3C O H..

H+

C

CH3

CH3

H3C +

Br -....: :

H Br....

:

H+Br -....: : +

C

CH3

CH3

H3C Br

CH

HC

H

H+

H Cl....

:

H+Cl -....: : + C C

H

H

H

H

H

+

CH3 C

O

O

: :

H....

Na+ ++CH3 C

O

O

: :.... :

-+ H+ NH2

-

sodium amide

:..

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REDOX REACTIONS IN ORGANIC CHEMISTRYThere are a variety of ways for calculating oxidation numbers for organic chemistry. In single C compounds, we use the same (rigorous) procedure used for inorganic chemistry, i.e., assign oxidation numbers (ox. #) to all atoms other than C based on their EN values and calculate the ox. # of C from these with the aid of a couple rules ...

1. The ox. # of a compound equals the charge on the compound

2. Certain atoms have fixed ox. #’s when bonded to C, i.e., H= +1, O = -2, halogens (X) = -1

Calculate the ox. # of C in each of the following compounds ...

Ox. # -4 -2 0 +2 +4 +4

CH4 CH3OH CO2H-C-OH

O

HO-C-OH

O

H-C-H

O

CH3Cl CH2Cl2 CHCl3 CCl4

Oxidation of an organic compound causes an increase in oxidation number and is also defined as:

any reaction that increases the number of atoms which are more EN than C, i.e., O, N, X or

any reaction that decreases the number of atoms which are less EN than C, i.e., H, Li, Na, Mg

Reduction is the opposite, i.e., a decrease in oxidation number and a reaction that decreases the number of EN atoms atoms.

Identify the following reactions as oxidation or reduction...

CH3-CH3 CH2=CH2 CH CH

CH3CH2OH CH3-C-OH

O

CH3-C-H

O

CH2=CH2 CH3-CH2OH+H2O

- H2O


Oxidizing Reagents for Organic ChemistryOxidant (media) Reduced Form Oxidant Reduced Form

(purple) MnO4- (OH- or neutral) MnO2 (brown ppte.) X2, e.g., Cl2 X-, e.g., Cl-

(purple) MnO4- (H+) Mn+2 (colorless, aq.) H2O2 2 O-2

hot conc. HNO3 NO2 (brown gas) O2 2 O-2

(orange Cr+6 ) H2CrO4 (aq. H2SO4) Cr+3 (blue-green) Pb(OAc)4 Pb(OAc)2

PCC (still Cr+6) (in CH2Cl2) Cr+3 (blue-green) Hg(OAc)2 HgOAc

HIO4 (colorless) HIO3 (colorless) NaOCl (bleach) Cl-

H2CrO4 is prepared by dissolving CrO3 (chromia or chromic anhydride) or K2Cr2O7 in aq. H2SO4.Jones reagent: CrO3 + H2SO4 (aq.) + acetone H2CrO4 Pyridinium Chlorochromate (PCC): CrO3 + C5H5N: + HCl C5H6NCrO3Cl Collins Reagent: CrO3 + 2 C5H5N: (C5H5N)2CrO3

Cr+6 in aqueous acid soln. are strong oxidants. Cr+6 in anhydrous solvent are mild oxidants.

Reducing Agents for Organic ChemistryReducing Reagent (conditions) Oxidized form

H2 (g) (Pt, Pd, Ni, etc. catalyst in EtOH) H+

LiAlH4 (in ether) liberates :H- H2

NaBH4 (in EtOH) liberates :H- H2

BH3 (g) H2

Li, Na, K, Zn, Hg, Mg, etc. Li+, Na+, K+, Zn+2, etc.

SnCl2 + 2HCl = H2SnCl4 H2SnCl6

Zn[Hg] + HCl ZnCl2 + H2


Balancing Redox Equations:

In order to calculate the theoretical and actual yield from a reaction, a balanced chemical equation is necessary. The following rules describe the Ion-Electron Half-Reaction Method for balancing redox equations ...

1. Break the equation into 2 half-reactions, i.e., oxidation and reduction2. Balance all atoms in each half-reaction other than H and O3. Balance O and H as follows ...

add H2O to balance O’s first, then balance H’s by adding H+

add sufficient electrons to balance the charges Note: e-'s are always added to the right side of the oxidation ½-reaction because ox. = loss of e-'sNote: e-'s are always added to the left side of the reduction ½-reaction because red. = gain of e-'s

if the reaction is carried out in alkaline media, add enough OH- to neutralize all H+

(combine them to make HOH) and be sure to add the same quantity of OH- to both sides to keep mass and charges balanced

4. Multiply each ½-reaction by a least common multiple so the number of e -'s transferred in each ½-reaction is equal.

5. Add the ½-reactions and cancel common terms from each side6. Check for charge and mass balance

Balance the following redox equations for practice ...

1. Cyclopentanol is cleaved by strong oxidants like hot, conc. HNO3 producing pentanedioic acid.

2. Cyclohexene is cleaved by strong oxidants like hot, acidic KMnO4 producing hexanedioic acid.

3. Cyclohexene is oxidized to a diol by cold, neutral or alkaline KMnO4 producing 1,2-cyclohexanediol.


Elimination Often Competes with Substitution: Strong dehydrating acids (H2SO4, H3PO4) favor elimination (dehydration) in alcohols. Because they

are strong acids, they readily protonate the alcohol thereby converting a poor leaving group (OH -) into a good leaving group (HOH), however, the anions produced after protonation of the alcohol (HSO4

- or H2PO4-) are very poor nucleophiles and can’t replace the leaving group.

CH3CH2-OH + H2SO4 (catalyst) CH2=CH2 + H2O (elimination)

(CH3)2CH-OH + H2SO4 (catalyst) CH3CH=CH2 + H2O (elimination)

(CH3)3C-OH + H2SO4 (catalyst) (CH3)2C=CH2 + H2O (elimination)

Strong non-dehydrating acids (like HI, HBr and HCl) also readily protonate an alcohol creating a good leaving group (HOH) but with the difference that the resulting Nu:-’s (like I-, Br-, and Cl-), are much better Nu:-’s and readily replace the leaving group which results in substitution. CH3CH2-OH + HBr CH3CH2-Br + H2O (substitution)

(CH3)2CH-OH + HBr (CH3)2CH-Br + H2O (substitution)

(CH3)3C-OH + HBr (CH3)3C-Br + H2O (substitution)

Very strong bases can cause elimination reactions with alkyl halides because strong hydrohalic acids (HX) produced by elimination react rapidly and completely with the excess strong base in a neutralization reaction, as per Le Chatalier. This is especially true with 2 and 3 alkyl halides which are bulky (hindered) and the Nu:- has difficulty contacting the reactive C atom.CH3CH2-Cl + CH3O-Na+ [CH3CH2-OCH3 + NaCl] & [CH2=CH2 + CH3O-H + NaCl]

1 (v. strong base) > 90 % sub. <10% elim.

(CH3)2CH-Cl + CH3O-Na+ [(CH3)2CH-OCH3 + NaCl] & [CH3CH=CH2 + CH3OH + NaCl]2 ~ 20 % sub. ~80% elim.

(CH3)3C-Cl + CH3O-Na+ [(CH3)3C-OCH3 + NaCl] & [(CH3)2C=CH2+ CH3OH + NaCl]3 ~3 % sub. ~ 97% elim.

(CH3)3C-Cl + Na+CN- (CH3)3C-CN + NaCl3 (v. good Nu:-) (almost all sub.)


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ORGANIC REACTION TYPES: - CHEM HOGchemrat.com/ChemHog2/Organic Chem_files/organicintro3.doc · Web viewAdd up the number of valence electrons in each atom to find the total number