lecture 6 resonance(1).pdf

8/15/2019 Lecture 6 Resonance(1).pdf

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Lecture 6Resonance Structures1. What is Resonance?

Resonance Structures and Resonance Hybrid

2. Using Curved Arrows (Arrow Pushing)

3. Rules in Writing Resonance Structures

Sample Problems

4. Can Lone pairs move? Sample Problems

5. Generalizations on Electron Movement?

Sample Problems

6. Criteria for Determining Major Resonance Contributor

Sample Problems

7. Chemical Bonding – The Complete Picture

Sample ProblemLesson 4|1


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What is Resonance?

Lesson 2|2


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» The structures of some molecules and ions are notadequately represented by a single Lewis structure.

Resonance is a method of writing or describing

these molecules and ions in terms of two or moreLewis structures.

» Individual Lewis structures are called resonance

structures or resonance forms or resonancecontributors.

Their relationship is shown by double‐headed

(resonance) arrows.

» The molecule or ion is a hybrid of the various

contributing structures.

Lesson 4|3


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» The unhybridized porbitals overlap toform the bond.

» The sigma and pi bondsmake a double bond.

sp2sp

2

sp2

sp2sp2

sp2

Lesson 4|4

» The sigma bonds between

the C atoms result from the

overlap between sp2 and

sp2 hybrid orbitals of the Catoms.


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The p orbitals can overlap

this way.

» This overlap corresponds

to the following Lewis

structure

The p orbitals can also

overlap this way.

» This overlap corresponds

to the following Lewis

structure

Lesson 4|5

• Individual structure is not

sufficient to represent the

benzene molecule.

• The two structures are needed

to represent benzene molecule.


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» Are NOT real» An individual resonance structure does NOT

accurately represent the structure of a molecule or

ion

» The resonance hybrid is a more accurate

representation of the molecule

The resonance hybrid is the combination of all the

various contributing resonance structures.

» Resonance structures are NOT in equilibrium

» Writing Resonance structures is NOT the same as

writing chemical equations

Lesson 4|6


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Resonance Structure is

a single Lewis

structure

Resonance Hybrid

is a singlestructure that is

constructed from

contributions

from all possible

resonance

structures.

Lesson 4|7


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Using Curved Arrows

Lesson 2|8


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» Single headed arrow denotes irreversible reaction

» Double arrow denotes reversible reaction at

equilibrium

» Double‐headed arrow denotes resonance structures

» Curved arrow denotes movement of electrons

Lesson 4|9


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» Curved arrow: A symbol used to show the movementof electrons

Tail start from theelectrons

• The electrons are

either (1) lone

pairs or (2) pi ()

electrons

Head lands either on(1) a sigma () bond

or (2) an atom

Lesson 4|10


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Lesson 4|11

Curved arrow (1) denotes

the movement of the lone

pair on N atom between

N and C(A) atoms, making

a pi bond. The new bond

between N and C(A)

atoms is now a double

bond.

Curved arrow (2)denotes the movement

of the pi electrons

between the C(A) and

C(B) atoms to move the

lone pair to the C(B)

atom.

After completing curved arrows (1) and (2), that is, after

moving the electrons, a new resonance structure is drawn as

shown below:

Note: the formal charge of N

changed from 0 to +1; while the

formal charge of C(B) changed

from 0 to ‐1.

Formal Charges of:C(B) = 0

C(A) = 0

N = 0


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The complete representation of the process in SLIDE #11 is shown below:

Lesson 4|12


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In drawing resonance structures, the

curved arrow is used to move pi electronsor lone pairs of electrons, and NOT to move

a charge or an atom or a sigma bond.

Lesson 4|13

Once the electrons have been

moved, the correct formal charges ofthe atoms involved in the electron

movement must be assigned.


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It is easy to make the following mistake:

1. Break the

bond

• The curve arrow is NOT permissible

because the arrow indicates the movement

of the

bond.

• In resonance, the

bond does NOT

delocalize.

Lesson 4|14


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It is easy to make the following mistakes:1. Put more than 8 electrons on B, C, N, O, and F

2. You make this mistake if you don’t count the Hatoms and the number of electrons around these

atoms.

The C atom with asterisk has

already 8 electrons:

• 1 bond with C• 1 bond with O and

• 2 bonds with 2 H atoms (not

drawn in skeletal structure

If the arrow were drawn

as shown, there would be

10 electrons on C atom

with asterisk

*

Lesson 4|15


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Electron pushing using

curved arrows is a survivalskill in organic chemistry.

Need to learn it well!

Lesson 4|16


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Rules in Writing

Resonance Structures

Lesson 2|17


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1. Delocalize or move Only electronsa) Move ONLY either electrons or lone pairs

b) Do not move atoms

The nuclei stays put, that is, the order of

bonding does not change

c) Do not move or break electrons

2. Do not exceed the octet electrons for second period

elements, such as B, C, N, O and F

3. Do not vary the total number of valence electrons

from resonance structure to another resonance

structure

Lesson 4|18


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Is not a valid resonance structure of

It violates Rule 1 (b) and (c); Broke bond between C

and H; moved atom H

Is not a valid resonance structure of

It violates Rule 2; C has 10 electrons

Movement of the electrons to the next C with

asterisk is not a valid resonance structure

It violates Rule 2; C with asterisk would end up with

10 electrons

*

Lesson 4|19


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Lesson 4|20


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» Determine if the pi electrons can be moved in thefollowing molecule:

Lesson 4|21


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» Determine if the pi electrons are delocalized in the followingmolecule:

Analysis:» As drawn, the pi electrons are found in C‐2 and C‐3.

Both of these atoms are sp2 hybridized. Both of themhave unhybridized p orbitals that form the pi bond.

For the pi electrons to move, the adjacent C‐1 musthave an unhybridized p orbital.

Answer: To answer this question, determine thehybridization of C‐1 atom. C‐1 is sp3 hybridized, whichmeans all its p orbitals are hybridized. In other words,there is no p orbital in C‐1 that can accommodate the pielectrons. Therefore, these pi electrons cannot move tobetween C(1) and C(2).

Lesson 4|22


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» Determine if the pi electrons can delocalize in thefollowing molecule:

Alternative Analysis:

If the electrons are delocalized towards C‐1

and C‐2 as shown below, C‐1 would have 10electrons, which is NOT acceptable for C:

Lesson 4|23


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» Determine if the pi electrons are delocalized in thefollowing molecule:

Analysis:

» If the pi electrons cannot be moved between

C(1) and C(2), then try moving the pi electrons toC(3) as shown below:

» Later, we will evaluate whether this resonancestructure is a good resonance contributor.

Lesson 4|24


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» Determine if the pi electrons can move in thefollowing molecule:

Lesson 4|25


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» Determine if the pi electrons can move in the following molecule:

Analysis:» As drawn, the pi electrons are found in C‐2 and C‐3. Bothof these atoms are sp2 hybridized. Both of them haveunhybridized p orbitals that form the pi bond. For the pi

electrons to delocalize, C‐1 must have an unhybridized porbital.

Answer: To answer this question, determine thehybridization of C‐1 atom. Unlike C‐1 in the previous

problem, this C‐1 is sp2

hybridized ( 3 sigma + 0 lonepair), which means that one p orbital is NOT hybridized.The + formal charge on C‐1 denotes that thisunhybridized p orbital is empty; hence, it can

accommodate the pi electrons from C‐2 and C‐3. Lesson 4|26


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» Determine if the pi electrons are delocalized in the following

molecule:

This movement of the pi electrons is represented by two

resonance structures shown below:

The hybrid resonance structure that shows the movement of thepi electrons from C(3) to C(1):

+

+

Lesson 4|27

The dots

represent the

movement of the

pi electrons.


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• When resonance is involved, different

resonance forms may suggest different

hybridization and bond angles,• But a real molecule can have only one

set of bond angles

• Bond angles must be compatible withall the important resonance forms.

• The bond angles (shape) imply the

hybridization of the atoms

• This hybridization must be the same in

all the resonance forms. Lesson 4|28


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Can Lone Pairs of

Electrons move?

Lesson 2|29


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» Can the lone pair on N atom move? Explain.

Lesson 4|31


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» Can the lone pair on N atom delocalize? Explain.

Step 1: Determine if the atom adjacent to the N atomthat contains a lone pair has unhybridized p orbital

» C‐2 is adjacent to N atom that contains the lone pair.» C‐2 is sp2 hybridized; hence, it contains unhybridized p

orbital

Step 2: Determine if the lone pair is in the unhybridizedp orbital

» N is sp2 hybridized; hence, it has unhybridized porbital. But this p orbital is already used for making

the pi bond with C‐2 atom. Therefore, the lone pairmust be in an sp2 hybrid orbital.

» Since the lone pair is NOT in an unhybridized p orbital,it CANNOT delocalize.

» This particular lone pair is localized on N atom. Lesson 4|32


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» Consider the molecule . The following

» resonance structures are valid.

» What is the hybridization on N? Explain.

Lesson 4|33


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If the following resonance structures are valid,

then the lone pair from N atom moved between N and the

adjacent C atom as shown below:

» Lone pair can move if it is located in p orbital of the N atom and

the adjacent C atom is sp2

hybridized, which means it has anavailable p orbital. Since the lone pair moved, it must be in the p

orbital of the N atom. If the N atom has ONE unhybridized p

orbital, then N must be sp2 hybridized.

Lesson 4|34


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Some Generalizations

about Electron

Movement

Lesson 2|35


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1. electrons are next to a positively charged atom

2. Lone pairs next to a positively charged atom

3. electrons are between two atoms with different

electronegativities

Lone Pair or pi electrons can move when:

Lesson 4|36


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4. electrons are conjugated in a bond system

Conjugated bonds involve alternating bonds

separated by a sigma bond

5. Lone pairs are next to a bond

Lone Pair or pi electrons can move when:

Lesson 4|37


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It is easy to make the mistake of

memorizing generalization number 5:

“Lone pair next to a bond” without

understanding the basic concept of

delocalization.

Lone Pairs are delocalized if they are

contained in unhybridized p orbital thatis adjacent to a bond

Lesson 4|38


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It is easy to forget to write

the formal charges of theatoms after moving or

delocalizing the lone pair orthe pi electrons.

Lesson 4|39


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» Consider the molecule . Determine if the

» following resonance structure is valid.

» Explain.

Lesson 4|40


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» Consider the molecule . Determine if the

» following resonance structure is valid.

» Explain.

» The N atom in is sp2 hybridized (2 sigma + 1lone pair); hence, it has unhybridized p orbital. Butthis p orbital is already used in forming a pi bond withthe C atom (double bond in N = C). Therefore, the

lone pair is not in unhybridized p orbital; rather it isfound in an sp2 hybrid orbital.

» Next side…please

Lesson 4|41


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» How can you tell if the lone pair is in the sp2

hybridorbital?

» The N atom has 3 sp2 hybrid orbitals. Two of these orbitals

are used to make the sigma bond with C‐1 and C‐5. Sincethe unhybridized p orbital of N is used to make the pi

bond with C‐1, the third sp2 must contain the lone pair.

» If the lone pair is NOT in an unhybridized p orbital, it

cannot move. Therefore, is NOT a valid

resonance structure.

Lesson 4|42


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» Draw resonance structures for the followingmolecule:

Lesson 4|43


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» The electrons between two atoms with differentelectronegativities can move if the other conditions

for electron movement are satisfied:

» Lone pair in a p orbital can move if it is next to a Cwith a positive formal charge as shown below:

Lesson 4|44


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» The complete process is shown below:

» The three resonance structures are arranged in order

of increasing importance: II, III, I

» The resonance hybrid is:

Lesson 4|45

The movement of the lone pairdenotes that N is sp2 hybridized

with the lone pair in the p orbital.


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» Draw resonance structures for the followingmolecule:

» Therefore, there are 3 resonance structures:

» These 3 resonance structures contribute to the

resonance hybrid shown below:

The delocalization of the lone

pair denotes that N is sp2

hybridized with the lone pair in

the unhybridized p orbital. +

‐

Next slide please…Lesson 4|46


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Criteria for Determiningthe Importance of the

Resonance Contributors

Lesson 4|47


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» There are resonance structures that are equivalent:

» Hence, their contributions to the resonance hybrid

are equal.

» However, some resonance structures are NOT

equivalent:

Lesson 4|48


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» If the resonance structures are not equivalent, theircontributions to the resonance hybrid are not equal.

» Which contribution is greater?

» The contribution of the resonance structure to theresonance hybrid depends on its stability.

» The more stable resonance structure contributesmore.

» This resonance structure is described as the majorcontributor or more significant contributor to theresonance hybrid.

Lesson 4|49


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If resonance structures are NOT equivalent their


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If resonance structures are NOT equivalent, their

contributions to the resonance hybrid vary

Factors Affecting the Stability of the Resonance

Structure

1. Octet electrons

2. Number of covalent bonds3. Negative Charge on the more electronegative

atom

4. Charge Separation

Lesson 4|51


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• A resonance structure of a molecule with

separated charges has a positive charge

and a negative charge.

• Resonance contributors with separated

charges are relatively unstable (relativelyhigh in energy) because energy is

required to keep the opposite charges

separated.

Lesson 4|53


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Major:• All octets filled

• Higher number of

covalent bonds

• Although O (an

electronegative

atom) has positive

charge, it has filled

octet (This is ok if

cannot be avoided!)

Minor:• C+ is not

octet filled

• Lowernumber of

covalent

bonds

Lesson 4|54


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N C O N C O

Which is the major resonance contributor?

Major:

• All octets filled

• Same number of

covalent bonds• The negative

charge is on O (an

electronegativeatom)

Minor:


• Same number of

covalent bonds

• The negative charge

is on the atom N

with the lower

electronegativity

Lesson 4|57


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Draw the resonance hybrid.

Lesson 4|58


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Equivalent:


• Same number of

covalent bonds• The negative

charge is on O (an

electronegativeatom)

Equivalent:


• Same number ofcovalent bonds

• The negative charge

is on O (an

electronegative

atom)

Resonance

hybrid is

‐

‐

Lesson 4|59


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Lesson 4|60


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Major:



Minor:



• But with Charge

separation and• The positive charge

is on O

Lesson 4|61


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Chemical Bonding –

The complete Picture

Lesson 4|62


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A. Draw resonance contributors for the followingspecies

B. Rank the resonance contributors in order of

decreasing contribution to the hybridC. Draw the resonance hybrid

D. Describe the chemical bonding in the molecule.

Lesson 4|63


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A. Draw resonance contributors for the followingspecies

Answer: Follow all the rules for writing valid resonance

structures. Use curved arrows to indicate the electronflow.

Since you are just getting started, draw all H atoms and

lone pairs to avoid any error in the formal charges:



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B. Rank the resonance contributors in order of decreasing

contribution to the hybrid

A is the most stable because it has complete octet and

does not have charge separation.

Both B and C have separated charges. But + and – in B

are very close to each other. Large energy is needed to

separate these charges. Therefore, B is less stable than

C. This structure is almost insignificant.

Decreasing Order of contribution: A > C > B



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D. Finally, describe the chemical bonding in the molecule:

1. Partial double bonds from O‐1 to C‐4.

2. O‐1, C‐2, C‐3 and C‐4 are all sp2

hybridized.3. Pi are electrons delocalized from O‐1 to C‐4

4. From the formal charges, C‐4 has a partial + charge and O‐1

has a partial – charge5. Single bonds; Sigma bonds from C‐4 to C‐6, and C‐2 to C‐7

6. All C atoms from C‐5 to C‐7 are sp3 hybridized.

7. The single bonds are longer than the partial double bonds8. The molecule is planar (~120o bond angle) from O‐1 to C‐4.

9. The molecule is tetrahedral (~about 109.5o) at C‐5, C‐6 and

C‐7. Lesson 4|67

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