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Home , Lone pair, Resonance (chemistry)

1 Lecture 5 1. next to empty 2p orbital

pair acceptors - lack of

C C CC

sp2 R+ is more common sp R+ is less common

R+ needs electrons, has to overlap with a. an adjacent 2p lone pair with electrons b. an adjacent

a. an adjacent 2p lone pair with electrons on a neutral atom (+ overall, delocalization of positive charge)

R R X = neutral atom with lone pair

R R C C R X 2D resonance R X C C R X R X

R R R R C R C R CX CX R N R N R 3D resonance R R R R R R R C R C R CX R O CX 3D resonance R O R R R R C better, more bonds, full octets C R F R F

b. an adjacent 2p lone pair with electrons on a negative atom (neutral overall, delocalization of electrons)

R R X =anion with lone pair

R R C C R X 2D resonance R X C C R X R X

R R R R C R C R CX CX R C R C R 3D resonance R R R R R R R C R C R CX R N CX 3D resonance R N R R R R C C better, more bonds, full octets R O R O

2 Lecture 5 Problem 1 – All of the following examples demonstrate delocalization of a lone pair of electrons into an empty 2p orbital. Usually in organic this is a site, but not always. There are many variations. Assume full octets on all nonhydrogen atoms below, unless you see a carbon with three bonds and a positive (this will be a carbocation) or an atom drawn with lone pairs explicitly drawn and a positive formal charge. Add in missing lone pairs and show proper formal charge (a number of examples need negative formal charge added). Use correct resonance arrows (curved and double headed) where appropriate. What is the hybridization of all nonhydrogen atoms? You should be able to draw a 3D picture of any of these .

R a. R R b. R C H C H C H C H R N R O R O R N H H better, more bonds, full octets better, more bonds, full octets 2 C, O are both sp C, N are both sp2

R H R H R H R H

CO CO CN CN

R R R H R H

c. R R R

H C H H C H H C H O N O N O N

H H H better, more bonds, full octets C, N and O are all sp2

d. R R R

C H C H C H O N O N O N

H H H better, more bonds, full octets, no charge C, N and O are all sp2 e. F F F F

H C H H C H H C H H C H O N O N O N O N

H H H H better, more bonds, full octets, less electronegative atom with positive formal charge (N>O>F) C, N, O and F are all sp2

3 Lecture 5 f. g.

R C N H R C N H R C O R C O

better, more bonds, full octets better, more bonds, full octets C and N are both sp C and O are both sp

R CN R CN R CO R CO

h. i. O O

R CN R CN C C R R R R better, more bonds, full better, more bonds, full octets, no formal charge, C and N are both sp octets, no formal charge, C and O are both sp2

j. k. O O H H H N N NN NN NN R R H O H H O H better, more bonds, full H O H octets, no formal charge, 2 C and O are both sp better, more bonds, full octets, less electronegative atom with positive formal charge (N>O) C, N and O are all sp2

l.

H H H

NN NN NN

H O H O H O

better, more bonds, full octets, no formal charge, C, N and O are all sp2

4 Lecture 5 m.

O O O O O O O O O

These 2 resonance structures are equivalent and better, more bonds, full octets, atoms on the ends are partially negative and the oxygen atom in the middle is positive, all O is sp2

n.

O O O

RN+2 RN RN

O O O

These 2 resonance structures are equivalent and better, more bonds, full octets, oxygen atoms on the ends are partially negative and the atom in the middle is positive, all N and O are sp2

Problem 2 – All of the following examples demonstrate delocalization of a lone pair of electrons into a pi bond. The variations are almost limitless. X, Y and Z below can be sp or sp2 hybridized and they can be carbon, nitrogen or oxygen. X and Z can also be fluorine. In this problem the sigma skeleton is sp2-sp2-sp2 for all parts. Assume full octets below on all nonhydrogen atoms. This means you will have to add in lone pairs, if missing. Include proper formal charge in your created resonance structures (some of the structures need additional positive formal charge). Use correct resonance arrows (curved and double headed) where appropriate. What is the hybridization of all nonhydrogen atoms? “R” represents an organic group or a atom. The generalized 3D example provided in this problem that can be modified to fit all of the parts of this problem.

R R No formal charge R R is indicated. X, Y and 2 X Y X Y Z are sp hybridized in these examples, but

R Z R R Z R there are many other possibilities. R R

5 Lecture 5 a. R R b. R R

R C R R C R R C R R C R C C C C N C N C

R R R R R R resonance structures are equivalent 2nd structure is better, N c. R R d. R R

C R C R R C R R C R O C O C C N C N

R 2nd structure is better, O R st R R 1 structure is better, N e. R R f. R R

R C R R C R C R C R N N N N O N O N

R 1st structure is better, no charge R 2nd structure is better, O

g. R R h. R R R C R C R C C O C O R C N O N O R 1st structure is better, O R 1st structure is better, O

R j. i. R N R N R O C O C C C O O O O R R 2nd structure is better, O resonance structures are equivalent

k. l.

N N R N R N O O N O N O O O

nd 2 structure is better, O resonance structures are equivalent

R m. R n.

N O N O O O O O O O O O

resonance structures are equivalent resonance structures are equivalent

R p. o. R R R

R C R R C R R C R R C R O C O C O N O N

nd R 2 structure is better, no charge R 2nd structure is better, no charge

6 Lecture 5 q. R r. R R R

R C R C R R C R R C O O O N O N O O 2nd structure is better, no charge 2nd structure is better, N R R R s. R R t. R

R C R R C R R C R R C R N N O O O O N N

R R R R resonance structures are equivalent resonance structures are equivalent

Problem 3 –The following examples also demonstrate delocalization of a lone pair of electrons into a pi bond, but with a different sigma skeleton (sp2-sp-sp). Again, there are numerous variations. Assume full octets below on all nonhydrogen atoms. This means you will have to add in lone pairs, if missing. Include proper formal charge in your resonance structures (some of the structures need additional positive formal charge). Use correct resonance arrows (curved and double headed) where appropriate. What is the hybridization of all nonhydrogen atoms? A generalized 3D example is provided that can be remade to fit all of the parts of the problem. Can you draw your own 3D structures?

X Y Z X Y Z

a. b. R

C CCR N CCR

R R c. d.

R O CCR C CN

R e. f.

N CN O CN

R

7 Lecture 5 g. h. R

C NN N NN

R R i. j.

R O NN C CNR

R k. l.

N CNR O CNR R m. n. R R N CNR O CNR R

Problem 4 –Azide and the can generate similar looking resonance structures, except for the formal charge. Azide is an anion and nitronium is a cation. First, draw an acceptable 2D for each of these ions. Draw two additional resonance structures for each and state which are the major and minor resonance contributors. Include all of the details of resonance: lone pairs, formal charge and curved and double headed arrows. Use your 2D resonance structures to determine the hybridization of each of the atoms and draw a 3D structure of the best resonance contributor. A generalized 3D skeleton is provided as a guide.

azide = N3 nitronium ion = NO 2 3D skeleton and resonance structures.

X Y Z X Y Z X Y Z

3. Pi bond is donated into an empty 2p orbital (usually a positive carbocation).

Problem 5 – Two additional resonance structures can be drawn for each of the following ? Draw them using proper resonance arrows and formal charge. The first resonance contributor is formed using pattern three resonance and the second resonance contributor is formed using pattern one. Which carbocation of the three (a, b or c) is more stable? Why? 8 Lecture 5 a.

N

CH2 b. O

CH2 c. F

CH2

Problem 6 – Which pi bond is more stabilizing to the carbocation in each part? Draw the additional resonance structure(s) using proper conventions of resonance. Are any of the resonance contributors unacceptable? a. R CH2 N CH2 O CH2

b.

R C C CH N CH 2 2 O CH2

Problem 7 – How many additional resonance structures can be drawn for each of the following carbocations? Draw them using proper resonance arrows and formal charge. Which carbocation is more stable, a or b? Why?

a. b.

H N H

C C

H H

4. Pi bond electrons are donated into an adjacent pi bond.

There can be as few as two adjacent pi bonds, or essentially infinite pi systems, such as graphite. The pi bonds can be C=C of , and aromatics, C=N, CN and many variations of C=O. One pi bond will donate pi electrons and the other pi bond will accept the pi electrons. Sometimes formal charge is created and sometimes formal charge is delocalized. In other examples neutral pi electrons are moved to new positions, but remain neutral. In all of these examples pi electrons are delocalized. We can only look at a few representative examples.

9 Lecture 5 These can be pi donor systems. These can be pi acceptor systems.

R R R R donate R R electrons CC CC many CN CO variations R R R R R R R

donate electrons many R CCR R CCR R CN variations

R R donate C C electrons many N O variations

10 Lecture 5 a. Electrons are delocalized and no formal charge is created. This is the best kind of resonance. All of the pi bonds in the ring have shifted over by one position. The electrons that were taken away with the first arrow are filled back in with the last arrow.

Benzene resonance

2D structures Pi bonds have shifted positions, but no formal charge is created.

3D structures

R R R R

CC CC

R C C R R C C R

C C C C

R R R R

Nonbenzenoid resonance

2D structures Pi bonds have shifted positions, but no formal charge is created.

3D structures

H H H H

C C C C H H H C C H C C C CH

C C C C C H C H H H C C C C

H H H H

11 Lecture 5 Problem 8 – How many atoms are in a plane in each of the two structures above? b. Electrons are delocalized in a neutral, conjugated pi system and formal charge is created. The additional resonance structures are minor resonance contributors, but are often informative about the chemistry of the functional group. Typically, an electronegative atom (nitrogen or oxygen) will be at the end receiving . i. R R R

R C O R C O R C O The pi bond shifts can be shown step wise or both CC CC resonance arrows can be CC used together to go from the first structure to the R R R R R R last structure. ii.

N R N R N R R CCC R CCC R CCC H H H iii.

C N C N C N etc. Five hydrogen atoms are not shown. iv. R R R R R R

R C C R C C R C C

CC R CC R CC R

R R R R R R Symbolically, this looks the same as "i" above. However, there is no electronegative atom to pull electrons to itself and the dipolar minor resonance contributors have much less importance than when an electronegative atom like oxygen or nitrogen is present. We don't usually write these, although they are valid contributors according to our rules. c. Electrons are delocalized in a positively charged, conjugated pi system and the positive formal charge is delocalized. The additional resonance structures can be minor or major resonance contributors, and are often informative about the chemistry of the functional group. Each of the structures could have been formed by protonation of the neutral heteroatom in strong . i. R R R

R C O R C O R C O

CC H CC H CC H

R R R R R R ii. 12 Lecture 5 H H H N R N R N R R CCC R CCC R CCC H H H iii.

C N H C N H C N H etc.

Problem 9 – What is the hybridization of all nonhydrogen atoms in parts b and c above?

Problem 10 – Which carbonyl bond (C=O) is able to get more electron density from the pi bonds of the aromatic ring? The nitrogen lone pair in the second ring does not participate in resonance. Why not? The nitrogen atom off the side of the third ring powerfully donates its electrons. Why? Use 2D and/or 3D resonance structures that explain your reasoning.

O O O

N N

Problem 11 – Write three additional resonance structures for the following cation. Order the resonance contributors from best (=1) to poorest. Where do you begin your ? Look at the charge on the structure. Remember, electrons are negatively charged. What is the hybridization of the nitrogen atom?

R R

N C R

R CC

R O H C D A B 13 Lecture 5 Problem 12 –Draw a 3D structure for each given representation below. Identify the best resonance structure among any other resonance structures. Drawing the 2D possibilities first may help you do this. Show sigma bonds as lines, wedges and dashes and the p orbitals in pi bonds, as well as any orbitals holding lone pairs of electrons. Draw two dots for pi bond electrons and two dots inside a circle for lone pair electrons. Indicate any formal charge present and give the hybridization, bond angles and shape of each nonhydrogen atom. Assume full octets, unless a carbocation is indicated. Include proper resonance arrow conventions. Rank your resonance structures from best (= 1) to poorest. It is helpful for drawing 3D structures to begin your drawing with a pi bond in the plane of the page and to include as many pi bonds and other atoms in the plane of the page or parallel to the page as possible. a. b. H c. H H3C H O C CCC N CO

CCN CH3 O C CH3 H C d. e. H O H N H O H H3C C C If you begin drawing where an arrow is N C C C N C CH3 NC H pointing, you can put more atoms in the H plane of the paper. H3C H

Problem 13 – Poorer resonance structures are sometimes used to reinforce obvious polar effects and indicate typical reactivity of a functional group. This is especially true for carbonyl bonds (C=O). The carbonyl functional feature shows up in , ketones, , esters, , acid chlorides, anhydrides, ureas, , urethanes…and more. Indicate in which polar pi bonds, below, such a resonance structure would be reasonable and show this resonance structure. 2D structures are fine. If not expected, point out why this is the case. Carbon-nitrogen pi bonds can be explained in a similar manner to the carbon-oxygen pi bonds.

R R Additional resonance is possible with the positively C O CO R R charged carbon, if it is C O C O connected to an atom with R R R R a lone pair of electrons or 2D resonance structures another pi bond. 3D resonance structures - carbonyl resonance

a. O b. c. d. e. O O CH2 NH H3C C H C C 3 H3C C H3C C H3C C H OH NH2 H H f, g. i. O h. N

O H3C C C H H3C C N

NH 2