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Resonance 1. Lone Pair Next to Empty 2P Orbital 1 Lecture 5 Resonance 1. Lone pair next to empty 2p orbital electron pair acceptors - lack of electrons 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 pi bond 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 chemistry this is a carbocation 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 formal charge (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 molecules. 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, oxygen 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 nitrogen 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 hydrogen 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 nd R 2 structure is better, O 1st structure is better, N R R 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 nd nd 2 structure is better, no charge R 2 structure is better, N 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 nitronium ion 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 Lewis structure 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.
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