A4-sp2Hybridization Definition In sphybridization,the s orbital is mixed with two of the 2p orbitals (e.g.2p,an 2p)to give three sp hybridized orbitals of equal energy.The remaining 2p,orbita is unaffected.The energy of each hybridized orbital is greater than the original s orbital but less than the original p orbitals.The remaining 2p orbital (in this case the 2p,orbital)remains at its original energy level (Fig.1). Energy 22py22 Original atomic orbitals sp2hybridized orbitals Fig.1.spHybridization. Electronic For carbon,there are four valence and the electrons to fit into the three hybridized sp of the hybrid ged ema ital.The firs st th ch that they are all half. filled.This le on I to een pamring or pl ng it into the vacant 2p,orb tals of equ ng to an o f h energy.Ho o fill up ort energy before mo if the en between orbit as higher energy 2p,orbital ting in thi Geometry The 2p,orbital has the usual dumbbell shape.Each of the sphybridized orbitals has a dofor med dumbbell sha similar to an sp'hybridized orbital.Howe the difference between the sizes of the major and minor lobes is larger for the sp hybridized orbital The hybridized orbitals and the2p,orbital occupy spaces as far apart from each ssible The lobes of th P of thex and z axes (Fig 2a).The thre ajor lobes she cupy ng sp ch that the 4 all placed in ng tor 1 are now ready to the ding of an sp2hybr idiz
A4 – sp2 Hybridization 9 Definition In sp2 hybridization, the s orbital is mixed with two of the 2p orbitals (e.g. 2px and 2pz) to give three sp2 hybridized orbitals of equal energy. The remaining 2py orbital is unaffected. The energy of each hybridized orbital is greater than the original s orbital but less than the original p orbitals. The remaining 2p orbital (in this case the 2py orbital) remains at its original energy level (Fig. 1). Electronic For carbon, there are four valence electrons to fit into the three hybridized sp2 configuration orbitals and the remaining 2p orbital. The first three electrons are fitted into each of the hybridized orbitals according to Hund’s rule such that they are all half- filled. This leaves one electron still to place. There is a choice between pairing it up in a half-filled sp2 orbital or placing it into the vacant 2py orbital. The usual principle is to fill up orbitals of equal energy before moving to an orbital of higher energy. However, if the energy difference between orbitals is small (as here) it is easier for the electron to fit into the higher energy 2py orbital resulting in three half-filled sp2 orbitals and one half-filled p orbital (Fig. 1). Four bonds are possible. Geometry The 2py orbital has the usual dumbbell shape. Each of the sp2 hybridized orbitals has a deformed dumbbell shape similar to an sp3 hybridized orbital. However, the difference between the sizes of the major and minor lobes is larger for the sp2 hybridized orbital. The hybridized orbitals and the 2py orbital occupy spaces as far apart from each other as possible. The lobes of the 2py orbital occupy the space above and below the plane of the x and z axes (Fig. 2a). The three sp2 orbitals (major lobes shown only) will then occupy the remaining space such that they are as far apart from the 2py orbital and from each other as possible. As a result, they are all placed in the x–z plane pointing toward the corner of a triangle (trigonal planar shape; Fig. 2b). The angle between each of these lobes is 120. We are now ready to look at the bonding of an sp2 hybridized carbon. sp2 2px 2py 2s 2pz Original atomic orbitals 2py sp2 hybridized orbitals Energy Fig. 1. sp2 Hybridization. x y z z b) 120o a) y x Fig. 2. (a) Geometry of the 2py orbital; (b) geometry of the 2 py orbital and the sp2 hybridized orbitals.
o Section A-Structure and bonding results in three half-filled sphybridized orbitals which form The us ne,for example ethe (H.C=CH far as the Hn h l to form a strong o b rong o bon n the two carbo toms of ethene due to the overlap of sphybridized orbitals from each carbon (Fig.3). 00%→&0D Fig.3.(a)Formation of a C-Habond;(formation of a C-Cabond. The full o bonding diagram for ethene is shown in Fig.4a and can be simplified as in Fig.4b.Eth 1 nar.We ha n he plains the t arbons but ha 11 ed why e the pre nt i na ation could nd th (Fig.5) efor be fu this planar shape.This bond involves H.UC- CL-H 可H Fig.4.(a)Bonding diagram for ethene;(b)simple representation ofbonds for ethene 4Hc-c心4 H Fig.5.Bond rotation around abond. each carbon which ov verlap side nd)with the 6)This nd nd the C-C bor the bond ould ha e to be br tation..Aπbond is we on,res ulting in ove ap.The presen ins wh ore reactive than al nes,since a i bond is more easily t en and is more
10 Section A – Structure and bonding Alkenes sp2 Hybridization results in three half-filled sp2 hybridized orbitals which form a trigonal planar shape. The use of these three orbitals in bonding explains the shape of an alkene, for example ethene (H2CCH2). As far as the C–H bonds are concerned, the hydrogen atom uses a half-filled 1s orbital to form a strong σ bond with a half filled sp2 orbital from carbon (Fig. 3a). A strong σ bond is also possible between the two carbon atoms of ethene due to the overlap of sp2 hybridized orbitals from each carbon (Fig. 3b). The full σ bonding diagram for ethene is shown in Fig. 4a and can be simplified as shown in Fig. 4b. Ethene is a flat, rigid molecule where each carbon is trigonal planar. We have seen how sp2 hybridization explains the trigonal planar carbons but we have not explained why the molecule is rigid and planar. If the σ bonds were the only bonds present in ethene, the molecule would not remain planar since rotation could occur round the C–C σ bond (Fig. 5). Therefore, there must be further bonding which ‘locks’ the alkene into this planar shape. This bond involves the remaining half-filled 2py orbitals on each carbon which overlap side-on to produce a pi (p) bond), with one lobe above and one lobe below the plane of the molecule (Fig. 6). This π bond prevents rotation round the C–C bond since the π bond would have to be broken to allow rotation. A π bond is weaker than a σ bond since the 2py orbitals overlap side-on, resulting in a weaker overlap. The presence of a π bond also explains why alkenes are more reactive than alkanes, since a π bond is more easily broken and is more likely to take part in reactions. H sp2 1s C + C H Fig. 3. (a) Formation of a C–H σ bond; (b) formation of a C–C σ bond. + C C sp2 C sp2 C Fig. 4. (a) σ Bonding diagram for ethene; (b) simple representation of σ bonds for ethene. C C H H H H C C H H H H Bond rotation Fig. 5. Bond rotation around a σ bond. C C H H H H H C C H H H
A4-sp2Hybridization Fig.6.Formation of a bond. Carbonyl groups The same theory explains the bonding within a carbonyl group(C=O)where both the carbon and oxygen atoms are sp'hybridized.The following energy level diagram (Fig.7)shows how the valence electrons of oxygen are arranged after sp hybridization.Two of the sp'hybridized orbitals are filled with lone pairs of electrons,which leaves two half-filled orbitals available for bonding.The sp orbital can be used to form a strong o bond,while the 2p,orbital can be used for the weaker bond.Figure 8 shows how the o and n bonds are formed in the carbonyl group and explains why carbonyl groups are planar with the carbon atom having a trigonal planar shape.It also explains the reactivity of carbonyl groups since the bond is weaker than the o bond and is more likely to be Energy 4420.2+2 什H4。 Original atomic orbitals sp2hybridized orbitals Fig.7.Energy level diagram for sphybridized oxygen. a "wC①①→."w sigma bond Fig.8 (a)Formation of the carbonyl o bond:(b)formation of the carbonyl x bond
A4 – sp2 Hybridization 11 Carbonyl groups The same theory explains the bonding within a carbonyl group (CO) where both the carbon and oxygen atoms are sp2 hybridized. The following energy level diagram (Fig. 7) shows how the valence electrons of oxygen are arranged after sp2 hybridization. Two of the sp2 hybridized orbitals are filled with lone pairs of electrons, which leaves two half-filled orbitals available for bonding. The sp2 orbital can be used to form a strong σ bond, while the 2py orbital can be used for the weaker π bond. Figure 8 shows how the σ and π bonds are formed in the carbonyl group and explains why carbonyl groups are planar with the carbon atom having a trigonal planar shape. It also explains the reactivity of carbonyl groups since the π bond is weaker than the σ bond and is more likely to be involved in reactions. Fig. 6. Formation of a π bond. sp2 hybridized orbitals 2px 2py 2pz 2s Energy 2py sp2 Original atomic orbitals Fig. 7. Energy level diagram for sp2 hybridized oxygen. C O H H C O H H sp2 a) sigma bond sp2 Fig. 8. (a) Formation of the carbonyl σ bond; (b) formation of the carbonyl π bond. C O H H 2py 2py b) C O H H H H H H 2py 2py π Bond
Section A-Structure and bonding Aromatic rings All the carbons in an aromatic ring are sp'hybridized which means that each carbon can form three c bonds and one n bond.In Fig.9a,all the single bonds are o while each double bond consists of one c bond and one x bond.However,this is an oversimplification of the aromatic ring.For example,double bonds are shorter than single bonds and if benzene had this exact structure,the ring would be deformed with longer single bonds than double bonds(Fig.9b). (a)Re ring:(b)' In fact,the C-Cbonds in ben e are all th to ,we nee ame lentich ta with all its o b awn such that we a ng into the zene ring.Since al the carbons are s here is a 2p over on ea carbon which can overlap with a 2p,orb ther s of it (Fig.10b).From this,it is clear that each 2p,orbital can overlap w vith its neigh bors right round the ring.This leads to a molecular orbital which involves all the 2p, orbitals where the upper and lower lobes merge to give two dough ut-like lobes above and below the plane of the ring(Fig.11a).The molecular orbital is symme r cal and the six it electrons are said to be delocalized around the aror natic ring since they are not localized between any two particular carbon atoms.The aromatic ring is often represented as shown in Fig.11b to represent this delocalization of the 8oeafmcingmoeou sentation of benzene to illustrate
Aromatic rings All the carbons in an aromatic ring are sp2 hybridized which means that each carbon can form three σ bonds and one π bond. In Fig. 9a, all the single bonds are σ while each double bond consists of one σ bond and one π bond. However, this is an oversimplification of the aromatic ring. For example, double bonds are shorter than single bonds and if benzene had this exact structure, the ring would be deformed with longer single bonds than double bonds (Fig. 9b). In fact, the C–C bonds in benzene are all the same length. In order to understand this, we need to look more closely at the bonding which takes place. Figure 10a shows benzene with all its σ bonds and is drawn such that we are looking into the plane of the benzene ring. Since all the carbons are sp2 hybridized, there is a 2py orbital left over on each carbon which can overlap with a 2py orbital on either side of it (Fig. 10b). From this, it is clear that each 2py orbital can overlap with its neighbors right round the ring. This leads to a molecular orbital which involves all the 2py orbitals where the upper and lower lobes merge to give two doughnut-like lobes above and below the plane of the ring (Fig. 11a). The molecular orbital is symmetrical and the six π electrons are said to be delocalized around the aromatic ring since they are not localized between any two particular carbon atoms. The aromatic ring is often represented as shown in Fig. 11b to represent this delocalization of the π 12 Section A – Structure and bonding H H H H H H b) = a) Fig. 9. (a) Representation of the aromatic ring; (b) ‘deformed’ structure resulting from fixed bonds. H H H H H H a) b) C C C C C = = C a) b) H H H H H H Fig. 11. Bonding molecular orbital for benzene; (b) representation of benzene to illustrate delocalization. Fig. 10. (a) σ Bonding diagram for benzene, (b) π Bonding diagram for benzene. a) b)
A4-sp?Hybridization electrons.Delocalization increases the stability of aromatic rings such that they are less reactive than alkenes(i.e.it requires more energy to disrupt the delocalized system of an aromatic ring than it does to break the isolated bond of an alkene). Conjugated Aromatic rings are not the only structures where delocalization of r electrons can systems take place.Delocalization occurs in conjugated systems where there are alternat- ing single and double bonds (e.g.1,3-butadiene).All four carbons in 1,3-butadiene are sp hybridized and so each of these carbons has a half-filled p orbital which can interact to give two r bonds(Fig.12a).However,a certain amount of overlap is also possible between the p orbitals of the middle two carbon atoms and so the bond connecting the two alkenes has some double bond character (Fig.12b) borne out by the observation that this bond is shorter in length than a typical single bond.This delocalization also results in increased stability.However,it is important to realize that the conjugation in a conjugated alkene is not as great as in the aromatic system.In the latter system,the electrons are completely delocalized round the ring and all the bonds are equal in length.In 1,3-butadiene, the r electrons are not fully delocalized and are more likely to be found in the ter- minal C-C bonds.Although there is a certain amount of character in the middle bond,the latter is more like a single bond than a double bond. Other examples of conjugated systems include a,B-unsaturated ketones and a,B- unsaturated esters (Fig.13).These too have increased stability due to conjugation. HC-CH -c-CH 1.3-Butadiene Fig.12.(a)Bonding in 1,3-butadiene;(b)delocalization in 1,3-butadien 9.9 Fg.13.x B-Uns
A4 – sp2 Hybridization 13 electrons. Delocalization increases the stability of aromatic rings such that they are less reactive than alkenes (i.e. it requires more energy to disrupt the delocalized π system of an aromatic ring than it does to break the isolated π bond of an alkene). Conjugated Aromatic rings are not the only structures where delocalization of π electrons can systems take place. Delocalization occurs in conjugated systems where there are alternating single and double bonds (e.g. 1,3-butadiene). All four carbons in 1,3-butadiene are sp2 hybridized and so each of these carbons has a half-filled p orbital which can interact to give two π bonds (Fig. 12a). However, a certain amount of overlap is also possible between the p orbitals of the middle two carbon atoms and so the bond connecting the two alkenes has some double bond character (Fig. 12b) – borne out by the observation that this bond is shorter in length than a typical single bond. This delocalization also results in increased stability. However, it is important to realize that the conjugation in a conjugated alkene is not as great as in the aromatic system. In the latter system, the π electrons are completely delocalized round the ring and all the bonds are equal in length. In 1,3-butadiene, the π electrons are not fully delocalized and are more likely to be found in the terminal C–C bonds. Although there is a certain amount of π character in the middle bond, the latter is more like a single bond than a double bond. Other examples of conjugated systems include α,β-unsaturated ketones and α,β- unsaturated esters (Fig. 13). These too have increased stability due to conjugation. C C H H H C C H H C C H H H H C C H H H 2py 2py 2py 2py 1,3-Butadiene a) Fig. 12. (a) π Bonding in 1,3-butadiene; (b) delocalization in 1,3-butadiene. C C H H3C H C O CH3 C C H H3C H C O OCH3 a) b) Fig. 13. (a) α,β-Unsaturated ketone; (b) α,β-unsaturated ester. C C H H H C H H C H 2py 2py 2py 2py b)