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• Notice that the molecular formula would be inadequate to distinguish between propanol and isopropanol.
• Practice converting from one type of representation to another with Skillbuilder 2.1
• Each corner or endpoint represents a carbon atom. There aresix carbon atoms in hexane and four in 2-butene and 2-butyne:
• The zigzag format is fairly accurate in representing the bondangles for sp3 and sp2 hybridized atoms– Linear geometry is shown for sp-hybridized atoms?
• Carbon atoms are not labeled, but a carbon is assumed to belocated at every corner or endpoint on the zigzag.
• Practice identifying the location of carbons and hydrogens in a skeletal structure. In your mind’s eye you should see the hydrogens and where they are located.
• When certain atoms are bonded together in specific arrangements, they undergo specific chemical reactions
• These characteristic groups of atoms/bonds are called functional groups. Every chemistry student needs to learn the term for each functional group (Table 2.1)
• Formal charge (section 1.4) affects the stability and reactivity of molecules, so you must be able to identify formal charges in bond-line representations
• The following structure is incomplete, because it doesn’t have formal charges correctly indicated.
• Formal charges must alwaysdrawn. Fix the structure by adding them
• Formal charge must be drawn, always, but drawing lone pairs is optional and they are often not included.
• By knowing the formal charge, the presence (or absence) of lone pairs is implied
• Oxygen is in 6th group of PTE, needs 6 valence electrons to be neutral, so an oxygen anion has 7. The negatively charged oxygen has one bond, so it must have 6 unshared electrons to total 7.
• The shape of a compound governs how it interacts biologically, and so it is important to accurately depict and interpret 3-D in bond-line structures.
… so the pi-bond is also delocalizedbetween carbons 2 and 3
• If all of the carbons have unhybridized p orbitals, then all 3 of them overlap side-on-side, and
• All three overlapping p orbitals allow the electrons to move throughout the overlapping area, and so we say the molecule has resonance (meaning it has delocalized electrons).
Resonance StabilizationWe developed the concept of resonance using the allyl cation as an example, and we saw that the two π electrons are spread out over the three carbon atoms of the allylic system. This spreading of electrons, called delocalization, is a stabilizing factor. That is, molecules and ions are stabilized by the delocalization of electrons. This stabilization is often referred to as resonance stabilization, and the allyl cation is said to be resonance stabilized. Resonance stabilization plays a major role in the outcome of many reactions, and we will invoke the concept of resonance in almost every chapter of this text-book. The study of organic chemistry therefore requires a thorough mastery of drawing resonance structures, and the following sections are designed to foster the necessary skills.
2.8 Curved ArrowsIn this section, we will focus on curved arrows, which are the tools necessary to draw resonance structures properly. Every curved arrow has a tail and head:
Tail Head
Curved arrows used for drawing resonance structures do not represent the motion of electrons—they are simply tools that allow us to draw resonance structures with ease. These tools treat the electrons as if they were moving, even though the electrons are actually not moving at all. In Chapter 3, we will encounter curved arrows that actually do represent the flow of electrons. For now, keep in mind that all curved arrows in this chapter are just tools and do not represent a flow of electrons.
It is essential that the tail and head of every arrow be drawn in precisely the proper location. The tail shows where the electrons are coming from, and the head shows where the electrons are going (remember, the electrons aren’t really going anywhere, but we treat them as if they were for the purpose of drawing the resonance structures). We will soon learn patterns for drawing proper curved arrows. But, first, we must learn where not to draw curved arrows. There are two rules that must be followed when drawing curved arrows for resonance structures:
1. Avoid breaking a single bond. 2. Never exceed an octet for second-row elements.
Let’s explore each of these rules:
1. Avoid breaking a single bond when drawing resonance structures. By definition, resonance struc-tures must have all the same atoms connected in the same order. Breaking a single bond would change this—hence the first rule:
+
Don’t break a single bond
⊕⊕
There are very few exceptions to this rule, and we will only violate it two times in this textbook (both in Chapter 8). Each time, we will explain why it is permissible in that case. In all other cases, the tail of an arrow should never be placed on a single bond.
2. Never exceed an octet for second-row elements. Elements in the second row (C, N, O, F) have only four orbitals in their valence shell. Each orbital can either form a bond or hold a lone pair. Therefore, for second-row elements the total of the number of bonds plus the number of lone pairs can never be more than four. They can never have five or six bonds; the most is four. Similarly, they can never have four bonds and a lone pair, because this would also require five orbitals. For the same reason, they can never have three bonds and two lone pairs. Let’s see some examples of curved arrows that violate this second rule. In each of these drawings, the central atom cannot form another bond because it does not have a fifth orbital that can be used.
• When multiple resonance structures can be drawn, we know a blend of all of them, the hybrid structure, is the actual structure of the compound
• Typically, not all of the resonance structures will contribute equally to the hybrid.
• The following rules, listed in order of importance, allow us to determine the most significant resonance form(s) for a given compound (i.e. the MAJOR resonance form)
It is fairly common to encounter a carbon atom with a positive charge, even though it lacks an octet (as seen in the first resonance form above). In contrast, oxygen is much more electronegative than carbon, so you should never draw a resonance form in which an oxygen atom lacks an octet. In the example below, the second resonance form is a minor contributor because the carbon atom lacks an octet, but the third resonance form shown is insignificant because the positively charged oxygen atom lacks an octet. Avoid drawing insignificant resonance forms.
Major contributor Minor contributor Insignificant
O
⊕
O
O
O⊝
O⊝
O⊕
2. The structure with fewer formal charges is more significant. Any resonance form that con-tains an atom bearing a +2 or −2 charge is highly unlikely. In the example below, the first resonance form is the best Lewis structure and the largest contributor to the hybrid because it has filled octets and no formal charges. The second resonance form is still a major con-tributor since it has filled octets, but it is less significant than the first because it has formal charges. The third resonance form is a minor contributor because it has a carbon atom that lacks an octet
Minor contributor
H C
NH2
NH⊝
⊕
Major contributor
H C
NH2
NH⊝
⊕
H C
NH2
NH
Largest contributor
In cases where there is an overall net charge, as seen in the example below, the creation of new charges is not favorable. For such charged compounds, the goal in drawing resonance forms is to delocalize the charge – relocate it to as many different positions as possible.
CH2CCH3
O
CH2CCH3
O
Delocalized negative charge
Insignificantresonance
⊝
⊝CH2CCH3
O⊝
⊝
⊕
3. Other things being equal, a structure with a negative charge on the more electronegative element will be more significant. To illustrate this, let’s revisit the previous example, in which there are two significant resonance forms. The first resonance form has a negative charge on oxygen, while the second resonance form has a negative charge on carbon. Since oxygen is more electronegative than carbon, the first resonance form is the major contributor:
Minor contributor
CH2CCH3
O⊝
Major contributor
CH2CCH3
O⊝
Similarly, a positive charge will be more stable on the less electronegative element. In the follow-ing example, both resonance forms have filled octets, so we consider the location of the positive charge. Nitrogen is less electronegative than oxygen, so the resonance form with N+ is the major contributor:
It is fairly common to encounter a carbon atom with a positive charge, even though it lacks an octet (as seen in the first resonance form above). In contrast, oxygen is much more electronegative than carbon, so you should never draw a resonance form in which an oxygen atom lacks an octet. In the example below, the second resonance form is a minor contributor because the carbon atom lacks an octet, but the third resonance form shown is insignificant because the positively charged oxygen atom lacks an octet. Avoid drawing insignificant resonance forms.
Major contributor Minor contributor Insignificant
O
⊕
O
O
O⊝
O⊝
O⊕
2. The structure with fewer formal charges is more significant. Any resonance form that con-tains an atom bearing a +2 or −2 charge is highly unlikely. In the example below, the first resonance form is the best Lewis structure and the largest contributor to the hybrid because it has filled octets and no formal charges. The second resonance form is still a major con-tributor since it has filled octets, but it is less significant than the first because it has formal charges. The third resonance form is a minor contributor because it has a carbon atom that lacks an octet
Minor contributor
H C
NH2
NH⊝
⊕
Major contributor
H C
NH2
NH⊝
⊕
H C
NH2
NH
Largest contributor
In cases where there is an overall net charge, as seen in the example below, the creation of new charges is not favorable. For such charged compounds, the goal in drawing resonance forms is to delocalize the charge – relocate it to as many different positions as possible.
CH2CCH3
O
CH2CCH3
O
Delocalized negative charge
Insignificantresonance
⊝
⊝CH2CCH3
O⊝
⊝
⊕
3. Other things being equal, a structure with a negative charge on the more electronegative element will be more significant. To illustrate this, let’s revisit the previous example, in which there are two significant resonance forms. The first resonance form has a negative charge on oxygen, while the second resonance form has a negative charge on carbon. Since oxygen is more electronegative than carbon, the first resonance form is the major contributor:
Minor contributor
CH2CCH3
O⊝
Major contributor
CH2CCH3
O⊝
Similarly, a positive charge will be more stable on the less electronegative element. In the follow-ing example, both resonance forms have filled octets, so we consider the location of the positive charge. Nitrogen is less electronegative than oxygen, so the resonance form with N+ is the major contributor:
It is fairly common to encounter a carbon atom with a positive charge, even though it lacks an octet (as seen in the first resonance form above). In contrast, oxygen is much more electronegative than carbon, so you should never draw a resonance form in which an oxygen atom lacks an octet. In the example below, the second resonance form is a minor contributor because the carbon atom lacks an octet, but the third resonance form shown is insignificant because the positively charged oxygen atom lacks an octet. Avoid drawing insignificant resonance forms.
Major contributor Minor contributor Insignificant
O
⊕
O
O
O⊝
O⊝
O⊕
2. The structure with fewer formal charges is more significant. Any resonance form that con-tains an atom bearing a +2 or −2 charge is highly unlikely. In the example below, the first resonance form is the best Lewis structure and the largest contributor to the hybrid because it has filled octets and no formal charges. The second resonance form is still a major con-tributor since it has filled octets, but it is less significant than the first because it has formal charges. The third resonance form is a minor contributor because it has a carbon atom that lacks an octet
Minor contributor
H C
NH2
NH⊝
⊕
Major contributor
H C
NH2
NH⊝
⊕
H C
NH2
NH
Largest contributor
In cases where there is an overall net charge, as seen in the example below, the creation of new charges is not favorable. For such charged compounds, the goal in drawing resonance forms is to delocalize the charge – relocate it to as many different positions as possible.
CH2CCH3
O
CH2CCH3
O
Delocalized negative charge
Insignificantresonance
⊝
⊝CH2CCH3
O⊝
⊝
⊕
3. Other things being equal, a structure with a negative charge on the more electronegative element will be more significant. To illustrate this, let’s revisit the previous example, in which there are two significant resonance forms. The first resonance form has a negative charge on oxygen, while the second resonance form has a negative charge on carbon. Since oxygen is more electronegative than carbon, the first resonance form is the major contributor:
Minor contributor
CH2CCH3
O⊝
Major contributor
CH2CCH3
O⊝
Similarly, a positive charge will be more stable on the less electronegative element. In the follow-ing example, both resonance forms have filled octets, so we consider the location of the positive charge. Nitrogen is less electronegative than oxygen, so the resonance form with N+ is the major contributor: