The methods reviewed above for drawing Lewis structures and determining formal charges on atoms are an essential starting point for a novice organic chemist, and work quite will when dealing with small, simple structures. But as you can imagine, these methods become unreasonably tedious and time-consuming when you start dealing with larger structures. It would be unrealistic, for example, to ask you to draw the Lewis structure below (of one of the four nucleoside building blocks that make up DNA) and determine all formal charges by adding up, on an atom-by-atom basis, the valence electrons.
And yet, as organic chemists, and especially as organic chemists dealing with biological molecules, you will be expected soon to draw the structure of large molecules such as this on a regular basis. Clearly, you need to develop the ability to quickly and efficiently draw large structures and determine formal charges. Fortunately, this ability is not terribly hard to come by - all it takes is a few shortcuts and some practice at recognizing common bonding patterns.
Let’s start with carbon, the most important element for organic chemists. Carbon is said to be tetravalent, meaning that it tends to form four bonds. If you look at the simple structures of methane, methanol, ethane, ethene, and ethyne in the figures from the previous section, you should quickly recognize that in each molecule, the carbon atom has four bonds, and a formal charge of zero.
This is a pattern that holds throughout most of the organic molecules we will see, but there are also exceptions.<style type="text/css"></style> In carbon dioxide, the carbon atom has double bonds to oxygen on both sides (O=C=O). Later on in this chapter and throughout this book we will see examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom bears a positive or negative formal charge, respectively. If a carbon has only three bonds and an unfilled valence shell (in other words, if it does not fulfill the octet rule), it will have a positive formal charge.
If, on the other hand, it has three bonds plus a lone pair of electrons, it will have a formal charge of -1. Another possibility is a carbon with three bonds and a single, unpaired (free radical) electron: in this case, the carbon has a formal charge of zero. (One last possibility is a highly reactive species called a ‘carbene’, in which a carbon has two bonds and one lone pair of electrons, giving it a formal charge of zero. You may encounter carbenes in more advanced chemistry courses, but they will not be discussed any further in this book).
You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection.
The pattern for hydrogens is easy: hydrogen atoms have only one bond, and no formal charge. The exceptions to this rule are the proton, H+, and the hydride ion, H-, which is a proton plus two electrons. Because we are concentrating in this book on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge.
Let us next turn to oxygen atoms. Typically, you will see an oxygen bonding in three ways, all of which fulfill the octet rule.
If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of-1. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of +1.
When we get to our discussion of free radical chemistry in chapter 17, we will see other possibilities, such as where an oxygen atom has one bond, one lone pair, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for virtually everything we see until chapter 17.
Nitrogen has two major bonding patterns, both of which fulfill the octet rule:
If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of +1. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs.
Two third row elements are commonly found in biological organic molecules: sulfur and phosphorus. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO4 3-), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that elements in the third row of the periodic table have d orbitals in their valence shell as well as s and p orbitals, and thus are not bound by the octet rule.
Finally, the halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of +1.
These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly.
Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear.
If you look ahead in this book at the way organic compounds are drawn, you will see that the figures are somewhat different from the Lewis structures we have been using in this section. In order to make structural drawings clearer and less crowded, organic chemists use an abbreviated drawing convention called ‘line structures’. The convention is quite simple and makes it easier to draw molecules, but line structures do take a little bit of getting used to. Carbon atoms are depicted not by a capital C, but by a ‘corner’ between two bonds, or a free end of a bond. Straight-chain molecules are usually drawn out in a 'zig-zig' shape. Hydrogens attached to carbons are generally not shown: rather, like lone pairs, they are simply implied (unless a positive formal charge is shown, all carbons are assumed to have a full octet of valence electrons). Hydrogens bonded to heteroatoms are shown, (to an organic chemist, a heteroatom is a nitrogen, oxygen, sulfur, halogen, or anything other than carbon or hydrogen) but are usually without showing the bond. The following examples illustrate the idea.
As you can see, the 'pared down' line structure for a molecule makes it much easier to see the basic structure. For larger, more complex biological molecules, it becomes impractical to use full Lewis structures.
Sometimes, one or more carbon atoms in a line structure will be depicted with a capital C, if doing so makes an explanation easier to follow. If you draw out a carbon with a C, however, you also need to draw in the hydrogens for that carbon.
Exercise 1.9: Give the molecular formula, (including overall charge, if any) of the molecules/ions whose structures are shown below, and add any non-zero formal charges at the appropriate atoms. All atoms (other than hydrogens) have a complete octet of valence electrons.
a) Draw a line structure for the DNA base 2-deoxycytidine (figure in section 1.3A)
b) Draw a line structure for the amino acid histidine.
Now that we have reviewed how to draw Lewis structures and learned the line structure shortcut, it is a good time to learn about the concept of constitutional isomers. Imagine if you were asked to draw a structure (Lewis or line) for a compound with the molecular formula C4H10. This would not be difficult - you could simply draw:
But when you compared your answer with that of a classmate, she may have drawn this structure:
Who is correct? The answer, of course, is that both of you are. A molecular formula only tells you how many atoms of each element are present in the compound, not what the actual atom-to-atom connectivity is. There are often many different possible structures for one molecular formula. Compounds that have the same molecular formula but different connectivity are called constitutional isomers (sometimes the term ‘structural isomer’ is also used). The Greek term ‘iso’ means ‘same’.
Fructose and glucose, two kinds of sugar molecules, are constitutional isomers with the molecular formula C6H12O6.
Later, we will see other types of isomers that have the same molecular formula and the same connectivity, but are different in other respects.
Exercise 1.11 : Draw all of the possible constitutional isomers with the given molecular formula.
The Index of Hydrogen Deficiency (IDH) concept is a helpful tool that we can use when trying to come up with possible Lewis structures when we know a molecular formula. The IHD is really just a simple formula that allows us to calculate how many units of unsaturation - in other words, how many multiple bonds or ring structures - exist in a molecule with a given molecular formula.
Here's how it works. To start off with, we will consider only simple hydrocarbons - molecules that contain only carbons and hydrogens. The IHD rule tells us that a saturated, straight-chain alkane will have the molecular formula CnHn+2, (n is the number of carbons in the molecule). The following alkanes are shown as examples.
The IHD for hydrocarbons, then, is defined as:
By this definition, the IHD value for saturated, straight-chain alkanes is of course equal to zero.
If we put a double bond or a ring structure into a hydrocarbon, the additional carbon-carbon bond means that there must be two fewer hydrogens in the molecule. Because of this, the IHD value is equal to 1.
If we put in a double bond and a ring, or a triple bond, then there are two extra carbon-carbon bonds, four fewer hydrogens, and IHD = 2.
If we are given a molecular formula and asked to come up with one or more possible Lewis structures that fit, figuring the IHD is a very good place to start. For example, if we are asked to draw a possible Lewis structure for a molecule with the molecular formula C7H8, we would first calculate that IHD = 4.
We now know that our molecule must have a combination of five multiple bonds or rings. A likely candidate for C7H9 would be a compound called toluene (a common solvent in the organic chemistry lab). There are, of course, other constitutional isomers with the same molecular formula.
We can easily modify the IHD formula to account the presence of oxygen, nitrogen, halogens and ionic charges:
Notice that we may ignore the number of oxygen atoms.
For example, given the molecular formula C4H9N, we calculate IHD = 1, so one possible structure is A below. Likewise, given the formula C7H8O2Cl- , we calculate IHD = 3 , for which one possible structure is B below.
Because sulfur and phosphorus are both in the third row of the periodic table and thus able to form more than four bonds, the IHD calculation can get more tricky when these elements are involved. In these cases, we will need to rely on our familiarity with the common organic functional groups (see next section) and use a little more trial and error when suggesting structures. Nevertheless, figuring the IDH will be particularly helpful to us in Chapter 5, when we learn how to use a technique called nuclear magnetic resonance to figure out the structures of unknown organic compounds.
Exercise 1.12 : Determine the IHD for
Online lecture from Kahn Academy:
Representing structures of organic molecules
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