Lone pairs, drawing

First, one of the n bonds must be broken to give the diradical and then, an electron is added to one of the nitrogen atoms. This nitrogen now has two lone pairs. Draw the structure of this anion. 

Challenge A nitrogen trifluoride molecule contains numerous lone pairs. Draw its Lewis structure. 

The geometry of NH3 is trigonal pyramidal (four electron groups, three bonding groups, one lone pair). Draw a three-dimensional picture of NH3 and imagine each bond as a rope that is being pulled. The pulls of the ropes do not cancel and the molecule is polar. 

Cyanide is a good anion, and the cation is stabilised by a lone pair of electrons on oxygen. Draw the disconnection again using the lone pair. 

Because the acylated product has a delocahsed lone pair and is less reactive than PhNHi. You may have been surprised that LiAlHi reduction completely removes the carbonyl oxygen atom. To help explain this, please draw the likely intermediate. 

Compare and contrast the electrostatic potential map of a typical detergent with that of a typical soap (stearate). Which part of each molecule will be most water soluble (hydrophilic) Draw a Lewis structure that describes each molecule s water-soluble group (make sure you indicate all necessary formal charges and lone pairs). Which part(s) of each molecule will be most grease soluble (lipophilic) What kinds of atoms and bonds are found in these groups ... 

Draw structures for the following molecules, showing lone pairs ... 

Just as there is a C=0 bond to the left of the lone pair, there is a second C=0 bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion. 

The following molecular models are representations of (a) adenine and (b) cytosine, constituents of DNA. Indicate the positions of multiple bonds and lone pairs for both, and draw skeletal structures (gray = C. red = O, blue = N, ivory = IT). 

First, look at the reaction and identify the bonding changes that have occurred. In this case, a C—Br bond has broken and a C-C bond has formed. The formation of the C-C bond involves donation of an electron pair from the nucleophilic carbon atom of the reactant on the left to the electrophilic carbon atom ol CH Br, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of CH3Br. At the same time the C—C bond forms, the C-Br bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the C-Br bond to Br. The bromine is now a stable Br- ion. 

Problem 24.20 Draw an orbital picture of thiazole. Assume that both the nitrogen and sulfur atoms are sp2-hybridized, and show the orbitals that the lone pairs occupy. 

Draw the most important Lewis structure for each of the following ring molecules (which have been drawn without showing the locations of the double bonds). Show all lone pairs and nonzero formal charges. If there are equivalent resonance... 

STRATEGY For the electron arrangement, draw the Fewis structure and then use the VSEPR model to decide how the bonding pairs and lone pairs are arranged around the central (nitrogen) atom (consult Fig. 3.2 if necessary). Identify the molecular shape from the layout of atoms, as in Fig. 3.1. 

Consider the bonding in CH2=CHCHO. (a) Draw the most important Lewis structure. Include all nonzero formal charges, (b) Identify the composition of the bonds and the hybridization of each lone pair—for example, by writing o(H ls,C2s/ 2). 

The monomer of the conducting polymer polyaniline is the compound aniline (aminobenzene). (a) Draw the structural formula of the aniline monomer. What is the hybridization of the N atom in (b) aniline (c) polyaniline (d) Indicate the locations of lone pairs, if any, in polyaniline, (e) Do the N atoms help to carry the current Explain your reasoning. See Box 19.1. 

We have the lone pair, because we can t use each of the five electrons to form a bond. Carbon can never have five bonds. Why not Electrons exist in regions of space called orbitals. These orbitals can overlap with orbitals from other atoms to form bonds, or the orbitals can contain two electrons (which is called a lone pair). Carbon has only four orbitals, so there is no way it could possibly form five bonds—it does not have five orbitals to use to form those bonds. This is why a carbon atom with a negative charge will have a lone pair (if you look at the drawing above, you will count four orbitals—one for the lone pair and then three more for the bonds). 

Now that we have established that formal charges must always be drawn and that lone pairs are usually not drawn, we need to get practice in how to see the lone pairs when they are not drawn. This is not much different from training yourself to see all the hydrogen atoms in a bond-line drawing even though they are not drawn. If you know how to count, then you should be able to figure out how many lone pairs are on an atom where the lone pairs are not drawn. 

EXERCISE 1.46 Draw aU lone pairs in the following structure ... 

PROBLEMS Review the common situations above, and then come back to these problems. For each of the following structures, draw all lone pairs. Try to recognize how many lone pairs there are without having to count. Then count to see if you were right. 

MORE PROBLEMS For each of the following structures, draw all lone pairs. 

The examples above are clear, bnt with bond-line drawings, it can be more difficult to see the violation because we cannot see the hydrogen atoms (and, very often, we cannot see the lone pairs either for now, we will continne to draw lone pairs to ease you into it). You have to train yonrself to see the hydrogen atoms and to recognize when yon are exceeding an octet ... 

Now that we know how to identify good arrows and bad arrows, we need to get some practice drawing arrows. We know that the tail of an arrow must come either from a bond or a lone pair, and that the head of an arrow must go to form a bond or a lone pair. If we are given two resonance structures and are asked to show the arrow(s) that get us from one resonance structure to the other, it makes sense that we need to look for any bonds or lone pairs that are appearing or disappearing when going from one structure to another. For example, consider the following resonance structures ... 

Answer Let s analyze the difference between these two drawings. We begin by looking for any double bonds or lone pairs that are disappearing. We see that oxygen is losing a lone pair, and the C=C on the bottom is also disappearing. This should automatically teU us that we need two arrows. To lose a lone pair and a double bond, we will need two tails. 

So we know we need two arrows. Let s start at the top. We lose a lone pair from the oxygen atom and form a C=0. Let s draw that arrow ... 

Notice that if we stopped here, we would be violating the second commandment. The central carbon atom is getting five bonds. To avoid this problem, we must immediately draw the second arrow. The C=C disappears (which solves our octet problem) and becomes a lone pair on carbon. 

In this example, we can see that one of the lone pairs on oxygen is coming down to form a bond, and the C=C double bond is being pushed to form a lone pair on a carbon atom. When both arrows are pushed at the same time, we are not violating either of the two commandments. So, let s focus on how to draw the resonance structure. Since we know what arrows mean, it is easy to follow the arrows. We just get rid of one lone pair on oxygen, place a double bond between carbon and oxygen, get rid of the carbon-carbon double bond, and place a lone pair on carbon ... 

There is one subtle point that must be mentioned. We said that you do not need to draw lone pairs—you only need to draw formal charges. There will be times when you will see arrows being pushed on structures that do not have the lone pairs drawn. When this happens, you might see an arrow coming from a negative charge ... 

The drawing on the left is the common way this is drawn. Just don t forget that the electrons are really coming from a lone pair (as seen in the drawing on the right). 

PROBLEMS For each of the structures below, draw the resonance structure that you get when you push the arrows shown. Be sure to include formal charges. (Hint In some cases the lone pairs are drawn and in other cases they are not drawn. Be sure to take them into account even if they are not drawn—you need to train yourself to see lone pairs when they are not drawn.)... 

We ask if there are any lone pairs that we can move to form a pi bond. So we draw an arrow that brings the lone pair down to form a pi bond ... 

---