Organic chemistry for dummies pdf download




















Additionally, you get the lowdown on how specific structural elements on a molecule affect its acidity; then you use that knowledge to compare the acidity of different acids. You work with a quantitative scale of acidity, called the pKa scale, which you can use to predict the direction of an acid-base reaction at equilibrium.

This section takes a closer look at these two types and shows you how to solve some basic problems associated with them. Bronsted-Lowry acids and bases Chemists care deeply about the mechanism of a reaction, which is organic-speak for how a reaction occurs. The complete mechanism of a reaction shows all the bond-making and bond-breaking steps in a reaction, including the order in which those steps take place.

You depict the mechanism on paper by using arrows to show the movement of the electrons in a reaction, taking the starting material into the product. Figure shows a schematic of the reaction between a Bronsted-Lowry acid and a Bronsted-Lowry base.

Here, the Bronsted base the proton acceptor pulls the proton off the Bronsted acid the proton donor. Acid-base reactions are the simplest organic reactions, so the mecha- nisms of these reactions are simple — a good place to start practicing your arrow-pushing skills.

In the example in Figure , the lone pair of electrons then becomes the bonding elec- trons in the base-H bond. Two pieces of organic-speak you should remember for Bronsted acids and bases: The deprotonated acid is referred to as the conjugate base, and the protonated base is called the conjugate acid. Chapter 4: Working with Acids and Bases 61 Q. Label the acid and base in the following reaction. Then show the mechanism arrow-pushing of the deprotonation reaction. Designate the conjugate acid and the conjugate base of this reaction.

After HCl gives up its proton, it becomes the chloride ion, the conjugate base. To draw the mechanism, start by moving the lone pair of electrons from the oxygen on water and pushing it to the proton on H-Cl.

This push represents a new bond forming between the oxygen and the hydrogen. The bond between H-Cl is then broken shown by another arrow and those two bonding electrons are then assigned to the chlorine as a lone pair. Then show the mechanism of the deprotona- tion reaction. Hint: Double-bond electrons can act as bases just like lone pairs. A more all-encompassing definition of acids and bases is the Lewis definition, in which acids are defined as molecules or ions that accept pairs of electrons in a reaction and bases are defined as molecules or ions that donate pairs of electrons in a reaction.

The result is the for- mation of a new covalent bond. Figure shows a general form of a Lewis acid and base reaction. Therefore, Bronsted acids and bases are also Lewis acids and bases. Label the Lewis acid and Lewis base in CH3NH2 acts as the Lewis base in this reaction because it donates a pair of the following reaction. Then show the mechanism of the acid-base reaction electrons to the Lewis acid BH3 to make a using arrows. Label the Lewis acid and Lewis base in the 4. Label the Lewis acid and Lewis base in the following reaction.

Then show the mecha- following reaction. Then show the mecha- nism of the acid-base reaction using nism of the acid-base reaction using arrows. Because acid and base reactions are so important in organic chemistry, being able to compare two acid struc- tures so you can say which acid is stronger than another is a really important skill. The secret to comparing the strength of two acids is this: Strong acids have stable conjugate bases. Therefore, the more stable the conjugate base of an acid, the stronger the acid, because an acid is more willing to give up a proton when doing so leads to a stable conjugate base.

Conversely, an acid with an unstable conjugate base is less willing to give up its proton because doing so leads to an unstable conjugate base. The question then becomes this: What features on a molecule stabilize a conjugate base? Because most acids are neutrally charged and upon deprotonation become negatively charged in the conjugate base form, any structural features that help to stabilize this nega- tive charge in the conjugate base leads to a stronger acid.

Contrasting atom electronegativity, size, and hybridization Charges are more stable on some atoms than others. Recall that elec- tronegativity increases as you go up and to the right on the periodic table.

The size of atoms increases as you go down the periodic table. Negative charges prefer to rest on larger atoms because larger atoms allow the negative charge to delocalize over a larger space electron delo- calization is always a stabilizing feature. Of course, you sometimes face a dilemma. In these cases, atom size trumps electronegativity.

Negative charges prefer to be placed in orbitals with more s character because s orbitals are closer to the atom nucleus. This means that a negative charge prefers to rest on an sp-hybridized atom over an sp2 atom and on an sp2 atom over an sp3 atom refer to Chapter 1 to brush up on determining atom hybridization.

Which acid is stronger, HF or HI? This A. HI is the stronger acid. The strength of presents a dilemma, because although negative charges prefer to rest on more an acid depends on the stability of the conjugate base, so the first thing to do is electronegative atoms, they also prefer to to deprotonate these acids and see which rest on bigger atoms.

However, atom size conjugate base is more stable. Fluorine is more elec- stronger acid than HF. Which acid is stronger, HCl or HF? Which acid is stronger, CH4 or NH3? Which acid is stronger, H2S or H2O? Solve It Solve It 9. Which of the following two acids is Which of the following two acids is stronger? A charge becomes more stable the more it can be delocalized over as many atoms as possible so that no one atom has to carry the full charge.

Therefore, nearby electronegative atoms that pull some of the negative charge away from the negatively charged atom and delocalize the charge will stabilize it.

Which of the two molecules shown The molecule with the CF3 group is more acidic because fluorine is a very elec- below is more acidic? Which of the two shown acids is more Which of the two shown acids is more acidic? As a general rule, the more resonance structures a mole- cule has, the more stable the structure. Which of the two shown acids is more First, draw the conjugate base of both acids and see which one is more stable.

The left-most conjugate base in this case O is less stable because this structure has no resonance structures, so the negative OH charge is localized on a single oxygen.

In OH the right-hand structure, on the other hand, the charge can delocalize through A. Therefore, the right- O hand acid is the more acidic because it C has the more stable conjugate base. The bottom line is that the more acidic a molecule is, the lower its pKa value. In general, a reaction equilibrium favors the side with the lower-energy molecules, and because strong acids and bases are high in energy, acid-base reactions favor the side with the weaker acids and bases.

For any acid-base reaction, if you know the pKa of the acid and the pKa of the conjugate acid, you can determine the direction of the equilibrium. The equilibrium lies in the direction of the side that has the weaker acid that is, the acid with the higher pKa value. You can find tables of pKa values in Organic Chemistry I For Dummies Wiley by yours truly or in most any introductory organic chemistry text.

Predict the direction of the equilibrium The equilibrium favors the products. The pKa of the acid and the conjugate acid in the following acid-base reaction. Predict the direction of the equilibrium in Predict the direction of the equilibrium in the following acid-base reaction.

Note: You the following acid-base reaction. Show the mechanism of this reaction by drawing an arrow from a lone pair on the oxygen and pushing it to the hydrogen on HI. Then draw a second arrow from the H-I bond to the iodine to show the breaking of this bond and the reassignment of these electrons as a lone pair on iodine.

However, the reaction works in the same way. Then use a second arrow to show the hydronium ion O-H bond being broken, with the bonding electrons placed onto the oxygen as a new lone pair. Therefore, a molecule with lone pairs of electrons a Lewis base readily adds to the alu- minum. Draw an arrow from an oxygen lone pair and push the arrow to the aluminum to show the formation of an O-Al bond.

Nitrogen and carbon are roughly the same size because both atoms are in the same row of the periodic table, but nitrogen is more electronegative than carbon, so NH2— is more stable than CH3—. Consequently, NH3 is a stronger acid than CH4. Oxygen and nitrogen are essentially the same size because these atoms are in the same row on the periodic table, but oxygen is more electronegative than nitrogen. H H C C C C H H sp sp2 Negative charges also called anions using organic-speak are more stable on sp-hybridized atoms than on sp2 hybridized atoms because sp orbitals have more s character 50 percent s character than sp2 orbitals 33 percent s character ; therefore, the sp orbital places the lone pair closer to the nucleus.

Electronegativity increases as you go up and to the right on the periodic table. Thus, the right-hand molecule with the closer chlorine is more acidic than the left-hand molecule. Keep in mind that the size of the neighboring atoms is unim- portant in terms of affecting the acidity: Size of the atom matters only when the charge is on that particular atom.

Therefore, the left struc- ture is more acidic than the right one. In the conjugate base of the right-hand structure, the charge is localized on a single atom. Therefore, the left structure is more acidic than the right. However, the base of the left structure has only two resonance structures, whereas the base of the right-hand structure has four resonance structures. Typically, the more resonance structures a molecule or ion has, the more stable it is. Therefore, the right structure is more acidic than the left.

However, the left structure has only two resonance structures, whereas the right-hand structure has three. Because the more reso- nance structures a molecule has, the more stable it is, the right structure is more acidic than the left. Therefore, the equilibrium lies in the direction of the products. Therefore, the equilibrium favors the reactants. I n this part, you get down to business by working with the simplest organic structures, those molecules consist- ing of just hydrogen and carbon atoms, better known as hydrocarbons.

These molecules form the bones of organic structures, so this part lays the brickwork that forms the foundation of your knowledge for the rest of the course. Here, you discover how to name hydrocarbons, study their 3-D properties, and get a close look at some introductory reactions with the reactions of alkenes, molecules contain- ing carbon-carbon double bonds. In a similar way, two molecules can have the same connections of atoms but differ from each other by the orientation of those atoms in three-dimensional space.

Such molecules are said to have different stereochemistry, or differing spatial arrangements of atoms. The study of stereochemistry is important for understanding the molecular basis of biology. Stereochemistry has far-reaching consequences because nature often treats two molecules with different stereochemistry quite differently.

For example, a molecule with one stereo- chemistry may smell like mint, while the same molecule with a different stereochemistry smells like cinnamon; the first molecule may treat a disease, while the other is poisonous. So how does stereochemistry manifest itself in molecules? Every carbon that has four differ- ent attachments to it called a chiral center can have two different configurations, much like a hand has two possible configurations, right or left.

But organic chemists have a thing for Latin, so they use the labels R and S, which stand for rectus right and sinister left. The goal of this chapter is to familiarize you with working with molecules of differing stereochemistry.

Part of that involves being able to assign whether a chiral center is of R or S configuration and also to distinguish the relationships between molecules with different stereochemistry, called stereoisomers. Also on the menu is working with a handy tool called a Fischer projec- tion, which aids in quickly comparing the stereochemistry of molecules. Identifying Chiral Centers and Assigning Substituent Priorities Because stereochemistry can be conceptually challenging, I take a gradual approach in this chapter to assigning the configuration of chiral centers as R or S.

Locate a chiral center. A chiral center is a carbon atom with four different attached atoms or groups. Assign the priorities of each of the four groups attached to the chiral center using the Cahn-Ingold-Prelog rules.

Prioritize the groups from 1 to 4 based on the atomic number of the first atom attached to the chiral center. The highest priority 1 goes to the group whose first atom has the highest atomic number; the lowest priority 4 goes to the group whose first atom has the lowest atomic number usually hydrogen. In the event of a tie — that is, two groups attach with the same atom type — continue down the chain of both groups until the tie is broken. Thus, a carbon attached to an oxygen has priority over a carbon attached to another carbon because oxygen has a higher atomic number than carbon.

Thus, a carbon attached to two other carbons has a higher priority than a carbon attached to just one other carbon. However, atom type trumps the number of groups, so a carbon attached to an oxygen has higher priority than a carbon attached to two or three other carbons. Figure shows some examples. For example, you treat a carbon-nitrogen triple bond as though the carbon were single-bonded to three nitrogens, with each of the nitrogens in turn bonded to a carbon. You treat a carbon double-bonded to nitrogen as though the carbon were two single bonds to nitrogen, each of which would be bonded back to the carbon.

The effect is that for identical atom types, triple bonds have higher priority than double bonds, and double bonds have higher priority than single bonds. Figure shows an example. Use a procedure for assigning the R or S designation to that chiral center using the group priorities you assigned in the second step. See the next section for more specifics. Keep in mind that, by convention, bonds coming out of the paper are indicated with a solid wedge, and bonds going back into the paper are indicated with a dashed line.

Normal bond lines indicate bonds in the plane of the paper. The easiest way to visualize a chiral center is to draw two bonds in the plane of the paper normal lines , one bond coming out of the paper a wedge , and one bond going back into the paper a dashed line. Figure bonds in the plane Using a bond going back into paper wedges of the paper C and dashes. Label each of the chiral centers in the fol- central carbon is tricky because it appears to be a chiral center because it lowing molecule with a star.

Prioritize the substituents off the chiral Compare the first atom of each of the four groups attached to the chiral carbon; the center by using the Cahn-Ingold-Prelog prioritizing scheme. The remaining two A. In this case, the top sub- HO CH2OH stituent is next attached to a carbon, 3 and the bottom substituent is attached 2 to an oxygen.

The substituent attached to the oxygen has the higher priority because oxygen has a higher atomic number than carbon. Label each of the chiral centers in the fol- 2. Label each of the chiral centers in the fol- lowing molecule with a star. Determine the substituent priority for each 4.

Determine the substituent priority for each chiral center by using the Cahn-Ingold- chiral center by using the Cahn-Ingold- Prelog rules.

Prelog rules. Rotate the molecule so the fourth-priority substituent is in the back on a dashed bond sticking back into the paper.

If you have difficulty rotating the molecule in 3-D, use a molecular modeling kit those plastic balls and sticks that represent atoms and bonds until you get the feel of doing this kind of transformation.

Draw a curve from the first- to second- to third-priority substituent. If the curve is clockwise, the configuration is designated R; if the curve is counter- clockwise, the configuration is S. Tip: Remember this step in terms of driving and pic- ture the curve as a steering wheel: If you steer clockwise, the car goes right R ; turn the wheel counterclockwise, and the car goes left S, which stands for sinister, or left, in Latin. If you have difficulty visualizing in 3-D — or want to save some time on an exam — I have a trick you can use: Swapping any two groups inverts the stereochemistry it makes the chiral center go from R to S or S to R.

Assign the chiral center in the following The next step is to point the fourth-prior- ity substituent to the back. In this case, molecule as R or S. Draw an arrow connecting H C the first- to second- to third-priority F Br substituents.

The chiral center is S. Bromine has the 4 Cl 2 highest priority because this atom has H C the highest atomic number; chlorine is F Br 1 second, fluorine is third, and hydrogen 3 as always is last, in order of descending atomic number. The curve is counterclockwise, so the configuration of this chiral center is S. Assign the chiral center in the following Rotate the molecule so the fourth-prior- ity substituent CH3 sticks in the back.

This step is the most difficult because it requires you to visualize the molecule in CH3 three dimensions. The chiral center is R. Prioritize the sub- stituents on the chiral center. Finally, the CH2CH3 group has higher pri- This curve is clockwise, so the configura- ority than CH3 because in the former tion for this chiral center is R. Identify any chiral centers in the following 6. Identify any chiral centers in the following molecule and then label each chiral center molecule and the label each chiral center as R or S.

Identify any chiral centers in the following 8. Identify any chiral centers in the following molecule and then label each chiral center molecule and then label each chiral center as R or S.

Identify any chiral centers in the following Fischer projections frequently represent stereochemistry, so become familiar with these projections. Even though Fischer projections look flat, by con- vention, you assume that the horizontal bars indicate bonds coming out of the plane of the paper like wedged bonds and the vertical bars indicate bonds going back into the paper like dashed bonds.

Figure shows how to convert a 3-D projection wedge and dash structure into a Fischer projection. You rotate a quarter turn around one of the bonds so that two of the bonds on the chiral center are coming out of the plane of the paper and two bonds are going back into the paper. The bonds coming out of the plane become the horizontal bars in the Fischer pro- jection, and the bonds going back into the plane become the vertical bars. However, trying a few problems using the quarter-turn method can help give you a better understanding of what the Fischer projection means.

Assign the priorities to each of the substituents using the Cahn-Ingold-Prelog rules. Draw the curve from the first- to second- to third-priority substituents. Assign R or S stereochemistry to the 4H chiral center in the following Fischer 3 projection. The first thing to wise, so the configuration is S. Convert the following molecule into its Keep in mind that your primary goal when performing this conversion into a Fischer projection.

Fischer projection is to make sure that the configuration of the chiral center H stays the same. This becomes the Fischer projection. Assign R or S stereochemistry to each Convert the following molecule into its Convert the following molecule into its Fischer projection. Fischer projection.

Just as the relationship of a right hand to a left hand is that of mirror images, a molecule with a chiral center of R configuration is the mirror image of the same molecule with the S configuration.

Chemists call molecules that are mirror images of each other enantiomers, so in orgo-speak, your right hand is the enantiomer of your left. The problem with this analogy is that molecules can have more than one chiral center. Now imagine an arm with two left hands attached weird, I know. The mirror image of this arm its enantiomer is an arm with two right hands attached.

Similarly, the enantiomer of a mole- cule with two R chiral centers is a molecule with two S chiral centers. The bottom line is that the enantiomer of a molecule is a molecule with all the configurations switched all R config- urations go to S, and all S configurations go to R.

If you were to have an arm with two right hands and another arm with one right hand and one left hand, the two arms would have a different relationship than that of mirror images.

One example of a pair of diastereomers is one molecule that has two R chiral centers and a molecule with one R and one S chiral center. With one exception, any molecule that has a chiral center is a chiral molecule.

A chiral mole- cule is a molecule that has a non-superimposable mirror image. Molecules that have chiral centers are chiral molecules — with one exception. Meso com- pounds are molecules that have chiral centers but are achiral not chiral as a result of having a plane of symmetry in the molecule. A plane of symmetry is an imaginary line that you can draw in a molecule for which both halves are mirror images of each other. Unlike chiral mole- cules, meso compounds have mirror images that are superimposable on the original mole- cule that is, the mirror images are identical to the original molecule.

Thus, being able to spot planes of symmetry in molecules that have chiral centers is important for determining whether the molecule is chiral or achiral. Watch for rotations around single bonds. Anytime you can rotate around a single bond to give a structure that has a plane of symmetry — in other words, when any conformation has a plane of symmetry — the molecule is meso and achiral.

Identify the relationship between the two the two molecules. In this case, the con- figurations are opposite, with the left shown molecules as identical molecules, enantiomers, or diastereomers. The two molecules are enantiomers of H C H C each other. State whether the following compound is the molecule in half. Therefore, the mirror image of this compound is identi- chiral.

This molecule is achiral and meso. This molecule has two chiral centers, so you may expect it to be chiral. However, close inspection of this molecule shows that it has a plane of symmetry that cuts plane of symmetry Identify the relationship between the two Identify the relationship between the two shown molecules as identical molecules, shown molecules as identical molecules, enantiomers, or diastereomers. State whether the following compound is shown molecules as identical molecules, chiral.

State whether the following compound is State whether the following compound is chiral. Perhaps the most difficult carbon to assign is the ring carbon attached to the chain. The ring carbon with the chain attached is a chiral center because the top side of the ring is different from the bottom side of the ring.

Unlike in problem 2, these two sides of the ring are considered different substituents. Chapter 5: Seeing Molecules in 3-D: Stereochemistry 93 Starting with the right chiral center, prioritizing the substituents is reasonably straightforward. The oxygen gets the highest priority because this atom has the highest atomic number. The hydrogen gets the lowest priority because it has the lowest atomic number. Finally, the methyl group and the complex substituent to the left both have first-atom carbons, but the methyl gets a lower priority than the complex substituent because the complex substituent is attached to an oxygen, breaking the tie.

Both of the adjacent atoms in the ring are carbons. The top carbon in the ring is attached to three carbons and the bottom carbon in the ring is attached to just one other carbon, so the top ring carbon has the higher priority note that the number of attached atoms matters only in the event of a tie. Hydrogen has the lowest priority as usual. Under the Cahn-Ingold-Prelog prioritizing scheme, carbon-carbon double bonds are treated as though they represent two single bonds from the first carbon to the second carbon; therefore, one of the second carbons has two single bonds to the first atom.

Triple bonds repre- sent three single bonds from the first carbon to the second carbon, so one of the second car- bons then has three single bonds back to the first atom. The basic idea to remember is that double bonds have higher priority over single bonds between the same atoms, and triple bonds have priority over double bonds between the same atoms.

Thus, the carbon-carbon triple bond has priority over the carbon-carbon double bond, and the carbon-carbon double bond has pri- ority over the carbon-carbon single bond. First, determine the substituent priorities. The OH is first because oxygen has the highest atomic number, and hydrogen is last. Because the fourth-priority substituent is already in the back, you can draw the curve from the first- to second- to third-priority substituent.

The curve is clockwise, so the chiral center is designated R as in b. Based on atomic numbers, bromine is the highest, NH2 is second, CH3 is next, and hydrogen is last as in a. Next, rotate the molecule so the fourth-priority substituent is in the back as in b. This curve is clockwise, so the chiral center is R.

Determine the substituent priorities. Based on atomic numbers, fluorine is the first, OCH3 is second, CH3 is third, and hydrogen is fourth as in a. Rotate the molecule so the fourth-priority substituent is in the back as in b. This curve is coun- terclockwise, so the configuration is S.

Starting with the left chiral center, prioritize the substituents. Because the fourth-priority substituent is in the back, you can draw the curve to obtain the configuration. In this case, the curve is clockwise, so the left chiral center is R see a.

Now do the right chiral center. You can draw the curve after assigning the substituent priorities because the fourth-priority substituent is already in the back. The curve is counterclockwise, so this chiral center is S see b. Starting with the left chiral center, the OH gets the high- est priority and the H gets the fourth priority based on atomic number.

The right side of the ring has a higher priority over the left side of the ring because the first carbon on the right side is attached to chlorine, which breaks the tie.

The fourth-priority substituent is in the back, so drawing the clockwise curve yields an R configuration. The chlorine gets the highest priority and the hydrogen, the fourth; the left-hand side of the ring gets higher priority over the right-hand side of the ring because the first carbon on the left side of the ring is attached to an OH, which breaks the tie.

Drawing the counterclockwise curve yields an S config- uration. The right-hand carbon drawn in a 3-D projection is a chiral center, however. Prioritizing the substituents and drawing the counterclockwise curve shows that this chiral center is S.

First, prioritize the substituents as in a. Then draw the curve from first- to second- to third-priority substituents as in b. Because the fourth-priority substituent is on a vertical bar and the curve is clockwise, the configuration is R. Begin by prioritizing the substituents as in a. Now draw the curve from first to second to third as in b. Because the fourth-priority substituent is on a horizontal bar, the configuration is exactly the opposite of what you may expect.

The curve is clockwise, so the configuration is S. First, find the configu- ration of the chiral center in the 3-D projection by assigning priorities to the substituents. The fourth-priority substituent is in the back, so draw the curve from the first- to second- to third- priority substituents; the curve is counterclockwise, so the configuration is S.

Next, assign the configuration to the Fischer projection. Because the fourth-priority substituent is on a horizon- tal bar and the curve from the first- to second- to third-priority substituents is clockwise, the configuration is S. As long as the configuration is correct S , placing substituents in alternative locations is perfectly fine. By convention, chiral centers in Fischer projections are aligned vertically, so you first rotate the molecule so that the bond connecting the two chiral centers is vertical.

Then perform two quarter turns to get the molecule into the right orienta- tion for the Fischer projection. Molecules with more than one chiral center are often easier to convert to the Fischer projection by bypassing the quarter turn and 3-D visualization business: Assign the R or S configurations to the chiral centers in the 3-D projection and then add the substituents to a Fischer projection to make sure that the configurations remain unchanged.

First, assign the R or S configuration for each chiral center. Both chiral centers are R. Next, make sure both chiral centers have R stereochemistry in the Fischer projection. After drawing the Fischer projection with two crosses, one for each chiral center, add the substituents so that each chiral center is R.

With these placements, the curve from first- to second- to third-priority substituents is counter- clockwise, which indicates the chiral center is R because the fourth-priority substituent is on a horizontal bar. Next, add substituents to the second chiral center on the Fischer projection.

With this placement, the priority curve is counterclockwise, indicating an R stereochemistry because the fourth-priority substituent is on a horizontal bar. Thus, both methods yield the same solution. You can use whichever one works best for you.

These Fischer projections are mirror images of each other you can draw a mirror plane between the two Fischer projections. Moreover, switching the position of any two substituents in a Fischer projection inverts the stereochemistry takes it from R to S or S to R. Assigning the stereochemistry to both structures shows that the chiral center is R in both cases. Therefore, the two molecules have to be identical but merely drawn in a different way.

Two chiral centers are in the molecule, but only the top one switches upon going from the left-hand structure to the right-hand structure. Both chiral centers need to be switched to make the molecules enantiomers mirror images. Thus, these molecules are diastereomers. Even though two chiral centers exist, a plane of symmetry bisects the ring so that the top half is a reflection of the bottom half. OH plane of symmetry OH s This molecule is chiral.

Therefore, the molecule is chiral. This problem is very tricky. Although the molecule as ini- tially drawn has no plane of symmetry, you can rotate around the central bond to give a struc- ture that does have a plane of symmetry, making the molecule meso and achiral.

Those are the names of the organic molecules in those products. And for chemists, knowing what to call different molecules is an important part of keeping everyone on the same page.

In order for you to get in on the conversation, though, you need to know how to name organic molecules yourself. The best place to start honing your skills is with the simplest of organic molecules, the alkanes. Alkanes, the bones of organic molecules, are organic compounds containing only carbon-carbon and carbon-hydrogen single bonds.

This chapter gives you practice with the naming, or nomenclature, of alkanes, and it lays the foundation for the nomenclature of all organic molecules.

In addition to giving proper names to molecules, you see how to draw a structure from a given name. Such naming can appear difficult at first, given that organic structures often have long, complex names — and frankly, somewhat difficult-to-pronounce names — but with a bit of practice, this process is actually quite straightforward and kind of fun, too. Understanding How to Name Alkanes When you name straight-chain alkanes, use the four following steps: 1. Find the parent chain. Identify the parent chain, which is the longest continuous chain of carbon atoms.

Table shows the parent names for compounds, based on the number of carbons in the parent chain. Number the parent chain. Number the parent chain starting from the side that reaches the first substituent sooner.

A substituent is an atom or group that attaches to the parent chain. Name each of the substituents. You can name straight-chain substituents using Table , but you should also know the names of the three common complex substituents shown in Figure The dashed line indicates where that substituent joins the parent chain. Order the substituents alphabetically, using a number to indicate the position of each substituent on the parent chain. The di- indicates two substituents of the same kind; for three like sub- stituents, you use the prefix tri-; for four, tetra-; for five, penta-; and so on.

Tert-butyl or t-butyl and sec-butyl or s-butyl go under the letter b. For example, dimethyl falls under m. Cyclopropyl goes under c, and isopropyl goes under i. Cycloalkanes, or alkanes in rings, are named in a similar way. The principal difference comes in selecting the parent chain. For a cycloalkane, the parent chain is either the ring or a substituent attached to the ring. To determine which of the two is the parent chain, you count the number of carbons in the ring and then count the longest continuous chain of car- bons in the substituent.

Whichever has more carbons is labeled the parent chain, and the other is considered a substituent. A three-carbon ring, for example, is called cyclopropane. A three-membered ring as a sub- stituent would be called a cyclopropyl substituent. Name the following alkane: Now name the substituents. Advanced embedding details, examples, and help! Publication date Usage Public Domain Mark 1. From models to molecules to mass spectrometry-solve organic chemistry problems with ease Got a grasp on the organic chemistry terms and concepts you need to know, but get lost halfway through a problem or worse yet, not know where to begin?

Have no fear — this hands-on guide helps you solve the many types of organic chemistry problems you encounter in a focused, step-by-step manner. Know how to solve the most common organic chemistry problemsWalk through the answers and clearly identify where you went wrong or right with each problemGet the inside scoop on acing your exams! Use organic chemistry in practical applications with confidenc. There are no reviews yet.

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