Organic Chemistry - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Chemistry 2017 (2016)

5 Steps to a 5: AP Chemistry 2017 (2016)


Review the Knowledge You Need to Score High


Organic Chemistry


Although organic chemistry is not specifically tested on the AP Chemistry exam, an overview of the subject will help you recognize and understand organic structures that appear in relation to other topics on the exam. In addition, this material is taught by some high school and college teachers. This is an enrichment chapter that should be considered a bonus chapter to this book.


Summary: Organic chemistry is the study of the chemistry of carbon. Almost all the compounds containing carbon are classified as organic compounds. Only a few—for example, carbonates and cyanides—are classified as inorganic. It used to be thought that all organic compounds had to be produced by living organisms, but this idea was proven wrong in 1828 when German chemist Friedrich Wöhler produced the first organic compound from inorganic starting materials. Since that time, chemists have synthesized many organic compounds found in nature and have also made many never found naturally. It is carbon”s characteristic of bonding strongly to itself and to other elements in long, complex chains and rings that gives carbon the ability to form the many diverse and complex compounds needed to support life.

Keywords and Equations

No keywords or equations specific to this chapter are listed on the AP exam.


Alkanes are members of a family of organic compounds called hydrocarbons, compounds of carbon and hydrogen. These hydrocarbons are the simplest of organic compounds, but are extremely important to our society as fuels and raw materials for chemical industries. We heat our homes and run our automobiles through the combustion (burning) of these hydrocarbons. Paints, plastics, and pharmaceuticals are often made from hydrocarbons. Alkanes are hydrocarbons that contain only single covalent bonds within their molecules. They are called saturated hydrocarbons because they are bonded to the maximum number of other atoms. These alkanes may be straight-chained hydrocarbons, in which the carbons are sequentially bonded; branched hydrocarbons, in which another hydrocarbon group is bonded to the hydrocarbon “backbone”; or they may be cyclic, in which the hydrocarbon is composed entirely or partially of a ring system. The straight-chained and branched alkanes have the general formula of C n H2n +2 , whereas the cyclic alkanes have the general formula of C n H2n . The n stands for the number of carbon atoms in the compound. The first 10 straight-chained alkanes are shown in Table 18.1 .

There can be many more carbon units in a chain than are shown in Table 18.1 , but these are enough to allow us to study alkane nomenclature—the naming of alkanes.

Alkane Nomenclature

The naming of alkanes is based on choosing the longest carbon chain in the structural formula, then naming the hydrocarbon branches while indicating onto which carbon that branch is attached. Here are the specific rules for naming simple alkanes:

  1. Find the continuous carbon chain in the compound that contains the most carbon atoms. This will provide the base name of the alkane.
  2. This base name will be modified by adding the names of the branches (substituent groups) in front of the base name. Alkane branches are named by taking the name of the alkane that contains the same number of carbon atoms, dropping the -aneending and adding -yl . Methane becomes methyl, propane becomes propyl, etc. If there is more than one branch, list them alphabetically.

Table 18.1 The First Ten Straight-Chained Alkanes

  1. The position where a particular substituent is attached to the chain is indicated by a location number. These numbers are assigned by consecutively numbering the carbons of the base hydrocarbon, starting at one end of the hydrocarbon chain. Choose the end that will result in the lowest sum of location numbers for the substituent groups. Place this location number in front of the substituent name and separate it from the name by a hyphen (for example, 2-methyl).
  2. Place the substituent names with their location numbers in front of the base name of the alkane in alphabetical order. If there are identical substituents (two methyl groups, for example), give the location numbers of each, separated by commas using the common Greek prefixes (di-, tri-, tetra-, etc.) to indicate the number of identical substituent groups (i.e., 2,3-dimethyl). These Greek prefixes are not considered in the alphabetical arrangement.
  3. The last substituent group becomes a part of the base name as a prefix.

    StudyingFigures 18.1 and 18.2 may help you learn the naming of substituted alkanes.

Figure 18.1 Naming an alkane.

Figure 18.2 Naming of another alkane.

Figure 18.3 Structural isomers of C5 H12 .

Structural Isomerism

Compounds that have the same molecular formulas but different structural formulas are called isomers. With hydrocarbons, this applies to a different arrangement of the carbon atoms. Isomers such as these are called structural isomers . Figure 18.3 shows the structural isomers of C5 H12 . Note that there are the same number of carbons and hydrogens in each structure. Only the way the carbons are bonded is different.

In writing structural isomers, or any other organic compounds, remember that carbon forms four bonds . One of the most common mistakes that a chemistry student makes is writing an organic structure with a carbon atom having fewer or more than four bonds.

Here is a practice problem. Name the following compound:

Answer: 5-ethyl-2,2-dimethylnonane


First, pick the longest chain. This is bold-faced in the diagram below. The carbons are attached by single bonds, so this is an alkane. Because the longest chain has nine carbons, it is a nonane.

Next, the longest chain should be numbered from one end to the other with the lowest number(s) going to the branches. For the preceding example the numbering of the chain (bold-face carbon atoms) would be:

Once these numbers have been assigned, do not alter them later.

All carbon atoms that are not part of the nine-atom main chain are branches. Branches have -yl endings. It may help you to circle the carbon atoms belonging in the branches. In the above example, there are three branches. Two consist of only one carbon and are called methyl groups. The remaining branch has two carbons, so it is an ethyl group. The branches are arranged alphabetically. If there is more than one of a particular type, use a prefix (di -, tri- , tetra -, etc.). The two methyl groups are designated dimethyl. The position of each branch is indicated with a number already determined for the main chain. Each branch must get its own number, even if it is identical to one already used.

In the above example this gives: 5-ethyl-2,2-dimethylnonane

  1. ethyl before methyl (alphabetical—prefixes are ignored)
  2. two methyl groups = dimethyl
  3. three branches = three numbers

Numbers are separated from other numbers by commas, and numbers are separated from letters by a hyphen.

Another type of isomerism is optical isomerism. These molecules are capable of rotating light to either the left or right and are said to be optically active. The presence of an asymmetric or chiral carbon (a carbon atom with four different groups attached to it) will make a compound optically active.

Common Functional Groups

If chemistry students had to learn the properties of each of the millions of organic compounds, they would face an impossible task. Luckily, chemists find that having certain arrangements of atoms in an organic molecule causes those molecules to react in a similar fashion. For example, methyl alcohol, CH3 -OH, and ethyl alcohol, CH3 -CH2 -OH, undergo the same types of reactions. The −OH group is the reactive part of these types of molecule. These reactive groups are called functional groups. Instead of learning the properties of individual molecules, one can simply learn the properties of functional groups.

In our study of the simple hydrocarbons, there are only two functional groups. One is a carbon-to-carbon double bond. Hydrocarbons that contain a carbon-to-carbon double bond are called alkenes. Naming alkenes is very similar to naming alkanes. The major difference is that the carbon base has an -ene ending instead of the -ane ending. The carbon backbone of the base hydrocarbon is numbered so the position of the double bond has the lowest location number.

The other hydrocarbon functional group is a carbon-to-carbon triple bond. Hydrocarbons that contain a triple bond are called alkynes. Alkynes use the -yne ending on the base hydrocarbon. The presence of a double or triple bond make these hydrocarbons unsaturated.

The introduction of other atoms (N, O, Cl, etc.) to organic compounds gives rise to many other functional groups. The major functional groups are shown in Table 18.2 , on the next page.

Table 18.2 Common Functional Groups


As we mentioned in the introduction to this chapter, carbon has the ability to bond to itself in long and complex chains. These large molecules, called macromolecules , may have molecular masses in the millions. They are large, complex molecules, but most are composed of repeating units called monomers . Figure 18.4 shows two macromolecules, cellulose and nylon, and indicates their repeating units.

Figure 18.4 Two macromolecules.

Macromolecules are found in nature. Cellulose, wool, starch, and DNA are but a few of the macromolecules that occur naturally. Carbon”s ability to form these large, complex molecules is necessary to provide the diversity of compounds needed to make up a tree or a human being. But many of the useful macromolecules that we use every day are created in the lab and industrial complex by chemists. Nylon, rayon, polyethylene, and polyvinyl chloride are all synthetic macromolecules. They differ by which repeating units (monomers) are joined together in the polymerization process. Our society has grown to depend on these plastics, these synthetic fabrics. The complexity of carbon compounds is reflected in the complexity of our modern society.


Any experiment would probably apply the concepts of organic chemistry in a synthesis situation.

Common Mistakes to Avoid

  1. When writing organic formulas, make sure that every carbon has four bonds.
  2. When naming alkanes, make sure to number the carbon chain so the sum of all location numbers is as small as possible.
  3. When naming branched alkanes, be sure to consider the branches when finding the longest carbon chain. The longest chain isn”t always the one in which the carbon atoms all lie in a horizontal line.
  4. In naming identical substituents on the longest carbon chain, be sure to use repeating location numbers, separated by commas (2,2-dimethyl).
  5. Be sure that every carbon has four bonds!

Review Questions

Use these questions to review the content of this chapter and improve your understanding of chemistry. Although organic chemistry won”t specifically be tested on the exam, this section contains questions of the same types as those found on the AP exam. First are five multiple-choice questions; following them is a multipart free-response question. Follow the time limitations given for practice pacing yourself.

Multiple-Choice Questions

Answer the following questions in 10 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.

1 . Cycloalkanes are hydrocarbons with the general formula C n H2n . If a 0.500 g sample of any alkene is combusted in excess oxygen, how many moles of water will form?

(A) 0.50

(B) 0.072

(C) 0.036

(D) 1.0

2 .

The organic compound shown above would be classified as:

(A) an organic base

(B) an ether

(C) an alcohol

(D) an aldehyde

3 .

The above compound would be classified as:

(A) an aldehyde

(B) a ketone

(C) an ester

(D) a carboxylic acid

4 . Which of the following compounds is optically active?




(D) CH3 CH2 CH2 CH2 OH

5 . A carboxylic acid may be represented as:



(C) R-O-R


Answers and Explanations for the Multiple-Choice Questions

1 . C —The general formula, C n H2n , means that 1 mol of H2 O will form per mole of empirical formula unit, regardless of the value of n . The moles of water formed are the mass of the alkene divided by the empirical formula mass. (0.500 g alkene)(1 mol alkene/14 g alkene)(1 mol H2 O/mol alkene) = 0.036 mol.

2 . C —Organic bases are, in general, amines (contain N). An ether would have an oxygen single-bonded to two carbons (R groups). An aldehyde has oxygen double-bonded to a carbon at the end of a chain. Aldehydes (RCHO) and alcohols (ROH) are often confused because of the similarity in their general formulas. Ketones have oxygen double-bonded to a carbon not at the end of a chain.

3 . D —Based on classification of organic compounds.

4 . A —Redrawing the structures may help you to recognize the correct answer. An optical isomer must be a carbon atom with four different groups attached to it. For A, the groups on the second carbon are: CH3 [, H, Cl, and [CH2CH2 CH3 . Answer C is misleading. It is similar to A, but two of the groups, the [CH2 CH3 groups, are the same.

5 . D —A = alcohol, B = aldehyde, C = ether

Free-Response Question

You have 20 minutes to answer the following question. You may use a calculator and the tables in the back of the book.

Question 1

The alkane hexane, C6 H14 , has a molecular mass of 86.17 g mol–1 .

(a) Like all hydrocarbons, hexane will burn. Write
a balanced chemical equation for the complete combustion of hexane. This reaction produces gaseous carbon dioxide, CO2 , and water vapor, H2 O.

(b) The complete combustion of 10.0 g of hexane produces 487 kJ. What is the molar heat of combustion (ΔH ) of hexane?

(c) Determine the pressure exerted by the carbon dioxide formed when 5.00 g of hexane is combusted. Assume the carbon dioxide is dry and stored in a 20.0 L container at 27°C.

(d) Hexane, like most alkanes, may exist in different isomeric forms. The structural formula of one of these isomers is pictured below. Draw the structural formula of any two other isomers of hexane. Make sure all carbon atoms and hydrogen atoms are shown.

Answer and Explanation for the Free-Response Question

Note that, while organic chemistry is not an AP topic, all of the materials in these questions depend upon basic AP chemistry knowledge, which is why this chapter can be valuable.

(a) 2 C6 H14 + 19 O2 → 12 CO2 + 14 H2 O

Give yourself 2 points for the answer shown above, or for the coefficients: 1, 9/2, 6, and 7. Give yourself 1 point if you have one or more, not all, of the elements balanced.

(b) (–487 kJ/10.0 g hexane)(86.17 g hexane/mol hexane) = –4.20 × 103 kJ/mol

Give yourself 2 points for the above setup and correct answer (this requires a negative sign in the answer). If the setup is partially correct, give yourself 1 point.

(c) The ideal gas equation should be rearranged to the form P = nRT /V .

This answer has an extra significant figure. The mole ratio should match the one given in your balanced equation. You will not be penalized again for an incorrectly balanced equation.

You will lose a point if you do not include a hexane-to-CO2 conversion. R = 0.08206 L atm mol–1 K–1 (This value is in your test booklet.)

T = 27°C + 273 = 300.0 K

In this case, there is no penalty if you forget to use the Kelvin temperature.

V = 20.0 L

P = (0.3481 mol CO2 )(0.08206 L atm mol–1 K–1 )(300.0 K)/(20.0 L) = 0.429 atm

Give yourself 2 points for the correct setup and answer. Give yourself 1 point if you did everything correctly, except the mole ratio or the Kelvin conversion.

(d) You may need to redraw one or more of your answers to match the answers shown below.

Give yourself 1 point for each correct answer, with a 2-point maximum. There are no bonus points for additional answers.

These compounds are 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane, respectively. These four, along with the original n -hexane, are the only isomers. If you think you have another isomer, you have simply redrawn one of these. Try naming your answer and see if it matches one of these names.

Total your points. The maximum is 8 points. Subtract one point if all your answers do not have the correct number of significant figures.

Rapid Review

  • Organic chemistry is the chemistry of carbon and its compounds.
  • Hydrocarbons are organic compounds of just carbon and hydrogen atoms.
  • Alkanes are hydrocarbons in which there are only single bonds.
  • Alkanes are named in a very systematic way. Review the rules for naming alkanes.
  • Isomers are compounds that have the same molecular formulas but different structural formulas. Review the writing of the various structural isomers of alkanes.Make sure that each carbon atom has four bonds.
  • A functional group is a group of atoms that is the reactive part of the molecule. Review the general functional groups.
  • Macromolecules are large molecules that may have molecular masses in the millions. Macromolecules are generally composed of repeating units called monomers.