RULE 1 -
Individual resonance forms are imaginary, not real. The real structure is a composite, or resonance hybrid, of the different forms. Species such as the acetate ion and benzene are no different from any other. They have single, unchanging structures, and they don't switch back and forth between resonance forms. The only difference between these and other substances is in the way they are represented in drawings.


RULE 2 -
Resonance forms differ only in the placement of their $\boldsymbol{\pi}$ or nonbonding electrons. Neither the position nor the hybridization of any atom changes from one resonance form to another. In the acetate ion, for instance, the carbon atom is $s p^{2}$-hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the $\pi$ electrons in the $\mathrm{C}=\mathrm{O}$ bond and the lone-pair electrons on oxygen differ from one form to another. This movement of electrons from one resonance structure to another can be indicated with curved arrows. A curved arrow always indicates the movement of electrons, not the movement of atoms. An arrow shows that a pair of electrons moves from the atom or bond at the tail of the arrow to the atom or bond at the head of the arrow.

The situation with benzene is similar to that with acetate. The $\pi$ electrons in the double bonds move, as shown with curved arrows, but the carbon and hydrogen atoms remain in place.

RULE 3
Different resonance forms of a substance don't have to be equivalent. As an example, we'll see in Chapter 22 that a compound such as acetone, which contains a $\mathrm{C}=\mathrm{O}$ bond, can be converted into its anion by reaction with a strong base. The resultant anion has two resonance forms. One form contains a carbon-oxygen double bond and has a negative charge on carbon; the other contains a carbon-carbon double bond and has a negative charge on oxygen. Even though the two resonance forms aren't equivalent, both contribute to the overall
resonance hybrid.

When two resonance forms are nonequivalent, the actual structure of the resonance hybrid resembles the more stable form. Thus, we might expect the true structure of the acetone anion to be more like that of the form that places the negative charge on the electronegative oxygen atom rather than on the carbon.

RULE 4
Resonance forms obey normal rules of valency. A resonance form is like any other structure: the octet rule still applies to second-row, main-group atoms. For example, one of the following structures for the acetate ion is not a valid resonance form because the carbon atom has five bonds and ten valence electrons:


RULE 2 -
RULE 5
The resonance hybrid is more stable than any individual resonance form. In other words, resonance leads to stability. Generally speaking, the larger the number of resonance forms a substance has, the more stable the substance is, because its electrons are spread out over a larger part of the molecule and are closer to more nuclei. We'll see in Chapter 15, for instance, that a benzene ring is more stable because of resonance than might otherwise be expected.


RULE 2 - 2.6 Drawing Resonance Forms
Look back at the resonance forms of the acetate ion and the acetone anion shown in the previous section. The pattern seen there is a common one that leads to a useful technique for drawing resonance forms. In general, any three-atom grouping with a $p$ orbital on each atom has two resonance forms:

The atoms $\mathrm{X}, \mathrm{Y}$, and Z in the general structure might be $\mathrm{C}, \mathrm{N}, \mathrm{O}, \mathrm{P}, \mathrm{S}$, or others, and the asterisk (*) might mean that the $p$ orbital on atom $Z$ is vacant, that it contains a single electron, or that it contains a lone pair of electrons. The two resonance forms differ simply by an exchange in position of the multiple bond and the asterisk from one end of the three-atom grouping to the other.

By learning to recognize such three-atom groupings within larger structures, resonance forms can be systematically generated. Look, for instance, at the anion produced when $\mathrm{H}^{+}$is removed from 2,4-pentanedione by reaction with a base. How many resonance structures does the resultant anion have?

The 2,4-pentanedione anion has a lone pair of electrons and a formal negative charge on the central carbon atom, next to a $\mathrm{C}=\mathrm{O}$ bond on the left. The $\mathrm{O}=\mathrm{C}-\mathrm{C}:^{-}$grouping is a typical one for which two resonance structures can be drawn.

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


Drawing Resonance Forms for an Anion -
Draw three resonance structures for the carbonate ion, $\mathrm{CO}_{3}{ }^{2-}$.


Strategy -
Look for three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital. Then exchange the positions of the multiple bond and the electrons in the $p$ orbital. In the carbonate ion, each singly bonded oxygen atom with three lone pairs and a negative charge is adjacent to the $\mathrm{C}=\mathrm{O}$ double bond, giving the grouping Ö: $=C-$ Ö:-


Solution -
Exchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures.


Drawing Resonance Forms for a Radical -
Draw three resonance forms for the pentadienyl radical, where a radical is a substance that contains a single, unpaired electron in one of its orbitals, denoted by a dot $(\cdot)$.


Strategy -
Find the three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital.


Solution -
The unpaired electron is on a carbon atom next to a $\mathrm{C}=\mathrm{C}$ bond, giving a typical three-atom grouping that has two resonance forms.

In the second resonance form, the unpaired electron is next to another double bond, giving another three-atom grouping and leading to another resonance form.
Three-atom grouping

Thus, the three resonance forms for the pentadienyl radical are:

PROBLEM Which of the following pairs of structures represent resonance forms, and which do not? Explain. 2-9
(a)

(b)
and

PROBLEM Draw the indicated number of resonance forms for each of the following substances:
2-10 (a) The methyl phosphate anion, $\mathrm{CH}_{3} \mathrm{OPO}_{3}{ }^{2-}$ (3)
(b) The nitrate anion, $\mathrm{NO}_{3}{ }^{-}$(3)
(c) The allyl cation, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}{ }^{+}$(2)
(d) The benzoate anion (2)


Solution - 2.7 Acids and Bases: The Brønsted-Lowry Definition
Perhaps the most important of all concepts related to electronegativity and polarity is that of acidity and
basicity. We'll soon see, in fact, that the acid-base behavior of organic molecules explains much of their chemistry. You may recall from a course in general chemistry that two definitions of acidity are frequently used: the Brønsted-Lowry definition and the Lewis definition. We'll look at the Brønsted-Lowry definition in this and the following three sections and then discuss the Lewis definition in Section 2.11.

A Brønsted-Lowry acid is a substance that donates a hydrogen ion, $\mathrm{H}^{+}$, and a Brønsted-Lowry base is a substance that accepts a hydrogen ion. (The name proton is often used as a synonym for $\mathrm{H}^{+}$because loss of the valence electron from a neutral hydrogen atom leaves only the hydrogen nucleus-a proton.) When gaseous hydrogen chloride dissolves in water, for example, a polar HCl molecule acts as an acid and donates a proton, while a water molecule acts as a base and accepts the proton, yielding chloride anion ( $\mathrm{Cl}^{-}$) and hydronium cation $\left(\mathrm{H}_{3} \mathrm{O}^{+}\right)$. This and other acid-base reactions are reversible, so we'll write them with double, forward-andbackward arrows.

Acid

Base

Conjugate base

Conjugate acid

Chloride ion, the product that results when the acid HCl loses a proton, is called the conjugate base of the acid, and hydronium ion, the product that results when the base $\mathrm{H}_{2} \mathrm{O}$ gains a proton, is called the conjugate acid of the base. Other common mineral acids such as $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HNO}_{3}$ behave similarly, as do organic acids such as acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$.

In a general sense,


Solution - 2.7 Acids and Bases: The Brønsted-Lowry Definition
In a general sense,

$$
\begin{array}{cc}
\mathrm{H}-\mathrm{A} & +\mathrm{B} \\
\text { Acid } & \text { Base }
\end{array}
$$

For example:

Notice that water can act either as an acid or as a base, depending on the circumstances. In its reaction with HCl , water is a base that accepts a proton to give the hydronium ion, $\mathrm{H}_{3} \mathrm{O}^{+}$. In its reaction with ammonia $\left(\mathrm{NH}_{3}\right)$, however, water is an acid that donates a proton to give ammonium ion $\left(\mathrm{NH}_{4}^{+}\right)$and hydroxide ion, $\mathrm{HO}^{-}$.

PROBLEM Nitric acid $\left(\mathrm{HNO}_{3}\right)$ reacts with ammonia $\left(\mathrm{NH}_{3}\right)$ to yield ammonium nitrate. Write the reaction, and
2-11 identify the acid, the base, the conjugate acid product, and the conjugate base product.


Solution - 2.8 Acid and Base Strength
Different acids differ in their ability to donate $\mathrm{H}^{+}$. Stronger acids, such as HCl , react almost completely with water, whereas weaker acids, such as acetic acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$, react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant ( $\boldsymbol{K}_{\mathbf{a}}$ ) for the acid-dissociation equilibrium. Recall from general chemistry that the concentration of solvent is ignored in the equilibrium expression and that brackets [ ] around a substance refer to the concentration of the enclosed species in moles per liter.

$$
\begin{gathered}
\mathrm{HA}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{~A}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \\
K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}
\end{gathered}
$$

Stronger acids have their equilibria toward the right and thus have larger acidity constants, whereas weaker acids have their equilibria toward the left and have smaller acidity constants. The range of $K_{\mathrm{a}}$ values for different acids is enormous, running from about $10^{15}$ for the strongest acids to about $10^{-60}$ for the weakest. Common inorganic acids such as $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HNO}_{3}$, and HCl have $K_{\mathrm{a}}$ 's in the range of $10^{2}$ to $10^{9}$, while organic acids generally have $K_{\mathrm{a}}$ 's in the range of $10^{-5}$ to $10^{-15}$. As you gain experience, you'll develop a rough feeling for which acids are "strong" and which are "weak" (always remembering that the terms are relative, not absolute).


Solution - 2.8 Acid and Base Strength
Acid strengths are normally expressed using $\mathrm{p} K_{\mathrm{a}}$ values rather than $K_{\mathrm{a}}$ values, where the $\mathbf{p} \boldsymbol{K}_{\mathbf{a}}$ is the negative common logarithm of the $K_{\mathrm{a}}$ :

$$
\mathrm{p} K_{\mathrm{a}}=-\log K_{\mathrm{a}}
$$

A stronger acid (larger $K_{\mathrm{a}}$ ) has a smaller $\mathrm{p} K_{\mathrm{a}}$, and a weaker acid (smaller $K_{\mathrm{a}}$ ) has a larger $\mathrm{p} K_{\mathrm{a}}$. TABLE 2.3 lists the $\mathrm{p} K_{\mathrm{a}}$ 's of some common acids in order of their strength, and a more comprehensive table is given in Appendix B .

Notice that the $\mathrm{p} K_{\mathrm{a}}$ value shown in TABLE 2.3 for water is 15.74 , which results from the following calculation. Because water is both the acid and the solvent, the equilibrium expression is


Solution - 2.8 Acid and Base Strength
$$
\begin{gathered}
\begin{array}{c}
\mathrm{H}_{2} \mathrm{O}
\end{array}+\underset{\text { (acid) }}{\mathrm{H}_{2} \mathrm{O}} \rightleftharpoons \mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \\
K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\begin{array}{c}
{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}
\end{array}}{\left[\mathrm{H}_{2} \mathrm{O}\right]}=\frac{\left[1.0 \times 10^{-7}\right]\left[1.0 \times 10^{-7}\right]}{[55.4]}=1.8 \times 10^{-16} \\
\mathrm{p} K_{\mathrm{a}}=15.74
\end{gathered}
$$

The numerator in this expression is the so-called ion-product constant for water, $K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.00 \times$ $10^{-14}$, and the denominator is the molar concentration of pure water, $\left[\mathrm{H}_{2} \mathrm{O}\right]=55.4 \mathrm{M}$ at $25^{\circ} \mathrm{C}$. The calculation is artificial in that the concentration of "solvent" water is ignored while the concentration of "acid" water is not, but it is nevertheless useful for making a comparison of water with other weak acids on a similar footing.


Solution - 2.8 Acid and Base Strength
Notice also in TABLE 2.3 that there is an inverse relationship between the acid strength of an acid and the base strength of its conjugate base. A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base. To understand this inverse relationship, think about what is happening to the acidic hydrogen in an acid-base reaction. A strong acid is one that loses $\mathrm{H}^{+}$easily, meaning that its conjugate base holds the $\mathrm{H}^{+}$ weakly and is therefore a weak base. A weak acid is one that loses $\mathrm{H}^{+}$with difficulty, meaning that its conjugate base holds the proton tightly and is therefore a strong base. The fact that HCl is a strong acid, for example, means that $\mathrm{Cl}^{-}$does not hold $\mathrm{H}^{+}$tightly and is thus a weak base. Water, on the other hand, is a weak acid, meaning that $\mathrm{OH}^{-}$holds $\mathrm{H}^{+}$tightly and is a strong base.


Solution - 2.8 Acid and Base Strength
|  | Acid | Name | $\mathrm{p} K_{\mathrm{a}}$ | Conjugate base | Name |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Weaker acid | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ | Ethanol | 16.00 | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ | Ethoxide ion | Stronger base |
|  | $\mathrm{H}_{2} \mathrm{O}$ | Water | 15.74 | $\mathrm{HO}^{-}$ | Hydroxide ion |  |
|  | HCN | Hydrocyanic acid | 9.31 | $\mathrm{CN}^{-}$ | Cyanide ion |  |
|  | $\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}$ | Dihydrogen phosphate ion | 7.21 | $\mathrm{HPO}_{4}{ }^{2-}$ | Hydrogen phosphate ion |  |
|  | $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ | Acetic acid | 4.76 | $\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}$ | Acetate ion |  |
|  | $\mathrm{H}_{3} \mathrm{PO}_{4}$ | Phosphoric acid | 2.16 | $\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}$ | Dihydrogen phosphate ion |  |
|  | $\mathrm{HNO}_{3}$ | Nitric acid | -1.3 | $\mathrm{NO}_{3}{ }^{-}$ | Nitrate ion |  |
| acid | HCl | Hydrochloric acid | -7.0 | $\mathrm{Cl}^{-}$ | Chloride ion | base |

PROBLEM The amino acid phenylalanine has $\mathrm{p} K_{\mathrm{a}}=1.83$, and tryptophan has $\mathrm{p} K_{\mathrm{a}}=2.83$. Which is the stronger $\mathbf{2 - 1 2}$ acid?


Solution - 2.8 Acid and Base Strength
Phenylalanine
( $\mathrm{p} K_{\mathrm{a}}=1.83$ )
Tryptophan
$\left(\mathrm{p} K_{\mathrm{a}}=2.83\right)$
PROBLEM Amide ion, $\mathrm{H}_{2} \mathrm{~N}^{-}$, is a much stronger base than hydroxide ion, $\mathrm{HO}^{-}$. Which is the stronger acid, $\mathrm{NH}_{3}$
2-13 or $\mathrm{H}_{2} \mathrm{O}$ ? Explain.


Solution - 2.9 Predicting Acid-Base Reactions from $\mathrm{p} K_{\mathrm{a}}$ Values
Compilations of $\mathrm{p} K_{\mathrm{a}}$ values like those in TABLE 2.3 and Appendix B are useful for predicting whether a given acid-base reaction will take place, because $\mathrm{H}^{+}$will always go from the stronger acid to the stronger base. That is, an acid will donate a proton to the conjugate base of a weaker acid, and the conjugate base of a weaker acid will remove a proton from a stronger acid. Since water ( $\mathrm{p} K_{\mathrm{a}}=15.74$ ) is a weaker acid than acetic acid ( $\mathrm{p} K_{\mathrm{a}}=4.76$ ), for example, hydroxide ion holds a proton more tightly than acetate ion does. Hydroxide ion will therefore react to a large extent with acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, to yield acetate ion and $\mathrm{H}_{2} \mathrm{O}$.

Another way to predict acid-base reactivity is to remember that the product conjugate acid in an acid-base reaction must be weaker and less reactive than the starting acid and the product conjugate base must be weaker and less reactive than the starting base. In the reaction of acetic acid with hydroxide ion, for example, the product conjugate acid $\left(\mathrm{H}_{2} \mathrm{O}\right)$ is weaker than the starting acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$, and the product conjugate base $\left(\mathrm{CH}_{3} \mathrm{CO}_{2}{ }^{-}\right)$is weaker than the starting base $\left(\mathrm{OH}^{-}\right)$.


Predicting Acid Strengths from $\mathbf{p} K_{\mathbf{a}}$ Values -
Water has $\mathrm{p} K_{\mathrm{a}}=15.74$, and acetylene has $\mathrm{p} K_{\mathrm{a}}=25$. Which is the stronger acid? Does hydroxide ion react to a significant extent with acetylene?

Acetylene


Strategy -
In comparing two acids, the one with the lower $\mathrm{p} K_{\mathrm{a}}$ is stronger. Thus, water is a stronger acid than acetylene and gives up $\mathrm{H}^{+}$more easily.


Solution -
Because water is a stronger acid and gives up $\mathrm{H}^{+}$more easily than acetylene, the $\mathrm{HO}^{-}$ion must have less affinity for $\mathrm{H}^{+}$than the $\mathrm{HC} \equiv \mathrm{C}$ : $^{-}$ion. In other words, the anion of acetylene is a stronger base than hydroxide ion, and the reaction will not proceed significantly as written.


Calculating $K_{\mathrm{a}}$ from $\mathrm{p} K_{\mathrm{a}}$ -
According to the data in TABLE 2.3, acetic acid has $\mathrm{p} K_{\mathrm{a}}=4.76$. What is its $K_{\mathrm{a}}$ ?


Strategy -
Since $\mathrm{p} K_{\mathrm{a}}$ is the negative logarithm of $K_{\mathrm{a}}$, it's necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the $\mathrm{p} K_{\mathrm{a}}(4.76)$, change the sign $(-4.76)$, and then find the antilog $\left(1.74 \times 10^{-5}\right)$.


Solution -
$K_{\mathrm{a}}=1.74 \times 10^{-5}$.

PROBLEM Will either of the following reactions take place to a significant extent as written, according to the 2-14 data in Table 2.3?

PROBLEM Ammonia, $\mathrm{NH}_{3}$, has $\mathrm{p} K_{\mathrm{a}} \approx 36$, and acetone has $\mathrm{p} K_{\mathrm{a}} \approx 19$. Will the following reaction take place to a 2-15 significant extent?

Acetone
PROBLEM What is the $K_{\mathrm{a}}$ of HCN if its $\mathrm{p} K_{\mathrm{a}}=9.31$ ?
2-16


Solution - 2.10 Organic Acids and Organic Bases
Many of the reactions we'll be seeing in future chapters, including practically all biological reactions, involve organic acids and organic bases. Although it's too early to go into the details of these processes now, you might keep the following generalities in mind:


Organic Acids -
Organic acids are characterized by the presence of a positively polarized hydrogen atom (blue in electrostatic potential maps) and are of two main kinds: acids such as methanol and acetic acid that contain a hydrogen atom bonded to an electronegative oxygen atom $(\mathrm{O}-\mathrm{H})$ and those such as acetone (Section 2.5) that contain a hydrogen atom bonded to a carbon atom next to a $\mathrm{C}=\mathrm{O}$ bond $(\mathrm{O}=\mathrm{C}-\mathrm{C}-\mathrm{H})$.

Methanol
$\left(\mathrm{p} K_{a}=15.54\right)$

Acetic acid
$\left(\mathrm{p} K_{\mathrm{a}}=4.76\right)$

Acetone
$\left(p K_{a}=19.3\right)$

Methanol contains an $\mathrm{O}-\mathrm{H}$ bond and is a weak acid, while acetic acid also contains an $\mathrm{O}-\mathrm{H}$ bond and is a somewhat stronger acid. In both cases, acidity is due to the fact that the conjugate base resulting from loss of $\mathrm{H}^{+}$ is stabilized by having its negative charge on a strongly electronegative oxygen atom. In addition, the conjugate base of acetic acid is stabilized by resonance (Section 2.4 and Section 2.5).

Anion is stabilized both by having negative charge on a highly electronegative atom and by resonance.

The acidity of acetone and other compounds with $\mathrm{C}=\mathrm{O}$ bonds is due to the fact that the conjugate base resulting from loss of $\mathrm{H}^{+}$is stabilized by resonance. In addition, one of the resonance forms stabilizes the negative charge by placing it on an electronegative oxygen atom.

Anion is stabilized both by resonance and by having negative charge on a highly electronegative atom.

Electrostatic potential maps of the conjugate bases from methanol, acetic acid, and acetone are shown in FIGURE 2.5. As you might expect, all three show a substantial amount of negative charge (red) on oxygen.
(a)


Organic Acids -
$\mathrm{CH}_{3} \mathrm{O}^{-}$
(b)

(c)

FIGURE 2.5 Electrostatic potential maps of the conjugate bases of (a) methanol, (b) acetic acid, and (c) acetone. The electronegative oxygen atoms stabilize the negative charge in all three.

Compounds called carboxylic acids, which contain the $-\mathrm{CO}_{2} \mathrm{H}$ grouping, occur abundantly in all living organisms and are involved in almost all metabolic pathways. Acetic acid, pyruvic acid, and citric acid are examples. You might note that at the typical pH of 7.3 found within cells, carboxylic acids are usually dissociated and exist as their carboxylate anions, $-\mathrm{CO}_{2}{ }^{-}$.

Acetic acid

Pyruvic acid

Citric acid


Organic Bases -
Organic bases are characterized by the presence of an atom (reddish in electrostatic potential maps) with a lone pair of electrons that can bond to $\mathrm{H}^{+}$. Nitrogen-containing compounds such as methylamine are the most common organic bases and are involved in almost all metabolic pathways, but oxygen-containing compounds can also act as bases when reacting with a sufficiently strong acid. Note that some oxygen-containing compounds can act both as acids and as bases depending on the circumstances, just as water can. Methanol and acetone, for instance, act as acids when they donate a proton but as bases when their oxygen atom accepts a proton.

Methylamine

Methanol

Acetone

We'll see in Chapter 26 that substances called amino acids, so-named because they are both amines $\left(-\mathrm{NH}_{2}\right)$ and carboxylic acids $\left(-\mathrm{CO}_{2} \mathrm{H}\right)$, are the building blocks from which the proteins in all living organisms are made. Twenty different amino acids go into making up proteins-alanine is an example. Interestingly, alanine and other amino acids exist primarily in a doubly charged form called a zwitterion rather than in the uncharged form. The zwitterion form arises because amino acids have both acidic and basic sites within the same molecule and therefore undergo an internal acid-base reaction.

Alanine
(uncharged form)

Alanine
(zwitterion form)


Organic Bases - 2.11 Acids and Bases: The Lewis Definition
The Lewis definition of acids and bases is more encompassing than the Brønsted-Lowry definition because it's not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond.


Lewis Acids and the Curved Arrow Formalism -
The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate $\mathrm{H}^{+}$(which has an empty $1 s$ orbital). Thus, the Lewis definition of acidity includes many species in addition to $\mathrm{H}^{+}$. For example, various metal cations, such as $\mathrm{Mg}^{2+}$, are Lewis acids because they accept a pair of electrons when they form a bond to a base. We'll also see in later chapters that certain metabolic reactions begin with an acid-base reaction between $\mathrm{Mg}^{2+}$ as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base.

Lewis acid
Lewis base
Acid-base complex (an organodiphosphate ion)
In the same way, compounds of group 3A elements, such as $\mathrm{BF}_{3}$ and $\mathrm{AlCl}_{3}$, are Lewis acids because they have
unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in FIGURE 2.6. Similarly, many transition-metal compounds, such as $\mathrm{TiCl}_{4}, \mathrm{FeCl}_{3}, \mathrm{ZnCl}_{2}$, and $\mathrm{SnCl}_{4}$, are Lewis acids.

| Boron
 trifluoride
 (Lewis acid) | Dimethyl
 ether
 (Lewis base) |
| :---: | :---: |

Acid-base complex

FIGURE 2.6 The reaction of boron trifluoride, a Lewis acid, with dimethyl ether, a Lewis base. The Lewis acid accepts a pair of electrons, and the Lewis base donates a pair of nonbonding electrons. Note how the movement of electrons from the Lewis base to the Lewis acid is indicated by a curved arrow. Note also how, in electrostatic potential maps, the boron becomes more negative (red) after reaction because it has gained electrons and the oxygen atom becomes more positive (blue) because it has donated electrons.


Lewis Acids and the Curved Arrow Formalism -
Look closely at the acid-base reaction in FIGURE 2.6, and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in $\mathrm{BF}_{3}$, a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.5. We'll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it.

Some further examples of Lewis acids follow:


Lewis Bases -
The Lewis definition of a base-a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid-is similar to the Brønsted-Lowry definition. Thus, $\mathrm{H}_{2} \mathrm{O}$, with its two pairs of nonbonding electrons
on oxygen, acts as a Lewis base by donating an electron pair to an $\mathrm{H}^{+}$in forming the hydronium ion, $\mathrm{H}_{3} \mathrm{O}^{+}$.

In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an $\mathrm{H}^{+}$but as bases when their oxygen atom accepts an $\mathrm{H}^{+}$.

Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. Reaction normally occurs only once in such instances, and the more stable of the two possible protonation products is formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms.

Acetic acid (base)


Using Curved Arrows to Show Electron Flow -
Using curved arrows, show how acetaldehyde, $\mathrm{CH}_{3} \mathrm{CHO}$, can act as a Lewis base.


Strategy -
A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid.


Solution -
Acetaldehyde

PROBLEM Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with 2-17 HCl , and show how the species in part (b) can act as Lewis acids in their reaction with $\mathrm{OH}^{-}$.
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{HN}\left(\mathrm{CH}_{3}\right)_{2}, \mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}$
(b) $\mathrm{H}_{3} \mathrm{C}^{+}, \mathrm{B}\left(\mathrm{CH}_{3}\right)_{3}, \mathrm{MgBr}_{2}$

PROBLEM Imidazole, which forms part of amino acid histidine, can act as both an acid and a base.
2-18

Imidazole

Histidine
(a) Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom.
(b) Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base.


Solution - 2.12 Noncovalent Interactions between Molecules
When thinking about chemical reactivity, chemists usually focus their attention on bonds, the covalent interactions between atoms within molecules. Also important, however, particularly in large biomolecules like proteins and nucleic acids, are a variety of interactions between molecules that strongly affect molecular properties. Collectively called either intermolecular forces, van der Waals forces, or noncovalent interactions, they are of several different types: dipole-dipole forces, dispersion forces, and hydrogen bonds.

Dipole-dipole forces occur between polar molecules as a result of electrostatic interactions among dipoles. The forces can be either attractive or repulsive depending on the orientation of the molecules-attractive when unlike charges are together and repulsive when like charges are together. The attractive geometry is lower in energy and therefore predominates (FIGURE 2.7).

FIGURE 2.7 Dipole-dipole forces cause polar molecules (a) to attract one another when they orient with unlike charges together, but (b) to repel one another when they orient with like charges together.

Dispersion forces occur between all neighboring molecules and arise because the electron distribution within molecules is constantly changing. Although uniform on a time-averaged basis, the electron distribution even in nonpolar molecules is likely to be nonuniform at any given instant. One side of a molecule may, by chance, have a slight excess of electrons relative to the opposite side, giving the molecule a temporary dipole. This temporary dipole in one molecule causes a nearby molecule to adopt a temporarily opposite dipole, resulting in a tiny attraction between the two (FIGURE 2.8). Temporary molecular dipoles have only a fleeting existence and are constantly changing, but their cumulative effect is often strong enough to hold molecules close together so that a substance is a liquid or solid rather than a gas.

FIGURE 2.8 Attractive dispersion forces in nonpolar molecules are caused by temporary dipoles, as shown in these models of pentane, $\mathrm{C}_{5} \mathrm{H}_{12}$.


Solution - 2.12 Noncovalent Interactions between Molecules
FIGURE 2.8 Attractive dispersion forces in nonpolar molecules are caused by temporary dipoles, as shown in these models of pentane, $\mathrm{C}_{5} \mathrm{H}_{12}$.

Perhaps the most important noncovalent interaction in biological molecules is the hydrogen bond, an attractive interaction between a hydrogen atom bonded to an electronegative O or N atom and an unshared electron pair on another O or N atom. In essence, a hydrogen bond is a very strong dipole-dipole interaction involving polarized $\mathrm{O}-\mathrm{H}$ or $\mathrm{N}-\mathrm{H}$ bonds. Electrostatic potential maps of water and ammonia clearly show the positively polarized hydrogens (blue) and the negatively polarized oxygens and nitrogens (red).

Hydrogen bonding has enormous consequences for living organisms. Hydrogen bonds cause water to be a liquid rather than a gas at ordinary temperatures, they hold enzymes in the shapes necessary for catalyzing biological reactions, and they cause strands of deoxyribonucleic acid (DNA) to pair up and coil into the double helix that stores genetic information.


A deoxyribonucleic acid segment -
One further point before leaving the subject of noncovalent interactions: biochemists frequently use the term hydrophilic, meaning "water-loving," to describe a substance that is attracted to water and the term hydrophobic, meaning "water-fearing," to describe a substance that is not strongly attracted to water.

Hydrophilic substances, such as table sugar, often have a number of -OH groups in their structure so they can form hydrogen bonds and dissolve in water, whereas hydrophobic substances, such as vegetable oil, do not have groups that form hydrogen bonds and do not dissolve in water.

PROBLEM Of the two vitamins A and C, one is hydrophilic and water-soluble while the other is hydrophobic 2-19 and fat-soluble. Which is which?

Vitamin A
(retinol)

Vitamin C (ascorbic acid)


Alkaloids: From Cocaine to Dental Anesthetics -
Just as ammonia $\left(\mathrm{NH}_{3}\right)$ is a weak base, there are a large number of nitrogen-containing organic compounds called amines that are also weak bases. In the early days of organic chemistry, basic amines derived from natural sources were known as vegetable alkali, but they are now called alkaloids. More than 20,000 alkaloids are known. Their study provided much of the impetus for the growth of organic chemistry in the nineteenth century and remains today an active and fascinating area of research.

FIGURE 2.9 The coca bush Erythroxylon coca, native to upland rain forest areas of Colombia, Ecuador, Peru, Bolivia, and western Brazil, is the source of the alkaloid cocaine. (credit: "Erythroxylum coca" by Danna Guevara/Wikimedia Commons, CC BY 4.0)

Many alkaloids have pronounced biological properties, and approximately $50 \%$ of pharmaceutical agents used today are derived from naturally occurring amines. As just three examples, morphine, an analgesic agent (painkiller), is obtained from the opium poppy Papaver somniferum. Ephedrine, a bronchodilator, decongestant, and appetite suppressant, is obtained from Ephedra sinica, an evergreen shrub native to Mongolia and northeastern China. Cocaine, both an anesthetic and a stimulant, is obtained from the coca bush Erythroxylon coca, endemic to the upland rain forest areas of central South America. (And yes, there really was a small amount of cocaine in the original Coca-Cola recipe, although it was removed in 1906.)

Morphine

Ephedrine

Cocaine


Alkaloids: From Cocaine to Dental Anesthetics -
Morphine

Ephedrine

Cocaine

Cocaine itself is rarely used medically because it is too addictive, but its anesthetic properties provoked a long search for related but nonaddictive compounds. This search ultimately resulted in the synthesis of the "caine" anesthetics that are commonly used today in dental and surgical anesthesia. Procaine, the first such compound, was synthesized in 1898 and marketed under the name Novocain. It was rapidly adopted and remains in use today as a topical anesthetic. Other related compounds with different activity profiles followed: Lidocaine, marketed as Xylocaine, was introduced in 1943, and mepivacaine (Carbocaine) in the early 1960s. More recently, bupivacaine (Marcaine) and prilocaine (Citanest) have gained popularity. Both are quick-acting, but the effects of bupivacaine last for 3 to 6 hours while those of prilocaine fade after 45 minutes. Notice some structural similarity of all the caines to cocaine itself.

Procaine
(Novocain)

Mepivacaine (Carbocaine)

Bupivacaine (Marcaine)

Lidocaine
(Xylocaine)

Prilocaine
(Citanest)
An estimate from the U.S. National Academy of Sciences is that less than $1 \%$ of all living species have been characterized. Thus, alkaloid chemistry remains an active area of research, and innumerable substances with potentially useful properties have yet to be discovered. Undoubtedly even the caine anesthetics will become obsolete at some point, perhaps supplanted by newly discovered alkaloids.


Key Terms -
- acidity constant $\left(K_{\mathrm{a}}\right)$
- alkaloid
- Brønsted-Lowry acid
- Brønsted-Lowry base
- conjugate acid
- conjugate base
- dipole moment ( $\mu$ )
- dispersion force
- electronegativity (EN)
- electrostatic potential map
- formal charge
- hydrogen bond
- hydrophilic
- hydrophobic
- inductive effect
- intermolecular force
- Lewis acid
- Lewis base
- noncovalent interaction
- $\mathrm{p} K_{\mathrm{a}}$
- polar covalent bond
- resonance form
- resonance hybrid
- van der Waals force


Summary -
Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we've reviewed some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation for understanding the specific reactions that will be discussed in subsequent chapters.

Organic molecules often have polar covalent bonds as a result of unsymmetrical electron sharing caused by differences in the electronegativity of atoms. A carbon-oxygen bond is polar, for example, because oxygen attracts the shared electrons more strongly than carbon does. Carbon-hydrogen bonds are relatively nonpolar. Many molecules as a whole are also polar, owing to the presence of individual polar bonds and electron lone pairs. The polarity of a molecule is measured by its dipole moment, $\boldsymbol{\mu}$.

Plus (+) and minus (-) signs are often used to indicate the presence of formal charges on atoms in molecules. Assigning formal charges to specific atoms is a bookkeeping technique that makes it possible to keep track of the valence electrons around an atom and offers some clues about chemical reactivity.


Summary -
Some substances, such as acetate ion and benzene, can't be represented by a single line-bond structure and must be considered as a resonance hybrid of two or more structures, none of which would be correct by themselves. The only difference between two resonance forms is in the location of their $\pi$ and nonbonding electrons. The nuclei remain in the same places in both structures, and the hybridization of the atoms remains
the same.
Acidity and basicity are closely related to the ideas of polarity and electronegativity. A Brønsted-Lowry acid is a compound that can donate a proton (hydrogen ion, $\mathrm{H}^{+}$), and a Brønsted-Lowry base is a compound that can accept a proton. The strength of a Brønsted-Lowry acid or base is expressed by its acidity constant, $\boldsymbol{K}_{\mathbf{a}}$, or by the negative logarithm of the acidity constant, $\mathbf{p} \boldsymbol{K}_{\mathbf{a}}$. The larger the $\mathrm{p} K_{\mathrm{a}}$, the weaker the acid. More useful is the Lewis definition of acids and bases. A Lewis acid is a compound that has a low-energy empty orbital that can accept an electron pair; $\mathrm{Mg}^{2+}, \mathrm{BF}_{3}, \mathrm{AlCl}_{3}$, and $\mathrm{H}^{+}$are examples. A Lewis base is a compound that can donate an unshared electron pair; $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ are examples. Most organic molecules that contain oxygen and nitrogen can act as Lewis bases toward sufficiently strong acids.

A variety of noncovalent interactions have a significant effect on the properties of large biomolecules. Hydrogen bonding-the attractive interaction between a positively polarized hydrogen atom bonded to an oxygen or nitrogen atom with an unshared electron pair on another O or N atom, is particularly important in giving proteins and nucleic acids their shapes.


Visualizing Chemistry -
PROBLEM Fill in the multiple bonds in the following model of naphthalene, $\mathrm{C}_{10} \mathrm{H}_{8}$ (black $=\mathrm{C}$, gray $=\mathrm{H}$ ). How 2-20 many resonance structures does naphthalene have? Draw them.

PROBLEM The following model is a representation of ibuprofen, a common over-the-counter pain reliever.
2-21 Indicate the positions of the multiple bonds, and draw a skeletal structure (black $=\mathrm{C}$, red = O, gray $=\mathrm{H})$.

PROBLEM cis-1,2-Dichloroethylene and trans-1,2-dichloroethylene are isomers, compounds with the same
$\mathbf{2 - 2 2}$ formula but different chemical structures. Look at the following electrostatic potential maps, and tell whether either compound has a dipole moment.

cis-1,2-Dichloroethylene

trans-1,2-Dichloroethylene

PROBLEM The following molecular models are representations of (a) adenine and (b) cytosine, constituents of
2-23 DNA (deoxyribonucleic acid). Indicate the positions of multiple bonds and lone pairs for both, and draw skeletal structures (black = C, red = O, blue = N, gray = H).
(a)

Adenine
(b)

Cytosine


Mechanism Problems -
PROBLEM Predict the product(s) of the following acid/base reactions. Draw curved arrows to show the 2-24 formation and breaking of bonds.
(a)

(b)

(c)

PROBLEM Use curved arrows to draw the protonated form of the following Lewis bases.
2-25 (a)

(b)

(c)

(d)

PROBLEM Use the curved-arrow formalism to show how the electrons flow in the resonance form on the left to 2-26 give the one on the right.

(a)

(b)

c)

PROBLEM Double bonds can also act like Lewis bases, sharing their electrons with Lewis acids. Use curved
2-27 arrows to show how each of the following double bonds will react with HCl and draw the resulting
carbocation.
(a) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$
(b)

(c)


Electronegativity and Dipole Moments -
PROBLEM Identify the most electronegative element in each of the following molecules:
2-28 (a) $\mathrm{CH}_{2} \mathrm{FCl}$
(b) $\mathrm{FCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$
(c) $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(d) $\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{Li}$

PROBLEM Use the electronegativity table given in Figure 2.3 to predict which bond in each of the following
2-29 pairs is more polar, and indicate the direction of bond polarity for each compound.
(a) $\mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}$ or $\mathrm{Cl}-\mathrm{Cl}$
(b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}$ or $\mathrm{H}-\mathrm{Cl}$
(c) $\mathrm{HO}-\mathrm{CH}_{3}$ or $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Si}-\mathrm{CH}_{3}$
(d) $\mathrm{H}_{3} \mathrm{C}-\mathrm{Li}$ or $\mathrm{Li}-\mathrm{OH}$

PROBLEM Which of the following molecules has a dipole moment? Indicate the expected direction of each.
2-30 (a)

(b)

(c)

(d)

PROBLEM (a) The $\mathrm{H}-\mathrm{Cl}$ bond length is 136 pm . What would the dipole moment of HCl be if the molecule
2-31 were $100 \%$ ionic, $\mathrm{H}^{+} \mathrm{Cl}^{-}$?
(b) The actual dipole moment of HCl is 1.08 D . What is the percent ionic character of the $\mathrm{H}-\mathrm{Cl}$ bond?


Electronegativity and Dipole Moments -
PROBLEM Phosgene, $\mathrm{Cl}_{2} \mathrm{C}=\mathrm{O}$, has a smaller dipole moment than formaldehyde, $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$, even though it
2-32 contains electronegative chlorine atoms in place of hydrogen. Explain.
PROBLEM Fluoromethane $\left(\mathrm{CH}_{3} \mathrm{~F}, \mu=1.81 \mathrm{D}\right)$ has a smaller dipole moment than chloromethane $\left(\mathrm{CH}_{3} \mathrm{Cl}, \mu=\right.$
2-33 1.87 D ) even though fluorine is more electronegative than chlorine. Explain.
PROBLEM Methanethiol, $\mathrm{CH}_{3} \mathrm{SH}$, has a substantial dipole moment ( $\mu=1.52$ ) even though carbon and sulfur
2-34 have identical electronegativities. Explain.


Formal Charges -
PROBLEM Calculate the formal charges on the atoms shown in red.
2-35 (a) ${ }_{\left(\mathrm{CH}_{3}\right)_{2}} \mathrm{OO}_{\mathrm{BF}}^{3}$
(b) $\mathrm{H}_{2} \ddot{\mathrm{C}}-\mathrm{N} \equiv \mathrm{N}$ :
(c) $\mathrm{H}_{2} \mathrm{C}=\mathrm{N}=\ddot{\mathrm{N}}$ :
(d) $: \ddot{0}=\ddot{0}-\ddot{0}:$

(f)

PROBLEM Assign formal charges to the atoms in each of the following molecules:
2-36 (a)

(b) $\mathrm{H}_{3} \mathrm{C}-\ddot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{N}$ :
(c) $\mathrm{H}_{3} \mathrm{C}-\ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{N}}$ :


Resonance -
PROBLEM Which of the following pairs of structures represent resonance forms?
2-37
(a)

and

(b)

(c)

(d)

PROBLEM Draw as many resonance structures as you can for the following species:
2-38 (a)

(b)

(c)

(e)

PROBLEM 1,3-Cyclobutadiene is a rectangular molecule with two shorter double bonds and two longer single 2-39 bonds. Why do the following structures not represent resonance forms?

Acids and Bases
PROBLEM Alcohols can act either as weak acids or as weak bases, just as water can. Show the reaction of
2-40 methanol, $\mathrm{CH}_{3} \mathrm{OH}$, with a strong acid such as HCl and with a strong base such as $\mathrm{Na}^{+}{ }^{-} \mathrm{NH}_{2}$
PROBLEM The $\mathrm{O}-\mathrm{H}$ hydrogen in acetic acid is more acidic than any of the $\mathrm{C}-\mathrm{H}$ hydrogens. Explain this result
2-41 using resonance structures.

Acetic acid

PROBLEM Draw electron-dot structures for the following molecules, indicating any unshared electron pairs.
2-42 Which of the compounds are likely to act as Lewis acids and which as Lewis bases?
(a) $\mathrm{AlBr}_{3}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(c) $\mathrm{BH}_{3}$
(d) HF
(e) $\mathrm{CH}_{3} \mathrm{SCH}_{3}$
(f) $\mathrm{TiCl}_{4}$

PROBLEM Write the products of the following acid-base reactions:
2-43 (a) $\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftarrows$ ? (b) $\mathrm{CH}_{3} \mathrm{OH}+\mathrm{NaNH}_{2} \rightleftarrows$ ?
(c) $\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}+\mathrm{NaOH} \rightleftarrows$ ?

PROBLEM Rank the following substances in order of increasing acidity:
2-44


Resonance -
PROBLEM Rank the following substances in order of increasing acidity:
2-44

Acetone ( $\mathrm{p} K_{\mathrm{a}}=19.3$ )

2,4-Pentanedione
$\left(\mathrm{p} K_{\mathrm{a}}=9\right)$

Phenol
$\left(p K_{a}=9.9\right)$

Acetic acid
$\left(p K_{a}=4.76\right)$

PROBLEM Which, if any, of the substances in Problem 2-44 is a strong enough acid to react almost completely
2-45 with NaOH ? (The $\mathrm{p} K_{\mathrm{a}}$ of $\mathrm{H}_{2} \mathrm{O}$ is 15.74.)
PROBLEM The ammonium ion $\left(\mathrm{NH}_{4}{ }^{+}, \mathrm{p} K_{\mathrm{a}}=9.25\right)$ has a lower $\mathrm{p} K_{\mathrm{a}}$ than the methylammonium ion $\left(\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right.$,
2-46 $\mathrm{p} K_{\mathrm{a}}=10.66$ ). Which is the stronger base, ammonia $\left(\mathrm{NH}_{3}\right)$ or methylamine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$ ? Explain.
PROBLEM Is tert-butoxide anion a strong enough base to react significantly with water? In other words,
2-47 can a solution of potassium tert-butoxide be prepared in water? The $\mathrm{p} K_{\mathrm{a}}$ of tert-butyl alcohol is approximately 18.


Potassium tert-butoxide -
PROBLEM Predict the structure of the product formed in the reaction of the organic base pyridine with the
2-48 organic acid acetic acid, and use curved arrows to indicate the direction of electron flow.

Pyridine Acetic acid
PROBLEM Calculate $K_{\mathrm{a}}$ values from the following $\mathrm{p} K_{\mathrm{a}}$ 's:
2-49 (a) Acetone, $\mathrm{p} K_{\mathrm{a}}=19.3$ (b) Formic acid, $\mathrm{p} K_{\mathrm{a}}=3.75$
PROBLEM Calculate $\mathrm{p} K_{\mathrm{a}}$ values from the following $K_{\mathrm{a}}$ 's:
2-50 (a) Nitromethane, $K_{\mathrm{a}}=5.0 \times 10^{-11}$
(b) Acrylic acid, $K_{\mathrm{a}}=5.6 \times 10^{-5}$

PROBLEM What is the pH of a 0.050 M solution of formic acid, $\mathrm{p} K_{\mathrm{a}}=3.75$ ?
2-51
PROBLEM Sodium bicarbonate, $\mathrm{NaHCO}_{3}$, is the sodium salt of carbonic acid $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right), \mathrm{p} K_{\mathrm{a}}=6.37$. Which of the
2-52 substances shown in Problem 2-44 will react significantly with sodium bicarbonate?


General Problems -
PROBLEM Maleic acid has a dipole moment, but the closely related fumaric acid, a substance involved in the
2-53 citric acid cycle by which food molecules are metabolized, does not. Explain.

Maleic acid

Fumaric acid

PROBLEM Assume that you have two unlabeled bottles, one of which contains phenol ( $\mathrm{p} K_{\mathrm{a}}=9.9$ ) and one of
2-54 which contains acetic acid ( $\mathrm{p} K_{\mathrm{a}}=4.76$ ). In light of your answer to Problem 2-52, suggest a simple way to determine what is in each bottle.

PROBLEM Identify the acids and bases in the following reactions:
2-55 (a)

(c)

(d)

PROBLEM Which of the following pairs represent resonance structures?
2-56
(a)

(b)

and

(c)

and

(d)

PROBLEM Draw as many resonance structures as you can for the following species, adding appropriate formal
2-57 charges to each:
(a) Nitromethane,

(b)
Ozone,

(c) Diazomethane, $\mathrm{H}_{2} \mathrm{C}=\stackrel{+}{\mathrm{N}}=\stackrel{\overline{\mathrm{N}}}{ }$ :

PROBLEM Carbocations, which contain a trivalent, positively charged carbon atom, react with water to give
2-58 alcohols:

How can you account for the fact that the following carbocation gives a mixture of two alcohols on reaction with water?

PROBLEM We'll see in the next chapter that organic molecules can be classified according to the functional
2-59 groups they contain, where a functional group is a collection of atoms with a characteristic chemical reactivity. Use the electronegativity values given in Figure 2.3 to predict the direction of polarization of the following functional groups.

Ketone
(b)

Alcohol
(c)

Amide
(d) $-\mathrm{C} \equiv \mathrm{N}$

Nitrile


General Problems -
Ketone
(b)

Alcohol
(c)

Amide
(d) $-\mathrm{C} \equiv \mathrm{N}$

Nitrile

PROBLEM The azide functional group, which occurs in azidobenzene, contains three adjacent nitrogen atoms.
2-60 One resonance structure for azidobenzene is shown. Draw three additional resonance structures, and assign appropriate formal charges to the atoms in all four.

Azidobenzene

PROBLEM Phenol, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$, is a stronger acid than methanol, $\mathrm{CH}_{3} \mathrm{OH}$, even though both contain an $\mathrm{O}-\mathrm{H}$ bond.
2-61 Draw the structures of the anions resulting from loss of $\mathrm{H}^{+}$from phenol and methanol, and use resonance structures to explain the difference in acidity.

Phenol $\left(\mathrm{p} K_{\mathrm{a}}=9.89\right) \quad$ Methanol $\left(\mathrm{p} K_{\mathrm{a}}=15.54\right)$
PROBLEM Thiamin diphosphate (TPP), a derivative of vitamin $B_{1}$ required for glucose metabolism, is a weak
2-62 acid that can be deprotonated by a base. Assign formal charges to the appropriate atoms in both TPP and its deprotonation product.

PROBLEM Which of the following compounds or ions have a dipole moment?
2-63
(a) Carbonate ion $\left(\mathrm{CO}_{3}{ }^{2-}\right)$
(b)

(c) $\oplus$
$\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}$

PROBLEM Use the $\mathrm{p} K_{\mathrm{a}}$ table in Appendix B to determine in which direction the equilibrium is favored.
2-64 (a)

(b)

(c)

$$
\overline{\mathrm{C}} \mathrm{H}_{3}+\mathrm{CH}_{3} \mathrm{NO}_{2} \rightleftarrows \mathrm{CH}_{4}+\overline{\mathrm{C}} \mathrm{H}_{2} \mathrm{NO}_{2}
$$


General Problems -
PROBLEM Which intermolecular force is predominantly responsible for each observation below?
$\mathbf{2 - 6 5}$ (a) $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{29} \mathrm{CH}_{3}$, a component found in paraffin wax, is a solid at room temperature while $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{3}$ is a liquid.
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$ has a higher boiling point than $\mathrm{CH}_{4}$.
(c) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$, which is found in vinegar, will dissolve in water but not in oil. Assume that oil is $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}$.
$702 \cdot$ Additional Problems


CHAPTER 3 -
Organic Compounds: Alkanes and Their Stereochemistry

FIGURE 3.1 The bristlecone pine is the oldest living organism on Earth. The waxy coating on its needles contains a mixture of organic compounds called alkanes, the subject of this chapter. (credit: modification of work "Gnarly Bristlecone Pine" by Rick Goldwaser/Flickr, CC BY 2.0)


CHAPTER CONTENTS - 3.1 Functional Groups
3.2 Alkanes and Alkane Isomers
3.3 Alkyl Groups
3.4 Naming Alkanes
3.5 Properties of Alkanes
3.6 Conformations of Ethane
3.7 Conformations of Other Alkanes

WHY THIS CHAPTER? The group of organic compounds called alkanes are simple and relatively unreactive, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we'll use alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules, a topic of particular importance in understanding biological organic chemistry.

According to Chemical Abstracts, the publication that abstracts and indexes the chemical literature, there are more than 195 million known organic compounds. Each of these compounds has its own physical properties, such as melting point and boiling point, and each has its own chemical reactivity.

Chemists have learned through years of experience that organic compounds can be classified into families according to their structural features and that the members of a given family have similar chemical behavior. Instead of 195 million compounds with random reactivity, there are a few dozen families of organic compounds whose chemistry is reasonably predictable. We'll study the chemistry of specific families throughout much of this book, beginning in this chapter with a look at the simplest family, the alkanes.


CHAPTER CONTENTS - 3.1 Functional Groups
The structural features that make it possible to classify compounds into families are called functional groups. A functional group is a group of atoms within a molecule that has a characteristic chemical behavior. Chemically, a given functional group behaves in nearly the same way in every molecule it's a part of. For example, compare ethylene, a plant hormone that causes fruit to ripen, with menthene, a much more complicated molecule found in peppermint oil. Both substances contain a carbon-carbon double-bond functional group, and both therefore react with $\mathrm{Br}_{2}$ in the same way to give a product in which a Br atom has added to each of the double-bond carbons (FIGURE 3.2). This example is typical: the chemistry of every organic molecule, regardless of size and complexity, is determined by the functional groups it contains.

Ethylene

Menthene

FIGURE 3.2 The reactions of ethylene and menthene with bromine. In both molecules, the carbon-carbon double-bond functional group has a similar polarity pattern, so both molecules react with $\mathrm{Br}_{2}$ in the same way. The size and complexity of the molecules are not important.

Look at TABLE 3.1, which lists many of the common functional groups and gives simple examples of their occurrence. Some functional groups have only carbon-carbon double or triple bonds; others have halogen atoms; and still others contain oxygen, nitrogen, or sulfur. Much of the chemistry you'll be studying is the chemistry of these functional groups.


Functional Groups with Carbon-Carbon Multiple Bonds -
Alkenes, alkynes, and arenes (aromatic compounds) all contain carbon-carbon multiple bonds. Alkenes have a double bond, alkynes have a triple bond, and arenes have alternating double and single bonds in a sixmembered ring of carbon atoms. They look different, but because of their structural similarities, they also have chemical similarities.

TABLE 3.1 Structures of Some Common Functional Groups


Functional Groups with Carbon-Carbon Multiple Bonds -
| Name | Structure* | Name ending | Example |
| :---: | :---: | :---: | :---: |
| Amine | <smiles>CN(C)C(C)(C)C</smiles> | -amine | $\mathrm{CH}_{3} \mathrm{NH}_{2}$
 Methylamine |
| Imine (Schiff base) | <smiles>CCC(=NC)C(C)(C)C</smiles> | None | <smiles>C=NC</smiles>
 Acetone imine |
| Nitrile | -C $\equiv \mathrm{N}$ | -nitrile | $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{~N}$
 Ethanenitrile |
| Thiol | <smiles>CC(C)(C)S</smiles> | -thiol | $\mathrm{CH}_{3} \mathrm{SH}$
 Methanethiol |
| Sulfide | <smiles></smiles> | sulfide | $\mathrm{CH}_{3} \mathrm{SCH}_{3}$
 Dimethyl sulfide |
| Disulfide | <smiles>CSC(C)(C)C</smiles> | disulfide | $\mathrm{CH}_{3} \mathrm{SSCH}_{3}$
 Dimethyl disulfide |
| Sulfoxide | <smiles>CCS+C(C)(C)C</smiles> | sulfoxide | <smiles>C[SeH]=[OH+]</smiles>
 Dimethyl sulfoxide |
| Aldehyde | <smiles>CC=O</smiles> | -al | Ethanal |
| Ketone | <smiles>CCC(=O)C(C)(C)C</smiles> | -one | <smiles>C=C=O</smiles>
 Propanone |
| Carboxylic acid | <smiles>CC(C)(C)C(=O)O</smiles> | -oic acid | <smiles>C[14CH2]O</smiles>
 Ethanoic acid |


Functional Groups with Carbon-Carbon Multiple Bonds -
| Ester | <smiles>CC(C)(C)C(=O)O</smiles> | -oate | <smiles>CC(C)=O</smiles>
 Methyl ethanoate |
| Thioester | <smiles>CC(C)(C)C(=O)S</smiles> | -thioate | <smiles>C[AsH]=O</smiles>
 Methyl ethanethioate |


Functional Groups with Carbon-Carbon Multiple Bonds -
TABLE 3.1 Structures of Some Common Functional Groups

| Name | Structure* | Name ending | Example |
| :---: | :---: | :---: | :---: |
| Amide | <smiles>CN(C)C(=O)C(C)(C)C</smiles> | -amide | <smiles>CC(N)=O</smiles>
 Ethanamide |
| Acid chloride | <smiles>CC(C)(C)C(=O)Cl</smiles> | -oyl chloride | <smiles>CC(=O)Cl</smiles>
 Ethanoyl chloride |
| Carboxylic acid anhydride | <smiles>CC(C)(C)C(=O)OC(=O)C(C)(C)C</smiles> | -oic anhydride | <smiles>COC(C)=O</smiles>
 Ethanoic anhydride |

*The bonds whose connections aren't specified are assumed to be attached to carbon or hydrogen atoms in the rest of the molecule.


Functional Groups with Carbon Singly Bonded to an Electronegative Atom -
Alkyl halides (haloalkanes), alcohols, ethers, alkyl phosphates, amines, thiols, sulfides, and disulfides all have a carbon atom singly bonded to an electronegative atom-halogen, oxygen, nitrogen, or sulfur. Alkyl halides have a carbon atom bonded to halogen ( -X ), alcohols have a carbon atom bonded to the oxygen of a hydroxyl group ( -OH ), ethers have two carbon atoms bonded to the same oxygen, organophosphates have a carbon atom bonded to the oxygen of a phosphate group $\left(-\mathrm{OPO}_{3}{ }^{2-}\right)$, amines have a carbon atom bonded to a nitrogen, thiols have a carbon atom bonded to the sulfur of an -SH group, sulfides have two carbon atoms bonded to the same sulfur, and disulfides have carbon atoms bonded to two sulfurs that are joined together. In all cases, the bonds are polar, with the carbon atom bearing a partial positive charge ( $\delta+$ ) and the electronegative atom bearing a partial negative charge ( $\delta-$ ).

Alkyl halide (haloalkane)

Amine

Alcohol

Thiol

Ether

Sulfide

Phosphate

Disulfide


Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) -
The carbonyl group, $\mathrm{C}=\mathrm{O}$ (pronounced car-bo-neel) is common to many of the families listed in TABLE 3.1. Carbonyl groups are present in a majority of organic compounds and in practically all biological molecules. These compounds therefore behave similarly in many respects but differ depending on the identity of the other atoms bonded to the carbonyl-group carbon. Aldehydes have at least one hydrogen bonded to the $\mathrm{C}=\mathrm{O}$, ketones have two carbons bonded to the $\mathrm{C}=\mathrm{O}$, carboxylic acids have an -OH group bonded to the $\mathrm{C}=\mathrm{O}$, esters have an ether-like oxygen bonded to the $\mathrm{C}=\mathrm{O}$, thioesters have a sulfide-like sulfur bonded to the $\mathrm{C}=\mathrm{O}$, amides have an amine-like nitrogen bonded to the $\mathrm{C}=\mathrm{O}$, acid chlorides have a chlorine bonded to the $\mathrm{C}=\mathrm{O}$, and so on. In all these functional groups, the carbonyl carbon atom bears a partial positive charge ( $\delta+$ ), and the oxygen bears a partial negative charge ( $\delta$-).

Acetone-a typical carbonyl compound

Aldehyde

Ketone

Carboxylic acid

Ester

Thioester

Amide

Acid chloride

PROBLEM Use Table 3.1 to identify the functional groups in each of the following molecules:
3-1 (a)
Methionine, an amino acid:
(b) Ibuprofen, a pain reliever:

(c) Capsaicin, the pungent substance in chili peppers:

PROBLEM Propose structures for simple molecules that contain the following functional groups:
3-2 (a) Alcohol (b) Aromatic ring
(c) Carboxylic acid (d) Amine
(e) Both ketone and amine (f) Two double bonds


Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) -
PROBLEM Identify the functional groups in the following model of arecoline, a veterinary drug used to control 3-3 worms in animals. Convert the drawing into a line-bond structure and a molecular formula (red = O , blue $=\mathrm{N}$, black $=\mathrm{C}$, gray $=\mathrm{H}$ ) .


Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers
Before beginning a systematic study of the different functional groups, let's look first at the simplest family of molecules to develop some general ideas that apply to all families. We saw in Section 1.7 that the carbon-carbon single bond in ethane results from $\sigma$ (head-on) overlap of carbon $s p^{3}$ hybrid orbitals. If we imagine joining three, four, five, or even more carbon atoms by $\mathrm{C}-\mathrm{C}$ single bonds, we can generate the large family of molecules called alkanes.

Methane

Ethane

Propane

Butane

Alkanes are often described as saturated hydrocarbons: hydrocarbons because they contain only carbon and hydrogen; saturated because they have only $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}-\mathrm{H}$ single bonds and thus contain the maximum possible number of hydrogens per carbon. They have the general formula $\mathrm{C}_{n} \mathrm{H}_{2 n+2}$, where $n$ is an integer. Alkanes are also occasionally called aliphatic compounds, a name derived from the Greek aleiphas, meaning "fat." We'll see in Section 27.1 that many animal fats contain long carbon chains similar to alkanes.

A typical animal fat
Think about the ways that carbon and hydrogen might combine to make alkanes. With one carbon and four hydrogens, only one structure is possible: methane, $\mathrm{CH}_{4}$. Similarly, there is only one combination of two carbons with six hydrogens (ethane, $\mathrm{CH}_{3} \mathrm{CH}_{3}$ ) and only one combination of three carbons with eight hydrogens (propane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ ). When larger numbers of carbons and hydrogens combine, however, more than one structure is possible. For example, there are two substances with the formula $\mathrm{C}_{4} \mathrm{H}_{10}$ : the four carbons can all be in a row (butane), or they can branch (isobutane). Similarly, there are three $\mathrm{C}_{5} \mathrm{H}_{12}$ molecules, and so on for larger alkanes.


Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers
Compounds like butane and pentane, whose carbons are all connected in a row, are called straight-chain alkanes, or normal alkanes. Compounds like 2-methylpropane (isobutane), 2-methylbutane, and 2,2-dimethylpropane, whose carbon chains branch, are called branched-chain alkanes.

Compounds like the two $\mathrm{C}_{4} \mathrm{H}_{10}$ molecules and the three $\mathrm{C}_{5} \mathrm{H}_{12}$ molecules, which have the same formula but different structures, are called Isomers, from the Greek isos + meros, meaning "made of the same parts." Isomers have the same numbers and kinds of atoms but differ in the way the atoms are arranged. Compounds like butane and isobutane, whose atoms are connected differently, are called constitutional isomers. We'll see shortly that other kinds of isomers are also possible, even among compounds whose atoms are connected in the same order. As TABLE 3.2 shows, the number of possible alkane isomers increases dramatically with the number of carbon atoms.

| TABLE 3.2 Number of Alkane Isomers |  |  |  |
| :--- | :--- | :--- | :--- |
| Formula | Number of isomers | Formula | Number of isomers |
| $\mathrm{C}_{6} \mathrm{H}_{14}$ | 5 | $\mathrm{C}_{10} \mathrm{H}_{22}$ | 75 |
| $\mathrm{C}_{7} \mathrm{H}_{16}$ | 9 | $\mathrm{C}_{15} \mathrm{H}_{32}$ | 4347 |
| $\mathrm{C}_{8} \mathrm{H}_{18}$ | 18 | $\mathrm{C}_{20} \mathrm{H}_{42}$ | 366,319 |
| $\mathrm{C}_{9} \mathrm{H}_{20}$ | 35 | $\mathrm{C}_{30} \mathrm{H}_{62}$ | $4,111,846,763$ |


Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers
Constitutional isomerism is not limited to alkanes-it occurs widely throughout organic chemistry. Constitutional isomers may have different carbon skeletons (as in isobutane and butane), different functional groups (as in ethanol and dimethyl ether), or different locations of a functional group along the chain (as in isopropylamine and propylamine). Regardless of the reason for the isomerism, constitutional isomers are always different compounds with different properties but with the same formula.


Different carbon skeletons $\mathrm{C}_{4} \mathrm{H}_{10}$ -
Different functional groups
$\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$

Different position of functional groups $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N}$

2-Methylpropane
(isobutane)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
Ethanol

Isopropylamine

$$
\text { and } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}
$$

Butane
$\mathrm{CH}_{3} \mathrm{OCH}_{3}$
Dimethyl ether
and

Propylamine

A given alkane can be drawn in many ways. For example, the straight-chain, four-carbon alkane called butane can be represented by any of the structures shown in FIGURE 3.3. These structures don't imply any particular three-dimensional geometry for butane; they indicate only the connections among atoms. In practice, as noted in Section 1.12, chemists rarely draw all the bonds in a molecule and usually refer to butane by the condensed structure, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ or $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}$. Still more simply, butane can be represented as $n-\mathrm{C}_{4} \mathrm{H}_{10}$, where $n$ denotes normal (straight-chain) butane.

$$
\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \quad \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}
$$


Different carbon skeletons $\mathrm{C}_{4} \mathrm{H}_{10}$ -
FIGURE 3.3 Some representations of butane, $\mathbf{C}_{\mathbf{4}} \mathbf{H}_{\mathbf{1 0}}$. The molecule is the same regardless of how it's drawn. These structures imply only that butane has a continuous chain of four carbon atoms; they do not imply any specific geometry.

Straight-chain alkanes are named according to the number of carbon atoms they contain, as shown in TABLE 3.3. With the exception of the first four compounds-methane, ethane, propane, and butane-whose names have historical roots, the alkanes are named based on Greek numbers. The suffix -ane is added to the end of each name to indicate that the molecule identified is an alkane. Thus, pentane is the five-carbon alkane, hexane is the six-carbon alkane, and so on. We'll soon see that these alkane names form the basis for naming all other organic compounds, so at least the first ten should be memorized.


TABLE 3.3 Names of Straight-Chain Alkanes -
| Number of carbons
 $(n)$ | Name | Formula
 $\left(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\right)$ | Number of carbons
 $(n)$ | Name | Formula
 $\left(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\right)$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | Methane | $\mathrm{CH}_{4}$ | 9 | Nonane | $\mathrm{C}_{9} \mathrm{H}_{20}$ |
| 2 | Ethane | $\mathrm{C}_{2} \mathrm{H}_{6}$ | 10 | Decane | $\mathrm{C}_{10} \mathrm{H}_{22}$ |
| 3 | Propane | $\mathrm{C}_{3} \mathrm{H}_{8}$ | 11 | Undecane | $\mathrm{C}_{11} \mathrm{H}_{24}$ |
| 4 | Butane | $\mathrm{C}_{4} \mathrm{H}_{10}$ | 12 | Dodecane | $\mathrm{C}_{12} \mathrm{H}_{26}$ |
| 5 | Pentane | $\mathrm{C}_{5} \mathrm{H}_{12}$ | 13 | Tridecane | $\mathrm{C}_{13} \mathrm{H}_{28}$ |

TABLE 3.3 Names of Straight-Chain Alkanes


TABLE 3.3 Names of Straight-Chain Alkanes -
TABLE 3.3 Names of Straight-Chain Alkanes

| Number of carbons
 $(n)$ | Name | Formula
 $\left(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\right)$ | Number of carbons
 $(n)$ | Name | Formula
 $\left(\mathrm{C}_{n} \mathrm{H}_{2 n+2}\right)$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | Hexane | $\mathrm{C}_{6} \mathrm{H}_{14}$ | 20 | Icosane | $\mathrm{C}_{20} \mathrm{H}_{42}$ |
| 7 | Heptane | $\mathrm{C}_{7} \mathrm{H}_{16}$ | 30 | Triacontane | $\mathrm{C}_{30} \mathrm{H}_{62}$ |
| 8 | Octane | $\mathrm{C}_{8} \mathrm{H}_{18}$ |  |  |  |


Drawing the Structures of Isomers -
Propose structures for two isomers with the formula $\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}$.


Strategy -
We know that carbon forms four bonds, nitrogen forms three, and hydrogen forms one. Write down the carbon atoms first, and then use trial and error plus intuition to put the pieces together.


Solution -
There are two isomeric structures. One has the connection $\mathrm{C}-\mathrm{C}-\mathrm{N}$, and the other has the connection $\mathrm{C}-\mathrm{N}-\mathrm{C}$.

PROBLEM Draw structures of the five isomers of $\mathrm{C}_{6} \mathrm{H}_{14}$.
3-4
PROBLEM Propose structures that meet the following descriptions:
3-5 (a) Two isomeric esters with the formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}$ :
(b) Two isomeric nitriles with the formula $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~N}$
(c) Two isomeric disulfides with the formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{~S}_{2}$

PROBLEM How many isomers are there with the following descriptions?
3-6 (a) Alcohols with the formula $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$ (b) Bromoalkanes with the formula $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}$
(c) Thioesters with the formula $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{OS}$


Solution - 3.3 Alkyl Groups
If you imagine removing a hydrogen atom from an alkane, the partial structure that remains is called an alkyl group. Alkyl groups are not stable compounds themselves, they are simply parts of larger compounds and are named by replacing the -ane ending of the parent alkane with an $-y l$ ending. For example, removal of a hydrogen from methane, $\mathrm{CH}_{4}$, generates a methyl group, $-\mathrm{CH}_{3}$, and removal of a hydrogen from ethane, $\mathrm{CH}_{3} \mathrm{CH}_{3}$, generates an ethyl group, $-\mathrm{CH}_{2} \mathrm{CH}_{3}$. Similarly, removal of a hydrogen atom from the end carbon of any straight-chain alkane gives the series of straight-chain alkyl groups shown in TABLE 3.4. Combining an alkyl group with any of the functional groups listed earlier makes it possible to generate and name many thousands of compounds. For example:

Methane

A methyl group

Methyl alcohol (methanol)

Methylamine
TABLE 3.4 Some Straight-Chain Alkyl Groups


Solution - 3.3 Alkyl Groups
Methane

A methyl group

Methyl alcohol (methanol)

Methylamine
TABLE 3.4 Some Straight-Chain Alkyl Groups

| Alkane |  | Name | Alkyl group |
| :--- | :--- | :--- | :--- |
| $\mathrm{CH}_{4}$ | Methane | $-\mathrm{CH}_{3}$ | Mame (abbreviation) |
| $\mathrm{CH}_{3} \mathrm{CH}_{3}$ | Ethane (Me) | $-\mathrm{CH}_{2} \mathrm{CH}_{3}$ | Ethyl (Et) |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Propane | $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Propyl (Pr) |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Butane | $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Butyl (Bu) |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Pentane | $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | Pentyl, or amyl |

Just as straight-chain alkyl groups are generated by removing a hydrogen from an end carbon, branched alkyl groups are generated by removing a hydrogen atom from an internal carbon. Two 3-carbon alkyl groups and four 4-carbon alkyl groups are possible (FIGURE 3.4).

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \frac{?}{2}$
Propyl

Isopropyl

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
Butane


Solution - 3.3 Alkyl Groups
Isopropyl

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
Butane

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}$ 之
Butyl

Isobutyl

~

sec-Butyl

tert-Butyl

FIGURE 3.4 Alkyl groups generated from straight-chain alkanes.
One further comment about naming alkyl groups: the prefixes sec- (for secondary) and tert- (for tertiary) used for the $\mathrm{C}_{4}$ alkyl groups in FIGURE 3.4 refer to the number of other carbon atoms attached to the branching carbon atom. There are four possibilities: primary $\left(1^{\circ}\right)$, secondary $\left(2^{\circ}\right)$, tertiary $\left(3^{\circ}\right)$, and quaternary $\left(4^{\circ}\right)$.

Primary carbon ( $1^{\circ}$ )
is bonded to one other carbon.

Secondary carbon ( $\mathbf{2}^{\circ}$ ) is bonded to two other carbons.

Tertiary carbon ( $3^{\circ}$ ) is bonded to three other carbons.


Quaternary carbon (40) <br> is bonded to four -
other carbons.The symbol $\mathbf{R}$ is used here and throughout organic chemistry to represent a generalized organic group. The $R$ group can be methyl, ethyl, propyl, or any of a multitude of others. You might think of $\mathbf{R}$ as representing the Rest of the molecule, which isn't specified.

The terms primary, secondary, tertiary, and quaternary are routinely used in organic chemistry, and their meanings need to become second nature. For example, if we were to say, "Citric acid is a tertiary alcohol," we would mean that it has an alcohol functional group $(-\mathrm{OH})$ bonded to a carbon atom that is itself bonded to three other carbons.

General class of tertiary alcohols, $\mathrm{R}_{3} \mathrm{COH}$

Citric acid-a specific tertiary alcohol

In addition to speaking of carbon atoms as being primary, secondary, or tertiary, we speak of hydrogens in the same way. Primary hydrogen atoms are attached to primary carbons $\left(\mathrm{RCH}_{3}\right)$, secondary hydrogens are attached to secondary carbons $\left(\mathrm{R}_{2} \mathrm{CH}_{2}\right)$, and tertiary hydrogens are attached to tertiary carbons $\left(\mathrm{R}_{3} \mathrm{CH}\right)$. There is, however, no such thing as a quaternary hydrogen. (Why not?)

PROBLEM Draw the eight 5-carbon alkyl groups (pentyl isomers). 3-7

PROBLEM Identify the carbon atoms in the following molecules as primary, secondary, tertiary, or quaternary:
3-8 (a)

(b)

(c)

PROBLEM Identify the hydrogen atoms on the compounds shown in Problem 3-8 as primary, secondary, or 3-9 tertiary.

PROBLEM Draw structures of alkanes that meet the following descriptions:
3-10 (a) An alkane with two tertiary carbons (b) An alkane that contains an isopropyl group
(c) An alkane that has one quaternary and one secondary carbon


Quaternary carbon (40) <br> is bonded to four - 3.4 Naming Alkanes
In earlier times, when relatively few pure organic chemicals were known, new compounds were named at the whim of their discoverer. Thus, urea $\left(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\right)$ is a crystalline substance isolated from urine; morphine $\left(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\right)$ is an analgesic (painkiller) named after Morpheus, the Greek god of dreams; and acetic acid, the primary organic constituent of vinegar, is named from the Latin word for vinegar, acetum.

As the science of organic chemistry slowly grew in the 19th century, so too did the number of known compounds and the need for a systematic method of naming them. The system of naming (nomenclature) we'll use in this book is that devised by the International Union of Pure and Applied Chemistry (IUPAC, usually spoken as eye-you-pac).

A chemical name typically has four parts in the IUPAC system: parent, prefix, locant, and suffix. The parent name identifies the main part of the molecule and tells how many carbon atoms are in that part. Prefixes identify the various substituent groups attached to the parent. Locants give the positions of the attached substituents. And the suffix identifies the primary functional group attached to the parent.

As we cover new functional groups in later chapters, the applicable IUPAC rules of nomenclature will be given. In addition, Appendix $A$ at the back of this book gives an overall view of organic nomenclature and shows how compounds that contain more than one functional group are named. (If preferred, you can study that appendix now.) For the present, let's see how to name branched-chain alkanes and learn some general rules that are applicable to all compounds.

All but the most complex branched-chain alkanes can be named by following four steps. For a very few compounds, a fifth step is needed.


Identify the parent hydrocarbon. -
(a) Find the longest continuous chain of carbon atoms in the molecule, and use the name of that chain as the parent name. The longest chain may not always be apparent from the manner of writing; you may have to "turn corners."

(b) If two different chains of equal length are present, choose the one with the larger number of branch points as the parent.

Named as a hexane with two substituents

NOT
as a hexane with one substituent

STEP 2


Number the atoms in the longest chain. -
(a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain.

NOT

The first branch occurs at C3 in the proper system of numbering, not at C4.
(b) If there is branching an equal distance away from both ends of the parent chain, begin numbering at the end nearer the second branch point.

NOT

STEP 3
Identify and number the substituents.
(a) Assign a number to each substituent to locate its point of attachment to the parent chain.

(b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents.

Named as a hexane

Substituents:
On C2, $\mathrm{CH}_{3}$
On C4, CH3
On C4, $\mathrm{CH}_{2} \mathrm{CH}_{3}$

$$
\begin{aligned}
& \text { (2-methyl) } \\
& \text { (4-methyl) } \\
& \text { (4-ethyl) }
\end{aligned}
$$


Write the name as a single word. -
Use hyphens to separate the different prefixes, and use commas to separate numbers. If two or more different substituents are present, cite them in alphabetical order. If two or more identical substituents are present on the parent chain, use one of the multiplier prefixes di-, tri-, tetra-, and so forth, but don't use these prefixes for alphabetizing. Full names for some of the examples we have been using are as follows:

STEP 5


Name a branched substituent as though it were itself a compound. -
In some particularly complex cases, a fifth step is necessary. It occasionally happens that a substituent on the main chain is itself branched. In the following case, for instance, the substituent at C 6 is a three-carbon chain with a methyl group. To name the compound fully, the branched substituent must first be named.

Named as a 2,3,6trisubstituted decane

A 2-methylpropyl substituent

Number the branched substituent beginning at the point of its attachment to the main chain, and identify it-in this case, a 2-methylpropyl group. The substituent is treated as a whole and is alphabetized according to the first letter of its complete name, including any numerical prefix. It is set off in parentheses when naming the entire molecule.

2,3-Dimethyl-6-(2-methylpropyl)decane
As a further example:

5-(1,2-Dimethylpropyl)-2-methylnonane

A 1,2-dimethylpropyl group

For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic, common names, as noted earlier.

Neopentyl
tert-Pentyl, also called
tert-amyl $(t-a m y l)$
Isopentyl, also called


5-Carbon alkyl groups -
The common names of these simple alkyl groups are so well entrenched in the chemical literature that IUPAC rules make allowance for them. Thus, the following compound is properly named either 4-(1-methylethyl)heptane or 4-isopropylheptane. There's no choice but to memorize these common names; fortunately, there are only a few of them.

4-(1-Methylethyl)heptane or 4-Isopropylheptane
When writing an alkane name, the nonhyphenated prefix iso- is considered part of the alkyl-group name for alphabetizing purposes, but the hyphenated and italicized prefixes sec- and tert- are not. Thus, isopropyl and isobutyl are listed alphabetically under $i$, but sec-butyl and tert-butyl are listed under $b$.


Naming Alkanes -
What is the IUPAC name for the following alkane?


Strategy -
Find the longest continuous carbon chain in the molecule, and use that as the parent name. This molecule has a chain of eight carbons-octane-with two methyl substituents. (You have to turn corners to see it.) Numbering from the end nearer the first methyl substituent indicates that the methyls are at C2 and C6.


Converting a Chemical Name into a Structure -
Draw the structure of 3-isopropyl-2-methylhexane.


Strategy -
This is the reverse of Worked Example 3.2 and uses a reverse strategy. Look at the parent name (hexane), and draw its carbon structure.

$$
\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C} \text { Hexane }
$$

Next, find the substituents (3-isopropyl and 2-methyl), and place them on the proper carbons.

Finally, add hydrogens to complete the structure.
Solution

3-Isopropyl-2-methylhexane

PROBLEM Give IUPAC names for the following compounds:
3-11 (a) The three isomers of $\mathrm{C}_{5} \mathrm{H}_{12}$
(b)

(c)

(d)

PROBLEM Draw structures corresponding to the following IUPAC names:
3-12 (a) 3,4-Dimethylnonane (b) 3-Ethyl-4,4-dimethylheptane (c) 2,2-Dimethyl-4-propyloctane (d) 2,2,4-Trimethylpentane

PROBLEM Name the eight 5-carbon alkyl groups you drew in Problem 3-7.

PROBLEM Give the IUPAC name for the following hydrocarbon, and convert the drawing into a skeletal $\mathbf{3 - 1 4}$ structure.


Strategy - 3.5 Properties of Alkanes
Alkanes are sometimes referred to as paraffins, a word derived from the Latin parum affinis, meaning "little affinity." This term aptly describes their behavior, for alkanes show little chemical affinity for other substances and are chemically inert to most laboratory reagents. They are also relatively inert biologically and are not often involved in the chemistry of living organisms. Alkanes do, however, react with oxygen, halogens, and a few other substances under appropriate conditions.

Reaction with oxygen occurs during combustion in an engine or furnace when an alkane is used as a fuel. Carbon dioxide and water are formed as products, and a large amount of heat is released. For example, methane (natural gas) reacts with oxygen according to the equation

$$
\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+890 \mathrm{~kJ} / \mathrm{mol}(213 \mathrm{kcal} / \mathrm{mol})
$$

The reaction of an alkane with $\mathrm{Cl}_{2}$ occurs when a mixture of the two is irradiated with ultraviolet light (denoted $h v$, where $v$ is the Greek letter $n u$ ). Depending on the time allowed and the relative amounts of the two reactants, a sequential substitution of the alkane hydrogen atoms by chlorine occurs, leading to a mixture of chlorinated products. Methane, for instance, reacts with $\mathrm{Cl}_{2}$ to yield a mixture of $\mathrm{CH}_{3} \mathrm{Cl}_{1}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}$, and $\mathrm{CCl}_{4}$. We'll look at this reaction in more detail in Section 6.6.

Alkanes show regular increases in both boiling point and melting point as molecular weight increases (FIGURE 3.5), an effect due to the presence of weak dispersion forces between molecules (Section 2.12). Only when sufficient energy is applied to overcome these forces does the solid melt or liquid boil. As you might expect, dispersion forces increase as molecular size increases, accounting for the higher melting and boiling points of larger alkanes.


Strategy - 3.5 Properties of Alkanes
FIGURE 3.5 A plot of melting and boiling points versus number of carbon atoms for the C1-C14 straight-chain alkanes. There is a regular increase with molecular size.

Another effect seen in alkanes is that increased branching lowers an alkane's boiling point. Thus, pentane has no branches and boils at $36.1^{\circ} \mathrm{C}$, isopentane (2-methylbutane) has one branch and boils at $27.85{ }^{\circ} \mathrm{C}$, and neopentane (2,2-dimethylpropane) has two branches and boils at $9.5^{\circ} \mathrm{C}$. Similarly, octane boils at $125.7^{\circ} \mathrm{C}$, whereas isooctane ( $2,2,4$-trimethylpentane) boils at $99.3^{\circ} \mathrm{C}$. Branched-chain alkanes are lower-boiling because they are more nearly spherical than straight-chain alkanes, have smaller surface areas, and consequently have smaller dispersion forces.


Strategy - 3.6 Conformations of Ethane
Up until now, we've viewed molecules primarily in a two-dimensional way and have given little thought to any consequences that might arise from the spatial arrangement of atoms in molecules. Now it's time to add a third dimension to our study. Stereochemistry is the branch of chemistry concerned with the three-dimensional aspects of molecules. We'll see on many occasions in future chapters that the exact three-dimensional structure of a molecule is often crucial to determining its properties and biological behavior.

We know from Section 1.5 that $\sigma$ bonds are cylindrically symmetrical. In other words, the intersection of a plane cutting through a carbon-carbon single-bond orbital looks like a circle. Because of this cylindrical symmetry, rotation is possible around carbon-carbon bonds in open-chain molecules. In ethane, for instance, rotation around the $\mathrm{C}-\mathrm{C}$ bond occurs freely, constantly changing the spatial relationships between the hydrogens on one carbon and those on the other (FIGURE 3.6).

FIGURE 3.6 Rotation occurs around the carbon-carbon single bond in ethane because of $\boldsymbol{\sigma}$ bond cylindrical symmetry.
The different arrangements of atoms that result from bond rotation are called conformations, and molecules that have different arrangements are called conformational isomers, or conformers. Unlike constitutional isomers, however, different conformers often can't be isolated because they interconvert too rapidly.

Conformational isomers are represented in two ways, as shown in FIGURE 3.7. A sawhorse representation views the carbon-carbon bond from an oblique angle and indicates spatial orientation by showing all $\mathrm{C}-\mathrm{H}$ bonds. A Newman projection views the carbon-carbon bond directly end-on and represents the two carbon atoms by a circle. Bonds attached to the front carbon are represented by lines to the center of the circle, and bonds attached to the rear carbon are represented by lines to the edge of the circle.

Sawhorse representation

Newman
projection


Strategy - 3.6 Conformations of Ethane
Sawhorse representation

Newman
projection

FIGURE 3.7 A sawhorse representation and a Newman projection of ethane. The sawhorse representation views the molecule from an oblique angle, while the Newman projection views the molecule end-on. Note that the molecular model of the Newman projection appears at first to have six atoms attached to a single carbon. Actually, the front carbon, with three attached green atoms, is directly in front of the rear carbon, with three attached red atoms.

Despite what we've just said, we actually don't observe perfectly free rotation in ethane. Experiments show that there is a small $(12 \mathrm{~kJ} / \mathrm{mol} ; 2.9 \mathrm{kcal} / \mathrm{mol})$ barrier to rotation and that some conformations are more stable than others. The lowest-energy, most stable conformation is the one in which all six $\mathrm{C}-\mathrm{H}$ bonds are as far away from one another as possible-staggered when viewed end-on in a Newman projection. The highestenergy, least stable conformation is the one in which the six $\mathrm{C}-\mathrm{H}$ bonds are as close as possible-eclipsed in a Newman projection. At any given instant, about $99 \%$ of ethane molecules have an approximately staggered conformation and only about $1 \%$ are near the eclipsed conformation.


Strategy - 3.6 Conformations of Ethane
The extra $12 \mathrm{~kJ} / \mathrm{mol}$ of energy present in the eclipsed conformation of ethane is called torsional strain. Its cause has been the subject of controversy, but the major factor is an interaction between $\mathrm{C}-\mathrm{H}$ bonding orbitals on one carbon and antibonding orbitals on the adjacent carbon, which stabilizes the staggered conformation relative to the eclipsed one. Because a total strain of $12 \mathrm{~kJ} / \mathrm{mol}$ arises from three equal hydrogen-hydrogen eclipsing interactions, we can assign a value of approximately $4.0 \mathrm{~kJ} / \mathrm{mol}(1.0 \mathrm{kcal} / \mathrm{mol})$ to each single interaction. The barrier to rotation that results can be represented on a graph of potential energy versus degree of rotation, in which the angle between $\mathrm{C}-\mathrm{H}$ bonds on the front and back carbons as viewed end-on (the dihedral angle) goes full circle from 0 to $360^{\circ}$. Energy minima occur at staggered conformations, and energy maxima occur at eclipsed conformations, as shown in FIGURE 3.8.

FIGURE 3.8 A graph of potential energy versus bond rotation in ethane. The staggered conformations are $12 \mathrm{~kJ} / \mathrm{mol}$ lower in energy than the eclipsed conformations.


Strategy - 3.7 Conformations of Other Alkanes
Propane, the next-higher member in the alkane series, also has a torsional barrier that results in hindered rotation around the carbon-carbon bonds. The barrier is slightly higher in propane than in ethane-a total of 14 $\mathrm{kJ} / \mathrm{mol}(3.4 \mathrm{kcal} / \mathrm{mol})$ versus $12 \mathrm{~kJ} / \mathrm{mol}$.

The eclipsed conformation of propane has three interactions-two ethane-type hydrogen-hydrogen interactions and one additional hydrogen-methyl interaction. Since each eclipsing $\mathrm{H} \longleftrightarrow \mathrm{H}$ interaction is the same as that in ethane and thus has an energy "cost" of $4.0 \mathrm{~kJ} / \mathrm{mol}$, we can assign a value of $14-(2 \times 4.0)=6.0$ $\mathrm{kJ} / \mathrm{mol}(1.4 \mathrm{kcal} / \mathrm{mol})$ to the eclipsing $\mathrm{H} \longleftrightarrow \mathrm{CH}_{3}$ interaction (FIGURE 3.9).

FIGURE 3.9 Newman projections of propane showing staggered and eclipsed conformations. The staggered conformer is lower in energy by $14 \mathrm{~kJ} / \mathrm{mol}$.


Strategy - 3.7 Conformations of Other Alkanes
FIGURE 3.9 Newman projections of propane showing staggered and eclipsed conformations. The staggered conformer is lower in energy by $14 \mathrm{~kJ} / \mathrm{mol}$.

The conformational situation becomes more complex for larger alkanes because not all staggered conformations have the same energy and not all eclipsed conformations have the same energy. In butane, for instance, the lowest-energy arrangement, called the anti conformation, is the one in which the two methyl groups are as far apart as possible $-180^{\circ}$ away from each other. As rotation around the $\mathrm{C} 2-\mathrm{C} 3$ bond occurs, an eclipsed conformation is reached where there are two $\mathrm{CH}_{3} \longleftrightarrow \mathrm{H}$ interactions and one $\mathrm{H} \leftrightarrows \mathrm{H}$ interaction. Using the energy values derived previously from ethane and propane, this eclipsed conformation is more strained than the anti conformation by $2 \times 6.0 \mathrm{~kJ} / \mathrm{mol}+4.0 \mathrm{~kJ} / \mathrm{mol}$ (two $\mathrm{CH}_{3} \longleftrightarrow \mathrm{H}$ interactions plus one $\mathrm{H} \longleftrightarrow \mathrm{H}$ interaction), for a total of $16 \mathrm{~kJ} / \mathrm{mol}(3.8 \mathrm{kcal} / \mathrm{mol})$.

Butane-anti conformation ( $0 \mathrm{~kJ} / \mathrm{mol}$ )


Strategy - 3.7 Conformations of Other Alkanes
Butane-anti conformation ( $0 \mathrm{~kJ} / \mathrm{mol}$ )

Butane-eclipsed
conformation
$(16 \mathrm{~kJ} / \mathrm{mol})$
As bond rotation continues, an energy minimum is reached at the staggered conformation where the methyl groups are $60^{\circ}$ apart. Called the gauche conformation, it lies $3.8 \mathrm{~kJ} / \mathrm{mol}(0.9 \mathrm{kcal} / \mathrm{mol})$ higher in energy than the anti conformation even though it has no eclipsing interactions. This energy difference occurs because the hydrogen atoms of the methyl groups are near one another in the gauche conformation, resulting in what is called steric strain. Steric strain is the repulsive interaction that occurs when atoms are forced closer together than their atomic radii allow. It's the result of trying to force two atoms to occupy the same space.

As the dihedral angle between the methyl groups approaches zero, an energy maximum is reached at a second eclipsed conformation. Because the methyl groups are forced even closer together than in the gauche conformation, both torsional strain and steric strain are present. A total strain energy of $19 \mathrm{~kJ} / \mathrm{mol}(4.5 \mathrm{kcal} /$ mol ) has been estimated for this conformation, making it possible to calculate a value of $11 \mathrm{~kJ} / \mathrm{mol}(2.6 \mathrm{kcal} /$ mol ) for the $\mathrm{CH}_{3} \longleftrightarrow \mathrm{CH}_{3}$ eclipsing interaction: total strain of $19 \mathrm{~kJ} / \mathrm{mol}$ minus the strain of two $\mathrm{H} \longleftrightarrow \mathrm{H}$ eclipsing interactions ( $2 \times 4.0 \mathrm{kcal} / \mathrm{mol}$ ) equals $11 \mathrm{~kJ} / \mathrm{mol}$.


Strategy - 3.7 Conformations of Other Alkanes
After $0^{\circ}$, the rotation becomes a mirror image of what we've already seen: another gauche conformation is reached, another eclipsed conformation, and finally a return to the anti conformation. A plot of potential energy versus rotation about the $\mathrm{C} 2-\mathrm{C} 3$ bond is shown in FIGURE 3.10.

FIGURE 3.10 A plot of potential energy versus rotation for the C2-C3 bond in butane. The energy maximum occurs when the two methyl groups eclipse each other, and the energy minimum occurs when the two methyl groups are $180^{\circ}$ apart (anti).

The notion of assigning definite energy values to specific interactions within a molecule is very useful, and we'll return to it in the next chapter. A summary of what we've seen thus far is given in TABLE 3.5.

The same principles just developed for butane apply to pentane, hexane, and all higher alkanes. The most favorable conformation for any alkane has the carbon-carbon bonds in staggered arrangements, with large substituents arranged anti to one another. A generalized alkane structure is shown in FIGURE 3.11.

| Interaction | Cause | Energy cost |  |
| :---: | :---: | :---: | :---: |
|  |  | ( $\mathrm{kJ} / \mathrm{mol}$ ) | ( $\mathrm{kcal} / \mathrm{mol}$ ) |
| $\mathrm{H} \leftrightarrow \mathrm{H}$ eclipsed | Torsional strain | 4.0 | 1.0 |
| $\mathrm{H} \leftrightarrow \mathrm{CH}_{3}$ eclipsed | Mostly torsional strain | 6.0 | 1.4 |
| $\mathrm{CH}_{3} \longleftrightarrow \mathrm{CH}_{3}$ eclipsed | Torsional and steric strain | 11.0 | 2.6 |
| $\mathrm{CH}_{3} \longleftrightarrow \mathrm{CH}_{3}$ gauche | Steric strain | 3.8 | 0.9 |


Strategy - 3.7 Conformations of Other Alkanes
FIGURE 3.11 The most stable alkane conformation is the one in which all substituents are staggered and the carbon-carbon bonds are arranged anti, as shown in this model of decane.
One final point: saying that one particular conformer is "more stable" than another doesn't mean the molecule adopts and maintains only the more stable conformation. At room temperature, rotations around $\sigma$ bonds occur so rapidly that all conformers are in equilibrium. At any given instant, however, a larger percentage of molecules will be found in a more stable conformation than in a less stable one.


Newman Projections -
Sight along the C1-C2 bond of 1-chloropropane, and draw Newman projections of the most stable and least stable conformations.


Strategy -
The most stable conformation of a substituted alkane is generally a staggered one in which large groups have an anti relationship. The least stable conformation is generally an eclipsed one in which large groups are as close as possible.


Solution -
Most stable (staggered)

Least stable (eclipsed)

PROBLEM Make a graph of potential energy versus angle of bond rotation for propane, and assign values to the 3-15 energy maxima.

PROBLEM Sight along the C2-C1 bond of 2-methylpropane (isobutane).
3-16 (a) Draw a Newman projection of the most stable conformation.
(b) Draw a Newman projection of the least stable conformation.
(c) Make a graph of energy versus angle of rotation around the $\mathrm{C} 2-\mathrm{C} 1$ bond.
(d) Assign relative values to the maxima and minima in your graph, given that an $\mathrm{H} \longleftrightarrow \mathrm{H}$ eclipsing interaction costs $4.0 \mathrm{~kJ} / \mathrm{mol}$ and an $\mathrm{H} \longleftrightarrow \mathrm{CH}_{3}$ eclipsing interaction costs $6.0 \mathrm{~kJ} / \mathrm{mol}$.

PROBLEM Sight along the C2-C3 bond of 2,3-dimethylbutane, and draw a Newman projection of the most
3-17 stable conformation.
PROBLEM Draw a Newman projection along the C2-C3 bond of the following conformation of 3-18 2,3-dimethylbutane, and calculate a total strain energy: