5.1: Algebraic Expressions
- Convert between written and symbolic algebraic expressions and equations.
- Simplify and evaluate algebraic expressions.
- Add and subtract algebraic expressions.
- Multiply and divide algebraic expressions.
Algebraic expressions are the building blocks of algebra. While a numerical expression (also known as an arithmetic expression) like 5 + 3 can represent only a single number, an algebraic expression such as 5𝑥+ can represent many different numbers. This section will introduce you to algebraic expressions, how to create them, simplify them, and perform arithmetic operations on them.
Algebraic Expressions and Equations
Xavier and Yasenia have the same birthday, but they were born in different years. This year Xavier is 20 years old and Yasenia is 23, so Yasenia is three years older than Xavier. When Xavier was 15, Yasenia was 18. When Xavier will be 33, Yasenia will be 36. No matter what Xavier’s age is, Yasenia’s age will always be 3 years more.
In the language of algebra, we say that Xavier's age and Yasenia's age are variable and the 3 is a constant. The ages change, or vary, so age is a variable . The 3 years between them always stays the same or has the same value, so the age difference is the constant . In algebra, letters of the alphabet are used to represent variables. The letters most often used for variables are \(x\), , , , , and . Suppose we call Xavier's age . Then we could use to represent Yasenia's age, as shown in the table below.
| Xavier’s Age | Yasenia’s Age |
|---|---|
| 15 | 18 |
| 20 | 23 |
| 33 | 36 |
| \(x\) |
To write algebraically, we need some symbols as well as numbers and variables. The symbols for the four basic arithmetic operations: addition, subtraction, multiplication, and division are summarized in Table 5.1, along with words we use for the operations and the result.
| Operation | Notation | Say: | The result is… |
|---|---|---|---|
| Addition | plus | The sum of and | |
| Subtraction | − | minus | The difference of and |
| Multiplication |
•
, (
)(
), (
)
,
(
),
|
times | The product of and |
| Division | ÷ , / | divided by | The quotient of and |
In algebra, the cross symbol
is normally not used to show multiplication because that symbol could cause confusion. For example, does
We perform these operations on two numbers. When translating from symbolic form to words, or from words to symbolic form, pay attention to the words of or and to help you find the numbers.
- The sum of 5 and 3 means add 5 plus 3, which we write as .
- The difference of 9 and 2 means subtract 9 minus 2, which we write as .
- The product of 4 and 8 means multiply 4 times 8, which we can write as .
- The quotient of 20 and 5 means divide 20 by 5, which we can write as .
Translate the following algebraic expressions from algebra into words.
1. \(12+14\)
2. \((30)(5)\)
3. \(64 \div 8\)
4. \(x-y\)
- Answer
-
- According to Table 5.1, this could be translated as 12 plus 14 OR the sum of 12 and 14.
- According to Table 5.1, this could be translated as 30 times 5 OR the product of 30 and 5.
- According to Table 5.1, this could be translated as 64 divided by 8 OR the quotient of 64 and 8.
- According to Table 5.1, this could be translated as \(x\) minus OR the difference of \(x\) and
Translate the following algebraic expressions from algebra into words.
\(18+11\)
\((27)(9)\)
\(84 \div 7\)
\(p-q\)
Translate the following phrases from words into algebraic expressions.
1. 43 plus 67
2. The product of 45 and 3
3. The quotient of 45 and 3
4. 89 minus 42
- Answer
-
1. According to Table 5.1, these words could be translated as \(47-19\).
2. According to Table 5.1, these words could be translated as \(72 \div 9\).
3. According to Table 5.1, these words could be translated as \(m+n\).
4. According to Table 5.1, these words could be translated as (13) (7).
Translate the following phrases from words into algebraic expressions.
43 plus 67
The product of 45 and 3
The quotient of 45 and 3
89 minus 42
What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. “Running very fast” is a phrase, but “The football player was running very fast” is a sentence. A sentence has a subject and a verb. In algebra, we have expressions and equations. Example 5.1 and Example 5.2 used expressions. An expression is like an English phrase. Notice that the English phrases do not form a complete sentence because the phrase does not have a verb. The following table has examples of expressions, which are numbers, variables, or combinations of numbers and variables using operation symbols.
| Expression | Words | English Phrase |
|---|---|---|
| 3 plus 5 | The sum of three and five | |
| minus one | The difference of and one | |
| 6 times 7 | The product of six and seven | |
|
\(x \div y\)
|
\(x\) divided by | The quotient of \(x\) and |
Translate the following phrases from words into algebraic expressions.
- Seven more than a number .
- A number times itself.
- Six times a number , plus two more.
- The cost of postage is a flat rate of 10 cents for every parcel, plus 34 cents per ounce \(x\).
- Answer
-
1. \(n+7\)
2. \(n \bullet n\) or \(n^2\)
3. \(6 n+2\)
4. \(10+34 x\)
Translate the following phrases from words into algebraic expressions.
Twenty less than a number \(n\). (Hint: you have a number \(n\) and you want 20 less than it.)
Add two to a number \(n\), then multiply it by six.
A number \(n\) to the third power minus five.
A plumber charges \(\$ 60\) per hour \(h\), plus a \(\$ 40\) flat fee for every job.
An equation is two expressions linked with an equal sign (the symbol =). When two quantities have the same value, we say they are equal and connect them with an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb. So, is read “ is equal to .” The following table has some examples of equations.
| Equation | English Sentence |
|---|---|
| The sum of three and five is equal to eight. | |
|
|
minus one equals fourteen. |
| The product of six and seven is equal to forty-two. | |
|
|
\(x\) is equal to fifty-three. |
| plus nine is equal to two times minus three. |
Translate the following sentences from words into algebraic equations.
- Two times \(x\) is 6.
- plus 2 is equal to times 3.
- The quotient of 35 and 7 is 5.
- Sixty-seven minus \(x\) is 56.
- Answer
-
1. \(2 x=6\)
2. \(n+2=3 n\)
3. \(35 \div 7=5\)
4. \(67-x=56\)
Translate the following sentences from words into algebraic equations.
1 . Five times y is 50.
2 . Half of a number n is 30.
3 . The difference of three times a number n and 7 is 2.
4 . Two times x plus 7 is 21.
French philosopher and mathematician René Descartes (1596–1650) is usually given credit for the use of the letters \(x\), , and to represent unknown quantities in algebra. He introduced these ideas in his publication of La Geometrie , which was printed in 1637. In this publication, he also used the letters , , and to represent known quantities. There is a (possibly fictitious) story that, when the book was being printed for the first time, the printer began to run short of the last three letters of the alphabet. So the printer asked Descartes if it mattered which of \(x\), , or were used for the mathematical equations in the book. Descartes decided it made no difference to him; so the printer decided to use \(x\) predominantly for the mathematics in the book, because the letters and would occur more often in the body of the text (written in French) than the letter \(x\) would! This might explain why the letter \(x\) is still used today as the most common variable to represent unknown quantities in algebra.
Simplifying and Evaluating Algebraic Expressions
To simplify an expression means to do all the math possible. For example, to simplify we would first multiply to get 8 and then add 1 to get 9. We have introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values. Consider . Do you add first or multiply first? Do you get different answers?
| Add first: | Multiply first: | Which one is correct? |
Early on, mathematicians realized the need to establish some guidelines when performing arithmetic operations to ensure that everyone would get the same answer. Those guidelines are called the order of operations and are listed in the table below.
| Step 1: Parentheses and Other Grouping Symbols | Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first. |
| Step 2: Exponents | Simplify all expressions with exponents. |
| Step 3: Multiplication and Division | Perform all multiplication and division in order from left to right. These operations have equal priority. |
| Step 4: Addition and Subtraction | Perform all addition and subtraction in order from left to right. These operations have equal priority. |
You may have heard about Please Excuse My Dear Aunt Sally or PEMDAS. Be careful to notice in Steps 3 and 4 in the table above that multiplication and division, as well as addition and subtraction, happen in order from LEFT to RIGHT. It is possible, for example, to have PEDMAS or PEMDSA. The PEMDAS trick can be misleading if not fully understood!
Use parentheses to make the following statements true.
1. \(17-10+3=10\)
2. \(2 \bullet 26-7=38\)
3. \(8+12 \div 5-3=14\)
4. \(5+2^3 \bullet 7=91\)
- Answer
-
1. Add the parentheses around the \(17-10\). Then you have \((17-10)+3=7+3=10\).
2. Add the parentheses around the \(26-7\). Then you have \(2 \bullet(26-7)=2 \bullet 19=38\).
3. Add the parentheses around the \(5-3\). Then you have\[8+12 \div(5-3)=8+12 \div 2=8+6=14 \nonumber \]
4. Add the parentheses around the \(5+2^3\). Then you have
\[\left(5+2^3\right) \bullet 7=(5+8) \bullet 7=13 \bullet 7=91 . \nonumber \]
Use parentheses and the order of operations to make each equation true.
\(24-17-6=13\)
\(3 \bullet 6+13=31\)
\(12-6 \div 5-3=3\)
\(5 \bullet 3^2+5=70\)
In the last example, we simplified expressions using the order of operations. Now we'll evaluate some expressions—again following the order of operations. To evaluate an expression means to find the value of the expression when the variable is replaced by a given number.
1. Evaluate \(3 x+5\) when \(x=2\).
2. Evaluate \(x^2+3 x+1\) when \(x=2\).
.
- Answer
-
1. Evaluate \(5 x-6\) when \(x=3\).
2. Evaluate \(x^2-6 x+3\) when \(x=3\).
Evaluate \(5 x-6\) when \(x=3\).
Evaluate \(x^2-6 x+3\) when \(x=3\).
Operations of Algebraic Expressions
Algebraic expressions are made up of
terms
. A term is a constant or the product of a constant and one or more variables. Examples of terms are 7,
, 5
, 9
, and
Some terms share common traits. When two terms are constants or have the same variable and exponent, we say they are like terms . If there are like terms in an expression, you can simplify the expression by combining the like terms. We add the coefficients and keep the same variable.
Add \(\left(x^2+4 x-9\right)+\left(3 x^2-x+12\right)\).
- Answer
-
Step 1 : Add the terms in any order and get the same result (think: \(2+3=3+2\) ) and drop the parentheses:
\[x^2+4 x-9+3 x^2-x+12 \nonumber \]
Step 2 : Group like terms together:
\[x^2+3 x^2+4 x-x-9+12 \nonumber \]
Step 3: Combine the like terms:
\[4 x^2+3 x+3 \nonumber \]
Add \(\left(2 x^2-4 x+5\right)+\left(3 x^2+x-12\right)\).
Subtract \((5 x 2+4 x-9)-(3 x 2-x+12)(5 x 2+4 x-9)-(3 \times 2-x+12)\).
- Answer
-
Step 1: Distribute the negative inside the parentheses (think: \(2-(3-4)=2-3+4=-1+4=3\), which is the correct answer). You cannot just drop the parentheses (for example, \(2-3-4=-1-4=-5\), which is not correct as we have already verified the answer is 3 ):
\[5 x^2+4 x-9-3 x^2+x-12 \nonumber \]
Step 2 : Group like terms together:
\[5 x^2-3 x^2+4 x+x-9-12 \nonumber \]
Step 3 : Combine the like terms:
\[2 x^2+x-21 \nonumber \]
Subtract \(\left(x^2-4 x+8\right)-\left(x^2+5 x+12\right)\).
Before looking at multiplying algebraic expressions we look at the Distributive Property, which says that to multiply a sum, first you multiply each term in the sum and then you add the products. For example, \(5(4+3)=5(4)+5(3)=20+15=35\) can also be solved as \(5(4+3)=5(7)=35\). If we use a variable, then \(5(x+3)=5 x+15\).
We can extended this example to \((5+2)(4+3)=(5)(4)+(5)(3)+(2)(4)+(2)(3)=20+15+8+6=49\), which can also be solved as \((5+2)(4+3)=(7)(7)=49\). If we use variables, then
\[(x+5)(x+4)=(x)(x)+(x)(4)+(5)(x)+(5)(4)=x^2+4 x+5 x+20=x^2+9 x+20 . \nonumber \]
Distributive Property: \(a(b+c)=a b+a c\)
Simplify each expression.
1. \((x-3) 5\)
2. \((-3)(x+y-2)\)
3. \(5^2(7+3)(x)\)
4. \(4+x \bullet 5\)
5. \((4+x) \bullet 5\)
- Answer
-
1. \(5 x-3 \bullet 5=5 x-15\)
2. \((-3) \bullet x+(-3) \bullet y-(-3) \bullet 2=-3 x-3 y+6\)
3. \(25(7+3)(x)=25(10)(x)=250 x\)
4. \(4+5 x\)
5. \((4) \bullet(5)+(x) \bullet(5)=20+5 x\)
Simplify each expression.
\(2(y+5)\)
\((-2)(a+b-4)\)
\(4^2(47-40+x)\)
\((18 \div 3)(x+7-4)\)
\(2(3 a+5)+(-3)(a+2)\)
Multiply \((4 x-9)(x+2)\).
- Answer
-
Step 1 : Use the Distributive Property:
\[(4 x)(x)+(4 x)(2)-(9)(x)-(9)(2) \nonumber \]
Step 2 : Multiply:
\[4 x^2+8 x-9 x-18 \nonumber \]
Step 3 : Combine the like terms:
\[4 x^2-x-18 \nonumber \]
Multiply \((x-4)(2 x-3)\).
You may have heard the term FOIL which stands for: First, Outer, Inner, Last. FOIL essentially describes a way to use the Distributive Property if you multiply a two-term expression by another two-term expression, but FOIL only works in that specific situation. For example, suppose you have a two-term expression multiplied by a three-term expression, such as . What terms qualify as inner terms and what terms qualify as outer terms? In this particular situation, FOIL cannot possibly work; the multiplication of should yield six terms, where FOIL is designed to only give you four! The Distributive Property works regardless of how many terms there are. FOIL can be misleading and applied inappropriately if not fully understood!
Divide \(\left(8 x^2+4 x-16\right) \div(4 x)\)
- Answer
-
Divide EACH term by \(4 x\) :
\[\left(8 x^2 \div 4 x\right)+(4 x \div 4 x)-(16 \div 4 x)=2 x+1-\dfrac{4}{x} \nonumber \]
Divide \(\left(16 x^2+4 x-8\right) \div(4)\).
Be careful how you divide! Sometimes students incorrectly divide only one term on top by the bottom term. For example, \(\dfrac{8 x^2+6 x-3}{2 x}\) might turn into \(4 x+3 x-3=7 x-3\) if done incorrectly. When we divide expressions, EACH term is divided by the divisor. So, \(\dfrac{8 x^2+6 x-3}{2 x}=\dfrac{8 x^2}{2 x}+\dfrac{6 x}{2 x}-\dfrac{3}{2 x}=4 x+3-\dfrac{3}{2 x}\). If you forget, it is always a good idea to check these rules by creating an example using numerical expressions. For example, \(\dfrac{9+6+3}{3}=\dfrac{18}{3}=6\). Dividing each term on top by 3 would yield \(\dfrac{9+6+3}{3}=\dfrac{9}{3}+\dfrac{6}{3}+\dfrac{3}{3}=3+2+1=6\), which is the correct answer. However, if you just divided the 9 on top by the 3 on the bottom, getting \(\dfrac{9+6+3}{3}=3+6+3=12\), this does not result in the correct answer.
Abu Ja’far Muhammad ibn Musa Al-Khwarizmi was born around 780 AD, probably in or around the region of Khwarizm, which is now part of modern-day Uzbekistan. For most of his adult life, he worked as a scholar at the House of Wisdom in Baghdad, Iraq. He wrote many mathematical works during his life, but is probably most famous for his book Al-kitab al-muhtasar fi hisab al-jabr w’al’muqabalah , which translates to The Condensed Book on the Calculation of al-Jabr ( completion ) and al’muqabalah ( balancing ). The word al-jabr would eventually become the word we use to describe the topic that he was writing about in this book: algebra . From another book of his, with the Latin title Algoritmi de numero Indorum ( Al-Khwarizmi on the Hindu Art of Reckoning ), our word algorithm is derived. In addition to writing on mathematics, Al-Khwarizmi wrote works on astronomy, geography, the sundial, and the calendar.
In 2012, Andrew Hacker wrote an opinion piece in the New York Times Magazine suggesting that teaching algebra in high school was a waste of time. Keith Devlin, a British mathematician, was asked to comment on Hacker's article by his students in his Stanford University Continuing Studies course "Mathematics: Making the Invisible Visible" on iTunes University. Devlin concludes that Hacker was displaying his ignorance of what algebra is.
Check Your Understanding
1. Juliette is 2 inches taller than her friend Vivian. Which algebraic equations represent their height? Use \(J\) for Juliette's height and \(V\) for Vivian's height.
\[
\begin{array}{l}
J=V+2 \\
V=J-2 \\
J+2=V \\
J=V-2
\end{array}
\nonumber \]
2. Which options represent algebraic expressions?
\[
\begin{array}{l}
2 x^2+3 x-1=0 \\
5 x+8 \\
2 n+3 m \\
5 x-7=3 x+1
\end{array}
\nonumber \]
3. Which expression equals \(10 x\) ?
\[
\begin{array}{l}
(8 x+12 x) \div 4 x-2 x \\
8 x+(12 x \div 4 x)-2 x \\
8 x+12 x \div(4 x-2 x) \\
(8 x+12 x) \div(4 x-2 x)
\end{array}
\nonumber \]
4. Using the expression \(3 x^2-7 x+2\), when a certain number is put in for \(x\), the result is 50 . What is the value of \(x\) ?
\(-2\)
\(-3\)
2
3
5. Which expression equals \((x-y)(x-y)\) ? Hint: Use the Distributive Property.
\[
\begin{array}{l}
x^2-y^2 \\
x^2+y^2 \\
x^2-2 x y-y^2 \\
x^2-2 x y+y^2
\end{array}
\nonumber \]
6. Given the expression \(9 x^3+3 x^2-6 x\), the Distributive Property allows it to be rewritten as:
\[
\begin{array}{l}
3 x\left(3 x^2+x-2\right) \\
3 x^2+x-2 \\
27 x^5-54 x^4 \\
27 x^6-54 x^3
\end{array}
\nonumber \]
7. Given the two algebraic expressions \((x+2)\) and \((x+y-5)\), the solution is \(x^2+x y-3 x+2 y-10\). What mathematical operation was performed on the two algebraic expressions?
8. Given the two algebraic expressions \(8 x^2-9 x+6\) and \(6 x\), the solution is \(3 x-1.5+\frac{1}{x}\). What mathematical operation was performed on the two algebraic expressions?


2.7: S07. Distances as Determined by Standard Candles - SOLUTIONS
Exercise 7.1.1
- Answer
-
Constant \(t\), \(r\), and \(\phi\) so we have
\[\begin{equation*}
\begin{aligned}
ds = a(t)r\,d\theta
\end{aligned}
\end{equation*}\]
Exercise 7.1.2
- Answer
-
Similar to 7.1.1 above, we now have constant \(t\), \(r\), and \(\theta\), which gives
\[\begin{equation*}
\begin{aligned}
ds = a(t)r\sin\theta \,d\phi
\end{aligned}
\end{equation*}\]
Exercise 7.1.3
- Answer
-
This is simply \(a^2(t)r^2 \sin \theta \, d\theta d\phi\).
Exercise 7.1.4
- Answer
-
Using the result from 7.1.3 above, we now just integrate over \(\theta\) and \(\phi\), so the area is
\[\begin{equation*}
\begin{aligned}
A &= \int_{0}^{2\pi}\int_{0}^{\pi} a(t)rsin\theta d\theta d\phi \\ \\ &= \int_{0}^{2\pi} 2a^2(t)r^2d\phi \\ \\ &= 4\pi a^2(t)r^2
\end{aligned}
\end{equation*}\]
Exercise 7.1.5
- Answer
-
We know that Luminosity = (Flux)\(\times\)(Surface Area). Making the appropriate substitutions we do indeed find that \(F = L/(4\pi d^2 a^2)\).
Exercise 7.2.1
- Answer
-
As we found in Chapter 6, the rate of arrival of the wave crests will be slower than the rate of emission by the factor of \(a(t_r)/a(t_e)\). The same argument applies to the rate of arrival of photons. We also saw that wavelength would be stretched out by a factor \(1+z \equiv \frac{\lambda_r}{\lambda_e} = a(t_r)/a(t_e)\). Therefore the rate of arrival of photons will be slowed down by a factor of \(1+z\).
Exercise 7.2.2
- Answer
-
The relationship between photon energy and wavelength is \(E = \frac{hc}{\lambda}\). Substituting this into the definition of redshift, we find that
\[\begin{equation*}
\begin{aligned}
E_r/E_e = \frac{1}{1 + z}
\end{aligned}
\end{equation*}\]
Exercise 7.3.1
- Answer
-
Solving Equation 7.4 for \(d_{\rm lum}\) we get
\[\begin{equation*}
\begin{aligned}
d_{\rm lum} = \sqrt{\frac{L}{4\pi F}}
\end{aligned}
\end{equation*}\]
Exercise 7.3.2
- Answer
-
Substituting in Equation 7.3 to our result above we get
\[\begin{equation*}
\begin{aligned}
d^2_{\rm lum} = \frac{L}{4\pi}\frac{4\pi d^2(1 + z)^2}{L} \quad \Longrightarrow \quad d_{\rm lum} = d(1 + z)
\end{aligned}
\end{equation*}\]


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5.1: Prelude to Discrete Random Variables
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Random Variable (RV) a characteristic of interest in a population being studied
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5.2: Probability Distribution Function (PDF) for a Discrete Random Variable
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A discrete probability distribution function has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one.
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5.3: Mean or Expected Value and Standard Deviation
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The expected value is often referred to as the "long-term" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ . In other words, after conducting many trials of an experiment, you would expect this average value.
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5.4: Binomial Distribution
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A statistical experiment can be classified as a binomial experiment if the following conditions are met: (1) There are a fixed number of trials. (2)There are only two possible outcomes: "success" or "failure" for each trial. (3) The trials are independent and are repeated using identical conditions. The outcomes of a binomial experiment fit a binomial probability distribution.
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5.5: Discrete Distribution (Playing Card Experiment)
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A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if an everyday experiment fits a discrete distribution. The student will demonstrate an understanding of long-term probabilities.
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5.6: Discrete Distribution (Lucky Dice Experiment)
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A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if a gambling game fits a discrete distribution. The student will demonstrate an understanding of long-term probabilities.
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5.7: Discrete Random Variables (Exercises)
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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.
Contributors and Attributions


8.2: The Genetic Code
-
- Last updated
- Save as PDF
- Ying Liu
- City College of San Francisco
Learning Objectives
By the end of this section, you will be able to do the following:
- Explain the “central dogma” of DNA-protein synthesis
- Describe the genetic code and how the nucleotide sequence prescribes the amino acid and the protein sequence
The cellular process of transcription generates messenger RNA (mRNA), a mobile molecular copy of one or more genes with an alphabet of A, C, G, and uracil (U). Translation of the mRNA template on ribosomes converts nucleotide-based genetic information into a protein product. That is the central dogma of DNA-protein synthesis. Protein sequences consist of 20 commonly occurring amino acids; therefore, it can be said that the protein alphabet consists of 20 “letters” (Figure 15.2). Different amino acids have different chemistries (such as acidic versus basic, or polar and nonpolar) and different structural constraints. Variation in amino acid sequence is responsible for the enormous variation in protein structure and function.
The Central Dogma: DNA Encodes RNA; RNA Encodes Protein
The flow of genetic information in cells from DNA to mRNA to protein is described by the central dogma (Figure 15.3), which states that genes specify the sequence of mRNAs, which in turn specify the sequence of amino acids making up all proteins. The decoding of one molecule to another is performed by specific proteins and RNAs. Because the information stored in DNA is so central to cellular function, it makes intuitive sense that the cell would make mRNA copies of this information for protein synthesis, while keeping the DNA itself intact and protected. The copying of DNA to RNA is relatively straightforward, with one nucleotide being added to the mRNA strand for every nucleotide read in the DNA strand. The translation to protein is a bit more complex because three mRNA nucleotides correspond to one amino acid in the polypeptide sequence. However, the translation to protein is still systematic and colinear , such that nucleotides 1 to 3 correspond to amino acid 1, nucleotides 4 to 6 correspond to amino acid 2, and so on.
The Genetic Code Is Degenerate and Universal
Each amino acid is defined by a three-nucleotide sequence called the triplet codon. Given the different numbers of “letters” in the mRNA and protein “alphabets,” scientists theorized that single amino acids must be represented by combinations of nucleotides. Nucleotide doublets would not be sufficient to specify every amino acid because there are only 16 possible two-nucleotide combinations (4 2 ). In contrast, there are 64 possible nucleotide triplets (4 3 ), which is far more than the number of amino acids. Scientists theorized that amino acids were encoded by nucleotide triplets and that the genetic code was “degenerate.” In other words, a given amino acid could be encoded by more than one nucleotide triplet. This was later confirmed experimentally: Francis Crick and Sydney Brenner used the chemical mutagen proflavin to insert one, two, or three nucleotides into the gene of a virus. When one or two nucleotides were inserted, the normal proteins were not produced. When three nucleotides were inserted, the protein was synthesized and functional. This demonstrated that the amino acids must be specified by groups of three nucleotides. These nucleotide triplets are called codons . The insertion of one or two nucleotides completely changed the triplet reading frame , thereby altering the message for every subsequent amino acid (Figure 15.5). Though insertion of three nucleotides caused an extra amino acid to be inserted during translation, the integrity of the rest of the protein was maintained.
Scientists painstakingly solved the genetic code by translating synthetic mRNAs in vitro and sequencing the proteins they specified (Figure 15.4).
In addition to codons that instruct the addition of a specific amino acid to a polypeptide chain, three of the 64 codons terminate protein synthesis and release the polypeptide from the translation machinery. These triplets are called nonsense codons , or stop codons . Another codon, AUG, also has a special function. In addition to specifying the amino acid methionine, it also serves as the start codon to initiate translation. The reading frame for translation is set by the AUG start codon near the 5' end of the mRNA. Following the start codon, the mRNA is read in groups of three until a stop codon is encountered.
The arrangement of the coding table reveals the structure of the code. There are sixteen "blocks" of codons, each specified by the first and second nucleotides of the codons within the block, e.g., the "AC*" block that corresponds to the amino acid threonine (Thr). Some blocks are divided into a pyrimidine half, in which the codon ends with U or C, and a purine half, in which the codon ends with A or G. Some amino acids get a whole block of four codons, like alanine (Ala), threonine (Thr) and proline (Pro). Some get the pyrimidine half of their block, like histidine (His) and asparagine (Asn). Others get the purine half of their block, like glutamate (Glu) and lysine (Lys). Note that some amino acids get a block and a half-block for a total of six codons.
The specification of a single amino acid by multiple similar codons is called "degeneracy." Degeneracy is believed to be a cellular mechanism to reduce the negative impact of random mutations. Codons that specify the same amino acid typically only differ by one nucleotide. In addition, amino acids with chemically similar side chains are encoded by similar codons. For example, aspartate (Asp) and glutamate (Glu), which occupy the GA* block, are both negatively charged. This nuance of the genetic code ensures that a single-nucleotide substitution mutation might specify the same amino acid but have no effect or specify a similar amino acid, preventing the protein from being rendered completely nonfunctional.
The genetic code is nearly universal. With a few minor exceptions, virtually all species use the same genetic code for protein synthesis. Conservation of codons means that a purified mRNA encoding the globin protein in horses could be transferred to a tulip cell, and the tulip would synthesize horse globin. That there is only one genetic code is powerful evidence that all of life on Earth shares a common origin, especially considering that there are about 10 84 possible combinations of 20 amino acids and 64 triplet codons.
Link to Learning
Link to Learning
Transcribe a gene and translate it to protein using complementary pairing and the genetic code at this site .
Scientific Method Connection
Scientific Method Connection
Which Has More DNA: A Kiwi or a Strawberry?
Question : Would a kiwi and strawberry that are approximately the same size (Figure 15.6) also have approximately the same amount of DNA?
Background : Genes are carried on chromosomes and are made of DNA. All mammals are diploid, meaning they have two copies of each chromosome. However, not all plants are diploid. The common strawberry is octoploid (8 n ) and the cultivated kiwi is hexaploid (6 n ). Research the total number of chromosomes in the cells of each of these fruits and think about how this might correspond to the amount of DNA in these fruits’ cell nuclei. What other factors might contribute to the total amount of DNA in a single fruit? Read about the technique of DNA isolation to understand how each step in the isolation protocol helps liberate and precipitate DNA.
Hypothesis : Hypothesize whether you would be able to detect a difference in DNA quantity from similarly sized strawberries and kiwis. Which fruit do you think would yield more DNA?
Test your hypothesis : Isolate the DNA from a strawberry and a kiwi that are similarly sized. Perform the experiment in at least triplicate for each fruit
- Prepare a bottle of DNA extraction buffer from 900 mL water, 50 mL dish detergent, and two teaspoons of table salt. Mix by inversion (cap it and turn it upside down a few times).
- Grind a strawberry and a kiwi by hand in a plastic bag, or using a mortar and pestle, or with a metal bowl and the end of a blunt instrument. Grind for at least two minutes per fruit.
- Add 10 mL of the DNA extraction buffer to each fruit, and mix well for at least one minute.
- Remove cellular debris by filtering each fruit mixture through cheesecloth or porous cloth and into a funnel placed in a test tube or an appropriate container.
- Pour ice-cold ethanol or isopropanol (rubbing alcohol) into the test tube. You should observe white, precipitated DNA.
- Gather the DNA from each fruit by winding it around separate glass rods.
Analyze your data : Did you notice an obvious difference in the amount of DNA produced by each fruit? Were your results reproducible?
Draw a conclusion : Given what you know about the number of chromosomes in each fruit, can you conclude that chromosome number necessarily correlates to DNA amount? Can you identify any drawbacks to this procedure? If you had access to a laboratory, how could you standardize your comparison and make it more quantitative?


2.4: Motivational Interviewing
Patient education and health promotion are core nursing interventions. Motivational interviewing (MI) is a communication skill used to elicit and emphasize a client’s personal motivation for modifying behavior to promote health. MI has been effectively used for several health issues such as smoking cessation, diabetes, substance use disorders, and adherence to a treatment plan. [1]
The spirit of motivational interviewing is a collaborative partnership between nurses and clients, focused on patient-centered care, autonomy, and personal responsibility. It is a technique that explores a client’s motivation, confidence, and roadblocks to change. During motivational interviewing, nurses pose questions, actively listen to client responses, and focus on where the client is now with a current health behavior and where they want to be in the future. [2]
Motivational interviewing uses these principles [3] :
- Express empathy. Use reflective listening to convey acceptance and a nonjudgmental attitude. Rephrase client comments to convey active listening and let clients know they are being heard.
- Highlight discrepancies. Help clients become aware of the gap between their current behaviors and their values and goals. Present objective information that highlights the consequences of continuing their current behaviors to motivate them to change their behavior.
- Adjust to resistance. Adjust to a client’s resistance and do not argue. The client may demonstrate resistance by avoiding eye contact, becoming defensive, interrupting you, or seeming distracted by looking at their watch or cell phone. Arguing can place the client on the defensive and in a position of arguing against the change. Focus on validating the client’s feelings.
- Understand motivations. Uncover a client’s personal reasons for making behavioral changes and build on them.
- Support self-efficacy. Encourage the client’s optimistic belief in the prospect of change and encourage them to commit to positive behavioral changes. Ask clients to elaborate on past successes to build self-confidence and support self-efficacy.
- Resist the reflex to provide advice. Avoid imposing your own perspective and advice.
When implementing motivational interviewing, it is important to assess the client’s readiness for change. Motivational interviewing is especially useful for clients in the contemplation stage who are feeling ambivalent about making change. Recall these five stages of behavioral change [4] :
- Precontemplation: Not considering change.
- Contemplation: Ambivalent about making change.
- Preparation: Taking steps toward implementing change.
- Action: Actively involved in the change process.
- Maintenance: Sustaining the target behavior.
Identify clients who are ambivalent about making a behavioral change or following a treatment plan by listening for the phrase, “Yes, but.” The “but” holds the key for opening the conversation about ambivalence. For example, a client may state, “I want to take my medication, but I hate gaining weight.” The content in the sentence after the “but” reveals the client’s personal roadblock to making a change and should be taken into consideration when planning outcomes and interventions. [5]
See the following box for an example of a nurse using motivational interviewing with a client.
Example of Motivational Interviewing [6]
Mr. L. had been in treatment for bipolar I disorder with medication management and supportive therapy for many years. He had a history of alcohol dependence but was in full recovery. Mr. L. was admitted to the intensive care unit with a toxic lithium level. He had been seen in the emergency room the preceding evening and was noted to have a very high blood alcohol level. The next day the nurse asked the client about his alcohol use using motivational interviewing.

Nurse (Using reflective listening): You seem pretty angry about the perception that you were hospitalized because you had been drinking.
Client: You better believe it! I am a man! I can have a few drinks if I want to!
Nurse: (Expressing empathy and acceptance): You want to be respected even when you are drinking.
Client: I have had some trouble in the past with drinking, but that is not now. I can quit if I want to! Compared to what I used to drink, this is nothing.
Nurse (Rolling with resistance): So you see yourself as having had drinking problems in the past, but the drinking you’ve done recently is not harmful for you.
Client: Well, I guess I did end up in the hospital.
Nurse (Using open-ended questioning): Tell me more about what happened.
Client: I was pretty angry after an argument with my girlfriend, and I decided to buy a bottle of whiskey.
Nurse (Exploring): And then?
Client: Well, I meant to have a couple of shots, but I ended up drinking the whole fifth. I really don’t remember what happened next. They said I nearly died.

Client: I guess that does sound like a problem…but I don’t want anyone else telling me whether or not I can drink!
Nurse (Emphasizing autonomy): Tell me how the choice to drink or not continues to support or oppose your health goals.
View the following supplementary YouTube videos about motivational interviewing:
- Introduction to Motivational Interviewing [7]
- Motivational Interviewing – Good Example – Alan Lyme [8]
Complete Western Region Public Health Training Center’s Motivational Interviewing course and receive a certificate of completion.
- Rubak, S., Sandbaek, A., Lauritzen, T., & Christensen, B. (2005). Motivational interviewing: A systematic review and meta-analysis. The British Journal of General Practice: The Journal of the Royal College of General Practitioners, 55 (513), 305–312. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1463134/ ↵
- Droppa, M., & Lee, H. (2014). Motivational interviewing: A journey to improve health. Nursing, 44 (3), 40–46. https://doi.org/10.1097/01.NURSE.0000443312.58360.82 ↵
- Droppa, M., & Lee, H. (2014). Motivational interviewing: A journey to improve health. Nursing, 44 (3), 40–46. https://doi.org/10.1097/01.NURSE.0000443312.58360.82 ↵
- Droppa, M., & Lee, H. (2014). Motivational interviewing: A journey to improve health. Nursing, 44 (3), 40–46. https://doi.org/10.1097/01.NURSE.0000443312.58360.82 ↵
- Droppa, M., & Lee, H. (2014). Motivational interviewing: A journey to improve health. Nursing, 44 (3), 40–46. https://doi.org/10.1097/01.NURSE.0000443312.58360.82 ↵
- Griffith, L. J. (2008). The psychiatrist's guide to motivational interviewing. Psychiatry (Edgmont, PA: Township) , 5 (4), 42–47. https://www.ncbi.nlm.nih.gov/pubmed/19727309 ↵
- Matulich, B. (2013, May 30). Introduction to motivational interviewing [Video]. YouTube. All rights reserved. https://youtu.be/s3MCJZ7OGRk ↵
- TheRETAchannel. (2013, July 18). Motivational interviewing - Good example - Alan Lyme [Video]. YouTube. All rights reserved. https://youtu.be/67I6g1I7Za o ↵


2.1: Who are the Salafis/Wahabis?
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Amongst the Hanbalis, there is a well-known movement called the Salafi movement. One of the most well-known Salafi groups are the wahabis, a movement with its origins in Saudi Arabia. Wahabis are vehemently against tomb worship, and many of the folk practices of Islam that have been tied to Sufism and/or reverence for particular mystics and holy men (often referred to as walis in Arabic). According to Hanbal’s Islamic theology, these practices are seen as a form of idol worship, defined as a partnering of mortal human beings with God. This is fundamentally in contradiction with one of the main tenets of Islam: monotheism, or tawhῑd .
Salaf means ancestors, and the Salafis, or followers of the Salaf , believe that the spiritual and temporal practices of the earliest Muslims and companions of the prophet provide a comprehensive guide for current-day life and government. Muslim intellectuals of the 19th and early 20th century, grappling with the prospect of modernization and Westernization, saw much potential in this approach because of the universal principles contained in the Qur’an, the prophet’s words and the way he and his companions conducted themselves. The ideal was that they would find a way to adopt Western technology and institutions while applying Islamic concepts to how they would be adopted. As the movement progressed, however, it became more of a fundamentalist movement, and currently one hears about Salafis in the news primarily in connection to radicalization.
1. The diversity of Middle Eastern identities in relation to religion are threefold:
• the Middle East is diverse in regard to multiple religious groups.
• Middle Eastern diaspora communities worldwide reflect a much different demographic make-up than in the Middle East, with a larger percentage of non-Muslims.
• Muslim-majority countries worldwide are very diverse, containing major world regions outside of the Middle East.
1. The Abrahamic faiths of Judaism, Christianity and Islam have been hugely influential, but share commonalities and differences.
2. The Abrahamic Faiths are not the only religions of cultural significance, and at the same time there is much diversity within them.


<IP_ADDRESS>: Passive Transport - Osmosis
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Osmosis is the diffusion of water through a semipermeable membrane according to the concentration gradient of water across the membrane. Whereas diffusion transports material across membranes and within cells, osmosis transports only water across a membrane and the membrane limits the diffusion of solutes in the water. Osmosis is a special case of diffusion. Water, like other substances, moves from an area of higher concentration to one of lower concentration. Imagine a beaker with a semipermeable membrane, separating the two sides or halves ( Figure \(\PageIndex{1}\) ). On both sides of the membrane, the water level is the same, but there are different concentrations on each side of a dissolved substance, or solute , that cannot cross the membrane. If the volume of the water is the same, but the concentrations of solute are different, then there are also different concentrations of water, the solvent, on either side of the membrane.
A principle of diffusion is that the molecules move around and will spread evenly throughout the medium if they can. However, only the material capable of getting through the membrane will diffuse through it. In this example, the solute cannot diffuse through the membrane, but the water can. Water has a concentration gradient in this system. Therefore, water will diffuse down its concentration gradient, crossing the membrane to the side where it is less concentrated. This diffusion of water through the membrane— osmosis —will continue until the concentration gradient of water goes to zero. Osmosis proceeds constantly in living systems.
Tonicity
Tonicity describes the amount of solute in a solution. The measure of the tonicity of a solution, or the total amount of solutes dissolved in a specific amount of solution, is called its osmolarity . Three terms—hypotonic, isotonic, and hypertonic—are used to relate the osmolarity of a cell to the osmolarity of the extracellular fluid that contains the cells. All three of these terms are a comparison between two different solutions (for example, inside a cell compared to outside the cell).
In a hypotonic solution, such as tap water, the extracellular fluid has a lower concentration of solutes than the fluid inside the cell, and water enters the cell. (In living systems, the point of reference is always the cytoplasm, so the prefix hypo – means that the extracellular fluid has a lower concentration of solutes, or a lower osmolarity, than the cell cytoplasm.) It also means that the extracellular fluid has a higher concentration of water than does the cell. In this situation, water will follow its concentration gradient and enter the cell. This may cause an animal cell to burst, or lyse.
In a hypertonic solution (the prefix hyper – refers to the extracellular fluid having a higher concentration of solutes than the cell’s cytoplasm), the fluid contains less water than the cell does, such as seawater. Because the cell has a lower concentration of solutes, the water will leave the cell. In effect, the solute is drawing the water out of the cell. This may cause an animal cell to shrivel, or crenate.
In an isotonic solution, the extracellular fluid has the same osmolarity as the cell. If the concentration of solutes of the cell matches that of the extracellular fluid, there will be no net movement of water into or out of the cell. The cell will retain its “normal” appearance. Blood cells in hypertonic, isotonic, and hypotonic solutions take on characteristic appearances ( Figure \(\PageIndex{2}\) ).
Remember that all three of these terms are comparisons between two solutions (i.e. inside and outside the cell). A solution can’t be hypotonic, that would be like saying that Bob is taller. That doesn’t make sense – you need to say that Bob is taller than Mike. You can say that the solution inside the cell is hypotonic to the solution outside the cell. That also means that the solution outside is hypertonic to the solution inside (just like Mike would be shorter than Bob).
Some organisms, such as plants, fungi, bacteria, and some protists, have cell walls that surround the plasma membrane and prevent cell lysis. The plasma membrane can only expand to the limit of the cell wall, so the cell will not lyse. In fact, the cytoplasm in plants is always slightly hypertonic compared to the cellular environment, and water will always enter the plant cell if water is available. This influx of water produces turgor pressure , which stiffens the cell walls of the plant ( Figure \(\PageIndex{3}\) ). In nonwoody plants, turgor pressure supports the plant. If the plant cells become hypertonic, as occurs in drought or if a plant is not watered adequately, water will leave the cell. Plants lose turgor pressure in this condition and wilt.
Query \(\PageIndex{1}\)
Query \(\PageIndex{2}\)
Query \(\PageIndex{3}\)
References
Unless otherwise noted, images on this page are licensed under CC-BY 4.0 by OpenStax.
Text adapted from: OpenStax, Concepts of Biology. OpenStax CNX. May 18, 2016 http://cnx.org/contents/b3c1e1d2-839...9a8aafbdd@9.10


14.5.2: Igneous Rocks
Igneous rocks form by the cooling of magma (molten rock material beneath the surface) or lava (molten rock material extruded onto the surface). Magma which originates at depths as great as 200 kilometers below the surface consists primarily of elements found in silicate minerals along with gases, notably water vapor. Because the molten material is less dense than the surrounding solidified rock, it works its way toward the surface where it flows out onto the surface as lava.
Cooling of the magma initiates the crystallization of elements contained within the molten rock. Usually, cooling does not take place at the same rate throughout the entire mass. As it cools, numerous small crystal develop and then ions are added around these centers of growth. Once the crystals grow enough for their edges to meet, growth ceases and crystallization commences in another part of the mass. Over time the magma cools into a solidified rock of interlocking crystals. The rate of cooling determines the degree to which crystals can grow. Those that form by slow cooling have large crystals and are described as being coarse-grained . Fine-gained igneous rocks form by more rapidly cooling when the molten material is exposed at the surface. As a result, crystals don't have a chance to grow very large.
Intrusive Igneous Rock
When magma intrudes into pre-existing rock it cools rather slowly because the surrounding host or country rock , as it is called, insulates the magma. As a result, crystals grow larger giving the rock mass a coarse texture. Such intrusive igneous, or plutonic rocks where the mineral grains are easily seen with the unaided eye are called porphyritic . The rock mass itself is called a pluton .
Physical geographers are especially interested in rocks that form beneath the surface after they have been exposed by erosion of the overlying host rock. A batholith is a huge intrusive igneous rock mass or pluton that when uncovered creates topographic highs in mountainous regions. The Idaho Batholith for example covers some 16,000 square miles. These landforms are massive in character lacking linear ridges and valleys.
When magma intrudes between the layers or rock and solidifies it creates a feature called a sill . Sills form a hogback or cuesta when the layer of rock is dipping, or the tops of mesas when lying horizontal. If the magma solidifies as a pocket of igneous rock that warps the overlying rock it will form a dome or laccolith . The Black Hills of South Dakota is such a domal feature. If magma cools in near vertical fractures a dike is formed . Dikes often form on the flanks of volcanoes. When exposed by erosion they take the form of a linear ridge. Shiprock, New Mexico a volcanic neck (Figure \(\PageIndex{4}\)), has several dikes radiating away from it.
Extrusive Igneous Rocks
Fine-grained rocks form if molten rock cools rapidly when it is extruded onto the surface. Basalt is a common extrusive igneous rock. Some fine grained rocks forming at the edge of a lava flow have small holes or vesicles which are void spaces left by escaping gasses. Very rapid cooling can produce rocks with a glassy texture such as obsidian.


2.1: Hydrocarbons
- Identify hydrocarbons and their subclasses: alkane, alkene, alkyne, and aromatic.
- Write IUPAC name for a skeletal structure and draw the skeletal structure from the IUPAC name for simple hydrocarbons.
- Understand some physical properties like melting points, boiling points, and solubilities of hydrocarbons.
- Predict the reactive sites, i.e., \(\delta\)- (nucleophilic) and \(\delta\)+ (electrophilic) regions in the structures of simple hydrocarbons.
- Identify primary, secondary, tertiary, or quaternary \(\ce{C's}\) and \(\ce{H's}\).
Classification of hydrocarbons
Organic compounds composed of \(\ce{C's}\) and \(\ce{H's}\) only are called hydrocarbons .
The hydrocarbons are divided into two major classes based on the absence or presence of an aromatic ring.
An aromatic ring is a cyclic chain of only sp 2 -hybridized atoms in planer geometry having \(\pi\)-electrons in the ring equal to 4n+2, where n is a positive integer, i.e., 0, 1, 2, 3, ...,. Benzene is an example of an aromatic ring that is a cycle of six sp 2 -hybridized \(\ce{C's}\) having six \(\pi\)-electrons, which is a number equal to 4n+2 for n = 1.
Benzene
- Hydrocarbons with at least one aromatic ring in their structure are called aromatic hydrocarbons .
-
Hydrocarbons that do not have an aromatic ring in their structure are called
aliphatic hydrocarbons
. The aliphatic hydrocarbons are subdivided into i) alkanes, ii) alkenes, and iii) alkynes. Figure \(\PageIndex{1}\) illustrates the classification of hydrocarbons.
- Alkanes have all \(\ce{C-C}\) single bonds, i.e., all the \(\ce{C's}\) are sp 3 -hybridized, e.g., propane ( \(\ce{\small{H}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\)).
- Alkenes have at least one \(\ce{C=C}\) double bond, e.g., propene ( \(\ce{\small{H}-{\underset{\underset{\Large{H}} |}{C}}={\underset{\underset{\Large{H}} |}{C}}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\)).
- Alkynes have at least one \(\ce{C≡C}\) triple bond, e.g., propyne ( \(\ce{\small{H}-{C}≡{C}-\overset{\overset{\Large{H}}|}{\underset{\underset{\Large{H}} |}{C}}\!-H}\)).
Alkanes
Nomenclature of alkanes
Organic compounds have trivial names as well as systematic names. Systematic names are based on the International Union of Pure and Applied Chemistry (IUPAC) recommendations, referred to as IUPAC nomenclature . Table 1 lists stem names that represent the number of \(\ce{C's}\) in a continuous chain of \(\ce{C's}\) (straight chain) or a cyclic chain part of an organic compound.
| # of \(\ce{C’s}\) | Stem name | # of \(\ce{C’s}\) | Stem name | # of \(\ce{C’s}\) | Stem name |
|---|---|---|---|---|---|
| 1 | Meth | 11 | Undec | 30 | Triacont |
| 2 | Eth | 12 | Dodec | 40 | Tetracont |
| 3 | Prop | 13 | Tridec | 50 | Pentacont |
| 4 | But | 14 | Tetradec | 60 | Hexacont |
| 5 | Pent | 15 | Pentadec | 70 | Heptacont |
| 6 | Hex | 16 | Hexadec | 80 | Octacont |
| 7 | Hept | 17 | Heptadec | 90 | Nonacont |
| 8 | Oct | 18 | Octadec | 100 | hect |
| 9 | Non | 19 | Nonadec | ||
| 10 | Dec | 20 | Icos |
Straight chain alkanes
To name a straight chain alkane, the first syllable, i.e., alk- is replaced with the stem name that tells the number of \(\ce{C's}\) in the chain, and the last syllable, i.e., -ane is retained as a suffix in which 'an' tells all bonds are single bonds, and 'e' tells it is a hydrocarbon, i.e., -ane means an alkane.
For example, \(\ce{CH4}\) is called methane, \(\ce{CH3-CH3}\) is called ethane, and \(\ce{CH3-CH2-CH3}\) is called propane, where meth-, eth-, and prop- are the stem names representing one, two, and three \(\ce{C's}\). Table 2 shows the formulas, IUPAC names of some straight-chain alkanes, and some of their physical properties. The physical properties will be discussed in a later section.
| Formula | IUPAC Name | Melting point ( o C) | Boiling point ( o C) | Density (g/ml) | Physical state at 20 o C |
|---|---|---|---|---|---|
| \(\ce{CH4}\) | Methane | −182 | −162 | 0.000656 | Gas |
| \(\ce{C2H6}\) | Ethane | −183 | −89 | 0.00126 | Gas |
| \(\ce{C3H8}\) | Propane | −188 | −42 | 0.00201 | Gas |
| \(\ce{C4H10}\) | Butane | −138 | 0 | 0.00248 | Gas |
| \(\ce{C5H12}\) | Pentane | −130 | 36 | 0.626 | Liquid |
| \(\ce{C6H14}\) | Hexane | −95 | 69 | 0.659 | Liquid |
| \(\ce{C7H16}\) | Heptane | −91 | 98 | 0.684 | Liquid |
| \(\ce{C8H18}\) | Octane | −57 | 126 | 0.703 | Liquid |
| \(\ce{C9H20}\) | Nonane | −54 | 151 | 0.718 | Liquid |
| \(\ce{C10H22}\) | Decane | −30 | 174 | 0.730 | Liquid |
| \(\ce{C11H24}\) | Undecane | 196 | 0.740 | Liquid | |
| \(\ce{C12H26}\) | Dodecane | −10 | 216 | 0.749 | Liquid |
| \(\ce{C13H28}\) | Tridecane | −5.4 | 235 | 0.756 | Liquid |
| \(\ce{C14H30}\) | Tetradecane | 5.9 | 253 | 0.763 | Liquid |
| \(\ce{C15H32}\) | Pentadecane | 10 | 270 | 0.769 | Liquid |
| \(\ce{C16H34}\) | Hexadecane | 18 | 287 | 0.773 | Liquid |
| \(\ce{C17H36}\) | Heptadecane | 22 | 303 | 0.777 | Liquid |
| \(\ce{C18H38}\) | Octadecane | 28 | 317 | 0.781 | Solid |
| \(\ce{C19H40}\) | Nonadecane | 32 | 330 | 0.785 | Solid |
| \(\ce{C20H42}\) | Icosane | 37 | 343 | 0.789 | Solid |
A regular zigzag line, also called a straight chain, usually represents the hydrocarbons, but it may be a wiggle waggle because of rotation allowed around each \(\ce{C-C}\) single bond. Figure \(\PageIndex{2}\) shows the semi-condensed formulas of decane presented in five different configurations and the corresponding skeletal formulas. The terminal \(\ce{C}\) in a chain is the carbon at the ends connected with only one other \(\ce{C}\). The internal carbon in a chain is the \(\ce{C}\) connected with at least two other \(\ce{C's}\).
- A continuous \(\ce{C}\) chain starts from a terminal carbon and reaches the other terminal \(\ce{C}\) through the connected continuous \(\ce{C-C}\) single bonds.
- A straight-chain hydrocarbon has all the internal \(\ce{C's}\) connected with only two other \(\ce{C's}\), no matter how much wiggle waggle the chain may be.
Branched alkanes
If at least one \(\ce{C}\) in an alkane is connected with three or four other \(\ce{C's}\), it is a branched alkane . The parent chain is the longest continues \(\ce{C}\) chain. The smaller chains branching from the parent chain are called branches, substituents , or alkyl groups . R- represents a general alkyl group. Rules for naming the branched alkanes are the following.
- Find the most extended continuous \(\ce{C}\) chain, called the parent chain. Name it by using the stem name that tells the number of \(\ce{C's}\) in it with the suffix -ane that means it is an alkane.
Not
- Name the branch chain by using the stem name that tells the number of \(\ce{C's}\) in the continues \(\ce{C}\) chain of the branch with the suffix -yl that tells it is a branch. Use the name of the branch as a prefix to the name of the parent chain.
- If there are two or more longest chains of equal length, choose the parent chain with the highest number of branches on it.
not
- If there is more than one branch, arrange their names alphabetically as prefixes to the name of the parent chain.
- If there is more than one branch with the same name, list it once with the prefix di-, tri-, tetra-, etc., added to the branch name to represent two, three, four, etc. of the same branches. The alphabetization of the branch names is based on the branch name without the prefixes di-, tri-, tetra-, etc. For example, dimethyl, trimethyl, and tetramethyl are all alphabetized based on the 'm' of methyl, not based on the 'd' or 't' first alphabets of the prefixes.
not
-
Number the \(\ce{C's}\) of the parent chain starting from the end that gives the lowest number to whichever branch appears first on the parent chain.
not
- If numbering from left or right has the same number for the first branch, start the numbering from the side that gives a lower number to the branch that comes alphabetically first.
not
-
List the number to which the branch is attached to the parent chain before the branch name, separated by a hyphen. This is the IUPAC name of the alkane.
,
, and
- If two or more branches have the same name, list their numbers in increasing order, separated by commas, and the final number separated by a hyphen from the branch name.
- If two branches are on the same \(\ce{C}\), use the same number twice, i.e., once for each branch.
- If a branch is further branched, name it as if it is a parent chain following the above rules, except that i) the suffix of the brach is -yl instead of -ane, ii) insert a number from which the beach is attached to the parent chain before -yl, iii) Enclose the branch name in small brackets and place it as a prefix to the name of the parent chain, iv) insert a number before the bracket to which the beach is attached on the parent change, as shown in the following example.
- The terminal \(\ce{C's}\), i.e., the \(\ce{C's}\) with only one single bond with another \(\ce{C}\) are primary carbon .
- The \(\ce{H's}\) bonded to primary \(\ce{C's}\) are primary hydrogen .
- The \(\ce{C's}\) with two single bonds with other \(\ce{C's}\) are secondary carbon .
- The \(\ce{H's}\) bonded to secondary \(\ce{C's}\) are secondary hydrogen .
- The \(\ce{C's}\) with three single bonds with other \(\ce{C's}\) are tertiary carbon .
- The \(\ce{H's}\) bonded to tertiary \(\ce{C's}\) are tertiary hydrogen .
- The \(\ce{C's}\) with four single bonds with other \(\ce{C's}\) are quaternary carbon .
The quaternary \(\ce{C's}\) do not have any \(\ce{H}\) bonded to them. The following figure labels the primary, secondary, and tertiary carbons and hydrogen with different colors.
Cycloalkanes
An alkane with one ring of \(\ce{C's}\) in its structure is called cycloalkane. IUPAC rules for naming cycloalkanes are the following.
- If there is one ring of \(\ce{C's}\) without any substituent on it, it is named by using the stem name that tells the number of \(\ce{C's}\) in the ring with prefix cyclo- and suffix -ane, as shown in the following examples.
- If there is only one branch or a substituent on the ring of \(\ce{C's}\), then the ring supplies the parent name, and the branch name is added as a prefix, as for alkanes but without any number preceding the branch name. Some examples of IUPAC names of branched cycloalkanes are shown below.
, , and
- If there are two substitutes, they are listed alphabetically, preceded by location number, as in the case of alkanes. The numbering of \(\ce{C's}\) in the chain starts from the point of attachment of the substituent that comes alphabetically first. The numbering goes clockwise or counterclockwise, whichever gives the second substitute the lowest number, as shown in the following examples.
, , and
- If there are more than two substituents, they are listed as prefixes in alphabetic order as for alkanes. The ring \(\ce{C's}\) attached to one of the substituents is assigned number 1, and the numbering is continued clockwise or counterclockwise so that the addition of numbers assigned to the substituents is the lowest possible. Some examples are shown below.
, , and
Sometimes the ring name is treated as a branch with the last syllable changed from -ane to -yl, e.g., cyclopropane becomes cyclopropyl as a branch. This change is applied particularly when the chain attached to the ring has more number of \(\ce{C's}\) than the number of \(\ce{C's}\) in the ring.
Cycloalkanes with two alkyl groups on different carbons have two configurations: i) both alkyl groups pointing in the same direction are called 'cis' orientation, and ii) the two bulky groups pointing in the opposite direction are called 'trans' orientation, as shown below for the case of 1,2-dimethyl cyclopropane and 1-ethyl-3-methylcyclohexane examples.
The cis- and trans- configuration can not inter-convert by rotation around \(\ce{C-C}\) bond as the rotation is restricted by the carbon bridge in the case of cycloalkanes. Therefore, the cis- and trans-configuration of cycloalkane are isomers of each other, i.e., the same formula but different compounds. For example, cis-1,2-dimethylcyclopropane has a boiling point of 37 C , and its isomer, i.e., trans-1,2-dimethylcylopropane has a boiling point of 28 o C. This phenomenon, i.e., cis-, trans- isomerism, is common in all cyclic compounds.
Bonding and physical properties of alkanes
Bonding
The general formula of straight chain and branched alkanes is \(C_{n}H_{(2n+2)}\), and that of cycloalkanes having one ring is \(C_{n}H_{(2n)}\) where n is a positive integer. All \(\ce{C's}\) in alkanes are sp 3 hybridized, i.e., tetrahedral geometry with all bond angles about 109.5 o , as illustrated in Figure \(\PageIndex{3}\) for the case of ethane molecule. The bond energy for \(\ce{C-C}\) bond is ~3.8 eV and for \(\ce{C-H}\) bond is ~4.4 eV.
Polarity
The electronegative difference for the \(\ce{C-H}\) bond is 0.4, which falls in the nonpolar bond category. The electrostatic potential map of a molecule shows \(\delta\)+ region in blue, \(\delta\)- region in red, and neutral or nonpolar region in green color. Alkanes are nonpolar, as shown in green in the electrostatic map of ethane in Figure \(\PageIndex{3}\). The strong \(\ce{C-C}\) and \(\ce{C-H}\) bonds make the alkanes stable and less reactive. The nonpolar nature of alkanes makes them more inert chemically.
Intermolecular forces and boiling points
Alkanes are nonpolar molecules where the only intermolecular interaction is London dispersion forces. Straight-chain alkanes are a homologous series.
As the molar mass, and, consequently, the surface area and the number of electrons in the electron cloud, gradually increase, the London dispersion forces increase, resulting in a gradual increase in melting and boiling points, as shown in Table 2.
The first four members for the alkane series, i.e., \(\ce{CH4}\) to \(\ce{C4H10}\) are gases, the next from \(\ce{C5H12}\) to \(\ce{C17H36}\) are liquid and the larger alkanes are wax or soft solids.
Natural gas is the primary source of methane and ethane. Petroleum, which in the crude oil form is a viscous liquid composed of thousands of organic compounds, primarily hydrocarbons, is the primary source of alkanes and also the main source of organic raw materials. The differences in the boiling points allow the segregation of crude oil into different fractions, as illustrated in Figure \(\PageIndex{4}\).
Solubility
Alkanes are not soluble in water because alkanes are nonpolar, and water is a polar solvent. Alkanes are soluble in each other or other nonpolar solvents. Alkanes float on water because they are less dense than water. The density of liquid alkanes varies in the range of 0.6 g/ml to 0.8 g/ml, which is less than the density of water, which is ~1 g/ml at room temperature. Petroleum and crude oil are mainly composed of alkanes, which is why oil spills make a layer of oil float on the water's surface.
Changes in physical properties with branching
The physical properties of branched alkanes are almost the same as that of straight chain alkanes, except that:
- the polling point increases as the molar mass increases, and
- for the same molar mass, the boiling point decreases as the branching increases, as shown in Table 1.
| Number of \(\ce{C's}\) | Name | Condensed formula | Boiling point |
|---|---|---|---|
| 5 | Pentane | \(\ce{CH3CH2CH2CH2CH3}\) | 36 o C |
| 6 | Hexane | \(\ce{CH3CH2CH2CH2CH2CH3}\) | 69 o C |
| 6 | 2-Methylpentane | \(\ce{(CH3)2CHCH2CH2CH3}\) | 60 o C |
| 6 | 2,2-Dimethylbutane | \(\ce{(CH3)3CCH2CH3}\) | 50 o C |
| 7 | Heptane | \(\ce{CH3CH2CH2CH2CH2CH2CH3}\) | 98 o C |
Straight-chain alkanes are long and cylindrical, as illustrated in Figure \(\PageIndex{5}\). Branching makes them spherical, resulting in less contact area between molecules, fewer London dispersion forces, and lower boiling points. Note that the effect of molar mass is dominant over that of branching.
Properties of cycloalkanes
The physical properties of cycloalkanes are almost similar to alkanes, except that bonds have to bend to acquire ring-shape that causes angle strain. Further, the molecule can not acquire the most stable configuration due to no or limited rotation around \(\ce{C-C}\) single bond. Therefore, the cycloalkanes, mainly the smaller three or four-member rings, are relatively unstable. Except for cyclopropane, there is a limited rotation possible around \(\ce{C-C}\) bond that allows the molecule to acquire bent-shape, rather than a planer shape that releases some of the strains, as shown by models in Figure \(\PageIndex{6}\). The strain decreases, and the stability increases in this order: cyclopropane > cyclobutane > cyclopentane > cyclohexane < cycloheptane > cyclooctane. Notably, the five and six-membered rings, e.g., cyclopentane and cyclohexane, have little or no strain, so they are the most common cyclic compounds found in nature.
Some uses of alkanes
The first four members, i.e., methane, ethane, propane, and butane, are gases used as fuels. The next four are components of gasoline, and higher than that are components of kerosene and diesel, used as fuels.
Alkanes and other hydrocarbons are part of almost all organic compounds but are not so common in pure form in living things. Some examples of uses in biological systems are the following. Long-alkanes having eighteen or more \(\ce{C,s}\) are semisolids found in waxes as in vaseline used in ointments and cosmetics. The waxes are found on surfaces of leaves and fruits to protect against water loss, prevent the leaching of essential minerals by rain, and protect against bacteria, fungi, and harmful insects. Alkanes are also found in some pheromones, e.g., sand bees use tricosane (\(\ce{C23H48}\)), pentacosane (\(\ce{C25H53}\)), and heptacosane (\(\ce{C27H56}\)) to identify a mate.
Alkenes
Nomenclature of alkenes
The procedure for IUPAC naming alkenes is the same as alkanes, except for the following changes.
- The suffix to the stem name of the parent chain is -ene, where 'en' tells its an alkene, and 'e' means it is a hydrocarbon, e.g., \(\ce{CH2=CH2}\) is ethene.
- A location number of the first \(\ce{C}\) of \(\ce{C=C}\) bond is inserted between the stem- name and -ene suffix, separated by hyphens, as in the following example, except for propene that does not need to show the location number.
3- ethy lpent- 2 -ene
-
The parent chain numbering starts from the end that gives the lowest number to the first \(\ce{C}\) of \(\ce{C=C}\) bond.
- If the first \(\ce{C}\) of \(\ce{C=C}\) bond receives the same number from either side, then the rule for alkane applies, i.e., start numbering from the side that gives the lower number to the branch that appears first, as shown in the following examples.
3-ethyl - 6- methyl hept- 3 -ene, \(\xcancel{\text{5-ethyl-2-methylhept-4-ene}}\)
4-ethyl - 2- methyl hex- 3 -ene, \(\xcancel{\text{3-ethyl-5-methylhept-3-ene}}\)
- If two or more double bonds exist, the suffix -ene is changed to -diene, triene-, etc. The rest of the nomenclature remains the same.
buta-1,3-diene
2- methyl buta-1,3-diene
-
In cycloalkenes, the \(\ce{C's}\) of \(\ce{C=C}\) are #1 and #2 and do not need to show the location number.
- If there is a branch on cycloalkene, the \(\ce{C'}\) of \(\ce{C=C}\) bond are #1 and #2, and the numbering is continued clockwise or counterclockwise to give the lowest number to the branch, as shown in the following examples.
cyclohexene
4- methyl cyclohex-1-ene
- there are two alkyl groups, one on each \(\ce{C}\) of \(\ce{C=C}\) bond (or third or fourth group different from the other group on the same \(\ce{C}\)) then a descriptor E indicating cis, or Z indicating the trans orientation of the bulkier groups, preceded by the location number of the first \(\ce{C}\) of the \(\ce{C=C}\) bond, is added as a prefix enclosed within brackets.
(2E)-pent-2-ene
(2Z)-pent-2-ene
- if the double bond is in a branch, the -ene suffix is replaced with -enyl, where 'en' tells it is an alkene and 'yl' means it is a branch.
In the older system:
- The parent chain may not be the longest, but the parent chain is the longest chain counting the double bond, and
- The location number of the double bond is inserted before the stem name, e.g., but-2-ene (\(\ce{CH3CH=CHCH3}\)) is written as 2-butene in the old IUPAC nomenclature.
Some common names are still being used in the chemical literature and industries. For example, ethylene, propylene, isobutylene, and isoprene are often used in place of IUPAC names ethene, propene, 2-methylpropene, and 2-methylbuta-1,3-diene.
| Structure and IUPAC name |
ethene |
prop-1-ene |
2-methylprop-1-ene |
2-methylbuta-1,3-diene |
| Common name | ethylene | propylene | isobutylene | isoprene |
Bonding and physical properties of alkenes
Alkenes are nonpolar compounds with only London dispersion forces as intermolecular forces, like alkanes. Therefore, the physical properties of alkenes are similar to alkanes with the same \(\ce{C}\) skeleton, i.e., they are insoluble in water, soluble in nonpolar solvents, and have densities lower than that of water. The significant difference between alkanes and alkenes is the \(\ce{C=C}\) bond in alkenes described below.
bonding
The general formula of alkenes with one double bond is \(C_{n}H_{(2n)}\), where n is a positive integer. The \(\ce{C's}\) in \(\ce{C=C}\) bond are sp 2 hybridized with trigonal planer geometry around each of the sp 2 \(\ce{C's}\) with bond angles around 120 o , as illustrated in Figure \(\PageIndex{7}\). Except for \(\ce{C=C}\), all other \(\ce{C's}\) in alkenes are sp 3 hybridized, i.e., tetrahedral geometry with all bond angles about 109.5 o . Although alkenes are nonpolar overall, the Electrostatic potential map in Figure \(\PageIndex{7}\) shows a red region where \(\pi\)-electrons are located above and below the \(\sigma\) bond and slightly blush region where \(\ce{H's}\) of sp2 \(\ce{C's}\) are located. These \(\delta {-}\) and \(\delta {+}\) regions impart reactivity to alkenes that will be described in later sections.
The general formula of an alkane is \(C_{n}H_{(2n+2)}\), where n is a positive integer. One degree of unsaturation is two less \(\ce{H's}\) than in an alkane of the same carbon structure. For example, one ring in cycloalkane or one double bond in alkenes reduces two \(\ce{H's}\), one unsaturation degree. The degree of unsaturation is also known as the index of hydrogen deficiency (IHD) . Remember, if a halogen with one bond is added to an alkane, it replaces one \(\ce{H}\), an \(\ce{O}\) with two bonds does not change the number of \(\ce{H's}\), and a \(\ce{N}\) with three bonds add one \(\ce{H}\). Based on this information, the degree of unsaturation, i.e., number of double bonds + rings, can be calculated by the following formula.
\[IHD = (C's + 1) - {\frac{H's-N's+X's}{2}}\nonumber \]
, where \(\ce{IHD}\) is an index of hydrogen deficiency, which is equivalent to the number of \(\pi\) bonds + rings, \(\ce{C's}\) is the number of \(\ce{C}\) atoms, \(\ce{H's}\) is the number of \(\ce{H}\) atoms, \(\ce{N's}\) is the number of \(\ce{N}\) atoms, and \(\ce{X's}\) is number of halogen atoms in the molecular formula.
If oxygen is in the molecule, it is not included in the IHD calculation.
What is the degree of saturation in cyclohexane (\(\ce{C6H10}\)?
Solution
\(\ce{C's}\) = 6, \(\ce{H's}\) = 10, \(\ce{N's}\) = 0, \(\ce{X's}\) = 0.
\(IHD = (C's + 1) - {\frac{H's-N's+X's}{2}}\) = \((6 + 1) - {\frac{10-0+0}{2}}\) = 2, which is one \(\pi\)-bond and one ring as shown below.
Substituents attached to sp 2 hybridized \(\ce{C's}\) of alkenes are also called vinylic. The bond energy for \(\ce{C=C}\) bond is ~7.4 eV which is less than two times of ~3.8 eV of \(\ce{C-C}\) bond, and the vinylic \(\ce{=C-H}\) bond is ~4.8 eV, which is stronger than alkane \(\ce{-C-H}\)~4.4 eV. Comparison ~7.4 eV for \(\ce{C=C}\) bond vs 3.8 eV for \(\ce{C-C}\) bond shows that, although a double bond ((\(\sigma\) + \(\pi\)) is stronger than a single bond, a \(\pi\) bond is weaker than a \(\sigma\) bond.
If there are two alkyl groups, one on each \(\ce{C}\) of a double bond, they can be pointing in the same direction, called cis orientation, or in the opposite direction, called trans orientation, as shown in the Figure below for the case of but-2-ne. The cis and trans alkenes of the same molecular formula are different compounds related as isomers of each other. This is because three is no rotation around \(\ce{C=C}\) bond without breaking and re-making the \(\pi\) bond.
Cis trans isomerism plays a vital role in biological systems. For example, fatty acids are significant components of fats and vegetable oils. Fatty acid contains a long alkyl chain attached to a carboxylic acid group. The alkyl chain is usually alkane in the cases of animal fats or includes one or more double bonds, almost always in the cis configuration in the cases of vegetable oils and fish oil. For example, models of steric acid with no double bonds, oleic acid with one cis double bond, and linoleic acid with two cis double bonds are shown below.
The cis-double bonds change the long cylindrical shape of fatty acids to a bent or more spherical shape with less contact area with the neighboring molecules. It results in lower intermolecular forces and melting points than the same chain with a trans or no double bond. That is why vegetable oils with more cis-double bonds are liquids, while animal fats with little to no double bonds are solids at room temperature. Another example is vitamin A having five trans double bonds, as shown in the model drawing below. Vitamin A leads to retinol synthesis, which plays a crucial role in our vision based on switching one of its double bonds between cis and trans configurations.
Model of vitamin A molecule with five trans double bonds.
Polarity and chemical reactivity
Although \(\ce{C=C}\) bond is nonpolar, the electrostatic potential map of ethene in Figure \(\PageIndex{7}\) shows there is an electron-rich region (red color in the map) above and below the axis joining the nuclei. This is because the \(\sigma\) bond is along the axis of nuclei, and the \(\pi\) bond places electrons above and below the \(\sigma\) bond. Molecules with negative charge or electron-rich (\(\delta\)-) regions, as in the case of \(\ce{C=C}\) bond in alkene, are nucleophiles . Molecules with positive charge or electron-deficient (\(\delta\)+) region are electrophiles. Chemical reactions between nucleophiles and electrophiles are facilitated by the attraction between the opposite charges that allow them to come closer to each other for bond-making and bond-breaking to happen.
The term acid is used for electrophile when \(\delta\)+ part is a proton, and the base is used in the place of nucleophile when \(\delta\)- region is a lone pair of electrons interacting with a proton.
Alkenes uses and importance in biological systems
Alkenes are among the raw materials for the synthesis of several organic chemicals. Mainly, ethene and propene are the raw material for synthesizing commercially important polymers: polyethylene and polypropylene.
Ethene is a ripening agent for fruits. It allows fruit growers to pick fruits while they are green and less susceptible to bruising and then treat them with ethene gas for ripening when ready for sale. Lycopene, -a red pigment in tomatoes, and carotene, -a yellow pigment in carrots, are polyenes, shown below.
Terpenes are a class of natural products having a general formula \(\ce{(C_{5}H_{8})n}\), where n is a positive integer. Isoprene (\(\ce{(C_{5}H_{8})}\), i.e., 2-methylbuta-1,3-diene is the building block of terpene.
isoprene, IUPAC name: 2-methylbuta-1,3-diene
Terpenoids are derivatives of terpenes in which heteroatoms/functional group is added. Natural rubber, essential oils, and steroids are examples of terpenes or terpenoids. They play important roles in living things, e.g., in defense against diseases. Some examples of terpenes and terpenoids are shown below, with the isoprene skeleton highlighted.
Alkynes
Nomenclature
The IUPAC nomenclature of alkynes followes the same rule as alkenes, except for the following changes.
- The suffix to the stem name is -yne where 'yn' tells it is an alkyne, and 'e' means it is a hydrocarbon, e.g., \(\ce{CH≡CH}\) is ethyne.
- If a double and a triple bond are in the same molecule, they are treated equally. One exception is when the double receives the same number as the triple bond counted from the other end of the parent chain; preference is given to the double bond, as shown in the following examples.
Note: in the second example, the double bond received a lower number than the triple bond. The common name of ethyne is acetylene, which is also accepted as its IUPAC name.
Bonding and physical properties of alkynes
Alkynes are nonpolar compounds with only London dispersion forces as intermolecular forces, like alkenes and alkanes. Therefore, the physical properties of alkynes are similar to alkanes with the same \(\ce{C}\) skeleton. They are insoluble in water, soluble in nonpolar solvents, and have lower densities than water's. The significant difference is the presence of a triple bond in alkynes described below.
Bonding
The general formula of alkyne with one triple bond is \(C_{n}H_{(2n-2)}\), where n is a positive integer. The \(\ce{C's}\) in \(\ce{C≡C}\) bond are sp hybridized with a linear geometry around each of the sp \(\ce{C's}\) with bond angles 180 o , as illustrated in Figure \(\PageIndex{8}\). Substituents attached to sp hybridized \(\ce{C's}\) of alkynes are also called acetylenic . The bond energy for \(\ce{C≡C}\) bond is ~10 eV which is about 2.5 times stronger than that of \(\ce{C-C}\) bond. Due to the cylindrical shape of two \(\pi\)-bond clouds, there is free rotation possible around a \(\ce{C≡C}\) bond, but due to the linear geometry, it means less.
polarity and chemical reactivity
Although \(\ce{C≡C}\) bond is nonpolar, the electrostatic potential map of ethyne in Figure \(\PageIndex{8}\) shows there is an electron-rich region (red color in the map) around the axis joining the nuclei. This is due to \(\pi\)-bonding electrons around the \(\sigma\) bond. The \(\pi\) bonds provide \(\delta\)- region in alkynes, making them nucleophiles, like alkenes. The \(\ce{H's}\) are also blue in the electrostatic potential map of ethyne, showing that the \(\ce{C-H}\) bond is polar with \(\delta {+}\) charge on \(\ce{H}\).
The \(\ce{C-H}\) bond is nonpolar with an electronegativity difference of 0.4 between \(\ce{C}\) and \(\ce{H}\). The green color in the electrostatic potential map of ethane reflects it. However, the \(\ce{C-H}\) bond becomes shorter, stronger, and polar with a change in the hybridization of the \(\ce{C}\) from sp 3 to sp 2 and sp. It means the electronegativity of \(\ce{C}\) increases with a change in the hybridization from sp 3 to sp 2 and sp. The reason is that the sp orbital is wider and shorter due to more s-orbital character (50% s-character) than sp 2 orbital (33% s-character), which, in turn is wider and shorter the sp 3 orbital (25% s-character), as illustrated in Fig. 7. The wide and shorter orbitals place the electron closer to the \(\ce{C}\) nucleus for a stronger attraction to the nucleus, which means higher electronegativity.
Uses of alkynes
Alkynes are less common in biological systems or the chemical industry than alkenes and alkanes. The first member, i.e., acetylene, is used in oxyacetylene torches for cutting and welding metals, as illustrated in the figure on the right. Acetylene is also a starting material for several chemicals used in the chemical industry.
Aromatic hydrocarbons
Aromatic hydrocarbons contain a planer cycle or ring (not bent or puckered) in their structure with all atoms sp 2 -hybridized and \(\pi\text{-electrons in the ring} = 4n +2\), where n is a positive integer. Benzene is an example of an aromatic ring that will be described here. Although aromatic hydrocarbons have systematic names like other hydrocarbons, some alternate and familiar names are so widely used that they are accepted as equivalent to IUPAC names.
Nomenclature of benzene derivatives
- The name of the six sp 2 hybridized \(\ce{C}\)-chain is benzene. It is used as the parent name for the substituted benzene rings.
-
If there is only one substituent, i.e., mono-substituted benzene, use the branch name as a prefix to the parent name 'benzene' without numbering.
- If the substituent is methyl (\(\ce{-CH3}\)), the parent name 'toluene' is often used for it, and if the substituent is ethenyl (\(\ce{-CH=CH2}\)), it is also called styrene. A few examples are given below.
-
If there are two substituents (di-substituted benzene), start numbering the chain from the first substituent and go clockwise or counterclockwise so that the second substituent receives the lower number. List the substitutes alphabetically, preceded by their number, as prefixes to the parent name 'benzene.'
- If one of the two substitutes is methyl (\(\ce{-CH3}\)), start numbering the chain from the point of attachment of methyl and go clockwise or counterclockwise so that the second substitute receives the lower number. List the second substituent preceded by its number as a prefix to the parent name 'toluene.'
- If there are two methyls (\(\ce{-CH3}\)) substituents, the parent name 'xylene' is often used. Numbering starts from one of the methyls and goes clockwise or counterclockwise, so the second methyl receives the lower number. List the numbers as prefixes to the parent name 'xylene,' numbers separated by commas from each other and by a hyphen from the parent name, as usual.
- Alternate to the numbers, the word 'ortho' or 'o' for a 1,2 relationship, 'meta' or 'm' for a 1,3 relationship, and 'para' or 'p' for a 1,3 relationship of two substituents on a benzene ring are also accepted. A few examples are the following.
-
If there are more than two substituents (poly-substituted benzene), start numbering the chain from the first substituent and go clockwise or counterclockwise so that the substituents receive an overall lower number. List the substitutes alphabetically, preceded by their number, as prefixes to the parent name 'benzene.'
- If one substituent is methyl (\(\ce{-CH3}\)), start numbering from the point of attachment of methyl (\(\ce{-CH3}\)) and go as usual to assign the lower overall number to the rest of the substituents, and use the parent name 'toluene.'
Bonding and physical properties of aromatic hydrocarbons
Aromatic hydrocarbons are nonpolar compounds with only London dispersion forces as intermolecular forces, like other hydrocarbons. Benzene has low solubility in water (~1.8 g/L at room temperature), melts at 5.53 o C, and boils at 80.1 o C, which are comparable to that of cyclohexane (immiscible in water, 6.5 o C, and boiling point is 80.7 o C). The primary difference is resonance resulting in significant stabilization of benzene compared to any other hydrocarbon, as described below.
Bonding
All \(\ce{C's}\) in the benzene ring are sp 2 hybridized, i.e., trigonal planer geometry with all bond angles about 120 o , as illustrated in Figure \(\PageIndex{10}\). All \(\ce{C's}\) in benzene ring and \(\ce{H's}\) or other substituents attached to them are in one plane. Substituents attached to the \(\ce{C's}\) of the benzene ring are also called phenylic . The bond energy for \(\ce{C\overset\cdots{-}C}\) bond in benzene ring is ~5.4 eV which is between a \(\ce{C-C}\) and a \(\ce{C=C}\) bond. the phenylic \(\ce{C-H}\) bond is ~4.9 eV, almost the same as in alkenes. The \(\ce{C\overset\cdots{-}C}\) bond in the benzene ring is (140 pm), i.e., about in the middle of a single and a double bond. This is explained by benzene having two equal resonance contributors in which a single and double bond alternates and the resonance hybrid has a 50% \(\pi\)-bond.
polarity and chemical reactivity
Although benzene is nonpolar, the electrostatic potential map of benzene in Figure \(\PageIndex{10}\) shows there is a circle of an electron-rich region (red color in the map) above and below the ring. This is due to the delocalized \(\pi\)-bond extending over all \(\ce{C's}\) of the benzene ring. The resonance in a cycle of aromatic rings imparts significantly more resonance stabilization, called aromatic stabilization, than in any other hydrocarbon.
Although the \(\delta^-\) region of delocalized elections makes benzene a nucleophile, its nucleophilic character is weak because donating these electrons causes a loss of aromatic stabilization. Therefore benzene needs stronger electrophiles to react with than an alkene and follows a different reaction route than an alkene does.
Benzene increases the risk of leukemia and other blood disorders. Polycyclic aromatic hydrocarbons (PAHs) contain two or more benzene rings fused edge to edge. Naphthalene has two benzene rings fused. Naphthalene is used in mothballs. Anthracene has there benzene rings linked together. Anthracene is used in the manufacture of dyes. Phenanthrene is also three benzene rings fused. Phenanthrene is known to cause cancer, i.e., it is a carcinogen . Benzo[a]pyrene has five benzene rings connected. Benzo[a]pyrene is a potent carcinogen found in tobacco smoke, barbecued meat, and automobile exhaust.
Uses of aromatic hydrocarbons
Benzene is used as a raw material for synthesizing several chemicals. Some familiar aromatic ring-containing compounds include explosives, e.g., trinitrotoluene (TNT), common medicines, e.g., aspirin, acetaminophen, ibuprofen, sulfanilamide, and flavoring agents, e.g., vanillin, shown below.
(Aspirin)
(Acetaminophene)
(Ibuprofen)
(Sulfanilamide)
(Vanilin)


30.1: Using Operations on Decimals to Solve Problems
Lesson
Let's solve some problems using decimals.
Exercise \(\PageIndex{1}\): Close Estimates
For each expression, choose the best estimate of its value.
-
\(76.2\div 15\)
- \(0.5\)
- \(5\)
- \(50\)
-
\(56.34\div 48\)
- \(1\)
- \(10\)
- \(100\)
-
\(124.3\div 20\)
- \(6\)
- \(60\)
- \(600\)
Exercise \(\PageIndex{2}\): Applying Division with Decimals
Your teacher will assign to you either Problem A or Problem B. Work together as a group to answer the questions. Be prepared to create a visual display to show your reasoning with the class.
Problem A:
A piece of rope is 5.75 meters in length.
- If it is cut into 20 equal pieces, how long will each piece be?
- If it is cut into 0.05-meter pieces, how many pieces will there be?
Problem B:
A tortoise travels 0.945 miles in 3.5 hours.
- If it moves at a constant speed, how many miles per hour is it traveling?
- At this rate, how long will it take the tortoise to travel 4.86 miles?
Exercise \(\PageIndex{3}\): Distance between Hurdles
There are 10 equally-spaced hurdles on a race track. The first hurdle is 13.72 meters from the start line. The final hurdle is 14.02 meters from the finish line. The race track is 110 meters long.
- Draw a diagram that shows the hurdles on the race track. Label all known measurements.
- How far are the hurdles from one another? Explain or show your reasoning.
-
A professional runner takes 3 strides between each pair of hurdles. The runner leaves the ground 2.2 meters
before
the hurdle and returns to the ground 1 meter
after
the hurdle.
About how long are each of the runner’s strides between the hurdles? Show your reasoning.
Exercise \(\PageIndex{4}\): Examining a Tennis Court
Here is a diagram of a tennis court.
The full tennis court, used for doubles, is a rectangle. All of the angles made by the line segments in the diagram are right angles.
- The net partitions the tennis court into two halves. Is each half a square? Explain your reasoning.
- Is the service line halfway between the net and the baseline? Explain your reasoning.
-
Lines painted on a tennis court are 5 cm wide. A painter made markings to show the length and width of the court, then painted the lines to the outside of the markings.
- Did the painter’s mistake increase or decrease the overall size of the tennis court? Explain how you know.
- By how many square meters did the court’s size change? Explain your reasoning.
Summary
Diagrams can help us communicate and model mathematics. A clearly-labeled diagram helps us visualize what is happening in a problem and accurately communicate the information we need.
Sports offer great examples of how diagrams can help us solve problems. For example, to show the placement of the running hurdles in a diagram, we needed to know what the distances 13.72 and 14.02 meters tell us and the number of hurdles to draw. An accurate diagram not only helped us set up and solve the problem correctly, but also helped us see that there are only nine spaces between ten hurdles.
To communicate information clearly and solve problems correctly, it is also important to be precise in our measurements and calculations, especially when they involve decimals.
In tennis, for example, the length of the court is 23.77 meters. Because the boundary lines on a tennis court have a significant width, we would want to know whether this measurement is taken between the inside of the lines, the center of the lines, or the outside of the lines. Diagrams can help us attend to this detail, as shown here.
The accuracy of this measurement matters to the tennis players who use the court, so it matters to those who paint the boundaries as well. The tennis players practice their shots to be on or within certain lines. If the tennis court on which they play is not precisely measured, their shots may not land as intended in relation to the boundaries. Court painters usually need to be sure their measurements are accurate to within \(\frac{1}{100}\) of a meter or one centimeter.
Practice
Exercise \(\PageIndex{5}\)
A roll of ribbon was 12 meters long. Diego cut 9 pieces of ribbon that were 0.4 meter each to tie some presents. He then used the remaining ribbon to make some wreaths. Each wreath required 0.6 meter. For each question, explain your reasoning.
- How many meters of ribbon were available for making wreaths?
- How many wreaths could Diego make with the available ribbon?
Exercise \(\PageIndex{6}\)
The Amazon rainforest covered 6.42 million square kilometers in 1994. In 2014, it covered only \(\frac{50}{59}\) as much. Which is closest to the area of the Amazon forest in 2014? Explain how you know without calculating the exact area.
- 6.4 million km 2
- 5.4 million km 2
- 4.4 million km 2
- 3.4 million km 2
- 2.4 million km 2
Exercise \(\PageIndex{7}\)
To get an A in her math class, Jada needs to have at least 90% of the total number of points possible. The table shows Jada’s results before the final test in the class.
| Jada's points | total points possible | |
|---|---|---|
| Homework | \(141\) | \(150\) |
| Test 1 | \(87\) | \(100\) |
| Test 2 | \(81\) | \(100\) |
| Test 3 | \(91\) | \(100\) |
- Does Jada have 90% of the total possible points before the final test? Explain how you know.
- Jada thinks that if she gets at least 92 out of 100 on the final test, she will get an A in the class. Do you agree? Explain.
Exercise \(\PageIndex{8}\)
Find the following differences. Show your reasoning.
- \(0.151-0.028\)
- \(0.106-0.0315\)
- \(3.572-2.6014\)
(From Unit 5.2.3)
Exercise \(\PageIndex{9}\)
Find these quotients. Show your reasoning.
- \(24.2\div 1.1\)
- \(13.25\div 0.4\)
- \(170.28\div 0.08\)
(From Unit 5.4.5)