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Learning Objectives
• Identify, given IUPAC names to the structural formula, and draw the structural formula from the IUPAC name of simple aldehydes, ketones, and imines.
• Understand the polarity and predict some physical properties, reactive sites, and relative reactivities of aldehydes, ketones, and imines based on the bond polarity.
Carbonyl group and its subclasses
A ($\ce{C=O}$) group is a carbonyl group. It has a $\sigma$-bond between sp2 orbitals and a $\pi$ between p orbitals of a $\ce{C}$ and an $\ce{O}$. Lone pairs of electrons occupy the remaining two sp2 orbitals of $\ce{O}$ as shown here: $\ce{-{\underset{|}{C}}=\overset{\Large{\cdot\cdot}}{O}\!:}$. The lone pairs are usually not shown. The carbonyl group is represented as $\ce{-{\underset{|}{C}}={O}}$, $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!{-}}$ or simply as $\ce{C=O}$. If the $\ce{O}$ is replaced with $\ce{N}$ it becomes an imine $\ce{C=NR}$ group.
The carbonyl group is subdivided into aldehydes, ketones, carboxylic acids, and carboxylic acid derivates based on what is bonded to the carbonyl carbon, i.e., what are the X and Y in this formula: $\ce{X-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Y}$, as listed in Table 1.
Table 1: Sub-groups of carbonyl group based on what is $\ce{X}$ and $\ce{Y}$ in the general formula of a carbonyl group: $\ce{X-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Y}$.
$\ce{X}$ $\ce{Y}$ Group name General formula
Hydrocrbon ($\ce{R}$) or $\ce{H}$ $\ce{H}$ Aldehyde $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-H}$
Hydrocarbon ($\ce{R}$) Hydrocarbon ($\ce{R'}$) Ketone $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R'}$
Hydrocrbon ($\ce{R}$) or $\ce{H}$ $\ce{-\!\!\!\!\!{\overset{\overset{\huge\;\enspace{NR}}|\!\!\!\!\!\!|\enspace\enspace}{C}}\!\!\!\!\!-R''}$ Imine $\ce{R'-\!\!\!\!\!{\overset{\overset{\huge\;\enspace{NR}}|\!\!\!\!\!\!|\enspace\enspace}{C}}\!\!\!\!\!-R''}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Alcohol group ($\ce{-OH}$) Carboxylic acid $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Oxy anion ($\ce{-O^{-}}$) Carboxylate anion $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Halogen ($\ce{-X}$, where $\ce{-X}$ = $\ce{-F}$, $\ce{-Cl}$, $\ce{-Br}$, or $\ce{-I}$) Acid halide $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-X}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Carboxyl ($\ce{-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}-R'}}$) Acid anhydride $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}-R'}}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Alkoxy ($\ce{-OR'}$) Ester $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OR'}$
Hydrocarbon ($\ce{R}$) or $\ce{H}$ Amine ($\ce{-NH2}$, $\ce{-NHR'}$, or $\ce{-NR'R''}$) Amide $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2}$
Hydrocarbon ($\ce{R}$) Cyano ($\ce{-C≡N}$) Nitrile $\ce{R-C≡N}$
The first three, i.e., aldehydes, ketones, and imines, are described in this section. The others, i.e., carboxylic acids and their derivatives, including carboxylate anion, acid anhydrides, acid halides, esters, and amides, are described in the next section. The nitrile group is classified as a carboxylic acid derivative.
Aldehydes
An aldehyde has either two $\ce{H's}$ or a $\ce{H}$ and a hydrocarbon $\ce{R}$ single bonded with the carbonyl carbon, i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-H}$ group. The condensed form of the aldehyde group is $\ce{-CHO}$.
Nomenclature of aldehydes
The IUPAC naming of aldehydes follows the rules of naming alcohols, with the following changes.
• The longest chain containing the aldehyde group is chosen as the parent name with the last 'e' of the suffix replaced with -al, e.g.,$\ce{CH3CHO}$ is ethanal.
• Start numbering from the aldehyde $\ce{C}$. If there is one aldehyde group, it is understood to be #1, i.e., no need to show the number. A few examples are shown below.
• If more than one functional groups are present, the order of preference of functional groups is the following: $\overleftarrow{\text{carboxylic acid> aldehyde>ketone>alcohol>amine>thiol}}$. A suffix represents the higher priority group, and prefixes represent the other groups. For example, 2-hydroxypropanal shown on the right.
• The suffix "carbaldehyde" represents an aldehyde group bonded to a cycloalkane. Numbering starts from the point of attachment of the aldehyde to the cycloalkane, as shown in the examples below.
• An aldehyde group bonded to a benzene ring has the parent name benzaldehyde, as shown in the following examples. Numbering, if needed, starts from the aldehyde attachment point.
The common name of methanal ($\ce{H2C=O}$) is formaldehyde, and ethanal ($\ce{CH3-CHO}$) is acetaldehyde. Aldehydes take their common names from the common names of carboxylic acids, described in a later section.
Example $1$
What is the IUPAC name of this compound shown in the figure on the right?
Solution
the longest chain containing aldehyde groups is four $\ce{C}$, i.e., butane.
Change the final 'e' with -al, and since there are two aldehyde groups, use dial.
Answer: butanedial
Note 1: the final 'e' is dropped when the following letter added starts with a vowel, e.g., -al; otherwise, it is retained as in butanedial.
Note 2: Location number is not needed even for this dialdehyde because it is understood that the aldehyde groups are at the end of the chain.
Example $2$
What is the IUPAC name of this compound shown in the figure on the right?
Solution
The longest chain is four $\ce{C}$, i.e., butane
There is an aldehyde group, so change the final 'e' with -al, i.e., butanal.
There is an alcohol group that is added as a hydroxy- prefix, i.e., hydroxybutanal.
The alcohol group needs the location number, start from aldehyde, and the alcohol group receives #4.
Answer: 4-hydroxybutanal.
Example $3$
Write the skeletal formula of 4-hydroxy-3-methylbenzaldehyde?
Solution
The parent name is benzaldehyde, i.e., an aromatic ring with an aldehyde group as shown on the right.
Start the number from the point of attachment of aldehyde and go either clockwise or counterclockwise and place a methyl group at #3 and an alcohol group at #4.
Answer:
Physical properties of aldehydes
The aldehydes have sp2-hybridized $\ce{C}$ and $\ce{O}$ with a double bond (a $\sigma$ and a $\pi$-bond) between them, two lone pairs occupying two sp2-orbital of an $\ce{O}$, and the $\ce{C}$ bonded with a $\ce{H}$ and a hydrocarbon $\ce{R}$ or another $\ce{H}$, i.e., $\ce{-{\underset{|}{C}}=\overset{\Large{\cdot\cdot}}{O}\!:}$. The $\ce{C=O}$ bond is polar, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, because $\ce{O}$ are more electronegative than $\ce{C}$ (3.3-2.1 = 0.9), as shown in Figure $1$. It makes the $\ce{C}$ a $\delta^{+}$, i.e., an electrophile in reactivity and the $\ce{O}$ a $\delta^{-}$, i.e., a nucleophile or a base in reactivity. The lone pair of electrons on $\ce{O}$ add to the base character of the carbonyl $\ce{O}$.
Note that the $\delta^{+}$ region extend from carbonyl $\ce{C}$ to the $\ce{C}$ attached to it, designated as $\alpha{C}$ and the $\ce{H's}$ on the $\alpha{C}$.
$\alpha$-, $\beta$-, $\gamma$-, $\delta$- designations of $\ce{C's}$
The $\ce{C}$ directly bonded to a functional group is designated as $\alpha$, the next one as $\beta$, the third as $\gamma$, and so on. A few examples are shown below for explanation.
, , and
Aldehyde group ($\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$) is polar. Aldehydes have boiling points higher than the alkanes of comparable molar mass due to the dipole-dipole interaction in addition to London dispersion forces. Aldehydes have boiling points lower than alcohols of the comparable mass because aldehydes do not have hydrogen bonding with each other, as compared below.
Name Condensed formula Molar mass Boiling point
Pentane $\ce{CH3CH2CH2CH2CH3}$ 72 g/mol 36 oC
Butanal $\ce{CH3CH2CH2CHO}$ 72 g/mol 76 oC
Butanol $\ce{CH3CH2CH2CH2OH}$ 74 g/mol 117 oC
Aldehydes can establish hydrogen bonding with water molecules through $\ce{\overset{\delta{-}}{O}}$ of carbonyl group with $\ce{\overset{\delta{+}}{H}}$ of water molecules. Therefore, aldehydes up to four $\ce{C's}$, i.e., methanal, ethanal, propanal, and butanal, are soluble in water. Pentanal with five $\ce{C's}$ is slightly soluble, and hexanal with six $\ce{C's}$ is insoluble. Aldehydes, except for formaldehyde, generally smell pleasant and are used in perfumes.
Ketones
A ketone has two hydrocarbon groups bonded with the carbonyl carbon, i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R'}$ group. The condensed form of the aldehyde group is $\ce{RCOR'}$.
Nomenclature of ketones
The IUPAC naming of ketones followed the rules of naming alcohols and aldehydes, summarized below.
• The longest chain containing the ketone group is chosen as the parent name, with the last 'e' of the suffix replaced with -one.
• The parent chain is numbered starting from the end, which gives the lower number to the ketone group, e.g., $\ce{CH3COCH3}$ is propan-2-one.
• For two ketone groups -dione and three -trione suffix is used.
• For cyclic ketones, the parent name of cycloalkane is used with the last 'e' replaced with -one. Some examples are shown below.
• If a group of higher precedence, e.g., an aldehyde, is also present, then ketone is represented by the prefix -oxo, as shown below.
• A benzene ring is represented by the prefix phenyl-, as shown below.
Some examples of the naming using the above rules are shown below.
Common names of ketones
Propan-2-one is acetone, and 1-phenylethan-1-one is acetophenone. Common names of other ketones are obtained by listing the alkyl groups bonded to the carbonyl group in the order of group size, followed by the word ketone, as shown below with common names in brackets.
Iso-Group
An alkyl group containing $\ce{-CH3}$ attached to the 2nd last $ce{C}$ takes iso- prefix, as shown in the examples below.
The group names with iso-prefix are often used in common names and are also accepted in IUPAC nomenclature.
Example $1$
What is the IUPAC name of this compound shown in the figure on the right?
Solution
the longest chain containing ketone groups is six $\ce{C}$, i.e., hexane.
Change the final 'e' with -one, i.e., hexanone.
Add the location of the ketone group. Start numbering the parent chain from the end, giving the ketone a lower number. Ketone receives #3
Answer: hexan-3-one
Example $2$
Write the skeletal formula of cyclopentanone.
Solution
The parent name is cyclopentane as shown in the figure on the right.
Add a ketone group to any carbon.
Answer:
Example $3$
Write the skeletal formula of 3-methylpentane-2-one?
Solution
Parent name is pentane:
Count from either side and add a carbonyl group to $\ce{C}$#2 and a methyl group to $\ce{C}$#3.
Answer:
Physical properties of Ketones
Ketones have sp2-hybridized $\ce{C}$ and $\ce{O}$ with a double bond (a $\sigma$ and a $\pi$-bond) between them, two lone pairs occupying two sp2-orbital of an $\ce{O}$, and the $\ce{C}$ bonded with two hydrocarbon groups. i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{:O:}}|\!\!\!\!|\enspace}{C}}\!\!-R'}$. The $\ce{C=O}$ bond is polar, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$ because $\ce{O}$ are more electronegative than $\ce{C}$ (3.3-2.1 = 0.9), as shown in Figure $2$. It makes the $\ce{C}$ a $\delta^{+}$, i.e., an electrophile in reactivity and the $\ce{O}$ a $\delta^{-}$, i.e., a nucleophile or a base in reactivity. The lone pair of electrons on $\ce{O}$ add to the base character of the carbonyl $\ce{O}$.
Comparison of carbonyl $\ce{C}$ of an aldehyde and a ketone
Figure $2$ compares the electrostatic potential maps of an aldehyde and a ketone. The following are the points to note:
• The carbonyl $\ce{C}$ is more $\delta{+}$ in an aldehyde than in a ketone. The reason is that the alkyl group of ketone partially neutralizes the $\delta{+}$ of the carbonyl $\ce{C}$ by hyperconjugation while the $\ce{H}$ of the aldehyde can not. (Note: hyperconjugation is a special form of resonace which involves $\sigma$-bonds. The details of it are out of the scope of this book)
• The carbonyl $\ce{C}$ of an aldehyde is more accessible to reagents in chemical reactions than the carbonyl $\ce{C}$ of a ketone. This is because $\ce{H}$ in an aldehyde is small compared to the hydrocarbon group in a ketone.
These two factors make an aldehyde a more reactive electrophile than a ketone.
Other than the reactivity difference between an aldehyde and a ketone described above, the physical characteristics, i.e., the trend in the boiling points, solubility in water, etc., are the same for aldehydes and ketones.
Some important aldehydes and ketones
Aldehydes and ketones are often part of the biochemical processes. They are often an intermediate in the conversion of food into energy.
Methanal or formaldehyde is a colorless gas with pungent order that has germicidal properties. A 40% mixture of formaldehyde in water called formalin is used to preserve biological specimens. Formaldehyde is a starting material of polymers used in fabrics, insulation, carpeting, and other products.
Propan-1-one, or acetone, is a colorless liquid with a mild odor used as a solvent in paints, nail polish removers, rubber cement, and cleaning fluids. Care must be taken in handling acetone as it is highly volatile and flammable.
Several natural products are aldehydes or ketones used to flavor foods or as components of fragrances. For example, muscone is a ketone used in musk perfumes. Oil of spearmint contains carvone which is a ketone; cinnamaldehyde is found in cinnamon, almonds contain benzaldehyde, and vanillin is found in vanilla beans. The structures of these aldehydes and ketones are shown below.
Formaldehyde
Acetone
Muscone
Carvone
Cinnamaldehyde
Benzaldehyde
Vanillin
Imines
Replacing carbonyl $\ce{O}$ with a $\ce{N}$ in a carbonyl ($\ce{C=O}$ makes an iminie group, i.e., $\ce{R'-\!\!\!\!\!{\overset{\overset{\huge\;\:\enspace{NR}}|\!\!\!\!\!\!\!|\enspace\enspace}{C}}\!\!\!\!\!-R''}$. Note that $\ce{N}$ has three bonds and one lone pair. The third bond of $\ce{N}$ is with a $\ce{H}$ or a hydrocarbon ($\ce{R}$) in this case. Imines are less common but important as reactive intermediates.
Nomenclature of imines
IUPAC rules for naming imines are the same as for the corresponding aldehydes or ketones, except that the -al or -one is replaced with the suffix -imine, as shown by the examples below.
Physical properties of imines
The $\ce{C=N}$ bond is polar, i.e., $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{N}}$ because $\ce{N}$ are more electronegative than $\ce{C}$ (3.0-2.5 = 0.5), as shown in Figure $4$. It makes the $\ce{C}$ a $\delta^{+}$, i.e., an electrophile in reactivity and the $\ce{N}$ a $\delta^{-}$, i.e., a nucleophile or a base in reactivity. The lone pair of electrons on $\ce{N}$ add to the base character of the $\ce{N}$ in an imine group.
A comparison of the electrostatic potential map in Figure $3$ shows that the $\ce{C}$ of an imine group is much less $\delta{+}$ than the corresponding carbonyl $\ce{C}$ (notice a less blue area in the case of imine compared to the corresponding ketone). This is because $\ce{C=O}$ bond is more polar (3.5-2.5 =1.0) than a $\ce{C=N}$ bond (3.0-2.5 = 0.5). Based on the less polar bond, one would expect imines to be more stable in chemical reactions than the corresponding aldehydes or ketones. Since $\ce{N}$ is less electronegative, it holds on to the lone pair less tightly than an $\ce{O}$, which makes imine more basic, i.e., they easily donate their lone pair to a proton. Once the $\ce{N}$ makes the bond with a proton, it becomes positive charge species that is more reactive than the carbonyl compound it is derived from. Imines are less common but more important as reactive intermediates in converting aldehydes and ketones, particularly in biochemical systems. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.04%3A__Functional_groups_containing_sp2-hybridized_heteroatom.txt |
Learning Objectives
• Identify, assign IUPAC name, and draw structure from the IUPAC name of carboxylic acids and their derivatives, including acid halides, acid anhydrides, esters, amides, and nitriles.
• Predict the changes in the polarity and its effect on the reactivity of carboxylic acids and their derivatives, including acid halides, acid anhydrides, esters, amides, and nitriles.
• Identify phosphoric acid, anhydrides of phosphoric acids, phosphate anions, and esters of phosphoric acids.
What are carboxylic acids and carboxylic acid derivatives?
Carboxylic acids have a carbonyl group ($\ce{C=O}$) and a hydroxyl group ($\ce{-OH}$) on the same carbon, i.e., $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$ group. The carboxylic acid group is represented as $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$, or as $\ce{-COOH}$. Carboxyl acids have some characteristics of the $\ce{C=O}$ group, some characteristics of the $\ce{-OH}$ group, and some additional characteristics due to the interaction of the two groups. In carboxylic acid derivates, the $\ce{-OH}$ group is replaced with another group, that includes acid halides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-X}$), acid anhydrides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}-R'}}$), easters ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OR'}$), and amides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2}$). Nitrile group that has the carbonyl $\ce{O}$ replaced with a $\ce{N}$ and the $\ce{-OH}$ group also replaced with the same $\ce{N}$, i.e., $\ce{R-C≡N}$ group is also classified as a carboxylic acid derivative. The nomenclature and physical characteristics of carboxylic acids and their derivatives are described in the following sections.
Carboxylic acids ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$)
Nomenclature of carboxylic acids
The IUPAC nomenclature of the carboxylic acids follows the following rules.
• The longest hydrocarbon chain containing the carboxylic acid group is chosen as the parent name, with the last letter 'e' of its suffix replaced with -oic acid. For example, $\ce{HCOOH}$ is methanoic acid, $\ce{CH3COOH}$ is ethanoic acid, and $\ce{CH3CH2COOH}$ is propanoic acid.
• If there are two carboxylic acid groups, the suffix changes to -dioic acid, e.g., $\ce{HOOC-COOH}$ is ethanedioc acid, and $\ce{HOOC-CH2-COOH}$ is propanedioic acid. (note that when the suffix begins with a consonant (the letter 'd' in this case), the last letter 'e' of the parent hydrocarbon name is not dropped)
• Start numbering from the $\ce{C}$ of the $\ce{-COOH}$ group. The $\ce{-COOH}$ group itself does not need a location number, as it is always at the end of the chain.
• If the $\ce{-COOH}$ group is bonded to a cyclic chain, the suffix -carboxylic acid is added to the name of the cyclic hydrocarbon. Numbering starts from the point of attachment of $\ce{-COOH}$ to the ring.
• If the $\ce{-COOH}$ group is bonded to a benzene ring, the parent name "benzoic acid is used.
• Numbering starts from the point of attachment of the $\ce{-COOH}$ group to the ring.
Example $1$
What is the IUPAC name of the compound shown on the right?
Solution
• The longest chain counting the $\ce{-COOH}$ group is three $\ce{C's}$ and a double bond, so the parent name is propene and replace the last letter 'e' with -oic acid, i.e., propenoic acid.
• There is a methyl group attached that becomes a prefix, i.e., methylpropenoic acid.
• A location number is needed for the methyl group and the double bond. Start numbering from the $\ce{C}$ of the $\ce{-COOH}$ group, double bond receives#2 and the methyl group receives #2.
Answer: 2-methylprop-2-enoic acid
The common name of 2-methylprop-2-enoic acid is methacrylic acid, and prop-2-enoic acid is acrylic acid, the monomers (the repeating units) in some polymers.
Example $2$
What is the IUPAC name of the compound shown on the right?
Solution
The $\ce{-COOH}$ group is attached to a five $\ce{C}$ cyclic chain, so the name of the cyclic chain becomes the parent name: cyclopentane.
Add the suffix -carboxylic acid to the parent name to indicate the carboxylic acid group attached to a cyclic chain.
Answer: cyclopentanecarboxylic acid
Example $3$
What is the IUPAC name of the compound shown on the right?
Solution
• A carboxylic acid bonded to a benzene ring takes "benzoic acid" as the parent name.
• A $\ce{-OH}$ group takes the prefix "hydroxy" in the presence of a $\ce{-COOH}$ group, i.e., hydroxybenzoic acid.
• Start numbering from the point of attachment of the $\ce{-COOH}$ group: the $\ce{-OH}$ group receives #4.
Answer: 4-hydroxybenzoic acid.
Common names of carboxylic acids
Common names of carboxylic acids are derived from the names of natural sources of these acids. Table 1 lists some of the common names of carboxylic acids.
Table 1: Common names and the sources of the common names of some of the carboxylic acids.
Condensed formula IUPACE name Common name Source of the common name
$\ce{HCOOH}$ methanoic acid formic acid Latin: formica, ant
$\ce{CH3COOH}$ ethanoic acid acetic acid Latin: acetum, vinegar
$\ce{CH3CH2COOH}$ propanoic acid propionic acid Greek: propion, first fat
(\ce{CH3(CH2)2COOH}\) butanoic acid butyric acid Latin: butyrum, butter
(\ce{CH3(CH2)4COOH}\) hexanoic acid caproic acid Latin: caper, goat
(\ce{CH3(CH2)14COOH}\) hexadecanoic acid palmitic acid Latin: palma, palm tree
(\ce{CH3(CH2)16COOH}\) octadecanoic acid stearic acid Greek: stear, solid fat
(\ce{CH3(CH2)18COOH}\) eicosanoic acid arachidic acid Greek: arachis, peanut
The first syllable of the common names, e.g., form-, acet-, prop-, etc., are also used as the first syllable of the common names of related compounds. For example, $\ce{HCOH}$ is formaldehyde, $\ce{CH3COH}$ is acetaldehyde, etc.
Physical properties of carboxylic acids
The carboxylic acid group has a $\ce{C=O}$ and a $\ce{-OH}$ groups, i.e., an sp2- and an sp3 hybridized $\ce{O}$ bonded to the same $\ce{C}$. Both $\ce{O's}$ have two lone pairs of electrons on them, i.e., $\ce{-\!\!{\overset{\overset{\huge\enspace\!{:O:}}|\!\!\!\!|\enspace}{C}}\!\!-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}-H}$. Lone pair of electrons are usually not shown except when needed, i.e., the carboxylic acid group is represented as $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$ or as $\ce{-COOH}$.
The carboxylic acid group has three polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ and $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O}-\overset{\delta{+}}{H}}$. This is because $\ce{O}$ are more electronegative than $\ce{C}$ (3.3-2.6 = 0.7) and $\ce{H}$ (3.3-2.2 = 1.17), as shown in Figure $1$. It makes $\ce\overset{\delta{+}}{C}$ an electrophile, $\ce\overset{\delta{-}}{O}$ a nucleophile or a base, and $\ce\overset{\delta{+}}{H}$ an acid in reactivity. Due to the acid protons, carboxylic acids are also classified as organic acids.
The polar $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$ and $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$ bonds allow dipole-dipole interactions and hydrogen bonding in addition to the London dispersion forces. Carboxylic acids have stronger intermolecular forces, higher melting points, higher boiling points, and higher solubilities in water compared to alcohols and aldehydes of comparable molar mass due to more intermolecular forces, as compared in Table 2.
Table 1: Compares boiling points and solubility of a carboxylic acid, alcohol, and aldehyde of comparable molar mass.
Condensed formula IUPAC name Molar mass (g/mol) Boiling point (oC) Solubility in water
$\ce{CH3(CH2)2COOH}$ Butanoic acid 88.1 163 Miscible
$\ce{CH3(CH2)3CH2OH}$ Pentan-1-ol 88.1 137 2.3 g/100 mL
$\ce{CH3(CH2)3CHO}$ Pentanal 86.1 103 Slightly soluble
Two carboxylic acids can make two hydrogen bonds with each other, as illustrated in Figure $2$, behaving as a dimer with two times higher molecular mass. It explains their higher boiling points than alcohols of the same molar mass. Carboxylic acids of up to five $\ce{C's}$, i.e., methanoic acid, ethanoic acid, propanoic acid, butanoic acid, and pentanoic acid, are soluble in water. Hexanoic acid is slightly soluble, and higher acids are insoluble.
Carboxylic acids have a sour taste because they are acids due to ionizable proton in their $\ce{-O-H}$ groups. For example, the sour taste of citrus fruits is due to citric acid, and the sour taste of vinegar is due to ethanoic acid.
Oxidation is i) loss of electrons, ii) gain of $\ce{O}$, or loss of $\ce{H}$; and the reduction is the opposite of these. Oxidation and reduction happen together and are collectively called Redox reactions. Most chemical reactions in biological systems are redox reactions, e.g., photosynthesis is a reduction of $\ce{CO2}$ to convert solar energy into potential chemical energy, and digestion of food is the opposite, i.e., oxidation to release the energy for the activities of life.
A major portion of organic compounds is hydrocarbon groups, gradually oxidized to alcohols, aldehydes or ketones, carboxylic acids, and finally, carbon dioxide and water, releasing energy. For example, methane ($\ce{CH4}$) oxidizes to methanol ($\ce{CH3OH}$), methanal ($\ce{CH2O}$), methanoic acid ($\ce{HCOOH}$), and finally to carbon dioxide ($\ce{CO2}$) that is exhaled, as shown below.
The alcohols, aldehydes, ketones, and carboxylic acids also serve as intermediates for synthesizing compounds the body needs. Carboxylic acids commonly appear in the metabolic process. For example, glucose, a six $\ce{C}$ compound, is first converted to two pyruvic acid molecules. Under low oxygen conditions (anaerobic), pyruvic acid is reduced to lactic acid.
In the presence of oxygen (aerobic), pyruvic acid releases a $\ce{CO2}$ and becomes a two $\ce{C}$ group that joins a four $\ce{C}$ compound oxalic acid to make a six $\ce{C}$ compound citric acid. Citric acid releases a $\ce{CO2}$ and becomes a five $\ce{C}$ compound $\alpha$-ketoglutaric acid, which releases another $\ce{CO2}$ and becomes four $\ce{C}$ compound succinic acid, as shown below. This process goes on through several intermediate carboxylic acids, and either all the $\ce{C's}$ of the starting compound convert to $\ce{CO2}$, or the intermediate is utilized to synthesize compounds needed by the body.
, , and
Note: The carboxylic acids are shown as neutral acids in the above example, but in the physiological conditions, they exist as anions, i.e., as a carboxylate group ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}$). These points will be discussed in a later chapter.
Acid halides and acid anhydrides -the most reactive acid derivatives
Nomenclature of acid halides
Acid halides contain the ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-X}$) group, where $\ce{X}$ can be $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$.
• IUPAC name of the acid halide takes the name of the corresponding carboxylic acid with the suffix -oic acid replaced with -oil halide, as shown in the following examples.
Acid chlorides and acid bromides are the most common.
Nomenclature of acid anhydrides
Acid anhydride contains two acyl $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$ groups bonded to a common $\ce{O}$ atom, i.e., ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}-}}$) group. A carboxylic acid anhydride is derived by condensing two carboxylic acids by losing a $\ce{H2O}$ molecule. The acid anhydride is symmetric anhydride if both the acids are the same and mixed or asymmetric anhydride if two different acids are condensed to form the anhydride,
• Symmetric acid anhydrides are named using the name of the corresponding acid with the last word 'acid' replaced with 'anhydride', as shown in the following examples.
• Mixed or asymmetric anhydrides are named by listing the names of the two acids in alphabetic order without the last word 'acid', followed by the word 'anhydride', as shown in the following examples.
Physical properties of acid halides and acid anhydrides
The acid halides group has two polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{X}}$ resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{X}}$. The acid anhydrides have four polar bonds i.e., two $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and two $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O}-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}-}$. This polarity of the group can be observed in the electrostatic potential maps of acid halide, acid anhydride, and an acid shown in Figure $3$
It is apparent from the comparison of the electrostatic potential maps shown in Figure $3$ that the carbonyl $\ce{C}$ is more bluish, i.e., higher $\delta{+}$ and stronger nucleophile, in the case of acid halide and acid anhydride than in carboxylic acid. The question is $\ce{Cl}$ in the acid halide is less electronegative than $\ce{O}$ in carboxylic acid, then why is the carbonyl $\ce{C}$ more $\delta{+}$ in the acid halide than in the acid? The answer is in the fact that the heteroatom not only draws the bonding electron away from the carbonyl $\ce{C}$, it also donates its lone pair through resonance that diminishes the $\delta{+}$ character on the carbonyl $\ce{C}$:
• The 2p-orbital of $\ce{O}$ and 2p-orbital of $\ce{C}$ of carboxylic acid are of similar size and overlap well for the reasonce to happen and diminish its $\delta{+}$ character.
• The 3p-orbital of $\ce{Cl}$ or 4p orbital of $\ce{Br}$ overlaps poorly with 2p-orbital of carbonyl $\ce{C}$ due to the size difference and does not diminish its $\delta{+}$ character.
• The (\ce{O}\) is shared between to $\ce{C=O}$ groups in the case of acid anhydride. So it diminish $\delta{+}$ character of $\ce{C=O}$ less than in carboxylic acids.
Due to the highly nucleophilic carbonyl $\ce{C}$, the acid halides and acid anhydrides are very reactive and primarily used as reactive intermediates in chemical synthesis. Again due to their high reactivity, they can not survive in biological systems. Biochemical systems use carboxylic acid derivatives containing $\ce{S}$ or phosphate groups as reactive intermediates, which will be described later.
Easters
Esters have an acyl group (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$) bonded with an alkoxy group (easters ($\ce{-OR'}$), i.e., an easter ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OR'}$) group.
Nomenclature of esters
• IUPAC name of an ester starts with the name of the alkyl group that is part of the alkoxy ($\ce{-OR'}$) group, followed by the name of the acid corresponding to the acyl (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$) group with the suffix -oic acid replaced with the suffix -oate, as shown in the following examples.
Example $1$
What is the IUPAC name of the compound shown on the right?
Solution
• Alkyl group in the alkoxy ($\ce{-OR'}$) group is four ($\ce{C's}$) i.e., butyl.
• The acyl group is three ($\ce{C's}$), i.e., propanoic acid; replace acid with -oate, i.e., propanoate.
Answer: butyl propanoate
Example $2$
What is the IUPAC of the compound shown on the right?
Solution
• It is an easter where the alkyl group in the alkoxy ($\ce{-OR'}$) group is one ($\ce{C's}$), i.e., methyl.
• The acyl group contains benzene, i.e., benzoic acid, and replace the -ic acid with -oate, i.e., benzoate.
Answer: methyl benzoate
Physical properties of esters
Esters group has a polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and two polar $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ groups, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O-R'}}$, as shown in Figure $4$. The reactivity of an ester's carbonyl $\ce{C}$ is almost the same as that of a carboxylic acid.
Easters are pretty common in nature. Small esters are volatile and soluble in water, making them easier to smell and taste. The fragrances of many perfumes and flavors of several fruits are due to esters, as shown in Table 3.
Table 3: Some esters and the flavor/taste of the fruit they are associated with are shown in the next row.
pentyl acetate pentyl methanoate ethyl heptanoate propyl acetate
Banana
Plum
Grape
Pear
octyl acetate ethyl butanoate pentyl butanoate propyl pentanoate
Orange
Pineapple
Apricot
Apple
Esters in fats
Fats are esters of carboxylic acids that contain a long chain hydrocarbon, an alkane or alkene with cis double bonds, called fatty acids, and propane-1,2,3-triol also called glycerol, as shown in one example in Figure $5$.
Aspirin -an ester in medical use
Aspirin is an ester of salicylic acid found in a willow tree's bark. Salicylic acid reduces pain and fever, but it irritates the stomach lining. Aspirin, an ester of salicylic acid, overcomes this problem and is commonly used to reduce pain and fever and as an anti-inflammatory agent. Methyl salicylate is another ester of salicylic acid found in wintergreen oil and used as skin ointments to soothe sore muscles.
Salicylic acid Aspirin Methyl salicylate
Amides
Amidess have an acyl group (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$) bonded with a nitrogen group ($\ce{-NRR'}$), i.e., an amide ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NRR'}$) group, where $\ce{R}$ and $\ce{R'}$ may be $\ce{H}$ or a hydrocarbon group.
Nomenclature of amides
• IUPAC name of an amide is the name of the acid corresponding to the acyl (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$) group with the suffix -oic acid replaced with the word amide.
• If a hydrocarbon group is bonded to nitrogen, its name appears as a prefix preceded by N-.
• If two identical hydrocarbon groups are bonded to nitrogen, the group name is preceded by N, N-di.
• If two different hydrocarbon groups are bonded to nitrogen, the group names, each preceded by N-, are listed alphabetically as prefixes. Some examples are shown below.
• An amide group attached to a benzene ring takes the base name 'benzamide,' as shown in the following examples.
Example $1$
What is the IUPAC name of the compound shown on the right?
Solution
• There is an amide group on a four $\ce{C}$ chain, i.e., butanoic acid, which changes to butanamide.
• There is a methyl group on nitrogen that becomes the prefix N-methyl.
Answer: N-methylbutanamide
Example $2$
What is the IUPAC name of the compound shown on the right?
Solution
• There are two amide groups, so it takes the suffix -diamide.
• The corresponding acid is a six $\ce{C}$ diacid named hexanedioic acid. Replace -oic acid with amide.
Answer: hexanediamid
Physical properties of amides
Amide group has a polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{N}}$, and $\ce{\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$ groups, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{N}-R'R"}$, as shown in Figure $5$.
Comparison of electrostatic potential maps in Figure $5$ shows that the carbonyl $\ce{C}$ of amide is less $\delta{+}$, i.e., less electrophilic than that of the corresponding carboxylic acid. This is because of two reasons i) $\ce{N}$ is less electronegative and draws electrons away less than $\ce{O}$ and ii) being less electronegative, $\ce{N}$ sends its lone pair of electrons more to the carbonyl $\ce{C}$ neutralizing its $\delta{+}$ by resonance than $\ce{O}$. Both of these factors make the carbonyl $\ce{C}$ less $\delta{+}$ and less electrophilic, which makes amides one of the lest reactive and the most stable carboxylic acid derivatives that are found commonly in nature. The resonance effect is illustrated in Figure $6$ below. Due to the resonance, the $\ce{C-N}$ bond has a significant double bond character.
Since the lone pair of electrons of amide are occupied in the resonance, they are less available to protons of acids. Therefore, amides are much less basic than amines. Amides have $\ce{\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$ bonds that allow them to make hydrogen bonds with water molecules. Therefore, amides containing up to five $\ce{C's}$ are soluble in water due the hydrogen bonding. Those with larger alkyl groups, i.e., with more than five $\ce{C's}$ are slightly soluble or insoluble due to the hydrophobic character of the alkyl groups dominating over the hydrophilic nature of the amide group.
Amides in nature and medicines
Proteins are made of small repeat units, called amino acids, joined through amide groups, as illustrated in Figure $7$, which makes amide one of the most common groups present in nature.
During the digestion of proteins, $\ce{N's}$ end up in urea excreted by kidneys. If kidneys malfunction, urea may build to a toxic level, resulting in uremia. Urea is also used as a fertilizer.
Barbiturates derived from barbituric acid have sedative and hypnotic effects and are used in medicines for these effects. Barbiturate drugs include phenobarbital and pentobarbital. Phenacetin and acetaminophen are amides used in Tylenol as alternatives to reduce fever and pain but with little anti-inflammatory effect. The structures of these amides are shown below.
urea
barbituric acid
phenobarbital pentobarbital phenacetin acetaminophen
cyclic amides are called lactams. A four-member lactam is a common feature in the structure of Penicillin and related synthetic antibiotics, as shown below.
A four-membered
lactam
Penicillin G Amoxicillin Cephalexin
Nitriles
Nitriles have cyano group that is a polar bond $\ce{-\overset{\delta{+}}{C}{≡}\overset{\delta{-}}{N}\!:}$ with a lone pair in one of the sp orbital of $\ce{N}$, as illustrated in Figure $5$.
IUPAC name of nitriles is composed of the name of the hydrocarbon skeleton, including the $\ce{C}$ in the nitrile group with the suffix -nitrile.
Another way of naming them is to take the name of the corresponding carboxylic acid but replace the suffix -oic acid with -onitrile, as shown in the following example.
Nitriles are not very common in nature but are important as intermediates in synthesizing organic compounds.
Phosphorous groups
Phosphorous ($\ce{P}$) is in the same group with $\ce{N}$ in periodic table. Like $\ce{N}$ in ammonia ($\ce{{H}-\overset{\bullet\bullet}{\underset{\underset{\huge{H}} |}{N}}\!-H}$), the $\ce{P}$ can have three bonds and a lone pair, as in phosphine ($\ce{{H}-\overset{\bullet\bullet}{\underset{\underset{\huge{H}} |}{P}}\!-H}$) with eight valence electrons, i.e., octet complete. Unlike $\ce{N}$, the $\ce{P}$ is in fourth row in the periodic table and, like other elements of the fourth and higher row, can have more than eight valence electrons in its compounds, e.g., ten valence electrons in phosphoric acid ($\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$). Phosphorous groups are important in biological systems, e.g., they are part of DNA molecules, phospholipids in cell membranes, and energetic molecules like adenosine triphosphate that are used as energy currency in biochemical reactions.
Phosphoric acid and phosphoric anhydrides
Phosphoric acid or orthphosphosporic acid has three $\ce{-OH}$ groups and one $\ce{=O}$ bonded to a $\ce{P}$ atom, i.e., $\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$. Like carboxylic acid anhydride, which is two carboxylic acids joined through a common $\ce{O}$, two phosphoric acids joined through a common $\ce{O}$ is a diphosphoric acid or a pyrophosphoric acid, and three phosphoric acids condensed in this way is a triphosphoric acid. Their corresponding anions formed after ionizing the acidic protons from the $\ce{-OH}$ groups are called phosphates, and alkoxy ($\ce{-OR}$ replacing one or more $\ce{-OH}$ groups are phosphate esters, as shown below.
$\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$
Phosphoric acid or orthphosphosporic acid
$\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$
diphosphoric acid or a pyrophosphoric acid
$\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$
triphosphoric acid
$\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O^{-}}$
Phosphate ion or orthphosphate ion
$\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O^{-}}$
diphosphate ion or a pyrophosphate ion
$\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-O^{-}}$
triphosphate ion
Phosphoric esters
Like carboxylic acid changes to an ester when its $\ce{-OH}$ group is replaced with an alkoxy ($\ce{-OR}$) group, phosphoric acid changes to mono phosphoric ester when one of its $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$), to diester when two $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$), and to triester when three $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$).
• The phosphoric esters are named by listing the names of the alkyl parts of the alkoxy groups in alphabetic order, followed by the world phosphate.
• In more complex phosphodiesters, it is common practice to name the organic compound followed by the word 'phosphate' or prefixed phospho-, as shown in the following examples.
dimethyl phosphate
ethyl methyl phosphate
dihydroxyacetone phosphate
adenosine triphosphate | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.05%3A_Functional_groups_containing_mix_of_sp3-_and_sp2-_or_sp-hybridized_heteroatom.txt |
• 3.1: Introduction to stereochemistry
Isomerism and the basic classification of isomers are described.
• 3.2: Confirmations
Representing conformations of simple organic molecules, including ethane and butane, by Sawhorse and Newman projections, drawing chair conformations of cyclohexane and substituted cyclohexane, and relative stability of the conformational isomers are described.
• 3.3: Configurations
Perspective drawing, Fisher projections, CIP rules for assigning R/S stereodescriptors, D/S stereodescriptors, and stereochemical relationships, including enantiomers, diastereomers, and meso are described.
03: Stereochemistry
Learning Objectives
• Identify isomers and differentiate constitutional isomers from sterioisomes.
• Identify subclasses of constitutional isomers, including skeletal isomers, functional group isomers, and positional isomers; and subclasses of sterioisomers, including conformers, enantiomers, diasteriomers.
What is stereochemistry?
Stereochemistry is the study of the relative arrangement of atoms in molecules and their manipulation. A major area of stereochemistry is the study of isomers which is introduced below.
Isomers
Isomers are species with the same atoms in the same numbers, i.e., the same molecular formula but are arranged differently in space.
Several types of isomers can be classified in different ways. Figure \(1\) presents one scheme of classifying isomers.
Constitutional isomers
Constitutional isomers have a different sequence of bonds that may result in another skeleton, i.e., skeletal isomers, different reactive features, i.e., functional groups isomers, or the same reactive features but placed at different locations, i.e., positional isomers.
Skeletal isomers
Skeletal isomers have different carbon skeletons. For example, hexane, 2-methylpentane, 3-methylpentane, and 2,2-dimethylbutane are skeletal isomers having the same molecular formula \(\ce{C6H14}\).
Functional group isomers
Functional group isomers have different reactive features, i.e., different functional groups. For example, propan-1-ol is alcohol, and methoxyethane is ether, but they are isomers having the same molecular formula \(\ce{C3H8O}\).
Positional isomers
Positional isomers have the same reactive features, i.e., the same functional groups, but the groups are attached at different positions. For example, propan-1-ol -a primary alcohol having an alcohol group linked to terminal \(\ce{C}\) and propan-2-ol -a secondary alcohol having an alcohol group attached to a non-terminal \(\ce{C}\), are isomers having the same molecule formula \(\ce{C3H8O}\).
Stereoisomers
• Stereoisomers have the same sequence of bonds but the atoms or groups of atoms are oriented differently in space.
• Conformers or conformational isomers are the stereoisomers in which the different orientations of atoms are a result of rotation around single bonds. The specific arrangement of atoms in a conformational isomer is also called conformation.
• Configurational isomers are stereoisomers that can be interconnected only by breaking and making some bonds. The specific arrangement of atoms in a configuration isomer is also called configuration.
Conformers or conformational isomers
Conformers are the isomers that are the result of rotation around single bonds. They are also called different conformations of the same molecule. For example, rotation around \(\ce{C-C}\) bond of ethane places a set of three \(\ce{H's}\) on one \(\ce{C}\) at different positions relative to \(\ce{H's}\) on the other \(\ce{C}\), as illustrated below.
Different arrangements of \(\ce{H's}\) in ethane due to rotation around \(\ce{C-C}\) bond
Illustration of rotation around \(\ce{C-C}\) bond in ethane (Copyright; mailto:[email protected], CC BY 2.5 via Wikimedia Commons)
Usually, the rotation around single bonds happens rapidly at room temperature. So, the conformers usually exist as a mixture and can not be easily separated.
Configurational isomers
Configurational isomers are stereoisomers that can be interconnected only by breaking and making some bonds. For example, cis-but-2-ene and trans-but-2-ene are configurational isomers in which \(\ce{-CH3}\) groups connected to two \(\ce{C's}\) of a double bond are oriented differently as shown in Figure \(1\). Another example is L-glyceraldehyde and D-glyceraldehyde in which four different groups are connected to the same \(\ce{C}\) but oriented differently as shown in Figure \(1\).
The configurational isomers are subdivided into enantiomers and diastereomers.
Enantiomers
A pair of stereoisomers that are related to each other as non-superimposable mirror images are Enantiomers.
For example, D-glyceraldehyde and L-glyceraldehyde shown above are enantiomers of each other. Imagine there is a mirror between D- and L-glyceraldehyde shown above. You will notice that they are mirror images of each other. If you try to overlap L-glyceraldehyde onto D-glyceraldehyde, two groups may overlap, but the other two will not overlap, no matter how you may rotate the molecule. D- and L- represent two different configurations of glyceraldehyde shown in Figure \(1\). Enantiomers are chiral molecules.
Chirality
An object or molecule that cannot be superimposed on its mirror image by any translation, rotational, or conformational changes is a chiral object. This geometric property is called chirality.
Achiral is not chiral, i.e., the objects or molecules that are identical to their mirror image are achiral.
For example, an amino acid with four different groups attached to the same carbon and hand are chiral, having a non-superimposable mirror image as illustrated in Figure \(2\). Like left and right hands that have a thumb and fingers in the same order, but are mirror images and not the same, chiral molecules have the same things attached in the same order, but are mirror images and not the same.
Diastereomers
Stereoisomers that are not enantiomers are diastereomers.
For example, cis-but-2-ene and trans-but-2-ene shown in Figure \(1\) are diastereomers because they have the same formula, and the same atom-connectivity, but methyl groups are oriented in the same direction in cis- and in opposite directions in trans-isomer, they do not mirror each other. The cis- and trans- represent two different configurations of but-2-ene as shown in Figure \(1\). | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/03%3A_Stereochemistry/3.01%3A_Introduction_to_stereochemistry.txt |
Learning Objectives
• Draw sawhorse projections and Newman projections of simple organic compounds.
• Name, predict relative stability, and energy barriers to interconversion of conformations of simple organic compounds, including ethane, butane, and cyclohexane.
• Draw chair conformations of cyclohexane, substituted cyclohexanes, and their flipped forms, and predict the stability of the bulky groups at the axial and equatorial positions.
Representing orientation of groups connected to a $\ce{C-C}$ bond
There are multiple ways of visualizing the orientation of atoms along a $\ce{C-C}$ bond, including sawhorse projection, Newman projection, and Fisher projection.
Sawhorse projection and Newman projection of a molecule
• Sawhorse projection looks at the $\ce{C-C}$ bond from an oblique angle, as shown in Figure $1$.
• Newman projection looks from front to back with the front atom represented as a dot with bonds originating from the dot and the back atom is represented as a wheel with bonds shown as lines originating from the wheel, as illustrated in Figure $1$.
The relative orientation of groups connected to the two $\ce{C's}$ of $\ce{C-C}$ bond is quantitatively shown by the dihedral angle.
Dihedral angle or torsion angel
Dihedral angle or torsion angle is the clockwise angle between half-planes through two sets of three atoms, e.g., the half-planes defined by $\ce{R-C-C}$ and $\ce{R'-C-C}$ in Figure $2$, having two atoms in common $\ce{C-C}$ in this case).
Conformational isomers
The torsion angle changes due to free rotation around the single bond. However, the rotation is not completely free; there are energy barriers. Therefore, as the torsion angle gradually changes from 0o to 360o, the relative potential energy of the molecule goes through local minima to local maxima, often at intervals of 60o.
Conformations
• Conformations are the different arrangements of atoms in a molecule resulting from rotation around a single bond.
• Eclipsed conformation is a conformation in which two substituents on adjacent atoms are at 0o torsion angle and the
• Staggered conformation is the confirmation in which the two substituents on adjacent atoms are at 60o torsion angle, as illustrated in Figure $3$.
Conformations of ethane
The eclipsed and staggered conformations of ethane are shown in the figure on the right. The eclipsed conformation of ethane is higher energy (unstable) relative to staggered confirmation by 12.5 kJ/mol. The eclipsed conformation changes to staggered, and vice versa, after every 60o rotation around the $\ce{C-C}$ bond as illustrated in Figure $4$.
Factors in stabilizing or destabilizing conformations
Factors that contribute to the stabilization or destabilization of the conformation are the following.
1. Destabilization due to electrostatic repulsion between the same charge dipoles in close proximity in eclipsed conformation,
2. destabilization due to steric repulsion, i.e., physical repulsion of the groups due to space requirement in close proximity in eclipsed conformation, and
3. stabilization due to hyperconjugation, i.e., an overlap of a $\sigma$-bond on one atom with an antibonding orbital on the neighboring atom in staggered conformation.
These factors are illustrated in Figure $5$ for the case of ethane.
Conformations of butane
The most stable conformation of butane ($\ce{CH3CH2CH2CH3}$) for a view of C2-C3 bond is illustrated on the right. This staggered conformation with the bulky ($\ce{-CH3}$ groups in this case) fathers apart is given a special name,'anti' and labeled A in the relative energy vs. tortion angle curve shown in Figure $6$.
A 60o rotation converts the anti-conformation to eclipsed conformation, labeled C in the figure, which is 16 kJ/mol higher energy relative to the anti-confirmation. A further 60o rotation converts the eclipsed conformation to another staggered conformation called gauche-conformation, labeled B, at a local energy minimum but 3.8 kJ/mol higher than anti-conformation. The gauche-conformation has the bulky ($\ce{-CH3}$ in this case) groups at 60o torsion angle where they are close enough to cause some steric repulsion between each other. Another 60o rotation converts the gauche-conformation to an eclipsed conformation called cis, labeled D, in which the bulky ($\ce{-CH3}$ in this case) groups are eclipsing each other. This cis-conformation is 3 kJ/mol higher in energy than the other eclipsed confirmation (labeled B) and it is the highest energy conformation of butane. These conformations keep repeating at an interval of 60o torsion angle as illustrated in Figure $6$.
Conformational isomers
• Conformations corresponding to local minima on the potential energy curve, e.g., anti- and gauche-conformations in the case of butane, are called conformational isomers, conformers, or rotamers.
• The conformations corresponding to the local maxima on the energy curve, e.g., eclipsed conformations in the case of butane, are the transition states between the conformational isomers.
Energy cost of eclipsing and gauche interaction of $\ce{-H}$ and $\ce{-CH3}$ groups
It can be calculated form Figure $4$ and Figure $6$ that each
• $\ce{-H}$ to $\ce{-H}$ eclipsing costs 4 kJ/mol,
• $\ce{-H}$ to $\ce{-CH3}$ eclipsing costs 6 kJ/mol,
• $\ce{-CH3}$ to $\ce{-CH3}$ eclipsing costs 11 kJ/mol, and
• $\ce{-CH3}$ to $\ce{-CH3}$ gauche interaction costs 3.8 kJ/mol.
Conformations and strains in cycloalkanes
If a cycloalkane is in a planar conformation, i.e., all the $\ce{C's}$ in the chain be in the same plane, there will be two kinds of strain:
1. angle strain, i.e., the strain caused by bending the bond angles from the ideal tetrahedal bond angles of 109.5o, and
2. eclipsing strains between a pair of $\ce{H}$ on each $\ce{C}$ with pairs of $\ce{H's}$ on the two neighboring carbons.
Cyclopropane has a planar ring with bond angles reduced to 60o and three pairs of eclipsing $\ce{H's}$, resulting in an overall ring strain of ~120 kJ/mol.
A planar cyclobutane would have bond angles reduced to 90o and four pairs of eclipsing $\ce{H's}$, but it changes to a puckered confirmation that increases angle strain a little but significantly reduces the eclipsing strain, resulting in an overall ring strain ~110 kJ/mol.
A planar cyclopentane would have bond angles of 108o, i.e., close to the regular value of 109.5o, but five pairs of eclipsing $\ce{H's}$. Cyclopentane is puckered, increasing angle strain slightly but significantly reducing the eclipsing strain, resulting in an overall strain of ~25 kJ/mol. Four of the $\ce{C's}$ of cyclopentane are almost in one plane and the fifth is raised, giving it an envelope shape, also called envelop conformation. Cyclopentane is one of the most common ring structures found in nature due to its low angle strain. For example, ribose -a sugar that is a component of DNA monomers, and fructose or fruit sugar are five-membered rings.
Cyclohexane acquires a bent conformation that looks like a chair -called chair conformation, in which all bond angles are close to the tetrahedral value of 109.5o and all $\ce{H's}$ are staggered with no ring strain. Cyclohexane is the most common ring structure due to its zero ring strain. For example, glucose -the most common monosaccharide (sugar) is six-membered rings. Starch, a carbohydrate component of food, and cellulose, a major component of wood, are made of glucose units. The conformations of cyclopropane, cyclobutane, cyclopentane, and cyclohexane are illustrated below.
Cyclopropane,
Ring strain ~120 kJ/mol
Cyclobutane
Ring strain ~110 kJ/mol
Cyclopentane
Ring strain ~25 kJ/mol
Cyclohexane
No ring strain
Medium-size rings (7-13 $\ce{C's}$) can pucker/bend to minimize angle strain but some transannular strain, i.e., steric repulsion between groups attached to non-adjacent ring atoms, appear between groups on some $\ce{C's}$ with other groups at a distance across the ring. For example, the figure on the right highlights the transannular strain between some $\ce{H's}$ with a red triangle in the lowest energy conformation of cyclodecane (Copyright; Chem540f09grp4, Public domain, via Wikimedia Commons). In larger rings (14 or more $\ce{C's}$), there is little or no ring strain. The order of strain in cyclic hydrocarbons is the following: cyclopropane > cyclobutane > cyclopentane > cyclohexane < cycloheptane <cyclooctane.
Conformations of cyclohexane
Cyclohexane which is the most common cyclic structure, exists primarily in a chair conformation in which all bonds are 111o, i.e., very close to tetrahedral value 109.5o, and all six pairs of $\ce{H's}$ are staggered as shown in Figure $7$.
Drawing chair conformation of cyclohexane
Follow the following 5 steps to draw the chair form of a cyclohexane ring.
1. Draw two parallel lines slanting down, separated by about half their length, and the bottom line starting from about the middle of the first as shown in Figure $8$.
2. Start a line from the lower end of the bottom line making a wide V shape, and extend it to about parallel to the top of the top line. Draw a line parallel to the first one in this step, starting from the top of the top line and extending to about the bottom of the bottom line. Figure $8$ shows these two lines in blue.
3. Connect the top of the right side line of step 2 to the bottom end of the top line of step 1 and do the same to the other line of step 2. These are shown in red in Figure $8$. The chair skeleton of the cyclohexane ring is complete at this stage.
4. Draw lines starting from the corners of the ring, going upwards from the corners pointing up, and going downwards from the corners pointing down. These bonds are approximately parallel to the axis of the ring called axial bonds, shown in green in Figure $8$.
5. Draw lines starting from the corners making a big V-shape to the axial bond and approximately parallel to the bond one bond away in the ring. These are approximately along the equator of the ring and called equatorial bonds, shown in black, blue, and red, corresponding to the color of the bonds one bond away to which they are drawn parallel in Figure $8$.
There are two chair conformations. They are flipped form of each other. To draw the flipped form of the first, follow the same five steps but start with two parallel lines slanting upwards, as illustrated in Figure $8$.
Ring flipping in cyclohexane
There is limited rotation around $\ce{C-C}$ in a cyclohexane chain that allows flipping of the cyclohexane ring that converts all the axial bonds into equatorial and all the equatorial bonds to axial as illustrated by animation and drawing in Figure $9$.
The ring flipping goes through twist boat conformation and half-chair and boat transition states with higher energy conformations as shown in Figure $10$. The two chair conformations are of equal energy in the case of cyclohexane and predominant in the mixture due to their low energy. For every 10,000 molecules in chair conformations, there is no more than one molecule in the twist boat conformation.
Conformations of monosubstituted cyclohexanes
Two chair conformations of cyclohexane are of equal energy. When one of the $\ce{H}$ is replaced with a bulky group like $\ce{-CH3}$ group, the bulky group is at an equatorial position in one and at an axial position in the other chair conformation and the two are not the same energy. The bulky group at the axial position has steric repulsion with the other two axial $\ce{H's}$ on the same face of the ring, called 1,3-diaxial interaction, as illustrated in Fig. 11. Therefore, the chair conformation with the bulky group at equatorial position is more stable and more predominant at equilibrium than its flipped form with the bulky group at axial position. This effect depends on the size of the group; the larger the group the stronger the effect.
Preference of equatorial groups over axial groups in nature
Six-membered rings are the most common. Among them, those with all or most of the bulky groups in equatorial positions are significantly more common because of the stability of the equatorial relative to the axial bulky groups. For example, several six-membered ring monosaccharides (sugars) have the most bulky groups in equatorial positions. Glucose with all the bulky groups at the equatorial positions is the most common among them. Glucose units combine through an equatorial group to form a polymer called cellulose, the major component of wood. Starch is also a polymer of glucose but with one connecting bond axial. This changing connecting bond from equatorial to axial changes cellulose, a hard structural material, to starch, a carbohydrate. Wood can survive centuries, but starch would get rotten and decompose in a few days if left out in the open. The structures of glucose, starch, and cellulose are shown in Figure $12$.
Fused ring systems also exist in nature; most have six- and five-membered rings. For example, the basic skeleton of steroid hormones is three six-membered rings designated A, B, and C, and one five-membered ring designated D, fused, as shown in Fig 13. The $\ce{C-C}$ bond shared between the two fused rings is the bridge. The two $\ce{C's}$ at the ends of the bridge are bridgehead $\ce{C's}$. Each bridgehead $\ce{C}$ has two $\ce{C-C}$ bonds, one in each of the fused rings. Trans ring fusion has all four bonds, from the bridgehead $\ce{C's}$ into the two fused rings, equatorial and cis ring fusion has one of the four bonds axial. Steroid hormones always have trans-fused rings B, C, and D, as shown in Fig. 13. Ring A may be cis-fused, as in example 5$\beta$ in Figure $1$, but in a majority of cases, it is also trans-fused as in example 5$\alpha$ in Figure $1$. This is because the cis fusion with one axial $\ce{C-C}$ bond at the ring connection is associated with steric strain. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/03%3A_Stereochemistry/3.02%3A_Confirmations.txt |
Learning Objectives
• understand the meaning of absolute configuration, chiral centers, and draw perspective drawings and Fisher projections to represent chiral compounds.
• Apply rules to assigning D/L or R/S stereodescriptors to the chiral compounds containing one or two chiral centers.
• Understand optical activity as a physical characteristic of chiral compounds and the factors that affect it.
• Recognize stereochemical relationships between organic compounds, including enantiomers, diastereomers, and meso.
What is a configuration?
A molecule's permanent geometry that results from its bonds' spatial arrangement is called configuration. Stereoisomers have different configurations for the same set of atoms and the same group of bonds. For example, cis-but-2-ene has a different configuration than its stereoisomer trans-2-butene, shown in Figure 3.1.1.
Representing a configurations
Representing configuration around a single atom will be described first. An sp-hybridized atom has a linear geometry, as in the case of hydrogen cyanide $\ce{H-C≡N}$ or carbon dioxide $\ce{O=C=O}$ molecules. An sp2-hybridized atom is a trigonal planer, i.e., there are three $\sigma$-bonds at 60o to each other in the same plane and one $\pi$-bond with one of the $\sigma$-bond, as in the case of formaldehyde $\ce{H2C=O}$ represented as . An sp3-hybridized atom is tetrahedral, i.e., four $\sigma$-bonds as two V's of 109o internal angle joined perpendicular to each other at vortexes, e.g., methane $\ce{CH4}$ modeled as . Tetrahedral geometry is a 3D geometry usually represented by perspective drawings or Fisher projections, described in the next sections.
Perspective drawing
Since the tetrahedral geometry is two V's perpendicular to each other, one V is placed in the plane of the screen (or page) and represented by solid lines. The other V is represented by a solid wedge representing the bonding coming out of the screen towards the viewer and a hashed wedge representing the bond going beyond the screen away from the viewer, as illustrated below for the case of two configurations of 2-chlorobutane. The view angle is from the direction of the top-left corner. The V in the plane of the screen appears larger and the other appears smaller. Usually, the solid wedge is near the viewer relative to the hashed wedge.
If the configuration of two or more connected $\ce{C's}$ needs to be shown, the V's in the plane of the screen is usually drawn pointing downwards or upwards, making a zig-zag line as illustrated below for the case of two configurations of 2-chlorobutane.
In the case of skeletal formulas, $\ce{C's}$ and $\ce{H's}$ are not shown. However, the configuration is still readable correctly from the bonds shown, as illustrated below for the case of two configurations of 2-chlorobutane.
Fisher projections
Fisher projections represent the configuration as a projection of the molecule in 2D with four bonds in the shape of a cross, as explained in Figure $1$.
Examples of Fisher projections of the two configurations of 2-chlorobutane are shown below, along with the perspective drawings.
When more than one consecutive $\ce{C's}$ needs to be shown by a Fisher projection, the $\ce{C}$ chain, i.e., the skeleton, is shown vertically. For this purpose, $\ce{C-C}$ single bond is rotated by 180o resulting in a semi-cyclic conformation. Then the $\ce{C's}$ of interest are viewed from the edge-on so that the horizontal bonds are pointing towards the viewer and drawn vertically from top to bottom, as illustrated Figure $2$.
Chiral center
A compound with sp-, sp2-hybridized central atom, or a sp3-hybridized central atom with two or more same groups have a superimposable mirror image, as illustrated in Figure $3$. Since the compound and its mirror image are identical in these cases, these compounds are achiral, i.e., not chiral.
A compound with four different groups attached to an sp3-hybridized central atom has two distinct configurations, which are related as non-superimposable mirror images of each other, i.e., chiral. Since the two configurations represent different compounds that are stereoisomers, making and breaking of bonds, i.e., a chemical change, is needed to inter-convert them, as illustrated in Figure $3$.
An atom, often a $\ce{C}$ with four different groups attached to it, is called a chiral center.
A chiral center is often a cause of the chirality of the molecule. A stereochemical descriptor, e.g., D or L and R or S, is used to distinguish the two configurations of a chiral center.
An absolute configuration is the spatial arrangement of the atoms of a chiral molecular entity specified by a stereochemical descriptor.
Stereocenter
A stereocenter is an atom, axis, or plane with at least three different groups attached to it, so interchanging any two groups creates a new stereoisomer. The chiral center is a sub-class stereocenter, as illustrated in Figure $3$. The $\ce{C's}$ of the $\ce{C=C}$-bond of cis-but-2-ene and trans-but-2-ene shown in Figure 3.1.1, have three different groups and they are sterocenters. For example, swapping $\ce{-CH3}$ and $\ce{-H}$ on any of the two $\ce{C's}$ converts cis-but-2-ene to its stereoisomer trans-but-2-ene and vice versa.
Stereochemical descriptors
Two sets of stereochemical descriptors, i.e., D or L and R or S, are in common use and described next.
D/L Stereochemical descriptors
The D and L stereochemical descriptors were developed for monosaccharides, i.e., sugars. The monosaccharides have an aldehyde or a ketone group placed on the top or near the top in Fisher projections. Every other $\ce{C}$ usually has an alcohol ($\ce{-OH}$) group on them, and they are chiral except the terminal $\ce{C's}$, which are achiral.
Looking at the bottom-most chiral $\ce{C}$ in the Fisher projection of a monosaccharide, if the $\ce{-OH}$ group is towards the right of the Fisher projection, it is designated D and if it is towards the left, it is designated L, as illustrated in Figure $4$.
A hyphen separates the name of the monosaccharide from the D or L stereodescrptor. The D or L defines the absolute configuration of the bottom-most chiral $\ce{C}$ and the name that follows it defines the absolute configuration of all other chiral $\ce{C's}$ in a monosaccharide.
Amino acids have a chiral (\ce{C}\) with an amine ($\ce{-NH2}$) and a carboxylic acid ($\ce{-COOH}$) attached to it. The $\ce{-COOH}$ group is placed on the top and the $\ce{-NH2}$ group is either on the left or right side of the Fisher projection.
If the $\ce{-NH2}$ group is towards the right of the Fisher projection of an amino acid it is designated D. If it is towards the left, it is designated L, as illustrated in Figure $4$.
• Natural monosaccharides (sugars) have D-configurations, and natural amino acids have L-configurations with few exceptions.
• Although the common names of monosaccharides and amino acids with a D or L descriptor are not IUPAC names, these are commonly used in biochemistry.
R/S Stereochemical descriptors
The R and S stereochemical descriptors are part of the IUPAC nomenclature. For this purpose, the four groups attached to the chiral center are assigned priority 1, 2, 3, and 4, where 1 is the highest priority and 4 is the lowest. Then R or S is assigned based on the groups' relative orientations, considering the priority order.
Sequence rules
The following is the sequence rules, also called Cahn-Ingold-Prelog (CIP) sequence rules for assigning priority.
1. A group attached to a stereocenter through an atom of a higher atomic number (Z) has higher priority. For example, the priority of the four groups in is: $\overleftarrow{\ce{-OH > -NH2 > -CH3 > -H}}$.
2. If there is a tie, look at atoms two bonds away from the stereocenter. List them according to their atomic number (Z) and compare them, one by one, starting from the highest atomic number. Stop at the first point of difference and assign higher priority to the higher atomic number at the first point of difference. For example, in there is a tie between $\ce{-CH3}$ and $\ce{-CH2-CH3}$ due to both having $\ce{C}$ at one bond away from chiral center. Comparison of atoms two bonds away: $\overset{\large{H, H, H}}{\small{C, H, H}}$ assigns higher priority to $\ce{-CH2-CH3}$ over $\ce{-CH3}$, i.e., the priority order is: $\overleftarrow{\ce{-OH > -CH2-CH3 > -CH3 > -H}}$.
3. If the tie does not break at the distance of two bonds, move to atoms at the distance of three bonds from the chiral center and repeat the process until the first point of difference arrives.
CIP rule for atoms with double or triple bond
Atom bonded by a double bond is treated as connected twice, and by a triple bond is treated as connected three times. For example, $\ce{-CH=CH2}$ and $\ce{-C#CH}$ are the same for the atom directly attached to the chiral center. For comparing atoms two bonds away, the groups are treated as explained in the illustration below, with the first point of difference highlighted in red.
Caution
The rule is to stop at the first point of difference, even if the order may change later, as illustrated below with the first point of difference highlighted in red.
Assigning R or S to the chiral center
After all four groups on the chiral center have been assigned priority numbers 1, 2, 3, and 4, two situations arise:
• Case 1: The lowest priority, i.e., #4, is pointing away from the viewer, i.e., it is on a hashed line in the perspective drawing or on a vertical line in Fisher projection. In this case, R (rectus, Latin for right) is assigned if an arrow drawn from #1 to #2 to #3 is clockwise or directed to the right, and S (sinister, Latin for left) is assigned if it is directed to the left or counterclockwise, as illustrated in Figure $5$.
• Case 2: The lowest priority, i.e., #4 is not pointing away from the viewer. In this case,
• swap #4 with the group pointing away from the viewers,
• since a swap of any two groups changes the configuration, reverse the assignments, i.e., assign R if an arrow drawn from #1 to #2 to #3 is clockwise or directed to the right and S if it is directed to the left or counterclockwise, as illustrated in Figure $5$.
The stereochemical descriptor R or S describes the absolute configuration of the chiral center. They are placed within small brackets at the beginning of the IUPAC name, separated by a hyphen, as shown in the examples below. If there is more than one chiral center in a compound, the locant number precedes each chiral center's R or S descriptor, as explained in the examples below.
Example $1$
Assign R/S to the compound shown on the right.
Solution
1. Assign priority based on the atomic number of atoms attached to the chiral center: $\ce{-Br}$ > $\ce{-NH2}$ > $\ce{-CH3}$ > $\ce{-H}$.
2. Swap #4 with the group away from the viewer (hashed wedge). No action needed as $\ce{-H}$ is already at the hashed wedge.
3. Draw an arrow from #1 to #2 to #3. The arrow is counterclockwise, i.e., the stereodescriptor is S.
Answer: S
Example $2$
Assign R/S to the compound shown on the right.
Solution
1. Assign priority based on the atomic number of atoms attached to the chiral center: $\ce{-OH}$ is #1, there is a tie between $\ce{-CH3}$ and $\ce{-COOH}$, and $\ce{-H}$ is #4.
1. Tiebreaker is $\ce{O}$ two bonds away $\overset{\large{O, O, O}}{\small{H, H, H}}$, assigning #2 to $\ce{-COOH}$ and #3 to $\ce{-CH3}$.
2. Swap #4 with the group away from the viewer (hashed wedge), i.e., $\ce{-OH}$ with $\ce{-H}$ in this case.
3. Draw an arrow from #1 to #2 to #3. The arrow is counterclockwise. Since two groups were swapped, reverse the assignment i.e., counterclockwise is R.
Answer: R
Trick for assigning R and S
When the lowest priority group #4 is on the vertical line in the Fisher projection, as in or , there is no need to swap to bring it on the vertical line. Draw the arrow from #1 to #2 to #3 and reverse the assignments, i.e., assign S if the arrow is clockwise and R if it is counterclockwise. This trick also works when the lowest priority group #4 is on the solid wedge in the perspective drawing, as in . This trick makes the R/S assignment easy, especially for complex organic compounds where redrawing the structure after the swap is not easy.
Caution
Do not try this trick when the lowest priority group #4 is in the plane of the page in the perspective drawing, as in or . These two structures have opposite configurations, but the arrow drawn from #1 to #2 to #3 is in the same direction for both. It will lead to an incorrect assignment in one of the two.
Example $3$
Assign R/S to the compound shown on the right.
Solution
1. Assign priority based on the atomic number of atoms attached to the chiral center: $\ce{-Cl}$ is #1, there is a tie between $\ce{-CH3}$ and $\ce{-CH2CH3}$, and $\ce{-H}$ is #4.
1. Tiebreaker is $\ce{C}$ two bonds away $\overset{\large{C, H, H}}{\small{H, H, H}}$, assigning #2 to $\ce{-CH2CH3}$ and #3 to $\ce{-CH3}$.
2. Swap #4 with the group away from the viewer (hashed wedge). Trick: No need for Fisher projection, as the assignment can be reversed in the last step.
3. Draw an arrow from #1 to #2 to #3. The arrow is clockwise. Since the lowest priority group is on vertical line, reverse the assignment i.e., clockwise is S.
Answer: S
Example $4$
Assign R/S to the compound shown on the right.
Solution
1. Assign priority based on the atomic number of atoms attached to the chiral center: $\ce{-NH2}$ is #1, there is a tie between $\ce{-C(CH3)2SH}$ and $\ce{-COOH}$, and $\ce{-H}$ is #4.
1. Tiebreaker is $\ce{S}$ two bonds away $\overset{\large{S, C, C}}{\small{O, O, O}}$, assigning #2 to $\ce{-C(CH3)2SH}$ and #3 to $\ce{-COOH}$.
2. Swap #4 with the group away from the viewer (hashed wedge). Trick: No need for perspective drawing with the lowest priority group on a solid wedge, as the assignment can be reversed in the last step.
3. Draw an arrow from #1 to #2 to #3. The arrow is clockwise. Since the lowest priority group is on the solid wedge, reverse the assignment i.e., clockwise is S.
Answer: S (This compound is (S)-penicillamine)
Example $5$
Assign R/S to the compound shown on the right.
Solution
1. Assign priority based on the atomic number of atoms attached to the chiral center: $\ce{-NH2}$ is #1, there is a tie between $\ce{-C(CH3)2SH}$ and $\ce{-COOH}$, and $\ce{-H}$ is #4.
1. Tiebreaker is $\ce{S}$ two bonds away $\overset{\large{S, C, C}}{\small{O, O, O}}$, assigning #2 to $\ce{-C(CH3)2SH}$ and #3 to $\ce{-COOH}$.
2. Swap #4 with the group away from the viewer (hashed wedge). Not needed as the lowest priority group is already away from the viewer on hashed wedge.
3. Draw an arrow from #1 to #2 to #3. The arrow is clockwise. Clockwise is R.
Answer: R (This compound is (R)-penicillamine)
Compounds with one chiral center
Compounds containing one chiral center are chiral. They have a stereoisomer called enantiomer, which is their non-superimposable mirror image. Some examples of compounds containing one chiral center, their enantiomers with their absolute configuration assigned by D/L system and R/S system, and their tetetrahederal geometry illustrated are shown in Figure $6$.
Enantiomers have the same physical and chemical properties, except when they react with or interact with other chiral objects or molecules.
For example, hands and gloves are chiral objects. The right-hand glove fits on the right hand, and the left-hand fits on the left. Similarly, enantiomer molecules react equally and when they are produced in a chemical reaction they are produced equally, except when one of the reagents is chiral. For example, enzymes are chiral reagents and usually react with one of the two enantiomers in the reactants and/or produce one of the two enantiomers.
Optical activity
The light has an electric field oscillating perpendicular to the direction of propagation of the light. Ordinary light has an electric field oscillating in all places perpendicular to the direction of propagation. When it passes through a polarizer, it transmits only the waves oscillating in one plane, i.e., plane polarized light, as illustrated in Figure $7$. The plane polarized light is composed of left-handed and right-handed circularly polarized light, which are non-superimposable mirror images of each other, i.e., chiral, as shown in Figure $7$.
When the plane-polarized light passes through a solution of chiral molecules, one enantiomer absorbs more right-handed and the other more left-handed component. Therefore, if the solution is of a pure enantiomer, the plane of the plane-polarized light rotates clockwise or counterclockwise. The enantiomer that rotates the plane polarized clockwise is called dextrorotatory (abbreviated as d or (+)), and the other that rotates it counterclockwise is called levorotatory (abbreviated as l or (-)).
• The ability of chiral compounds to rotate the plane of plane-polarized light is called optical activity.
• Chiral compounds are optically active, and achiral compounds are optically inactive.
• An instrument used to measure the ability of a compound to rotate the plane of a polarized light is called a polarimeter, illustrated in Figure $8$.
The rotation of the plane of plane-polarized light by an enantiomer is called observed rotation, measured in degrees. The observed rotation depends on the concentration of the sample, the path length of the light through the sample, the wavelength of the light used, and temperature.
Specific rotation
When the concentration is fixed to 1 g/mL (or density of the pure sample in g/mL) and path length to 1 dm, the observed rotation is specific rotation [$\alpha$].
Usually, the wavelength ($\lambda$) is 589 nm, i.e., sodium D light and temperature (T) is room temperature and shown as subscript and superscrpt: $[\alpha]^{T}_{\lambda}$, as in the following formula.
$[\alpha]^{T}_{\lambda} = \frac{\text{Observed rotation (in degree)}}{\text{Path length (dm)}\times\text{Concentration (g/mL)}} \nonumber$
The specific rotation is a characteristic physical property of chiral compounds. If one enantiomer is dextrorotatory, the other is leavorotatory to the same degree, as shown in the following examples.
S and R define the absolute configuration of the chiral center, and dextrorotatory (+) or levorotatory (-) is an experimental property of the compound. A compound with S configuration may be dextrorotatory (+) or levorotatory (-), the same for R configuration. For example, (S)-lactic acid is dextrorotatory (+), and its salt (S)-sodium lactate is levorotatory (-), as shown below.
Racemic mixture
Mixing a chiral compound with its enantiomer diminishes its optical activity. A 50:50 mixture of enantiomers is optically inactive called a racemic mixture.
Compounds with more than one chiral centers
The maximum number of stereoisomers possible for chiral compounds is given by the formula 2n, where n is the number of chiral centers. For example, glyceraldehyde shown in Figure 3.1.1 has one chiral center and (21= 2) two stereoisomers: D-glyceraldehyde and L-glyceraldehyde. Erythrose, i.e., another monosaccharide, has two chiral centers and (22= 4) four stereoisomers, as shown in Figure $9$.
D-erythrose and L-erythrose are enantiomers, and D-threose and L-threose are also enantiomers, but erythrose is not an enantiomer of threose.
Diastereomers
Stereoisomers that are not enantiomers are diastereomers of each other. For example, D-erythrose is a diastereomer of D-threose and L-threose, as illustrated in Figure $9$.
The formula 2n tells the maximum possible number of stereoisomers of compounds having n number of chiral centers. The actual number may be less than the maximum. For example, tartaric acid has two chiral centers and three stereoisomers: (+) tartaric acid and (-) tartaric acid, which are enantiomers and achiral stereoisomers that is diastereomers of the first two, as illustrated in Figure $10$. The third isomer is achiral because its mirror image is identical, simply rotated by a 180o.
Meso compound
A meso compound is an achiral compound with two or more chiral centers. A meso compound has a plane of symmetry that divides into two mirror-image halves, as illustrated in Figure $10$.
Enantiomers have the same physical and chemical properties, except when interacting with chiral objects or reagents. Diastereomers have different physical and chemical properties, as can be observed in Table 1.
Table 1: Some of the physical and chemical properties of enantiomers ((+)-tartaric acid and (-)-tartaric acid) and their diastereomer (meso tartaric acid)
Stereoisomer (+)-Tartaric acid (-)-Tartaric acid Meso tartaric acid
Specific rotation +12.7 -12.7 0
Melting point (oC) 171-174 171-174 146-148
Density at 20oC (g/mL) 1.7598 1.7598 1.660
Solubility in water at 20oC (g/100 mL) 139 139 125
$\ce{pK_{1}}$ at 25oC 2.98 2.98 3.23
$\ce{pK_{2}}$ at 25oC 4.34 4.34 4.82
Chirality in life
A large number of compounds in living things are chiral. For example, amino acids and carbohydrates are chiral. Macro-molecules like proteins and nucleic acids are also chiral. Chirality is also observed on a macro-scale, e.g., our hands, foot, and ears are chiral. Similarly, horns and spots in axis deer are chiral, as shown in Figure $11$.
Stereospecific reactions in living things
Most chemical reactions in living things are stereospecific, i.e., one stereoisomer selectively reacts, and one is selectively produced. For example, the human digestion enzyme chymotrypsin has 268 chiral centers and 2268 possible stereoisomers, but only one isomer is produced. This is because amino acids that constitute chymotrypsin and other enzymes have L-configuration. The enzyme-catalyzed reactions are often stereospecific because their binding sites are chiral, and only one enantiomer reactant fits in them nicely for the reaction, as illustrated in Figure $12$. Similarly, one specific surface of enzyme-bound substrates is exposed for the reaction that results in the selective production of one enantiomer.
Chirality manifests in taste, flavor, odor, and drug action as their receptors have chiral binding sites, as illustrated in Figure $12$. Examples include:
• L-aspartame tastes sweet and is used as an artificial sweetener but D-aspartame is tasteless,
• R-(-)-carvone smells like spearmint, and S-(+)-carvone smells like caraway,
• R-(+)-limonene smells like orange and lemon, and S-(-)-limonene smells like a spruce tree,
• (S)-penicillamine has antiarthritic activity, and (R)-penicillamine is toxic, and
• S-ibuprofen has anti-inflammatory action, and R-ibuprofen is inactive. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/03%3A_Stereochemistry/3.03%3A_Configurations.txt |
• 4.1: What is a reaction mechanism
Homolytic and heterolytic bond breaking and making elementary steps are described, and the reaction mechanism is defined.
• 4.2: Free Radical Reactions
Free radical reaction mechanism and its application examples in halogenation of alkanes, combustion, thermal cracking, polymerization, and the aging process are described.
• 4.3: Acid-base reactions
Proton exchange reactions, i.e., acid-base reactions, factors that affect the strength of acids, direction of acid-base equilibrium based on the pKa of acids involved, and some simple organic acid-base elementary reaction steps are described. The ionization of amino acids and phosphate groups under physiological conditions is discussed.
• 4.4: Nucleophilic substitution and elimination reactions
Nucleophilic substitution and elimination reaction mechanisms and examples of alcohols, ethers, amines, and thiols reactions are described. Effects of factors that affect these reactions and define the conditions of these mechanisms are described
• 4.5: Nucleophilic acyl substitution reactions
Nucleophilic acyl substitution reaction mechanisms and their applications to the reactions of carboxylic acid and their derivatives are described. Applications of these reactions for the synthesis of condensation polymers are also described.
• 4.6: Nucleophilic Addition Reactions
Nucleophilic addition reaction mechanisms under base-promoted and acid-catalyzed conditions are described and applied to cyanohydrin, hydrates, hemiacetals, and acetals formation.
• 4.7: Electrophilic addition reactions
The electrophilic addition mechanism and its application to adding halogen acids and water to akenes and alkynes are described. The concepts of regioselectivity, stereoselective and steriospecific reactions, and tautomerization are described with examples of electrophilic addition reactions.
• 4.8: Electrophilic aromatic substitution reactions
The electrophilic aromatic substitution mechanism is described and applied to some reactions of benzene, including halogenation, nitration, sulfonation, alkylation, and acylation of benzene examples.
• 4.9: Reduction and oxidation (redox) reactions
The redox process is defined, and examples of organic redox reactions, including reduction of alkenes, carbonyl compounds, thiols, and oxidation of alcohols, aldehydes, and disulfide are described.
• 4.10: Reactions with cyclic transition state
Examples of reactions involving cyclic five- or six-member transition states, including cyclic hemiacetal formation of monosaccharides, Diels-Alder reactions, and decarboxylation of beta-keto acids, are described.
04: Organic reactions
What is a chemical reaction?
Chemical reactions involve rearranging atoms in a substance or substances called reactants resulting in new substances called products. In other words, chemical reactions involve breaking and/or making bonds. There are two major ways of breaking and making bonds, i.e., hemolytic and heterolytic bond breaking and making.
Hemolytic bond breaking and making
A covalent bond is a shared pair of electrons. In hemolytic bond breaking or hemolytic cleavage, the two elections are divided equally between the products as represented by the following reaction of a chlorine molecule (\(\ce{Cl2}\)) splitting into two chlorine atoms.
Recall that the half-headed curvy arrow represents the movement of one electron, pair of dots represents lone pair of electrons, a line represents a pair of bonding electrons, and a single dot represents an unpaired valance electron. A specie with an unpaired valence electron is called a free radical, e.g., \(\underset{\bullet\bullet}{_•^•\stackrel{\bullet}{Cl}{_•^•}}\) in the above reaction is a free radical. Free radicals are usually reactive species.
The Reverse of the above reaction is a hemolytic bond making where two reactants contribute one electron each to make a covalent bond, as shown in the following reaction between two chlorine atoms.
Heterolytic bond breaking and making
Heterolytic bond breaking happens so that a shared pair of electrons in the covalent bond are retained by one of the bonded atoms decreasing the charge by one, and the other bonded atom loses the shared electron increasing the charge by one, as shown in the following reaction.
Note that a regular curvy arrow represents the movement of two electrons, in this case, a bonding pair of electrons ending up as the fourth lone pair on the chlorine atom. Usually, the more electronegative atom receives the bonding pair of electrons and the less electronegative atom loses it in the heterolytic bond breaking. The reverse of the above reaction is heterolytic bond making, where the bonding pair of electrons are donated by one of the reactants, as shown in the following reaction.
Reaction mechanism
Chemical reactions often involve more than one bond-making and/or bond-breaking event. Further, reactions often happen in a sequence of steps that add up to yield the overall reaction equation. Individual reaction steps in the sequence are called elementary reactions.
Description of the step-by-step sequence of elementary reaction by which the overall chemical change occurs is called a reaction mechanism. There are two major reaction mechanisms, i) free radical reaction mechanisms that involve hemolytic bond breaking and making and ii) polar reaction mechanisms that involve heterolytic bond breaking and making. Some of the important reaction mechanisms are described in the following sections. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.01%3A_What_is_a_reaction_mechanism.txt |
Learning Objectives
• Understand free radical reaction mechanisms, including elementary reactions of its three main steps: initiation, propagation, and termination.
• Learn exams of free radical reactions in industry and daily life: converting alkanes to haloalkanes, cracking process, combustion, polymerization, and the aging process.
Free radical reaction mechanism
Hemolytic bond breaking requires energy equal to the bond dissociation energy, and hemolytic bond making releases energy equal to the bond dissociation energy. When a molecule absorbs energy in the form of heat or a photon of UV light, some of the weak bonds, like $\ce{Cl-Cl}$ bond (239 kJ/mol) or $\ce{O-O}$ bond (146 kJ/mol) breaks initiating free radicals, e.g.:
,
where hv represents a photon of UV light and ∆ represent heat energy. The reaction in which free radicals are created from neutral species, as in the above example, is called the free radical initiation step in a free radical reaction mechanism.
Free radicals are very reactive species. They usually abstract an atom, e.g., an H atom, from a hydrocarbon molecule, e.g.:
This reaction happens easily because the energy needed to break an $\ce{H-C}$ bond (413 kJ/mol) is compensated by the energy released by making $\ce{H-Cl}$ bond (427 kJ/mol). The $\ce{\stackrel{\bullet\;\;\;\;}{CH3}}$ repeats similar process when it collides with a $\ce{Cl2}$ molecule as shown in the second reaction above. Again, the energy released by making $\ce{C-Cl}$ bond (339 kJ/mol) compensates for the energy needed in breaking $\ce{Cl-Cl}$ bond (239 kJ/mol). Reactions in which one radical converts into a neutral specie and create another free radical that repeats the process, as in the above two reactions, are called propagation reactions. The propagation reactions often happen easily as the energy released in the bond-making compensates fully or partially for the energy needed in bond-making. The propagation reactions shown above happen in a cycle as the free radical produced in one is the reactant in the other and vice versa. The two propagation steps add up to the following overall reaction:
Overall reaction: $\ce{CH4 + Cl2 -> CH3Cl + HCl}$
This cycle of propagation reaction may repeat hundreds of times until one of the reactant exhaust or a free radical may collide with another free radical and terminate each other, as in the following reactions.
The homolytic bond-making between two free radicals that terminate the two radicals is called the termination reaction. The initiation, propagation, and termination are typical elementary reactions in a free radical reaction mechanism.
A $\ce{\stackrel{\bullet\;\;\;\;}{CH3}}$ may collide with $\ce{CH4}$ molecule and abstract a $\ce{H}$ atom and a $\underset{\bullet\bullet}{_•^•\stackrel{\bullet}{Cl}{_•^•}}$ may collide with a $\ce{Cl2}$ molecule and abstract a $\ce{Cl}$ atom, but there is no net chemical change in these elementary steps as shown below.
Examples of free radical reactions
Conversion of alkanes to haloalkanes
Exposing a mixture of alkane and halogen (chlorine or bromine) to UV light or heat at ~100oC converts alkanes to haloalkanes, as described in the free radical reaction mechanism for the case of $\ce{CH4}$ to $\ce{CH3-Cl}$ conversion.
$\ce{Cl2 + CH4 ->[\text{UV or heat}]CH3Cl + HCl}\nonumber$
This reaction is useful in industrial organic synthesis for converting alkanes found in petroleum to alkyl halides that serve as intermediates in synthesizing organic compounds. The reaction is not very useful for laboratory organic synthesis as the product of the reaction competes with the initial alkane resulting in a mixture of products, as shown in the following reactions.
$\ce{Cl2 + CH3Cl ->[\text{UV or heat}]CH2Cl2 + HCl}\nonumber$
$\ce{Cl2 + CH2Cl2 ->[\text{UV or heat}]CHCl3 + HCl}\nonumber$
$\ce{Cl2 + CHCl3 ->[\text{UV or heat}]CCl4 + HCl}\nonumber$
The haloalkane product can not compete effectively when the initial alkane concentration is high. So, the side products are minimized in the industrial process by employing a higher concentration of alkane. This reaction converts relatively less reactive alkanes to more reactive haloalkanes having polar $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Cl}}$ or $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Br}}$ bond. The haloalkanes are used in the polar reactions described in later sections.
Combustion
Hydrocarbons found in petroleum, coal, and natural gas are primarily consumed in combustion processes for producing heat. In a combustion process, ground state $\ce{O2}$ is first converted to an excited state $\ce{O2^{\ast}}$, that strips a $\ce{H}$ off of an organic molecule in the free radical initiation step.
$\ce{R-CH3 + O2^{\ast} -> R-\!\!\stackrel{\bullet\;\;\;\;}{CH2}+ HO\!\!\stackrel{\bullet}{O}}\nonumber$
The free radicals, i.e., $\ce{R-\!\!\stackrel{\bullet\;\;\;\;}{CH2}}$ and $\ce{HO\!\!\stackrel{\bullet}{O}}$ react further producing more radicals and, ultimately, convert the organic compounds to $\ce{CO2}$, $\ce{H2O}$, and heat. The combustion reaction of octane and ethanol are shown below as examples.
$\ce{2CH3CH2CH2CH2CH2CH2CH2CH3 + 25O2 -> 16CO2 + 18H2O + {Heat}}\nonumber$
$\ce{CH3CH2OH + 3O2 -> 2CO2 + 3H2O + {Heat}}\nonumber$
Cracking
Although $\ce{C-C}$ (347 kJ/mol) and $\ce{C-H}$ (413 kJ/mol) bonds are stable at room temperature, at 450 oC to 900 oC there is enough thermal energy to break them homolytically. The resulting radical species react further and usually end up in smaller chain alkanes and alkenes, as illustrated in the figure on the right (Pengeldi, CC0, via Wikimedia Commons). This process is called thermal cracking. For example, one of the many reactions involved in the thermal cracking of octane is the following.
Catalysts and steam may be added to accelerate the sections and/or modify the nature of products. Cracking convert long-chain alkanes to more valuable small-chain alkanes and alkenes.
Polymerization
Polymers are long molecules composed of hundreds or thousands of repeat units called monomers. One polymer synthesis mechanism is free radical polymerization, involving initiation, propagation, and termination steps. The process initiates by homolytic cleavage of a weak bond, like $\ce{O-O}$ bond in benzoyl peroxide ($\ce{C6H5COO-OOCC6H5}$, by heat or UV-photon. The radical produced by the homolytic cleavage adds to a $\pi$ bond of an alkene producing a new radical. The new radical adds to another alkene repeating the process in propagation steps, as illustrated below.
,
where $\ce{X}$ is $\ce{H}$ in polyethylene, $\ce{CH3}$ in polypropylene, benzene ring $\ce{C6H5}$ in polystyrene, $\ce{Cl}$ in polyvinylchloride, etc., as illustrated below.
Ultimately, the radicals react with each other by hemolytic bond =-making that terminates the radicals.
Polyethylene is used in plastic bags, bottles, toys, piping, etc. Polypropylene is used in ropes, carpets, pipes, furniture, food containers, etc. Polystyrene is used in foam packing, building insulation foam, plastic cutlery, foam containers for food and drinks, etc. Polyvinylchloride is used in pipes, window and door frames, bank cards, cable insulation, etc. In polytetrafluorethylene or Teflon, all $\ce{H's}$ and $\ce{X}$ in the monomer are $\ce{F's}$. Teflon is used in non-stick coatings, lubricants, electrical insulation for wires, etc. Various objects of common use made of these polymers are illustrated in Figure $1$.
Free radical reactions in biological processes
Free radical reactions are involved in the aging of food products. For example, fatty acids have long alkane chains, many of which have $\ce{C=C}$ bonds. The allylic hydrogens in fatty acid chains are susceptible to hemolytic bond cleavage, as shown below.
Oxygen is a diradical that adds to the free radical in a hemolytic bond-making step, producing a new radical. The new radical abstracts allylic hydrogen from another fatty acid producing a radical that repeats the cycle, i.e., propagation steps. The propagation reactions install hydroperoxyl groups ($\ce{-O-OH}$) on the fatty acids. The hydroperoxides are unstable and decompose to short-chain aldehydes and carboxylic acids responsible for the aged fats' unpleasant "rancid" smell. A similar process happens with the low-density lipoproteins deposited in arteries that lead to cardiovascular diseases. The aging process in organisms is also related to similar free radical reactions.
Living things have a mechanism of getting rid of unwanted free radicals by reacting them with radical scavengers. For example, vitamin C is a radical scavenger in the blood, and vitamin E is a radical scavenger in fats. The radical-scavengers neutralize the toxic radicals by donating $\ce{H}$ atoms from their $\ce{-OH}$ groups that break the propagation chain reaction. The new radicals generated in the scavenging reactions are less reactive and easily excreted before they can do more damage. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.02%3A_Free_Radical_Reactions.txt |
Learning Objectives
• Recognize acids and bases, understand the strength of acids and based, and predict the strength based on the structural features of the molecules.
• Understand simple acid-base reactions that are elementary steps in organic reaction mechanisms.
• Predict the direction of acid-base equilibrium based on the acid/base strength.
• Predict the ionization of acid/base functional groups in biochemicals under physiological conditions.
What is an acid-base reaction?
An acid-base reaction is a reaction in which a proton ($\ce{H^{+}}$) is exchanged between reactants. For example, when acetic acid ($\ce{CH3COOH}$) is mixed with water, a proton is transferred from acetic acid to water, as shown in the reaction equation below.
$\ce{CH3COOH + H2O <=> CH3COO^{-} + H3O^{+}}$
An acid donates a proton, and a base accepts a proton. $\ce{CH3COOH}$ is an acid and $\ce{H2O}$ is a base in the forward reaction. $\ce{CH3COO^{-}}$ is a base and $\ce{H3O^{+}}$ is an acid in the reverse reaction. Two species that are related by the addition or removal of a proton are conjugate acid-base pairs. For example, $\ce{CH3COOH}$ and $\ce{CH3COO^{-}}$ are conjugate acid-base pair. Similarly, $\ce{H3O^{+}}$ and $\ce{H2O}$ are conjugate acid-base pair. Since a proton has a +1 charge, its removal from an acid decreases the charge by one, and its addition to a base increases the charge by one. So, a general formula of an acid-conjugate acid-base pair could be represented as $\ce{HB}$ and $\ce{B^{-}}$, e.g., $\ce{CH3COOH}$ and $\ce{CH3COO^{-}}$ or as $\ce{HB^{+}}$ and $\ce{B}$, e.g., $\ce{H3O^{+}}$ and $\ce{H2O}$. Acid-base reactions are generally fast and reversible, establishing equilibrium. That is why equilibrium arrows are shown separating reactants and products.
Strength of an acid and a base
Strength of an acid
The strength of an acid is a measure of its ability to donate a proton to a base. Often the solvent is water, and the reference base is $\ce{H2O}$, as in the following general reaction.
$\ce{HA + H2O <=> A^{-} + H3O^{+}}\nonumber$
The relationship of molar concentrations of reactants and products in an equilibrium reaction is expressed by the following formula:
$\ce{K_{eq} = \frac{[A^{-}][H3O^{+}]}{[HA][H2O]}}\nonumber$
,where square brackets indicate the molar concentration of the specie within the brackets and $\ce{K_{eq}}$ is a constant called the equilibrium constant. Since water is a solvent, its concentration $\ce{[H2O]}$ is almost a constant and merged with the constant to define a new constant, as shown below.
$\ce{ K_{a} = K_{eq}[H2O] = \frac{[A^{-}][H3O^{+}]}{[HA]}}\nonumber$
, where $\ce{K_{a}}$ is a constant called acid dissociation constant that expresses the strength of the acid, the higher the value of $\ce{K_{a}}$; the stronger the acid. Since $\ce{K_{a}}$ is often a large negative number, it is expressed on a log scale with its signed reversed, as shown in the following formula.
$\ce{pK_{a} = -Log_{10}(K_{a})}\nonumber$
The $\ce{pK_{a}}$ is a quantitative measure of the acid strength, i.e., the smaller the $\ce{pK_{a}}$ means the large the $\ce{K_{a}}$ and stronger the acid.
Strength of a base
The strength of a base is a measure of its ability to accept a proton from an acid. Often the solvent is water, and the reference base is
$\ce{H2O}$, as in the following general reaction.
$\ce{A^{-} + H2O <=> HA+ HO^{-}}\nonumber$
The relationship between molar concentrations of reactants and products in an equilibrium reaction is expressed by the following formula:
$\ce{K_{eq} = \frac{[HA][HO^{-}]}{[A^{-}][H2O]}}\nonumber$
Again $\ce{[H2O]}$ is assumed to be a constant and merged with the constant to define a new constant, as shown below.
$\ce{ K_{b} = K_{eq}[H2O] = \frac{[HA][HO^{-}]}{[A^{-}]}}\nonumber$
, where $\ce{K_{b}}$ is a constant called base dissociation constant that expresses the strength of the base. the higher the value of $\ce{K_{b}}$ the stronger the base. again take a log and reverse the sign to arrive at a constant called $\ce{pK_{b}}$ as shown in the following formula.
$\ce{pK_{b} = -Log_{10}(K_{b})}\nonumber$
The $\ce{pK_{b}}$ is a quantitative measure of the base strength, i.e., the smaller the $\ce{pK_{b}}$ means the large the $\ce{K_{b}}$ and stronger the base.
Relationship between $\ce{pK_{a}}$ and $\ce{pK_{b}}$ of conjugate acid-base pair
The $\ce{K_{a}}$ of an acid $\ce{HA}$ and $\ce{K_{b}}$ of a conjugate acid-base $\ce{A^{-}}$ pair are reciprocal of each other as proven below.
$\ce{ {K_{a}} \times {K_{b}} = \frac{[A^{-}][H3O^{+}]}{[HA]}\times \frac{[HA][HO^{-}]}{[A^{-}]} = {[HO^{-}][H3O^{+}]} = K_{w} = 10^{-14}}\nonumber$
That rearranges to,
$\ce{ {K_{b}} = \frac{10^{-14}}{K_{a}}}\nonumber$
It shows that if acid is a strong acid (large $\ce{K_{a}}$), its conjugate base is a weak base ((small $\ce{K_{b}}$), and vice versa. Taking the log and changing the sign on both sides leads to a relationship between $\ce{pK_{a}}$ and $\ce{pK_{b}}$ of conjugate acid-base pair, as shown below.
$\ce{ -Log_{10}{K_{b}} = -Log_{10}(\frac{10^{-14}}{K_{a}})}\nonumber$
$\ce{ pK_{b} = 14 - pK_{a}}\nonumber$
Therefore, $\ce{ pK_{a}}$ values are tabulated in the reference books, and $\ce{ pK_{b}}$ values, if needed, are calculated from the $\ce{ pK_{a}}$ values of their conjugate acids using the above formula.
Factors affecting the strength of acids
The $\ce{pK_{a}}$ is a quantitative measure of acid strength, i.e., the smaller the $\ce{pK_{a}}$, the stronger the acid. For example, acetic acid ($\ce{CH3COOH, pK_{a} = 4.76}$) is a stronger acid than ethanol ($\ce{CH3CH2OH, pK_{a} = 15.9}$). Approximate $\ce{pK_{a}}$ of acids commonly encountered in organic chemistry are listed in Table 1.
Table 1: Approximate $\ce{pKa}$ values of acids commonly encountered in organic chemistry (Acidic protons are shown in the formula, and $\ce{R{-}}$ is an alkyl group or hydrogen)
$\ce{pK_a}$: <0 2 5 10 15
Compounds:
Strong acids like sulfuric acid
$\ce{HO-\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\!{O}}|\!\!|\enspace}{S\!}}}\!\!\!-OH}$
Phosphate esters $\ce{RO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$, $\ce{RO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OR}} |}{P}}}\!\!\!\!-OH}$
Carboxylic acids $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$
Phosphate esters $\ce{RO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-O^{-}}$
Phenols
Ammonium ions $\ce{R-NH3^{+}}$
Thiols $\ce{R-SH}$
Water $\ce{H2O}$
Alcohols $\ce{R-OH}$
$\ce{pK_a}$: 20 25 35 45 >50
Compounds: Aldehydes and ketones $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-CH2-R}$ Terminal alkynes $\ce{R-C≡CH}$
Hydrogen $\ce{H-H}$
Amines $\ce{R-NHR}$
Alkenes $\ce{R2C=CHR}$ Alkanes $\ce{R-CH2-R}$
The major factors affecting the acid strength are the following.
Electronegativity
Within the same row of the periodic table, the more electronegative the atom to which the proton is bonded, the more acidic the proton is. Electronegativity (EN) increases from left to right in the same row, e.g., $\ce{C}$, EN 2.6 < $\ce{N}$, EN 3.0 < $\ce{O}$, EN 3.4 < $\ce{F}$, EN 4.0.
The acid strength increases as the electronegativity of the atom carrying acidic proton increases within the same row of the periodic table, e.g., $\ce{CH4, pK_{a}}$ ~60 < $\ce{NH3, pK_{a}}$ ~38 < $\ce{H2O, pK_{a}}$ 14.0 < $\ce{HF, pK_{a}}$ 3.2.
This effect of electronegativity explains why alcohols ($\ce{R-OH}, pK_a \text ~16$) are more acidic than amines ($\ce{R-NHR}, pK_a \text ~40$) and amines are more acidic than alkanes ($\ce{R-CH3}, pK_a \text ~60$). This trend related to electronegativity does not hold when comparing elements from different rows.
Size of atom
The electronegativity decreases from top to bottom of a periodic table, e.g., for halogens: $\ce{F}$, EN 3.98 > $\ce{Cl}$, EN 3.16 > $\ce{Br}$, EN 2.69 > $\ce{I}$, 2.66. . The size of atoms has an opposite trend, i.e., the size increases from top to bottom of the periodic table, e.g., the size of halogens varies in this order: $\ce{F}$ < $\ce{Cl}$ < $\ce{Br}$ < $\ce{I}$.
The acid strength increases as the size of the atom carrying the acidic proton increases in the same period, e.g., $\ce{HF, pK_{a}}$ 3.2 < $\ce{HCl, pK_{a}}$ -7 < $\ce{HBr, pK_{a}}$ -8 < $\ce{HI, pK_{a}}$ -9.9.
This is because as the size of the atom increases, its bond with a proton becomes weaker and easier to break. Further, the negative charge a proton leaves behind is more easily stabilized when spread over a larger atom than a smaller atom. The effect of the size of an atom on acid strength is opposite to that of electronegativity within the same period.
The effect of the size of the atom bearing acidic proton on acidity is dominant over the effect of electronegativity.
The effect of size explains why thiols ($\ce{R-SH}, pK_a \text ~10$) are stronger acids than alcohols ($\ce{R-OH}, pK_a \text ~16$).
Resonance
When a proton leaves an acid $\ce{HA}$, it usually leaves a negative charge on the conjugate base $\ce{A^{-}}$. Resonance can stabilize the negative charge by distributing it on more than one atom.
An acid with conjugated base resonance stability can let go of its acidic proton more easily. It is stronger than an equivalent acid with no resonance stabilization of its conjugate base.
For example, both carboxylic acids and alcohols have acidic proton on an $\ce{-OH}$ group, but carboxylic acids are stronger acids ($\ce{pK_{a}, \text ~5}$) than alcohols ($\ce{pK_{a}, \text ~16}$) due to resonance stabilization of the negative charge on their conjugate bases, as illustrated below.
Carboxylic acid:
Alcohol:
Phenols are also stronger acids ($\ce{pK_{a}, \text ~10}$) than alcohols for the same reason, as shown below.
However, phenols are weaker acids than carboxylic acids because the negative charge is shared by more electronegative $\ce{O}$ in carboxylic acids than by less electronegative $\ce{C}$ in phenols.
Protons on $\ce{\alpha C}$ to a $\ce{C=O}$ group, as in aldehydes, ketones, and carboxylic acid derivatives, are more acidic ($\ce{pK_{a}, \text ~20}$) than protons on alkanes ($\ce{pK_{a}, \text ~60}$) due to the resonance effect, as shown below.
Hybridization
Electrons in an s-orbital are nearer to and more attracted by the nucleus than in a p-orbital of the same shell. This is because of the spherical shape of the s-orbital placing electrons nearer to the nucleus versus the dumbbell shape of the p-orbital in the same shell. Therefore, the valence electrons in an sp-orbital having 50% s-character are attracted more to the nucleus than an sp2 orbital with 33% s-character and an sp3 orbital with 25% s-character. Recall that electronegativity is a measure of the ability of a nucleus to attract valence electrons.
The electronegativity of an atom and, consequently, the acidity of a proton attached to it increase with a change in hybridization in this order:sp3 < sp2 < sp.
For example, the acidity of protons of sp3-hybridized ethane, sp2-hybridized ethene, and sp-hybridized ethyne increases in this order: $\ce{CH3CH3}$ $pK_a \;51$ < $\ce{CH2=CH2}$ $pK_a \;44$ < $\ce{CH≡CH}$ $pK_a \;25$.
Inductive effect
The electron-withdrawing effect of an electronegative atom, i.e., the inductive effect increases the acidity of the proton adjacent to it.
• Near the electronegative atom stronger, the effect, as illustrated in the example below where $\ce{Cl}$ is the electronegative atom affecting the acidity of $\ce{-COOH}$ proton.
• The more electronegative the atom stronger the effect, as illustrated in the example below, where the electronegativity increases in this order: $\overrightarrow{\ce{ H < I < Br < Cl < F}}$.
Amino acids are more acidic than carboxylic acids due to the inductive effect of ammonium ion, e.g., glycine is more acidic than acetic acid as shown below.
Charge
A proton $\ce{H^{+}}$ on a positively charged specie is easier to remove and, consequently, more acidic than on the same species in a neutral state.
The acidity of the following species reflects this: $\overrightarrow{\ce{H2O, pK_a \;\text15.7 < H3O^{+}, pK_a \;\text-1.74}}$ and $\overrightarrow{\ce{NH3, pK_a \;\text ~38 < NH4^{+}, pK_a \;\text 9.24}}$. The effect of the charge is reduced but still exists when the charge is on an atom other than the atom bearing the acidic proton. Negatively charges species, conversely, are less acidic than the same specie in a neutral or positively charged state, as illustrated below for the case of phosphoric acid species.
$\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$, $\ce{pK_a \;\text 2.14}$ > $\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OH}} |}{P}}}\!\!\!\!-OH}$, $\ce{pK_a \;\text 7.20}$ > $\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} |}{P}}}\!\!\!\!-OH}$, $\ce{pK_a \;\text 12.37}$
Examples of organic acid-base reactions
Reactions of organic acids
Carboxylic acids are organic acids. The following example shows that when a carboxylic acid is dissolved in water, it dissociates by donating its proton to water.
$\ce{CH3COOH + H2O <=> CH3COO^{-} + H3O^{+}}$
The reaction is written as an equilibrium because it goes in both directions, forward and reverse. acetic acid $\ce{CH3COOH}$ is the acid in reactants and hydronium ion $\ce{H3O^{+}}$ is the acid in products.
The general rule is that a stronger acid goes to a weaker one in an acid-base reaction. It is equally correct to say a stronger base goes to a weaker base.
For example, it is more accurate to say that the above reaction is reverse directed, or there is more concentration of reactants than the products at equilibrium, because $\ce{H3O^{+}}$ is a stronger acid than $\ce{CH3COOH}$. This fact is represented by a longer equilibrium arrow in the direction of weaker acid in the equilibrium, as shown below.
$\ce{\underset{pK_a \;\text 4.76}{CH3COOH} + H2O <<=> CH3COO^{-} + \underset{pK_a \;\text -1.74}{H3O^{+}}}\nonumber$
This reaction is made strongly forward directed by using a strong base like $\ce{^{-}OH}$ that produces a weaker acid $\ce{H2O}$, making the reaction almost irreversible, as shown below.
$\ce{\underset{pK_a \;\text 4.76}{CH3COOH} + ^{-}OH -> CH3COO^{-} + \underset{pK_a \;\text 14.0}{H2O}}\nonumber$
This is an example of an acid-base neutralization reaction. The $\ce{^{-}OH}$ ion is obtained by dissolving sodium hydroxide $\ce{NaOH}$ in water, but $\ce{Na^{+}}$ is a spectator ion, i.e., it does not take part in the reaction and is usually not shown in the equation. As shown below, other organic acids, like phenols and thiols, do similar acid-base neutralization reactions.
$\ce{\underset{pK_a \;\text 9.95}{C6H5OH} + ^{-}OH -> C6H5O^{-} + \underset{pK_a \;\text 14.0}{H2O}}\nonumber$
$\ce{\underset{pK_a \;\text 10.4}{CH3SH} + ^{-}OH -> CH3S^{-} + \underset{pK_a \;\text 15.7}{H2O}}\nonumber$
Terminal alkyne has acidic protons but they are weak acids ($\ce{pK_a \;\text ~25}$). A base like $\ce{^{-}OH}$ is not a strong enough base to remove all of the alkyne protons, as shown by the following equilibrium reaction.
$\ce{\underset{pK_a \;\text 25}{R-C≡CH} + ^{-}OH <<=> R-C≡C^{-} + \underset{pK_a \;\text 14.0}{H2O}}\nonumber$
An amide anion ($\ce{^{-}NH2}$ in the form of $\ce{NaNH2}$) is a sufficiently strong base to neutralize alkyne, as shown below.
$\ce{\underset{pK_a \;\text 25}{R-C≡CH} + ^{-}NH2 -> R-C≡C^{-} + \underset{pK_a \;\text 38}{NH3}}\nonumber$
Alcohols ($\ce{R-OH, pK_a \;\text ~25}$) are also weak acids and not fully neutralized with bases like $\ce{^{-}OH}$, as shown below.
$\ce{\underset{pK_a \;\text 16}{R-OH} + ^{-}OH <=> R-O^{-} + \underset{pK_a \;\text 14}{H2O}}\nonumber$
A hydride ion ($\ce{H^{-}}$ in the form of $\ce{NaH}$) is a sufficiently strong base to neutralize alcohols, as shown below.
$\ce{\underset{pK_a \;\text 16}{R-OH} + H^{-} -> R-O^{-} + \underset{pK_a \;\text 35}{H2}}\nonumber$
Reactions of organic bases
Amines ($\ce{R-NH2}$) are organic bases that produce $\ce{^{-}OH}$ ions in water, as shown below.
$\ce{\underset{pK_a \;\text 15.7}{H2O} + R-NH2 <<=> ^{-}{OH} + \underset{pK_a \;\text 10}{R-NH3^{+}}}\nonumber$
Amines neutralize strange acids like ($\ce{HCl}$) and produce ammonium salts, as shown below.
$\ce{\underset{pK_a \;\text -7}{HCl} + R-NH2 -> R-NH3^{+}{Cl^{-}}}\nonumber$
Ammine.HCl drugs -increasing shelf-life and solubility in body fluids
Many drugs contain amine functional groups. Amines are usually unstable and insoluble in water. Neutralizing the amine ($\ce{R-NH2}$) with $\ce{HCl}$ converts them to ammonium chloride salts ($\ce{R-NH3^{+}C^{-}}$), also called amine.HCl salts ($\ce{R-NH2.HCl}$), which are water-soluble and more stable. Therefore, drugs containing amine functional groups are usually sold in ammonium chloride form, making them soluble in body fluids like blood plasma and increasing their shelf-life. For example, methadone, a narcotic analgesic, and procaine, a local anesthetic, are marketed as hydrochloride salts, as shown below.
Alcohols are amphoteric substances like water, i.e., they can donate a proton as acid and accept a proton as bases. Like alcohols, the oxygen of carbonyl ($\ce{C=O}$ group also has two lone pairs and can accept a proton. A strong acid like $\ce{HCl}$ or $\ce{H2SO4}$can protonate an alcohol or a carbonyl group, as shown below.
$\ce{\underset{pK_a \;\text -7}{HCl} + R-OH <=>> Cl^{-} + \underset{pK_a \;\text ~-3}{R-OH2^{+}}}\nonumber$
$\ce{\underset{pK_a \;\text -9}{H2SO4} + R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R <=>> Cl^{-} + \underset{pK_a \;\text ~-7}{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{C}}\!\!\!\!\!-R}}\nonumber$
The acid-base reactions that produce cations by accepting protons, like $\ce{R-OH2^{+}}$ or $\ce{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{C}}\!\!\!\!\!-R}$ and anions by donating protons, like $\ce{R-O^{-}}$, $\ce{R-C≡C^{-}}$, $\ce{R-S^{-}}$, etc. are important intermediates in organic reactions described in the later sections.
Ionization of acidic and basic functional groups at physiological pH
The Henderson-Hasselbach equation for an acid ($\ce{HA}$) is: $\color{blue}\text {Henderson-Hasselbach equation: } \ce{ pH = pK_{a} + Log_{10}(\frac{[A^{-}]}{[HA]})}\nonumber$, where $\ce{pH = -Log_{10}[H3O^{+}]}$, and $\ce{pK_{a} = -Log_{10}K_{a}}$. For a base ($\ce{B}$) is: $\color{blue}\ce{ pH = pK_{a} + Log_{10}(\frac{[HB^{+}]}{[B]})}\nonumber$, where $\ce{pK_a}$ is that of the conjugate acid $\ce{HB^{+}}$. The Henderson-Hasselbach equation helps estimate the ratio of acid ($\ce{HA}$) to its conjugate base (($\ce{A^{-}}$) and of base ($\ce{HB}$) to its conjugate acid (($\ce{B^{+}}$) at the pH of the medium. According to Henderson-Hasselbach equaiton,
• for an acid ($\ce{HA}$), if $\ce{pH}$ of medium is two units lower than the $\ce{pK_a}$ of the acid, it will be present substantially in its de-protonated form $\ce{A^{-}}$, and
• for a base ($\ce{B}$), if $\ce{pH}$ of medium is two units higher than the $\ce{pK_a}$ of its conjugated acid $\ce{HB^{+}}$, it will be present substantially in its protonated form $\ce{HBA^{+}}$.
Following conclusions are drawn by applying the above rules to the functional groups commonly found in cells.
• The $\ce{pH}$ of cells is generally between 7 and 8.5. Carboxylic acids ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$) have $\ce{pK_a}$ 4 to 5. So, carboxylic acids exist in an ionized form $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}$ in a physiological medium.
• The conjugate acid forms of amines $\ce{R-NH3^{+}}$ have $\ce{pK_{a}}$ 9 to 11. So, amines exist in the ionized form $\ce{R-NH3^{+}}$ in a physiological medium.
• Amino acids $\ce{R-\!\!\!\!\!\!\!\!\!{\underset{\underset{\huge\enspace\enspace\enspace{NH2}} |}{C}}\!\!\!\!\!\!\!\!\!H-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$ have amine and carboxylic acid groups. Amino acids exist as $\ce{R-\!\!\!\!\!\!\!\!\!{\underset{\underset{\huge\enspace\enspace\enspace\underset{\small{+}}{N}{H3}} |}{C}}\!\!\!\!\!\!\!\!\!H-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}$. A specie with separate positive and negative charge groups on it is called a zwitterion. Amino acids exist as a zwitterions.
• Phosphodiesters $\ce{RO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OR}} |}{P}}}\!\!\!\!-OH}$ present in DNA and RNA molecules have $\ce{pK_{a}}$ 1 to 3. So, phosphodiester also exist in an ionized form $\ce{RO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}|\!\!|\enspace}{\underset{\underset{\huge\enspace\,{OR}} |}{P}}}\!\!\!\!-O^{-}}$ in a physiological medium.
• The conjugate acid forms of amines, i.e., ammonium ion $\ce{R-NH3^{+}}$ have $\ce{pK_{a}}$ 9 to 11. So, amines exist as ammonium ions $\ce{R-NH3^{+}}$ in a physiological medium.
Derevation of Henderson-Hasselbalch equation
The equation is derived by performing the following steps: i) re-arranging $\ce{ K_{a}}$ expression:
$\ce{ K_{a} = \frac{[A^{-}][H3O^{+}]}{[HA]}}\nonumber$
$\ce{ [H3O^{+}] = K_{a}\times\frac{[HA]}{[A^{-}]}}\nonumber$
Taking the log of both sides and reversing the sign:
$\ce{ -Log_{10}[H3O^{+}] = -Log_{10}(K_{a}\times\frac{[HA]}{[A^{-}]})}\nonumber$
$\ce{ -Log_{10}[H3O^{+}] = -Log_{10}K_{a}-Log_{10}(\frac{[HA]}{[A^{-}]})}\nonumber$
making a substitution for $\ce{ pK_{a}}$, and $\ce{ pH}$:
$\boxed{\text {Henderson-Hasselbach equation: } \ce{ pH = pK_{a} + Log_{10}(\frac{[A^{-}]}{[HA]})}}\nonumber$
, where $\ce{pH = -Log_{10}[H3O^{+}]}$, and $\ce{pK_{a} = -Log_{10}K_{a}}$.
A similar derivation of a general base $\ce{B}$ and its conjugate acid $\ce{HB^{+}}$ leads to this equation:
$\boxed{\ce{ pH = pK_{a} + Log_{10}(\frac{[HB^{+}]}{[B]})}}\nonumber$
, where $\ce{pK_a}$ is that of the conjugate acid $\ce{HB^{+}}$. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.03%3A_Acid-base_reactions.txt |
Learning Objectives
• Identify nucleophiles and electrophiles, their similarities, and differences from acids and bases.
• Understand SN2, E2, SN1, or E1 reaction mechanisms, factors that affect them, and conditions that dictate which one will take place.
• Learn some examples of SN2, E2, SN1, and E1 reactions of alcohols, ethers, amines, and thiols.
Nucleophile and electrophile
Nucleophile is a neutral or anionic specie that can donate a lone pair or $\pi$ bonding electrons to make a covalent bond.
Examples of nucleophile are negative or partial negative ($\delta{-}$) atoms with lone pair or $\pi$ bond in the following $\ce{H3{N}{_{\bullet}^{\bullet}}}$, $\ce{H2\overset{\!\bullet\bullet}{O}{_{\bullet}^{\bullet}}}$, $\ce{H\overset{\!\!\!\bullet\bullet}{\underset{\bullet\bullet}{O}{_{\bullet}^{\bullet}}}^{-}}$, and $\ce{H2C=CH2}$.
Electrophile is an electron-deficient atom of a neutral or cationic specie that can receive an electron pair to make a covalent bond.
Examples of electrophile are positive charge or partial positive ($\delta{+}$) charge $\ce{C's}$ or $\ce{H's}$ in the following species: $\ce{(CH3)3\overset{+}{C}}$, $\ce{\overset{\delta{+}}{C}H3{-}\overset{\delta{-}}{Cl}}$, $\ce{(CH3)2\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{Cl}}$.
Differences between nucleophile-electrophile and acid-base
The base is a substance that donates a pair of electrons to a proton to make a covalent bond. So, a base is a sub-class of nucleophiles and an acidic proton is a subclass of electrophiles. Acid-base reactions are fast reactions. In acid-base reactions, emphasis is on thermodynamic, specifically on the equilibrium constant K quantified in terms of pKa, i.e., smaller the pKa means larger K and stronger acid. Nucleophiles donate electrons to an electron-deficient carbon or some atom usually other than a proton. Emphasis in nucleophilic-electrophilic reaction is on the kinetics, a good nucleophile reacts faster and a poor nucleophile reacts slower.
Similarities and differences in nucleophilicity and basicity
The similarity is that the stronger bases are stronger nucleophiles when compared within the same row of the periodic table. For example, basicity follows the order $\overrightarrow{\ce{HO^{-}<NH2^{-}<F^{-}}}$ and their nucleophilicity follows the same order. Similarly, anionic species, e.g., $\ce{NH2^{-}}$, $\ce{HO^{-}}$ and $\ce{RO^{-}}$ are stronger bases and good nucleophiles compared to the corresponding neutral species of the same element, i.e., $\ce{NH3}$, $\ce{H2O}$ and $\ce{ROH}$.
The difference is that the basicity and nucleophilicity are affected differently when comparing atoms of different rows. For example, hydride ion ($\ce{H^{-}}$) from the 1st row is a stronger base but a poor nucleophile. the opposite is true for the 3rd-row and higher elements, e.g., $\ce{SH^{-}}$, $\ce{Cl^{-}}$, $\ce{Br^{-}}$. and $\ce{I^{-}}$ are week bases but usually good nucleophiles. Unlike basicity, nucleophilicity is affected by steric factors and solvents. Most of these differences are related to the fact that in acid-base reactions, the electron recipient is the 1s orbital of a proton, and in nucleophilic-electophilic reactions, the electron recipient is 2s, 2p, or larger orbital of carbon or some other element. The details of these factors are out of the scope of this book. Base strength is described as stronger or weak, and nucleophilic strength is described as good or poor.
Nucleophilic substitution mechanisms
In a nucleophilic substitution reaction, a nucleophile ( $\ce{Nu{_{\bullet}^{\bullet}}^{-}}$ ) attacks and makes a covalent bond with $\delta{+}$ atom of the target molecule, called substrate ($\ce{\overset{\delta{+}}{R-}\overset{\delta{-}}{X}}$). The polar bond of the target molecule breaks heterolytically, leaving the bonding electrons with the more electronegative end called leaving group ($\ce{X{_{\bullet}^{\bullet}}^{-}}$), as shown in the following generalized reaction.
$\ce{{Nu}{_{\bullet}^{\bullet}}^{-} + \overset{\delta{+}}{R-}\overset{\delta{-}}{X} -> {\overset{\delta{+}}{R-}\overset{\delta{-}}{Nu}} + X{_{\bullet}^{\bullet}}^{-} }\nonumber$
Since one nucleophile, i.e., $\ce{Nu{_{\bullet}^{\bullet}}^{-}}$, replaces another nucleophile, i.e., $\ce{X{_{\bullet}^{\bullet}}^{-}}$ on the substrate, it is called a nucleophilic substitution reaction. The nucleophilic substitution reactions are based on the fact the incoming nucleophile has a stronger tendency to make a covalent bond with the electrophilic center than the leaving group. In other words, the incoming nucleophile is stronger and the leaving group is a weaker nucleophile. Mechanisms of the nucleophilic substitution reactions are described below.
Nucleophilic substitution bimolecular (SN2)
One mechanism for nucleophilic substitution reaction is concerted bond-making and breaking in a single step, as shown below.
The incoming nucleophile approaches the electrophilic $\ce{\overset{\delta{+}}{C}}$ from the side opposite to the leaving group. The incoming nucleophile starts making the bond and the leaving group starts breaking the bond simultaneously. The other three groups pointing away from the leaving group, start moving to the other side, away from the incoming nucleophile. In the transition state, the three groups acquire a trigonal planar arrangement perpendicular to the bond being broken and formed. This reaction mechanism is called SN2, where S is for substitution, N is for nucleophilic, and 2 is for bimolecular.
• If the electrophilic $\ce{\overset{\delta{+}}{C}}$ is a chiral center, its configuration is inverted in the product of SN2 reaction.
• This elementary step of SN2 reaction involves two reactants in its rate-determining step, the incoming nucleophile, and the substrate, and it is a bimolecular reaction.
The rate of an SN2 reaction is directly proportional to the substrate concentration and the nucleophile concentration. The rate depends on the substrate's structure, the nature of the leaving group, the nature of the nucleophile, and the solvent, as described next.
Effect of the structure of the substrate
The rate follows the following order with respect to the nature of the electrophilic $\ce{\overset{\delta{+}}{C}}$: $\overleftarrow{\text{ methyl > primary > secondary > tertiary}}$. The steric hindrance posted by the substrate to the nucleophile explains it, which follows the same order, as illustrated below.
Type of electrophile carbon Methyl Primary Secondary Tertiary
Model
Structures with mechanism arrows
The relative SN2 reaction rate 145 1 0.008 no reaction
Effect of the nucleophile
Good nucleophiles react faster. Particularly, anionic nucleophiles, like $\ce{HO^{-}}$ or $\ce{RO^{-}}$ are employed for SN2 as they react much faster than their corresponding neutral counterparts, i.e., $\ce{H2O}$ or $\ce{ROH}$, as shown in the following example.
Effect of the leaving group
Leaving group propensity has a trend opposite to the basicity of the leaving group, i.e., good leaving groups are weaker bases. For example, the basicity of halide ions follows this order: $\overleftarrow{\ce{F^{-} > Cl^{-} > Br^{-} > I^{-}}}$ but the rate of reaction of alkyl halides, under the same conditions, follows the opposite trend, i.e., $\overrightarrow{\ce{R-F < R-Cl < R-Br < R-I}}$. The effect of the leaving group on the rate of the reaction is shown in the following table.
Reaction Relative rate
$\ce{HO^{-} + RCH2-F -> RCH2-OH + F^{-}}$ 1
$\ce{HO^{-} + RCH2-Cl -> RCH2-OH + Cl^{-}}$ 200
$\ce{HO^{-} + RCH2-Br -> RCH2-OH + Br^{-}}$ 10,000
$\ce{HO^{-} + RCH2-I -> RCH2-OH + I^{-}}$ 30,000
Effect of the solvent
The solvent is needed to dissolve both the substrate and the nucleophile. The substrate is a polar compound like $\ce{\overset{\delta{+}}{CH3-}\overset{\delta{-}}{Br}}$ and the nucleophile is usually in the form of an ionic solid like $\ce{Na^{+}OH^{-}}$ or $\ce{CH3O^{-}Na^{+}}$. Nonpolar solvents do not work because polar and ionic substances do not dissolve in nonpolar solvents. Polar solvents can dissolve polar and ionic compounds. Polar solvents fall into two categories:
• polar protic solvent that have an acidic proton like water ($\ce{H2O}$) and methanol ($\ce{CH3-OH}$), and
• polar aprotic like acetone ( $\ce{(CH3)2\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$ ) and dimethylsulfoxide ( $\ce{(CH3)2\overset{\delta{+}}{S}{=}\overset{\delta{-}}{O}}$ }\)) which do not have an acidic proton. Their $\delta{+}$ end is at the center of the molecule, surrounded by bulky groups.
Polar protic solvent dissolves ionic compounds by ion-dipole interactions forming a layer of solvent around cation and anion. They are not suitable for SN2 reactionas the solvent layer prevents nucleophiles from approaching electrophilic $\ce{\overset{\delta{+}}{C}}$.
Aprotic solvents dissolve the ionic compound by the ion-dipole interaction but leave the anion, i.e., nucleophile, almost free. This is because the anion is prevented from approaching the $\delta{+}$ pole of the solvent due to steric hindrance, as illustrated in Figure $1$. Recall that electrostatic force is inversely proportional to the square of the distance between the +ve charge and the -ve charge, i.e., the longer the distance, the weaker the force.
Polar aprotic solvents are suitable for SN2 reaction as the nucleophiles are relatively free to approach electrophilic $\ce{\overset{\delta{+}}{C}}$ of the substrate.
Examples of SN2 reactions
Alkyl halides are derived from alkanes by free radical reactions. The alkyl halides are then converted to a number of other classes of organic compounds by SN2 reaction, e.g., ethyl bromide can be converted to:
alcohol: $\ce{CH3CH2Br + OH^{-} -> CH3CH2OH + Br^{-}}$,
a thiol: $\ce{CH3CH2Br + HS^{-} -> CH3CH2SH + Br^{-}}$,
an ether: $\ce{CH3CH2Br + CH3CH2O^{-} -> CH3CH2OCH2CH3 + Br^{-}}$,
a thioether: $\ce{CH3CH2Br + CH3CH2S^{-} -> CH3CH2SCH2CH3 + Br^{-}}$,
an amine: $\ce{CH3CH2Br + ^{-}NH2 -> CH3CH2NH2 + Br^{-}}$,
a nitrile: $\ce{CH3CH2Br + ^{-}C≡N -> CH3CH2C≡N + Br^{-}}$, and
and an alkyne: $\ce{CH3CH2Br + ^{-}C≡CCH3 -> CH3CH2C≡CCH3 + Br^{-}}$.
Elimination bimolecular (E2)
Recall that the $\ce{C}$ connected to a functional group is $\alpha\ce{C}$, and the one adjacent to it is $\beta\ce{C}$. Also, recall that $\ce{\overset{\delta{+}}{H-}\overset{\delta{-}}{Br}}$ is a strong acid because the acidic polar leaves electron on elecronegative $\ce{Br}$. Similarly, protons on $\beta\ce{C}$ are acidic because they can send the bonding electrons to $\ce{Br}$ through the following mechanism.
Nucleophiles are bases at the same time. The base $\ce{HO^{-}}$ attacks $\beta\ce{H}$, the $\ce{H-C}$ bonding electrons establish a $\pi$-bond between $\alpha\ce{C}$ and $\beta\ce{C}$, and the $\alpha\ce{C}$ let go the leaving group $\ce{Br^{-}}$. Since a $\beta\ce{H}$ is eliminated along with the leaving group, This reaction mechanism is called $\beta$-elimination. It is a bimolecular reaction because of two species involved in this elementary step: the base and the substrate. This specific $\beta$-elimination is called elimination bimolecular (E2), where E stands for elimination and 2 for bimolecular.
SN2 vs. E2
SN2 and E2 reactions compete because the same reagent is nucleophile in SN2 and base in the E2 reaction. However, the effect of the structure of the substrate is opposite in SN2 and E2. The structure of substrate affects SN2 in this order: $\overleftarrow{\text{ methyl > primary > secondary > tertiary}}$. Tertiary substrates do not react by SN2 mechanics because the nucleophile is prevented from approaching $\alpha\ce{C}$ by steric hindrance. However, $\beta\ce{H}$ is still exposed and easily approachable by the base, as shown below.
Further, E2 is facilitated by lowering steric crowding around the $\alpha\ce{C}$. Therefore, tertiary substrates yield E2 product under the SN2 conditions. The steric crowding around $\alpha\ce{C}$ of secondary substrates is less than tertiary. So, E2 and SN2 compete, and elimination and substation products are formed from the secondary substrates. There is no steric crowding in the case of primary or methyl substrates. So, SN2 dominated over E2, and substitution product is formed almost exclusively in the case of primary and methyl substrates. The effect of the structure of the substrate on E2 reaction is in this order: $\overrightarrow{\text{ methyl < primary < secondary < tertiary}}$, which is the opposite of SN2. E2 reactions synthesize alkenes from alkyl halides, as in the above example, and form similar functional groups that will be described later.
Substitution nucleophilic unimolecular SN1
Ionic compounds like $\ce{Na^{+}Cl^{-}}$ dissociate into ions in polar protic solvents like $\ce{H2O}$ better than in polar aprotic solvents like. This is because unlike polar aprotic solvents solvating only cations, polar protic solvents solvent both cations and anions. Polar covalent bonded compounds like $\ce{\overset{\delta{+}}{H-}\overset{\delta{-}}{Br}}$ also dissociate in polar portic solvent by the same mechanism. Polar organic compounds like $\ce{\overset{\delta{+}}{H3C-}\overset{\delta{-}}{Br}}$ have less tendency to dissociate. However, if the substrate is tertiary like $\ce{(H3C)3\overset{\!\!\!\!\delta{+}}{C-}\overset{\delta{-}}{Br}}$, the effect of steric crowding on $\alpha\ce{C}$ and the dissociation ability of the polar protic solvent together are strong enough to dissociate polar organic compounds, as in 1st step of the mechanism shown in Figure $2$.
The carbonation produced in the 1st step, e.g., $\ce{(H3C)3C^{+}}$ in this case, is a strong electrophile and reacts with any nucleophile around, including solvent molecules, e.g., ($\ce{H2O}$ in this case. Since the 2nd step in non-selective, the solvent wins over any other nucleophile in the system because of its higher concentration. The neutral nucleophile becomes a cationic group in the product of the 2nd step, e.g., $\ce{R-OH2^{+}}$ in this case. The strong acid like $\ce{R-OH2^{+}}$ donates its proton to any base ($\ce{{_{\bullet}^{\bullet}}B}$) in the medium, including the solvent, e.g., $\ce{H2O}$ in this case.
The overall reaction is substitution nucleophilic. The overall reaction is unimolecular proportional to the substrate concentration in the 1st step. So, this mechanism is called substitution nucleophilic unimolecular SN1, where S stands for substitution, N or nucleophilic, and 1 for unimolecular.
• SN1 is unimolecular.
• The carbonation intermediate is trigonal planar and the nucleophile can attack it from either side to make a new covalent bond. Therefore, if SN1 reaction happens on a chiral carbon, around 50% of the product retains the configuration of the reactant, and the other 50% has the inverted configuration, i.e., the product is a racemic mixture.
• SN1 Reaction takes place on tertiary alkyl halides. Secondary and primary alkyl halides do not undergo SN1 reactions.
Elimination unimolecular (E1)
The carbocation formed in SN1 reaction has acidic protons on $\beta\ce{C}$ because the electrons left on the conjugate base are stabilized as a $\pi$-bond, as in the mechanism shown below.
This mechanism eliminates a leaving group and a proton $\beta$ to the leaving group. The 1st step, which is the rate-determining step, is unimolecular. This mechanism is called elimination unimolecular (E1), where E stands for elimination and 1 for unimolecular. E1 always competes with SN1 because they have a common first step.
What decides the reaction will happen by SN2, E2, SN1, or E1 mechanism?
All of the SN2, E2, SN1, or E1 involve two nucleophiles, an incoming nucleophile and a leaving group which is a nucleophile attached to the substrate. Incoming nucleophile substitutes the leaving group in substitution reactions and eliminates $\beta$$\ce{H}$ and the leaving group in elimination reactions. The question is what decides which mechanism the reaction will follow? This section addresses the answers to this question. The solvent and the nucleophile determine SN2 or SN1 condition.
Strong nucleophiles, usually anionic like $\ce{HO^{-}}$, $\ce{RO^{-}}$, and polar aprotic solvent like acetone ($\ce{(CH3)2C=O}$) or dimethylsulfoxide ($\ce{(CH3)2S=O}$) define SN2 and E2 condition.
The substrate dictates SN2, E2, or both will happen: SN2 happens on the primary or methyl substrate, E2 happens on the tertiary substrate, and both happen simultaneously on a secondary substrate. SN2 and E2 reactions do not take place in polar protic solvents.
Polar protic solvents like water ($\ce{H2O}$) or methanol ($\ce{CH3-OH}$) and neutral nucleophiles like $\ce{H2O}$, $\ce{ROH}$ define SN1 and E1 condition.
Anionic nucleophiles can not exist in protic sovlents due to acid-base neutralization reactions between them. Neutral species are poor nucleophiles but are OK in SN1 or E1 reactions because they are not involved in the rate-determining step. The following data on the dissociation rate of tertiary butyl chloride shows the effect of polar protic solvent on the rate-determining step of SN1 and E1 reactions.
Solvent Polarity (Dielectric constant) Protic or aprotic Relative rate
Water ($\ce{H2O}$) 78 Polar protic 40
Ethanol ($\ce{CH3CH2-OH}$) 24 Polar protic 1
Acetone ($\ce{(CH3)2C=O}$ 21 Polar aprotic 0.005
SN1 and E1 usually compete as both have the same rate-determining step. Tertiary substrate easily react by SN1 and E1 mechanism. Secondary substrate may or may not take place by SN1 and E1. For example, secondary alkyl halides do not react, but secondary alcohols and ethers react by SN1 and E1. Primary substrates usually do not react by SN1 and E1 mechanisms.
Effect of leaving group on SN2, E2, SN1, and E1 mechanisms
Recall that the stronger the acid, the weaker the conjugate base, and voice versa. For example, $\ce{HI}$ is a strong acid and $\ce{I^{-}}$ is a weak base, while $\ce{HF}$ is a weak acid and $\ce{F^{-}}$ is a strong base. In other words, strong bases do not tend to protons, and weak bases easily leave protons in acid-base reactions. The same applies to leaving groups in SN2, E2, SN1, and E1 reactions.
Strong bases are poor leaving groups, and weak bases are good leaving groups in SN2, E2, SN1, and E1 reactions.
The basicity and hence the leaving propensity of leaving groups follows these trends:
1. Basicity decreases from top to bottom in a group of periodic table. For example, the base strength of halides follows this order: $\overleftarrow{\ce{F^{-} > Cl^{-} > Br^{-} > I^{-}}}$. Leaving propensity follows the opposite trend i.e., $\ce{-I}$ is the best-leaving group, while $\ce{-F}$ is the worst leaving group. $\ce{-F}$ does not act as a leaving group.
2. Basicity decreases from right to left in a row of the periodic table. For example, the base strength of 2nd row elements follows this order: $\overleftarrow{\ce{F^{-} < OH^{-} < NH2^{-} < CH3^{-}}}$. Leaving propensity follows the opposite trend, i.e., $\ce{-OH}$ $\ce{-NH2}$. $\ce{-CH3}$ are the worst leaving group than $\ce{-F}$ and do not act as leaving group.
3. Basicity decreases upon protonation in the cases of amphoteric species. For example, $\ce{H2O}$ is a weaker base than $\ce{OH^{-}}$. That is why, $\ce{-OH}$ is not a leaving group, but $\ce{-\overset{+}{O}H2}$ is a good leaving group that leaves as a weak base $\ce{H2O}$.
Nucleophilic substitution and elimination reactions of alcohols, ethers, amines, and sulfur compounds
Reactions of alcohols
Substitution reactions
The $\ce{O}$ of an alcohol ($\ce{R-OH}$) is protonated by a strong acid to convert it from not leaving group (($\ce{-OH}$) to a good leaving group (($\ce{-\overset{+}{O}H2}$). The protonated alcohols under go SN1 or E1 reactions, except when the substrate is primary. that undergoes SN2 or E2 reactions. For example, primary, secondary, and tertiary alcohols undergo nucleophilic substitution with $\ce{HCl}$, $\ce{HBr}$, or $\ce{HI}$ as shown below.
$\ce{(CH3)3C-OH + HBr <=> (CH3)3C-{\overset{+}{O}}H2 + Br^{-} ->[\Delta] (CH3)3C-Br + H2O}\nonumber$ by SN1 mechanism
$\ce{(CH3)2CH-OH + HBr <=> (CH3)2CH-{\overset{+}{O}}H2 + Br^{-} ->[\Delta] (CH3)2CH-Br + H2O}\nonumber$ by SN1 mechanism
$\ce{CH3CH2-OH + HBr <=> CH3CH2-{\overset{+}{O}}H2 + Br^{-} ->[\Delta] CH3CH2-Br + H2O}\nonumber$ by SN2 mechanism
Elimination reactions may compete with substitution reactions described above, but the alkene products of elimination add $\ce{HCl}$, $\ce{HBr}$, or $\ce{HI}$ and end in the same final product as the substitution products. These reactions of alkens will be described in a later section.
Elimination reactions
Inorder to perform elimination reactions, the $\ce{O}$ of alcohols ($\ce{R-OH}$) is protonated by sulfuric acid ($\ce{H2SO4}$) in water solution. The conjugate base of sulfuric, i.e., $\ce{[HSO4]^{-}}$ is a weak nucleophile and does not cause the substitution reaction. Water is a nucleophile, but its creation does not change the intermediate. However, water acts as a base picking up $\beta$$\ce{H}$, causing elimination reactions, as shown below.
$\ce{(CH3)3C-OH <=>[H2SO4 + H2O] (CH3)3C-{\overset{+}{O}}H2 ->[\Delta +HSO4^{-} + H2O] (CH3)2C=CH2 + H2O}\nonumber$ by E1 mechanism
$\ce{(CH3)2CH-OH <=>[H2SO4 + H2O] (CH3)2CH-{\overset{+}{O}}H2 ->[\Delta +HSO4^{-} + H2O] CH3CH=CH2 + H2O}\nonumber$ by E1 mechanism
$\ce{CH3CH2-OH <=>[H2SO4 + H2O] CH3CH2-{\overset{+}{O}}H2 ->[\Delta +HSO4^{-} + H2O] CH2=CH2 + H2O}\nonumber$ by E2 mechanism
These reactions are called dehydration of alcohols that convert alcohols to alkenes.
Alcohols are common intermediates in biochemical reactions. However, unlike in a chemical laboratory, strong acids like $\ce{HBr}) or \(\ce(H2SO4) needed to activate alcohol groups do not survive under physiological conditions. Phosphoric acid and its anhydrides, i.e., pyrophosphoric acid and triphosphoric acids are weak acids and their conjugate bases, i.e., phosphate, pyrophosphate, and triphosphate, are weak bases and good leaving groups. Therefore, biological systems convert alcohols into phosphate or pyrophaste esters by reacting with adenosine triphosphate (ATP). It allows them to participate in the nucleophilic substitution reactions under physiological conditions as shown below. Reactions of ethers Like alcohols, the \(\ce{O}$ of ethers ($\ce{R-O-R'}$) is protonated by a strong acid to convert it from not leaving group (($\ce{-OR'}$) to a good leaving group (($\ce{-\overset{+}{O}HR}$). For example, ethers undergo nucleophilic substitution reactions with $\ce{HCl}$, $\ce{HBr}$, or $\ce{HI}$ producing alcohols and an alkyl halides, as shown below.
$\ce{(CH3)3C-O-C(CH3)3 + HBr <=> (CH3)3C-{\overset{+}{O}}HC(CH3)3 + Br^{-} ->[\Delta] (CH3)3C-Br + (CH3)3C-OH}\nonumber$ by SN1 mechanism
$\ce{CH3CH2-O-CH3 + HBr <=> CH3CH2-{\overset{+}{O}}HCH3 + Br^{-} ->[\Delta] CH3CH2-OH + CH3-Br}\nonumber$ by SN2 mechanism
Ethylene oxide: A sterilant based on SN2 reactions
Ethylene oxide is a three-membered cyclic ether. It is a colorless gas with a boiling point of 11 oC. The $\ce{C-O}$ bonds in ethylene oxide are unstable due to angle strain. Therefore, the ether groups in ethylene oxide act as excellent leaving groups because angle strain is released. Ethylene oxide reacts fast by SN2 mechanism with amino ($\ce{-NH2}$) and sulfhydryl ($\ce{-SH}$) groups that are commonly present in biochemicals, as shown below.
These reactions modify the biochemicals, leading to the death of microorganisms. Ethylene oxide is used as a fumigant in foods and textiles and to sterilize surgical instruments in hospitals.
Reactions of amines
Amine $\ce{-NH2}$ and also its protonated form $\ce{-NH3^{+}}$ are poor as leaving groups and do not act as leaving groups. However, amines are good nucleophiles and act as incoming nucleophiles in various reactions, e.g., in SN2 reactions with alkyl halides, as shown below.
$\ce{CH3CH2-Br + CH3-NH2 -> CH3CH2-{\overset{+}{N}}H2-CH2CH3 + Br^{-}}$ an SN2 reaction
Reactions of thiols
Thiolate ion is a good nucleophile for SN2 reactions as shown in the following example.
$\ce{CH3-\overset{\bullet\bullet}{\underset{\bullet\bullet}{S}}^{-} + CH3-I -> CH3-{\overset{\bullet\bullet}{S}}-CH3 + I^{-}}$ an SN2 reaction
Methylating agent in lab and in biochemical systems
Methyl ($\ce{\overset{\delta{+}}{C}H3{-}\overset{\delta{-}}{X}}$ is the best electrophilic site, and iodide ($\ce{-I}$) is the best-leaving group that makes methyliodide ($\ce{\overset{\delta{+}}{C}H3{-}\overset{\delta{-}}{I}}$) the best substrate for methylating any nucleophile in an SN2 reactions, as shown in the following examples.
$\ce{CH3-\overset{\bullet\bullet}{\underset{\bullet\bullet}{S}{_{\bullet}^{\bullet}}}^{-} + CH3-I -> CH3-\overset{\bullet\bullet}{\underset{\bullet\bullet}{S}}-CH3 + I^{-}}$ an SN2 reaction
Sulfur is a larger atom that can accommodate three alkyl groups. Therefore, dimethyl sulfide produced in the above reaction can be methylated one more time, producing a solfonium salt, as shown below.
$\ce{CH3-\overset{\bullet\bullet}{\underset{\bullet\bullet}{S}}-CH3 + CH3-I -> (CH3)3{\overset{\bullet\bullet}{S}}^{+} I^{-}}$ an SN2 reaction
Dimethyl sulfide ($\ce{(CH3)2{\overset{\bullet\bullet}{S}}^{+}{-}}$) is an excellent leaving group that makes trimethylsulfonium ion ($\ce{(CH3)3{\overset{\bullet\bullet}{S}}^{+}}$) an excellent methylating agent, as shown in an example below.
$\ce{HO^{-} + (CH3)3{\overset{\bullet\bullet}{S}}^{+} -> CH3-OH + CH3-\overset{\bullet\bullet}{\underset{\bullet\bullet}{S}}-CH3}$ an SN2 reaction
methyliodide ($\ce{H3C-I}$) is a common methylating agent in laboratory,- but it is not available in biological systems. S-adenosylmethionine (SAM), which is similar to trimethylsulfonium ion ($\ce{(CH3)3{\overset{\bullet\bullet}{S}}^{+}}$) is a common methylating agent in biological systems, as shown in an example below.
In the above example of a biochemical reaction, noradrenaline hormone is converted into a more potent adrenaline hormone by methylation reaction using SAM as a methylating agent. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.04%3A_Nucleophilic_substitution_and_elimination_reactions.txt |
Learning Objectives
• Understand nucleophilic acyl substitution mechanisms, and factors that affect them.
• Apply nucleophilic acyl substitution to reactions of acid halides, anhydrides, carboxylic acids, esters, and amides.
• Wright the reactions with reagents to convert carboxylic acids into acid halides and phosphate esters in living things.
• Apply nucleophilic acyl substitution reactions to synthesize condensation polymers.
What is a nucleophilic acyl substitution reaction?
A carbon double bonded with oxygen, i.e., $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$ is a carbonyl group. The $\ce{C}$ of the carbonyl group is bonded to two other groups. If one of the group bonded to the carbonyl $\ce{C}$ is an alkyl ($\ce{R{-}}$), or hydrogen ($\ce{H{-}}$), it become an acyl group, i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$. The acyl group has a polar double bond, i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-}$ where the carbonyl $\ce{C}$ is partial positive ($\delta{+}$), i.e., it is an electrophile. If the acyl group is attached to a nucleophile that can act as a leaving group ($\ce{-Lv}$), a stronger nucleophile ($\ce{Nu^{-}}$) can substitute it from the acyl group in reactions called nucleophilic acyl substitution reactions, as shown below in a generalized reaction.
$\ce{Nu^{-} + R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Lv -> R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Nu + Lv^{-}}\nonumber$
Mechanisms of nucleophilic acyl substitution reactions
In nucleophilic acyl substitution reactions, there are two reactants, the nucleophile and the acyl group containing substrate. Nucleophiles can exist in anionic base form $\ce{Nu^{-}}$ in a basic medium and neutral acid form $\ce{HNu}$ in a neutral or acidic medium. The anionic form is a better nucleophile than its neutral acid form. Similarly, the substrate can exist as neutral $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Lv}$ in neutral or basic medium or as protonated $\ce{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{\overset{\delta{+}}{C}}}\!\!\!\!\!\!-\overset{\delta{-}}{Lv}}$ form in an acidic medium. The protonated form has more $\delta{+}$ charge and is a better electrophile. The anionic nucleophile cannot coexist with protonated substrate, because of their acid-base reaction. The other three combinations, i.e., basic nucleophile + neutral substrate, neutral nucleophile + protonated substrate, and neutral nucleophile + neutral substrate, can coexist as described below.
Base-promoted mechanism
The nucleophile in its more reactive basic form $\ce{Nu^{-}}$ and neutral substrate $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-}$ can coexist in a basic medium. The nucleophile $\ce{Nu^{-}}$ attacks the electrophilic carbonyl $\ce{\overset{\delta{+}}{C}}$ and, simultaneously, the $\ce{\overset{\delta{+}}{C}}$ breaks the weakest $\pi$-bond, as shown in step#1 of the mechanism below. The electrophilic $\ce{C}$ changes hybridization from sp2 to sp3, i.e., a tetrahedral intermediate.
The nucleophilic $\ce{O^{-}}$ created in the first step attacks the $\ce{\overset{\delta{+}}{C}}$ to re-establish the $\pi$-bond, and either $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Nu}}$ or $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Lv}}$-bond breaks. This step is also called the collapse of the tetrahedral intermediate. Breakage of $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Nu}}$-bond reverses the first step and breakage of $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{Lv}}$-bond leads to the products of step#2. Later will likely happen when the $\ce{-Lv}$ is a good leaving group. The second step is usually not reversible because $\ce{Lv^{-}}$ is usually a poor nucleophile. The leaving group ultimately picks up a proton from any acid molecule ($\ce{HB^{+}}$) in the medium, as shown in step#3.
The rate of reaction is increased by converting the neutral or acid form of the nucleophile ($\ce{HNu}$) to its more reactive conjugate base form ($\ce{Nu^{-}}$) in the basic medium in this mechanism. Therefore, it is called base promoted nucleophilic acyl substitution mechanism.
Acid-catalyzed mechanism
In this mechanism, the nucleophile is in less reactive neutral $\ce{HNu}$ form, but the substrate is in a more reactive protonated $\ce{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{\overset{\delta{+}}{C}}}\!\!\!\!\!\!-\overset{\delta{-}}{Lv}}$ form, in an acidic medium. Acid in the medium protonate the carbonyl $\ce{O}$ in step#1, as shown in the mechanics below.
Nucleophile $\ce{HNu}$ attacks the protonated electrophilic carbonyl $\ce{\overset{\delta{+}}{C}}$ and, simultaneously, the $\pi$-bond breaks in step#2, leading to a tetrahedral intermediate-I. The neutral nucleophile becomes +ve charged after donating its lone pair of electrons. It carries an acid proton. Although the medium is acidic in this case, it is usually amphoteric, with some basic groups still present. The incoming nucleophile donates its acidic proton to any basic molecule ($\ce{:B}$ in the medium in step#3, leading to a tetrahedral intermediate-II. Then the acidic medium protonates either $\ce{-Nu}$ or $\ce{-Lv}$-groups. Protonation of $\ce{-Nu}$ reverses the reaction but protonation of $\ce{-Lv}$ makes it a better-leaving group in tetrahedral intermediate-III of step#4. The lone pair on $\ce{O}$ re-establishes the $\pi$-bond with the electrophilic $\ce{C}$, and at the same time, the leaving group leaves, leading to a protonated acyl product in the step#5. The protonated acyl donates its proton to any base $\ce{:B}$ in the medium and becomes the product in step#6.
Acid is a catalyst in this mechanism as it is consumed in step#1 but re-generated in step#6, and accelerates two slow steps in this mechanism, i.e., step#2 and step#5. That is why it is called the acid-catalyzed nucleophilic substitution mechanism. Specifically, the protonation of the substrates $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-}$ converts it to a more reactive $\ce{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{\overset{\delta{+}}{C}}}\!\!\!\!\!\!-\overset{\delta{-}}{Lv}}$ form that accelerates step#2, and protonation of $\ce{-Lv}$ in tetrahedra intermediate-II converts it to a good leaving $\ce{-Lv^{+}H}$ group in tetrahedral intermediate-III that accelerates step#5. The other steps in the mechanism are acid-base reactions which are inherently fast.
All the steps in this mechanism are reversible. It means there is an equilibrium between reactants and products. Three situations can arise, which are the following.
1. If the incoming nucleophile ($\ce{HNu}$) is a poor nucleophile compared to the leaving group ($\ce{HLv}$), the reactants dominate,
2. If the incoming nucleophile ($\ce{HNu}$) is a good nucleophile compared to the leaving group ($\ce{HLv}$), the products dominate, and
3. If the nucleophilicity of the incoming nucleophile ($\ce{HNu}$) is comparable to that of the leaving group ($\ce{HLv}$), about an equal-equal mixture of reactants and products exists.
In situation#3, the equilibrium can be manipulated to favor products or reactants based on Le Chatelier’s principle. Employing one of the reactants in excess or removing one of the products drives the reaction forward. Similarly, adding one of the products in excess or removing one of the reactants drives the reaction in the reverse direction.
Neutral nucleophile and neutral substrate reaction mechanism
In this mechanism, both the nucleophile $\ce{HNu}$ and the substrate $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-}$ are in their less reactive neutral forms in a neutral medium. Nucleophile $\ce{HNu}$ attacks the electrophilic carbonyl $\ce{\overset{\delta{+}}{C}}$ and, simultaneously, the $\pi$-bond breaks in step#1, leading to tetrahedral intermediate-I. The acidic proton on the incoming nucleophile in intermediate-I is removed by any base molecule ($\ce{:B}$ in the medium in step#2, leading to tetrahedral intermediate-II, as shown below.
The nucleophilic $\ce{O^{-}}$ in tetrahedral intermediate-II re-establishes the $\pi$-bond, and, simultaneously, either $\ce{Nu^{-}}$ or $\ce{Lv^{-}}$ leaves. Departure of $\ce{Nu^{-}}$ reverses step#2 and departure of $\ce{Lv^{-}}$ leads to the product in step#3. $\ce{Lv^{-}}$ is neutralized by any acid molecule present in the medium in step#4.
Since both the nucleophile and the substrate are in their less reactive neutral forms, it works only in situations where a good leaving group ($\ce{-Lv}$) is attached to the acyl substrate. Step#3 is usually irreversible because the $\ce{-Lv}$ groups chosen for these reactions are usually good leaving groups and poor nucleophiles.
Effect of leaving group on nucleophilic acyl substitution reactions
Good leaving groups ($\ce{Lv^{-}}$) are usually weak bases and poor nucleophiles at the same time. Weak bases do not fully share their electrons with acidic protons, and poor nucleophiles do not fully share their electrons with electrophilic $\ce{C's}$. Poor nucleophiles increase the rates of slow steps in the following ways:
1. their bond with $\ce{C}$ is strongly polar, making the acyl $\ce{\overset{\delta{+}}{C}}$ a more reactive electrophile, and
2. they leave easily increasing the rate of collapse of tetrahedral intermediates.
Leaving groups commonly encountered in nucleophilic acyl substitution reactions are halogens like chlorine ($\ce{-Cl}$) or bromine ($\ce{-Br}$) , carboxylate ($\ce{-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R}$), hydroxyl ($\ce{-OH}$), alkoxy ($\ce{-OR}$), and amine ($\ce{-NH2}$) groups. Their basicity and nucleophilicity increases in this order: $\overrightarrow{\ce{Cl^{-} < R-COO^{-} < HO^{-} ≈ RO^{-} < ^{-}NH2}}$. Therefore, their ability to leave increases in the opposite order $\overrightarrow{\ce{^{-}NH2 < HO^{-} ≈ RO^{-} < R-COO^{-} < Cl^{-}}}$, i.e., halogens like chlorine ($\ce{-Cl}$) are the best-leaving groups and amines ($\ce{-NH2}$) are the worst-leaving group.
These leaving groups are found in the following subclasses of acyl substrates: amides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2}$), carboxylic acids ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH}$), easters ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OR}$), acid anhydrides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R}$), and acid halides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Cl}$). These substrates are carboxylic acids and their derivatives. Their reactivity as acyl substrates in nucleophilic acyl substitution reactions increases in this order: $\overrightarrow{\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2} < \ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH} ≈ \ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OR} < \ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R} < \ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-Cl }}$. It means a group that is higher in the reactivity order can be easily converted to a group that is lower in the order, but the reverse does not happen normally. For example, the following reaction is feasible:
,
but the following reaction, i.e., the reverse of it, does not happen:
.
Examples of nucleophilic acyl substitution reactions
Reactions of acid halides
The acid halides, such as acid chlorides or bromides are the most reactive class of acyl substrates that can react with neutral nucleophiles and neutral substrates in a neutral medium. For example, acid chlorides can react with carboxylic acid, water, alcohol, or amines to form acid anhydrides, carboxylic acids, esters, and amines, respectively, as shown below.
Reactions of acid anhydrides
The acid anhydrides are reactive acyl substrates, second only to acid halides. Anhydride can react with water to produce two equivalents of carboxylic acid; with alcohols to produce one equivalent of ester and one equivalent of carboxylic acid; and with two amines to produce one equivalent of amide and one equivalent of an amine salt of carboxylic acid, as shown below.
Two moles of amine are needed in the last reaction because initially formed carboxylic acid reacts with the amine base producing an amine salt, as shown below.
Reactions of carboxylic acids
Carboxylic acids react with alcohols to produce ester but slowly. These reactions are reversible because the reactivity of the incoming nucleophiles is about the same as that of leaving nucleophoiles. Acid catalysis is applied to accelerate these reactions, as shown in the example below.
Since there is an equilibrium, excess alcohol is used, or water is removed to drive the reaction forward. These relations are called alcoholysis reactions because the reactant alcohol is also solvent. If a reverse reaction is desired, water is employed in excess as a solvent and the reaction is called hydrolysis of esters.
Carboxylic acids do not react with halides and another carboxylic acid because the incoming nucleophiles are poor relative to the leaving group. Base can not be added to accelerate the reactions of carboxylic acids because the base neutralizes the substrate, as shown below.
The carboxylate anion ( $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}$ is the least reactive acyl substrate that is below amides in the reactivity order. Carboxylic acids do not react with amines for the same reason, i.e., amine base neutralizes the carboxylic acid producing ammonium salts, as shown below.
Protonated amines in the ammonium salts are not nucleophiles as they do not have lone pair of electrons on the $\ce{N}$.
Reactions of esters
Esters are comparable in reactivity with carboxylic acids but without the acidic protons. Therefore, both acid-catalysis and base-promoted reactions can be carried out with esters. For example, water hydrolyzes esters under acidic conditions, as shown below.
It is a reversible reaction that can be driven forward by employing excess water or by removing the alcohol product. These reactions are called acid-catalyzed hydrolysis of esters.
The alkoxy group of esters can be substituted by an alkoxy group of another alcohol, as shown below.
It is a reversible reaction that can be driven forward by employing excess reactant alcohol or by removing the product alcohol. These reactions are called acid-catalyzed transesterification reactions.
Base-promoted reactions of esters
Transesterification of esters can be performed by employing a conjugated base of the alcohos, as shown in the following example.
Similarly, the example below shows that esters can be hydrolyzed using alkali, i.e., hydroxide ions.
The base-promoted hydrolysis of esters is called saponification which is used to hydrolyze fats, as shown in Figure $1$. Fats are tri-esters of glycerol and fatty acids. Saponification of fasts produces sodium salts of fatty acids and glycerol. Sodium salts of fatty acids are soaps.
Reactions of amides
Amides are the least reactive among the acyl substrates. Amides do not react with halides, water, or alcohols under neutral conditions because the incoming nucleophiles are poor than the leaving group. However, water hydrolyzes amides under acid catalysis and heating, as shown below.
Acid converts poor leaving $\ce{-NRR'}$ group to a good leaving $\ce{-\overset{+}{N}HRR'}$ group, and also removes the product amine $\ce{NRR'}$ by acid-base reaction, as illustrated below.
The ammonium ion ($\ce{RR'N^{+}HCl^{-}}$ does not have a lone pair on $\ce{N}$ and is not a nucleophile, that makes the reaction irreversible.
As shown below, alcohols react with amides under acid catalysis and heating.
Activating carboxylic acids in the laboratory and in biochemical systems
Carboxylic acids are common raw materials but must be converted into reactive derivatives, like acid halide, before conversion to other carboxylic acid derivates. Reagents like thionyl chloride ($\ce{SOCl2}$), phosphorous trichloride ($\ce{PCl3}$), or phosphorous pentachloride ($\ce{PCl5}$) are used to convert carboxylic acids into acid chlorides for this purpose, as shown below.
The above reagents are not available in biochemical systems. Adenosine triphosphate, shown below is a common reagent available in biochemical systems.
As illustrated below, adenosine triphosphate is used in biochemical systems to convert acylates into acyl phosphates or acyl adenylates.
The phosphate and adenylate are good leaving groups. For example, coenzyme-A ($\ce{CoASH}$) containing nucleophilic thiol ($\ce{-SH}$) group is common in biochemical systems.
$\ce{CoASH}$ displaces adenylate group from acyls through nucleophilic acyl substitution mechanism, as illustrated below.
Thioesters, like $\ce{-SCoA}$, are also good leaving groups. For example, the neurotransmitter acetylcholine is synthesized by reacting acetyl$\ce{-SCoA}$ with choline through a nucleophilic acyl substitution mechanism, as shown below.
Condensation polymerization reactions
When two molecules combine to form a single molecule, usually with a loss of a small molecule like water or ammonia, the reaction is called a condensation reaction. For example, nucleophilic acyl substitution reactions between carboxylic acids and alcohols are condensation reactions, as shown below.
Polymers called polyesters are produced when molecules with two carboxylic acids and others with two alcohols condense to form esters. For example, benzene-1,4-dicarboxylic acid, commonly known as terephthalic acid, and ethane-1,2-diol, known as ethylene glycol, condense to produce a polyester called Polyethylene terephthalate (PET), as shown in Figure $2$. Polyethylene terephthalate is used to make drink bottles and polyester fabrics.
Nucleophilic acyl substitution reactions between carboxylic acids and amines are another example of condensation reactions that can produce polyamides. For example, condensation of hexanedioic acid and hexane-1,6-diamine, known as hexamethylene diamine, produces a polyamide called nylon 66, as illustrated in Figure $3$.
The first digit of 66 in the name nylon 66, tells number of carbons in one monomer (hexanedioic acid) and the second digit for the other monomer (hexane-1,6-diamine). Nylone 66 is commonly used in clothing, fishing lines, and guitar strings. Kevlar used to make bulletproof jackets is another example of polyamide, shown in Figure $4$
Condensation reaction between diisocyanates ($\ce{O=C=N-R-N=C=O}$) and alcohols ($\ce{HO-R-OH}$) produces polyurethane where ureththane groups ( $\ce{-{\overset{\Large{H}}{N}}-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O\bond{-}}$ ) link the monomer units in the polymer molecule, as illustrated in Figure $5$.
Polyurethane makes foam, e.g., mattresses foam.
Nucleic acids, i.e., DNA and RNA are biopolymers in which phosphate diester linkages hold the monomer units together, as illustrated in the figure on the right (copyright: Public domain). Nucleic acids are described in detail in a separate chapter. Proteins are polyamide bipolymers. For example, cinnamycin (lanthiopeptin), shown in the figure on the left (copyright: Public domain), is an antibacterial peptide produced by Streptomyces cinnamoneus containing 19 amino acids. Proteins are also described in detail in a separate chapter. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.05%3A_Nucleophilic_acyl_substitution_reactions.txt |
Learning Objectives
• Understand and be able to draw nucleophilic addition mechanisms in basic and acidic mediums.
• Apply nucleophilic addition reactions to simple examples, like the addition of cyanide, water, and alcohols to aldehydes and ketones.
The carbonyl group of an aldehyde or a ketone is a polar double bond just like the acyl group of a carboxylic acid derivative, i.e., $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-R'}$ where the carbonyl $\ce{\overset{\delta{+}}{C}}$ is an electrophile. However, this electrophile does not have a leaving group on it. It can add a nucleophile but usually needs ether base-promoted or acid catalysis conditions. These two mechanisms are described next.
Base-promoted nucleophilic addition reactions
A nucleophile in its more reactive basic form $\ce{Nu^{-}}$ in basic medium attacks the electrophilic carbonyl $\ce{\overset{\delta{+}}{C}}$. Simultaneously, the $\pi$-bond bond break, as shown in step#1 of the mechanism below. The $\ce{O^{-}}$ either re-establish $\pi$-bond expelling the nucleophile, or it is neutralized by an acid-base reaction with any acid group present in the medium in step#2 of the mechanism shown below.
Remember, the medium is basic but amphoteric and has some acidic groups. If there is no sufficiently acidic group in the medium, an acid ($\ce{HB^{+}}$) is added after mixing the reactants for step#2 to happen. The carbonyl $\ce{C}$ changes hybridization from sp2 to sp3, and $\ce{C=O}$ group converts to an alcohol ($\ce{-OH}$) in this reaction. It is called nucleophilic addition reaction because the overall result is the addition of a nucleophile ($\ce{HNu}$) to a carbonyl ($\ce{C=O}$) group.
Cyanohydrin formation
The addition of cyanide ($\ce{^{-}CN}$) to aldehydes or ketones is an example of base-promoted nucleophilic addition. It converts the $\ce{C=O}$ into cyanohydrin, i.e., a $\ce{C}$ with a cyanide ($\ce{-CN}$) and hydroxyl ($\ce{-OH}$ groups on it. Two examples of cyanohydrin formation are shown below.
Acid-catalyzed nucleophilic addition reactions
The acid in the medium converts less electrophilic $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}|\!\!|\enspace}{\overset{\delta{+}}{C}}}\!\!-R'}$ to a more electrophilic $\ce{R-\!\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{\overset{\Large{+}}{O}H}}|\!\!\!\!\!\!\!|\enspace\enspace}{\overset{\delta{+}}{C}}}\!\!\!\!\!\!-{R'}}$ form by protonating its $\ce{O}$ in step#1. Neutral nucleophile can exist only in neutral ($\ce{HNu}$) form in the acidic medium. The neutral nucleophile ($\ce{HNu}$) attacks the carbonyl $\ce{C}$ making a $\sigma$-bond and, simultaneously, the $\ce\pi$-bond breaks in step#2. Although the medium is acidic, it is amphoteric and has some basic groups that receive acidic proton on the incoming $\ce{-{\overset{+}Nu}H}$ in step#3, as illustrated below.
The acidic medium converts bad leaving $\ce{-OH}$-group to a good leaving $\ce{-\overset{+}{O}H2}$-group in step#4. The $\ce{-\overset{+}{O}H2}$-group leaves as $\ce{H2O}$, pushed by lone pair on $\ce{-Nu}$ in step#5. Second $\ce{HNu}$ attacks the electrophilic $\ce{C}$ in step#6, and the acidic proton on the second incoming $\ce{-\overset{+}{Nu}H}$ is removed by any base ($\ce{:\!\!B}$) in the medium in step#7. All the steps of the mechanism are reversible. Two situations can arise:
1. if $\ce{HNu}$ is in limited supply, equilibrium exists between the substrate and tetrahedral intermediate-II of step#3, and
2. if $\ce{HNu}$ is in excess, equilibrium exists between the substrate and the product of reation#7.
The reaction can be made irreversible by removing $\ce{H2O}$ from the products. The reaction can be reversed by mixing the acid catalyst and excess water with the product.
Hydration of aldehydes and ketones
An example of nucleophilic addition is the hydration of aldehydes and ketones when diluted with water, as shown below.
The hydrates are usually unstable and exist in a very small proportion relative to the initial aldehyde or ketone, e.g., the hydrate of acetone, shown below.
A few exceptions exist where one or both of the groups attached with the carbonyl $\ce{C}$ are $\ce{H's}$, hydrate form is dominant, e.g., the hydrated acetaldehyde and formaldehyde, as shown below.
Hemiacetals and acetals
The addition of alcohol ($\ce{R-OH}$) to an aldehyde or a ketone is another example of a nucleophilic addition reaction. If $\ce{R-OH}$ is in limited supply, the reaction stops after the addition of one $\ce{R-OH}$ molecule resulting in a product called hemiacetal, as in the following example.
Hemiacetal has an $\ce{-O-R}$-group and an $\ce{-OH}$ bonded to the same $\ce{C}$. Hemiacetals are usually unstable and exist in small proportion at equilibrium. Exceptions are five-membered and six-membered hemiacetals formed by the reaction of ($\ce{-OH}$)-group and the $\ce{C=O}$-group on the same molecule. For example, more than 99% of glucose molecules exist in six-membered hemiacetal forms in equilibrium with open-chain aldehyde form in a water solution, as shown in Figure $1$.
When alcohol is in excess, a second $\ce{R-OH}$ adds to the hemiacetal resulting in an acetal. An acetal has two $\ce{-O-R}$-groups bonded to the same $\ce{C}$, as shown in the example below.
The acetal groups in carbohydrates are called glycoside linkages. Glycoside linkages connect monosaccharide units together in polysaccharides like starch and cellulose and in disaccharides, e.g., lactose, maltose, and sucrose, as shown in Figure $2$ | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.06%3A_Nucleophilic_Addition_Reactions.txt |
Learning Objectives
• Draw an electrophilic addition mechanism with curly arrows showing the movement of electrons.
• Apply the electrophilic addition to reactions of alkenes and alkynes with halogen acids and water.
• Define the concepts: regioselectivity, stereoselective and stereospecific reactions, and tautomerism with examples of electrophilic addition relations of alkenes and alkynes.
Electrophilic addition mechanism
The $\pi$-bond of alkenes distributes electrons above and below the $\sigma$-bond that creates a local electron-rich ($\delta{-}$) region, as shown by red color region in the following electrostatic potential map of ethene.
Therefore, the alkenes are nucleophiles ($\delta{-}$) attracted to electrophiles ($\delta{+}$) and capable of donating their $\pi$-electrons to make a bond with them, as shown in the following reaction mechanism.
In first step of this mechanism, $\pi$-bond acts as an incoming nucleophile making a $\sigma$-bond with the electrophilic $\delta{+}$ end of the polar reagent $\ce{\overset{\delta{+}}{A}{-}\overset{\delta{-}}{B}}$ while the ($\ce{\overset{\delta{-}}{B}}$) acts as a leaving group, leaving as a nucleophile $\ce{B^{-}}$. One of the $\pi$-bonded carbon that receives the electrophile $\ce{\overset{\delta{+}}{A}}$ becomes sp3 hybridized, and the other carbon stays sp2 hybridized with an empty p-orbital as a carbocation. In the second step, the nucleophile ($\ce{B^{-}}$) attacks the electrophilic carbocation and makes a new bond. As a result, the reagent $\ce{A-B}$ add to the $\ce{C=C}$-bond. Since the process begins with the addition of an electrophile $\ce{\overset{\delta{+}}{A}}$, it is called an electrophilic addition reaction.
Examples of electrophilic addition reactions of alkenes
Halogen acids ($\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{Cl}}$, $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{Br}}$, and $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{l}}$) are among many reagents that add to alkenes by electrophilic addition mechanism, as shown in the following example.
Water ($\ce{H2O}$) has more polar $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{O}}$ bonds than the halogen acids, but $\ce{-OH}$ is a bad leaving group and usually do not act as a leaving group. The bad leaving $\ce{-OH}$ is first converted to a good leaving $\ce{-\overset{+}{O}H2}$ by adding stronger acid like sulfuric acid ($\ce{H2SO4}$) in the water. Then, the hydronium ion ($\ce{H-\overset{+}{O}H2}$) enters the first step of the electrophilic addition reaction, as shown below.
$\ce{H2O}$ acts as an incoming nucleophile in step#2, another water molecule removes its acidic proton in step#3. This reaction is called hydration of alkene because the result is the addition of a water molecule $\ce{\overset{\delta{+}}{H}{-}\overset{\delta{-}}{OH}}$ to a $\ce{C=C}$-bond of an alkene.
Sulfuric acid is used as an acid catalyst because its conjugate base $\ce{HSO4^{-}}$ is a poor nucleophile. $\ce{H2O}$ is also a poor nucleophile, but being in higher concentration wins the completion in step#2. Even if $\ce{HSO4^{-}}$ add to the carbocation in the second step, it is substituted by $\ce{H2O}$ under the reaction conditions.
Regioselectivity of electrophilic addition reactions
In the cases of unsymmetrical alkenes, two constitutional isomers are possible, but one is produced exclusively or as a major product. The reason is that the two potential carbonation intermediates have unequal stability. The stability of carbocations increases in this order: $\overrightarrow{\text{primary (\(\ce{R{\overset{+}{C}}H2}$) < secondary ($\ce{R2{\overset{+}{C}}H}$) < tertiary ($\ce{R3{\overset{+}{C}}}$) }}\). The following examples show that the more stable carbocation is formed exclusively or predominantly when there is a choice.
Chemical reactions in which two constitutional isomers are possible, but one is formed exclusively or predominantly are called regioselective reactions. Electrophilic additions are regioselective reactions.
Stereoselectivity of electrophilic addition reactions
When reactants and catalysts are achiral, but the product is chiral, it is produced equally in both (R) and (S) configurations. For example, the carbocation intermediate in electrophilic addition reactions is trigonal planer. The nucleophile can attack it from either side with about equal probability resulting in a racemic mixture, as illustrated in the following example.
However, suppose one of the reactants or the catalyst is chiral and a pure enantiomer. In that case, one of the enantiomers (R or S) is usually formed exclusively or as a major product. Enzymes in biochemical systems are chiral catalysts that usually produce one enantiomer exclusively.
Chemical reactions that produce one stereoisomer exclusively or preferentially relative to the others are called stereoselective reactions.
Enzyme-catalyzed reactions are stereoselective reactions. For example, hydration fumarate, catalyzed by an enzyme fumarase, produces (S)-malate, as shown below.
The R-enantiomer is not produced in the above reaction.
Enzymes usually also react with one stereoisomer in the reactant exclusively. For example, fumarase catalyzes the hydration of fumarate but does not react with its enantiomer maleate, as illustrated below.
Chemical reactions that preferentially react with one stereoisomer among reactant and preferentially produce one stereoisomer in the products are called stereospecific reactions.
Enzyme-catalyzed reactions are often stereospecific reactions. For example, the fumarase enzyme selectively reacts with fumarate (and not with its stereoisomer maleate) and selectively produces (S)-malate (and not its stereoisomer (R)-malate).
Electrophilic addition reactions of alkynes
Alkynes have a triple $\ce{C≡C}$-bond, i.e., one $\sigma$-bond and two $\pi$-bonds. The electrophilic addition reactions happen on $\pi$-bonds of alkynes. If the reagent is in a one-to-one mole ratio, it adds to one of the $\pi$-bonds. If the reagent is in excess, a second addition reaction happens on the product of the first addition, as shown below.
In step#2, the nucleophile ($\ce{Br} in this case) adds to the carbon carrying the nucleophile of step#1. This is because the carbonation intermediate of step#2 is stabilized by resonance with the lone pair of electrons on the nucleophile added in the first step. Hydration of alkyne follows the same step#1 as for alkenes, but the intermediate enol containing \(\ce{C=C}$ and $\ce{-OH}$ together exists in equilibrium with its structural isomer ketone, as shown in the following example.
Tautomers are constitutional isomers that are readily interconvertible. Tauomerization is the chemical reaction that interconverts the tautomers.
Enol and its isomer ketone, e.g., but-2-ene-2-ol and but-2-one in the above example, are called tautomers, and the equilibrium between the two in step#2 of the above reaction is tautomerization.
4.08: Electrophilic aromatic substitution reactions
Learning Objectives
• Understand the difference in the electrophilicity of $\pi$-bond of benzene and alkenes.
• Draw the electrophilic aromatic substitution mechanism with curly arrows showing the flow of electrons.
• Apply the electrophilic aromatic substitution to some reactions of benzene, including halogenation, nitration, sulfonation, alkylation, and acylation reactions.
Which electrophiles can react with an aromatic substrate?
The $\pi$-bonds in a benzene ring of aromatic compounds are weaker nucleophiles than the $\pi$-bonds in alkenes. This is because breaking a $\pi$-bond of alkene costs about 260 kJ/mole energy, but breaking a $\pi$-bond in an aromatic substrate costs an additional 208 kJ/mol because the aromatic stabilization is lost. Unlike alkenes, the aromatic substrates do not react with partial positive ($\ce{\overset{\delta{+}}{A}{-}\overset{\delta{-}}{B}}$) electrophiles. The aromatic substrates react with electrophiles in their most reactive cation $\ce{E^{+}}$ form. The cation electrophiles are generated in situ by acid-base or Lewis acid-Lewis base reactions. For example, halogens ($\ce{X-X}$ react as Lewis bases with Lewis acids like $\ce{AlX3}$ or $\ce{FeX3}$, where $\ce{X}$ is a halogen atom ($\ce{Cl}$ or $\ce{Br}$). The Lewis acid receives a lone pair from one halogen atom causing a heterolytic breaking of $\ce{X-X}$. The other halogen leaves as $\ce{X^{+}}$, as shown below.
Similar reaction happens when an alkylhalide ($\ce{R-X}$ or an acyl halide ( $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-X}$) reacts with Lewis acid, as shown below.
An $\ce{-OH}$ bonded with a potential electrophile $\ce{E^{+}}$ can be converted into a better leaving $\ce{-\overset{+}{O}H2}$ group by adding a stronger acid to the substrate. The $\ce{-{\overset{+}{O}}H2}$ leaves as neutral neucleophile $\ce{H2O}$, leaving behind the $\ce{E^{+}}$. For example, protonation of nitric acid with sulfuric acid generated nitronium ion ($\ce{\overset{+}{N}O2}$), as shown below.
Like the auto-ionization of water, the autoionization of sulfuric acid followed by elimination of $\ce{H2O}$ generates protonated sulfur trioide ($\ce{\overset{+}{S}O3H}$), as shown below.
Electrophilic aromatic substitution mechanism
The nucleophilic $\pi$-bond of an aromatic compound attacks the cation electrophile ($\ce{E^{+}}$), as shown in step#1 in the mechanism illustrated below. Any base group in the medium removes the acidic proton that re-establishes the $\pi$-bond in Step#2.
Removal of the proton by a base is preferred over electrophile attacking the carbonation intermediate in step#2, because aromatic stabilization decreases the energy barrier for the former. It is called electrophilic aromatic substitution reaction because an electrophile $\ce{E^{+}}$ substitutes another electrophile $\ce{H^{+}}$ from an aromatic substrate.
Examples of electrophilic aromatic substitution reactions
Some fo the important electrophilic aromatic substitution reactions of benzene are listed below.
• Halogenation of benzene substitutes a $\ce{-H}$ with a halogen ($\ce{-Cl}$ or $\ce{-Br}$), as shown below.
• Nitration of benzene substitutes a $\ce{-H}$ with nitro group ($\ce{-NO2}$) by the following reaction.
• Sulfonation of benzene substitutes a $\ce{-H}$ with sulfonic acid group ($\ce{-SO3H}$) by the following reaction.
• Alkylation of benzene substitutes a $\ce{-H}$ with alkyl acid group ($\ce{-R}$), e.g.,:
• Acylation of benzene substitutes a $\ce{-H}$ with acyl group ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R}$), e.g.,: | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.07%3A_Electrophilic_addition_reactions.txt |
Learning Objectives
• Define the redox process and calculate the oxidation number of an atom in an organic compound.
• Learn catalytic reduction of alkenes and carbonyl compounds.
• Learn reduction of carbonyl compounds by $\ce{NaBH4}$ or $\ce{LiAlH4}$, and by $\ce{NADH}$ in living things.
• Learn oxidation of alcohols and aldehydes by oxidizing agents in laboratories and by $\ce{NAD^{+}}$ in living things.
• Learn oxidation of thiol ($\ce{-SH}$) to disulfide ($\ce{-S-S{-}}$) and its revers, in proteins.
What are oxidation and reduction?
Oxidation is i) the loss of electrons, ii) the gain of $\ce{O}$, or iii) the loss of $\ce{H}$. Reduction is the opposite, i.e., i) the gain of electrons, ii) the loss of $\ce{O}$, or iii) the gain of $\ce{H}$. The reduction and oxidation reactions are coupled, called redox reactions. Electrons' loss or gain is easily recognized in inorganic chemical reactions. For example, consider the following reaction:
$\ce{ 4Fe + 3O2 -> 2Fe2O3}$,
$\ce{Fe}$ lost three three electrons to become $\ce{Fe^{3+}}$ and $\ce{O}$ gained two electrons to become $\ce{O^{2-}}$ in this reaction.
Oxidation number
Organic chemicals usually have covalent bonds where the loss or the gain of electrons from an atom is accounted for by oxidation number. An increase in the oxidation number is oxidation, and a decrease in the oxidation number is reduction. The following is a simple procedure for assigning an oxidation number to an atom in a neutral organic compound. Start with oxidation number zero for a free atom,
1. a bond with a more electronegative atom increases the oxidation number by one, e.g., a $\ce{C-O}$ bond increases the oxidation number of $\ce{C}$ by +1;
2. a bond with a less electronegative atom decreases the oxidation number by one, e.g., a $\ce{C-H}$ bond decreases the oxidation number of $\ce{C}$ by -1; and
3. a bond with the same atom does not change the oxidation number, e.g., a $\ce{C-C}$-bond does not change the oxidation number of $\ce{C}$.
For example, in $\ce{CH4}$ four bonds of $\ce{C}$with less electronegative $\ce{H's}$ decrease its oxidation number to -4. $\ce{H's}$. In $\ce{CH3-OH}$ three bonds of $\ce{C}$with less electronegative $\ce{H's}$ decrease its oxidation number to -3 and one bond with more electronegative $\ce{O}$ increases to +1, with the overall oxidation number of the $\ce{C}$ = -3 +1 = -2. So, conversion of methane ($\ce{CH4}$) to methanol ( $\ce{CH3-OH}$) is oxidation.
The following chart illustrates the oxidation states of $\ce{C}$ in organic compounds and their changes with organic transformations.
Examples of organic reduction reactions
Catalytic hydrogenation of alkenes
When an alkene and hydrogen are mixed, no reaction happens, as shown in the following example.
This is because alkene is an electrophile but $\ce{H-H}$ is a nonpolar bond that is not a nucleophile. Finely divided nickel ($\ce{Ni}$), platinum or palladium supported on carbon, i.e. $\ce{Pd/C}$ or $\ce{Pt/C}$ catalyze the reaction, as shown below.
Mechanism of catalytic hydrogen of alkenes
In step#1, the catalyst absorbs $\ce{H-H}$ and $\ce{RR'C=CR"R'"}$ on the surface, as illustrated in Figure $1$. In step#2, $\ce{H-H}$-bond breaks. $\ce{H's}$ and the $\pi$-bond of alkene become bonded with the catalyst surface. The surface-bonded species can migrate along the surface. In step#3, the $\pi$-bond attacks and makes a bond with $\ce{H}$ at the expense of breakage of $\ce{H{-}}$catalyst bond. The second $\ce{C}$ of the $\pi$-bond stays bonded with the catalyst. In step#4, $\ce{C{-}}$catalyst surface bonding electrons make a bond with the second $\ce{H}$ at the expense of breakage of $\ce{H{-}}$catalyst and $\ce{C{-}}$catalyst bonds. The alkane product of step#4 is no more attached to the catalyst and departs. The refreshed catalyst surface repeats the process.
Alkenes are planer around $\ce{C=C}$ bond. Both $\ce{H's}$ add from the same face of alkene facing the catalyst surface.
Syn addition is the addition of two substituents from the same side (same face) of the $\pi$-bond. Addition of $\ce{H-H}$ to $\ce{C=C}$ of an alkene is an example of syn addition.
Catalytic hydrogenation of vegetable oils
Vegetable oils are fats of tri-ester of glycerol with long chain fatty acids containing one or more $\ce{C=C}$ bonds. The $\ce{C=C}$ bonds create kinks in the long, preventing the chains' packing and resulting in lower melting points. Partial hydrogen of $\ce{C=C}$ bonds in vegetable oil is carried out to convert it to semisolid margarine, as illustrated in Figure $2$. Hydrogenation makes them straight-chain alkanes that pack nicely and increases their melting points.
Reduction of aldehydes and ketones
Catalytic hydrogenation of aldehydes and ketones
Hydrogen ($\ce{H2}$) reduces aldehydes and ketones to alcohols in the presence of a catalyst like finely divided $\ce{Ni}$, $\ce{Pd/C}$, or $\ce{Pt/C}$. as shown in the following examples.
If $\ce{C=C}$-bond and a $\ce{C=O}$-bond are present in the same molecule, catalytic hydrogen reduces both, as shown below.
Reduction of aldehydes and ketones by nucleophilic addition reaction
The $\ce{C=O}$ group can be selectively reduced to $\ce{C-OH}$ group by nucleophilic addition of hydride ($\ce{H^{-}}$) ion to the electrophilic $\ce{C}$ of $\ce{\overset{\delta{+}}{C}=\overset{\delta{-}}{O}}$ group, by simplified mechanics shown below.
The $\ce{H^{-}}$ is not employed in free form, but it is donated by a reducing agent like sodium borohydride $\ce{NaBH4}$ or lithium aluminum hydride $\ce{LiAlH4}$. Note that the $\ce{H^{-}}$ do not survive in an acid medium, so the $\ce{H^{-}}$ reacts in the first step, and then acid is added to donates a proton to $\ce{O}$ in the second step.
The nucleophilicity of $\ce{H^{-}}$ ion depends on the donor reagent. For aldehydes and ketones, $\ce{NaBH4}$ and $\ce{LiAlH4}$ work, but usually $\ce{NaBH4}$ is employed as it is tolerant to water. $\ce{NaBH4}$ reduces aldehydes, ketones, and acid halides but does not reduce other carboxylic acid derivatives and $\ce{C=C}$, as shown in the example below.
Reduction of aldehydes and ketones in biological systems
$\ce{NaBH4}$ and $\ce{LiAlH4}$ are used as $\ce{H^{-}}$ donors in laboratories. Nicotinamide adenine dinucleotide ($\ce{NADH}$) coenzyme in its reduced form $\ce{NADH}$ is used as a hydride donor in biological systems. $\ce{NADH}$ is oxidised to $\ce{NAD^{+}}$ after donating $\ce{H^{-}}$, as illustrated below.
Enzyme-catalyzed conversion of acetaldehyde to ethanol during fermentation is an example of reduction using $\ce{NADH}$.
Conversion of pyruvate to lactate during glycolysis under anaerobic (absence of oxygen) conditions is another example of reduction using $\ce{NADH}$, as shown below.
Cumulation of lactate after thought exercises causes muscle fatigue.
Reduction of carboxylic acid their derivatives
The $\ce{NaBH4}$ is less reactive and reduces only acid halids among carboxylic acids and their derivatives. $\ce{LiAlH4}$ is more reactive and reduces carboxylic acids and their derivatives. The mechanism is nucleophilic acyl substitution by $\ce{H^{-}}$ followed by nucleophilic addition, as shown below in the simplified form.
The reduction of propanoyl chloride by $\ce{NaBH4}$ is shown below.
$\ce{NaBH4}$ does not reduce the rest of the carboxylic acids and their derivatives. $\ce{LiAlH4}$ reduces carboxylic acids and their derivatives, as illustrated below with examples of reduction of acid halide, carboxylic acid, ester, and amide.
Reduction of disulfide $\ce(S-S}$) bond in biological systems
The $\ce{S-S}$ bond commonly occurs in the protein tertiary structure. It is responsible for straight or curly hair shapes. The $\ce{S-S}$ bond can be reduced to $\ce{-SH}$ group by a variety of reducing agents, including $\ce{NaBH4}$ and $\ce{Na}$ metal. A thiol-disulfide exchange reaction often achieves this conversion in biological systems, as in the following example.
$\ce{R-S-S-R + 2HOCH2CH2SH <=> 2R-SH + HOCH2CH2S-SCH2CH2OH}\nonumber$
Oxidation of alcohols and aldehydes
Primary and secondary alcohols are oxidized by several oxidizing agents, including Jones reagent (a mixture of chromium trioxide ($\ce{CrO3}$) and sulfuric acid ($\ce{H2SO4}$)), chromic acid (($\ce{H2CrO4}$), or potassium dichromate ($\ce{KCr2O7}$) in sulfuric acid ($\ce{H2SO4}$) acid. The mechanism with Jones reagent is shown below.
Primary alcohols are oxidized to aldehydes. Aldehydes convert to hydrates by adding water, and the hydrates are oxidized to carboxylic acids, as illustrated below.
Secondary alcohols are oxidized to a ketone, as shown in the following example. Ketone is not oxidized further as they do not have hydrogen on the carbonyl carbon needed to carry out step#4 of the mechanism.
Tertiary alcohols are not oxidized for the same reason, i.e., there is no hydrogen on the carbon to which alcohol is attached to carry out step#4.
Tests for aldehydes and reducing sugar based on easy oxidation of aldehydes
The fact that an aldehyde is oxidized further easily while ketone is not was used as a test to distinguish between aldehydes and ketones. In this test Tollens' reagent ($\ce{[Ag(NH3)2]OH}$) is mixed with the test sample. Ketones do not react but aldehydeis oxidized by the following reaction:
$\ce{2[Ag(NH3)2]OH + R-CHO -> 2Ag(s) + 4NH3 + R-COOH + 2H2O}\nonumber$
$\ce{Ag}$ produced in the above reaction deposits and forms silver mirror on the wall of the test tube as shown in Figure $3$
Benedict's reagent is a mixture of sodium carbonate ($\ce{Na2CO3}$), sodium citrate ($\ce{Na2C6H5O7}$), and copper(II) sulfate pentahydrate ($\ce{CuSO4.5H2O}$). Benedict's reagent produces red-precipitate ($\ce{Cu2O}$) as a positive test of aldehyde by the following reaction.
$\ce{2Cu(C6H5O7)^{-} + 5OH^{-} + R-CHO -> Cu2O(s, red) + 2C6H5O7^{3-} + 3H2O}\nonumber$
This test distinguishes aldehydes from ketones and is also positive for reducing sugars (carbohydrates) that contain an aldehyde group, as shown in Figure $4$.
Alcohol exhaled in breath is oxidized by dichromate $\ce{Cr2O7^{2-}}$ ions along with producing green color chromium(III) ions (\ce{Cr^{3+}}\) by the following reaction.
$\ce{CH3CH2OH + Cr2O7^{2-}(reddish orange) ->[H2SO4 + H2O] CH3COOH + Cr^{3+}(green)}\nonumber$
This reaction is used to test alcohol in drunk-and-drive cases by law enforcement.
Oxidation of alcohols in living systems
Alcohols are oxidized by nicotinamide adenine dinucleotide ($\ce{NAD^+}$) to aldehydes and ketones in the cells. For example, ethanol, methanol, and lactate are oxidized to ethanal, methanal, and pyruvate, respectively, as shown below.
Unpleasant effects of drinking alcohol, i.e., flushing, nausea, dizziness, sweating, headache, and low blood pressure, are caused by ethanol (acetaldehyde). Antabuse drugs prohibit aldehyde dehydrogenase from converting acetaldehyde to acetic acid, causing the unpleasant effects of acetaldehyde to persist. Methanal produced by the oxidation of methanol is a poisonous substance that damages many tissues and causes blindness. It is used to poison ethanol meant to be used as a solvent. In methanol poison medical cases, ethanol is injected intravenously into the patient to compete with methanol for alcohol dehydrogenase enzyme. This way, methanol is excreted before converting to poisonous methanal in the body.
Oxidation of thiol ($\ce{-SH}$) to disulfide ($\ce{-S-S{-}}$) linkage
Thiol ($\ce{-SH}$) is present in aminoacid cysteine. Two cysteine units in the same or different protein chains can oxidize their thiol groups to form disulfide ($\ce{-S-S{-}}$) linkage. The disulfide linkage plays an important part in defining the tertiary structure of proteins. This oxidation reaction can be carried out easily by a variety of oxidizing agents, including iodine ($\ce{I2}$), bromine ($\ce{I2}$), and oxygen ($\ce{O2}$), as shown in an example below.
$\ce{2R-SH + 1/2O2 -> RS-SR + H2O}\nonumber$
4.10: Reactions with cyclic transition state
Learning Objectives
• Learn examples of reactions that involve five- or six-member transition state, including cyclonic hemiacetal formation of monosaccharides, Diels-Alter reactions producing six-membered cyclic products, and decarboxylation of $\beta$-keto acids.
Intramolecular reactions happen if the two reacting groups are on the same molecule and can come to a bonding distance through a five- or six-member cyclic transition state. Some examples of it are described in the next sections.
Cyclic hemiacetal formation of monosaccharides
Monosaccharides, like glucose, fructose, galactose, etc., have a $\ce{C=O}$-group on one $\ce{C}$ and $\ce{-OH}$-group on every other $\ce{C}$. $\ce{-OH}$-group can add to $\ce{C=O}$-group forming a hemiacetal. Monosaccharides exist primarily in a five- or six-membered hemiacetal form because one of their $\ce{-OH}$-group can form a five- or six-membered transition state for the reaction, as shown in Figure $1$ for the case of D-glucose and D-fructose.
Diels–Alder reaction
A conjugated diene, e.g., butadiene, and an alkene, e.g., ethene, make a cyclic six-member transition state. They react by the Diels-Alder reaction mechanism and produce a six-member cyclic product. This reaction can be intermolecular, e.g., between butadiene and then, or intramolecular, e.g., in the biosynthesis of antibiotic lovastatin, illustrated in Figure $2$.
Decarboxylation
Decarboxylation is the removal of carbon dioxide ($\ce{CO2}$) from a carboxylic acid ($\ce{R-COOH}$), as in this example: $\ce{R-COOH ->[\Delta] R-H + CO2}$.
This reaction requires high temperatures, such as in the thermal decomposition process. However, if there is a second carbonyl ($\ce{C=O}$)) group $\beta\}$ to the $\ce{-COOH}$ group, it can easily acquire a six-member transition state and decarboxylate at moderate temperatures, as illustrated in Figure $3$.
Ketone bodies and diabetes mellitus
Acetoacetic acid and its reduced product $\beta$-hydroxybutyric acid, shown below, are produced in the liver as a result of the metabolism of fatty acids and some amino acids.
Acetoacetic acid and $\beta$-hydroxybutyric acid are called ketone bodies. Their concentration in the blood of healthy persons is about 0.01 mmol/L but in persons suffering from starvation or diabetes mellitus may be up to 500 times higher.
Carboxylic acids exist as carboxylate anions under physiological conditions. Decarboxylation of the $\beta$-keto carboxylates happens spontaneously under physiological conditions. For example, acetoacetate decarboxylates and produces carbon dioxide and acetone, as illustrated in Figure $4$.
Carbon dioxide leaves under moderate conditions in this case because the anion left behind is in resonance with the $\beta$-$\ce{C=O}$ group. The body does not metabolize acetone but exhales through the lungs. Acetone is responsible for its characteristic sweet smell in the breath of people with swear diabetes. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/04%3A_Organic_reactions/4.09%3A_Reduction_and_oxidation_%28redox%29_reactions.txt |
• 5.1: What are carbohydrates
Carbohydrates, monosaccharides, and polysaccharides are defined, and their general formula and functional groups and drawing Fisher projections are described.
• 5.2: General class names and Common names of monosaccharides
Class names, common names, D/L stereodescriptors, and drawing Fisher projections of important monosaccharides are described.
• 5.3: Cyclic structures of monosaccharides
Cyclic hemiacetal forms of monosaccharides and the related terms, like Howarth projections, anomeric carbon, alpha- and beta-configurations of anomeric carbon, furanose, and pyranose forms, and mutarotation are described.
• 5.4: Reactions of monosaccharides
Reactions of monosaccharides, including conversion to glycosides, oxidation to aldonic acids, and reduction to alditols, are described along with defining and testing reducing sugars.
• 5.5: Disaccharides
Disaccharides, including maltose, cellobiose, lactose, and sucrose, and the nomenclature of the glycosidic linkage are described
• 5.6: Oligosaccharides
Blood types based on the oligosaccharides attached to the red blood cell surfaces and blood compatibility based on the blood type are described.
• 5.7: Polysaccharides
Structures and some characteristics of polysaccharides, including starches, cellulose, and chitin, are described.
05: Carbohydrates
Learning Objectives
• Define carbohydrates, monosaccharides, polysaccharides, and write their general formulas.
• Draw and interpret fisher projections of monosaccharides.
What are carbohydrates
Carbohydrates are primary energy storage compounds, e.g., glucose (\(\ce{C6H12O6}\)) synthesized by using carbon dioxide (\(\ce{CO2}\)) from the air, water (\(\ce{H2O}\)) from the soil, and energy from sunlight, along with the release of oxygen (\(\ce{O2}\)), as illustrated in Figure \(1\).
Oxidation of carbohydrates, e.g., the reverse of the photosynthesis reaction, release energy that plant and animals use for activities. Carbohydrates are the support structure material in plants, e.g., cell-wall and wood, shell material of crustaceans, and connective tissues in animals.
Glucose is a polyhydroxy aldehyde, and fructose, i.e., another carbohydrate found in honey, is a polyhydroxy ketone, as illustrated in Figure \(2\).
Carbohydrates
Carbohydrates are polyhydroxy aldehydes, polyhydroxy ketones, or other compounds that hydrolyze to polyhydroxy aldehydes or polyhydroxy ketones.
The general formula of simple carbohydrates is \(\ce{C_{n}H_{2n}O_{n}}\), which can also be written as \(\ce{C_{n}.(H2O)_{n}}\) which is the origin of the name carbohydrates, i.e., hydrates of carbon.
Fisher projections
Fisher projections are two-dimensional representations of molecules for showing the configuration of chiral centers:
• the chiral center is in the plane of the page,
• horizontal lines represent bonds projecting toward the viewers,
• and vertical lines represent bonds projecting away from the viewer.
• The parent \(\ce{C}\) chain is placed on the vertical line, with the most oxidized \(\ce{C}\), i.e., the \(\ce{C=O}\) in carbohydrates, at the top or near the top end, and the numbering starts from the top most \(\ce{C}\), as illustrated in Figure \(2\).
Monosaccharides and polysaccharides
Monosaccharides
Simple carbohydrates that can not be hydrolyzed to more simple ones are called monosaccharides. For example, D-glucose, D-fructose, and D-threose shown in previous figures are monosaccharides.
Polysaccharides
Linear or branched chain polymers comprised of monosaccharide repeat units (monomers) are called polysaccharides. For example, starch is a polysaccharide with a D-glucose monomer, illustrated in Figure \(3\).
Polysaccharides hydrolyze to monosaccharides, as illustrated in the following general hydrolysis reaction.
5.02: General class names and Common names monosaccharides
Learning Objectives
• Assign and interpret class names of monosaccharides.
• Assign D/L stereodescriptors to the common names and define epimers.
• Draw structure from the name and vice versa for important monosaccharides, including D-ribose, D-glucose, D-mannose, and D-galactose.
General class names of monosaccharides
Monosaccharides are either polyhydroxy aldehydes that take aldo as prefixes or polyhydroxy ketones that take keto as prefixes in their general name. The general formula of monosaccharides is \(\ce{C_{n}H_{2n}O_{n}}\) where n can be 3, 4, 5, 6, 7, or 8 representing triose, tetrose, pentose, hexose, heptose, or octose, respectively, in the general name. For example, D-glucose belongs to aldohexose, where the aldo- prefix tells it is an aldehyde, -hex- in the middle of the name tells it has six \(\ce{C's}\), and -ose suffix denotes it is a carbohydrate. D-fructose belongs to ketohexose, i.e., it is a monosaccharide with a ketone group and six \(\ce{C's}\). D-glyceraldehyde is an aldotriose, i.e., a monosaccharide with an aldehyde group and three \(\ce{C's}\).
Common names of monosaccharides
Common names are specific for each monosaccharide. All \(\ce{C's}\) in a monosaccharide are chiral centers except the two terminal \(\ce{C's}\) and the \(\ce{C}\) of ketone group if it is a ketose. The absolute configuration of the penultimate \(\ce{C}\), i.e., the second-last \(\ce{C}\), is explicitly expressed by D- or L-stereochemical descriptors and the absolute configuration of all other chiral centers is implicit in the common name of the monosaccharide. There is one set of common names for all D-isomers, and their mirror images (enantiomers) have the same common name with D- replaced with L-. For example, D-glyceraldehyde and L-glyceraldehyde, D-glucose, and L-glucose are enantiomer pairs shown in Figure \(1\).
D/L Stereochemical descriptors
If the \(\ce{-OH}\) group on the second-last carbon (penultimate \(\ce{C}\) or the second \(\ce{C}\) from the bottom end) in Fisher projection of a monosaccharide is on the right side, it is assigned D- and if it is on the left side, it is assigned L-configuration. Monosaccharides that are enantiomer pairs have the same common name, but D- is replaced with L- or vice versa. For example, D-glyceraldehyde and L-glyceraldehyde, D-glucose and L-glucose enantiomer pairs are shown in Figure \(1\). with the penultimate \(\ce{C}\) defining D- or L-configuration shown in red color.
D-configuration of monosaccharides is commonly found in nature. The D/L stereodescriptors do not indicate the rotation of the plane polarized light, i.e., the enantiomer's dextro/levo rotatory nature. However, if one enantiomer is dextro (d-), the other is levo (l-) to the same degree, and vice versa.
The structures as Fisher projections and common names of D-aldoses are shown in Figure \(2\). and those of D-ketoses are shown in Figure \(3\). These are the most common monosaccharides found in nature.
Some important monosaccharides
D-Glucose is the most abundant monosaccharide in nature. Plants produce it in a photosynthesis process. D-Galactose and D-mannose are two important diastereomers of D-glucose that differ from D-glucose in the configuration of only one chiral center.
Epimers
Epimers are diastereomers that differ in absolute configuration of only one chiral center. For example, D-galactose configuration is different from D-glucose only at \(\ce{C}\)#4, i.e., D-galactose is a \(\ce{C}\)4-epimer of D-glucose. Similarly, D-mannose is \(\ce{C}\)2-epimer of D-glucose.
D-Fructose is another important monosaccharide that differs at \(\ce{C}\)#1 and \(\ce{C}\)#2 from glucose. That is, the \(\ce{C=O}\) is an aldehyde group at \(\ce{C}\)#1 in D-glucose, but it is a ketone at \(\ce{C}\)#2 in D-fructose. Fisher projections of D-glucose, D-galactose, D-mannose, and D-fructose are shown in Figure \(4\). with the differences from D-glucose highlighted by red-color fonts. D-ribose is another important monosaccharide present in RNA.
Drawing structures of important monosaccharides
D-ribose is aldopentose, i.e., an aldehyde with five \(\ce{C's}\). All chiral \(\ce{C's}\) have \(\ce{-OH}\) groups oriented towards the right in the Fisher projection. D-allose is aldohexose with the same structural features as D-ribose, i.e., all chiral \(\ce{C's}\) have \(\ce{-OH}\) groups oriented towards the right in the Fisher projection. D-glucose is \(\ce{C}\)3 epimer of D-allose. The other three important monosaccharides can be drawn by relating them to D-glucose, i.e., D-galactose is a \(\ce{C}\)4 epimer, D-mannose is a \(\ce{C}\)2 epimer of D-glucose, and D-fructose has a ketone group at \(\ce{C}\)2 in the place of aldehyde group of D-glucose at \(\ce{C}\)1. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/05%3A_Carbohydrates/5.01%3A_What_are_carbohydrates.txt |
Learning Objectives
• Draw and inter-convert open-chain and cyclic hemiacetal forms of compounds containing carbonyl and alcohol groups in the same molecule.
• Convert Fisher projections to Howarth projections and to chair conformation in the cases of pyranose forms of monosaccharides.
• Define Howarth projection, anomeric carbon, $\alpha$ and $\beta$ configuration of the anomeric carbon, furanose, and pyranose forms, and mutarotation of monosaccharides.
Cyclic hemiacetal
Alcohol ($\ce{R-OH}$) and a carbonyl ($\ce{C=O}$) groups react with each other and form a hemiacetal group. The hemiacetals are usually unstable and revert to the reactants. However, suppose the alcohol and the carbonyl groups are on the same molecule and can react to form a five- or six-member cyclic hemiacetal. The product is more stable than the reactants, as shown in the following example reaction.
Note that the hemiacetal $\ce{C}$ is a new chiral center that is produced as a racemic mixture, which is represented by a wavy line in the $\ce{-OH}$ group. Monosaccharides are polyhydroxy aldehydes or polyhydroxy ketones that often exist as hemiacetal as the dominant form in equilibrium with the open-chain form, as described in the next section.
Fisher projections to Haworth projections
Fisher projections represent the open-chain forms of monosaccharides where the $\ce{C}$-chain is a vertical line. Hemiacetals are five or six-member ring structures represented by pentagon or hexagon shapes.
Figure $1$ explains the steps to re-draw Fisher projection in cyclic conformation.
1. Rotate the Fisher projection by 90o clockwise from vertical to horizontal orientation. The groups on the right of Fisher's projection end up above, and those to the left end below in the horizontal orientation.
2. Move $\ce{C's}$#1, 4, 5, and 6 up by 60o each to arrive at the cyclic open-chain conformation#1.
3. Rotate the $\ce{C}$4-$\ce{C}$5 bond by 60o counterclock wise to bring the $\ce{C}$#5$\ce{-OH}$ group in the cycle of $\ce{C's}$ and $\ce{C}$#6 pointing above, i.e., cyclic conformation#1 in Figure $1$.
4. $\ce{C}$#5$\ce{-OH}$ and a carbonyl ($\ce{C=O}$) of $\ce{C}$#1 react to produce the hemiacetal group with the new $\ce{-OH}$ group on $\ce{C}$#1 either pointing away from the hemiacetal $\ce{O}$, i.e., $\alpha$- or point towards the hemiacetal $\ce{O}$, i.e., $\beta$-orientation in the cyclic hemiacetal.
Anomeric $\ce{C}$
Anomeric $\ce{C}$ is the $\ce{C}$ of carbonyl ($\ce{C=O}$) group in the open chain conformation that becomes the $\ce{C}$ of the hemiacetal group in the cyclic structure of a monosaccharide. For example, $\ce{C}$#1 of D-glucose is the anomeric $\ce{C}$, as highlighted in the figure on the right.
Haworth projection
Haworth projection is five or six-membered cyclic hemiacetals of monosaccharides, presented as pentagon or hexagon shape. It is viewed through its edge perpendicular to the plane of the page, such that the anomeric $\ce{C}$ s the right corner in the plane of the page and ether group ($\ce{R-O-R}$) is the top right corner in the rear, as shown in the figure on the right. Bold lines usually show the ring bonds toward the viewer (above the page) and those below the page (away from the viewer) as solid lines. Bonds outside the ring are shown as solid lines pointing above or below it.
Assigning $\alpha$- or $\beta$- to the cyclic structures of monosaccharides
Conversion of monosaccharides from open chain to cyclic hemiacetal produced a new chiral center at the anomeric $\ce{C}$. It is a mixture of two conformations: the $\ce{-OH}$ group on anomeric $\ce{C}$ going away from the ether ($\ce{R-O-R}$) group is $\alpha$- and towards it is $\beta$-configuration, as illustrated in the figure on the right with the help of D-glucose example.
D-Fructose exists in five-membered and six-membered hemiacetal forms, as shown in Figure $2$.
Furanose and pyranose forms of monosaccharides
Furanose is a five-membered, and pyranose is a six-membered cyclic hemiacetal of monosaccharide. These names originate from the names of cyclic ethers, furan, and pyran, shown in the figure on the right. Based on this nomenclature, the class name of $\alpha$-D-Glucose is $\alpha$-D-glucopyranose and of $\beta$-D-glucose is $\beta$-D-glucopyranose. D-Fructose exists in pyranose and furanose forms, as shown in Figure $2$.
Chair conformations of pyranose forms
Haworth projections assume a flat ring structure. It is close to the structure of furanose forms (five-membered rings), but pyranose forms (six-membered rings) exist in chair conformations. The figure on the right shows a chair conformation of a six-membered cyclic structure. The points to note are the following.
1. Each $\ce{C}$ has two bonds (shown in black) in the cycle, one axial bond (shown in red) pointing along the axis of the cycle and one equatorial bond (shown in blue) pointing approximately along the equator of the cycle.
2. The axial bonds point in opposite directions, up and down, on neighboring $\ce{C'}$. The same applies to equatorial bonds.
3. Bulky group, i.e., any group other than $\ce{-H}$, is more stable at the equatorial than axial position because of steric strain (push away) from the other two axial positions on the same face of the ring., as illustrated in the figure on the left.
How to draw chair conformations of monosaccharides
Alpha ($\alpha$) and axial start with the letter a. Remember $\alpha$- means an axial $\ce{-OH}$ at anomeric $\ce{C}$ in the case of pyranose chair conformation.
Since $\alpha$$\ce{-OH}$ is axial, $\beta$$\ce{-OH}$ is equatorial at anomeric $\ce{C}$. The Howarth projection of $\beta$-D-glucose in the figure on the right shows that the bulky groups alternate between up and down positions on $\ce{C}$#1 to $\ce{C}$#5. It means all of them are either axial or equatorial. Since $\beta$$\ce{-OH}$ is equatorial, all bulky groups in $\beta$-D-glucose are equatorial, as shown in Figure $3$.
After learning how to draw $\beta$-D-glucopyranose, the other monosaccharides can be drawn easily based on the knowledge of the difference from $\beta$-D-glucopyranose. For example, $\beta$-D-manopyranose is $\ce{C}$#2 epimer, $\beta$-D-galactopyranose is $\ce{C}$#4 epimer, and their $\alpha$ forms have an axial $\ce{-OH}$ at anomeric $\ce{C}$, as shown in Figure $3$.
Mutarotation
D-Glucose is a chiral compound. A freshly prepared aqueous solution of $\beta$-D-glucopyranose has a specific rotation of +112.2. A freshly prepared solution of $\alpha$-D-glucopyranose has a specific rotation of +18.7. However, the specific rotation of $\beta$-D-glucopyranose and $\alpha$-D-glucopyranose gradually changes to +52.5. This is because the open chain and the two cyclic forms establish equilibrium in the solution, as illustrated in Figure $4$.
Open chain and the two cyclic forms of D-glucose exist in equilibrium in solution where $\beta$-D-glucose has all bulky group in equatorial position is the most stable, $\alpha$-D-glucose with one bulky group in axial position is the second most stable and, and open chain for is the least stable, as reflected by their proportions at equilibrium shown in Figure $4$.
Mutarotation is the change in the optical rotation of carbohydrate solution over time due to the change in the composition of different compound forms to achieve equilibrium composition. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/05%3A_Carbohydrates/5.03%3A_Cyclic_structures_of_monosaccharides.txt |
Learning Objectives
• Write and understand chemical reactions of monosaccharides, including conversion of cyclic hemiacetal to glycosides and N-glycosides, oxidation of open-chain aldehyde form to aldonic acids, and reduction to alditols.
• Define reducing sugars with the examples of D-glucose and D-fructose.
Glycosidic bond formation
Aldehydes and ketones react with alcohols to form hemiacetals. If the alcohol reagent is in excess, a second molecule of the alcohol reacts and converts hemiacetal to acetal. Acetals can be isolated. Acetals convert to alcohol and the aldehyde or ketone when their aqueous solution is acidified, as shown in the general reaction below.
Cyclic (pyranose and furanose) forms of monosaccharides are hemiacetals that react with alcohols to form acetals.
Glycosides are the acetals of monosaccharides. The glucose formed by a reaction with alcohol is also called O-glycoside.
For example, $\beta$-D-glucopyranose reacts with methanol to form glycosides, shown in Figure $1$.
Glycoside and glycosidic bond
An acetal of a monosaccharide is a glycoside. The bond between anomeric $\ce{C}$ to the new ether $\ce{-OR}$ group in the glycoside is called the glycosidic bond. Anomeric $\ce{C}$ and glycosidic bonds are indicated by red arrows in Figure $1$.
Glycoside is named by replacing the last e of the carbohydrate name with ide and adding the name of the alkyl part of the new ether $\ce{-OR}$ group as a prefix separated by a hyphen. For example, $\beta$-D-glucopyranose reacts with methanol to form methyl-$\beta$-D-glucopyranoside and methyl-$\alpha$-D-glucopyranoside, as shown in Figure $1$.
Since monosaccharides have many alcohol groups, two monosaccharides can react with each to form a glycoside, called a disaccharide, three can form a trisaccharide, and many can form a polysaccharide. The di-, tri-, and polysaccharides can be hydrolyzed to monosaccharides in an acidic aqueous solution.
Monosaccharide, disaccharide, oligosaccharide, and polysaccharide
A carbohydrate that can not be hydrolyzed to a simpler carbohydrate is a monosaccharide, e.g., D-glucose. A glycoside of two monosaccharides is a disaccharide, e.g., cellobiose. A glycoside of three to ten monosaccharides is an oligosaccharide, e.g., a fragment of cellulose. A glycoside of more than ten monosaccharides is a polysaccharide, e.g., cellulose. The figure below illustrates the structures of D-glucose -a monosaccharide; cellobiose -a disaccharide; and cellulose, -a polysaccharide.
Amines react with hemiacetals forms of monosaccharides the same way as alcohols and produce N-glycosides. For example, $\beta$-D-2-deoxyribose reacts with thymine and produced an N-glycoside called deoxythymidine, a part of a monomer of DNA molecule, as shown in the figure below.
Exceptions to the general formula of carbohydrates
The general formula of carbohydrates is $\ce{C_{m}(H2O)_{n}}$, and usually for monosaccharides m is equal to n, i.e., $\ce{C._{n}(H2O)_{n}}$ or $\ce{C_{n}H_{2n}O_{n}}$. However, there are exceptions.
• Carbohydrates do not necessarily conform to this formula. For example, $\beta$-D-2-Deoxyribose $\ce{C5H10O4}$ has less $\ce{O's}$ than what is dictated by the general formula.
• Compounds following this general formula are not necessarily always carbohydrates. For example, acetic acid $\ce{C2H4O2}$ follows the general formula, but it is not a carbohydrate.
Oxidation to aldonic acids
Aldehydes are oxidized by a variety of reagents to carboxylic acids. Similarly, the aldehyde group ($\ce{-CHO}$) of the open-chain form of aldoses is oxidized to a carboxylic acid ($\ce{-COOH}$) group by oxygen $\ce{O2}$ in the presence of enzymes called oxidases. The acid is named by replacing the suffix -ose of the class name aldose with --onic acid. For example, D-glucose is oxidized to D-gluconic acid. Since, $\ce{-COOH}$ exists in ionized ($\ce{-COO^{-}}$) form, the suffix -onic acid is replaced with -onate for the ionized form. For example, D-glucose is oxidized to D-gluconate, as shown in the figure below.
Reducing sugars
A carbohydrate that reacts with a mild oxidizing agent under alkaline conditions to form an aldonic acid (or aldonate anion in physiological conditions) is called a reducing sugar. For example, D-glucose is a reducing sugar, as shown in the figure above. The carbohydrate reduces the oxidizing agent.
Common reagents used to test the presence of a reducing sugar include Tollen's reagent in which $\ce{Ag^{+}}$ is reduced to $\ce{Ag}$ that forms a silver mirror on the glass, and Benedict's reagent in which $\ce{Cu^{2+}}$ is reduced to $\ce{Cu^{+}}$ that forms red color precipitate $\ce{Cu2O}$, as shown in Figure $2$.
Although open-chain form is in small concentration at equilibrium, it is continuously formed from the hemiacetal forms as the oxidation reaction removes it. The 2-ketoses are also reducing sugars. This is because 2-ketoses exist in equilibrium with the corresponding aldoses under the basic conditions. The aldose form is the actual reducing agent in this mixture. For example, D-fructose is a reducing sugar. D-fructose exists in equilibrium with D-glucose in basic conditions, where D-glucose is the reducing agent, as illustrated in the figure below.
Diabetes and blood-sugar test
D-Glucose is also called blood sugar, as it is normally present in blood at about 70 mg/dL to 130 mg/dL. It may rise up to 140 mg/dL after eating food, but it returns to the normal range in healthy persons. Enzyme insulin controls blood sugar. If blood glucose control is not functioning properly, the blood glucose may stay higher than normal -a condition called hyperglycemia or diabetes; or it may stay below the normal level -a condition called hypoglycemia.
Blood glucose is usually tested based on enzyme-catalyzed reactions. Glucose oxidase enzyme oxidizes $\beta$-D-glucose using $\ce{O2}$ to D-gluconate and hydrogen peroxide $\ce{H2O2}$. $\alpha$-D-glucose does not react directly, but it converts to $\beta$-D-glucose as the latter is consumed. A second enzyme, peroxidase, causes $\ce{H2O2}$ to react with 2-methylaniline and produces a colored product that is monitored to measure blood glucose, as illustrated in the figure below.
Reduction to alditols
Aldehydes are reduced to alcohols by a variety of reducing agents. Aldoses in the open-chain form are reduced by reducing agents like sodium borohydride ($\ce{NaBH4}$), $\ce{H2}$ in the presence of $\ce{Pt}$, $\ce{Pd}$, or $\ce{Ni}$. Oxidases enzymes are reducing agents in biochemical systems. Although open-chain aldehyde form is in small concentration in the equilibrium mixture, cyclic hemiacetal forms convert to open-chain form as the latter is consumed. The product alcohol is named by replacing -ose suffix of the aldose name with -itol. For example, D-glucose is reduced to D-glucitol, as shown below.
D-glucitol is found in berries, cherries, plums, pears, seaweed, and algae. It is commonly known as D-sorbitol and is a sugar substitute for diabetes. Other alditol examples include D-erythritol, D-mannitol, and D-xylitol. D-xylitol is used in cereals, sugarless candies, and gums. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/05%3A_Carbohydrates/5.04%3A_Reactions_of_monosaccharides.txt |
Learning Objectives
• Be able to draw and name the glycosidic linkage in disaccharides.
• Know common disaccharides' structures, sources, and properties, including maltose, cellobiose, lactose, and sucrose.
Nomenclature of glycosidic linkage
Disaccharides are glycosides of two monosaccharides. Glycosidic linkage is formed by a reaction between the hemiacetal group of one monosaccharide and an alcohol group of the other, as illustrated in the figure below.
The glycosidic linkage is named by listing the number of the $\ce{C's}$ linked, separated by a comma. In the case of anomeric carbon, $\alpha$/$\beta$ orientation is also indicated, separated by a hyphen from the number of the anomeric carbon. For example, $\beta$-cellobiosee is linked by $\beta$-1,4 glycosidic linkage and $\beta$-maltose by $\alpha$-1,4 glycosidic linkage. The $\beta$ in the name $\beta$-cellobiose and $\beta$-maltose indicates the orientation of the free hemiacetal group in the second saccharide unit. It is not related to the orientation of the glycosidic linkage, as indicated by the black arrow in the figure. $\beta$-Cellobiose and $\beta$-maltose are configurational isomers having different physical and chemical properties.
Examples of disaccharides include maltose, cellobiose, lactose, and sucrose, described next.
Maltose
Maltose is a disaccharide of two D-glucose units connected through $\alpha$-1,4-glycosidic linkage, as shown below.
The second D-glucose unit with free hemiacetal group can be either in $\alpha$- or $\beta$-configuration, as indicated by $\alpha$/$\beta$ prefix to the name. An aqueous solution exists in equilibrium between $\alpha$, $\beta$, and open-chain aldose forms. Since common oxidizing agents can reduce the open-chain aldose form, maltose is a reducing sugar. Maltose is found in the juice of sprouted barley grains and other grains.
Cellobiose
Cellobiose is a disaccharide of two D-glucose units connected by $\beta$-1,4-glycosidic linkage, as shown below. Hemiacetal forms of cellobiose are shown below.
Cellulobiose is a reducing sugar because, like maltose, the D-glucose unit with free hemiacetal group can exist in $\alpha$, $\beta$, or open-chain aldose forms in solution. Cellobiose is obtained by hydrolysis of cellulose.
Lactose
Lactose is a disaccharide of D-galactose joined by $\beta$-1,4-glycosidic linkage with D-glucose, as shown below.
Lactose is a reducing sugar because, like maltose, the D-glucose unit with free hemiacetal group can exist in $\alpha$, $\beta$, or open-chain aldose forms in solution. Lactose is present in milk, up to 5% to 8% in human milk and 4% to 6% in cow milk. Lactase is the enzyme that hydrolysis lactose in the digestion process. Many adults develop lactose intolerance due to a lack of or insufficient lactase production. Without lactase, lactose enters the colon undigested and is fermented by bacteria causing bloating and abdominal cramps. Therefore, some products that use lactose also add lactase to avoid the problem.
Sucrose
Sucrose is a disaccharide of D-glucose (in pyranose form) and D-fructose (in furanose form) joined by $\alpha$-1,$beta)-2-glycosidic linkage, as shown below. Both anomeric \(\ce{C,s}$ in the glycosidic linkage can not exist in open-chain aldose or ketose forms. Therefore, sucrose is a non-reducing sugar. Sucrose is commonly used as table sugar. It is the most abundant disaccharide found in sugar cane and sugar beats.
Relative sweetness
Monosaccharides and disaccharides are sweet, but sweetness varies. Fructose is the sweetest monosaccharide, and lactose is the least sweet, as shown in Table 1. Honey is a mixture of fructose and glucose and has a sweetness about the same as table sugar (sucrose). Relative sweetness in the table is based on the results of a group of people tasting and ranking them in order of taste relative to sucrose (table sugar), rated 1. Some artificial sweeteners listed in the table are commonly used as sugar-substitute in diet foods.
Table 1: Relative sweetness of some carbohydrates and artificial sweeteners (Based on the results of a group of people tasting and ranking them in order of taste relative to sucrose (table sugar) rated 1 as reference).
Carbohydrates Sweetness relative to sucrose Sugar alcohols Sweetness relative to sucrose Artificial sweeteners Sweetness relative to sucrose
Fructose 1.74 Xylitol 1.00 Advantame 2,000
Sucrose (table sugar) 1.00 Maltitol 0.80 Saccharin 450
Honey 0.97 Sorbitol 0.60 Acesulfame-K 200
Glucose 0.74 Aspartame 180
Maltose 0.33 Stevia 150
Galactose 0.65 Sucralose 60
Lactose 0.16 | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/05%3A_Carbohydrates/5.05%3A_Disaccharides.txt |
Learning Objectives
• Differentiate between blood types based on the oligosaccharides attached to the red blood cell surface and blood compatibility related to the blood type.
Oligosaccharides are polymers containing a small number, typically three to ten of monosaccharides. Oligosaccharides include dextrins which are microbial breakdown products of long polysaccharides like starch or cellulose and other glycans linked to lipids or proteins through N- or O-glycosidic bonds. The primary role of the last class is cell recognition and cell adhesion. For example, oligosaccharides attached to red-blood cells define the blood type A, B, AB, or O, depending on the linked oligosaccharides, as described next.
A, B, AB, or O blood types
The membrane of animal plasma cells typically has oligosaccharides comprising 4 to 17 monosaccharides attached to them. Their purpose is cell recognition, i.e., if there is a foreign cell or an object in the biochemical system, it is identified through the surface oligosaccharides and either destroyed or flagged for destruction by the body's immune system. For example, red blood cells have different types based on the oligosaccharides attached to their cell membrane, as shown in Figure \(1\).
Blood type O has a combination of N-acetylgalactoseamine-galactose-Fucose. This combination is present in all other blood types also. So, Blood type O is not recognized as foreign in any blood type. A person with blood type O can donate blood to every type, i.e., type O is a universal donor. Any class other than type O has some additional monosaccharide attached which is recognized as foreign by blood type O. So, a person with blood type O can accept only type O blood.
Blood type B has an additional monosaccharide, i.e., galactose attached to galactose of type O oligosaccharide. Type B can receive blood from type O or type B and donate to type B or type AB, as shown in Figure \(2\).
Blood type A has an additional monosaccharide, N-acetyl galactosamine, attached to galactose of type O oligosaccharide. Type A can receive blood from type O or type A and can donate to type A or type A.B
Blood type AB has a mix of type A and type B. Type AB can accept all types, i.e., it is a universal recipient but can donate only to type AB.
5.07: Polysaccharides
Learning Objectives
• Learn the structures and some characteristics of polysaccharides, including starches, cellulose, and chitin.
Polysaccharides are long polymers from ten to thousands of monosaccharides joined by glycosidic linkages. The most abundant polysaccharides are starch, glycogen, cellulose, and chitin. Except for chitin, all others are composed of D-glucose. Chitin is composed of a modified form of glucose. These are described next.
Starch
Starch is a storage form of D-glucose in plants. It is found in potatoes, beans, rice, wheat, and other grains and roots, as illustrated in Figure $1$. Starch is a mixture of two forms, 20% to 25% amylose and 75 to 80% amylopectin.
Amylose
Amylose is an unbranched chain of up to 4,000 D-glucose units joined by $\alpha$-1,4-glycosidic linkage, as shown below.
The orientation of bonds with $\alpha$-1,4-glycosidic linkage makes the amylose chain exist in a helical structure, as shown in Figure $2$, that allows water molecules to access and establish hydrogen bonds with $\ce{-OH}$ groups. Starch is a water-soluble polymer. It can dissolve in water due to extensive hydrogen bonding with the water molecules when mixed with water.
Amylopectin
Amylopectin is a branched chain polymer of D-glucose, as shown below.
The main chain comprises 10,000 D-glucose units joined by $\alpha$-1,4-glycosidic linkage. The main chain exists in the form of a helix. Branches originate at every 24 to 30 unit intervals on the main chain connected to the main chain by a (\alpha\)-1,6-glycosidic linkage, as shown in Figure $2$. This helix with branches going outwards makes amylopectin more accessible to water for hydrogen bonding and easier to digest.
Glycogen
Glycogen is an energy-storage polysaccharide in animals with the same structure as amylopectin. it has up to 106 D-glucose units joined by $\alpha$-1,4-glycosidic linkages and branching through (\alpha\)-1,6-glycosidic linkages. The main difference from amylopectin is that glycogen has more frequent branching at 10 to 15 D-glucose units interval, as illustrated in Figure $3$. More branching makes it more soluble in water and easier to hydrolyze multiple D-glucose units when the body needs D-glucose.
Cellulose
Cellulose is the most abundant polysaccharide in nature that makes up about 50% of the cell wall of plant cells. Cotton, shown in Figure $4$, is almost pure cellulose.
Cellulose is a linear polysaccharide of about 2200 D-glucose units joined by $\beta$-1,4-glycosidic linkage, as shown below.
The $\beta$-1,4-glycosidic linkage allowes glucose units to adopt a linear structure with intra-molecule hydrogen bonding. The linear polymer packs nicely with inter-molecular hydrogen bonding and the London-dispersion forces, as shown in Figure $5$. It gives cellulose its mechanical strength and makes it insoluble in water. Due to the extensive inter- and intra-molecular hydrogen bonding within cellulose, water can not establish enough hydrogen-bonds to dissolve it.
Animals have $\alpha$-glucosidase enzymes that allow them to hydrolyze starch and glycogen to D-glucose, but they do not have $\beta$-glucosidase enzymes needed to hydrolyze cellulose. Grazing animals and termite hose bacteria in their stomach that have the $\beta$-glucosidase enzymes. Therefore, grazing animals and termites can digest cellulose.
Chitin
Chitin is the second most abundant polysaccharide, second only to cellulose. It is found in the exoskeleton of crustaceans and insects, as shown in Figure $6$.
Chitin has the same structure as cellulose except that it is composed of N-acetylglucosamine, an amide derivative of D-glucose, in place of D-glucose in cellulose, as illustrated below Figure $7$. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/05%3A_Carbohydrates/5.06%3A_Oligosaccharides.txt |
• 6.1: What are lipids?
Lipids are defined, their basic structural features are described, and their categories are listed.
• 6.2: Fatty acyls
Structural features of fatty acids, prostaglandins, and waxes, their nomenclature, importance, and uses are described.
• 6.3: Glycerolipids
Glycerolipids are defined, and the difference in composition and properties of fats and oils and their chemical reactions, including hydrolysis, saponification, and hydrogenation is described.
• 6.4: Glycerophospholipids
Glycerophospholipids are defined, and classification, structural features, and their role in snake poisoning and infant respiratory distress syndrome are described that allow them to organize into lipid bilayers in the watery environment of cells.
• 6.5: Sphingolipids
Sphingolipids, their subclasses, their role in signal conduction in the nervous system, and their role in multiple sclerosis are described.
• 6.6: Saccharolipids, polyketides, and prenols
Structurals of saccharolipids, polyketides, and prenols and terms related to terpenes, including mono-, sesqui-, di-, and tri-terpenes are described.
• 6.7: Sterols
Sterols, particularly cholesterol, bile salts, and steroid hormones, and their functions, production, consumption, and associated medical problems are described.
• 6.8: Cell membrane
Cell membrane, its composition and control of fluidity, and three modes of transport are described: diffusion, facilitated transport, and active transport.
06: Lipids
Learning Objectives
• Define lipids and understand the basic characteristics of their structural components.
Lipids
Lipids can be defined as hydrophobic or amphiphilic biochemicals, where hydrophobic means water-hating or non-polar and amphiphilic means having both hydrophilic, i.e., water-loving or polar and lipophilic, i.e., fat-loving or non-polar components within the same molecule, but overall water insoluble.
Lipids have a long alkyl group that imparts hydrophobic or lipophilic character to the lipids and, usually, a polar and/or ionic group that is a hydrophilic component of the molecule, as illustrated in the figure on the right.
Lipids are a diverse group of biomolecules, including waxes, fats, fat-soluble vitamins (such as vitamins A, D, E, and K), phospholipids, steroids, terpenes, and others. They have a common property of being water insoluble. They can be dissolved in nonpolar solvents like ether or chloroform. Lipids have different functions in living things, including energy storage, signaling, hormonal activities, acting as structural components of cell membranes, etc.
Classification of lipids
Lipids are divided into eight categories: 1) fatty acyls; 2) glycerolipids; 3) glycerophospholipids; 4) sphingolipids; 5) saccharolipids; 6) polyketides; 7) steroids; and 8) prenol lipids. Six of the lipid categories are illustrated in Figure $1$.
Five lipid categories contain long alkyl chain carboxylic acids or carboxylic acid derivatives. These include fatty acyls, glycerolipids, glycerophospholipids, sphingolipids, and saccharolipids. Polyketides are derived from condensation reactions of $\beta$-ketoacyl subunits. Steroids and prenols are derived from condensation reactions of isoprene subunits. These categories of lipids are described in the following sections.
6.02: Fatty acyls
Learning Objectives
• Understand the basic structural features of fatty acids, the notations reflecting them, and their relationship to the melting points and health effects.
• Define essential fatty acids, omega fatty acids, and their importance.
• Understand the basic structure features, nomenclature, and functions of prostaglandins.
• Understand the structures and uses of important waxes.
Fatty acyls are a group of lipids that contain a fatty acid or its derivative.
Fatty acids
Fatty acids
Fatty acids are carboxylic acids with an alkyl chain that is usually unbranched and containing an even number of $\ce{C's}$, usually between 8 to 20 $\ce{C's}$.
Figure $1$ shows some examples of fatty acids that have 18 $\ce{C's}$ in the chain. Carboxylic acid group ($\ce{-COOH}$ is hydrophilic, and the alkyl chain is hydrophobic, making fatty acids amphiphilic. The hydrophobic property of the long alkyl chain dominates, making fatty acids insoluble in water.
Unsaturated, monounsaturated, and polyunsaturated fatty acids
• The alkyl chain can be saturated with all $\ce{C-C}$ $\sigma$-bonds, called saturated fatty acid (SFA), e.g., steric acid;
• or may contain some double bonds ($\ce{C=C}$), called unsaturated fatty acid (USFA)., e.g., oleic acid and linoleic acid Figure $1$.
• The unsaturated fatty acid may have one $\ce{C=C}$ bond called monounsaturated fatty acid (MUFA), e.g., oleic acid;
• or more than one $\ce{C=C}$ bonds called polyunsaturated fatty acid (PUFA), e.g., linoleic acid.
Notations for and labelling $\ce{C's}$ in fatty acids
Short notations
Carboxylic acids are usually referred to by their trivial names. Short notation is used to indicate # of $\ce{C's}$:# of $\ce{C=C}$ bonds, e.g., steric acid Figure $1$ is 18:0 meaning eighteen carbons and zero double bonds. Similarly, oleic acid is 18:1 meaning eighteen carbons and one double bond, and linoleic acid is 8:2 meaning eighteen carbons and two double bonds.
Labelling $\ce{C's}$
Systematic IUPAC nomenclature starts numbers from carbonyl-carbon ($\ce{C=O}$) as shown in Figure $2$, red numbers. Trivial nomenclature uses Greek alphabets starting from $\ce{C}$#2 as $\alpha$, next one as $\beta$, than $\gamma$, and so on, as shown in Figure $2$, black numbers. IUPAC short notation shows the position labels of the double bonds' first $\ce{C's}$ in a bracket, separated by commas, next to the regular short notation. The numbering begins from ($\ce{-COOH}$ end. For example, steric acid 18:0, i.e., eighteen carbons and zero double bonds, oleic acid is 18:1(9), i.e., eighteen carbons and one double bond at $\ce{C}$# 9, and linoleic acid is 18:2(9, 12), i.e., eighteen carbons and bonds, one at $\ce{C}$# 9 and the other at $\ce{C}$# 12.
Omega-minus ($\omega$-#) labeling
In omega-minus ($\omega$-#) labeling the last $\ce{C}$, i.e., the $\ce{-CH3}$ at the end opposite to carboxyl ($\ce{-COOH}$) group is labeled $\omega$-1, the $\ce{C}$ next to it as $\omega$-2, than $\omega$-3 and so on, as as shown in Figure $2$, in blue fonts. Recall that $\omega$ is the last alphabet in Greek, so, $\omega$-1 is the last carbon. The "omega-#" ($\omega$-#) label is usually used to refer to the position of the first $\ce{C}$ of the first $\ce{C=C}$ bond from the $\omega$-end, i.e., from the $\ce{CH3-}$ end, opposite to the $\ce{-COOH}$ end. For example, $\alpha$-linolenic acid in Figure $2$ is an $\omega$-3 fatty acid.
Table 1 lists common fatty acids with their IUPAC short notations, $\omega$-#, and melting points.
Table 1: Some of the common fatty acids
Common name Condensed formula IUPAC notation* Melting point (oC) $\omega$-# Found in
Saturated fatty acids
Caprylic acid CH3(CH2)6COOH 8:0 17 Milk
Capric acid CH3(CH2)8COOH 10:0 32 Coconut
Lauric acid CH3(CH2)10COOH 12:0 44 Coconut
Myristic acid CH3(CH2)12COOH 14:0 55 Nutmeg
Palmitic acid CH3(CH2)14COOH 16:0 63 Palm
Stearic acid CH3(CH2)16COOH 18:0 69 Animal fat
Arachidic acid CH3(CH2)18COOH 20:0 75 Corn
Monounsaturated fatty acids
Myristoleic acid CH3(CH2)3CH=CH(CH2)7COOH 14:1(9) 0 $\omega$−5 Butter
Oleic acid CH3(CH2)7CH=CH(CH2)7COOH 18:1(9) 14 $\omega$-9
Olives,
pecan,
grapeseed
Elaidic acid CH3(CH2)7CH=CH(CH2)7COOH 18:1(9t) 45 $\omega$-9 Milk
Erucic acid CH3(CH2)7CH=CH(CH2)11COOH 22:1(13) 34 $\omega$-9 Rapeseed
Polyunsaturated fatty acids
Linoleic acid CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH 18:2(9,12) -5 $\omega$−6
Soybean,
safflower,
sunflower
$\alpha$-Linolenic acid CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH 18:3(9,12,15) -11 $\omega$−3 Corn
Arachidonic acid CH3(CH2)4CH=CHCH2CH=CHCH2CH=CHCH2CH=CH(CH2)3COOH 20:4(5,8,11,14) -50 $\omega$−6
Meat, eggs,
fish
Eicosapentaenoic acid CH3CH2CH=CHCH2CH=CHCH2CH=CHCH2CH=CHCH2CH=CH(CH2)3COOH 20:5(5,8,11,14,17) -65 $\omega$−3 Fish
Docosahexaenoic acid CH3CH3CH=CHCH2CH=CHCH2CH=CHCH2CH=CHCH2CH=CHCH2CH=CH(CH2)2COOH 22:6(4,7,10,13,16,19) -44 $\omega$−3 Fish
* IUPAC notation shows Numbers of $\ce{C's}$:number of double bonds(position of cis-double bond, except when there is symbol t for trans-configuration)
Essential fatty acids
$\alpha$-Linolenic acid (18:3(9,12,15)) -an $\omega$−3 fatty acid, and linoleic acid (18:2(9,12)) - an $\omega$−6 fatty acid are essential fatty acids (EFAs) because humans and other animals cannot synthesize them. They must be included in the diet as they are needed for good health. Although eicosapentaenoic acid (20:5(5,8,11,14,17)) and docosahexaenoic acid (22:6(4,7,10,13,16,19)), called long-chain unsaturated fatty acids, can be synthesized from $\alpha$-linolenic acid, the conversion efficiency is low, and it is recommended that these should also be included in the diet. Arachidonic acid (20:4(5,8,11,14)) is another long-chain unsaturated fatty acid that can be synthesized from linoleic acid.
Shapes and physical properties of fatty acids
Shapes
The saturated fatty acids are cylindrical that pack nicely in the solid state with a large contact area between the molecules. The double bonds in unsaturated fatty acids are usually in the cis configuration. A cis-$\ce{C=C}$ introduces a kink in the structure that does not let molecules pack with a large contact area between neighboring molecules, as illustrated in Figure $3$. Some unsaturated fatty acids have trans-double bonds and are cylindrical, like unsaturated fatty acids, but they are rare.
Melting points
The melting point of fatty acids is affected by two factors, as shown in Table 1:
1. an increase in the number of $\ce{C's}$ increases the melting point because it increases London dispersion forces holding the molecules together, and
2. the presence of cis-$\ce{C=C}$ bond decreases the melting point due to the kink in the structure that reduces the contact area between molecules and, consequently, London dispersion forces are less.
Fats containing saturated fatty acids are considered unhealthy as they tend to deposit in arteries forming plaque, i.e., atherosclerosis. It can lead to high blood pressure, rupture of arteries, or heart attack. Saturated fatty acids are more common in animal fats, while unsaturated fatty acids, considered healthier, are more common in vegetable oils.
It has been observed that people from Alaska eat more unsaturated fats and have a low occurrence of atherosclerosis and heart attack. Fats in Alaskan food is primarily from fish, such as salmon, tuna, and herring, as shown in Figure $4$. Vegetable oils have higher contents of $\omega$-6 fatty acids, such as linoleic acid and arachidonic acid. Fish oil mainly contains $\omega$-3 fatty acids, such as linolenic acid, eicosapentaenoic acid, and docosahexaenoic acid.
Prostaglandins
Prostaglandins are lipids containing 20 $\ce{C's}$ that are derived from arachidonic acid (20:4(5, 8, 11, 14)) and act like hormones. They are found in small concentrations in almost all body tissues and cells. Their activities include blood pressure regulation, blood clotting, inflammation, gastric secretions, kidney functions, and reproductive activities. They are usually short-lived and produced locally where they perform their function.
Prostaglandins synthesis
Prostaglandins synthesis starts from arachidonic acid using enzymes like cyclooxygenase (COX-1 and COX-2) and prostaglandin synthase, like PGE-synthase and PGF-synthase, as illustrated in Figure $5$. The prostaglandins have a five-membered ring and two side chains. The prostaglandin PGG2 is synthesized from arachidonate and quickly converts to PGH2. PGH2 regulates constriction and dilation of blood vessels and stimulates platelet aggregation. PGH2 converts to PGE2 and PGF2 which induce labor by stimulating uterine contraction. PGF2 also lowers blood pressure by relaxing smooth muscle cells in blood vessels. PGH2 is also a precursor of several other Prostaglandins and related biomolecules.
Prostaglandins names
The names of prostaglandins are abbreviated as three letters followed by a subscript, where the first two letters, i.e., PG, stands for prostaglandin, the 3rd letter indicates the substitution pattern, and the subscript tells the number of $\ce{C=C}$ bonds in the side chains. For example, the 3rd letter G has endoperoxide in the ring and a hydroperoxide group at $\ce{C's}$ 15, H has endoperoxide in the ring, and a hydroxyl group at $\ce{C's}$ 15, E has a $\beta$-hydroxyketone, and F has a 1,3-diol groups in the ring. The subscript 2 shows two $\ce{C=C}$ bonds in the side chains. The letter E in PGE2 also indicates that it is soluble in ether and F in PGF2 indicates it is soluble in phosphate buffer.
Nonsteroidal anti-inflammatory drugs (NSAIDs) action on prostaglandin synthesis
When tissues are injured, arachidonic acid is converted into prostaglandins that cause inflammation and pain. Nonsteroidal anti-inflammatory drugs, like aspirin, ibuprofen, and naproxen, shown below, inhibit the conversion of arachidonic acid into prostaglandins. Ibuprofen and naproxen inhibit the cyclooxygenase (COX) enzyme by binding with it. Aspirin deactivates the cyclooxygenase (COX) by transferring an acetyl group to the hydroxyl group of the active site of the cyclooxygenase. Care must be taken using NSAIDs as their long-term use can cause gastrointestinal, liver, or kidney damage.
Waxes
Lipid waxes are esters of one long-chain fatty acid with unbranched long-chain alcohol containing 14 to 30 $\ce{C's}$ per chain. For example, myricyl palmitate found in beeswax can be obtained by the esterification of palmitic acid and myricyl alcohol:
$\ce{CH3(CH2)14{-}\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-OH +HO{-}(CH2)29CH3 -> CH3(CH2)14{-}\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O{-}(CH2)29CH3 + H2O}\nonumber$
Functions of waxes
Waxes form a waterproof coating on plants' leaves, fruits, and stems to minimize water evaporation and protect against parasites. Examples include palm trees and jojoba wax from jojoba bushes. The wax coating on animals' skin, fur, and feathers protects water birds from water wetting and rainwater and shelters them from rainwater. Bees use wax to construct protection and housing for honey, as illustrated below. Sperm whales use spermaceti wax in their heads as a part of their eco analysis, i.e., navigation system.
Beeswax and carnauba wax protective coatings on cars, furniture, floors, etc. Jojoba wax and spermaceti are used to make candles and cosmetics. A mixture of waxes obtained from wool is used in lotions for face and skin softening.
Petroleum waxes and earwax
Petroleum-derived waxes or paraffin are also called waxes, but they are not esters but mixtures of long-chain hydrocarbons. Earwax is a mixture of glycerol esters, cholesterol, phospholipids, etc., from dead skin and secretions of cerumen glands. Like other lipid waxes, it protects the ear canal against bacteria, fungi, and water. Excess earwax can cause blockage in the ear canal and hearing loss. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.01%3A_What_are_lipids.txt |
Learning Objectives
• Define glycerolipids and triglycerides and understand their structures.
• Understand the difference in composition and properties of fats and oils and their primary role as energy storage molecules.
• Understand the chemical properties of fats and oils, including hydrolysis, saponification, and partial or complete hydrogenation.
What is a glycerolipid?
Glycerolipids have two components: propane-1,2,3-triol, also called glycerol, and one, two, or three fatty acids. Recall that a condensation reaction between an alcohol and a carboxylic acid forms an ester by eliminating a water molecule. The following example shows the reaction of glycerol with three molecules of stearic acid, creating a triester.
A mono-, di-, or tri-ester of glycerol is called glycerolipid.
The name of a glycerolipid begins with glyceryl and is followed by carboxylate. For example, the triester of glycerol shown above is glyceryl tristearate, which is commonly known as tristearin.
Glycerolipids that are triesters of glycerol with three fatty acids are called triacylglycerol or triglycerides. For example, glyceryl tristearate shown above is a triglyceride.
Triglycerides may be esters of three molecules of the same fatty acids, e.g., tristearin. Still, often the triglycerides found in nature are esters of two or three different fatty acids. Fatty acids usually found in fats and vegetable oils include lauric, myristic, palmitic, stearic, oleic, linoleic, and linolenic acids. An example of a mixed triglyceride of stearic, oleic, and palmitic acid is shown below.
Fats and oils
Fats and oils are triglycerides used as energy storage molecules in animals and plants.
Energy storage is essential for hibernating animals that live in icy environments. They have plenty of food available during summer but no food and below-freezing temperatures in winter. They eat plants, seeds, and nuts and accumulate fat that becomes their energy source when hibernating in winter. For example, the polar bear shown on the right can accumulate up to 14 kg of fat per week in summer when the food is plenty and uses it as an energy source when the animal hibernates for 4 to 7 months in winter.
Although glycogen is a quick energy source, fats and oils are more energy dense for two reasons:
• first, fats and oils are in a more reduced form than glycogen and release more energy per mass upon oxidation, and
• second, glycogen is hydrophilic and absorbs water, increasing its mass.
Physical properties of fats and oils
Pure fats and oils are colorless, odorless, and tasteless. The color, taste, and odor associated with butter -an animal fat, and olive oil -a vegetable oil, is due to a small number of other substances mixed with the triglycerides.
Fats come from animal sources, such as meat, whole milk, lard, butter, and cheese. Vegetable oils come from plant sources, e.g., soybeans, peanuts, sunflowers, etc.
Fats are usually solid at room temperature because they have a higher proportion of long-chain unsaturated fatty acids with higher melting points, e.g., lard and butter shown in Figure \(1\). Vegetable oils are usually liquid at room temperature because they contain more unsaturated or short-chain saturated fatty acids.
For example, canola oil, safflower oil, flaxseed oil, sunflower oil, corn oil, olive oil, soyabean oil, peanut oil, and cottonseed oil are vegetable oils that are liquid at room temperature and have a higher proportion of unsaturated fatty acids as shown in Figure \(2\). Palm and coconut oils are vegetable oils with a higher balance of saturated fatty acids. Still, they are liquid at room temperature because they have short-chain saturated fatty acids. For example, coconut mainly comprises lauric acid (12:0), a short-chain fatty acid.
Chemical properties of fats and oils
Three important reactions of triglycerides are hydrolysis by water in the presence of acid or lipase enzyme, saponification, i.e., base-promoted hydrolysis, and hydrogenation of double bonds in the fatty acids.
Hydrolysis
Easters are hydrolyzed (split) by water in the presence of an acid catalyst into alcohol and carboxylic acid. The same reaction happens with triglycerides, i.e., fats and vegetable oils—for example, tripalmitin hydrolyzes into glycerol and three palmitic acid molecules, as shown below.
Lipase enzymes do the same reaction during the digestion of triglycerides, e.g., trilaurin is hydrolyzed by lipase to glycerol and three molecules of lauric acid, as shown in the reaction below.
The digestion of lipids starts in the mouth with lingual lipases secreted by glands in the tongue and continues in the stomach with lingual and gastric lipases. Glycerol in the hydrolysis product is soluble in water, but fatty acids are not. Fatty acids are emulsified in the stomach and later mixed with bile and pancreatic juice to continue the process of digestion.
Saponification
Saponification is base-promoted hydrolysis of triglycerides that produces glycerol and salts of the fatty acids, a soap, as shown below for the case of saponification of tripalmitin.
Different varieties of soaps are shown in the figure on the right. Sodium salts of fatty acids make hard soaps, while potassium salts make soft soaps. Three of the most common soaps are sodium stearate, oleate, and linoleate. Salts of saturated fatty acids make rigid, and those of polyunsaturated fatty acids make soft soap. Perfumes are added for scent and dyes for color, sand is added in scouring soap, and the air is blown into the soap to make it float on water.
Hydrogenation
Alkenes add \(\ce{H2}\) and covert to alkane in the presence of \(\ce{Ni}\), \(\ce{Pt}\), or \(\ce{Pd}\) catalyst. The same reaction happens with the \(\ce{C=C}\) in unsaturated fatty acids found in triglycerides. The conversion of unsaturated fatty acids to saturated fatty acids increases the melting point of triglycerides. Therefore, vegetable oils with more unsaturated fatty acids become semisolid or solid after partial or complete hydrogenation, as illustrated by the reaction in Figure \(3\). Hydrogenation of vegetable oils is a commercial process to convert vegetable oils to semisolid products like margarine and shortening.
The driving force for the industrial process of hydrogenation of vegetable oils is that margarine or shortenings are used as cheaper alternatives to butter with a longer shelf-life. Complete hydrogenation of vegetable oils converts them into hard margarine. Often partial hydrogenation is performed as the partially hydrogenated product is semisolid and mimics butter better than hard shortening. Dyes and flavors are mixed with it to mimic butter's color, taste, and odor.
Unwanted trans-fatty acids are by-products of the hydrogenation of vegetable oils.
Like all other processed food products, hydrogenated vegetable oil products are associated with health problems. Partial hydrogenation converts some of the cis-\(\ce{C=C}\) bonds to trans-\(\ce{C=C}\) bonds, as illustrated in Figure \(3\). Trans-\(\ce{C=C}\) bonds are rare in some fatty acids. Research reports indicate that trans-fatty acids affect blood cholesterol like unsaturated fatty acids. Research reports also indicated that trans-fatty acids increase the level of low-density lipids (LDL), considered bad cholesterol, and decrease the level of high-density lipids (HDL), regarded as good cholesterol. Health organizations force the manufacturers to show the trans-fatty acid contents on the product labels. They are trying to educate the consumers about trans-fatty acids, as shown in Figure \(4\) from U.S. Food and Drug Administration. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.03%3A_Glycerolipids.txt |
Learning Objectives
• Define glycerophospholipids and understand the structural features that allow them to organize into lipid bilayers.
• Understand the classification of glycerophospholipids and their role in infant respiratory distress syndrome.
What are glycerophospholipids?
Glycerophospholipids are triesters of propane-1,2,3-triol (glycerol): 1st and 2nd ester bonds with fatty acids, and the 3rd ester bond with phosphate, as illustrated in Figure \(1\). The phosphate group is often a diester: an ester with the primary alcohol of glycerol and 2nd ester bond with a small molecule, such as choline, ethanolamine, serine, or inositol.
Phospholipids have two hydrophobic tails, which are the hydrocarbon groups of two fatty acids and a polar head comprising of ester groups with fatty acids, phosphate having a -ve charge, and often also have a +ve charge on the nitrogen of the small molecule attached with the phosphite, as shown in the figure on the right. The 1st fatty acid attached to the primary alcohol group is usually a saturated fatty acid, and the 2nd fatty acid attached to the secondary alcohol group of glycerol is usually an unsaturated fatty acid.
Classes of glycerophospholipids
The glycerol esterified with two fatty acids and one phosphoric acid is also called phosphatidyl. Phosphatidyls with phosphate esterified with choline are called phosphatidylcholines (lecithin). Phosphatidyl esterified with ethanolamine, or sometimes with serine, are called phosphatidylethanolamines or phosphatidylserine (common name cephalins). Another category is phosphatidylinositols with inositol-esterified phosphate, as shown in Figure \(2\).
Lipid bilayer
Recall the general solubility rule "like dissolves like". Oil droplets that are nonpolar floating on the surface of the water, which is polar, ultimately coalesce, making bigger droplets. Glycerophospholipids are amphiphilic, having an overall cylindrical shape with nonpolar tails of fatty acid chains and polar heads of ester groups on the glycerol part and ionic groups on phosphate and amine groups. Glycerophospholipids form a bilayer in the watery environment of living things based on the principle "like dissolves like", as illustrated in Figure \(3\).
The hydrophobic tails of fatty acids form the interior of the bilayer, and the hydrophilic polar heads make the outer layers in contact with water. The molecules are in a fluid (dynamic) state due to thermal energy (Figure \(3\) middle) but held together by intermolecular interactions: dipole-dipole interaction in the polar heads and London dispersion forces in the hydrophobic tails. Saturated fatty acids pack tightly together, making a rigid bilayer. However, adding unsaturated fatty acids with kinks makes the packaging lose, resulting in a more fluid bilayer, as illustrated in Figure \(3\) right.
The glycerophospholipid bilayer is a significant component of the cell membrane, the membranes around the nucleus, and some cell organelles. The hydrophobic nature of the interior of the bilayer becomes a barrier to the movement of ions and water in and out of the cell. Still, the bilayer's fluid nature allows the diffusion process to move some particles across the membrane. Other components of the cell membrane, such as cholesterol, adjust the rigidity, and proteins control the movement of particles in and out of the cell as needed. The structure of cell membranes is described in a later section.
Snake venom
Snakes inject their venom through fangs when they bite their prey, as shown in the figure on the right. The venom of eastern diamondback rattlesnakes and Indian cobra contains phospholipase enzymes that catalyze the hydrolysis of fatty acid on the secondary hydroxyl group of glycerol. The resulting phospholipids with one less carboxylate group, called lysophospholipids, is no more cylindrical molecule and do not fit correctly in the lipid-bilayer, causing the red blood cell membrane to rupture. This poses a significant health risk and may kill humans. The structure of the lysophospholipid example is shown in the figure on the left.
Infant respiratory distress syndrome (IRDS)
In the breathing process, the exchange of \(\ce{O2}\) and \(\ce{CO2}\) takes place in air sacs called alveoli within the lungs, as illustrated in Figure \(4\). Pulmonary surfactants, composed of a mixture of lecithin and sphingomyelin lipids, are released into the lungs of newborn babies that reduce the surface tension and ease the inflation of alveoli. Lecithins are mixtures of glycerophospholipids, including phosphatidylcholine, phosphatidylethanolamine, phosphatidylinositol, phosphatidylserine (shown in Figure \(2\)) and phosphatidic acid shown in the figure on the right.
In the cases of premature birth, before 28 weeks of gestation, the level of the surfactant is insufficient, causing the air sacs to collapse and have to reopen with each breathing, a condition called infant respiratory distress syndrome. This makes breathing difficult, may damage the alveoli, and lead to hypoxia and acidosis due to less oxygen inhalation and less carbon dioxide exhalation.
The lecithin to sphingomyelin ratio in mature fetal lungs is ~2.5, the ratio 2.4 to 1.6 poses a low risk, and less than 1.5 poses a high risk of infant respiratory distress syndrome. Treatment includes steroids given to the mother or infant to assist the development of lungs, the application of surfactants, and the use of a ventilator to help breathing. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.04%3A_Glycerophospholipids.txt |
Learning Objectives
• Define and understand the structure of sphingolipids, their subclasses, and their role in the nervous system.
• Understand the role of sphingolipids in multiple sclerosis.
Sphingolipids and their subclasses
Sphingolipids are a class of lipids having a sphingoid base that is a set of aliphatic amino alcohols, including sphingosine, shown in Figure \(1\). When the amino (\(\ce{-NH2}\) group is attached to a fatty acid by an amide bond, it is called ceramide. When the primary alcohol of ceramide is linked to a phosphorylcholine or phosphorylethanolamine group, it is called sphingomyelin. When the primary alcohol of ceramide is bound to glucose or galactose by a glycosidic bond, it is called cerebroside. When a glycosidic bond connects the primary alcohol of ceramide to an oligosaccharide with one or more sialic acids, it is called ganglioside.
Figure \(2\) illustrates two examples of sphingomyelin. Like glycerophospholipids, these sphingolipids are an essential component of the lipid bilayer. Mainly, they are abundant in the brain and nerves. They are abundant in the white matter of myelin sheath, i.e., a coating surrounding the nerve cells. Myelin sheath increases the speed of nerve impulses and is essential in protecting nerve cells, signal transduction, and cell recognition.
Multiple sclerosis
Multiple sclerosis is a nervous system disease in which the myelin sheaths wrapped around axons of nerve cells are damaged, as illustrated in Figure \(3\). Symptoms are related to the retardation of signal conduction by the nerves that, in turn, reduces sensation, coordination, movement, cognition, muscle weakness, blindness, and other functions involving nerves. The severity of the effects depends on the amount of damage. Studies indicate that vitamin D may lessen the severity of the disease. Nearly 1 million people in the US and about 2.8 million worldwide have multiple sclerosis.
6.06: Saccharolipids polyketides and prenols
Learning Objectives
• Define and understand the basic structural features of saccharolipids, polyketides, and prenols.
• Define mono-, sesqui-, di-, and tri-terpenes and recognize isoprene units in terpenes, terpenoids, and steroids.
Saccharolipids
Saccharolipids are fatty acids linked to a saccharide (carbohydrate) backbone by linkages other than glycosidic linkages. They are compatible with membrane bilayers. For example, saccharolipid lipid A found in E. Coli is shown in the figure on the right.
Polyketides
Polyketides are a broad class of natural products derived from $\beta$-polyketones, i.e., compounds containing $\ce{[-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-CH2{-}]_{n}}$ repeat units in their backbone or their reduced forms, such as $\ce{[-\!\!\!\!\!{\overset{\overset{\huge\enspace\enspace{OH}}|}{C}}\!\!\!\!\!-CH2{-}]_{n}}$, and $\ce{[-CH2-CH2{-}]_{n}}$. Polyketides are important in the pharmaceutical industry. They are used as antimicrobial, antiparasitic, and anticancer agents; some are toxins. About 20% of the top-selling medicines are polyketides. Some examples of medically important polyketide products are shown below.
Prenols
Prenols are synthesized from five-carbon isoprene precursors: isopentenyl pyrophosphate (IPP) and dimethylallyl pyrophosphate (DMAPP), as shown in Figure $1$ (left). The five-carbon skeleton of isoprene units in these products can often be easily distinguished, as shown in by different color sections of some examples in Figure $1$ (right). Since they contain a multiple of five $\ce{C's}$ in their skeleton, they are also classified as monoterpens (10 $\ce{C's}$), sesquiterpenes (15 $\ce{C's}$), diterpenes (20 $\ce{C's}$), triterpenes ((30 $\ce{C's}$), and tetraterpenes (40 $\ce{C's}$). Terpenoids are modified terpenes that contain additional functional groups, usually oxygen functional groups as those shown in $1$ (right).
Steroids -other lipids described later are derived from triterpene squalene, as illustrated in Figure $2$. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.05%3A_Sphingolipids.txt |
Learning Objectives
• Understand the core structure of steroids and sterols.
• Understand the structural features of cholesterol, its production, consumption, transport, and medical problems.
• Understand bile acids' structure features, their functions, and medical problems.
• Understand the structural features of steroid hormones, functions, abuses, and medical problems.
What are steroids?
Steroids are biologically active compounds that have a core structure composed of four fused rings in a specific configuration: three six-membered rings designated A, B, and C and one five-membered ring D, as Figure \(1\) right. Gonane has this core structure of steroids. Other steroids have gonane skeletons with some groups attached. For example, cholestane has methyl groups at positions 10 and 13 and an aliphatic chain at position 17. Cholesterol has the structure of cholestane. It is a biomarker found in rocks and petroleum deposits.
What are sterols?
Sterols are organic compounds derived from gonane with \(\ce{H}\) #3 replaced with an alcohol (\(\ce{-OH}\)) group. The sterols are a sub-class of steroids. The simplest sterol is the alcohol gonane shown in the figure on the right. Other sterols have other groups attached to the gonane structure.
Cholesterol
Cholesterol is the most abundant sterol found in animals. It is a structural component of animal cell membranes, a precursor for steroid hormones, bile acids, and vitamin D. Cholesterol has the \(\ce{C}\) skeleton of cholestane, with an \(\ce{-OH}\) group at position #3 and a \(\ce{C=C}\) bond at position 5 and 6, as shown in Figure \(2\).
Cholesterol sources and consumption
Cholesterol is biosynthesized from lanosterol, which, in turn, is synthesized from squalene, as described in the previous section. Cholesterol is synthesized in the liver using fats, proteins, and carbohydrates as raw materials. An average human weighing 68 kg synthesizes about 1000 mg of cholesterol daily. Food is also a source of cholesterol intake, as shown in the figure on the right. Typically about 307 mg of cholesterol intake is through food per day in the US. The liver reduces cholesterol production to adjust for the intake from food sources.
Cholesterol is amphiphilic like phospholipids and sphingolipids, with \(\ce{-OH}\) as a polar head and cholestane structure as a hydrophobic tail. The cholesterol molecule is stiff due to restricted movements around \(\ce{C-C}\) bonds locked in the cyclic structure.
Cholesterol is used in the cell membrane to impart rigidity to the ring -the higher the proportion of cholesterol in the cell membrane, the more rigid the membrane.
Usually, cholesterol is 30% of the cell membrane. It is involved in signaling and is found in the brain, plasma membrane, and myelin sheath of nerve cells. Cholesterol is a precursor to vitamin D and steroid hormones, including corticosteroids, such as cortisol and aldosterone, and sex hormones, such as progesterone, estrogens, and testosterone.
The liver recycles the excess cholesterol by converting it into bile acids stored in the gallbladder and excreted into the digestive tract. Bile acids emulsify fats in the food, which helps digest the food. About 50% of the excreted cholesterol is re-absorbed into the bloodstream by the small intestine.
Transport of cholesterol
Cholesterol is insoluble in water like other lipids. It is transported by the blood, which is water rich medium. Like soap carries greases in water during washing, lipoprotein transport glycerolipids and cholesterol in the body. The lipoproteins exist as particles suspended in blood. Glycerides and cholesteryl esters are in the core, and phospholipids, proteins (apolipoprotein), and cholesterol form the outer layer having their hydrophobic parts oriented outwards in contact with water, as illustrated in Figure \(3\).
Lipoproteins are classified based on their density, as shown in the figure on the right (Copyright: Modified from İnfoCan at Turkish Wikipedia., FAL, via Wikimedia Commons, Public domain). Their density is based on the proportion of proteins in them -the higher the proportion of proteins, the higher the density. They also have roles in the transportation process of cholesterol and lipids:
• Chylomicrons – Transport triglycerides from the small intestine to the liver.
• Very low-density lipoproteins (VLDL) - transport newly synthesized triglycerides from the liver to the fatty tissues.
• Intermediate-density lipoproteins (IDL) are between VLDL and LDL in density. They are not usually seen in the blood.
• Low-density lipoproteins (LDL) - carry cholesterol from the liver to other tissues. They are also sometimes referred to as "bad cholesterol."
• High-density lipoproteins (HDL) - collect cholesterol from other tissues and return it to the liver. They are sometimes referred to as "good cholesterol."
Atherosclerosis and heart attack
Excessive fats and cholesterol in the body are associated with diabetes; cancers of the breast, pancreas, and colon; heart attack; and stroke. Lipoproteins transport cholesterol and fats in the body. LDLs transport cholesterol from the liver to the cells where they are needed, and HDLs transport surplus cholesterol back to the liver, where it is converted to other steroids and bile acids. If the LDL level is too high, some cholesterol is deposited in the artery as plagues, as shown in Figure \(4\). That is why LDLs are sometimes called bad cholesterol. The plaque causes inflammation and death of the tissue in that area and restricts blood flow. It may lead to a heart attack if the blood flow to the heart is inadequate or a stroke if the blood flow to the brain is insufficient.
Since HDLs transport the surplus cholesterol back to the liver for recycling and avoid plaque formation, HDLs are also called good cholesterol. The HDL levels must be sufficiently high to prevent a heart attack. A lipid panel test is prescribed to assess the risk of a heart attack. It measures the levels of total cholesterol, LDLs, HDLs, and triglycerides in the blood. The normal levels are the following.
Total cholesterol About 150 mg/dL, less is better
LDL ("bad") cholesterol About 100 mg/dL, less is better
HDL ("good") cholesterol At least 40 mg/dL in men and 50 mg/dL in women; higher is better
Triglycerides Less than 150 mg/dL, less is better
Bile salts
Bile salts are synthesized from cholesterol in the liver. First, additional \(\ce{-OH}\) groups are incorporated, and the alkyl side chain is shortened and oxidized to a carboxylic acid (\(\ce{-COOH}\)) group. Then the \(\ce{-COOH}\) group is esterified with either glycine or taurine that imparts a negative charge group, as illustrated in Figure \(5\).
Bile salts act like soaps with an ionic head and the remaining hydrophobic part. They are stored in the gallbladder and released into the small intestine, where they emulsify fats, i.e., convert the larger fat drops into smaller droplets, just like greases and dirt are emulsified by soaps. The smaller droplets have a larger surface area, increasing the fats' reaction rate with the pancreatic lipase enzymes, as illustrated in Figure \(6\).
Since bile salts are excreted with the feces, they remove cholesterol in two ways: 1) they are themselves the products of cholesterol, and 2) they dissolve cholesterol in the form of bile salt-cholesterol particles that are eliminated.
Gallstones and jaundice
When a large amount of cholesterol accumulates in the gallbladder, it turns into solid particles, i.e., gallstones. The gallstones are almost 100% cholesterol with a small amount of calcium salts, glycerophospholipids, and fatty acids. Small gallstones pass through the bile duct and enter the duodenum, i.e., the initial part of the small intestine, and are excreted. However, large gallstones may get stuck in the bile duct and cause pain, as illustrated in Figure \(7\). If a gallstone block the bile duct, the bile can not be excreted. Then the bile pigment, known as bilirubin, will back up into the liver and be passed via blood. It causes jaundice which gives a yellow color to the skin and whiteness to the eyes.
Steroid hormones
Steroid hormones have structural similarities with cholesterol and are derived from it. Two significant subclasses of steroid hormones are sex and adrenocortical hormones, described in the following sections.
Sex hormones
The male sex hormones are called androgens which include testosterone and androsterone. The female sex hormones are estrogens, e.g., estradiol and estrone, and progestins, e.g., progesterone. These five of the most essential sex hormones are shown in the figure below. All five are present in males and females, but androgens are produced more in males, and estrogens and progestins are produced more in females.
Functions of sex hormones
Testosterone and androsterone are the most potent male sex hormones. They control the development of male secondary sex characteristics, which include the growth of muscles, facial hair, maturation of sex organs, and sperms. Estrogens and progestins control the development of female secondary sex characteristics. Estrogens increase the size of the uterus, deposit fat in the breasts, and broaden the pelvis of females. Progestins prepare the uterus for nurturing the fertilized egg. If the egg is not fertilized, the levels of estrogens and progestins drop sharply, and menstruation begins.
Birth control
The release of estrogens and progestins during pregnancy prevents ovulation. This is mimicked by synthetic estrogens, such as ethynyl estradiol, and synthetic progestins, such as norethindrone, shown below, used in birth control formulations. As with the use of any steroid hormones, there are risks associated with synthetic hormones for birth control, including weight gain and a greater chance of forming blood clots.
Anabolic steroids
One physiological function of testosterone is increasing muscle mass and decreasing body fat. Several synthetic equivalents, called anabolic steroids, have been developed for use as appearance and performance-enhancing drugs, as shown in the figure below. Although there are some benefits of anabolic steroids, their harmful effects are numerous and outweigh their benefits. They may not produce a euphoric high but develop substance use disorder. They can cause early heart attacks, strokes, kidney failure, liver tumors, make pimples pop up, hair fallout, grow breasts, and reduced testicles in males, grow beards in females, and cause extreme mood changes that may lead to committing suicide. Some of these harmful effects are related to the changes in cholesterol levels, causing an increase in low-density lipoprotein (bad cholesterol), a decrease in high-density lipoprotein (good cholesterol), high blood pressure, and liver damage. The harmful effects are long-lasting and may be irreversible.
Adrenocortical hormones
Adrenocortical hormones are produced by adrenal glands on the top of the kidneys, as shown in the figure on the right (Copyright; Alan Hoofring (Illustrator), Public domain, via Wikimedia Commons). There are two major subclasses: mineralocorticoids which regulate \(\ce{Na^{+}}\) and \(\ce{K^{+}}\) ions, and glucocorticoids which control carbohydrate metabolism.
Mineralocorticoid
Aldosterone is an important mineralocorticoid that enhances the reabsorption of \(\ce{Na^{+}}\) and \(\ce{K^{+}}\) ions in the kidney tubules. The control of \(\ce{Na^{+}}\) and \(\ce{K^{+}}\) concentrations controls water absorption and swelling of tissues.
glucocorticoid
Cortisone is a glucocorticoid that increases blood glucose levels and glycogen synthesis in the liver. Cortisol is another glucocorticoid released under stress to increase blood glucose levels at the expense of the metabolism of carbohydrates, fats, and proteins.
These hormones or their synthetic derivative, such as prednisone, shown below, are used to reduce inflammation, asthma, and arthritis. As with the use of steroids in general, health problems can result if these steroids are used for the long term. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.07%3A_Sterols.txt |
Learning Objectives
• Define and understand the composition of the cell membrane and how the composition adjusts the fluidity fo the membrane.
• Understand three modes of transport through the cell membrane: diffusion, facilitated transport, and active transport.
What is a cell membrane?
The cell membrane, also known as the plasma membrane, cytoplasmic membrane, and plasmalemma, is a lipid bilayer with proteins dispersed in it that separates the cell interior from the extracellular space, as illustrated in the figure on the right (Copyright; National Human Genome Research Institute, Public domain).
Composition of the cell membrane
A cell membrane is a complex structure with several components, as shown in Figure \(1\). and described here.
• Phospholipid bilayer that has polar heads on the outside in contact with water and nonpolar tails inside the bilayer. Unsaturated fatty acids in the lipid hydrophobic tails increase the membrane fluidity -the more proportion of unsaturated fatty acids, the higher the fluidity.
• Cholesterol interspersed between phospholipids controls the rigidity of the membrane -the more the proportion of cholesterol, the more rigid the membrane.
• Two types of proteins: integral proteins that span the membrane and serve as transporters of species, and peripheral proteins that are loosely attached to the outer side of the membrane that act as enzymes to facilitate the interaction with the cell's environment.
• Glycoproteins and glycolipids have carbohydrate oligomers attached to the outer lipid layer and serve the purpose of cell-to-cell recognition.
The cell membrane controls the movement of substances in and out of the cells and organelles. It is selectively permeable to ions and organic molecules. It plays a role in cell adhesion, cell signaling, and attachment surface for the cytoskeleton to shape the cells and attach to the extracellular matrix to hold them together in tissues.
Transport through the cell membrane
The cell membrane is a partition between intracellular and extracellular spaces. Still, some substances needed by the cell need to enter, and some products or wastes need to exit the cell. The cell membrane allows a selective movement of substances in and out of the cell in several ways.
Diffusion (passive) transport
The molecules in the lipid bilayer are vibrating due to thermal energy. Therefore, some molecules, such as \(\ce{O2}\), \(\ce{CO2}\), urea, water, etc., can move across the membrane from a higher concentration region to a lower concentration region through the process of diffusion, as illustrated in Figure \(2\) left.
Facilitated transport
Integral proteins form channels through which certain substances can diffuse more rapidly than by simple diffusion. The proteins have a channel size that matches the substance's size or changes the shape to adjust to the size of the substance that needs to be selectively transported through the facilitated transport, as illustrated in Figure \(2\) right. Particularly, water-soluble substances such as chloride ion (\(\ce{Cl^-}\)), bicarbonate ion \(\ce{HCO3^{-}}\), and glucose molecules do not move fast enough through simple diffusion and are transported by the facilitate transport process to meet the need of the cells.
Active transport
Sometimes substances need to be moved against the concentration gradient. It takes place at the expense of energy in the form of ATP, just like pumping substances at the cost of electricity in everyday life. For example, \(\ce{K^+}\) concentration is greater inside the cell, and that of \(\ce{Na^+}\) is greater outside the cell. In the conduction of nerve impulses and contraction of muscles, \(\ce{K^+}\) moves into the cell, and \(\ce{Na^+}\) moves out of the cell by active transport process, as illustrated in Figure \(3\). | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/06%3A_Lipids/6.08%3A_Cell_membrane.txt |
• 7.1: What are proteins?
Proteins are defined, and their basic classification and the importance of their structure in their functions are described.
• 7.2: Amino acids
Amino acids found in proteins, their classification, acid-base properties, and sources are described.
• 7.3: Primary structure of proteins
Peptide bond, disulfide bond, their nomenclature, and their characteristics, along with the primary structure of proteins and their importance, are described.
• 7.4: Secondary structure of proteins
Secondary structures of proteins, including alpha-helix, beta-pleated sheets, random coils, and triple helix, are described.
• 7.5: Tertiary structure of proteins
The tertiary structure of proteins and interaction, including disulfide bonds, salt bridges, coordinate covalent bonds, hydrogen bonding, and hydrophobic interaction, are described as responsible for forming and maintaining the tertiary structure of proteins.
• 7.6: Quaternary structure of proteins
The quaternary structure of proteins and its need are described.
• 7.7: Summary of protein structure levels
The structure levels of proteins are described and illustrated.
• 7.8: Protein misfolding and denaturation
Misfolding of proteins, medical problems associated with the misfolding, and denaturation of proteins are described.
• 7.9: Enzymes
Terminology related to enzymes, their nomenclature, and classification are described along with the models of enzyme action, factors that affect enzyme activity, and inhibitors that retard or destroy the enzyme activity.
07: Proteins
Learning Objectives
• Define proteins and understand their classification and the importance of their structure in their functions.
Proteins
Proteins are bio-polymers containing one or more polymer chains composed of amino acid monomers linked together by amide bonds, i.e., proteins are polyamide biochemicals.
Classification of proteins based on their functions
Proteins are considered the most abundant biochemicals that perform various functions in living things. The major types of proteins, classified according to their role in biochemical systems, are the following.
• Structural: Structural proteins are the primary structural materials in animals. For example, collagen constitutes tendons and cartilage, and keratin comprises hair, skin, wool, and nails.
• Contractile: These proteins are responsible for movements; e.g., myosin and actin are proteins responsible for the contraction of muscles.
• Transport: These proteins move the molecules in the body. For example, hemoglobin transports oxygen, and lipoproteins carry lipids.
• Storage: They store nutrients or essential chemicals, e.g., casein in milk and ovalbumin in eggs store nutrients, and ferritin stores iron in the spleen and liver.
• Hormones: These proteins regulate metabolism and the nervous system; e.g., insulin regulates blood glucose levels, and growth hormones regulate growth.
• Enzymes: They catalyze biochemical reactions in the body of living things, e.g., sucrase catalyzes the hydrolysis of sucrose, and digestive enzymes like trypsin catalyze the hydrolysis of proteins.
• Protection: These proteins protect living things, e.g., antibodies destroy harmful foreign substances, and fibrinogen coagulates blood when needed to avoid blood loss through injuries.
Importance of protein structure
Many proteins are present in living things' bodies to perform different functions. For example, the human body contains about 100,000 different proteins. The structure of proteins, i.e., the exact sequence of amino acids and how they fold and interact with other protein molecules in the physiological environment, is critically important for their functions. For example, hemoglobin in red blood cells is needed to transport oxygen. It comprises more than 800 amino acids arranged in a specific sequence. Sickle cell anemia is caused by a change of one out of more than 800 amino acids in hemoglobin. It causes red blood cells to change from regular rounded shapes to sickle shapes, as shown in the figure to the right, which can not function properly and causes other medical problems.
To understand the proteins, their functions, and their malfunction, it is essential to understand the structure and properties of amino acids and proteins described in the next section. Proteins have levels of structures, i.e., i) primary -the sequence of amino acids, ii) secondary -the folding of sections of the protein chains, iii) tertiary -the overall shape of the protein polymer, and iv) quaternary -a combination of more than one proteins in a unit which is described in a later section. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.01%3A_What_are_proteins.txt |
Learning Objectives
• Define and distinguish amino acids, $\alpha$-amino acids, and proteinogenic amino acids.
• Draw Fisher projections and assign D/L or R/S stereodescriptors to proteinogenic amino acids.
• Understand the classification of proteinogenic amino acids based on the characteristics of the side chain.
• Define isoelectric point and understand the ionization states of amino acids under physiological conditions.
• Define essential amino acids and their sources.
What are amino acids?
Amino acids are the building blocks of proteins, i.e., they are the monomers of proteins. Amino acids are organic compounds that contain both an amine ($\ce{-NH2}$) and a carboxylic acid ($\ce{-COOH}$) group in the same molecule. For example, alanine, shown on the right, is an amino acid. Proteins contain a subclass of amino acids called $\alpha$-Amino acids.
Alpha ($\alpha$)-amino acids
$\alpha$-Amino acids have an amine ($\ce{-NH2}$) on the $\alpha\ce{C}$ to $\ce{-COOH}$ group, i.e., both the amine ($\ce{-NH2}$) group and the carboxylic acid ($\ce{-COOH}$) group are attached to the same $\ce{C}$. For example, alanine is an $\alpha$-amino acid.
Proteinogenic amino acids
Proteinogenic amino acids are a subclass of $\alpha$-amino acids incorporated into proteins during biosynthesis. Twenty proteinogenic amino acids are usually present in proteins, and two additional are included in exceptional cases. This chapter's word "amino acid" refers to the 20 standard proteinogenic amino acids.
Configuration of $\alpha\ce{C}$ of an $\alpha$-amino acids
the $\alpha\ce{C}$ of $\alpha$-amino acid have four different groups attached to it: amine ($\ce{-NH2}$), carboxylic acid (($\ce{-COOH}$), hydrogen ($\ce{-H}$), and alkyl side chain ($\ce{-R}$, as shown in Figure $1$. There is one exception "glycine" that has two $\ce{H's}$ at the $\alpha\ce{C}$
Three groups, i.e., $\ce{-NH2}$, ($\ce{-COOH}$, and $\ce{-H}$ are present in all $\alpha$-amino acids, and the fourth group, i.e., the side chain ($\ce{-R}$) varies in different $\alpha$-amino acids.
The $\ce{-COOH}$ is acidic (pKa 2-5) and the $\ce{-NH2}$ is basic (pKa ~10). The $\ce{-COOH}$ loses its proton and exists as a carboxylate ion ( $\ce{-COO^{-}}$; the $\ce{-NH2}$ gains a proton and exists as ammonium ion ( $\ce{-NH3^{+}}$ at physiological pH of 7.4. These groups ionize the same way if present in the side chain in addition to the $\alpha\ce{C}$.
Zwitterions
Compounds that have a positive charge on one atom and a negative charge on another atom in the same molecule are called zwitterions. Most of the $\alpha$-amino acids usually exist as zwitterions at physiological pH, as shown in Figure $1$.
A $\ce{C}$ with four different groups is a chiral center. $\alpha\ce{C}$ of $\alpha$-amino acids is a chair center at $\alpha\ce{C}$, with one exception, "glycine" that has two $\ce{H's}$ at the $\alpha\ce{C}$. $\alpha$-Amino acids are presented in Fisher projections with the $\ce{C}$-chain placed vertically and the $\ce{-NH2}$ and $\ce{-H}$ bonds shown horizontally, as shown in Figure $1$.
D- and L-configurations of $\alpha$-amino acids
The $\alpha$-amino acids that have their $\ce{-NH2}$ group placed to the left side of their Fisher projections are called D-amino acids, and those having $\ce{-NH2}$ group placed to the left side are called L-amino acids. The D- and the corresponding L-amino acids are enantiomers, as shown in the figure below for the case of alanine.
Proteinogenic amino acids are L-amino acids, with the exception of glycine, which is not chiral. In the R/S system, proteinogenic amino acids are (S) at the $\alpha\ce{C}$, with the exception of "cysteine" being (R) and glycine not chiral.
The $\alpha$-amino acids are known by their common names, which are also abbreviated in three letters and one letter, as shown in Figure $2$. For example, Alanine is abbreviated as Ala in three letters or as A in one letter form. The proteinogenic amino acids are classified based on the nature of the side chain.
Classification of proteinogenic amino acids
$\alpha$-Amino acids are classified based on the hydrophobicity of their side chain.
Hydrophobic and hydrophilic
Hydrophobic means "water fearing", i.e., molecules or entities that tend to repel water, not dissolve, or not wetted by water. Hydrophilic means "water-loving", i.e., molecules or entities that tend to mix with, dissolve, or be wetted by water.
Hydrocarbons and other non-polar compounds are hydrophobic compounds that do not dissolve in water. Polar or ionic compounds are usually soluble in water. Based on these criteria, amino acids are classified into the following classes, shown in Figure 7.2.2.
• Nonpolar amino acids with an aliphatic side chain are hydrophobic. Methionine having a thioether group in the side chain is non-polar and placed in this group.
• Nonpolar amino acids with an aromatic side chain are hydrophobic. Tyrosine has phenol as part of its side chain, but it is hydrophobic and placed in this group.
• Polar neutral amino acids with a polar neutral side chain are hydrophilic. This class of side chain contains alcohol ($\ce{-OH}$) or amide $\ce{-CONH2}$ groups in their side chain that hydrogen bond with water but do not ionize.
• Amino acids in a special class include cysteine, glycine, and proline.
• Speciality of cysteine is that its thiol {$ce{-SH}$ group is easily oxidized forming is a disulfide ($\ce{S-S}$ bond, which is the only covalent bond in protein besides the amide bonds. Cystein is classified as nonpolar and hydrophobic because {$ce{-SH}$ is a nonpolar group.
• Glycine is the only proteinogenic acid that has no chiral center. The $\ce{-H}$ side chain places it at a borderline between hydrophilic and hydrophobic categories, it is considered neutral.
• Proline is the only amino acid with a secondary $\alpha$-amine group. The side chain is a five member ring with $\ce{N}$ of $\alpha$-amine as part of the ring. It is classified as a hydrophilic. The ring structure puts restrictions on the allowed configurations when it is incorporated in proteins. It creates a bent or kink in the protein backbone structure.
• Acidic amino acids with an acidic side chain are hydrophilic. These amino acids has carboxylic acid ($\ce{-COOH}$) group in their side chain that ionize to anion $\ce{-COO^{-}}$ under physiological conditions.
• Basic amino acids with a basic side chain are hydrophilic. These amino acids have basic primary or secondary amine groups in their side chain that ionize to cation $\ce{-NH3^{+}}$ or $\ce{-NRH2^{+}}$. Histidine has an amine group that has pKa 6.04 and is not ionized at pH 7.4, but it is placed in this group as it ionizes at pH below 6.0.
Acid-base nature of $\alpha$-amino acids
The pKa is a measure of the strength of an acid, i.e., the lower the pKa stronger the acid. Amino acids have $\ce{-COOH}$ group that is acidic with pKa 2-3 and $\ce{-NH2}$ on adjacent $\ce{C}$ that is basic with pKa ~40. Conjugate acid of $\ce{-NH2}$, i.e., $\ce{-NH3^{+}}$ has pKa ~10. Acids lose their proton when they are in a medium with a pH higher than the pKa of the acid. For example $\ce{-COOH}$ exist as $\ce{-COO^{-}}$ in physiological medium with pH ~7.4. Bases gain protons in a medium with a pH lower than the pKa of their conjugate acid. For example, $\ce{-NH2}$ exist as $\ce{-NH3^{+}}$ in physiological medium with pH ~7.4. Amino acids exist as zwitterions, i.e., have both cation $\ce{-NH3^{+}}$ and anion $\ce{-COO^{-}}$ in the same molecule. Some amino acids have an additional acid or base group in their side chain that also ionizes depending on the pH of the medium.
The gain or loss of protons is an equilibrium process. In a strongly acidic medium, basic groups gain more protons than the protons lost by their acid groups. So, the amino acids have an overall positive charge in a strong acid medium. In a strongly basic medium basic groups gain fewer protons than those lost by their acid groups. So, the amino acids have an overall negative charge in a strongly basic medium. At a certain pH in the middle, an amino acid has an equal positive and negative charge and is neutral overall.
Isoelectric point (pI)
An amino acid's isoelectric point (pI) is the pH at which it has equal positive and negative charges and carries no net charge, i.e., it is neutral overall.
The pKa value of $\ce{-COOH}$ and $\ce{-NH3^{+}}$ on the $\alpha$-$\ce{C}$, pKa values of acidic or basic groups in the side chain, and pI values of 20 $\alpha$-amino acids found in proteins are listed in Table 1 below.
Table 1: Names, three-letters, and one-letter abbreviations, pKa values of alpha-carboxylic acid and ammonium groups, pKa values of acid groups in the side chain, and isoelectric point (pI) of 20 amino acids found in proteins. (Reference: D.R. Lide, Handbook of Chemistry and Physics, 72nd Edition, CRC Press, Boca Raton, FL, 1991)
Amino acid Three letters abbreviations One letter abbreviations pKa of $\alpha$$\ce{-COOH}$ pKa of $\alpha$$\ce{-NH3^{+}}$ pKa of side chain group Isoelectric point (pI)
Alanine Ala A 2.34 9.69 6.00
Arginine Arg R 2.17 9.04 12.48 10.76
Asparagine Asn N 2.02 8.80 5.41
Aspartic acid Asp D 1.88 9.60 3.65 2.77
Cysteine Cys C 1.96 10.28 8.18 5.07
Glutamic acid Glu E 2.19 9.67 4.25 3.22
Glutamine Gln Q 2.17 9.13 5.65
Glycine Gly G 2.34 9.60 5.97
Histidine His H 1.82 9.17 6.00 7.59
Isoleucine Ile I 2.36 9.60 6.02
Leucine Leu L 2.36 9.60 5.98
Lysine Lys K 2.18 8.95 10.53 9.74
Methionine Met M 2.28 9.21 5.74
Phenylalanine Phe F 1.83 9.13 5.48
Proline Pro P 1.99 10.60 6.30
Serine Ser S 2.21 9.15 5.68
Threonine Thr T 2.09 9.10 5.60
Tryptophan Trp W 2.83 9.39 5.89
Tyrosine Tyr Y 2.20 9.11 10.07 5.66
Valine Val V 2.32 9.62 5.96
Essential amino acids
Nine amino acids are essential because humans can not synthesize them fast enough to meet their demands.
The essential amino acids are valine, isoleucine, leucine, methionine, phenylalanine, tryptophan, threonine, histidine, and lysine.
The essential amino acids are obtained from foods. Foods from animal sources, e.g., eggs, milk, fish, meat, etc., are complete foods with all the essential amino acids. Foods from plant sources, e.g., wheat, rice, corn, etc., are usually deficient in one or more essential amino acids. So, vegetarians have to eat various vegetarian foods to obtain all the essential amino acids.
Deficiency of essential amino acids in different foods
• Wheat, rice, and oats are deficient in lysine.
• Corn is deficient in lysine and tryptophan.
• Soy is deficient in methionine.
• Beans are deficient in methionine and tryptophan.
• Peas and peanuts are deficient in methionine
• Almonds and walnuts are deficient in lysine, tryptophan
• Foods from animal sources, e.g., milk, eggs, meat, fish, etc., have all the essential amino acids. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.02%3A_Amino_acids.txt |
Learning Objectives
• Understand a peptide bond and disulfide, their nomenclature, and their characteristics.
• Define and write proteins' primary structure, importance, and related terminologies.
Peptide
An amine reacts with a carboxylic acid and makes an amide by eliminating water. Amino acids have both an amine and a carboxylic acid group. If the amine group of one amino acid reacts with the carboxylic acid of another, the two amino acids become bonded through an amide bond. For example, alanine and glycine become bonded through an amide bond, as illustrated below.
Peptide bond
An amide bond that links two amino acids is called a peptide bond or peptide linkage. For example, a peptide bond that links alanine and glycine is highlighted in a brown box in the above illustration.
A single amino acid is also called a monopeptide, and two amino acids linked by a peptide bond is called a dipeptide. Two amino acids, alanine, and glycine, are monopeptides; their product, alanylglycine, shown above, is a dipeptide. One amino acid in a dipeptide has free ammonium ($\ce{-NH3^{+}}$) group that can make a peptide with $\ce{-COO^{-}}$ of another amino acid. Similarly, one amino acid in a dipeptide has a free carboxylate ( $\ce{-COO^{-}}$) group that can make a peptide bond with $\ce{-NH3^{+}}$ group of another amino acid. Three amino acids linked by peptide bonds is called a tripeptide, four are called tetrapeptide, five are pentapeptide, six are hexapeptide, seven are heptapeptide, and so on. A tripeptide of glycine, histidine, and lysine is shown below.
Shorter chains of amino acids linked by peptide bonds are called peptides; longer ones are called polypeptides. Polypeptides and proteins are used interchangeably, but more than 50 amino acid chains are usually called proteins. Amino acids in peptides are usually called residues.
Structure of a peptide backbone
The backbone of a peptide chain is $\ce{-C-C-N{-}}$ where the middle $\ce{C}$ is the carbonly $\ce{C=O}$ and $\ce{C-N}$ is the peptide bond. The peptide bond has two resonance contributors, as shown below.
Due to the resonance, the peptide bond has about 40% double bond character. There is no free rotation around the peptide bond due to its double bond character. Therefore, the four groups around the peptide bond (shown in blue color in the above structure) exist in the same plane as in the case of alkenes. Free rotation is around the other $\sigma$-bonds in the peptide backbone. Therefore, the structure of the peptide backbone is like cards connected by a swivel at the opposite corners, as shown in Figure $1$.
The peptide bond's rigidity limits the peptide backbone's possible orientations, affecting its secondary and tertiary structure. The $\ce{N-H}$ groups can establish hydrogen bonds with the $\ce{C=O}$ groups within the same chain or between the neighboring chains that play an essential role in determining the secondary and tertiary structures of proteins that are described in a later section.
Disulfide bond
Amino acid cysteine has a thiol ($\ce{-SH}$) group that can easily be oxidized to disulfide ($\ce{-S-S{-}}$) bond or disulfide linkage linking two cysteines into a dimer called cystine, as illustrated below.
When a cysteine residue makes a disulfide bond with another cysteine residue in the same chain or another chain, it provides a covalent linkage that binds parts of the same chain or two different chains, as shown in Figure $2$. Examples of both types are found in the structure of insulin, which is composed of two polypeptides joined by sulfide linkage, as shown in Figure $2$.
Naming peptides
When amino acids combine by peptide bonds, one amino acid on one terminal has free ammonium ($\ce{-NH3^{+}}$) group, an amino acid on the other terminal has free carboxylate ($\ce{-COO^{-}}$) group, as highlighted in the structures of dipeptide and tripeptides shown above.
N-Terminus and C-terminus of a peptide
• An amino acid in a peptide that has free ammonium ($\ce{-NH3^{+}}$) group is called N-terminus. For example, glycine in the tripeptide shown above is N-terminus.
• An amino acid that has a free carboxylate ($\ce{-COO^{-}}$) group is called C-terminus. For example, lysine in the tripeptide shown above is C-terminus.
Amino acids in a peptide are written horizontally from left to right, where N-terminus is the leftmost amino acid, and C-terminus is the rightmost amino acid.
A peptide is named by listing the names of its constituent amino acids in a sequence from N-terminus to C-terminus, with the last syllable changed to yl, except for the C-terminus. For example, the dipeptide of alanine and glycine is alanylglycine, and the tripeptide of glycine, histidine, and lysine is glycylhistidyllysine.
Often three-letter abbreviations of the amino acids in a peptide are written in a sequence from N-terminus to C-terminus, separated by hyphens. For example, the dipeptide alanylglycine can be written as Ala-Gly, and the tripeptide glycylhistidyllysine as Gly-His-Lys. For polypeptides, one-letter abbreviations of the amino acid residues are usually written in a sequence from N-terminus to C-terminus. For example, dipeptide Ala-Gly is AG, and tripeptide Gly-His-Lys is GHL.
What is the primary structure of proteins?
Primary structure of proteins
The primary structure of peptides or proteins is the sequence of amino acids linked together by peptide bonds. For example, the primary structures of the dipeptide and tripeptides shown above are Ala-Gly and Gly-His-Lys.
The primary structure of a protein is shown as a sequence of amino acids written from the N-terminus to the C-terminus. When the sequence of amino acids is known, three-letter abbreviations are separated by hyphens, e.g., Gly-His-Lys. When the sequence of amino acids in a peptide is not known, the three-letter abbreviations of the constituent amino acid are listed, separated by commas. For example, Ala, Gly could mean Ala-Gly or Gly-Ala, which are different compounds with different properties related to each other as constitutional isomers. Similarly, Gly, His, Lys could mean any one of the following six constitutional isomers: Gly-His-Lys, Gly-Lys-His, His-Lys-Gly, His-Gly-Lys, Lys-Gly-His, or Lys-His-Gly.
The number of constitutional isomers increases exponentially as the number of amino acids in the peptide increases. Constitutional isomers of a polypeptide of n amino acids chosen from 20 amino acids commonly found in proteins are given by 20n. For example, a polypeptide containing 60 amino acids selected from 20 amino acids found in proteins may have 2060, i.e., 1078, which is an enormous number of possibilities. This analysis shows that the number of proteins that can be synthesized using 20 amino acids is enormously large. An analog is the entire English language composed of letters from 26 different alphabets.
When the primary structure of a polypeptide is modified, its function is affected. The extent of the effect depends on the number of amino acids replaced and their nature. For example, human insulin comprises two peptides: chain A of 21 amino acids and chain B of 30 amino acids, two chains joined by disulfide linkages, as shown in Figure 7.3.1. The amino acids at positions 8, 9, and 10 in chain A (-Thr-Ser-Ile-) and position 30 in chain B (-Thr) in human insulin are replaced with -Ala-Ser-Val- and -Ala, respectively, in bovine insulin, but the two perform the same function. Humans can use bovine insulin, though it is less effective in humans and sometimes causes an allergic reaction.
As shown below, vasopressin and oxytocin are two nonapeptides that differ in two amino acid residues at positions 3 and 8. Their cysteine residues form a disulfide bond, and the carboxylate group ($\ce{-COO^{-}}$) on C-terminus is converted to a primary amide (($\ce{-CONH2}$).
Both vasopressin and oxytocin are hormones the pituitary gland produces but have different functions. Vasopressin is an antidiuretic hormone that regulates blood pressure by adjusting the water reabsorbed by the kidneys. Oxytocin stimulates uterine contractions in labor.
Another example where a slight change in the primary structure of a protein alters its function significantly is sickle cell anemia. Hemoglobin in the red blood cell is responsible for carrying oxygen. Hemoglobin comprises four polypeptides, two alpha chains, and two beta chains containing 574 amino acid residues. A change of glutamic acid (a hydrophilic amino acid) with valine (a hydrophobic amino acid) in the sixth position of the two beta-chains changes the structure of hemoglobin so much that it causes the red blood cells to change from a rounded shape to a sickle body. Sickled cells do not function properly and block the blood flow -a medical condition called sickle cell anemia, illustrated in The Figure on the right.
Some examples and uses of peptides
Peptides have different functions in the body of living things. Some of them are commercially important also. For example, dipeptide Ala-Gly shown is a dietary supplement. Aspartame, 200 times sweeter than sucrose and used as a sugar substitute, is a methyl ester of a dipeptide Asp-Phe, shown below.
The tripeptide Gly-His-Lys shown before is a human copper-binding peptide with wound healing and skin remodeling activity. Met-enkephalin (Tyr-Gly-Gly-Phe-Met) is a pentapeptide related to pain signals in the body.
Vasopressin and oxytocin are two nonapeptide hormones produced by the pituitary gland. Vasopressin is an antidiuretic hormone that regulates blood pressure by adjusting the water reabsorbed by the kidneys. Oxytocin stimulates uterine contractions in labor.
Substance P (SP) is an undecapeptide (Arg-Pro-Lys-Pro-Gln-Gln-Phe-Phe-Gly-Leu-Met) that is a neuropeptide acting as a neurotransmitter and as a neuromodulation. Insulin is a combination of two polypeptides linked by disulfide linkage that regulates glucose in the blood. Malfunctioning of insulin causes diabetes. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.03%3A_Primary_structure_of_proteins.txt |
Learning Objectives
• Define the secondary structure of proteins and understand the structural features of major secondary structures, including $\alpha$-helix, $\beta$-pleated sheet, random coil, and triple helix structures.
The polypeptide backbone is composed of repeated $\ce{-C-C-N{-}}$ units where the side chain ($\ce{-R}$) is hanging on the first $\ce{C}$. The second carbon is the carbonyl ($\ce{C=O}$) group and the nitrogen is the amine ($\ce{-NH{-}}$) group in amide. The polymer chain acquires confirmation to achieve:
1. maximize hydrogen bonding between the $\ce{C=O}$ and $\ce{-NH{-}}$ of the same chain (intramolecular) or among the neighboring chains (intermolecular), and
2. minimize the steric-strain due to the $\ce{-R}$ groups.
This way, the polymer backbone acquires repetitive patterns in portions of the backbone.
Secondary structure of proteins
The secondary structure of proteins comprises organized regions of polypeptide backbone stabilized by hydrogen bonds between atoms.
The two common secondary structures encountered in proteins are ($\alpha$-helix and $\beta$-pleated sheet. The other portions of the polymer backbone that are regular but not repetitive are called random coils. Triple-helix is another common secondary structure found in collagen proteins in connective tissues. These secondary structures are illustrated in Figure $1$. below and described in the following sections.
$\alpha$-Helix
In the $\alpha$-helix portion, the polypeptide chain acquires a coil shape spiraling clockwise from N-terminus to C-terminus, held by hydrogen bonds between the carbonly groups and amine groups in the backbone and side chains protruding outwards, as illustrated in the model shown on the left
Each $\ce{N-H}$ is hydrogen bonded with a $\ce{C=O}$ group four units away, shown by hashed lines in the model. Each side chain ($\ce{-R}$) group on an amino acid reside is protruding outwards from the helix.
The $\alpha$-helix is represented by a right-handed spiral shape ribbon, as shown by the red-color ribbon in the structure of glucagon shown in Figure $2$. The side chain on each amino acid can be seen protruding out from the helix. The random coil portion of glucagon is shown by a blue-color tube. Since the side chains hang out in free space, short and long-side chain amino acids can easily be accommodated in an $\alpha$-helix without steric strain. Keratin, a fibrous protein of hair, fingernails, horns, and wool, is composed of a major portion of $\alpha$-helix.
$\beta$-Pleated sheet
The $\beta$-Pleated sheet is a portion of polypeptide chains in which sections of polypeptide chains are aligned parallel or antiparallel and held together by hydrogen bonds between the carbonyl groups and amine groups of the neighboring polypeptide chains. The $\beta$-Pleated sheet is usually represented as a ribbon with an arrowhead pointing toward the N-terminus, as shown on the right.
Since the polypeptide backbone is in a zig-zag shape, multiple chains running parallel to each other in a $\beta$-pleated sheet is a pleated shape as shown in Figure $3$. The two neighboring chains are parallel if both have N-terminus on the same side, as shown on the right, and antiparallel if the two adjacent chains have N-terminus on the opposite sides, as shown on the left. The neighboring chains may be different polypeptides or the same polypeptide chain with hairclip-like bents at the ends.
The alternate $\ce{C=O}$ and $\ce{N-H}$ on the backbone of neighboring chains point towards each other and hydrogen bond with each other, as shown in Figure $3$. The other set of alternate $\ce{C=O}$ and $\ce{N-H}$ point to the other side and establish hydrogen bonds with the chains on the other sides. These hydrogen bonds give mechanical strength to the proteins. Silk is made of a fibrous protein fibroin with a significant portion as $\beta$-pleated sheets.
The alternate side chains ($\ce{-R}$) point above and below in a $\beta$-pleated sheet. The amino acids in $\beta$-pleated sheets usually have short side chains, such as glycine, valine, alanine, and serine, as the long chains cause more steric strain.
Random coil
Most of the proteins have a mix of $\alpha$-helix and $\beta$-pleated sheets and organized but not repeated structures between the two that are called random coils as illustrated in Figure $4$. The random coins are represented by tubes.
Triple helix
The triple helix is a secondary structure found in collagen. Collagen is a structural protein that is strong and elastic, present in connective tissues of the tendon, cartilage, blood vessels, skin, and bone. It is the most abundant protein of vertebrates making up 30% to 35% of their proteins by weight.
The triple helix is made of three collagen peptides which are left-handed helices that are placed together on a common axis and displaced (translated) along the axis, forming a right-handed triple helix, as shown in Figure $1$ and Figure $5$.
The collagen peptide is composed of repeat units of Gly-X-Y, where X is usually proline, and Y is generally hydroxyproline - a proline with a hydroxyl ($\ce{-OH}$ group installed on the side chain. Proline has a secondary $\alpha$-amine group that causes bent in the peptide chain, which is needed for the helical structure of collagen peptides. The extra ($\ce{-OH}$) in the side chain of hydroxyproline allows more hydrogen bonding in the collagen peptide, which makes connective tissues stronger. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.04%3A_Secondary_structure_of_proteins.txt |
Learning Objectives
• Define the tertiary structure of proteins and understand the interactions that form and maintain the tertiary structure.
What is the tertiary structure of proteins?
The secondary structure of proteins represents the folding of portions of the polypeptide held primarily by hydrogen bonding between $\ce{C=O}$ and $\ce{N-H}$ groups of the polymer backbone. The overall protein chain also folds in a configuration that causes more stabilizing interaction primarily between the groups on the side chains ($\ce{-R}$) and the between side chain groups and the groups on the polymer backbone.
Tertiary structure
The three-dimensional arrangement of all the atoms of a single polypeptide chain in space, held together by stabilizing interactions between the side chain and the backbone groups, is called the tertiary structure of proteins.
The interactions stabilizing the tertiary structure include disulfide linkage, salt bridge, coordinate bonds with metal ions, hydrogen bonding, and hydrophobic interaction, as shown in Figure $1$. and explained below.
Disulfide linkage
When thiol ($\ce{-SH}$) groups of two cysteine residues come close to each other in a folded protein chain, they can make disulfide linkage (($\ce{-S-S{-}}$), i.e., the only covalent bond, besides peptide bond, holding amino acid residues together in proteins.
$\ce{R-SH + HS-R' ->[Oxidation] R-S-S-R + 2H^{+} + 2e^{-}}$
Salt bridge
The side chain of acidic amino acids, e.g., Asp, loses protons and becomes an anion, and that of basic amino acids, e.g., Lys, gains a proton and becomes a cation. When the opposite ions come close to each other, they make an ionic bond, like that in salts, called salt-bridge, e.g., $\ce{...-CH2-COO^{-} ^{+}NH3-(CH2)4-...}$) can form between Asp and Lys residues. This is an ionic bond holding opposite charges together.
Coordinate bond
Some metals can act as Lewis acids and make a coordinate covalent bond by accepting a lone pair of electrons from Lewis bases. Oxygen and nitrogen atoms in proteins have lone pairs of electrons and can act as Lewis bases. For example, oxygen atoms of two carboxylate $\ce{-COO^{-}}$ groups can make coordinate covalent bonds with $\ce{Mg^{2+}}$, resulting in a coordinate covalent linkage ($\ce{-COO^{-}\!\!\!\!\!->\!\!Mg^{2+}\!\!\!\!\!\!\!\!<-\!\!\!\!\! ^{-}OOC{-}}$) between the amino acid residues.
Hydrogen bonding
Hydrogen atoms bonded to oxygen or nitrogen atoms, e.g., in $\ce{-OH}$ and $\ce{-NH2}$, can make hydrogen bonds with any oxygen or nitrogen atom. These bonds could be between side chains, backbone groups, or side chains and backbone groups. For example, $\ce{-OH}$ in the side chain of serine can make hydrogen with the carbonyl oxygen of the amide group ($\ce{-CONH2}$) of asparagine residue, as shown below.
Hydrophobic interactions
As shown below, amino acids with aliphatic or aromatic hydrophobic side chains, e.g., valine and phenylalanine, establish hydrophobic interaction.
Hydrophobic interaction is London dispersion forces between nonpolar groups. The physiological medium is predominantly water. As a nonpolar liquid like oil, when mixed with a polar liquid like water, the oil drops merge, expelling water. Similarly, hydrophobic groups in proteins come close to each other due to the hydrophobic interaction and tend to stay in the interior of the protein, expelling water away from those regions. Ionic and polar compounds dissolve in water. Similarly, ionic and polar groups on the side chains stay on the proteins' outer surfaces, interacting more with water.
The tertiary structure of proteins may be predominantly $\alpha$-helix, $\beta$-pleated sheets, or random coil, but often it is a mix of these, for example, Carboxypeptidase A from bovine pancreas, shown in Figure $2$. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.05%3A_Tertiary_structure_of_proteins.txt |
Learning Objectives
• Define and understand the quaternary structure of proteins and its need.
Several proteins function as a single polypeptide with primary, secondary, and tertiary structures. Some proteins require more than one polypeptide with their primary, secondary, and tertiary structures combined to function.
Quaternary structure of proteins
The association of more than one protein into a closely packed arrangement is called the quaternary structure of proteins.
The individual proteins in the assembly are called monomers or subunits. The submit are held together by the same interactions that hold the tertiary structure of individual subunits, i.e., hydrogen bonds, hydrophobic interactions, salt bridges, etc.
An association of two proteins is called a dimer; three proteins is a trimer; four is a tetramer;five is a pentamer. For example, hemoglobin, shown in Figure $1$. is a tetramer of two $\alpha$ subunits of 141 amino acid residues each and two $\beta$ subunits of 146 amino acid residues each. These subunits are structurally similar, about the same size, and work cooperatively in the hemoglobin tetramer.
7.07: Summary of protein structure levels
Learning Objectives
• Understand the structure levels of proteins.
The summary of protein structural levels is illustrated in Figure $1$ and described in a video. The protein structure in summary form is the following.
1. The primary structure is the sequence of amino acids in a polypeptide chain.
2. The secondary structure is the organized sections of the polypeptide chain, e.g., $\alpha$ helix and $\beta$ pleated sheets.
3. The tertiary structure is the overall folded 3D structure of the entire polypeptide chain, held by disulfide linkages, salt bridges, hydrogen bonding, and hydrophobic interactions.
4. Quaternary structure is the association of more than one polypeptide chain (subunit) held together by the same type of interactions holding a subunit's tertiary structure.
7.08: Protein misfolding and denaturation
Learning Objectives
• Understand misfolding of proteins and the medical problems associated with it.
• Understand the denaturation of proteins and what causes the denaturation.
Protein misfolding and associated diseases
Newly synthesized proteins fold in a specific way, i.e., into secondary, tertiary, and quaternary structures, to be able to perform their function, as illustrated in Figure $1$. Some proteins can fold in only one way, but others can fold in multiple ways. There are proteins in the cell, called chaperones, that help newly formed proteins to fold in the way needed for their function.
Sometimes normal proteins misfold and become pathological. Often, these proteins are in soluble $\alpha$ helix forms that re-assembles into $\beta$-pleated sheet forms that are sticky and aggregate into plaques or amyloid structures, as illustrated in Figure $2$, with the example of plaque formation in Alzheimer's affected brain.
Prions are small proteins found in nerve tissue. Their exact functions are unknown, but when they misfold, they can cause more normal proteins to misfold. This protein misfolding is related to diseases such as mad cow disease in cows, Creutzfeldt–Jakob disease in humans, Alzheimer's disease, and familial amyloid cardiomyopathy or polyneuropathy, as well as intracellular aggregation diseases such as Huntington's and Parkinson's disease.
Denaturation of proteins
Proteins usually keep their primary, secondary, tertiary, and quaternary structures under physiological conditions. However, some physical processes and chemical agents can beak the interactions, i.e., hydrogen bonding, disulfide linkages, salt bridges, and hydrophobic interactions.
Denaturation
Loss of the secondary, tertiary, and quaternary structures of proteins by a physical process or a chemical agent while maintaining the primary structure almost intact is called denaturation of proteins.
Proteins unfold and become almost linear polypeptide chains upon denaturation. Denatured proteins can not perform their functions. Denaturation can be caused by heat, acids or bases, organic compounds and solvents, heavy metal ions, and agitation, as explained below.
• Heating above 50 oC disrupts hydrogen bonding and hydrophobic interactions, causing protein denaturation. For example, cooking food and sterilizing surgical instruments by autoclave treatment. In laser surgery, laser, i.e., the light of a single wavelength, is focused on a spot, causing heat that denatures proteins. Heating by laser cauterizes incisions, i.e., burns the site or the wound. It helps prevent blood loss.
• Acids and bases disrupt hydrogen bonding and salt bridges, e.g., by neutralizing some ions involved in the salt bridges.
• Organic compounds like urea disrupt hydrogen bonding by replacing them with stronger hydrogen bonding with the compound, and organic solvents disrupt hydrogen bonding and hydrophobic interactions. For example, 75% alcohol sterilizes skin by denaturing and coagulating proteins.
• Heavy metals ions like $\ce{Pb^{2+}}$ and $\ce{Hg^{2+}}$ attack $\ce{-SH}$ groups making salt bridges, like $\ce{-S^{-}Pb^{2+} ^{-}S{-}}$. Egg white or milk is an antidote to heavy metal poisoning as they precipitate the metal ions in the stomach, based on the salt bridge forming reaction. Vomiting is induced to remove the metal before the precipitate is dissolved, releasing the metal in the later parts of the digestive system.
• Agitation physically disrupts hydrogen bonding and hydrophobic interactions. Whipped cream and whipped egg white are prepared based on the agitation process. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.06%3A_Quaternary_structure_of_proteins.txt |
Learning Objectives
• Understand enzyme-related terminology, nomenclature, and classification of enzymes.
• Understand the mode of action of enzymes, the factors that affect them, and the inhibitors that retard or damage the enzyme activity.
Enzymes and the related terminology
Three conditions are necessary for a chemical reaction to happen:
1. Reactants must collide -the more frequent the collisions faster the reaction,
2. the reactant must have proper orientation at the time of the collision -the higher the probability of proper orientation at the time of the collision, the faster the reaction, and
3. there must be enough energy at collision to surpass the energy barrier, i.e., the energy of activation for the reaction -the lower the activation energy, the faster the reaction.
Often chemical reactions are possible but so slow that they are practically useless. For example, a reaction between hydrogen $\ce{H2}$ and nitrogen $\ce{N2}$ producing ammonia $\ce{NH3}$ is possible, but to make it practically useful, high-pressure, high-temperature, and catalysts are needed.
$\ce{N2(g) + H2(g) <=>[heat, pressure, catalyst] 2NH3(g)\;\; -92.4 kJ mol^{-1}}$
Catalyst
A catalyst is a reagent that increases the rate of a chemical reaction without itself being altered in the process.
The catalysts usually increase the rate of chemical reactions by improving the last two factors, i.e., increasing the probability of proper orientation and providing an alternate route for the reaction with lower activation energy. There are more constraints for chemical reactions in living things, e.g., the reaction has to occur under physiological conditions of pH ~7.4 and body temperature ~37 oC. Special catalysts called enzymes are used to regulate chemical reactions in living things.
Enzyme
An enzyme is a substance that regulates the rate of chemical reaction in living things without itself being altered in the process. In other words, enzymes are biological catalysts.
Like other catalysts, the enzymes usually increase the rate of chemical reactions by improving the last two factors, i.e., increasing the probability of proper orientation and providing an alternate route for the reaction with lower activation energy. The enzymes achieve the appropriate orientation of the reactants by binding them in a specific region within the enzyme, which has the geometry of its interacting groups right for securing the reactant in a particular orientation.
Substrate
A reactant in an organic or biochemical reaction is a substrate.
Active site
The enzyme's active site is the region within an enzyme where the substrate binds for the reaction.
Enzymes are usually proteins having primary, secondary, and tertiary structures. The active site is usually a small region (10% to 20%) within the enzyme. A few side chains of amino acid residues within the active site participate in the catalytic action, as shown in Figure $1$. The rest of the amino acid residues define and hold the secondary and tertiary structures.
Some enzymes need a non-protein part to combine with them for their function.
Apoenzyme, cofactors, and coenzyme
The enzymes that need a non-protein portion to combine with them for their function are called apoenzymes. The non-protein portion of the enzymes is called the cofactor, as illustrated in Figure $2$. The cofactor could be a metallic ion, .e.g., $\ce{Zn^2+}$ or $\ce{Mg^2+}$, or an organic compound. Organic cofactor is called coenzyme. The apoenzyme and cofactors together are called holoenzymes, as illustrated in Figure $2$.
Enzymes usually make the reaction happen millions of times faster. For example, one molecule of carbonic anhydrase enzyme, shown in Figure $1$, can catalyze about one million molecules in one second in human blood by the following reaction.
Copyright: Synpath, CC0, via Wikimedia Commons.
Overall reaction: $\ce{CO2 + H2O <=> HCO3^{-} + H^{+}}$
Selective
A catalyst or a reagent is selective if it produces one product preferentially or exclusively when more than one product is possibly formed.
Enzymes are not only selective; they usually react with one compound or stereoisomer.
Stereospecific
Stereospecific catalysts or reagents are those that, when reacting with one stereoisomer, selectively produce a stereoisomer and either do not react with the isomers of the reactant or selectively produce other stereoisomers from them. Enzymes are usually stereospecific, i.e., they are selective among reactants and selective among products.
For example, the enzyme arginase is stereospecific, hydrolysis amino acid L-arginine to L-ornithine and urea, but does not react with D-arginine.
Usually, enzymes react with one compound or a specific bond of one compound. For example, enzyme ureas catalysis hydrolysis of urea ($\ce{(NH2)2C=O}$) and does not hydrolyze any other amide.
$\ce{(NH2)2C=O + H2O ->[urease] 2NH3 + CO2}\nonumber$
Trypsin is a digestive enzyme that cleaves peptide bonds of proteins but not every peptide bond, only those on the C-side of lysine and arginine residue.
Some enzymes are specific for a class of compounds, e.g., lipases catalyzing triglyceride hydrolysis, but do not react with carbohydrates or proteins.
Names and classification of enzymes
Names of enzymes are derived by replacing the end of the name of a reactant or reaction with the suffix ase. For example, sucrase hydrolysis sucrose, lipase hydrolyzes lipids, oxidase catalyzes oxidation reactions, dehydrogenase removes hydrogen atoms, etc. Some old names of enzymes have the suffix in, e.g., digestive enzymes pepsin and trypsin. In the recent classification of enzymes, the name or the class name indicates the type of reaction it catalyzes. There are six major classes of enzymes, as described in the following.
Oxidoreductases
Oxidoreductases do oxidation-reduction reactions, i.e., oxidases oxidize a substance, reductases reduce a substance, and dehydrogenases remove hydrogen. For example, lactate dehydrogenase reduces pyruvate and oxidizes lactate, as shown below.
Transferases
Transferases transfer a group between two compounds. For example, kinases transfer phosphate groups, and transaminases transfer amino groups, as shown below.
Hydrolases
Hydrolases do hydrolysis reactions: lipases hydrolyze lipids, carbohydrates hydrolyze carbohydrates, proteases hydrolyze proteins, phosphatases hydrolyze phosphate esters, and nucleases hydrolyze nucleic acids. For example, acetylcholinesterase hydrolysis acetylcholine, as shown below.
Lyases
Lyases add two groups to double bond or remove two groups from adjacent atoms to create a double bond or a ring structure by means other than hydrolysis and oxidation, e.g., carboxylases add or remove CO2, and deaminases add or remove NH3. For example, actonitase, shown in Figure $3$, adds or removes water. double bond or add a new ring structure.
Isomerases
Isomerases rearrange atoms in a molecule: isomerases convert cis to trans or trans to cis isomer, and epimerases convert D to L or L to D streoisomers. Phosphohexose isomerizes shown below isomerizes glucose-6-phosphate to fructose-6-phosphate
Ligases
Ligases or synthetases catalyze the joining of two molecules, e.g., glutamate to glutamine conversion by glutamine synthetase shown below.
(Copyright; Hbf878, CC0, via Wikimedia Commons)
How do enzymes catalyze the reactions?
The chemical reactions that enzymes catalyze can occur without enzymes, but the reactions' rates are usually prolonged due to high activation energies. Enzymes provide alternate routes to the reactions with lower energy barriers (activation energies) and proper orientations of substrates that result in fast reactions, as illustrated in Figure $3$
In a generalized enzyme-catalyzed reaction, enzyme (E) and substrate (S) bind to make an enzyme-substrate complex (ES). The intermolecular interactions between the enzyme and the substrate usually loosen the bonds of the substrate that need to be broken, resulting in a lower energy barrier for the catalyzed reaction that leads to the product (P) at a faster rate than otherwise. The product leaves the enzyme so it can bind with another substrate.
$\ce{E + S <=> ES -> E + P}\nonumber$
For example, carbonic anhydrase catalyzes the reaction of water ($\ce{H2O}$) with carbon dioxide ($\ce{CO2}$) in blood as illustrated in Figure $4$. The carbonic anhydrase enzyme is bound with cofactor $\ce{Zn^{2+}}$ through three histidine residues. The $\ce{Zn^{2+}}$ is also bound with $\ce{H2O}$ that bond makes water ready for the proton transfer. After the proton transfer, a strong electrophile ${-\overset{-}{O}H}$ is generated that attacks $\ce{CO2}$, which is placed in the proper location in the hydrophobic pocket nearby. The product is displaced with another water molecule that repeats the process. One molecule of carbonic anhydrase can convert about million $\ce{CO2}$ molecules per second in this reaction.
Models of enzyme action
Enzymes are usually specific for a particular substrate or a class of reaction. This specificity was explained first by models of enzyme action explained below.
Lock-and-key model
The lock and key model assumes enzymes have a fixed shape with active sites similar to a lock in which a particular substrate with a fixed shape similar to a key can fit, as shown in the figure on the right.
This model explains the specificity of enzymes but ignores the dynamic nature of molecules. Further, according to this model, the products would be expected to be snug-fit in the active site and not easy to release. In reality, all the bonds in a molecule are vibrating and stretching, and the whole molecule is jiggling due to thermal energy, as illustrated in Figure $5$.
Induced-fit model
The induced-fit model of enzyme action accounts for the flexibility of enzyme and substrate molecules. According to this model, the flexibility of the enzyme molecule allows the active site to adapt to the shape of the substrate and, at the same time, the substrate adopts the shape of the enzyme to acquire the best possible orientation for the reaction to occur, as illustrated in Figure $6$. Then the active site shape re-adjusts to let the products be released to allow the next cycle of the enzyme action. The observation from experiments during the actual catalysis reaction supports the view that not only does the active site of the enzyme change shape, the backbone and the side chains of the enzyme molecule remain in constant motion during the enzyme action.
Experimental data shows that the active site is usually a tiny portion (10% to 20%) of the enzyme. Within the active site, two or a few side chains of amino acid residues usually catalyze the reaction. Usually, the catalytically active residue is one of the following: His, Cys, Asp, Arg, and Glu. These amino acids have acidic or basic or thiol functional groups, which are not only capable of hydrogen bonding but also capable of acid, base, electrophile, or nucleophile catalysis. For example, the pepsin enzyme breaks peptide bonds by using histidine and cystine, as illustrated in Figure $7$.
The following video from RCSBProteinDataBank explains how enzymes work.
Isoenzymes as a medical diagnostic tool
Isoenzymes are different enzymes that perform the same function in different body parts. They usually have a tertiary structure and differ in only a few amino acid residues. For example, lactate dehydrogenase (LDA) catalyzes the conversion between pyruvate and lactate, as shown below.
LDA is a tetramer made of two sub-units, the H-form, and the M-form, in five combinations found in different tissues, as shown in Table 1. Similarly, creatine phosphokinase (CPK) catalyzes the interconversion of phosphocreatine to creatine. CPK is a dimer in 3 isoenzyme forms shown in Table 2.
These isoenzymes usually function within cells. However, when some disease damages a tissue, the cells die, and the isoenzymes are released into the blood. Analysis of blood serum is used to diagnose the location of the damage. For example, an elevated level of LDH5 indicates liver damage and myocardial infarction (heart damage) is characterized by a high level of LDH1 isoenzyme. The heart damage will also elevate CK2.
Table 1: Different forms of lactate dehydrogenase (LDA), their subunit composition the location in the body.
Type Subunits Illustration Location
LDH1 HHHH Heart and Erythrocyte
LDH2 HHHM Heart and Erythrocyte
LDH3 HHMM Brain and Kidney
LDH4 HMMM Skeletal Muscle and Liver
LDH5 MMMM Skeletal Muscle and Liver
Table 2: Isoenzymes of creatine phosphokinase (CPK) dimer, composition, and location in the body.
Isoenzyme Subunit Illustration Tissue of Origin
CPK1 BB Brain
CPK2 MB Heart
CPK3 MM Skeletal muscle
Factors that affect enzyme activity
The enzyme activity is related to how much it increases the reaction rate compared to the same reaction without the enzyme. The factors that affect enzyme activity include concentration, temperature, pH, and the presence of inhibitors.
Concentration
Usually, the enzyme is in very low concentration relative to the concentration of the substrate. Therefore, an increase in the enzyme concentration increases the reaction rate linearly, i.e., doubling the enzyme concentration doubles the reaction rate, and tripling it triples the rate. An increase in the substrate concentration increases the reaction rate but not linearly; it follows a curvilinear curve. The rate increase reaches a saturation level and does not increase after that with a further increase in the substrate concentration, as illustrated in the figure on the left. This is because all the enzyme's active sites become occupied with the substrate at the saturation point, and the reaction proceeds at its maximum rate, as illustrated in the figure on the right.
Temperature
Generally, an increase in temperature increases the rate of a chemical reaction. This is because more molecules have energy than activation energy at higher temperatures. The same applies to enzyme-catalyzed reactions at lower temperature ranges. Still, the rate becomes optimum at around body temperature and then starts to fall off, as illustrated in the figure on the right. The enzymes have secondary, tertiary, and quaternary structures optimized to perform best at the body temperature of about 37 oC. Denaturation inactivates the enzymes. It is reversible if the temperature is slightly above body temperature, but enzymes are denatured beyond repair at a much higher temperature.
pH
The enzymes work best at the pH of the tissue or organ where they work. This is because the structure of enzymes is optimized for the pH at which they usually function. An increase or a decrease in pH from the optimum value disrupts the secondary, tertiary, and quaternary structures and causes a reduction in enzyme activity. The effect of a slight change in pH is reversible, but a significant difference in pH denatures the enzymes permanently.
The optimum pH for most enzymes is around the physiological pH of 7.4. Still, some enzymes have different optimum pH, depending on the pH in their natural environment, as illustrated in the figure on the left. For example, the optimum pH of the starch-splitting amylase is pH 7-7.5 in the mouth. Pepsin breaks down proteins at around pH 2 in the stomach. Trypsin breaks down proteins at pH 8 in the intestine.
Enzyme inhibition
Inhibitors are substances that make enzyme lose their catalytic activity. Inhibitors prevent the substrate from binding with the active site of the substrate. The inhibition may be reversible or irreversible. There are two subclasses of reversible inhibitors; competitive inhibitors and noncompetitive inhibitors.
Competitive inhibitor
The competitive inhibitor has a shape similar to the enzyme's natural substrate. So, they compete with the substrate for the active site but do not react, as illustrated in Figure $8$. Since there is competition between the inhibitor and the substrate for the active site, increasing the concentration of the substrate wins by outnumbering the inhibitor in the completion, and the enzyme regains its activity.
Noncompetitive inhibitor
A noncompetitive inhibitor does not have a shape similar to the substrate and does not bind to the active site. It binds with the enzyme outside the active area but changes the folding patterns of the protein such that the active site can not acquire the proper shape. So, the substrate can not fit into the active site properly, as illustrated in Figure $8$.
Examples of noncompetitive inhibitors are toxic metals, like $\ce{Pb^{2+}}$, $\ce{Hg^{2+}}$, and $\ce{Ag^{+}}$. Since the inhibitor does not compete with the substrate, increasing the substrate concentration does not recover the enzyme activity. However, some chemical agents can remove the inhibitor from the enzyme, and the catalytic activity can be recovered.
Irreversible inhibitor
Irreversible inhibitors bind with enzymes and destroy their catalytic activity permanently. They usually bind with enzymes by covalent bonds that are not easily broken. Irreversible inhibitors include antibiotics, insecticides, and never gases. For example, penicillin is an irreversible inhibitor of an enzyme needed to form cell walls in bacteria. The bacteria without a complete cell wall can not survive. It does not affect the cell membranes of humans. Similarly, a diisopropyl fluorophosphate (DIPF) which is an organophosphate insecticide, binds with the $\ce{-OH}$ of a serine residue in the enzyme acetylcholinesterase, as shown in the figure on the right. The enzyme-catalyzed reaction of acetylcholine, shown below, is needed for nerve impulse transmission. Inhibition of acetylcholinesterase blocks nerve impulses, causing paralysis. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/07%3A_Proteins/7.09%3A_Enzymes.txt |
Learning Objectives
• Understand the function of nucleic acids.
• Learn the structures and names of nucleotides -the building blocks of nucleic acids, and their constituents, including ribose, deoxyribose, nitrogen bases, and phosphate groups.
Nucleic acids are biopolymers that carry the codes for synthesizing proteins and pass them on from generation to generation, i.e., they are genetic materials. In other words, nucleic acids are the instruction manual for biochemical reactions in living things.
Nucleotides are the building blocks, i.e., the repeat units or monomers of nucleic acids.
Nucleotides are composed of three sub-units:
1. a 5-carbon carbohydrate,
2. a base that is an aromatic compound containing nitrogen, and
3. an anion of phosphoric acid, i.e., phosphate ($\ce{PO4^{3-}}$).
5-Carbon carbohydrate
Two 5-carbon carbohydrates, i.e., ribose and deoxyribose, are found in nucleic acids, as shown in the figure below. Both are D-sugars in five-membered cyclic form with hydroxyl ($\ce{-OH}$) group at anomeric ($\ce{C}$#1') pointing towards $\ce{O}$ in the ring, i.e., in a $\beta$ configuration. The number labels have prime symbols, i.e., 1', 2', 3', etc., to distinguish them from regular numbers 1, 2, 3, etc., used for the nitrogen bases of the nucleic acids. The only difference between ribose and deoxyribose is that hydroxyl ($\ce{-OH}$) group at $\ce{C}$#2' is missing in the latter, which gives the deoxy- prefix to its name.
Types of nucleic acids
There are two types of nucleic acids: ribose in its nucleotides, called ribonucleic acid (RNA), and deoxyribose in its nucleotides, called deoxyribonucleic acid (DNA).
Nitrogen bases
Nitrogen bases in nucleic acids are derivatives of two aromatic compounds, purine and pyrimidine, shown below.
Cyclic compounds that contain atoms other than $\ce{C's}$ in the cycle are called heterocyclic compounds. Purine and pyrimidine are heterocyclic aromatic compounds because they contain $\ce{N's}$ in the cycles. Purine is bicyclic, containing a six-membered ring with two $\ce{N's}$ fused with a five-membered ring with two $\ce{N's}$. Pyrimidine is a six-membered cycle with two $\ce{N's}$. Both purine and pyrimidine are planer molecules like benzene.
There are five nitrogen bases found in nucleic acids: two are purines, i.e., adenine (A) and guanine (G), and three are pyrimidines, i.e., cytosine (C), thymine (T), and uracil (U), as shown below. To remember what are pyrimidines, remember "cut pyramid" where 'c' is cytosine, 'u' for uracil, 't' for thymine, and 'pyramid' for pyrimidines
Purines Pyrimidines
DNA contains four nitrogen bases, i.e., adenine, guanine, cytosine, and thymine. RNA also has four nitrogen bases, i.e., adenine, guanine, cytosine, and uracil. Note that the first three, i.e., adenine, guanine, and cytosine, are common in DNA and RNA, but the fourth, i.e., ti.e.ine in DNA, is replaced with uracil in RNA.
Nucleosides
When a monomer sugar like ribose or deoxyribose reacts with an amine, the $\ce{-OH}$ group at $\ce{C}$#1' is replaced with a $\ce{N}$ of the amine. The product is called N-glycoside, and the $\ce{C-N}$ bond in the N-glycoside is called an N-glycosidic bond, as shown in the following example.
Purines connect through $\ce{N}$#9 and pyrimidine through $\ce{N}$#1 with anomeric carbon ($\ce{C}$#1') of ribose or deoxyribose to form the N-glycoside.
The N-glycosides are named by using the name of purine but ending with -osine e.g., adenine becomes adenosine and guanine becomes gunosine, or by using the name of pyrimidine ending with -idine, e.g., cytosine becomes cytidine and uracil become uridine. One letter abbreviation of the nitrogen bases remains the same for the corresponding nucleoside, e.g., adenine (A) and adenosine (A). The four nucleosides found in RNA are shown below with their names and one-letter abbreviations.
In DNA there is deoxyribose in place of ribose. So the names of nucleosides found in DNA begin with the deoxy- prefix. The four nucleosides found in DNA are shown below with their names and one-letter abbreviations.
Phosphate
Phosphoric acid ($\ce{H3PO4}$) is an oxyacid with three acidic protons, i.e., it is triprotic acid. Just as two carboxylic acids condense with each other and form a carboxylic acid anhydride with the elimination of water molecule, two phosphoric acids condense and form diphosphoric acid, also called pyrophosphoric acid, with the elimination of water.
Similarly, three phosphoric acids can condense to form triphosphoric acid. Phosphoric acid, diphosphoric acid, and triphosphoric acid are shown below.
Under physiological conditions at pH ~7.4, the phosphoric acids lose proton and exist as phosphate anions, as shown below.
Phosphoric acids also condense with alcohols ($\ce{R-OH}$) and form mono-, di-, or triesters, as illustrated below.
Nucleotide
Phosphate esters of nucleosides are called nucleotides. The nucleotides found in nucleic acids are formed when $\ce{-OH}$ group at $\ce{C}$#5' of ribose or deoxyribose makes ester with a monophosphate, diphosphate, or triphosphate, as shown below.
The names of the nucleotides are derived by using the name of the corresponding nucleoside with monophosphate, diphosphate, or triphosphate added at the end. Abbreviation of the nucleotide starts with the one-letter abbreviation of the nitrogen base followed by MP, DP, or TP depending on whether the phosphate group is monophosphate, diphosphate, or triphosphate, as shown in the example structures above.
The nucleotides like ADP and ATP are high-energy molecules used as energy currency in biochemical systems. For example, when ATP converts to ADP by releasing a phosphate, energy is released to do work or synthesize other compounds. Synthesis of nucleic acids begins with triphosphate esters that convert into monophosphate esters when incorporated in the nucleic acid polymer. The energy released in converting triphosphate to monophosphate drives the reaction forward.
The four nucleotides found in RNA are shown below with their names and abbreviations.
, , , and
The nucleotides found in DNA are named similarly to those of RNA: start with the name of the corresponding nucleoside and end with monophosphate, diphosphate, or triphosphate. Remember that the names of DNA nucleosides begin with deoxy. The abbreviations of the nucleotides found in DNA start with a small alphabet d representing the deoxy- prefix of the nucleotides. The four nucleotides found in DNA are shown below with their names and abbreviations.
, , , | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.01%3A_Nucleotides_-the_building_blocks_of_nucleic_acids.txt |
Learning Objectives
• Define nucleic acids and understand the mechanism of nucleic acid synthesis and the related terminologies.
• Define and be able to write the primary structure of nucleic acids.
Nucleic acid
Nucleic acids are long unbranched strands (polymers) of nucleotide subunits (monomers) that primarily store and express genomic information in living things.
Nucleic acids are long unbranched strands (polymers) of nucleotide subunits (monomers). A nucleotide triphosphate is the starting material for nucleic acid synthesis. The synthesis of DNA is controlled by an enzyme called DNA polymerase. A nucleoside triphosphate containing adenine (A), guanine (G), cytosine (C), or thymine (T) uses $\ce{-OH}$ group at 3' position of deoxyribose to attack $\alpha$-phosphorous of triphosphate group at 5' position of the next nucleoside in an SN2 mechanism. It results in a phosphodiester linkage between the 3' position of the first and 5' position of the second nucleotide, and a pyrophosphate (diphosphate) is eliminated as shown in Figure $1$. The pyrophosphate breaks down into two phosphate units which make the reaction irreversible.
Next, nucleotide triphosphate repeats the process and links at the 3' end of the growing strand; the cyclic process keeps repeating this way, as shown in Figure $1$. The first nucleotide has unlinked triphosphate at the 5' position, called the free 5' end. The last nucleotide has an unlinked $\ce{-OH}$ group at the 3' position, and it is called the free 3' end. DNA synthesis occurs from 5' to 3' as shown in Figure $2$. Deoxyribose and phosphodiester linkage alternate in the polymer's backbone, and the nitrogen bases hang on the side (1' position) of the deoxyribose units. Two nucleotides linked by a phosphodiester linkage are called a dinucleotide, 3 to 10 linked nucleotides are called an oligonucleotide, and many nucleotides linked are called a polynucleotide.
The synthesis of RNA takes place the same way as that of DNA, except for the following differences:
1. the enzyme carrying out the synthesis is RNA polymerase,
2. the sugar (5-carbon carbohydrate) is ribose, and
3. uracil (U) is used in place of thymine (T).
Primary structure of nucleic acid
The sequence of nucleotides in the nucleic acid is called the primary structure of nucleic acid.
The primary structure is written from the 5' to 3' direction, where the 5'-end is on the left end, and the one-letter abbreviation of the nitrogen base represents the nucleotides. For example, the primary structure of the oligonucleotide product shown in Figure $2$ is ACGT. It is the sequence of nucleotides that carries the genetic information.
8.03: Secondary structure and replication of DNA
Learning Objectives
• Understand the double helix model and the replication process of DNA.
Secondary structure of DNA
Double helix model of DNA
It was observed from experimental data that although the composition of DNA varies from specie to specie, the amount of adenine (A) is always equal to the amount of thymine (T), and the amount of guanine (G) is always equal to the amount of cytosine (C). In 1953, Watson and Crick proposed a model of DNA with the help of Franklin's X-ray analysis. The Model, illustrated in Figure \(1\), is:
1. a DNA molecule is made up of two strands wound around each other like a spiral stair ladder,
2. the two strands run antiparallel, i.e., one from a 5' to 3' and the other from a 3' to 5' direction,
3. each strand has a backbone of alternating deoxyribose and phosphate, groups like the rails of the stair ladder, and
4. the two strands are connected at each nucleotide through hydrogen bonds between nitrogen bases: adenine associated with thymine of the second strand, and guanine connected with cytosine of the second strand, or vice versa, like a closed zip.
The logic of complementary base pairing
Two strands of DNA are complementary to each other as adenine (A) always faces thymine (T), and guanine (G) meets cytosine (C) in the other strand. The AT and GC are called complementary base pairs. The logic of this complementing is the following:
1. Purines, i.e., adenine (A) and guanine (G), are larger, and pyrimidines, i.e., cytosine (C) and thymine (T), are smaller. So, a purine in one strand pairs with a pyrimidine in the other strand to keep a nearly uniform gap between the two strands.
2. Adenine (A) makes two hydrogen bonds with thymine (T), and guanine (G) makes three hydrogen bonds with cytosine (C) in the other strand. The other possible combinations of purine and pyrimidine are unstable: A can have no hydrogen bond with C, and G can have only one with T.
The base pairing, i.e., AT and GC, complements the two strands. The multiple hydrogen bonding between each nucleotide of DNA holds the two strands together like a closed zip.
Replication of DNA
When a cell divides into two, each daughter cell needs a complete set of genetic information from the parent cell. The genetic information is in the form of the sequence of nucleotides in the DNA.
DNA replication is the process by which the DNA is copied in a cell at the time of cell division.
The process begins when an enzyme helicase catalyzes breaking hydrogen bonds, causing the DNA double helix to unzip, as illustrated by simulation in Figure \(2\) and explained in a video from yourgenome. Each strand acts as a temple for the synthesis of a new strand. As the complementary nucleotides come together, DNA polymerase joins the nucleotide by forming phosphodiester linkages from a 5' to 3' direction, as simulated in Figure \(3\). The leading strand is one strand of the initial DNA that unzips in a 3' to 5' direction. The leading strand's complementary strand is continually synthesized from the 5' to 3' direction. The other stand of the initial DNA that unzips in a 5' to 3' direction is called the lagging strand. The complementary strand of the lagging strand is synthesized in portions from 5' to 3' direction, called Okazaki fragments. An enzyme, called DNA ligase joins the fragments by forming phosphodiester linkages between them. This way, almost perfect copies of the entire DNA are synthesized. At this time, the cell can divide, and each daughter cell receives a replica of the parent DNA, which in many cases is identical to the DNA of the parent cell. The complementary base pairing in DNA ensures the correct placement of the nucleotides in the new DNA strands. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.02%3A_Primary_structure_of_deoxyribonucleic_acid_%28DNA%29.txt |
Learning Objectives
• Understand the basic structural features and functions of RNAs involved in protein synthesis, including mRNA, tRNA, and rRNA.
• Understand the transcription process, i.e., synthesis of mRNA from the DNA template, codons on DNA and mRNA, and anticodons on tRNA that specify amino acid to be incorporated in the protein.
Ribonucleic acid (RNA)
Ribonucleic acid (RNA) is a type of nucleic acid present in all cell types. It is structurally similar to DNA, as shown in Figure \(1\), but differs from it concerning the following.
1. RNA is often single-stranded.
2. The backbone of RNA is made of ribose units (rather than deoxyribose in DNA) connected by phosphodiester linkages.
3. Three nitrogen bases in RNA, i.e., adenine (A), Guanine (G), and cytosine (C), are the same as in DNA, but the fourth one is uracil (U) in RNA in the place of thymine (T) in DNA.
Types of RNAs Involved in protein synthesis
Three types of RNA are involved in protein synthesis, i.e., messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA), as illustrated in Figure \(1\). Some other RNAs have functions other than protein synthesis. Some viruses use RNA as genetic material rather than DNA.
Messenger RNA (mRNA)
Messenger RNA (mRNA) is a single-stranded RNas as illustrated in Figure \(1\). It is synthesized as a complementary strand of a section of one of the two strands of DNA that carry information for synthesizing a protein cell needs. The message RNA is synthesized in the nucleus and travels to the cytoplasm (watery interior of the cell outside the nucleus), where protein-synthesis machinery, i.e., ribosomes, reads the information and synthesizes the protein.
Transfer RNA (tRNA)
The transfer RNA (tRNA) is a small RNA comprising 70 to 90 nucleotides. Its job is to bring an amino acid to the protein synthesis location and incorporate it at a specific place in the protein, dictated by the codons on the mRNA. Hydrogen bonding in some complimentary bases in the tRNA makes it fold, producing loops and double-stranded portions, as illustrated in Figure \(2\). The 3D shape of the tRNA is twisted, appearing like L-shape, but it is usually depicted as a cloverleaf-like shape in 2D presentation. The loops include a D loop, a T loop, an anticodon loop, a variable loop that is a slight bulge between the T loop and the anticodon loop, and a receptor stem that contains the 3' end and 5' end of the tRNA molecule. The anticodon makes the tRNA specific for one amino acid. The anticodon is complementary to the codon on the mRNA.
Activation of tRNA
Attaching an amino acid to tRNA is called activation of tRNA. The 3' end of tRNA always has adenosine (A) followed by two cytosines (C), i.e., ACC sequence. The free 3' \(\ce{-OH}\) group of adenosine nucleotide makes an ester bond with an amino acid through the following reaction.
Anticodon
There is a different amino acyl-tRNA synthetase for each of the tRNA. Each tRNA is specific for one of the 20 amino acids. The particular shape of tRNA dictates this specificity. The tRNA and the amino acid corresponding to it fit in the active site of the corresponding amino acyl-tRNA synthetase, as illustrated in Figure \(2\). Each tRNA has a separate anticodon, a sequence of three nucleotides complementary to the corresponding codon on the mRNA, i.e., A pairs with U, G pairs with C, and vice versa. For example, mRNA codon GCA is complementary to the anticodon UGC on a tRNA shown in Figure \(2\) that specifies amino acid alanine in the protein.
Ribosomal RNA (rRNA)
Ribosomal RNA (rRNA) is the most abundant type of RNA that combines with proteins and makes ribosomes. Ribosome has two subunits, a large subunit, and a small subunit, as illustrated in Figure \(1\). Ribosome is the site for protein synthesis. There are many ribosomes in the cells.
Transcription
Gene is a section of DNA that is a basic unit of heredity passed from parent to child. Gene is a sequence of nucleotides in the DNA that encodes information for making specific proteins that lead to the expression of a particular physical character or trait, such as hair color or eye color, or some other specific function in the cell. Humans have about 20,000 protein-coding genes. Some genes encode the synthesis of RNAs that have functions other than protein synthesis.
Transcription is the process of mRNA synthesis using a protein synthesis gene portion of a DNA strand as a template, as illustrated in Figure \(3\).
The process begins when the gene portion of the DNA unwinds, and RNA polymerase catalyzes mRNA synthesis from a 3' to 5' direction, using a strand of the unwound DNA going from a 5' to 3' direction as a template. The unwound gene portion of the DNA is called the transcription bubble. The process stops when the RNA polymerase reaches the sequence of nucleotides on the template DNA that is a stop single. The mRNA is released from the DNA, and the unwound portion of the DNA returns to its regular double helix structure. The mRNA is processed further in the nucleus. Then it leaves from the nucleus to the cytoplasm (watery interior of the cell outside the nucleus), where protein-synthesis machinery, i.e., ribosomes, reads the information and synthesizes the protein in a process called translation.
The DNA strand used as a temple for mRNA synthesis s called the template strand, and the other is called the information strand. The mRNA complements the template strand, i.e., A in the template couples with U and G with C in the mRNA. The sequence of nucleotides in the mRNA is complementary to that of the template strand and the same as in the information strand, except that T in the information strand is replaced with U in the mRNA. Therefore, the primary structure of the information strand is shown as the primary structure of the DNA.
Codons
The gene encodes information about synthesizing a protein with the correct primary structure.
• A code called a codon is a sequence of three nucleotides (a trinucleotide) read from a 5' to 3' direction from mRNA, as illustrated in Figure \(4\).
• The codons:
• start from the first appearance of AUG, i.e., adenine-uracil-guanine, reading from 5' to 3' direction on a mRNA,
• are successive sets of frames of three nucleotides,
• non-overlapping,
• Since there are four different nucleotides, there are 64 ways to spell out three-letter codons, as shown in Figure \(5\).
• Three codons, i.e., UGA, UAA, and UAG, are stop codons signaling the termination of protein synthesis. The remaining 61 specify amino acids.
• Since there are only 20 amino acids, amino acids have more than one codon, except tryptophan and methionine have one codon each. For example, four codons: GGU, GGC, GGA, and GGG, specify glycine.
• Codon AUG has two roles: i) in the middle of a series of codons, it specifies methionine, and ii) at the beginning of an mRNA, it signals the start of protein synthesis by bringing in methionine as the first amino acid, but the first methionine is usually removed later from the beginning (N-terminus) of the protein.
8.05: Translation-Protein Synthesis
Learning Objectives
• Define and understand the mechanism of biological translation.
Translation
Translation in biology is the process of protein synthesis using the information encoded in mRNA.
Translation process
The process starts when the ribosome binds to mRNA in the cytoplasm, as illustrated in Figure \(1\). The ribosome moves along the mRNA from 5' to 3" direction until it reaches the stats codon AUG. The AUG is also the codon for methionine. So, the methionine-tRNA (methionine-loaded aminoacyl-tRNA) with the complementary anticodon UAC arrives and aligns opposite the codon AUG. The aminoacyl-tRNA with a complementary anticodon to the codon next to AUG on the mRNA arrives and aligns opposite the codon.
Ribosome catalyzes a peptide bond formation between \(\ce{-NH2}\) group of the amino acid on the second tRNA with the carboxylate group of the first by nucleophilic acyl substitution mechanism. The first tRNA becomes empty, and the second tRNA becomes peptidyl-tRNA. The ribosome moves on to the next codon -a process called translocation. The empty-tRNA leaves and a new aminoacyl-tRNA with a complementary anticodon to the third codon align on the mRNA and repeat the above process, as illustrated in Figure \(2\). The cycle repeats, and the peptide keeps elongating until the ribosome reaches a stop codon. The translation process stops at this point, and the newly formed peptide is released. The first methionine is usually removed from the peptide. The empty tRNA are re-loaded with the amino acid later on, as illustrated in Fig. 8.4.2. The hydrogen bonding and the other intramolecular interaction, like salt bridges, disulfide bonds, etc., make peptides acquire the secondary, tertiary, and sometimes quaternary structure. This is how functional proteins with the appropriate forms are synthesized.
Figure \(3\) presents a simulation that summarizes the process of transcription and translation and the trinucleotide codons in DNA, mRNA, anticodons on tRNA, and the corresponding amino acids in the polypeptide that are shown in the simulation. Watch the DNA and RNA video summary by clicking this link or the video below.
Antibiotics that interrupt protein synthesis in bacteria
Antibiotics that interrupt protein synthesis in bacteria but not in humans are clinically useful. Some examples and their actions are shown in Table 1.
Table 1: Some antibiotics act by interrupting protein synthesis in bacteria but not in humans.
Antibiotic Effect on protein synthesis in bacteria
Tetracycline Prevents the aminoacyl-tRNA from binding to the ribosome
Erythromycin Prevent the translocation of the ribosome along the mRNA
Streptomycin Inhibits the initiation of protein synthesis
Chloramphenicol Prevents the new peptide bond from being formed | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.04%3A_Ribonucleic_acid_%28RNA%29_and_transcription.txt |
Learning Objectives
• Understand mutation, its causes, types, and effects, including genetic diseases and cancer.
What is mutation?
A mutation is any change in an organism's DNA nucleotide sequence. It may occur due to an error during DNA replication, exposure to mutagens, or a viral infection. A mutagen is a physical or chemical agent that can cause a permanent change in an organism's DNA. Physical mutagens include exposure to radioactivity, X-rays, UV light, etc. Chemical mutagens are chemical agents that react with and change the DNA, e.g., polycyclic aromatic hydrocarbons (PAH) found in smoke and barbecued foods. Some mutagens that may lead to cancer are called carcinogens. Mutations may be harmful, beneficial, or may have no effect. Somatic mutations in body cells affect daughter cells but are not passed on to the offspring. Germline mutations occur in eggs or sperm and are passed to the offspring. When a mutation alters proteins or enzymes severely, the cells may not survive, and the person may have a genetic disease.
Types of mutation
A mutation may be due to i) replacement, ii) deletion, or iii) insertion of one or more nucleotides, as illustrated in Figure $1$. There are many types of mutations. One subclass of mutation, i.e., related with change in one nucleotide is described here to introduce the basic terminologies related with mutaions.
Point mutation
Point mutation is the addition, deletion, or change (substitution) of one nucleotide with another. It changes the codon in which the nucleotide is located.
Substitution mutation
A replacement of one nucleotide with another in the DNA is substitution mutation. There are three sub-type of substitution mutations.
1. If the initial codon and the new codon represent the same amino acid, the mutation is called a silent mutation.
2. If a point mutation changes the codon to a different amino acid, it is called a missense mutation. If missense mutation replaces an amino acid with significantly different properties, it may cause disease. For example, sickle cell disease is caused by a point mutation of codon GAG for glutamic acid (acidic) to GTG for valine (nonpolar) in the hemoglobin gene. On the other hand, if the new amino acid is similar to the previous one, there may be little or no change rendered in the proteins. For example, a change from AAA for lysine to AGA for arginine may not affect the protein as both amino acids are basic amino acids.
3. If a point mutation changes the codon for an amino acid to a stop codon, it is called a nonsense mutation. It results in premature termination of protein synthesis rendering nonfunctional protein. For example, $\beta$-thalassemia is caused by a change from CAG for glutamine to UAG, i.e., stop signal. The missense and nonsense mutations are illustrated in Figure $2$
Deletion mutation
Deletion mutation is the loss of one or more nucleotides from a segment of DNA. Since the codon is three consecutive nucleotides, the loss of one nucleotide will not only change the codon from which the nucleotide is lost, but all the codons following it will be changed. Deletion mutation is related to a significant number of genetic diseases, e.g., two-thirds of cystic fibrosis cases are due to the loss of three nucleotides that results in the loss of amino acid phenylalanine in the protein involved, and cat cry syndrome is due to a partial chromosome deletion.
Insertion mutation
In insertion mutation, one or more nucleotides are inserted into the typical sequence of nucleotides in the DNA. So, the codon where insertion happens and all the codons following it are changed.
Effects of mutations
Some mutations do not cause a significant change in protein structures, allowing the protein to perform its function. Others may change an amino acid vital to the structure of the proteins with an amino acid of significantly different properties. It results in the proteins not being able to function correctly. For example, sickle cell disease is caused by replacing glutamic acid with a nonpolar valine that prevents the hemoglobin protein from working correctly. If the protein is an enzyme, it may no longer catalyze the reaction. In this situation, the reactants may accumulate and become poisonous, or the product may be vital for survival and not synthesized. For example, an enzyme required to metabolize galactose-1-phosphate is absent in galactosemia. It results in the accumulation of galactose-1-phosphate, which may cause cataracts and mental retardation.
Genetic disorder
Genetic disorders are health conditions caused by mutations in the genetic material. For example, phenylketonuria (PKU) results when DNA can not carry the correct codes for the synthesis of the enzyme phenylalanine hydroxylase that hydroxylates the phenyl ring of phenylalanine to convert it to tyrosine. When phenylalanine cumulates, other enzymes convert it to phenylpyruvate—cumulation of phenylalanine and phenylpyruvate cause mental retardation. If diagnosed early on, a diet can be prescribed to the child that avoids foods containing phenylalanine. It can prevent the buildup of phenylpyruvate.
Similarly, if the enzyme that converts tyrosine to melanin is not functioning, melanin is not produced, causing albinism. Melanin is a pigment that gives color to skin and hair. People with albinism do not have skin, eye, or hair pigments. This disease also happens in animals. When a genetic disorder is inherited from one or both parents, it is called a hereditary disease. A few more genetic diseases are listed below.
• Cystic fibrosis results from a defective gene. It affects the lungs and digestive system. People with this disease can not digest food properly and have repeated lung infections.
• Down syndrome is a genetic disease in which peoples have cognitive impairment that may be mild or severe.
• Hemochromatosis is an inherited condition where the body absorbs and stores so much iron that it can lead to organ damage.
• Haemophilia is a genetic disorder in which blood does not clot properly. It makes bleeding challenging to control.
• Huntington's disease is a genetic disorder that affects the nervous system, and the effect worsens over time.
• Tay-Sachs disease is a genetic disorder that causes brain damage.
• Duchenne muscular dystrophy is a condition that causes gradual loss of muscle function.
• Thalassemia is a genetic disorder that causes less hemoglobin to be produced, making the blood cells small and pale.
• Tourette syndrome is a genetic disorder related to neurological problems. Peoples with this disorder make involuntary vocal sounds and movements. Relaxation and exercise may reduce the symptoms.
Cancer
Cancer is a disease in which some of the body's cells grow and multiply uncontrolled. Mutation in a cell may cause this uncontrolled division of cells. The uncontrolled division and growth of cells appear as a tumor, as illustrated in Figure $3$. It is benign if the tumor remains limited in growth without harming the neighboring tissue. However, the tumor is cancer if it invades other issues and interferes with their normal functions.
The most common types of cancer are lung cancer, prostate cancer, breast cancer, colorectal cancer, stomach cancer, cervical cancer, and skin cancer. Melanoma and skin cancers account for about 40% of cancer cases. Cancer types common in children include acute lymphoblastic leukemia, brain tumors, and non-Hodgkin lymphoma.
The mutations causing cancers can result from errors during DNA replication or exposure to carcinogens, i.e., the substances or radiations that cause cancer. The carcinogenic substances include some compounds in tobacco smoke, automobile exhaust fumes, processed meat; asbestos; benzene; toxic metals like nickel, arsenic, beryllium, chromium, and cadmium and their compounds; nitrosamine; ethylene oxide, etc. About 22% of cancer-related deaths occur due to smoking. Cancer-causing radiations include radioactivity resulting in ionizing radiations, X-rays, and UV-radiations. About 15% of cancers are due to cancer-causing viral infections like HIV, hepatitis B, hepatitis C, Helicobacter pylori, Epstein-Bari virus, papillomavirus, etc. Sometimes inherited defects in genes are the cause of cancer. These factors cause or are at least partially involved in gene mutations that result in cancer. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.06%3A_Mutations_and_genetic_diseases.txt |
Learning Objectives
• Define viruses and understand how they infect and multiply, including reverse transcription.
• Understand the basic structural features, infection ways, and treatment of two viral infections, including covid-19 and HAIDS.
What is a virus?
A virus is a tiny infectious microbe that consists of a genetic material (DNA or RNA) surrounded by a protein coat called the capsid that covers and protects the genetic material.
A single virus outside the cell that has genetic material and capsid surrounding it is also called a virus particle or a virion. As Figure \(1\) illustrates, viruses have various shapes.
A virus does not have the machinery to replicate and synthesize proteins, so it injects its genetic material into a host cell and hijacks its nucleic acid and protein synthesis machinery to reproduce. Viruses infect nearly all life forms, including humans, animals, plants, bacteria, and fungi. Figure \(2\). illustrates some of the human viral infections.
How do viruses infect and multiply?
The virus infection of cells begins with the attachment of the virus to the receptors at the host cell surface, followed by penetrating the cell membrane or cell wall. The protein coat (capsid) is removed, and the genetic material is injected into the host cell. The host cell replicates the viral DNA producing copies of it, and also translates it, making the proteins that the newborn virus needs. New viruses are assembled using the viral DNA and the protein coat produced by the host cell. They ultimately kill the host cell by bursting out of it in a lysis process, as illustrated in Figure \(3\). Some viruses take a portion of the host cell membrane to form an envelope around the capsid during lysis.
Reverse transcription
Viruses that contain viral RNA as their genetic material are called retroviruses. Once inside the host cell, the retroviruses first synthesize DNA using the viral RNA as a template, the reverse transcriptase enzyme in the virus, and the nucleotides in the host cell. The single DNA strand forms a double-strand viral DNA called a provirus. The provirus integrates with the DNA of the host cell using enzyme ligase, which is also present in the virus. Then the viral DNA directs the synthesis of viral RNA and the proteins needed to assemble new virus particles, as illustrated in Figure \(4\). Finally, the newborn virus particles burst out of the host cell and infect other cells.
Examples of viral infections
Viruses are responsible for various human infections, as illustrated in Figure \(2\), and also in animals, plants, fungi, and bacteria. Two well-known example diseases caused by viruses, COVID-19 and AIDS, are described briefly here.
COVID-19
Coronavirus disease 2019, commonly known as COVID-19, is caused by SARS-CoV-2, discovered in Wuhan, China, in December 2019 and spread worldwide. SARS-CoV2 is a member of a family of viruses that cause various diseases, from head or chest colds to more severe conditions like severe acute respiratory syndrome (SARS). SARS-CoV-2, also called coronavirus, spreads through droplets expelled or projected out of the mouth or nose of an infected person during breathing, coughing, sneezing, or speaking.
The common name coronavirus refers to the crown-shaped spike proteins sticking out of the virus, as illustrated in Figure \(5\). Coronavirus uses these spike proteins to attach to the host cells to infect it. Like other viruses, it replicates by reverse transcription inside the host cell, as illustrated in Figure \(6\). Genetic changes do happen to the virus over time that result in variants that have different attributes regarding how fast the virus speeds or how severe illness it causes. The virus and its variants are constantly monitored to update or improve its treatment as it changes over time.
According to CDC, some antibodies can protect from coronavirus by targeting these spike proteins, but the best protection is vaccines. Several COVID-19 vaccines have been developed. The most common COVID-19 vaccines, i.e., Pfizer and Moderna vaccines, use either self-replicating RNA or mRNA, which cause the cells to produce SARS-CoV-2 spike protein. This process teaches the body's immune system how to identify and destroy the pathogen. RNA vaccines usually use modified nucleotides in mRNA.
AIDS
Acquired immunodeficiency syndrome (AIDS) is the late stage of the human immunodeficiency virus (HIV) when the virus badly damages the body's immune system. HIV infection is transmitted through sexual activity, blood transfusion, mother to child, etc., and primarily destroys CD4+ T cells, components of the human immune system.
HIV is a spherical retrovirus, as illustrated in Figure \(7\). It has two copies of RNA enclosed by a nucleocapsid and bound to enzymes needed for its development surrounded in a conical capsid. It also has an envelope of lipid bilayer taken from the cell membrane of the host cell. It also has proteins from the host cell, like glycoproteins, gp120 and gp41, that allow it to attach to the target cell to begin the infection. Destruction of CD4+ T cells by HIV infection makes the person's immune system ineffective. It increases the risk of common diseases such as tuberculosis, tumors, pneumonia, skin cancer, etc., that are rare, as illustrated in Figure \(8\).
There is no cure for AIDS, it stays for life long, but its progress can be slowed down by treatments that interfere with the life cycle of HIV at different stages. These include drugs that inhibit entry, reverse transcription, and translation. Entry inhibitors include enfuvirtide (Fuzeon) and maraviroc (Selzentry). Transcription inhibitors include nucleosides analogous to natural nucleosides, e.g., 3'-azido-2'deoxythymidine (AZT), 2',3'-dideoxyinosine (ddC), 2',3'-dideoxycytidine (ddC) and 2',3'-didehydro-2'3'-dideoxythymidine (d4T), illustrated below. They incorporate into viral DNA and prevent the formation of a sugar-phosphate backbone as they do not have \(\ce{-OH}\) groups at the 3' position of the sugar needed for this purpose. The translation inhibitors include saquinavir (Invirase), ritonavir (Norvir), fosamprenavir (Lexiva), etc.
3'-azido-2'deoxythymidine (AZT)
2',3'-dideoxyinosine (ddC)
2',3'-dideoxycytidine (ddC)
2',3'-didehydro-2'3'-dideoxythymidine (d4T) | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.07%3A_Viruses.txt |
Learning Objectives
• Understand recombinant DNA technology and its applications in biomedical technologies.
• Understand PCR and its applications in genetic testing, fingerprinting, and human genome project.
Recombinant DNA
DNA can be isolated from the cell and cut at a specific place using restriction enzymes. Fragments of DNA from different sources, e.g., an insulin gene from a human and a DNA from Escherichia coli (E. coli) cut using the same restriction enzyme, can be combined into a new recombinant DNA. The recombinant DNA can be introduced into the cell, e.g., back into E. coli, where it propagates using the cellular machinery of the host cell. It is then used to produce proteins of interest, e.g., human insulin used to treat diseases like diabetes.
How is recombinant DNA produced?
The DNA of host cells, e.g., small circular plasmids of DNA from E. coli, are isolated first. The host cells are soaked in a detergent solution that disrupts the plasma membrane and releases the plasmids. A restriction enzyme is used to cut the DNA of the host cell at a specific location. The same enzyme is used to cut a piece of donor DNA, e.g., an insulin gene from a human. The cut ends of the DNA are called sticky ends. When the cut donor and host DNA are mixed, they join at the sticky ends, as illustrated in Figure \(1\). The resultant recombinant DNA is introduced to a new culture of E. Coli. The host cells that have taken up the recombinant DNA are selected and grown. When these bacteria grow and divide, they produce the protein encoded in the inserted gene, i.e., insulin in this example. Insulin is extracted, purified, and used to treat diabetes.
Some applications of recombinant DNA technology
The recombinant DNA technology is used in food production, medicine, agriculture, and bioengineering. Some examples are listed below.
• Chymosin is an enzyme used to manufacture chees. Currently, ~60% of hard cheese is produced in the US using genetically engineered chymosin.
• Insulin is usually synthesized by inserting the human insulin gene into E. Coli, or yeast, produces insulin processed and used to treat diabetes.
• Human growth hormone (HGT) is needed for patients whose pituitary glands do not produce sufficient hormones. Recombinant HGT is commonly used for this purpose.
• Blood clotting factor VIII is needed for patients suffering from the bleeding disorder hemophilia. It is produced these days by recombinant DNA technology.
• The Hepatitis B vaccine needed to treat hepatitis B infection is produced by recombinant DNA technology.
• The recombinant HIV protein tests the presence of antibodies that may have been produced in response to an HIV infection.
• Herbicide-resistant crops, including soy, corn, canola, alfalfa, cotton, etc., have been developed that contain recombinant genes resistant to the herbicide glyphosate found in Roundup. It allowed the use of Roundup to control weeds without affecting the crop.
• Insect-resistant corps: Bacillus thuringeiensis is a bacteria that naturally produces a protein with insecticidal properties. Crops containing the recombinant gene from the bacteria hold promise to control insect predators without using an insecticide.
• Interferon produced by recombinant DNA technology is used to treat cancer and viral disease.
• The influenza vaccine is used to prevent influenza.
Polymerase chain reaction (PCR)
polymerase chain reaction (PCR) quickly produces millions of copies of a DNA segment of interest. The selected DNA is heated to denature and separate the two strands and mixed with DNA polymerase, a short synthetic DNA called primer to select the segment to be amplified, and nucleotides, to produce a complementary strand. The process is repeated several times to make millions of copies of the desired DNA fragment, as illustrated in Figure \(2\).
Genetic testing
Genetic testing is done in a laboratory to study an individual's DNA to diagnose a defective gene that may cause a congenital disease, ancestry studies, or forensic studies. For example, breast cancer may be related to defects in breast cancer genes BRCA1 and BRCA2. Patients are screened for these defects by using blood or saliva samples to extract the DNA, and the PCR technique is applied to amplify and study the defective genes. Genetic testing is also used to study the DNA of tumors or cancer cases.
Fingerprinting
Fingerprinting is a technique based on PCR that allows determining the nucleotide sequence of some areas of human DNA that are unique to individuals. A small sample from flood, skin, saliva, or semen is used to extract the DNA for amplification by PCR. Fluorescent or radioactive isotopes are incorporated in the amplified DNA for easy monitoring. The DNA is cut into pieces by restriction enzymes, placed on a gel, separated by electrophoresis, and studied to identify the individual, as illustrated in Figure \(3\). This technique is used for paternity tests, criminal investigations, and forensics.
Human genome
The human genome project meant to study all human DNA (human genome) comprehensively, was launched in October 1990 and completed in April 2003. It shows about 20,000 protein-coding genes in humans, representing only about 1.5% of the human genome. The role of the rest of the genome is still being explored. It is mainly related to regulating genes and serving as a protein recognition site. These studies provide fundamental information to study human biology, defects in DNA that lead to genetic diseases, and find cures for conditions. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.08%3A_Genetic_engineering.txt |
• 9.1: Basics of metabolism
Metabolism, its subclasses: catabolism, anabolism; stages of food catabolism, and some important compounds involved in food catabolism, including ATP/ADP, NAD+/NADH, FAD/FADH2 pairs are described.
• 9.2: Digestion of food
Stage 1 of food catabolism, i.e., hydrolysis of carbohydrates, fats, and proteins by various digestive enzymes, is described.
• 9.3: Oxidation of glucose -the glycolysis
Glycolysis -the metabolic pathway of glucose oxidation and the fate of its end product pyruvate under aerobic and anaerobic conditions are described.
• 9.4: Citric acid cycle
The third stage of food catabolism, i.e., the citric acid cycle and its eight reactions are described.
• 9.5: Oxidative Phosphorylation
Forth stage of food catabolism, i.e., i) electron transport chain through complexes I through IV, higher energy molecules involved in the process, including flavin mononucleotide, coenzyme Q, iron-sulfur clusters, and their redox chemistry, and ii) ATP synthesis mechanism in complex V, along with accounting for ATP yield, body heating by brown fat, and reactive oxygen species produced in the process are described.
• 9.6: Oxidation of fatty acids
Stage two of fat's catabolism, including beta-oxidation of fatty acids, ATP yield per fatty acid, ketogenesis causing ketosis or ketoacidosis, and catabolism of glycerol, are described.
• 9.7: Degradation of amino acids
Stage two of catabolism of proteins, i.e., catabolism of amino acids, including transamination, oxidative amination, and urea cycle that takes care of the N and processing of the carbon skeleton of the amino acid to intermediates that enter into citric acid cycle for energy production are described.
09: Food to energy metabolic pathways
Learning Objectives
• Define metabolism, its subclasses: catabolism and anabolism, and stages of food catabolism.
• Understand the structure of a typical cell and mitochondrion -the sites of catabolism.
• Understand the basic structural features and functions of some common compounds involved in the catabolism of food, including ATP/ADP, NAD+/NADH, FAD/FADH2 pairs.
• Understand the basic structural features of enzymes that are involved in their catalytic activity.
What is metabolism
A large number of chemical reactions take place in living things almost all the time. Some of these reactions synthesize the substances living things need from the raw materials available. Usually, these reactions convert simple molecules or compounds to more complex molecules or compounds at the expense of energy. For example, photosynthesis converts carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$) into glucose ($\ce{C6H12O6}$) by utilizing energy from sunlight.
$\ce{6CO2 + 6H2O + Energy (2808 kJ/mol) -> C6H12O2}$
Other reactions break down the complex molecules to release the energy needed as heat, to do work, to provide energy for synthetic reactions, to dispose of waste byproducts, etc. For example, the reverse of photosynthesis happens in the digestion of glucose.
$\ce{6CO2 + 6H2O + Energy (2808 kJ/mol)C6H12O2 -> 6CO2 + 6H2O + Energy (2808 kJ/mol)}$
Metabolism
Metabolism is chemical reactions taking place in living things needed to sustain life.
Metabolism can be subdivided into two categories of reactions.
Catabolism and anabolism
• Catabolism is a set of chemical reactions in living things that breaks down molecules to release energy or obtain intermediates needed in other reactions.
• Anabolism, called biosynthesis, is a set of reactions in living things constructing molecules from smaller units.
The figure on the right summarizes metabolism and its sub-categories: catabolism and anabolism (Copyright: Linares-Pastén, J. A. (2018), CCBYSA 4, via Wikimedia Commons)
Metabolism usually happens through a series of interconnected chemical reactions called metabolic pathways. Metabolic pathways have a lot of similarities in different species. It indicates their common origin in the early stages of the evolution of species and their retention due to their efficiency. Some diseases, such as type II diabetes and cancer, disrupt normal metabolism. The difference in metabolic pathways from the normal allows scientists to find therapeutic interventions. Some of those metabolic pathways, particularly the catabolic pathways related to the digestion of food and its conversions to obtain energy, will be described in this chapter.
Stages of catabolism of food
The catabolism of food starts from the digestion of food. It can be divided into three major stages, as illustrated in the figure on the right (Copyright: modified from Tim Vickers, vectorized by Fvasconcellos, Public domain, via Wikimedia Commons).
Stage 1
It is mainly the hydrolysis of food molecules in the digestive system, shown in the figure on the left (Copyright: National Cancer Institute, Public domain, via Wikvia Wikimedians).
Polysaccharides are hydrolyzed to monosaccharides, fats are hydrolyzed to glycerol and fatty acids, and proteins are hydrolyzed to amino acids. The products of the stage 1 reactions diffuse into the bloodstream and are transported to cells, entering the next phase of food catabolism.
Stage 2
It happens in the cells, which is the degradation of monosaccharides, fatty acids, and amino acids. It yields smaller groups, usually a two-carbon acetyl group or, in the case of some amino acids, four-carbon carboxylate groups, which enter the next stage.
Stage 3
It happens in mitochondria in the cells. Stage 3 can be divided into three sub-stages described below.
Stage 3i: It is the citric acid cycle. In this stage, the two-carbon acetyl or four-carbon carboxylate groups are oxidized to carbon dioxide ($\ce{CO2}$) at the expense of the reduction of coenzymes $\ce{NAD^+}$ and $\ce{FAD}$ to $\ce{NADH}$ and $\ce{FADH2}$.
Stage 3ii: It is electron transport where coenzymes $\ce{NADH}$ and $\ce{FADH2}$ are reduced back to their oxidized forms $\ce{NAD^+}$ and $\ce{FAD}$ at the expense of reduction of oxygen ($\ce{O2}$) into water ($\ce{H2O}$) via electron transport process with release of energy.
Stage 3iii: It is an oxidative phosphorylation process where the energy released during the electron transport stage is used to synthesize adenosine triphosphate (ATP), which is a high-energy molecule from adenosine diphosphate (ADP), and phosphate (Pi), which are lower-energy molecules. The energy is temporarily stored in the form of ATP and released wherever it may be needed by the reverse reaction, i.e., conversion of ATP into ADP and Pi.
Cell structure related to food catabolism
Stages 2 and 3 of food catabolism happen in the cells. A cell membrane surrounds a typical animal cell. Organelles are organized or specialized structures within the cells. A typical animal cell is illustrated in the right figure with some organelles labeled (Copyright; National Human Genome Research Institute, Public domain).
The nucleus in the cell contains hereditary material, i.e., DNA. The space between the cell membrane and the nucleus is called the cytoplasm. The cytosol is the fluid part of the cytoplasm containing an aqueous solution of electrolytes and enzymes that catalyze many of the cell's chemical reactions. Within the cytoplasm are organelles that perform specialized functions. For example, ribosomes are the sites for protein synthesis, and mitochondria are the cell's energy factory where stage 3 of food catabolism occurs.
Mitochondrion
The mitochondrion (plural mitochondria) has an outer membrane and inner membrane and an inter-membrane space between the two, as shown in the figure on the left (Copyright; LadyofHats, Public domain, via Wikimedia Commons). The fluid section surrounded by the inner membrane is called the matrix. The enzymes that catalyze the chemical reactions of stage 3 of food catabolism are located in the matrix and along the inner membrane. These reactions ultimately convert the food molecules into $\ce{CO2}$, $\ce{H2O}$, and energy. The enzymes that catalyze stage 4 reactions that use this energy to produce high-energy ATP also occur in the matrix along the inner membrane.
The principal compounds involved in the common metabolic pathways
The principal compounds involved in the common metabolic pathways are adenosine triphosphate ($\ce{ATP}$), which is the agent for the temporary storage of energy and transfer of phosphate ($\ce{PO4^{3-}}$ or $\ce{Pi}$) group; nicotinamide adenine dinucleotide ($\ce{NAD^+}$) and flavin adenine dinucleotide ($\ce{FAD}$) which are agents for the transfer of electrons during the biological oxidation-reduction sections; and coenzyme A ($\ce{CoA}$) which is the agent for the transfer of acetyl ($\ce{CH3CO{-}}$) group. These compounds are described here.
ATP -the energy currency and agent for the transfer of phosphate groups
Adenosine triphosphate ($\ce{ATP}$) is a nucleotide composed of adenine base bonded to ribose sugar by an N-glycosidic bond and triphosphate bonded to ribose by an ester bond, as shown in Figure $1$. The N-glycoside of adenine with ribose is a nucleoside, i.e., adenosine. When adenosine is bonded to diphosphate, it makes adenosine diphosphate ($\ce{ADP}$).
Hydrolysis of $\ce{ATP}$ splits one phosphate ( $\ce{PO4^{3-}}$ or $\ce{Pi}$) and convert it to $\ce{ADP}$ with the release of 30.5 kJ/mol energy.
$\ce{ATP + H2O -> ADP + Pi}$ $\delta$H = -30.5 kJ/mol
This energy is used in the processes that require energy, such as muscle contraction, nerve signal conduction, and biosynthesis. Catabolism of food releases energy temporarily stored in the form of $\ce{ATP}$ by reversing the above reaction.
$\ce{ADP + Pi -> ATP + H2O}$ $\delta$H = +30.5 kJ/mol
The cycle of reactions between $\ce{ATP}$ and $\ce{ADP}$ shown in Figure $1$ happen rapidly, producing one to two million $\ce{ATP's}$ per second. An average human body produces $\ce{ATP}$ about equal to the human body mass per day, but it contains about 1 g of $\ce{ATP}$ at a time.
$\ce{ADP}$ can also hydrolyze with water releasing energy and converting to adenosine monophosphate ($\ce{AMP}$
$\ce{ADP + H2O -> AMP + Pi}$ $\delta$H = -30.5 kJ/mol
$\ce{ATP}$ is also a phosphorylation agent in metabolic reactions. For example, D-glucose is phosphorylated by the reaction shown below.
In summary the main roles of $\ce{ATP}$/$\ce{ADP}$ pair are:
1. energy transfer, i.e., releases energy when $\ce{ATP}$ converts not $\ce{ADP}$ + $\ce{P_{i}}$ and absorbs energy in the reverse reaction,
2. transfer of phosphate ($\ce{P_{i}}$), i.e., releases phospahte when $\ce{ATP}$ converts not $\ce{ADP}$ + $\ce{P_{i}}$ and consumes phosphate in the reverse reaction.
$\ce{NAD^+}$ and $\ce{FAD}$ -agents for the transform of electrons
Oxidation and reduction
Oxidation is:
• loss of electron ($\ce{e^-}$),
• gain of oxygen ($\ce{O}$), or
• loss of hydrogen ($\ce{H}$).
Reduction is:
• gain of electron ($\ce{e^-}$),
• loss of oxygen ($\ce{O}$), or
• gain of hydrogen ($\ce{H}$).
Generally, oxidation reactions release energy, and reduction reactions gain energy.
Clarifications on oxidation-reduction processes
Oxidation and reductions are coupled, i.e., if one reagent is oxidized, another is reduced simultaneously. For example, in the reaction shown below, pyruvate is oxidized by losing hydrogen coenzyme $\ce{NAD^{+}}$ is reduced by gaining hydrogen. So, oxidation-reduction couples are collectively called redox reactions.
$\ce{\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + NADH + H^{+} <->[\text{Lactate dehydrogenase}] \underbrace{CH3-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{\enspace\;{CH}}}\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Lactate} + NAD^{+}}$
Oxygen may be added to the compound, i.e., make a single or double bond with the substrate, or it may change from a single bond to a double bond in the substrate, e.g., $\ce{-C-O}$ to $\ce{-C=O}$, both ways it is oxidation. That is, an increase in the bond with oxygen is oxidation.
In metabolic reactions, $\ce{H}$ is represented as a proton $\ce{H^+}$ and an electron $\ce{e^-}$. So, the addition of $\ce{H^+}$ + $\ce{e^-}$ is reduction, and their removal is oxidation. Usually, $\ce{2H^{+}}$ + $\ce{2e^{-}}$ are transferred from or to a coenzyme in metabolic reactions.
If bonds with hydrogen added or removed are counterbalanced by bonds with oxygen removed or added, the overall reaction is not a redox. For example, an alkene's hydration, shown below, is not a redox reaction.
$\ce{CH3CH2CH=CHCH2CH3 + H2O ->[H2SO4] CH3CH2CH2-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{\enspace\;{CH}}CH2CH3}}$
Broader definition of oxidation and reduction
Oxidation and reduction are not limited to making or breaking a bond with oxygen or hydrogen; broader definitions are the following.
Adding a bond with a more electronegative atom is oxidation, and its removal is reduction. The opposite is true for a less electronegative atom, i.e., The addition of a bond with a less electronegative atom is reduction, and its removal is oxidation. For example, conversion of $\ce{R-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{\enspace\;{CH}}-R'}}$ into $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R'}$ replaces bond on $\ce{C}$ from less electronegative $\ce{H}$ with more electronegative $\ce{O}$ is oxidation. Similarly, conversion of $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-CH2R'}$ into $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-SCH2R'}$ replaces bond on $\ce{C}$ from a $\ce{C}$ with more electronegative $\ce{S}$ is oxidation. The reverse of these is reductions.
Nicotinamide adenine dinucleotide ($\ce{NAD^{+}}$)
Nicotinamide adenine dinucleotide is a coenzyme composed of two nucleotides, an adenosine diphosphate (ADP) and a second nucleotide in which the nitrogen base is nicotinamide provided by vitamin niacin. The two nucleotides are linked by diphosphate linkage, as shown in Figure $2$. The oxidized form is represented as $\ce{NAD^{+}}$ and its reduced form is represented as $\ce{NADH}$. The $\ce{NAD^{+}}$ is reduced to $\ce{NADH}$ by reacting with two hydrogen ($\ce{2H^{+}}$ + $\ce{2e^{-}}$) leaving one $\ce{H^{+}}$ in the products, as shown in Figure $2$. Oxidation of $\ce{NADH}$ to $\ce{NAD^{+}}$ is the reverse reaction.
The $\ce{NADH}$/$\ce{NAD^{+}}$ redox is coupled with $\ce{\underbrace{R-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{\enspace\;{CH}}}\!-R'}}_{Alcohol}$ / $\ce{\underbrace{R-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R'}}_{Carbonyl}$, redox reactions or it involves carbonyl ($\ce{C=O}$) group in food catabolism .
An example of an oxidation reaction in metabolism is the oxidation of the alcohols ($\ce{-OH}$) group to carbonyl ($\ce{C=O}$) group. For example, ethanol is oxidized to ethanal in the liver at the expense of reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$ as shown below.
Flavin adenine dinucleotide ($\ce{FAD}$)
Flavin adenine dinucleotide is a coenzyme composed of two nucleotides, an adenosine diphosphate (ADP) and a second nucleotide which is riboflavin (vitamin B2). Riboflavin is composed of flavin and ribitol, which is a sugar alcohol. The two nucleotides are linked by an ester linkage between ribitol and diphosphate, as shown in Figure $3$. The oxidized form is represented as $\ce{FAD}$, and its reduced form is represented as $\ce{FADH_{2}}$. The oxidation-reduction happens in the $\ce{N}$ containing rings of the flavin part. The $\ce{FAD}$ is reduced to $\ce{FADH_{2}}$ by reacting with two hydrogen ($\ce{2H^{+}}$ + $\ce{2e^{-}}$), as shown in Figure $2$. Oxidation of $\ce{FADH_{2}}$ to $\ce{FAD}$ is the reverse reaction.
The $\ce{FADH_{2}}$/$\ce{FAD}$ redox is coupled with $\ce{\underbrace{RR'\!\!\!\!\!{\overset{\;\;\;\overset{\huge\;{H}}|}{\enspace\;{C}}}\!-\!\!\!\!\!{\overset{\;\;\;\overset{\huge\;{H}}|}{\enspace\;{C}}}\!RR'}}_{Single bond}$ / $\ce{\underbrace{RR'C=CRR'}_{Double bond}}$, redox reactions, i.e., it involves $\ce{C=C}$ bond in food catabolism.
An example is the oxidation of a $\ce{C-C}$ bond to $\ce{C=C}$ bond of fumarate at the expense of reduction of $\ce{FAD}$ to $\ce{FADH_{2}}$ as shown below.
Another example is the following step in the $\beta$-oxidation of fatty acids catalyzed by acyl-$\ce{CoA}$ dehydrogenase.
Coenzyme A ($\ce{CoA}$) -the agent for transfer of acetyl ($\ce{CH3CO{-}}$) group
Coenzyme A ($\ce{CoA}$) is composed of several components, as illustrated in Figure $4$.
The reactive part of $\ce{CoA}$ is the thiol ($\ce{-SH}$) group which forms a high-energy thioester bond with acyl groups. Free coenzyme is usually represented as $\ce{HS-CoA}$, and when it is as a thioester of the acyl group, it is represented as $\ce{Ac-S-CoA}$, where $\ce{Ac{-}}$ is an acyl group. For example, pyruvate transfers its acetyl ($\ce{CH3CO{-}}$) groups to $\ce{HS-CoA}$ in the following reaction, which is a step in the catabolism of carbohydrates.
The $\ce{-S-CoA}$ group in $\ce{Ac-S-CoA}$ is a good leaving group that easily transfers the acyl group to other compounds during biosynthesis. For example, the acetyl group is transferred from $\ce{Ac-S-CoA}$ to an acyl carrier protein (ACP) which is a step in the synthesis of fatty acids.
Role of enzymes in metabolism
Nearly all metabolic reactions in living things are catalyzed by enzymes. Unlike chemical reactions in laboratories where different solvents, extreme temperature, and pressure conditions can be applied, and strong acids or bases can be used to catalyze the reactions, biomedical reactions in living things must occur under physiological conditions. Enzymes are specialized catalysts that catalyze the reactions under physiological conditions. The major factors responsible for the catalytic activity of the enzymes are the following.
1. Enzymes have functional groups in their active sites that act as handles to bind with and lock the substrate in a proper orientation such that the reagent can easily approach the reactive site on the substrate. This way, any steric hindrance to the reaction is minimized.
2. Enzymes and their co-factors also usually neutralize the ionic groups on the substrate that otherwise repel the charged or partially charged electrophile or nucleophile from approaching the substrate. This way, the electrostatic factors hindering the reaction are minimized.
3. Enzymes usually have functional groups in their active sites that catalyze the reaction of the substrate by activating it as a nucleophile, electrophile, acid, or base catalyst.
The following two examples from glucose catabolism illustrate the factors mentioned above.
The enzyme malate dehydrogenase catalyzes the oxidation of malate to oxaloacetate in the citric acid cycle. The substrate is bound by the enzyme in such a way that i) the charge of its two carboxylates ($\ce(-COO^{-}}$) groups are neutralized by arginine side chains of the enzyme, ii) histidine side acts as a base catalyst and removes the proton from the $\ce{-OH}$ group of the substrate, and at the same time, iii) the $\ce{H}$ on the $\ce{C}$ carrying the $\ce{-OH}$ group is exposed to nicotine amide ring of $\ce{NAD^{+}}$ that removes the proton, as illustrated in Figure $1$,
Another example is the formation of citrate from oxaloacetate and acetyl-$\ce{CoA}$ in the citric acid cycle, catalyzed by the enzyme citrate synthase as illustrated in Figure $2$.
1. The substrate is bonded by the enzyme in such a way that the two substrates are at a bonding distance. The $\ce{-CH3}$ hydrogens of acetyl-$\ce{CoA}$ are acidic protons due to being $\alpha$ to the carbonyl group. The asparagine side chain acts as a base catalyst and removes the acidic proton from $\ce{-CH3}$ group of acetyl-$\ce{CoA}$. The proton transfer to asparagine is facilitated by another proton transfer from histidine (His274) to the $\ce{C=O}$ group converting it into its enol form -a nucleophile.
2. The enol returns its proton to the histidine side chain in the second step activating the $\ce{C=C}$ bond of the enol for a nucleophilic attack on the electrophilic ketone ($\ce{C=O}$ group of oxaloacetate substrate, in the second step. This step is facilitated by the transfer of another proton from another histidine ((His320), the ($\ce{C=O}$) group involved in the reaction. The result is the conversion of the $\ce{C=O}$ group into an ($\ce{-O-H}$ group of citroyl-$\ce{CoA}$ ubterneduate.
3. Citroyl-$\ce{CoA}$ intermediate is hydrogen-bonded to a histidine (His274). This hydrogen bond activates the thioester group for a hydrolysis reaction. Water molecule hydrolyzes the thioester group releasing the citrate product from the enzyme in this reaction step.
Nearly every metabolic reaction is catalyzed by an enzyme in a similar way as described above. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.01%3A_Basics_of_metabolism.txt |
Learning Objectives
• Understand what happens in stage 1 of food catabolism.
Food digestion, stage 1 of food catabolism, happens in the digestive system shown in the figure on the right (Copyright: public domain, via Wikimedia Common). It is primarily the hydrolysis of biopolymers or larger molecules in food into smaller molecules that can transfer from the digestive system to the bloodstream and transport to cells. Carbohydrates are hydrolyzed into monosaccharides, fats into fatty acids and mono- or di-glycerides, proteins into amino acids, and DNA and RNA into mononucleotides by different enzymes.
Digestion of carbohydrates
The carbohydrates in human food are starch, sucrose, and lactose. Their digestion begins in the mouth through enzymes in the saliva and continues in the small intestine by pancreatic amylase. The enzymes degrade the starch into smaller and smaller fragments that ultimately result in glucose and maltose, which the small intestine can absorb.
Sucrose (table sugar) is hydrolyzed into glucose and fructose by the enzyme sucrase. Lactose found in milk is hydrolyzed into glucose and galactose by the enzyme lactase. The majority of the adult population have problems digesting unfermented dairy because they can not produce sufficient amounts of the enzyme lactase -a condition called lactose intolerance.
Digestion of fats
Digestion of fats begins in the mouth through lingual lipase, but it mainly happens in the small intestine. Bile acids emulsify the fats, and the pancreatic enzyme lipase hydrolysis them into free fatty acids and mono- and diacyl-glycerides.
Digestion of proteins
Digestion of proteins happens in the stomach and the duodenum, i.e., the first portion of the small intestine. Three enzymes: pepsin, secreted by the stomach, and trypsin and chymotrypsin, secreted by the pancreas, hydrolyze proteins into smaller peptides which are then hydrolyzed into amino acids by various exopeptidases and dipeptidases.
9.03: Oxidation of glucose -the glycolysis
Learning Objectives
• Understand glycolysis and the essential reactions in this metabolic pathway responsible for glucose oxidation to two pyruvate molecules.
• Understand the reactions that pyruvate undergoes in the absence and presence of oxygen before entering the third stage of catabolism.
Glycolysis
Oxidation of glucose is the 2nd stage of the catabolism of carbohydrates. It happens in the cytoplasm of the cell. It is a metabolic pathway consisting of ten glycolysis reactions, as illustrated in Figure $1$.
The ten reactions of glycolysis are the following (copyright: modified from a Public Domain resource at Wikipedia.)
1. Glucose is converted to glucose-6-phosphate by an SN2 reaction mechanism between a primary alcohol acting as a nucleophile and a $\ce{P}$ of $\ce{ATP}$ as an electrophile, where $\ce{ADP}$ acts as a good leaving group, as shown below.
2. Glucose-6-phosphate, which is an aldohexose, isomerizes to fructose-6-phosphate, which is a ketohexose.
3. The same mechanism phosphorylates the primary alcohol group of fructose-6-phosphate as in the first step producing fructose-1,6-bisphosphate.
4. The open-chain form of fructose-1,6-bisphosphate, which is in equilibrium with the cyclic form, splits into two compounds, glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.
5. This reaction is an isomerization reaction between glyceraldehyde-3-phosphate and its isomer dihydroxyacetone phosphate. The aldehyde or ketone in these two isomers tautomerizes to their enol forms, which either revert to the initial compound or to its isomers when the enol form changes back to an aldo or keto form.
6. The aldehyde ($\ce{-HC=O}$) group of glyceraldehyde-3-phosphate is oxidized in three steps: activated by the addition of $\ce{HS-CoA}$ (step 6i), followed by oxidation of $\ce{-OH}$ to a carbonyl ($\ce{C=O}$ group at the expense of reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$ (step 6ii)); and finally displacement of $\ce{HS-CoA}$ by a phosphate ($\ce{-PO4^{2-}}$) group (step 6iii). The overall reaction is the following.
The ketone group in dihydroxyacetone phosphate does not oxidize directly but converts to glyceraldehyde-3-phosphate due to the equilibrium creation#5. So, reaction#6 and reactions after this happen twice for each glucose molecule.
7. A phosphate group is transferred from 1,3-bisphosphoglycerate to ADP producing an ATP and a 3-phosphoglycerate.
Since this step happens twice for each glucose molecule processed, it compensates for the two ATP consumed, one in step#1 and the other in step#3.
8. 3-Phosphoglycerate is isomerized to 2-phosphoglycerate. In this step, the enzyme phosphoglycerate mutase first transfers its phosphate group to the $\ce{-OH}$ at position#2 and then receives the phosphate from position#3 of the intermediate, resulting in the isomerization of 3-Phosphoglycerate to 2-phosphoglycerate.
9. 2-Phosphoglycerate is dehydrated, i.e., $\ce{H2O}$ is eliminated form it, producing phosphoenolpyruvate.
10. Phosphoenolpyruvate transfers its phosphate group to ADP producing pyruvate and ATP.
Since the reaction happens two times for each glucose, two ATP are produced in this step.
In summary, glycolysis is the oxidation of glucose without the need for oxygen, i.e., anaerobic oxidation, consisting of two phases. The first phase comprises reactions#1 to 5. It is the preparatory phase, which is also energy consuming phase in which a six $\ce{C's}$ glucose ($\ce{C6H12O6}$) is phosphorylated twice at the expense of two ATP's and converts to two glyceraldehyde-3-phosphate molecules having three $\ce{C's}$ each. The second phase, comprising reactions#6 to 10, is the payoff phase in which four ATP's are produced, i.e., two more than the ATP consumed in the first phase. Two $\ce{NAD^{+}}$ are reduced to two $\ce{NADH}$ along with the production of two pyruvates $\ce{CH3COCOO^{-}}$ in the second phase, as shown in the following reaction.
$\ce{\underbrace{C6H12O6}_{Glucose} + 2NAD^{+} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + 2NADH + 2ATP + 2H^{+} + 2H2O}$
Fate of pyruvate
Two $\ce{NAD^{+}}$ are reduced to two $\ce{NADH}$ which need to be oxidized back to $\ce{NAD^{+}}$ for the process to continue. It happens in different ways depending on whether sufficient oxygen is present, i.e., aerobic condition or aerobic respiration, or if there is no oxygen, i.e., anaerobic condition. The fate of pyruvate also depends on whether the condition is aerobic or anaerobic, as described below.
Aerobic condition
In aerobic conditions, $\ce{NADH}$ is not oxidized at this stage, it is oxidized to $\ce{NAD^{+}}$ at the expense of oxygen in mitochondria. The energy released by the oxidation of $\ce{NADH}$ is used to produce more $\ce{ATP's}$ in the third stage of catabolism. Pyruvate is also transferred to mitochondria and undergoes oxidation at the cost of $\ce{NAD^{+}}$ reduced to $\ce{NADH}$ through a series of reactions catalyzed by a complex of three enzymes and five coenzymes, collectively called the pyruvate dehydrogenase complex. The overall reaction is the decarboxylation of pyruvate, i.e., carbon dioxide ($\ce{CO2}$) eliminated. The acetyl ($\ce{CH3-CO-}$) group is transferred to $\ce{HS-CoA}$ producing acetyl-$\ce{CoA}$, as shown in the following reaction.
$\ce{\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + HS-CoA + NAD^{+} ->[\text{Pyruvate dehydrogenase complex}] \underbrace{CH3-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-CoA}_{Acetyl\;CoA} + CO2 + NADH + H^{+}}$
Acetyl-$\ce{CoA}$ enters into the citric acid cycle, also called the Krebs cycle, in mitochondria, the third stage of catabolism.
The processing of pyruvate under aerobic conditions is also called oxidative decarboxylation or link reaction, which links glycolysis to the citric acid cycle, as explained in the video below.
Two pyruvates are produced per glycolysis of one glucose molecule. The overall reaction of one glucose molecule under aerobic conditions before entry into the citric acid cycle is the following.
$\ce{\underbrace{C6H12O6}_{Glucose} + 2HS-CoA + 4NAD^{+} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-CoA}_{Acetyl\;CoA} + 2CO2 + 4NADH + 2ATP + 4H^{+} + 2H2O}$
There is a net gain of six energetic molecules: $\ce{2ATP}$ and $\ce{4NADH}$ from glycolysis and link reaction under aerobic conditions.
Anaerobic condition
During vigorous exercises, as shown in the figure on the right, or strenuous physical work, oxygen depletes in muscles creating anaerobic (oxygen-free) conditions. Under anaerobic conditions, $\ce{NADH}$ is oxidized to $\ce{NAD^{+}}$ in the cytoplasm at the expense of the reduction of pyruvate to lactate, as shown in the reaction below.
$\ce{\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + NADH + H^{+} ->[\text{Lactate dehydrogenase}] \underbrace{CH3-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{\enspace\;{CH}}}\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Lactate} + NAD^{+}}$
Overall, glycolysis of glucose under anaerobic conditions is the following reaction.
$\ce{\underbrace{C6H12O6}_{Glucose} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{C}}\!\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Lactate} + 2ATP}$
The two $\ce{NADH}$ produced in the glycolysis of one glucose are consumed in the anaerobic conversion of two pyruvates into two lactates.
• There is a net gain of two energetic molecules, i.e., $\ce{2ATP}$ in glycolysis of one glucose followed by anaerobic conversion of two pyruvates into two lactates.
• The accumulation of lactate makes muscles tire and sour. The person keeps beating heavily after the exercise to pay the oxygen debt. Most of the lactate is transported to the liver, where it is re-oxidized to pyruvate.
Fermentation
Instead of converting pyruvate to lactate, as in humans and animals, yeast has an enzyme called pyruvate decarboxylase that decarboxylates pyruvate to acetaldehyde. Then $\ce{NADH}$ is oxidized to $\ce{NAD^{+}}$ at the expense of the reduction of acetaldehyde to ethanol in a process called fermentation, as shown in the reaction below.
Copyright; User: Cwernert at en.Wikipedia, vectorized by User: GKFX., CC BY-SA 3.0, via Wikimedia Commons.
The overall reaction of glycolysis of glucose in the fermentation process by yeast.
$\ce{\underbrace{C6H12O6}_{Glucose} + 2ADP + 2Pi -> 2\underbrace{CH3-CH2-OH}_{Ethanol} + 2CO2 + 2ATP}$ | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.02%3A_Digestion_of_food.txt |
Learning Objectives
• Understand the citric acid cycle, its reactions, and the yield of energetic molecules.
What is the citric acid cycle?
The citric acid cycle is the first part of the third stage of food catabolism, including carbohydrates, fats, and proteins. It is called the citric acid cycle because a $\ce{2C's}$ acetyl-$\ce{CoA}$ produced in the second stage of catabolism of foods reacts with a $\ce{4C's}$ oxaloacetate and produces $\ce{6C's}$-citrate in the first reaction. The citrate is oxidized through a series of eight reactions producing two carbon dioxide ($\ce{CO2}$) and re-generates the oxaloacetate to repeat the next round of the cycle, as shown in Figure $1$. It is also called the tricarboxylic acid cycle because citrate has three carboxylates ($\ce{-COO^{-}}$) groups, i.e., triacid. Another name for it is the Krebs cycle in honor of Hans Krebs, who discovered this metabolic pathway.
The oxidation reactions in the citric acid cycle release energy, which is coupled to the reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$, $\ce{FAD}$ to $\ce{FADH}$, or conversions of $\ce{ADP}$ to high-energy $\ce{ATP}$. Guanosine triphosphate ($\ce{GTP}$) -another high-energy molecule is also produced from guanosine diphosphate ($\ce{GDP}$), but this reaction ultimately reverses to transfer its energy to make $\ce{ATP}$.
Reactions of the citric acid cycle
The following equations of the reactions are modified from a public domain at Wikipedia.
1. To understand the first reaction, recall that the $\ce{-CH3}$ group of acetyl-$\ce{CoA}$ has relatively acidic protons as it is $\alpha$ to a carbonyl ($\ce{C=O}$) group. An enzyme, acting as a base, removes a proton from the $\ce{-CH3}$ group making it a carbanion which, being a strong nucleophile, attacks the ketone-$\ce{C}$ of oxaloacetate, which is an electrophile. This nucleophilic addition reaction converts the $\ce{C=O}$ into an $\ce{-OH}$ group. It is followed by the hydrolysis of the thioester by $\ce{H2O}$ through nucleophilic acyl substitution mechanism producing citrate and $\ce{HS-CoA}$, as shown in the following overall reaction.
2. To understand the second reaction, recall that tertiary $\ce{-OH}$ do not oxidize but are easily eliminated through E2-dehydration mechanics. The tertiary $\ce{-OH}$ of citrate is eliminated, and then $\ce{H2O}$ adds to the alkene intermediate, but it installs a secondary $\ce{-OH}$ group in isocitrate product, as shown below.
3. The third reaction has three steps. In the first step, the secondary $\ce{-OH}$ in isocitrate is oxidized to a ketone group at the expense of reduction of a $\ce{NAD^{+}}$ into $\ce{NADH}$. In this step, one $\ce{-H}$ is picked up by an enzyme ($\ce{A^{-}}$) and the second by the nicotinamide ring (shown in blue color) of the coenzyme $\ce{NAD^{+}}$. The $\ce{COO^{-}}$ group in the intermediate-I, being $\beta$ to the $\ce{C=O}$ group, eliminates (decarboxylate) easily as $\ce{CO2}$. The decarboxylation is facilitated by cofactor $\ce{Mg^{2+}}$ that binds with and draws electrons from the $\ce{O's}$. The enzyme $\ce{H-A}$ protonates the enole $\ce{C=C}$ bond of the intermediate-II that produces $\alpha$-ketoglutarate, as shown below.
4. The fourth reaction resembles the previous link reaction's oxidative decarboxylation of pyruvate to acetyl-$\ce{CoA}$. In this reaction, $\ce{NAD^{+}}$ is reduced into $\ce{NADH}$ at the expense of oxidative decarboxylation of $\alpha$-ketoglutarate and the resulting acyl-group is transferred to $\ce{HS-CoA}$ resulting in succinyl-$\ce{CoA}$ product as shown in the reaction below.
5. The fifth reaction has two steps. In the first step, a phosphate ($\ce{Pi}$) group substitutes $\ce{-S-CoA}$ group from succinyl-$\ce{CoA}$. In the second step, the phosphate group is transferred to $\ce{GDP}$ that converts to a $\ce{GTP}$ by the same mechanism as in steps #7 and step#10 of glycolysis. $\ce{GTP}$ reverts back to $\ce{GDP}$ and transfers its phosphate group to $\ce{ADP}$ that converts to one $\ce{ATP}$. So, production of one $\ce{GTP}$ is equal to production of one $\ce{ATP}$. The overall reaction five is shown below.
6. In the sixth reaction, one $\ce{FAD}$ is reduced into $\ce{FADH_{2}}$ at the expense of oxidation of succinate to fumarate by the following overall reaction.
7. In the seventh reaction $\ce{H2O}$ hydrates the $\ce{C=C}$ of fumarate producing malate, as shown below.
Recall that the enzymes are stereospecific. In this case, only fumarate (not its cis-isomer) reacts with the enzyme, and only (S)-malate (not its enantiomers (R)-malate) is produced.
8. In the eighth step, one more $\ce{NAD^{+}}$ is reduced into $\ce{NADH}$ at the expense of oxidation of malate into oxaloacetate, as shown in the reaction below.
The oxaloacetate is the reactant of step#1 and begins the cycle again by reacting with another acetyl-. One citric acid cycle has the following overall reaction with one acetyl-$\ce{CoA}$.
$\ce{ \underbrace{CH3-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-CoA}_{Acetyl\;CoA} + 3NAD^{+} + FAD + GDP + P_{i} + 2H2O -> HS-CoA + 3NADH + 3H^{+} + FADH_{2} + GTP + 2CO2}$
Recall that the catabolism of one glucose molecule produces two pyruvates that, intern, produce two acetyl-$\ce{CoA}$ in the oxidative decarboxylation of pyruvate under aerobic conditions. So, the citric acid cycle runs twice for the catabolism of one glucose molecule. One $\ce{GTP}$ ultimately converts into one $\ce{ATP}$.
Two citric acids from one glucose molecule yield ten energetic molecules from two turns citric acid cycles: $\ce{6NADH}$, $\ce{2FADH_{2}}$, and $\ce{2ATP}$
Predicting the production or consumption of energetic molecules in food catabolism
The information in this note is meant to help predict which energetic molecule will be produced or consumed during food catabolism reactions. Recall that oxidation generally release energy (exothermic), and reduction consumes energy (endothermic). The energy released from oxidation reactions in food catabolism is utilized to produce one of the two energetic coenzymes, i.e., either $\ce{NADH}$ from the reduction of $\ce{NAD^{+}}$ or $\ce{FADH2}$ from the reduction of $\ce{FAD}$. Recall that $\ce{C=O}$ is stronger bond (~800 kJ/mol bond dissociation energy) compared to $\ce{C=C}$ bond (~600 kJ/mole bond dissociation energy). $\ce{NADH}$ is more energetic than $\ce{FADH2}$. Therefore, any oxidation reaction involving carbonyl ($\ce{C=O}$) group produces $\ce{NADH}$ and any oxidation involving $\ce{C=C}$ bond produces $\ce{FADH2}$. There is only one reaction described in food catabolism in this book, i.e., reaction#6 in the citric acid cycle that involved oxidation of $\ce{C-C}$ to $\ce{C=C}$ bond and produced $\ce{FADH2}$. All other oxidation reactions in the citric acid cycle and glycolysis involve $\ce{C=O}$ group, i.e., either alcohol ($\ce{CH-OH}$) to carbonyl ($\ce{C=O}$) conversion, or ketone ($\ce{R-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-R'}$ to thioester ($\ce{R-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-R"}$) conversions produce $\ce{NAD}$. Conversionerisons of pyruvate to lactate and ethanol under anaerobic conditions are reduction reactions involving $\ce{C=O}$ group and consume $\ce{NADH}$.
Reactions that are not redox reactions, e.g., acyl ($\ce{ R-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-}$) transfer, hydration (addition of $\ce{H2O}$) , dehydration (elimination of $\ce{H2O}$), and isomerization, involve a small amount of energy. They are usually not coupled with the formation or consumption of energetic molecules like $\ce{NADH}$, $\ce{FADH2}$, or $\ce{ATP}$.
Exceptions include phosphate ester or thioester group, which are high-energy groups. The addition or removal of phosphate ester or thioester groups without redox reaction is usually coupled with $\ce{ATP}$/ $\ce{ADP}$ conversion. Two reactions, i.e., reaction#1 and reaction#3 in glycolysis (Figure 9.3.1) add phosphate ester bond and consume $\ce{ATP}$, and two reactions, i.e., reaction#& and reaction#10, remove phosphate ester and produce $\ce{ATP}$. One reaction, i.e., reaction#5 in the citric acid cycle (Figure 9.4.1), involves the replacing a thioester group with a carboxylate group without redox. It is coupled with the production of $\ce{GTP}$, which, in turn, is associated with the output of $\ce{ATP}$.
In summary:
• redox reactions involving $\ce{C=O}$ group relate to $\ce{NADH}$,
• redox reactions involving $\ce{C=C}$ group relate to $\ce{FADH2}$, and
• all other reactions that involve energetic groups like phosphate esters or thioesters relate to $\ce{ATP}$ in food catabolism.
$\ce{NADH}$ and $\ce{FADH}$ are high-energy molecules that are oxidized by $\ce{O2}$ in part two of the third stage of catabolism of food. The energy released from the oxidation of $\ce{NADH}$ and $\ce{FADH}$ is used to produce more $\ce{ATPs}$, which is described in the next section.
The following video summarizes the citric acid cycle, also called the Krebs cycle. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.04%3A_Citric_acid_cycle.txt |
Learning Objectives
• Understand parts 2 and 3 of the third stage of food catabolism, i.e., oxidative phosphorylation, and the basics of the redox chemistry of the molecules involved, including flavin mononucleotide, ubiquinone, and cytochrome c.
• Understand the electron transport process basics, including basic redox reactions and proton pumping in complexes I through IV.
• Understand the ATP synthesis process, accounting for the ATP yield, body heating, and reactive oxygen species produced.
What is oxidative phosphorylation?
In food catabolism up to this point, the energy is transferred from food to high-energy molecules, such as energetic electron carriers $\ce{NADH}$ and $\ce{FADH2}$. Recall that these electron carrier molecules do not deliver elections alone; they deliver protons and electrons, i.e., $\ce{2H^{+} + 2e^{-}}$. The electrons in $\ce{NADH}$ and $\ce{FADH}$ are transported through a series of enzymes located in the inner membrane of mitochondria, and ultimately, the $\ce{2H^{+} + 2e^{-}}$ reduce $\ce{O}$ into $\ce{H2O}$ by the reaction shown below.
$\ce{4H^{+} + 4e^{-} + O2 -> 2H2O}$
In each step, the electrons are transferred from a substance with a higher reduction potential to one with a lower reduction potential. Energy is released in each step proportional to the difference in the reduction potential. The energy is converted to electrochemical potential energy by pumping protons ($\ce{H^{+}}$) from the matrix to intermembrane spaces through sets of enzymes called complexes I, III, & IV. The electrochemical energy is harnessed by $\ce{ATP}$ synthase, which allows protons to flow back from the intermembrane space to the matrix and couples the energy released to synthesize higher energy $\ce{ATP}$ by phosphorylating lower energy $\ce{ADP}$ with inorganic phosphate ($\ce{P_i}$). These interconnected processes of electron transport and $\ce{ATP}$ synthesis are collectively called oxidative phosphorylation. It happens in the mitochondria of cells as illustrated in Figure $1$ and is explained in the following sections.
Electron and proton transport molecules
In addition to $\ce{NAD^{+}}$ and $\ce{FAD}$, the following molecules are involved in the transport of electrons and protons in the electron transport chain. The first electron acceptor from $\ce{NADH}$ or $\ce{FADH2}$ in the electron transport chain is flavin mononucleotide which changes from its oxidized form represented as $\ce{FMN}$ to its reduced form represented as $\ce{FMNH2}$ as illustrated in the figure on the right. Structure of $\ce{FMN}$ comprises three parts, phosphate, ribitol, and flavin. It is the same as one of the two nucleotides in flavin adenine dinucleotide. Nitrogen atoms in flavin part of $\ce{FMN}$ take part in the redo process, as illustrated in the figure on the right. $\ce{FMNH2}$ pass on the electrons to iron-sulfur clusters, which include $\ce{[2Fe-2S]}$ and $\ce{[4Fe-4S]}$ clusters. The redox couple in iron-sulfur clusters is $\ce{Fe^{3+}/Fe^{2+}}$, i.e., the following redox reaction:
$\ce{Fe^{3+} + e^{-} <=>[Reduction][Oxidation] Fe^{2+}}$.
Finally, the electrons are transferred to coenzyme-Q, which in oxidized form is ubiquinone represented as $\ce{Q}$ and in its reduced form is ubiquinol represented as $\ce{QH2}$, as illustrated in the figure above on the left. Coenzyme-Q is hydrophobic, i.e., lipid-soluble. Coenzyme-Q moves freely in the hydrophobic environment within the biliary inner membrane to pass on electrons from complex I and II to complex III. Coenzyme-Q not only can accept two electrons and two protons, but it can also accept one electron to become semiubiquinone, a resonance-stabilized radical anion.
Cytochrome c (Cyt c) is another electron carrier protein with a heme group attached. Heme group is illustrated in the figure on the right. Iron in the heme group of cytochrome c is the redox couple ($\ce{Fe^{3+}/Fe^{2+}}$) as in the case of iron-sulfur clusters. $\ce{Fe^{3+}}$ changes to $\ce{Fe^{2+}}$ when an electron is added to it, and the reverse happens when an electron is removed. Cytochrome c is water soluble and travels between complex III and complex IV in the watery environment along the inner membrane's surface facing intermembrane space.
Electron transport chain
In this part 2 of stage 3, electrons are removed from $\ce{NADH}$ and $\ce{FADH2}$ and passed on to reduce $\ce{O2}$ into $\ce{H2O}$ through a series of enzyme complexes, i.e., complexes I through IV and mobile electron carriers like ubiquinone ($\ce{Q}$) and cytochrome c ($\ce{Cyt \; c}$) by a process called electron transport chain. At each step, electrons pass from a substance with a higher reduction potential to one with a lower one. Energy is released proportional to the difference in the reduction potentials.
The complex I, III, and IV extend from the matrix to the intermembrane space through the inner membrane, as illustrated in Figure $1$. These three complexes utilize the energy released to pump protons ($\ce{H^{+}}$) from the matrix to the intermembrane space. Complex II extends to the matrix from the inner membrane but does not cross the membrane to intermembrane space. Therefore complex II transports electrons but does not pump protons. Coenzyme-Q ($\ce{Q}$) is a hydrophobic electron carrier molecule that moves in a hydrophobic environment inside the lipid bilayer of the inner membrane from complexes I and II to complex III. Cytochrome c ($\ce{Cyt \; c}$) is a hydrophilic electron carrier molecule that moves along the surface of the inner membrane facing the intermembrane space from complex III to complex IV. The enzyme complexes and the mobile electron carriers involved in the electron transport chain and $\ce{ATP}$ synthesis are described in the following sub-sections, where the chemical equations and figures are taken from the public domain source via Wikipedia.
Complex I
Complex I, also known as NADH-coenzyme Q oxidoreductase or NADH dehydrogenase, is the first enzyme complex in the electron transport chain. Complex I is illustrated in the figure on the right. The electron transport begins when an NADH binds with and donates two electrons to complex I. The electrons are received by flavin mononucleotide ($\ce{FMN}$) attached to the complex, which is reduced to $\ce{FMNH2}$ form. Then the electrons are transferred through a series of iron-sulfur clusters within complex I to coenzyme-Q ($\ce{Q}$) that, in turn, reduces to its reduced form $\ce{QH2}$. $\ce{QH2}$ travels through the lipid bilayer to and delivers the electrons to complex III.
As the electrons move through complex I and release part of their energy, four protons ($\ce{4H^{+}}$) are pumped from the matrix to intermembrane space by utilizing the energy released. It generated polarity across the inner membrane, with the positive side (P-side) facing the intermembrane area and the negative side (N-side) facing the matrix. The exact mechanics of proton pumping is not yet clearly understood. Still, the research indicates that it happens through conformational changes in complex I such that the protein bind protons on the N-side and releases them on the P-side of the membrane. The overall redox reaction in complex I is the following.
$\ce{NADH + Q +5H^{+}_{matrix} -> NAD^{+} + QH_{2} + 4H^{+}_{intermembrane}}$
$\ce{NAD^{+}}$ liberated in this reaction becomes available to oxidize more substrates in the metabolic pathways.
Complex II
Complex II, also known as succinate-Q oxidoreductase or succinate dehydrogenase, is a second entry point to the electron transport chain. It is part of the citric acid cycle and the electron transport chain. It contains protein subunits bonded with flavin adenine dinucleotide $\ce{FAD}$, iron-sulfur clusters, and a heme group, as illustrated in the figure on the right. Succinate is oxidized to fumarate in the citric acid cycle at the expense of oxidation of $\ce{FAD}$ to $\ce{FADH2}$ in complex II. The electrons and protons transfer from $\ce{FADH2}$ through iron-sulfur clusters to coenzyme-Q in complex II by the following overall reaction.
$\ce{Succinate + Q -> Fumarate + QH2}$
Coenzyme-Q passes on the electrons from complex II to complex III as from complex I to complex III.
Recall that (\ce{FADH2}\) is less energetic than (\ce{NADH}\). Since oxidation $\ce{FADH2}$ to $\ce{FAD}$ at complex II releases less energy than oxidation of $\ce{NADH}$ to $\ce{NAD^{+}}$ at complex I, complex II do not pump protons from matrix to intermembrane space. The energy released by electron transport is utilized to pump protons in complex III and IV, as described in the next section.
Complex III
Complex III, also known as cytochrome C reductase or cytochrome bc1 complex, contains protein subunits, an iron-sulfur cluster, and cytochromes. A cytochrome is a protein with at least one heme group that transfers one electron by accepting and releasing an electron based on the following reaction:
$\ce{Fe^{3+} + e^{-} <=> Fe^{2+}}$
Complex III passes on electrons from coenzyme-Q ($\ce{QH2}$) to cytochrome c ($\ce{Cyt~ c}$). Since $\ce{QH2}$ delivers two electrons ($\ce{e^{-}}$) while $\ce{Cyt~ c}$ can accept only one, the process happens in two steps, as illustrated in the figure on the right. In the first step, $\ce{QH2}$ transfers one $\ce{e^{-}}$ to $\ce{Cyt~ c}$ and one to another coenzyme-Q in quinone ($\ce{Q}$) form and releases $\ce{2H^{+}}$. $\ce{Q}$ converts to a semiubiquinone ($\ce{Q^{\bullet -}}$). In the second step, another $\ce{QH2}$ transfers one electron to a second $\ce{Cyt~ c}$ and one to $\ce{Q^{\bullet -}}$ converting it to $\ce{Q}$ which is released from the complex and repeats the above two steps. This way, two electrons are transferred from a $\ce{Q}$ to two $\ce{Cyt~ c}$ and $\ce{4H^{+}}$ are pumped out into intermembrane space by complex III as shown in the following reaction.
$\ce{QH2 + 2Cyt \; c_{ox} + 2H^{+}_{matrix} -> Q + 2Cyt \; c_{red} + 4H^{+}_{intermembrane}}$
$\ce{Cyt~ c}$ carries the electron to complex IV.
Complex IV
Complex IV, also known as cytochrome c oxidase, is the final complex in the electron transport chain. It contains several subunits, two heme groups, and several metal ion cofactors, including three atoms of copper, one magnesium, and one zinc. Complex IV reduces oxygen into water at the expense of oxidation of $\ce{Cyt~c}$ and pumps $\ce{4H^{+}}$ from matrix to intermembrane space as illustrated in the figure on the right and shown in the following overall reaction.
$\ce{4Cyt \; c_{red} + O2 + 8H^{+}_{matrix} -> 4Cyt \; c_{ox} + 2H2O + 4H^{+}_{intermembrane}}$
$\ce{ATP}$-synthase
$\ce{ATP}$-synthase, also known as complex V is the final complex in the oxidative phosphorylation process, i.e., part 3 of stage 3 of food catabolism. It comprises two parts in the shape of a mushroom, as illustrated in the figure on the left. The first part, FO, is the hydrophobic part embedded in the inner membrane (a blue and purple region in the model image on the right (Copyright; Alex. X, CC BY-SA 3.0, via Wikimedia Commons). The FO acts as a pore for the flow of protons ($\ce{H^{+}}$) across the membrane. The second part is F1 which is hydrophilic and spheroidal and protrudes into the matrix (red area in the image on the right). $\ce{ATP}$ synthesis takes place in the F1 part of $\ce{ATP}$-synthase. It operates on the principle of chemiosmosis, which is described below.
Chemiosmosis
Water movement from the direction of higher to lower concentration across a semipermeable membrane is called osmosis. When protons ($\ce{H^{+}}$) are pumped from the matrix to intermembrane space by complex I, III, & IV, a proton concentration gradient is developed across the membrane. Since protons are positive charge ions ($\ce{H^{+}}$), there is also an electrical gradient; collectively, it is called an electrochemical gradient. Chemiosmosis is the flow of ions ($\ce{H^{+}}$) across a semipermeable membrane down their electrochemical gradient.
An electrochemical gradient developed due to protons ($\ce{H^{+}}$) is a form of electrochemical potential energy. $\ce{ATP}$-Synthase harnesses the electrochemical potential energy to synthesize $\ce{ATP}$ as the protons flow through it, just as electricity is generated as water from a dam flows through a turbine. The higher energy $\ce{ATP}$ is synthesized by condensing lower energy $\ce{ADP}$ and phosphate $\ce{P_{i}}$ in a process called oxidative phosphorylation. It takes three to four $\ce{H^{+}}$ to synthesize one $\ce{ATP}$ by the following overall reaction.
$\ce{ADP + P_{i} +4H^{+}_{intermembrane} <=> ATP + H2O + 4H^{+}_{matrix}}$
This is an equilibrium reaction, i.e., when the electrochemical gradient is low, $\ce{ATP}$-synthase consumes $\ce{ATP}$ and returns protons from the matrix to the intermembrane space.
Structure and mechanisms of operation of $\ce{ATP}$-synthase
$\ce{ATP}$-synthase comprises several protein sub-units. It is in the shape of a mushroom with two major parts: a stem-like part which is a hydrophobic portion embedded in the inner membrane, is called FO, and a mushroom head-like part that protrudes into the matrix, is called F1, as illustrated in Figure $2$ a. Note: the subscript in FO is the letter O, not the number zero.
Mechanism of FOc ring rotation in $\ce{ATP}-synthase The steps of the rotation of the FOc ring are illustrated below. Steps in the rotation mechanism of FOc ring in \(\ce{ATP}$-synthase (Copyright; Asw-hamburg at German Wikipedia., CC BY-SA 3.0, via Wikimedia Commons)
Step a: FOc subunits are arranged in a circle. All of them have an amino acid asp residue with an acidic side chain but in neutral ($\ce{peptide-COOH}$) form except one (labeled #1 in step a), which, being near a cation group ($\ce{peptide-NH3^{+}}$ of arg residue of FOb subunit, is an anion, i.e., $\ce{peptide-COO^{-}}$. Electrostatic interaction between opposite charges builds mechanical tension in the FOc ring like a spiral spring.
1. Step b: A $\ce{H^{+}}$ from outside (intermembrance space) neutralizes the anion ($\ce{ peptide-COO^{-} + H^{+} -> peptide-COOH}$).
2. Step c: The strained structure relaxes as the electrostatic tension is removed, causing the peptide to rotate.
3. Step d: Twisting motion bring FOc#1 back to the other side of the stator FOb $\ce{peptide-NH3^{+}}$ group causing the rotor FOc to rotate by 30o.
4. Step e: FOc#1 subunit is neutral like other subunits in the ring.
5. Step f: The rotor FOc is rotated by 30o, and FOc#2 is under the spell of the positive charge of FOb $\ce{peptide-NH3^{+}}$ group.
6. Step g: The asp residue of FOc#2 ionizes the $\ce{-COOH) (\(\ce{ peptide-COOH -> peptide-COO^{-} + H^{+}}$) and the spring comes under tension again.
7. Step h: The asp residue releases the $\ce{H^{+}}$ that moves inside (the matrix).
8. Step i: The initial position of step a is restored and the following has happed in the process:
1. FOc#2 has taken the place of FOc#1,
2. electrochemical energy of one $\ce{H^{+}}$ movement transferred to kinetic energy through 30o rotation of the rotor, and
3. one $\ce{H^{+}}$ is smuggled from the intermembrane space to the matrix.
The simulation of the movement described above is illustrated in Figure $2$ c.
Mechanism of catalysis of $\ce{ATP}$ synthesis in F1($\alpha$$\beta$)3 complex
The F1($\alpha$$\beta$)3 complex catalyzes $\ce{ATP}$ synthesis. It has three $\alpha$ and three $\beta$ subunits arranged alternately as three $\alpha$$\beta$ dimers in the shape of carpels in an orange. It is the $\beta$ subunits that catalyze the $\ce{ATP}$ synthesis. Rotation of F1$\gamma$ subunit inside the F1($\alpha$$\beta$)3 complex causes conformational changes in the F1$\beta$ subunits that are linked with the mechanism of $\ce{ATP}$ synthesis.
F1$\beta$ switched between three states with each 360o rotation of F1$\gamma$ subunit inside the F1($\alpha$$\beta$)3 complex, as illustrated in Figure $2$ d.
1. First is the open-state, shown in brown, where $\ce{ADP}$ and $\ce{P_{i}}$ enter the active state.
2. Second is the loose-state in which the F1$\beta$ closes up around the substrate $\ce{ADP}$ and $\ce{P_{i}}$ binding them loosely, shown in red.
3. Third, is the tight-state in which F1$\beta$ tightens around the substrate $\ce{ADP}$ and $\ce{P_{i}}$ forcing them to condense into an $\ce{ATP}$ product.
4. Finally, F1$\beta$ reverts to an open-state that releases the $\ce{ATP}$ and binds $\ce{ADP}$ and $\ce{P_{i}}$ to repeat the process.
Since one $\ce{H^{+}}$ cause a 30o turn, a complete (360o) turn of F1$\gamma$ subunit inside F1($\alpha$$\beta$)3 complex needs $\ce{12H^{+}}$ and produces three $\ce{ATP's}$ from three F1$\beta$ subunits in it, that gives the following overall reaction.
$\ce{ADP + P_{i} + 4H^{+}_{intermembrane} <=> ATP + H2O + 4H^{+}_{matrix}}$
The next video presents the oxidative phosphorylation of glucose in a summary form.
Heating the body
$\ce{NADH}$ is reduced in the electron transport chain at the expense of reduction of $\ce{O2}$ by the following overall reaction.
$\ce{1/2O2 + NADH + H^{+} -> H2O + NAD^{+}}$
The potential difference between these redox pairs is 1.14 volts, equivalent to -218 kJ/mol. Reduction of one $\ce{NADH}$ can produce three $\ce{ATP}$. Production of $\ce{ATP}$ costs 30.5 kJ/mole, which is equivalent to 30.5 kJ/mole x 3 = 91.5 kJ/mole of $\ce{NADH}$. So the percentage of energy conserved as $\ce{ATP}$, i.e., the energy efficiency is: $\frac{91.5}{218}\times{100} = 42\text{%}$. The remaining 58% of energy ends up heating the body.
Some compounds, known as uncouplers, uncouple electron transport from $\ce{ATP}$ synthase, i.e., the electron transport pumps $\ce{H^{+}}$ as usual but $\ce{H^{+}}$ return to the matrix without producing $\ce{ATP's}$. So, all energy released during the electron transport process becomes heat energy. Examples of uncouplers are 2,4-dinitrophenol and dicumarol that combine with $\ce{H^{+}}$ and, being hydrophobic, carry them through the inner membrane. Some compounds, like oligomycin A prevent $\ce{ATP}$ synthesis by blocking $\ce{H^{+}}$-channels in $\ce{ATP}$ synthase. Salicylic acid (aspirin), if taken in extreme excess, also blocks $\ce{H^{+}}$-channels in $\ce{ATP}$ synthase. When the electron transport chain operates without $\ce{ATP}$ synthesis, all of the electron transport energy is used as heat.
Certain animals adapted to the cold environment have developed uncoupling systems to generate more heat for heating their body. They contain a large amount of brown fat, a tissue with many mitochondria, and are brown due to iron in the cytochrome in the mitochondria. The electron transport works in the brown fat. Still, the mitochondria have certain proteins embedded in the inner membrane that allow $\ce{H^{+}}$ to return to the matrix without $\ce{ATP}$ production. The body of newborn babies loses more heat per unit mass because of the larger surface area to mass ratio. Newborn babies have brown fat deposits on arteries that heat the blood circulating in their bodies. Adults usually do not have brown fat except those who live and work in a cold climate.
Reactive oxygen species
The electron transport chain uses four electrons and four protons to reduce oxygen by the following overall reaction,
$\ce{O2 + 4e^{-} + 4H^{+} -> 2H2O}$.
However, some electrons, particularly during the reduction of coenzyme-$\ce{Q}$ in the complex-III, leak and cause the following reaction.
$\ce{O2 ->[e^{-}] \underset{Superoxide}{O2^{\bullet-}} ->[e^{-}] \underset{Peroxide}{O2^{2-}}}$
Species like superoxide ($\ce{O2^{\bullet{-}}}$), peroxide ($\ce{O2^{2-}}$) and their product hydroxyl radical ($\ce{HO^{\bullet}}$) are called reactive oxygen species which are harmful. They damage proteins, cause mutations by reacting with DNA, and are responsible for aging, as illustrated in the figure on the right (taken from https://www.hiclipart.com/free-trans...-clipart-npvtr). The body has mechanisms to suppress the production of reactive species and destroy them if formed. Production of reactive oxygen species increases with an increase in the inter-member potential. The reactive oxygen species, oxidants, activate uncoupling proteins that reduce the membrane potential. Some substances in the body, e.g., vitamins C, E and antioxidant enzymes, react with and destroy the reactive oxygen species.
Summary of glucose catabolism
Aerobic catabolism of a glucose molecule converts a $\ce{6C's}$ glucose molecule to six carbon dioxide along with $\ce{ATP's}$, $\ce{GTP's}$, and reduced coenzyme $\ce{NADH's}$, and $\ce{FADH2{'s}}$, as shown in the figure on the right.
Glucose first goes through glycolysis that produces two pyruvate and overall two $\ce{ATP}$ and two $\ce{NADH}$. Two pyruvates go through oxidative decarboxylation producing two acetyl-$\ce{CoA}$ and two $\ce{NADH}$. Each acetyl-$\ce{CoA}$ goes through one turn of the citric acid cycle. So, the two turns of the citric acid cycles convert two acetyl-$\ce{CoA}$ into four $\ce{CO2}$, six $\ce{NADH}$, two $\ce{FADH2}$, and two $\ce{GDP}$. These steps are summarized in Figure $3$ and Figure $4$ shown below.
Figure $3$ illustrates the summary of aerobic catabolism of glucose and production of $\ce{ATP's}$, $\ce{GTP's}$, and reduced coenzyme $\ce{NADH's}$, and $\ce{FADH2{'s}}$ at different stages.
Insulin -the blood glucose-regulating hormone
Insulin is a hormone that regulates blood glucose levels. The glucose level increases after a meal, and the insulin promotes its absorption into the liver, fat, and muscle cells, converting it into glycogen or fats. High levels of insulin also inhibit the production of glucose by the liver. Low insulin levels have the opposite effects, which happen when a person has diabetes.
Comparison of $\ce{ATP}$ yield in aerobic and anaerobic catabolism of glucose
Anaerobic catabolism, i.e., the two forms of fermentation shown in Figure $3$ yields a net two $\ce{ATP's}$ from a glucose molecule.
Aerobic catabolism yields about thirty $\ce{ATP's}$. Two $\ce{ATP's}$ are produced before the citric acid cycle and two $\ce{GTP's}$, ten $\ce{NADH's}$, and two $\ce{FADH2{'s}}$ are produced during citric acid cycle. $\ce{GTP's}$ later convert to $\ce{ATP's}$. the reduced coenzymes, i.e., $\ce{NADH's}$ and $\ce{FADH2{'s}}$ enter the electron transport chain and produce heat and $\ce{ATP's}$ as explained in the next section.
Account of $\ce{ATP}$ production from complete oxidation of glucose
Two $\ce{ATP's}$ are produced before the citric acid cycle along with four $\ce{NADH}$. The citric acid cycle yields $\ce{GTP's}$, six $\ce{NADH's}$, and two $\ce{FADH2{'s}}$. $\ce{GTP's}$ convert to $\ce{ATP's}$, i.e., one $\ce{GTP}$ is equivalent to one $\ce{ATP}$. $\ce{NADH's}$ and $\ce{FADH2{'s}}$ enter the electron transport chain and produce heat and $\ce{ATP's}$. Recall that $\ce{NADH}$ is more energetic and pumps more proton through complex I, III, and IV than $\ce{FADH2}$ which is less energetic and pumps less protons through complex III and IV only. Therefore, $\ce{NADH}$ yields more $\ce{ATP's}$ than $\ce{FADH2}$. Earlier books state that one $\ce{NADH's}$ is equivalent to three $\ce{ATP's}$, but currently accepted average values are the following.
$\ce{GTP}~ \approx~ 1\ce{ATP}$
$\ce{NADH} ~\approx~ 2.5\ce{ATP}$
$\ce{FADH2} ~\approx~ 1.5\ce{ATP}$
Based on these conversions, complete aerobic respiration of glucose produces approximately $\ce{32ATP's}$, as shown in the following equation and detailed in Table 1.
$\ce{\underbrace{C6H12O6}_{Glucose} + 6O2 + 10NAD^{+} + 2FAD + + 2GTP + 2ADP + 4Pi -> 6CO2 + 6H2O + 10NADH + 2FADH2 + 2GTP + 2ATP}$
$\ce{\underbrace{C6H12O6}_{Glucose} + 6O2 +32ADP + 32Pi -> 6CO2 + 6H2O + 32ATP}$
Table 1: $\ce{ATP}$ Production from complete aerobic catabolism of one glucose molecule.
Reaction $\ce{ATP}$ or reduced coenzymes Total $\ce{ATP's}$ output
Glycolysis
glucose → glucose-6-phosphate -1
fructose 6-phosphate → fructose 1,6-bisphosphate -1
2 glyceraldehyde-3-phosphate → 2 1,3-bisphosphoglycerat $\ce{2NADH}$ 5
2 1,3-bisphosphoglycerate → 2 3-phosphoglycerate $\ce{2ATP}$ 2
2 phosphoenolpyruvate → 2 pyruvate $\ce{2ATP}$ 2
Oxidative phosphorylation of two pyruvate
2 pyruvate → 2 acetyl-CoA $\ce{2NADH}$ 5
Two turns citric acid cycles to consume to 2 acetyl-$\ce{CoA}$
2 isocitrate → 2 $\alpha$-ketoglutarate $\ce{2NADH}$ 5
2 α-ketoglutarate → 2 succinyl-$\ce{CoA}$ $\ce{2NADH}$ 5
2 succinyl-CoA → 2 succinate $\ce{2GTP}$ 2
2 succinate → 2 fumarate $\ce{2FADH2}$ 3
2 malate → 2 oxaloacetate $\ce{2NADH}$ 5
Total 32 | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.05%3A_Oxidative_Phosphorylation.txt |
Learning Objectives
• Understand the catabolism of fatty acids, i.e., $\beta$-oxidation of fatty acids, the reactions involved in the process, the number of cycles needed, and the calculation of ATP yield per fatty acid.
• Understand obesity, ketogenesis, ketosis, ketoacidosis, and catabolism of glycerol, associated with $\beta$-oxidation of fatty acids.
$\beta$-Oxidation of fatty acids
Although glucose is a quick energy source that animals need, fatty acids have higher energy per unit mass. Therefore, fats are the main energy storage compounds in animals. Fatty acids ($\ce{R-COOH}$) in fats are a major source of energy for animals. Hydrolysis of fats, i.e., the 1st stage of catabolism of fats, begins in the digestive tract and completes in the cytosol of the cells. The second stage of fatty acid catabolism starts with the conversion of ($\ce{R-COOH}$) to the thioester group of coenzyme A, ($\ce{R-CO-S-CoA}$), i.e., an acyl-$\ce{CoA}$, at the cost of energy of $\ce{ATP}$ to $\ce{AMP}$ conversion. The thioester is called an activated fatty acid. Some of the important terms used in this section are illustrated in the figure below.
Acyl-$\ce{CoA}$ is transported from the cytosol into mitochondria for the second stage of its catabolism. In the second stage, the Acyl-$\ce{CoA}$ is fragmented into two $\ce{C's}$ fragments in the form of acetyl-$\ce{CoA}$. The activated fatty acid goes through a set of four reactions in which the $\beta\ce{C}$ is oxidized to a carbonyl $\ce{C=O}$ group. Another coenzyme A makes thioester group with the $\beta$-$\ce{C=O}$, and an acetyl-$\ce{CoA}$ group is split off, leaving behind an acyl-$\ce{CoA}$ with two $\ce{C's}$ less than the initial acyl-$\ce{CoA}$. This process is called $\beta$-Oxidation of fatty acids. The process repeats on the fragment acyl-$\ce{CoA}$ again and again until the last acyl-$\ce{CoA}$ fragment left is acetyl-$\ce{CoA}$. The acetyl-$\ce{CoA}$ enters the citric acid cycle, which is the third stage of the catabolism, as in the case of the catabolism of glucose. These reactions are explained next.
Activation of fatty acids
The activation of a fatty acid begins with an SN2 reaction of acylate anion as a nucleophile with $\ce{ATP}$. It produces acyl-adenylate and pyrophosphate, as shown below.
Pyrophosphate is removed by hydrolysis, making the reaction irreversible: $\ce{PP_i + H2O -> 2P_i + 2H^{+}}$.
Acyl-adenylate goes through a nucleophilic acyl substitution reaction with coenzyme A ($\ce{HS-CoA}$) attacking as a nucleophile and AMP as leaving group.
As described next, acyl-CoA product is transported from the cytosol into mitochondria for the $\beta$-oxidation process.
Reaction 1 -oxidation
Reaction 1 is the dehydrogenation of of acyl-$\ce{CoA}$ by enzyme acyl-$\ce{CoA}$-dehydrogenase that removes $\ce{H}$ from $\alpha$ and $\beta$ $\ce{C's}$ generating a trans $\ce{C=C}$ bond. An $\ce{FAD}$ is reduced to $\ce{FADH2}$ as the acyl-$\ce{CoA}$ is oxidized in this reaction. Resonance stabilization of the $\ce{C=C}$ bond by conjugated $\ce{C=O}$ group makes the product stable, lowering the activation energy for the reaction.
Reaction 2 -hydration
The $\ce{C=C}$ is hydrated by addition of $\ce{H2O}$ that selectively installs an $\ce{-OH}$ group at $\beta$$\ce{C}$, creating a new chiral center in L-configuration. This is an electrophilic addition reaction. The $\ce{C=O}$ makes the $\beta$$\ce{C}$ more nucleophilic by withdrawing electrons from it by resonance. That is why the $\beta$$\ce{C}$ selectively reacts with the incoming $\ce{H2O}$ electrophile.
Reaction 3 -oxidation
The secondary $\ce{-OH}$ group is oxidized to a ketone ($\ce{C=O}$) group at the expense of reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$ by the following reaction.
Reaction 4 -cleavage by thiolysis
The fourth reaction is a nucleophilic acyl substitution reaction in which an anion of coenzyme A ($\ce{^{-}S-CoA}$) acts as a nucleophile, the ketone ($\ce{C=O}$) as electrophile and $\ce{^{-}CH2-CO-S-CoA}$ as the leaving group. $\ce{^{-}CH2-CO-S-CoA}$ is a good leaving group because the negative charge is resonance stabilized and mainly resides on $\ce{O}$ of carbonyl ($\ce{C=O}$) group. A proton later neutralizes the negative charge.
The second product of the above reaction is an acyl-$\ce{CoA}$ with two $\ce{2C's}$ less than the initial acyl-$\ce{CoA}$.
Number of $\beta$-oxidation cycles
The product acyl-$\ce{CoA}$ of the first $\beta$-oxidation cycle goes through the process of four reactions again and again until the last acyl-$\ce{CoA}$ product is acetyl-$\ce{CoA}$, i.e., $\ce{CH3CO-S-COA}$. The process is illustrated in Figure $1$, with the help of an example of $\beta$-oxidation of stearic acid which is a typical $\ce{18C's}$ saturated fatty acid.
If the starting fatty acid has $\ce{nC's}$, there are: $\frac{n}{2}$ acetyl-$\ce{CoA}$ produced. Since the last cycle produces two acetyl-$\ce{CoA}$, the number of cycles for a fatty acid containing $\ce{nC's}$ is: $\frac{n}{2}-1$. For example, stearic acid has $\ce{18C's}$ and it goes through: $\frac{18}{2}-1 = 8~\beta\text{-oxidation cycles}$.
$\ce{ATP}$ Yield from fatty acid oxidation
Each $\beta$-oxidation cycle of a fatty acid yields one $\ce{FADH2}$, one $\ce{NADH}$ and one acetyl-$\ce{CoA}$. These molecules produce $\ce{ATP's}$ when they enter the citric acid cycle and oxidative phosphorylation, as described earlier in the catabolism of glucose. One $\ce{FADH2}$ ≈ 1.5 $\ce{ATP's}$, one $\ce{NADH}$ ≈ 2.5 $\ce{ATP's}$, and one acetyl-$\ce{CoA}$ ≈ 10 $\ce{ATP's}$ that makes 14 $\ce{ATP's}$ per $\beta$-oxidation cycle. Since there are ($\frac{n}{2}-1$) or ($0.5\times{n}-1$) cycles for a fatty acid containing $\ce{nC's}$, there are: ($(0.5\times{n}-1)\times{14}))~\ce{ATP's}$ produced. The last cycle yields two acetyl-$\ce{CoA}$, so there are addition $\ce{10ATP's}$ from the the last acetyl-$\ce{CoA}$. The activation step converts one $\ce{ATP}$ to one $\ce{AMP}$ which is equivalent to conversion of $\ce{2ATP}$ to $\ce{2ADP}$. So, after adding ten from the last acetyl-$\ce{CoA}$ to the formula ( $(0.5\times{n}-1)\times{14})~\ce{ATP's}$ and subtracting 2 for the $\ce{2ATP's}$ consumed in the activation step, the formula for the number of $\ce{ATP's}$ produced per fatty acid containing $\ce{nC's}$ become the following:
$\ce{ATP's}~\text{produced per fatty acid containing}~\ce{nC's} = ((0.5\times{n}-1)\times{14} + 10 -2)~\ce{ATP's}$.
It simplifies to:
$\ce{ATP's}~\text{produced per fatty acid containing}~\ce{nC's} = (7\times{n}-6)~\ce{ATP's}$.
For example, stearic acid contains $\ce{18C's}$ and produces $(7\times18-6) = 120~\ce{ATP's}$. So, one mole of stearic acid (284.48 g/mol) produces 120 mol $\ce{ATP's}$, which is significantly higher than 32 mol $\ce{ATP's}$ produced by one mole of glucose (180.156 g/mol). When converted to moles of $\ce{ATP's}$ produced per unit mass, stearic acid produces 0.42 mol $\ce{ATP's}$/g, which is more than two times higher than 0.18 mol $\ce{ATP's}$/g of glucose. This is because the average oxidation state of glucose $\ce{C's}$ is higher than those of fatty acid $\ce{C's}$. Glucose is a quick energy source, but animals use fats as relatively longer-term energy storage molecules due to their higher energy density.
Unsaturated fatty acids
Unsaturated fatty acids have $\ce{C=C}$ bond in their alkyl chain. $\beta$-Oxidation of unsaturated fatty acids happens the same way as for saturated fatty acids except for the following changes. If $\ce{C=C}$ bond is a trans-$\ce{C=C}$ bond between $\alpha$ and $\beta$ $\ce{C'}$, reaction 1 in the $\beta$-oxidation is not needed and $\ce{FADH2}$ from reaction 1 is not produced. If $\ce{C=C}$ bond is a cis-$\ce{C=C}$ bond or it is between $\beta$ and $\gamma$ $\ce{C'}$, It goes through isomerization reactions catalyzed by isomerase enzymes to convert the $\ce{C=C}$ bond into a trans $\ce{C=C}$ bond between $\alpha$ and $\beta$ $\ce{C'}$ before reaction 3 happens on it. Unsaturated fatty acids yield a little less $\ce{ATP's}$ due to one $\ce{FADH2}$ less produced pre $\ce{C=C}$ bond, but the difference is small and the formula: $(7\times{n}-6)~\ce{ATP's}$, still gives a reasonably accurate estimate of $\ce{ATP}$ yield.
Fat and obesity
Although carbohydrates and glucose are quick energy sources, fat has more energy per mass or volume. Storage of fats for long-term energy supply is an important survival feature for several animals. For example, hibernating animals like polar bear converts food into fats during summer when food is plenty and utilize stored fats to survive for months without food during hibernation in winter. Whales and penguins are kept warm by a layer of body fat called blubber and use it as an energy source to survive when food is unavailable. Camel store fat in their hump and can survive for months without food and water by utilizing the fat reserves. Migrator birds also store fat and use it as an energy source during migration.
Humans can store fats. In earlier times, major human food was vegetables, and fats in food accounted for about 20% of food calories, but these days more than 60% of food calories are from fats. The cumulation of too much fat is associated with overweight and obesity. Obesity is an excess accumulation of body fat that negatively affects health. It is measured in terms of body mass index (BMI), which is the ratio of a person's mass to the square of a person's height. BMI 18.5 kg/m2 to 24.9 kg/m2 is normal, less is underweight, more is overweight, and over 30 kg/m2 is obese, as shown in Figure $2$. Obesity is associated with medical conditions like diabetes, high blood pressure, stroke, heart disease, gallstones, arthritis, and some cancers. According to NIH, nearly 3 out of 4 adults aged 20 or above in the US are overweight or obese. It is a preventable condition, as described in the following video message from NIH.
Like insulin regulates blood glucose, leptin is a hormone adipose tissues produce. Leptin governs the long-term balance between food intake and energy expenditure. When there is more adipose tissue (fatty tissues), more leptin is released, which decreases hunger by giving a feeling of fullness, and less fatty tissue means less fat and less leptin, which has the opposite effect. Like diabetes disturbs the function of insulin, obesity may be related to the malfunctioning of leptin. This is an active area of research these days for finding obesity treatment.
Ketone bodies
Acetyl-$\ce{CoA}$ produced during the $\beta$-oxidation of fatty acids enter the citric acid cycle. When many fatty acids degrade, the citric acid cycle can not take all of the acetyl-$\ce{CoA}$. The excess acetyl-$\ce{CoA}$ accumulates in the liver and converts into ketone bodies by a pathway known as ketogenesis, as shown in the figure on the right (Copyright; Sav vas, CC0, via Wikimedia Commons). In this figure, the enzymes are colored red, and the substrates/products are colored blue. The three ketone bodies, i.e., acetoacetate, acetone, and $\beta$-hydroxybutyrate, are marked within orange boxes.
Ketone bodies are transported from the liver to other tissues where acetoacetate and $\beta$-hydroxybutyrate are converted back to acetyl-$\ce{CoA}$ and enter the citric acid cycle to produce energy. Normally there is constant production and consumption of acetone bodies maintaining ~1 mg/dL concentration in blood.
Acetone cannot be converted back into acetyl-$\ce{CoA}$ except in the liver, where it can be converted into lactate and then to pyruvate.
Ketosis
When the synthesis rate of ketone bodies exceeds their consumption rate, they start accumulating and may start excreting in urine and via breathing -a condition called ketosis. Since two of the three ketone bodies are acids, their accumulation in the blood lowers the blood pH -a condition called ketoacidosis. The smell of acetone, fruity or like nail polish remover smaller, is detectable in the breath of persons suffering from ketosis. Ketone bodies can be tested in blood, urine, or breath, e.g., by using test strips with a color chart to read the results, as shown in the figure on the left. In ketosis, the ketone bodies in the blood are between 0.5 to 3.0 millimole/L (mmol/L). In ketoacidosis, the concentration may go above 10 mmol/L.
Ketosis can happen during fasting, starvation, prolonged low carbohydrate diets, prolonged intense exercises, alcoholism, or due to uncontrolled diabetes.
Diabetes and ketone bodies
The liver and pancreas play important roles in maintaining blood glucose levels, illustrated in the figure on the right (copyright: C. Muessig, CC BY-SA 3.0, via Wikimedia Commons) and described next. The normal glucose level in the blood is 4.5 to 5.5 mM. When glucose level is above normal, usually after a meal, the pancreas senses it and secrets insulin that increases the flow of glucose into muscles and fatty tissues, where it is converted into glycogen, lowering blood glucose. When the blood glucose level is low, the pancreas secretes another hormone, glucagon, into the bloodstream. Glucagon triggers glycogen breakdown, particularly in the liver, releasing glucose into the bloodstream.
In diabetes, there is either insufficient insulin or not functioning properly. Less glucose is sent to muscles and, as a result, muscles cause $\beta$-oxidation of fatty acids to meet the energy needs. Higher levels of acetyl-$\ce{CoA}$ from the $\beta$-oxidation produce more ketone bodies than their consumption, accumulating ketone bodies. It causes ketosis or ketoacidosis. Low pH of blood plasma due to ketoacidosis triggers kidneys to excrete urine with high acid levels. Glucose and ketone bodies also spill into the urine. High glucose concentration in the blood causes higher osmotic pressure, increasing dehydration. The symptoms include frequent urination and excessive thirst. These conditions can be treated with low carbohydrate diets, insulin therapy, or other medications to control diabetes.
Catabolism of glycerol
Glycerol, a part of triglycerides (fats), accounts for about 5% of their energy. Glycerol is first converted into glycerol-3-phosphate by an SN2 reaction of a primary alcohol group of glycerol acting as a nucleophile, phosphorus in the phosphate of ATP as a nucleophile, and ADP as a leaving group. Then, the secondary alcohol of glycerol is oxidized to a ketone group at the expense of reduction of a $\ce{NAD^{+}}$ coenzyme into $\ce{NADH}$ as shown below.
Dihydroxyacetone phosphate produced by the second reaction is an intermediate in the glycolysis of glucose and enters the same catabolic pathway. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.06%3A_Oxidation_of_fatty_acids.txt |
Learning Objectives
• Understand the catabolism of amino acids, including transamination, oxidative amination, and urea cycle that takes care of the $\ce{N}$ and processing of the $\ce{C}$ skeleton of the amino acid to intermediates that enter into citric acid cycle for energy production.
Amino acids are the products of stage 1 of protein catabolism. Usually, amino acids are used to synthesize proteins and other substances that need nitrogen, e.g., nucleotides. Only about 10% of our energy is usually derived from amino acids. Still, when there is a shortage of carbohydrates and fats, e.g., during fasting or starvation, amino acids are used as an alternate energy source. Prolonged use of amino acids as an energy source may lead to the destruction of essential tissues.
Degradation of amino acids has two sets of catabolic pathways: one that deals with the processing of $\ce{N}$ and the other that deals with the processing of the remaining $\ce{C}$ skeleton of the amino acids.
Processing of $\ce{N}$ of amino acids
Processing of $\ce{N}$ usually happens in the liver and has three major stages: i) transamination, ii) oxidative deamination, and iii) urea cycle, as illustrated with some examples in Figure $1$.
Transamination
Transamination is the process of exchange of an ammonium ($\ce{-NH3^{+}}$) group of amino acids with ketone ($\ce{C=O}$) group of an $\alpha$-ketoacid. Usually $\alpha$-amino acids exchange $\ce{-NH3^{+}}$ group with ketone ($\ce{C=O}$) group of an $\alpha$-ketoglutarate. For example, alanine exchanges its $\ce{-NH3^{+}}$ group with a $\ce{C=O}$ of an $\alpha$-ketoglutarate producing a new $\alpha$-keto acid (pyruvate in this example) and a new $\alpha$-amino acids (L-glutamate in this example), as shown in reaction 1 in Figure $1$. Other $\alpha$-keto acids can also receive $\ce{-NH3^{+}}$ of amino acids also. For example, L-glutamate can transfer its $\ce{-NH3^{+}}$ to oxaloacetate that regenerates $\alpha$-ketoglutarate and produces aspartate, as shown in reaction 3 in Figure $1$.
Oxidative deamination
Oxidative deamination reaction replaces $\ce{-NH3^{+}}$ with a $\ce{C=O}$, producing an $\alpha$-keto acid and an ammonium ($\ce{NH4^{+}}$) ion, as shown in reaction 2 in Figure $1$. This is an oxidation reaction that is coupled with the reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$. It happens in mitochondria where $\ce{NADH}$ enters oxidative phosphorylation pathway to produce $\ce{ATP's}$. The other product, i.e., $\ce{NH4^{+}}$, is a toxic substance that needs to be disposed off. Some of $\ce{NH4^{+}}$ may be used in anabolic reactions, e.g., reactions 4 and 5 in Figure $1$, but most of it enters the urea cycle.
Urea cycle
Ammonium ion $\ce{NH4^{+}}$ is a toxic substance that is converted to less toxic urea in the urea cycle by the following overall reaction.
$\ce{2NH4^{+} + CO2 -> \underbrace{H2N-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2}_{Urea} + 2H^{+} + H2O}$.
Before entry into the urea cycle, $\ce{NH4^{+}}$ reacts with two $\ce{ATP's}$ and bicarbonate ($\ce{HCO3^{-}}$), producing carbamoyl phosphate, two $\ce{ADP's}$ and phosphate ($\ce{P_i}$, as shown below.
This reaction is catalyzed by an enzyme carbamoyl phosphate synthetase I. Blood contains $\ce{HCO3^{-}}$ which is the product of dissolution of $\ce{CO2}$. The product of the above reaction, i.e., Carbamoyl phosphate, goes through four reactions called the urea cycle illustrated in Figure $2$.
• The first reaction of the urea cycle is catalyzed by ornithine transcarbamylase. The $\ce{-COO^{-}}$ group of ornithine substitutes phosphate ($\ce{P_{i}}$ from carbamoyl phosphate producing citrulline in mitochondria. Citrulline moves out from the matrix into the cytoplasm.
• The second reaction is a condensation reaction between the carbonyl ($\ce{C=O}$) group of citrulline and $\ce{-NH3^{+}}$ group of aspartate that produces argininosuccinate. This reaction takes place in the cytosol and is catalyzed by argininosuccinate synthetase.
• In the third reaction, argininosuccinate is cleaved by the enzyme argininosuccinase to form arginine, which stays in the cycle, and fumarate, which leaves the cycle.
• In the fourth reaction, arginase cleaves arginine to form urea and ornithine. Ornithine is transported back to the matrix in mitochondria to start the next urea cycle, and urea is released into the blood.
Since $\ce{HCO3^{-}}$ is produced by the dissolution of $\ce{CO2}$ in water, the $\ce{NH4^{+}}$ + $\ce{HCO3^{-}}$ are equivalent to $\ce{NH3}$ + $\ce{CO2}$ + $\ce{H2O}$. So, the overall equation of the urea cycle becomes:
$\ce{NH3 + CO2 + H2O + aspartate + 3ATP + 3H2O -> urea + fumarate + 2ADP + 2P_{i} + AMP + PP_{i} + H2O}$
Since $\ce{NH3}$ is removed to convert aspartate into fumarate along with $\ce{PP_{i} + H2O -> 2P_{i}}$, substituting these for aspartate and fumarate simplifies the above equation into:
$\ce{2NH3 + CO2 + 3ATP + 3H2O -> urea + 2ADP + 4P_{i} + AMP}$
One $\ce{NADH}$ is produced during oxidative deamination of L-glutamate (reaction 2 in Figure $1$). Another $\ce{NADH}$ is produced when fumarate from reaction 3 of the urea cycle is processed in the citric acid cycle, i.e., fumarate is converted into malate by enzyme fumarase and then malate is oxidized to oxaloacetate at the expense of reduction of $\ce{NAD^{+}}$ to $\ce{NADH}$ by enzyme malate dehydrogenase. Adding these reactions to the above reaction of the urea cycle results in the following overall reaction.
$\ce{CO2 + glutamate + aspartate + 3ATP + 2NAD^{+} + 3H2O -> urea + \alpha{-ketoglutarate} + oxaloacetate + 2ADP + 2P_{i} + AMP + 2PP_{i} + 2NADH}$
Three $\ce{ATP}$ are consumed, one of them to $\ce{AMP}$ that makes total equivalent to four $\ce{ATP's}$ consumed but two $\ce{NADH}$ enter oxidative phosphorylation and yield ~5$\ce{ATP}$, i.e., there is net production of an $\ce{ATP}$ in the process of $\ce{N}$ of an amino acid.
The urea released into the blood is later filtered out by the kidneys and excreted with urine. An adult passes about 25 to 30 g of urea in urine per day. If urea is not eliminated properly, it builds to toxic levels and needs medical treatment, like dialysis. Reducing protein intake also lowers urea output.
The link between the urea cycle and the citric acid cycle
The urea cycle and citric acid cycle are two separate cycles but are linked with each other. Aspartate that provides one $\ce{N}$ in the urea cycle is produced by oxaloacetate transamination, which is an intermediate in the citric acid cycle. Fumarate is a product of the urea cycle that is an intermediate of the citric acid cycle and returns to it.
processing of $\ce{C}$ skeleton of amino acids
$\alpha$-Ketoacids are left as $\ce{C}$ skeleton of amino acid after transamination or oxidative deamination. For example, pyruvate is left after transamination of alanine and $\alpha$-ketoglutarate is left after oxidative deamination of glutamate, as shown in reactions 1 and 2, respectively, in Figure $1$. $\alpha$-Ketoacids are either used as precursors for the synthesis of other compounds, or they enter the citric acid cycle to produce $\ce{CO2}$, $\ce{H2O}$, and energy. $\alpha$-Ketoglutarate and oxaloacetate are $\alpha$-keto acids that are intermediates in the citric acid cycle and directly enter the cycle. Other $\alpha$-keto acids go through a series of reactions to convert into one of the intermediates in the citric acid cycle, or they convert to pyruvate or acetyl-(\ce{CoA}\) that enter the citric acid cycle, as illustrated in Figure $3$. Details of the conversation of $\alpha$-keto acids into intermediates of the citric acid cycle or pyruvate or acetyl-(\ce{CoA}\) are not described here.
Amino acids that can degrade to pyruvate or oxaloacetate are called glucogenic because these products can form glucose through the glucogenesis pathway. For example, alanine yields pyruvate, and aspartate yield oxaloacetate, as shown in reactions 1 and 3 in Figure $1$. Amino acids that degrade to acetyl-(\ce{CoA}\) or acetoacetic acid, which can not form glucose but can be converted into ketone bodies, are called ketogenic. Lysine and leucine are ketogenic amino acids. Several amino acids can catabolize to produce both glucogenic and ketogenic intermediates, and they fall into both classes, as shown in Figure $3$. | textbooks/chem/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.07%3A_Degradation_of_amino_acids.txt |
• 1.1: Sand and Water
• 1.2: Chemicals Compose Ordinary Things
Chemistry is the branch of science dealing with the structure, composition, properties, and the reactive characteristics of matter. Matter is anything that has mass and occupies space. Thus, chemistry is the study of literally everything around us—the liquids that we drink, the gases we breathe, the composition of everything from the plastic case on your phone to the earth beneath your feet. Moreover, chemistry is the study of the transformation of matter.
• 1.3: The Scientific Method - How Chemists Think
Science is a process of knowing about the natural universe through observation and experiment. Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method. Science is broken down into various fields, including chemistry, and is both qualitative and quantitative.
• 1.4: Analyzing and Interpreting Data
• 1.5: A Beginning Chemist - How to Succeed
Most people can succeed in chemistry, but it often requires dedication, hard work, the right attitude and study habits!
• 1.6: Hypothesis, Theories, and Laws
Although many have taken science classes throughout the course of their studies, people often have incorrect or misleading ideas about some of the most important and basic principles in science. Most students have heard of hypotheses, theories, and laws, but what do these terms really mean? Prior to reading this section, consider what you have learned about these terms before. What do these terms mean to you? What do you read that contradicts or supports what you thought?
• 1.7: The Scope of Chemistry
Chemistry is the study of matter and the ways in which different forms of matter combine with each other. You study chemistry because it helps you to understand the world around you. Everything you touch or taste or smell is a chemical, and the interactions of these chemicals with each other define our universe. Chemistry forms the fundamental basis for biology and medicine.
• 1.E: Exercises
Exercises for Chapter 1 of Tro's Introductory Chemistry textmap.
01: The Chemical World
Chemistry is the branch of science dealing with the structure, composition, properties, and the reactive characteristics of matter. Matter is anything that has mass and occupies space. Thus, chemistry is the study of literally everything around us—the liquids that we drink, the gases we breathe, the composition of everything from the plastic case on your phone to the earth beneath your feet. Moreover, chemistry is the study of the transformation of matter. Crude oil is transformed into more useful petroleum products, such as gasoline and kerosene, by the process of refining. Some of these products are further transformed into plastics. Crude metal ores are transformed into metals, that can then be fashioned into everything from foil to automobiles. Potential drugs are identified from natural sources, isolated and then prepared in the laboratory. Their structures are systematically modified to produce the pharmaceuticals that have led to vast advances in modern medicine. Chemistry is at the center of all of these processes; chemists are the people that study the nature of matter and learn to design, predict, and control these chemical transformations. Within the branches of chemistry you will find several apparent subdivisions. Inorganic chemistry, historically, focused on minerals and metals found in the earth, while organic chemistry dealt with carbon-containing compounds that were first identified in living things. Biochemistry is an outgrowth of the application of organic chemistry to biology and relates to the chemical basis for living things. In the later chapters of this text we will explore organic and biochemistry in a bit more detail and you will notice examples of organic compounds scattered throughout the text. Today, the lines between the various fields have blurred significantly and a contemporary chemist is expected to have a broad background in all of these areas.
In this chapter, we will discuss some of the properties of matter and how chemists measure those properties. We will introduce some of the vocabulary that is used throughout chemistry and the other physical sciences.
Let’s begin with matter. Matter is defined as any substance that has mass. It is important to distinguish here between weight and mass. Weight is the result of the pull of gravity on an object. On the Moon, an object will weigh less than the same object on Earth because the pull of gravity is less on the Moon. The mass of an object, however, is an inherent property of that object and does not change, regardless of location, gravitational pull, or anything else. It is a property that is solely dependent on the quantity of matter within the object.
Contemporary theories suggests that matter is composed of atoms. Atoms themselves are constructed from neutrons, protons and electrons, along with an ever-increasing array of other subatomic particles. We will focus on the neutron, a particle having no charge; the proton, which carries a positive charge; and the electron, which has a negative charge. Atoms are incredibly small. To give you an idea of the size of an atom, a single copper penny contains approximately 28,000,000,000,000,000,000,000 atoms (that’s 28 sextillion). Because atoms and subatomic particles are so small, their mass is not readily measured using pounds, ounces, grams or any other scale that we would use on larger objects. Instead, the mass of atoms and subatomic particles is measured using atomic mass units (abbreviated amu). The atomic mass unit is based on a scale that relates the mass of different types of atoms to each other (using the most common form of the element carbon as a standard). The amu scale gives us a convenient means to describe the masses of individual atoms and to do quantitative measurements concerning atoms and their reactions. Within an atom, the neutron and proton both have a mass of one amu; the electron has a much smaller mass (about 0.0005 amu).
Atomic theory places the neutron and the proton in the center of the atom in the nucleus. In an atom, the nucleus is very small, very dense, carries a positive charge (from the protons) and contains virtually all of the mass of the atom. Electrons are placed in a diffuse cloud surrounding the nucleus. The electron cloud carries a net negative charge (from the charge on the electrons) and in a neutral atom there are always as many electrons in this cloud as there are protons in the nucleus (the positive charges in the nucleus are balanced by the negative charges of the electrons, making the atom neutral).
An atom is characterized by the number of neutrons, protons and electrons that it possesses. Today, we recognize at least 116 different types of atoms, each type having a different number of protons in its nucleus. These different types of atoms are called elements. The neutral element hydrogen (the lightest element) will always have one proton in its nucleus and one electron in the cloud surrounding the nucleus. The element helium will always have two protons in its nucleus. It is the number of protons in the nucleus of an atom that defines the identity of an element. Elements can, however, have differing numbers of neutrons in their nucleus. For example, stable helium nuclei exist that contain one, or two neutrons (but they all have two protons). These different types of helium atoms have different masses (3 or 4 amu) and they are called isotopes. For any given isotope, the sum of the numbers of protons and neutrons in the nucleus is called the mass number. All elements exist as a collection of isotopes, and the mass of an element that we use in chemistry, the atomic mass, is the average of the masses of these isotopes. For helium, there is approximately one isotope of Helium-3 for every one million isotopes of Helium-4, hence the average atomic mass is very close to 4 (4.002602).
As different elements were discovered and named, abbreviations of their names were developed to allow for a convenient chemical shorthand. The abbreviation for an element is called its chemical symbol. A chemical symbol consists of one or two letters, and the relationship between the symbol and the name of the element is generally apparent. Thus helium has the chemical symbol He, nitrogen is N, and lithium is Li. Sometimes the symbol is less apparent but is decipherable; magnesium is Mg, strontium is Sr, and manganese is Mn. Symbols for elements that have been known since ancient times, however, are often based on Latin or Greek names and appear somewhat obscure from their modern English names. For example, copper is Cu (from cuprum), silver is Ag (from argentum), gold is Au (from aurum), and iron is Fe (from ferrum). Throughout your study of chemistry, you will routinely use chemical symbols and it is important that you begin the process of learning the names and chemical symbols for the common elements. By the time you complete General Chemistry, you will find that you are adept at naming and identifying virtually all of the 116 known elements. Table $1$ contains a starter list of common elements that you should begin learning now!
Table $1$: Names and Chemical Symbols for Common Elements
Element Chemical Symbol Element Chemical Symbol
Hydrogen H Phosphorus P
Helium He Sulfur S
Lithium Li Chlorine Cl
Beryllium Be Argon Ar
Boron B Potassium K
Carbon C Calcium Ca
Nitrogen N Iron Fe
Oxygen O Copper Cu
Fluorine F Zinc Zn
Neon Ne Bromine Br
Sodium Na Silver Ag
Magnesium Mg Iodine I
Aluminum Al Gold Au
Silicon Si Lead Pb
The chemical symbol for an element is often combined with information regarding the number of protons and neutrons in a particular isotope of that atom to give the atomic symbol. To write an atomic symbol, begin with the chemical symbol, then write the atomic number for the element (the number of protons in the nucleus) as a subscript, preceding the chemical symbol. Directly above this, as a superscript, write the mass number for the isotope, that is, the total number of protons and neutrons in the nucleus. Thus, for helium, the atomic number is 2 and there are two neutrons in the nucleus for the most common isotope, making the atomic symbol $\ce{^{4}_{2}He}$. In the definition of the atomic mass unit, the “most common isotope of carbon”, $\ce{^{12}_{6}C}$, is defined as having a mass of exactly 12 amu and the atomic masses of the remaining elements are based on their masses relative to this isotope. Chlorine (chemical symbol Cl) consists of two major isotopes, one with 18 neutrons (the most common, comprising 75.77% of natural chlorine atoms) and one with 20 neutrons (the remaining 24.23%). The atomic number of chlorine is 17 (it has 17 protons in its nucleus), therefore the chemical symbols for the two isotopes are $\ce{^{35}_{17}Cl}$ and $\ce{^{37}_{17}Cl}$.
When data is available regarding the natural abundance of various isotopes of an element, it is simple to calculate the average atomic mass. In the example above, ${\displaystyle {}_{17}^{35}{\text{Cl}}}$$\ce{^{35}_{17}Cl}$ was the most common isotope with an abundance of 75.77% and ${\displaystyle {}_{17}^{37}{\text{Cl}}}$$\ce{^{37}_{17}Cl}$ had an abundance of the remaining 24.23%. To calculate the average mass, first convert the percentages into fractions; that is, simply divide them by 100. Now, chlorine-35 represents a fraction of natural chlorine of 0.7577 and has a mass of 35 (the mass number). Multiplying these, we get (0.7577 × 35) = 26.51. To this, we need to add the fraction representing chlorine-37, or (0.2423 × 37) = 8.965; adding, (26.51 + 8.965) = 35.48, which is the weighted average atomic mass for chlorine. Whenever we do mass calculations involving elements or compounds (combinations of elements), we always need to use average atomic masses. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.02%3A_Chemicals_Compose_Ordinary_Things.txt |
Learning Objectives
• Identify the components of the scientific method.
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments; which leads to additional observations, hypotheses, and experiments in repeated cycles (Figure \(1\)).
Step 1: Make observations
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: "the outside air temperature is cooler during the winter season," "table salt is a crystalline solid," "sulfur crystals are yellow," and "dissolving a penny in dilute nitric acid forms a blue solution and a brown gas." Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: "the melting point of crystalline sulfur is 115.21° Celsius," and "35.9 grams of table salt—the chemical name of which is sodium chloride—dissolve in 100 grams of water at 20° Celsius." For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal.
Step 2: Formulate a hypothesis
After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either one of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun.
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it.
Step 3: Design and perform experiments
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is—under conditions in which a single variable changes.
Step 4: Accept or modify the hypothesis
A properly designed and executed experiment enables a scientist to determine whether or not the original hypothesis is valid. If the hypothesis is valid, the scientist can proceed to step 5. In other cases, experiments often demonstrate that the hypothesis is incorrect or that it must be modified and requires further experimentation.
Step 5: Development into a law and/or theory
More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply states what happens; it does not address the question of why.
One example of a law, the law of definite proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered.
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
Example \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, an observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice were added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter to ever play professional basketball.
5. Limestone is relatively insoluble in water, but dissolves readily in dilute acid with the evolution of a gas.
Answer a
experiment
Answer b
law
Answer c
theory
Answer d
hypothesis
Answer e
observation
Summary
The scientific method is a method of investigation involving experimentation and observation to acquire new knowledge, solve problems, and answer questions. The key steps in the scientific method include the following:
• Step 1: Make observations.
• Step 2: Formulate a hypothesis.
• Step 3: Test the hypothesis through experimentation.
• Step 4: Accept or modify the hypothesis.
• Step 5: Develop into a law and/or a theory.
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.03%3A_The_Scientific_Method_-_How_Chemists_Think.txt |
Examples of the practical applications of chemistry are everywhere (Figure \(1\)). Engineers need to understand the chemical properties of the substances needed to design biologically compatible implants for joint replacements; or to design roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement. Archeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, DNA matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books.
Hopefully at this point you are fully convinced of how important and useful the study of chemistry can be. You may, however, still be wondering exactly what it is that a chemist does. Chemistry is the study of matter and the changes that matter undergoes. In general, chemists are interested in both characteristics that you can test and observe, like a chemical's smell or color, and characteristics that are far too small to see, like what the oxygen you breathe in or the carbon dioxide you breath out looks like under a microscope 1,000 times more powerful than any existing in the world today.
Wait a minute… how can a chemist know what oxygen and carbon dioxide look like under a microscope that doesn't even exist? What happened to the scientific method? What happened to relying on observations and careful measurements? In fact, because chemists can't see the underlying structure of different materials, they have to rely on the scientific method even more! Chemists are a lot like detectives. Suppose a detective is trying to solve a murder case—what do they do? Obviously, the detective starts by visiting the site of the crime and looking for evidence. If the murderer has left enough clues behind, the detective can piece together a theory explaining what happened.
Even though the detective wasn't at the crime scene when the crime was committed and didn't actually see the murderer kill the victim, with the right evidence, the detective can be pretty sure of how the crime took place. It is the same with chemistry. When chemists go into the laboratory, they collect evidence by making measurements. Once chemists have collected enough clues from the properties that they can observe, they use that evidence to piece together a theory explaining the properties that they cannot observe—the properties that are too small to see.
What kinds of properties do chemists actually measure in the laboratory? Well, you can probably guess a few. Imagine that you go to dinner at a friend's house and are served something that you don't recognize, what types of observations might you make to determine exactly what you've been given? You might smell the food. You might note the color of the food. You might try to decide whether the food is a liquid or a solid because if it's a liquid, it's probably soup or a drink. The temperature of the food could be useful if you wanted to know whether or not you had been served ice cream! You could also pick up a small amount of food with your fork and try to figure out how much it weighs—a light dessert might be something like an angel cake, while a heavy dessert is probably a pound cake. The quantity of food you have been given might be a clue too. Finally, you might want to know something about the food's texture—is it hard and granular like sugar cubes, or soft and easy to spread, like butter?
Believe it or not, the observations you are likely to make when trying to identify an unknown food are very similar to the observations that a chemist makes when trying to learn about a new material. Chemists rely on smell, color, state (whether it is a solid or liquid or gas), temperature, volume, mass (which is related to weight—as will be discussed in a later section), and texture. There is, however, one property possibly used to learn about a food, but that should definitely not be used to learn about a chemical—taste!
In the sections on the Atomic Theory, you will see exactly how measurements of certain properties helped early scientists to develop theories about the chemical structure of matter on a scale much smaller than they could ever hope to see. You will also learn how these theories, in turn, allow us to make predictions about new materials that humankind has not yet created.
The video below gives you some important tips on how to study chemistry in this class. With practice, you too can learn to think like a chemist, and you may even enjoy it! | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.05%3A_A_Beginning_Chemist_-_How_to_Succeed.txt |
Learning Objectives
• Describe the difference between hypothesis and theory as scientific terms.
• Describe the difference between a theory and scientific law.
Although many have taken science classes throughout the course of their studies, people often have incorrect or misleading ideas about some of the most important and basic principles in science. Most students have heard of hypotheses, theories, and laws, but what do these terms really mean? Prior to reading this section, consider what you have learned about these terms before. What do these terms mean to you? What do you read that contradicts or supports what you thought?
What is a Fact?
A fact is a basic statement established by experiment or observation. All facts are true under the specific conditions of the observation.
What is a Hypothesis?
One of the most common terms used in science classes is a "hypothesis". The word can have many different definitions, depending on the context in which it is being used:
• An educated guess: a scientific hypothesis provides a suggested solution based on evidence.
• Prediction: if you have ever carried out a science experiment, you probably made this type of hypothesis when you predicted the outcome of your experiment.
• Tentative or proposed explanation: hypotheses can be suggestions about why something is observed. In order for it to be scientific, however, a scientist must be able to test the explanation to see if it works and if it is able to correctly predict what will happen in a situation. For example, "if my hypothesis is correct, we should see ___ result when we perform ___ test."
A hypothesis is very tentative; it can be easily changed.
What is a Theory?
The United States National Academy of Sciences describes what a theory is as follows:
"Some scientific explanations are so well established that no new evidence is likely to alter them. The explanation becomes a scientific theory. In everyday language a theory means a hunch or speculation. Not so in science. In science, the word theory refers to a comprehensive explanation of an important feature of nature supported by facts gathered over time. Theories also allow scientists to make predictions about as yet unobserved phenomena."
"A scientific theory is a well-substantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experimentation. Such fact-supported theories are not "guesses" but reliable accounts of the real world. The theory of biological evolution is more than "just a theory." It is as factual an explanation of the universe as the atomic theory of matter (stating that everything is made of atoms) or the germ theory of disease (which states that many diseases are caused by germs). Our understanding of gravity is still a work in progress. But the phenomenon of gravity, like evolution, is an accepted fact.
Note some key features of theories that are important to understand from this description:
• Theories are explanations of natural phenomena. They aren't predictions (although we may use theories to make predictions). They are explanations as to why we observe something.
• Theories aren't likely to change. They have a large amount of support and are able to satisfactorily explain numerous observations. Theories can, indeed, be facts. Theories can change, but it is a long and difficult process. In order for a theory to change, there must be many observations or pieces of evidence that the theory cannot explain.
• Theories are not guesses. The phrase "just a theory" has no room in science. To be a scientific theory carries a lot of weight; it is not just one person's idea about something
Theories aren't likely to change.
What is a Law?
Scientific laws are similar to scientific theories in that they are principles that can be used to predict the behavior of the natural world. Both scientific laws and scientific theories are typically well-supported by observations and/or experimental evidence. Usually scientific laws refer to rules for how nature will behave under certain conditions, frequently written as an equation. Scientific theories are more overarching explanations of how nature works and why it exhibits certain characteristics. As a comparison, theories explain why we observe what we do and laws describe what happens.
For example, around the year 1800, Jacques Charles and other scientists were working with gases to, among other reasons, improve the design of the hot air balloon. These scientists found, after many, many tests, that certain patterns existed in the observations on gas behavior. If the temperature of the gas is increased, the volume of the gas increased. This is known as a natural law. A law is a relationship that exists between variables in a group of data. Laws describe the patterns we see in large amounts of data, but do not describe why the patterns exist.
What is a Belief?
A belief is a statement that is not scientifically provable. Beliefs may or may not be incorrect; they just are outside the realm of science to explore.
Laws vs. Theories
A common misconception is that scientific theories are rudimentary ideas that will eventually graduate into scientific laws when enough data and evidence has accumulated. A theory does not change into a scientific law with the accumulation of new or better evidence. Remember, theories are explanations and laws are patterns we see in large amounts of data, frequently written as an equation. A theory will always remain a theory; a law will always remain a law.
Summary
• A hypothesis is a tentative explanation that can be tested by further investigation.
• A theory is a well-supported explanation of observations.
• A scientific law is a statement that summarizes the relationship between variables.
• An experiment is a controlled method of testing a hypothesis.
Contributions & Attributions
• Henry Agnew (UC Davis) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.06%3A_Hypothesis_Theories_and_Laws.txt |
Learning Objectives
• To recognize the breadth, depth, and scope of chemistry.
• Define chemistry in relation to other sciences.
• Identify the main disciplines of chemistry.
Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—the screen on which you are reading this book, for example. Others are not so obvious. Because we move so easily through air, we sometimes forget that it, too, is matter. Because of this, chemistry is a science that has its fingers in just about everything. Being able to describe the ingredients in a cake and how they change when the cake is baked, for example, is chemistry!
Chemistry is one branch of science. Science is the process by which we learn about the natural universe by observing, testing, and then generating models that explain our observations. Because the physical universe is so vast, there are many different branches of science (Figure \(1\)). Thus, chemistry is the study of matter, biology is the study of living things, and geology is the study of rocks and the earth. Mathematics is the language of science, and we will use it to communicate some of the ideas of chemistry.
Although we divide science into different fields, there is much overlap among them. For example, some biologists and chemists work in both fields so much that their work is called biochemistry. Similarly, geology and chemistry overlap in the field called geochemistry. Figure \(1\) shows how many of the individual fields of science are related. At some level, all of these fields depend on matter because they all involve "stuff"; because of this, chemistry has been called the "central science", linking them all together.
There are many other fields of science, in addition to the ones (biology, medicine, etc.) listed here.
Example \(1\): Science Fields
Which fields of study are branches of science? Explain.
1. sculpture
2. astronomy
Solution
1. Sculpture is not considered a science because it is not a study of some aspect of the natural universe.
2. Astronomy is the study of stars and planets, which are part of the natural universe. Astronomy is therefore a field of science.
Exercise \(1\)
Which fields of study are branches of science?
1. physiology (the study of the function of an animal’s or a plant’s body)
2. geophysics
3. agriculture
4. politics
Answer a:
yes
Answer b:
yes
Answer c:
yes
Answer d:
no
Areas of Chemistry
The study of modern chemistry has many branches, but can generally be broken down into five main disciplines, or areas of study:
• Physical chemistry: Physical chemistry is the study of macroscopic properties, atomic properties, and phenomena in chemical systems. A physical chemist may study such things as the rates of chemical reactions, the energy transfers that occur in reactions, or the physical structure of materials at the molecular level.
• Organic chemistry: Organic chemistry is the study of chemicals containing carbon. Carbon is one of the most abundant elements on Earth and is capable of forming a tremendously vast number of chemicals (over twenty million so far). Most of the chemicals found in all living organisms are based on carbon.
• Inorganic chemistry: Inorganic chemistry is the study of chemicals that, in general, are not primarily based on carbon. Inorganic chemicals are commonly found in rocks and minerals. One current important area of inorganic chemistry deals with the design and properties of materials involved in energy and information technology.
• Analytical chemistry: Analytical chemistry is the study of the composition of matter. It focuses on separating, identifying, and quantifying chemicals in samples of matter. An analytical chemist may use complex instruments to analyze an unknown material in order to determine its various components.
• Biochemistry: Biochemistry is the study of chemical processes that occur in living things. Research may cover anything from basic cellular processes up to understanding disease states so that better treatments can be developed.
In practice, chemical research is often not limited to just one of the five major disciplines. A particular chemist may use biochemistry to isolate a particular chemical found in the human body such as hemoglobin, the oxygen carrying component of red blood cells. He or she may then proceed to analyze the hemoglobin using methods that would pertain to the areas of physical or analytical chemistry. Many chemists specialize in areas that are combinations of the main disciplines, such as bioinorganic chemistry or physical organic chemistry.
History of Chemistry
The history of chemistry is an interesting and challenging one. Very early chemists were often motivated mainly by the achievement of a specific goal or product. Making perfume or soaps did not need a lot of theory, just a good recipe and careful attention to detail. There was no standard way of naming materials (and no periodic table that we could all agree on). It is often difficult to figure out exactly what a particular person was using. However, the science developed over the centuries by trial and error.
Major progress was made toward putting chemistry on a solid foundation when Robert Boyle (1637-1691) began his research in chemistry (Figure \(3\)). He developed the basic ideas about the behavior of gases. He could then describe gases mathematically. Boyle also helped form the idea that small particles could combine to form molecules. Many years later, John Dalton used these ideas to develop the atomic theory.
The field of chemistry began to develop rapidly in the 1700's. Joseph Priestley (1733-1804) isolated and characterized several gases: oxygen, carbon monoxide, and nitrous oxide. It was later discovered that nitrous oxide ("laughing gas") worked as an anesthetic. This gas was used for that purpose for the first time in 1844 during a tooth extraction. Other gases discovered during that time were chlorine, by C.W. Scheele (1742-1786) and nitrogen, by Antoine Lavoisier (1743-1794). Lavoisier has been considered by many scholars to be the "father of chemistry". Among other accomplishments, he discovered the role of oxygen in combustion and definitively formulated the law of conservation of matter.
Chemists continued to discover new compounds in the 1800's. The science also began to develop a more theoretical foundation. John Dalton (1766-1844) put forth his atomic theory in 1807. This idea allowed scientists to think about chemistry in a much more systematic way. Amadeo Avogadro (1776-1856) laid the groundwork for a more quantitative approach to chemistry by calculating the number of particles in a given amount of a gas. A lot of effort was put forth in studying chemical reactions. These efforts led to new materials being produced. Following the invention of the battery by Alessandro Volta (1745-1827), the field of electrochemistry (both theoretical and applications) developed through major contributions by Humphry Davy (1778-1829) and Michael Faraday (1791-1867). Other areas of the discipline also progressed rapidly.
It would take a large book to cover developments in chemistry during the twentieth century and up to today. One major area of expansion was in the area of the chemistry of living processes. Research in photosynthesis in plants, the discovery and characterization of enzymes as biochemical catalysts, elucidation of the structures of biomolecules such as insulin and DNA—these efforts gave rise to an explosion of information in the field of biochemistry.
The practical aspects of chemistry were not ignored. The work of Volta, Davy, and Faraday eventually led to the development of batteries that provided a source of electricity to power a number of devices (Figure \(4\)).
Charles Goodyear (1800-1860) discovered the process of vulcanization, allowing a stable rubber product to be produced for the tires of all the vehicles we have today. Louis Pasteur (1822-1895) pioneered the use of heat sterilization to eliminate unwanted microorganisms in wine and milk. Alfred Nobel (1833-1896) invented dynamite (Figure \(5\)). After his death, the fortune he made from this product was used to fund the Nobel Prizes in science and the humanities. J.W. Hyatt (1837-1920) developed the first plastic. Leo Baekeland (1863-1944) developed the first synthetic resin, widely used for inexpensive and sturdy dinnerware.
Today, chemistry continues to be essential to the development of new materials and technologies, from semiconductors for electronics to powerful new medicines, and beyond.
Summary
• Chemistry is the study of matter and the changes it undergoes and considers both macroscopic and microscopic information.
• Matter is anything that has mass and occupies space.
• The five main disciplines of chemistry are physical chemistry, organic chemistry, inorganic chemistry, analytical chemistry and biochemistry.
• Many civilizations contributed to the growth of chemistry. A lot of early chemical research focused on practical uses. Basic chemistry theories were developed during the nineteenth century. New materials and batteries are a few of the products of modern chemistry. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.07%3A_The_Scope_of_Chemistry.txt |
1.1: Soda Pop Fizz
1.2: Chemicals Compose Ordinary Things
1.3: All Things Are Made of Atoms and Molecules
1.4: The Scientific Method: How Chemists Think
Use the following paragraph to answer the first two questions. In 1928, Sir Alexander Fleming was studying Staphylococcus bacteria growing in culture dishes. He noticed that a mold called Penicillium was also growing in some of the dishes. In Figure 1.13, Petri dish A represents a dish containing only Staphylococcus bacteria. The red dots in dish B represent Penicillium colonies. Fleming noticed that a clear area existed around the mold because all the bacteria grown in this area had died. In the culture dishes without the mold, no clear areas were present. Fleming suggested that the mold was producing a chemical that killed the bacteria. He decided to isolate this substance and test it to see if it would kill bacteria. Fleming grew some Penicillium mold in a nutrient broth. After the mold grew in the broth, he removed all the mold from the broth and added the broth to a culture of bacteria. All the bacteria died.
1. Which of the following statements is a reasonable expression of Fleming’s hypothesis?
1. Nutrient broth kills bacteria.
2. There are clear areas around the Penicillium mold where Staphylococcus doesn't grow.
3. Mold kills bacteria.
4. Penicillium mold produces a substance that kills Staphylococcus.
5. Without mold in the culture dish, there were no clear areas in the bacteria.
2. Fleming grew Penicillium in broth, then removed the Penicillium and poured the broth into culture dishes containing bacteria to see if the broth would kill the bacteria. What step in the scientific method does this represent?
1. Collecting and organizing data
2. Making a hypothesis
3. Testing a hypothesis by experiment
4. Rejecting the old hypothesis and making a new one
5. None of these
A scientific investigation is NOT valid unless every step in the scientific method is present and carried out in the exact order listed in this chapter.
1. True
2. False
Which of the following words is closest to the same meaning as hypothesis?
1. fact
2. law
3. formula
4. suggestion
5. conclusion
Why do scientists sometimes discard theories?
1. the steps in the scientific method were not followed in order
2. public opinion disagrees with the theory
3. the theory is opposed by the church
4. contradictory observations are found
5. congress voted against it
Gary noticed that two plants which his mother planted on the same day, that were the same size when planted, were different in size after three weeks. Since the larger plant was in the full sun all day and the smaller plant was in the shade of a tree most of the day, Gary believed the sunshine was responsible for the difference in the plant sizes. In order to test this, Gary bought ten small plants of the same size and type. He made sure they had the same size and type of pot. He also made sure they had the same amount and type of soil. Then Gary built a frame to hold a canvas roof over five of the plants while the other five were nearby but out in the sun. Gary was careful to make sure that each plant received exactly the same amount of water and plant food every day.
1. Which of the following is a reasonable statement of Gary’s hypothesis?
1. Different plants have different characteristics.
2. Plants that get more sunshine grow larger than plants that get less sunshine.
3. Plants that grow in the shade grow larger.
4. Plants that don’t receive water will die.
5. Plants that receive the same amount of water and plant food will grow the same amount.
2. What scientific reason might Gary have for insisting that the container size for the all plants be the same?
1. Gary wanted to determine if the size of the container would affect the plant growth.
2. Gary wanted to make sure the size of the container did not affect differential plant growth in his experiment.
3. Gary want to control how much plant food his plants received.
4. Gary wanted his garden to look organized.
5. There is no possible scientific reason for having the same size containers.
3. What scientific reason might Gary have for insisting that all plants receive the same amount of water everyday?
1. Gary wanted to test the effect of shade on plant growth and therefore, he wanted to have no variables other than the amount of sunshine on the plants.
2. Gary wanted to test the effect of the amount of water on plant growth.
3. Gary's hypothesis was that water quality was affecting plant growth.
4. Gary was conserving water.
5. There is no possible scientific reason for having the same amount of water for each plant every day.
4. What was the variable being tested in Gary's experiment?
1. the amount of water
2. the amount of plant food
3. the amount of soil
4. the amount of sunshine
5. the type of soil
5. Which of the following factors may be varying in Gary’s experimental setup that he did not control?
1. individual plant variation
2. soil temperature due to different colors of containers
3. water loss due to evaporation from the soil
4. the effect of insects which may attack one set of plants but not the other
5. All of the above are possible factors that Gary did not control.
When a mosquito sucks blood from its host, it penetrates the skin with its sharp beak and injects an anti-coagulant so the blood will not clot. It then sucks some blood and removes its beak. If the mosquito carries disease-causing microorganisms, it injects these into its host along with the anti-coagulant. It was assumed for a long time that the virus typhus was injected by the louse when sucking blood in a manner similar to the mosquito. But apparently this is not so. The infection is not in the saliva of the louse, but in the feces. The disease is thought to be spread when the louse feces come in contact with scratches or bite wounds in the host's skin. A test of this was carried out in 1922 when two workers fed infected lice on a monkey, taking great care that no louse feces came into contact with the monkey. After two weeks, the monkey had NOT become ill with typhus. The workers then injected the monkey with typhus and it became ill within a few days. Why did the workers inject the monkey with typhus near the end of the experiment?
1. to prove that the lice carried the typhus virus
2. to prove the monkey was similar to man
3. to prove that the monkey was not immune to typhus
4. to prove that mosquitoes were not carriers of typhus
5. the workers were mean
Eijkman fed a group of chickens exclusively on rice whose seed coat had been removed (polished rice or white rice). The chickens all developed polyneuritis (a disease of chickens) and died. He fed another group of chickens unpolished rice (rice that still had its seed coat). Not a single one of them contracted polyneuritis. He then gathered the polishings from rice (the seed coats that had been removed) and fed the polishings to other chickens that were sick with polyneuritis. In a short time, the birds all recovered. Eijkman had accurately traced the cause of polyneuritis to a faulty diet. For the first time in history, a food deficiency disease had been produced and cured experimentally. Which of the following is a reasonable statement of Eijkman’s hypothesis?
1. Polyneuritis is a fatal disease for chickens.
2. White rice carries a virus for the disease polyneuritis.
3. Unpolished rice does not carry the polyneuritis virus.
4. The rice seed coat contains a nutrient that provides protection for chickens against polyneuritis.
5. None of these is a reasonable statement of Eijkman's hypothesis.
The three questions below relate to the following paragraphs.
Scientist A noticed that in a certain forest area, the only animals inhabiting the region were giraffes. He also noticed that the only food available for the animals was on fairly tall trees and as the summer progressed, the animals ate the leaves high and higher on the trees. The scientist suggested that these animals were originally like all other animals but generations of animals stretching their necks to reach higher up the trees for food, caused the species to grow very long necks.
Scientist B conducted experiments and observed that stretching muscles does NOT cause bones to grow longer nor change the DNA of animals so that longer muscles would be passed on to the next generation. Scientist B, therefore, discarded Scientist A's suggested answer as to why all the animals living in the area had long necks. Scientist B suggested instead that originally many different types of animals including giraffes had lived in the region but only the giraffes could survive when the only food was high in the trees, and so all the other species had left the area.
1. Which of the following statements is an interpretation, rather than an observation?
1. The only animals living in the area were giraffes.
2. The only available food was on tall trees.
3. Animals which constantly stretch their necks will grow longer necks.
4. A, B, and C are all interpretations.
5. A, B, and C are all observations.
2. Scientist A's hypothesis was that
1. the only animals living in the area were giraffes.
2. the only available food was on tall trees.
3. animals which constantly stretch their necks will grow longer necks.
4. the animals which possess the best characteristics for living in an area, will be the predominant species.
5. None of the above are reasonable statements of Scientist A's hypothesis.
3. Scientist A's hypothesis being discarded is
1. evidence that the scientific method doesn’t always work.
2. a result achieved without use of the scientific method.
3. an example of what happened before the scientific method was invented.
4. an example of the normal functioning of the scientific method.
5. an unusual case.
When a theory has been known for a long time, it becomes a law.
1. True
2. False
During Pasteur's time, anthrax was a widespread and disastrous disease for livestock. Many people whose livelihood was raising livestock lost large portions of their herds to this disease. Around 1876, a horse doctor in eastern France named Louvrier, claimed to have invented a cure for anthrax. The influential men of the community supported Louvrier's claim to have cured hundreds of cows of anthrax. Pasteur went to Louvrier's hometown to evaluate the cure. The cure was explained to Pasteur as a multi-step process during which: 1) the cow was rubbed vigorously to make her as hot as possible; 2) long gashes were cut into the cows skin and turpentine was poured into the cuts; 3) an inch-thick coating of cow manure mixed with hot vinegar was plastered onto the cow and the cow was completely wrapped in a cloth. Since some cows recover from anthrax with no treatment, performing the cure on a single cow would not be conclusive, so Pasteur proposed an experiment to test Louvrier's cure. Four healthy cows were to be injected with anthrax microbes, and after the cows became ill, Louvrier would pick two of the cows (A and B) and perform his cure on them while the other two cows (C and D) would be left untreated. The experiment was performed and after a few days, one of the untreated cows died and one of them got better. Of the cows treated by Louvrier's cure, one cow died and one got better. In this experiment, what was the purpose of infecting cows C and D?
1. So that Louvrier would have more than two cows to choose from.
2. To make sure the injection actually contained anthrax.
3. To serve as experimental controls (a comparison of treated to untreated cows).
4. To kill as many cows as possible.
A hypothesis is
1. a description of a consistent pattern in observations.
2. an observation that remains constant.
3. a theory that has been proven.
4. a tentative explanation for a phenomenon.
A number of people became ill after eating oysters in a restaurant. Which of the following statements is a hypothesis about this occurrence?
1. Everyone who ate oysters got sick.
2. People got sick whether the oysters they ate were raw or cooked.
3. Symptoms included nausea and dizziness.
4. The cook felt really bad about it.
5. Bacteria in the oysters may have caused the illness.
Which statement best describes the reason for using experimental controls?
1. Experimental controls eliminate the need for large sample sizes.
2. Experimental controls eliminate the need for statistical tests.
3. Experimental controls reduce the number of measurements needed.
4. Experimental controls allow comparison between groups that are different in only one independent variable.
A student decides to set up an experiment to determine the relationship between the growth rate of plants and the presence of detergent in the soil. He sets up 10 seed pots. In five of the seed pots, he mixes a precise amount of detergent with the soil and the other five seed pots have no detergent in the soil. The five seed pots with detergent are placed in the sun and the five seed pots with no detergent are placed in the shade. All 10 seed pots receive the same amount of water and the same number and type of seeds. He grows the plants for two months and charts the growth every two days. What is wrong with his experiment?
1. The student has too few pots.
2. The student has two independent variables.
3. The student has two dependent variables.
4. The student has no experimental control on the soil.
A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but does not put fertilizer on row 2. Both rows receive the same amount of sun and water. She checks the growth of the corn over the course of five months. What is acting as the control in this experiment?
1. Corn without fertilizer.
2. Corn with fertilizer.
3. Amount of water.
4. Height of corn plants.
If you have a control group for your experiment, which of the following is true?
1. There can be more than one difference between the control group and the test group, but not more three differences, or else the experiment is invalid.
2. The control group and the test group may have many differences between them.
3. The control group must be identical to the test group except for one variable.
4. None of these are true.
If the hypothesis is rejected by the experiment, then:
1. the experiment may have been a success.
2. the experiment was a failure.
3. the experiment was poorly designed.
4. the experiment didn't follow the scientific method.
A well-substantiated explanation of an aspect of the natural world is a:
1. theory.
2. law.
3. hypothesis.
4. None of these.
1.5: A Beginning Chemist: How to Succeed | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/01%3A_The_Chemical_World/1.E%3A_Exercises.txt |
Chemistry, like all sciences, is quantitative. It concerns quantities, things that have amounts and units. Dealing with quantities and relating them to one another is very important in chemistry. In this chapter, we will discuss how we deal with numbers and units, including how they are combined and manipulated.
• 2.1: Taking Measurements
Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5.
• 2.2: Scientific Notation - Writing Large and Small Numbers
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator requires a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes.
• 2.3: Significant Figures - Writing Numbers to Reflect Precision
Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation.
• 2.4: Significant Figures in Calculations
To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
• 2.5: The Basic Units of Measurement
Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable. The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry.
• 2.6: Problem Solving and Unit Conversions
During your studies of chemistry (and physics also), you will note that mathematical equations are used in a number of different applications. Many of these equations have a number of different variables with which you will need to work. Note also that these equations will often require the use of measurements with their units. Algebra skills become very important here!
• 2.7: Solving Multi-step Conversion Problems
Sometimes you will have to perform more than one conversion to obtain the desired unit.
• 2.8: Units Raised to a Power
Conversion factors for area and volume can also be produced by the dimensional analysis method. Remember that if a quantity is raised to a power of 10, both the number and the unit must be raised to the same power of 10.
• 2.9: Density
Density is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant.
• 2.E: Measurement and Problem Solving (Exercises)
Exercises for Chapter 2 of Tro's Introductory Chemistry textmap.
02: Measurement and Problem Solving
Learning Objectives
• Express quantities properly, using a number and a unit.
A coffee maker’s instructions tell you to fill the coffee pot with 4 cups of water and to use 3 scoops of coffee. When you follow these instructions, you are measuring. When you visit a doctor’s office, a nurse checks your temperature, height, weight, and perhaps blood pressure (Figure \(1\)); the nurse is also measuring.
Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far they walk from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 kilometers, 12 miles, 12 furlongs, or 12 yards. Both a number and a unit must be included to express a quantity properly.
To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. The next two sections examine the rules for expressing numbers.
Example \(1\)
Identify the number and the unit in each quantity.
1. one dozen eggs
2. 2.54 centimeters
3. a box of pencils
4. 88 meters per second
Solution
1. The number is one, and the unit is a dozen eggs.
2. The number is 2.54, and the unit is centimeter.
3. The number 1 is implied because the quantity is only a box. The unit is box of pencils.
4. The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds.
Key Take Away
• Identify a quantity properly with a number and a unit. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.01%3A_Taking_Measurements.txt |
Learning Objectives
• Express a large number or a small number in scientific notation.
• Carry out arithmetical operations and express the final answer in scientific notation
Chemists often work with numbers that are exceedingly large or small. For example, entering the mass in grams of a hydrogen atom into a calculator would require a display with at least 24 decimal places. A system called scientific notation avoids much of the tedium and awkwardness of manipulating numbers with large or small magnitudes. In scientific notation, these numbers are expressed in the form
$N \times 10^n \nonumber$
where N is greater than or equal to 1 and less than 10 (1 ≤ N < 10), and n is a positive or negative integer (100 = 1). The number 10 is called the base because it is this number that is raised to the power $n$. Although a base number may have values other than 10, the base number in scientific notation is always 10.
A simple way to convert numbers to scientific notation is to move the decimal point as many places to the left or right as needed to give a number from 1 to 10 (N). The magnitude of n is then determined as follows:
• If the decimal point is moved to the left n places, n is positive.
• If the decimal point is moved to the right n places, n is negative.
Another way to remember this is to recognize that as the number N decreases in magnitude, the exponent increases and vice versa. The application of this rule is illustrated in Example $1$.
Example $1$: Expressing Numbers in Scientific Notation
Convert each number to scientific notation.
1. 637.8
2. 0.0479
3. 7.86
4. 12,378
5. 0.00032
6. 61.06700
7. 2002.080
8. 0.01020
Solution
Solutions to Example 2.2.1
Explanation Answer
a
To convert 637.8 to a number from 1 to 10, we move the decimal point two places to the left: 637.8
Because the decimal point was moved two places to the left, n = 2.
$6.378 \times 10^2$
b
To convert 0.0479 to a number from 1 to 10, we move the decimal point two places to the right: 0.0479
Because the decimal point was moved two places to the right, n = −2.
$4.79 \times 10^{−2}$
c This is usually expressed simply as 7.86. (Recall that 100 = 1.) $7.86 \times 10^0$
d Because the decimal point was moved four places to the left, n = 4. $1.2378 \times 10^4$
e Because the decimal point was moved four places to the right, n = −4. $3.2 \times 10^{−4}$
f Because the decimal point was moved one place to the left, n = 1. $6.106700 \times 10^1$
g Because the decimal point was moved three places to the left, n = 3. $2.002080 \times 10^3$
h Because the decimal point was moved two places to the right, n = -2. $1.020 \times 10^{−2}$
Addition and Subtraction
Before numbers expressed in scientific notation can be added or subtracted, they must be converted to a form in which all the exponents have the same value. The appropriate operation is then carried out on the values of N. Example $2$ illustrates how to do this.
Example $2$: Expressing Sums and Differences in Scientific Notation
Carry out the appropriate operation and then express the answer in scientific notation.
1. $(1.36 \times 10^2) + (4.73 \times 10^3) \nonumber$
2. $(6.923 \times 10^{−3}) − (8.756 \times 10^{−4}) \nonumber$
Solution
Solutions to Example 2.2.2.
Explanation Answer
a
Both exponents must have the same value, so these numbers are converted to either
$(1.36 \times 10^2) + (47.3 \times 10^2) = (1.36 + 47.3) \times 10^2 = 48.66 × 10^2$
or
$(0.136 \times 10^3) + (4.73 \times 10^3) = (0.136 + 4.73) \times 10^3) = 4.87 \times 10^3$.
Choosing either alternative gives the same answer, reported to two decimal places.
In converting 48.66 × 102 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$4.87 \times 10^3$
b
Converting the exponents to the same value gives either
$(6.923 \times 10^{-3}) − (0.8756 \times 10^{-3}) = (6.923 − 0.8756) \times 10^{−3}$
or
$(69.23 \times 10^{-4}) − (8.756 \times 10^{-4}) = (69.23 − 8.756) \times 10^{−4} = 60.474 \times 10^{−4}$.
In converting 60.474 × 10-4 to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$6.047 \times 10^{−3}$
Multiplication and Division
When multiplying numbers expressed in scientific notation, we multiply the values of $N$ and add together the values of $n$. Conversely, when dividing, we divide $N$ in the dividend (the number being divided) by $N$ in the divisor (the number by which we are dividing) and then subtract n in the divisor from n in the dividend. In contrast to addition and subtraction, the exponents do not have to be the same in multiplication and division. Examples of problems involving multiplication and division are shown in Example $3$.
Example $3$: Expressing Products and Quotients in Scientific Notation
Perform the appropriate operation and express your answer in scientific notation.
1. $(6.022 \times 10^{23})(6.42 \times 10^{−2}) \nonumber$
2. $\dfrac{ 1.67 \times 10^{-24} }{ 9.12 \times 10 ^{-28} } \nonumber$
3. $\dfrac{ (6.63 \times 10^{−34})(6.0 \times 10) }{ 8.52 \times 10^{−2}} \nonumber$
Solution
Solution to Example 2.2.3
Explanation Answer
a
In multiplication, we add the exponents:
$(6.022 \times 10^{23})(6.42 \times 10^{−2})= (6.022)(6.42) \times 10^{[23 + (−2)]} = 38.7 \times 10^{21} \nonumber$
In converting $38.7 \times 10^{21}$ to scientific notation, $n$ has become more positive by 1 because the value of $N$ has decreased.
$3.87 \times 10^{22}$ b
In division, we subtract the exponents:
${1.67 \times 10^{−24} \over 9.12 \times 10^{−28}} = {1.67 \over 9.12} \times 10^{[−24 − (−28)]} = 0.183 \times 10^4 \nonumber$
In converting $0.183 \times 10^4$ to scientific notation, $n$ has become more negative by 1 because the value of $N$ has increased.
$1.83 \times 10^3$ c
This problem has both multiplication and division:
${(6.63 \times 10^{−34})(6.0 \times 10) \over (8.52 \times 10^{−2})} = {39.78 \over 8.52} \times 10^{[−34 + 1 − (−2)]} \nonumber$
$4.7\times 10^{-31}$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.02%3A_Scientific_Notation_-_Writing_Large_and_Small_Numbers.txt |
Learning Objectives
• Identify the number of significant figures in a reported value.
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit. In the ruler illustration below, the bottom ruler gave a length with 2 significant figures, while the top ruler gave a length with 3 significant figures. In a correctly reported measurement, the final digit is significant but not certain. Insignificant digits are not reported. With either ruler, it would not be possible to report the length at $2.553 \: \text{cm}$ as there is no possible way that the thousandths digit could be estimated. The 3 is not significant and would not be reported.
Measurement Uncertainty
Some error or uncertainty always exists in any measurement. The amount of uncertainty depends both upon the skill of the measurer and upon the quality of the measuring tool. While some balances are capable of measuring masses only to the nearest $0.1 \: \text{g}$, other highly sensitive balances are capable of measuring to the nearest $0.001 \: \text{g}$ or even better. Many measuring tools such as rulers and graduated cylinders have small lines which need to be carefully read in order to make a measurement. Figure $1$ shows two rulers making the same measurement of an object (indicated by the blue arrow).
With either ruler, it is clear that the length of the object is between $2$ and $3 \: \text{cm}$. The bottom ruler contains no millimeter markings. With that ruler, the tenths digit can be estimated and the length may be reported as $2.5 \: \text{cm}$. However, another person may judge that the measurement is $2.4 \: \text{cm}$ or perhaps $2.6 \: \text{cm}$. While the 2 is known for certain, the value of the tenths digit is uncertain.
The top ruler contains marks for tenths of a centimeter (millimeters). Now the same object may be measured as $2.55 \: \text{cm}$. The measurer is capable of estimating the hundredths digit because he can be certain that the tenths digit is a 5. Again, another measurer may report the length to be $2.54 \: \text{cm}$ or $2.56 \: \text{cm}$. In this case, there are two certain digits (the 2 and the 5), with the hundredths digit being uncertain. Clearly, the top ruler is a superior ruler for measuring lengths as precisely as possible.
Example $1$: Reporting Measurements to the Proper Number of Significant Figures
Use each diagram to report a measurement to the proper number of significant figures.
a.
b.
Solutions
Solutions to Example 2.3.1
Explanation Answer
a. The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it’s at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. The symbol psi stands for “pounds per square inch” and is a unit of pressure, like air in a tire. The measurement is reported to three significant figures. 4.33 psi
b. The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle’s width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. The measurement is reported to three significant figures. 1.25 cm
Exercise $1$
What would be the reported width of this rectangle?
Answer
1.25 cm
When you look at a reported measurement, it is necessary to be able to count the number of significant figures. The table below details the rules for determining the number of significant figures in a reported measurement. For the examples in the table, assume that the quantities are correctly reported values of a measured quantity.
Table $1$: Significant Figure Rules
Rule Examples
1. All nonzero digits in a measurement are significant.
• 237 has three significant figures.
• 1.897 has four significant figures.
2. Zeros that appear between other nonzero digits (middle zeros) are always significant.
• 39,004 has five significant figures.
• 5.02 has three significant figures.
3. Zeros that appear in front of all of the nonzero digits are called leading zeros. Leading zeros are never significant.
• 0.008 has one significant figure.
• 0.000416 has three significant figures.
4. Zeros that appear after all nonzero digits are called trailing zeros. A number with trailing zeros that lacks a decimal point may or may not be significant. Use scientific notation to indicate the appropriate number of significant figures.
• 1400 is ambiguous.
• $1.4 \times 10^3$ has two significant figures.
• $1.40 \times 10^3$ three significant figures.
• $1.400 \times 10^3$ has four significant figures.
5. Trailing zeros in a number with a decimal point are significant. This is true whether the zeros occur before or after the decimal point.
• 620.0 has four significant figures.
• 19.000 has five significant figures.
Exact Numbers
Integers obtained either by counting objects or from definitions are exact numbers, which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures:
Example $2$
Give the number of significant figures in each. Identify the rule for each.
1. 5.87
2. 0.031
3. 52.90
4. 00.2001
5. 500
6. 6 atoms
Solution
Solution to Example 2.3.2
Explanation Answer
a All three numbers are significant (rule 1). 5.87, three significant figures
b The leading zeros are not significant (rule 3). The 3 and the 1 are significant (rule 1). 0.031, two significant figures
c The 5, the 2 and the 9 are significant (rule 1). The trailing zero is also significant (rule 5). 52.90, four significant figures
d The leading zeros are not significant (rule 3). The 2 and the 1 are significant (rule 1) and the middle zeros are also significant (rule 2). 00.2001, four significant figures
e The number is ambiguous. It could have one, two or three significant figures. 500, ambiguous
f The 6 is a counting number. A counting number is an exact number. 6, infinite
Exercise $2$
Give the number of significant figures in each.
1. 36.7 m
2. 0.006606 s
3. 2,002 kg
4. 306,490,000 people
5. 3,800 g
Answer a
three significant figures
Answer b
four significant figures
Answer c
four significant figures
Answer d
infinite (exact number)
Answer e
Ambiguous, could be two, three or four significant figures.
Accuracy and Precision
Measurements may be accurate, meaning that the measured value is the same as the true value; they may be precise, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise. The video below demonstrates the concepts of accuracy and precision.
Example $3$
The following archery targets show marks that represent the results of four sets of measurements.
Which target shows
1. a precise, but inaccurate set of measurements?
2. a set of measurements that is both precise and accurate?
3. a set of measurements that is neither precise nor accurate?
Solution
1. Set a is precise, but inaccurate.
2. Set c is both precise and accurate.
3. Set d is neither precise nor accurate.
Summary
Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.03%3A_Significant_Figures_-_Writing_Numbers_to_Reflect_Precision.txt |
Learning Objectives
• Use significant figures correctly in arithmetical operations.
Rounding
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
Consider the measurement $207.518 \: \text{m}$. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table $1$.
Number of Significant Figures Rounded Value Reasoning
Table $1$: Rounding examples
6 207.518 All digits are significant
5 207.52 8 rounds the 1 up to 2
4 207.5 2 is dropped
3 208 5 rounds the 7 up to 8
2 210 8 is replaced by a 0 and rounds the 0 up to 1
1 200 1 is replaced by a 0
Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.
It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division.
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer.
Multiplication and Division
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows:
The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up.
Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 102, whereas 450.0 has four significant figures and would be written as 4.500 × 102. In scientific notation, all significant figures are listed explicitly.
Example $1$
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 23.096 × 90.300
2. 125 × 9.000
Solution
a
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. $2.0856 \times 10^3$
b
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
The calculator gives 1,125 as the answer, but we limit it to three significant figures. $1.13 \times 10^3$
Addition and Subtraction
How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column.
We drop the last digit—the 1—because it is not significant to the final answer.
The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5.
Example $2$
1. 13.77 + 908.226
2. 1,027 + 611 + 363.06
Solution
a
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place, we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 $922.00 = 9.2200 \times 10^2$
b
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. $2,001.06 = 2.001 \times 10^3$
Exercise $2$
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
1. 217 ÷ 903
2. 13.77 + 908.226 + 515
3. 255.0 − 99
4. 0.00666 × 321
Answer a:
$0.240 = 2.40 \times 10^{-1}$
Answer b:
$1,437 = 1.437 \times 10^3$
Answer c:
$156 = 1.56 \times 10^2$
Answer d:
$2.14 = 2.14 \times 10^0$
Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator.
Calculations Involving Multiplication/Division and Addition/Subtraction
In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.
In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step.
Example $3$
1. 2(1.008 g) + 15.99 g
2. 137.3 s + 2(35.45 s)
3. ${118.7 g \over 2} - 35.5 g$
Solution
a.
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation for multiplication first. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
2(1.008 g) + 15.99 g =
Perform multiplication first.
2 (1.008 g 4 sig figs) = 2.016 g 4 sig figs
The number with the least number of significant figures is 1.008 g; the number 2 is an exact number and therefore has an infinite number of significant figures.
Then, perform the addition.
2.016 g thousandths place + 15.99 g hundredths place (least precise) = 18.006 g
Round the final answer.
Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise).
18.01 g (rounding up)
b.
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation for multiplication first. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
137.3 s + 2(35.45 s) =
Perform multiplication first.
2(35.45 s 4 sig figs) = 70.90 s 4 sig figs
The number with the least number of significant figures is 35.45; the number 2 is an exact number and therefore has an infinite number of significant figures.
Then, perform the addition.
137.3 s tenths place (least precise) + 70.90 s hundredths place = 208.20 s
Round the final answer.
Round the final answer to the tenths place based on 137.3 s.
208.2 s
c.
Table with two columns and 1 row. The first column on the left is labeled, Explanation, and underneath in the row is an explanation for division first. The second column is labeled, Answer, and underneath in the row is an answer.
Explanation Answer
${118.7 g \over 2} - 35.5 g$ =
Perform division first.
${118.7 g \over 2}$ 4 sig figs = 59.35 g 4 sig figs
The number with the least number of significant figures is 118.7 g; the number 2 is an exact number and therefore has an infinite number of significant figures.
Perform subtraction next.
59.35 g hundredths place − 35.5 g tenths place (least precise) = 23.85 g
Round the final answer.
Round the final answer to the tenths place based on 35.5 g.
23.9 g (rounding up)
Exercise $3$
Complete the calculations and report your answers using the correct number of significant figures.
1. 5(1.008s) - 10.66 s
2. 99.0 cm+ 2(5.56 cm)
Answer a
-5.62 s
Answer b
110.2 cm
Summary
• Rounding
• If the number to be dropped is greater than or equal to 5, increase the number to its left by 1 (e.g. 2.9699 rounded to three significant figures is 2.97).
• If the number to be dropped is less than 5, there is no change (e.g. 4.00443 rounded to four significant figures is 4.004).
• The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures.
• The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.04%3A_Significant_Figures_in_Calculations.txt |
Learning Objectives
• State the different measurement systems used in chemistry.
• Describe how prefixes are used in the metric system and identify how the prefixes milli-, centi-, and kilo- compare to the base unit.
How long is a yard? It depends on whom you ask and when you asked the question. Today we have a standard definition of the yard, which you can see marked on every football field. If you move the ball ten yards, you get a first down and it does not matter whether you are playing in Los Angeles, Dallas, or Green Bay. But at one time that yard was arbitrarily defined as the distance from the tip of the king's nose to the end of his outstretched hand. Of course, the problem there is simple: new king, new distance (and then you have to re-mark all of those football fields).
SI Base Units
All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science because of the difficulty in converting from one unit to another. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités. The SI was adopted by international agreement in 1960 and is composed of seven base units in Table $1$.
Quantity SI Base Unit Symbol
Table $1$: SI Base Units of Measurement
Length meter $\text{m}$
Mass kilogram $\text{kg}$
Temperature kelvin $\text{K}$
Time second $\text{s}$
Amount of a Substance mole $\text{mol}$
Electric Current ampere $\text{A}$
Luminous Intensity candela $\text{cd}$
The first units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these seven base units.
Unfortunately, the Metric System is Not Ubiquitous
The map below shows the adoption of the SI units in countries around the world. The United States has legally adopted the metric system for measurements, but does not use it in everyday practice. Great Britain and much of Canada use a combination of metric and imperial units.
Prefix Multipliers
Conversions between metric system units are straightforward because the system is based on powers of ten. For example, meters, centimeters, and millimeters are all metric units of length. There are 10 millimeters in 1 centimeter and 100 centimeters in 1 meter. Metric prefixes are used to distinguish between units of different size. These prefixes all derive from either Latin or Greek terms. For example, mega comes from the Greek word $\mu \varepsilon \gamma \alpha \varsigma$, meaning "great". Table $2$ lists the most common metric prefixes and their relationship to the central unit that has no prefix. Length is used as an example to demonstrate the relative size of each prefixed unit.
Prefix Unit Abbreviation Meaning Example
Table $2$: SI Prefixes
giga $\text{G}$ 1,000,000,000 1 gigameter $\left( \text{Gm} \right)=10^9 \: \text{m}$
mega $\text{M}$ 1,000,000 1 megameter $\left( \text{Mm} \right)=10^6 \: \text{m}$
kilo $\text{k}$ 1,000 1 kilometer $\left( \text{km} \right)=1,000 \: \text{m}$
hecto $\text{h}$ 100 1 hectometer $\left( \text{hm} \right)=100 \: \text{m}$
deka $\text{da}$ 10 1 dekameter $\left( \text{dam} \right)=10 \: \text{m}$
1 1 meter $\left( \text{m} \right)$
deci $\text{d}$ 1/10 1 decimeter $\left( \text{dm} \right)=0.1 \: \text{m}$
centi $\text{c}$ 1/100 1 centimeter $\left( \text{cm} \right)=0.01 \: \text{m}$
milli $\text{m}$ 1/1,000 1 millimeter $\left( \text{mm} \right)=0.001 \: \text{m}$
micro $\mu$ 1/1,000,000 1 micrometer $\left( \mu \text{m} \right)=10^{-6} \: \text{m}$
nano $\text{n}$ 1/1,000,000,000 1 nanometer $\left( \text{nm} \right)=10^{-9} \: \text{m}$
pico $\text{p}$ 1/1,000,000,000,000 1 picometer $\left( \text{pm} \right)=10^{-12} \: \text{m}$
There are a couple of odd little practices with the use of metric abbreviations. Most abbreviations are lowercase. We use "$\text{m}$" for meter and not "$\text{M}$". However, when it comes to volume, the base unit "liter" is abbreviated as "$\text{L}$" and not "$\text{l}$". So we would write 3.5 milliliters as $3.5 \: \text{mL}$.
As a practical matter, whenever possible you should express the units in a small and manageable number. If you are measuring the weight of a material that weighs $6.5 \: \text{kg}$, this is easier than saying it weighs $6500 \: \text{g}$ or $0.65 \: \text{dag}$. All three are correct, but the $\text{kg}$ units in this case make for a small and easily managed number. However, if a specific problem needs grams instead of kilograms, go with the grams for consistency.
Example $1$: Unit Abbreviations
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kiloliter
2. microsecond
3. decimeter
4. nanogram
Solutions
Solutions to Example 2.5.1
Explanation Answer
a The prefix kilo means “1,000 ×,” so 1 kL equals 1,000 L. kL
b The prefix micro implies 1/1,000,000th of a unit, so 1 µs equals 0.000001 s. µs
c The prefix deci means 1/10th, so 1 dm equals 0.1 m. dm
d The prefix nano means 1/1000000000, so a nanogram is equal to 0.000000001 g. ng
Exercise $1$
Give the abbreviation for each unit and define the abbreviation in terms of the base unit.
1. kilometer
2. milligram
3. nanosecond
4. centiliter
Answer a:
km
Answer b:
mg
Answer c:
ns
Answer d:
cL
Summary
• Metric prefixes derive from Latin or Greek terms. The prefixes are used to make the units manageable.
• The SI system is based on multiples of ten. There are seven basic units in the SI system. Five of these units are commonly used in chemistry. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.05%3A_The_Basic_Units_of_Measurement.txt |
Learning Objectives
• To convert a value reported in one unit to a corresponding value in a different unit using conversion factors.
During your studies of chemistry (and physics also), you will note that mathematical equations are used in many different applications. Many of these equations have a number of different variables with which you will need to work. You should also note that these equations will often require you to use measurements with their units. Algebra skills become very important here!
Converting Between Units with Conversion Factors
A conversion factor is a factor used to convert one unit of measurement into another. A simple conversion factor can convert meters into centimeters, or a more complex one can convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. Always remember that a conversion factor has to represent a fact; this fact can either be simple or more complex. For instance, you already know that 12 eggs equal 1 dozen. A more complex fact is that the speed of light is $1.86 \times 10^5$ miles/$\text{sec}$. Either one of these can be used as a conversion factor depending on what type of calculation you are working with (Table $1$).
Table $1$: Conversion Factors from SI units to English Units
English Units Metric Units Quantity
1 ounce (oz) 28.35 grams (g) *mass
1 fluid once (oz) 29.6 mL volume
2.205 pounds (lb) 1 kilogram (kg) *mass
1 inch (in) 2.54 centimeters (cm) length
0.6214 miles (mi) 1 kilometer (km) length
1 quarter (qt) 0.95 liters (L) volume
*Pounds and ounces are technically units of force, not mass, but this fact is often ignored by the non-scientific community.
Of course, there are other ratios which are not listed in Table $1$. They may include:
• Ratios embedded in the text of the problem (using words such as per or in each, or using symbols such as / or %).
• Conversions in the metric system, as covered earlier in this chapter.
• Common knowledge ratios (such as 60 seconds $=$ 1 minute).
If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter.
$1\; \rm{cm} = \dfrac{1}{100} \; \rm{m} = 10^{-2}\rm{m} \nonumber$
or
$100\; \rm{cm} = 1\; \rm{m} \nonumber$
Suppose we divide both sides of the equation by $1 \text{m}$ (both the number and the unit):
$\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}} \nonumber$
As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1:
$\dfrac{ \text{100 cm}}{\text{1 m}} = \dfrac{ \text{1000 mm}}{\text{1 m}}= \dfrac{ 1\times 10^6 \mu \text{m}}{\text{1 m}}= 1 \nonumber$
We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units.
Performing Dimensional Analysis
Dimensional analysis is amongst the most valuable tools that physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others. The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis.
Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as $\mathrm{\dfrac{100\:cm}{1\:m}}$ and multiply:
$3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}} \nonumber$
The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out:
$\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}} \nonumber$
The final step is to perform the calculation that remains once the units have been canceled:
$\dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \nonumber$
In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows:
quantity (in old units) × conversion factor = quantity (in new units)
You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you encounter will not always be so simple. If you master the technique of applying conversion factors, you will be able to solve a large variety of problems.
In the previous example, we used the fraction $\dfrac{100 \; \rm{cm}}{1 \; \rm{m}}$ as a conversion factor. Does the conversion factor $\dfrac{1 \; \rm m}{100 \; \rm{cm}}$ also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten:
$3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}} \nonumber$
For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out. Figure $1$ shows a concept map for constructing a proper conversion.
General Steps in Performing Dimensional Analysis
1. Identify the "given" information in the problem. Look for a number with units to start this problem with.
2. What is the problem asking you to "find"? In other words, what unit will your answer have?
3. Use ratios and conversion factors to cancel out the units that aren't part of your answer, and leave you with units that are part of your answer.
4. When your units cancel out correctly, you are ready to do the math. You are multiplying fractions, so you multiply the top numbers and divide by the bottom numbers in the fractions.
Significant Figures in Conversions
How do conversion factors affect the determination of significant figures?
• Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact.
• Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.)
• Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact.
• In contrast, conversion factors that come from measurements (such as density, as we will see shortly) or that are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer.
Example $1$
Steps for Problem Solving for Example 2.6.1 and 2.6.2
Example $1$ Example $2$
Steps for Problem Solving The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters? A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms?
Identify the "given" information and what the problem is asking you to "find." Given: 4.7 L
Find: mL
Given: 18 ms
Find: s
List other known quantities. $1\, mL = 10^{-3} L$ $1 \,ms = 10^{-3} s$
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$4.7 \cancel{\rm{L}} \times \dfrac{1 \; \rm{mL}}{10^{-3}\; \cancel{\rm{L}}} = 4,700\; \rm{mL}$
or
$4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL}$
or
4.7 x 103 2SF, not ambiguous
$18 \; \cancel{\rm{ms}} \times \dfrac{10^{-3}\; \rm{s}}{1 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
or
$18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s}$
Think about your result. The amount in mL should be 1000 times larger than the given amount in L. The amount in s should be 1/1000 the given amount in ms.
Exercise $1$
Perform each conversion.
1. 101,000 ns to seconds
2. 32.08 kg to grams
3. 1.53 grams to cg
Answer a:
$1.01000 x 10^{-4} s$
Answer b:
$3.208 x 10^{4} g$
Answer c:
$1.53 x 10^{2} cg$
Summary
• Conversion factors are used to convert one unit of measurement into another.
• Dimensional analysis (unit conversions) involves the use of conversion factors that will cancel unwanted units and produce the appropriate units. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.06%3A_Problem_Solving_and_Unit_Conversions.txt |
Multiple Conversions
Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. We will set up a series of conversion factors so that each conversion factor produces the next unit in the sequence. We first convert the given amount in km to the base unit, which is meters. We know that 1,000 m =1 km.
Then we convert meters to mm, remembering that $1\; \rm{mm}$ = $10^{-3}\; \rm{m}$.
Concept Map
Calculation
\begin{align*} 54.7 \; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1\; \cancel{\rm{mm}}}{\cancel{10^{-3} \rm{m}}} & = 54,700,000 \; \rm{mm} \ &= 5.47 \times 10^7\; \rm{mm} \end{align*} \nonumber
In each step, the previous unit is canceled and the next unit in the sequence is produced, each successive unit canceling out until only the unit needed in the answer is left.
Example $1$: Unit Conversion
Convert 58.2 ms to megaseconds in one multi-step calculation.
Solution
Solution for Example 2.7.1
Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 58.2 ms
Find: Ms
List other known quantities
$1 ms = 10^{-3} s$
$1 Ms = 10^6s$
Prepare a concept map.
Calculate.
\begin{align} 58.2 \; \cancel{\rm{ms}} \times \dfrac{10^{-3} \cancel{\rm{s}}}{1\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} & =0.0000000582\; \rm{Ms} \nonumber\ &= 5.82 \times 10^{-8}\; \rm{Ms}\nonumber \end{align}\nonumber
Neither conversion factor affects the number of significant figures in the final answer.
Example $2$: Unit Conversion
How many seconds are in a day?
Solution
Solution for Example 2.7.2
Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 1 day
Find: s
List other known quantities.
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
Prepare a concept map.
Calculate. $1 \: \text{d} \times \dfrac{24 \: \text{hr}}{1 \: \text{d}}\times \dfrac{60 \: \text{min}}{1 \: \text{hr}} \times \dfrac{60 \: \text{s}}{1 \: \text{min}} = 86,400 \: \text{s} \nonumber$
Exercise $1$
Perform each conversion in one multi-step calculation.
1. 43.007 ng to kg
2. 1005 in to ft
3. 12 mi to km
Answer a
$4.3007 \times 10^{-11} kg$
Answer b
$83.75\, ft$
Answer c
$19\, km$
Career Focus: Pharmacist
A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school. Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities. Curiously, an outdated name for pharmacist is chemist, which was used when pharmacists formerly did a lot of drug preparation, or compounding. In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health.
Summary
In multi-step conversion problems, the previous unit is canceled for each step and the next unit in the sequence is produced, each successive unit canceling out until only the unit needed in the answer is left.
2.08: Units Raised to a Power
Learning Objectives
• To convert a value reported in one unit raised to a power of 10, to a corresponding value in a different unit raised to the same power of 10, using conversion factors.
Conversion factors for area and volume can also be produced by the dimensional analysis method. Just remember that if a quantity is raised to a power of 10, both the number and the unit must be raised to the same power of 10. For example, to convert $1500 \: \text{cm}^2$ to $\text{m}^2$, we need to start with the relationship between centimeter and meter. We know that 1 cm = 10-2 m or 100 cm =1 m, but since we are given the quantity in 1500 cm2, then we have to use the relationship:
$1\, cm^2 = (10^{-2}\, m)^2 = 10^{-4}\, m^2 \nonumber$
CONCEPTMAP
CALCULATION
$1500 \: \cancel{\text{cm}}^2 \times \left( \dfrac{10^{-2} \: \text{m}}{1 \: \cancel{\text{cm}}} \right)^2 = 0.15 \: \text{m}^2 \nonumber$
or
$1500 \: \cancel{\text{cm}}^2 \times \left( \dfrac{1 \: \text{m}}{100 \: \cancel{\text{cm}}} \right)^2 = 0.15 \: \text{m}^2 \nonumber$
or
$1500 \: \cancel{\text{cm}}^2 \times \dfrac{1 \: \text{m}^2}{10,000 \: \cancel{\text{cm}^2}} = 0.15 \: \text{m}^2 \nonumber$
Example $1$: Volume of a Sphere
What is the volume of a sphere (radius 4.30 inches) in cubic cm (cm3)?
Solution
Solution for Example 2.8.1
Steps for Problem Solving What is the volume of a sphere (radius 4.30 inches) in cubic cm (cm3)?
Identify the "given” information and what the problem is asking you to "find."
Given: radius = 4.30 in
Find: cm3 (volume)
Determine other known quantities.
Volume of a sphere: V = $\dfrac{4}{3} \times \pi \times r^3$
= $\dfrac{4}{3} \times 3.1416 \times (4.3\underline{0}in)^3$
= $33\underline{3}.04 in^3$
Prepare a concept map.
Calculate. $33\underline{3}.04 \cancel{in^3} \left(\dfrac{2.54cm}{1 \cancel{in}}\right)^3 = 5.46 \times10^3 cm^3$
Think about your result. A centimeter is a smaller unit than an inch, so the answer in cubic centimeters is larger than the given value in cubic inches.
Exercise $1$
Lake Tahoe has a surface area of 191 square miles. What is the area in square km (km2)?
Answer
495 km2 | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.07%3A_Solving_Multi-step_Conversion_Problems.txt |
Learning Objectives
• Define density.
• Use density as a conversion factor.
Density ($\rho$) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows:
$\text{Density}=\dfrac{\text{Mass}}{\text{Volume}} \nonumber$
or just
$\rho =\dfrac{m}{V} \label{eq2}$
Based on this equation, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relationship of mass and volume. Pure water, for example, has a density of 0.998 g/cm3 at 25° C. The average densities of some common substances are in Table $1$. Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table $1$: Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Density can be measured for all substances—solids, liquids and gases. For solids and liquids, density is often reported using the units of g/cm3. Densities of gases, which are significantly lower than the densities of solids and liquids, are often given using units of g/L.
Example $1$: Ethyl Alcohol
Calculate the density of a 30.2 mL sample of ethyl alcohol with a mass of 23.71002 g
Solution
This is a direct application of Equation \ref{eq2}:
$\rho = \dfrac{23.71002\,g}{30.2\,mL} = 0.785\, g/mL \nonumber$
Exercise $1$
1. Find the density (in kg/L) of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
2. If you have a 2.130 mL sample of acetic acid with mass 0.002234 kg, what is the density in kg/L?
Answer a
$0.274 \,kg/L$
Answer b
$1.049 \,kg/L$
Density as a Conversion Factor
Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
$\dfrac{13.6\:g}{1\:mL} = 1 \nonumber$
and
$\dfrac{1\:mL}{13.6\:g} = 1 \nonumber$
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 2.0 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top, so that our final answer has a unit of mass:
$\mathrm{2.0\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}=27.2\:g=27\:g} \nonumber$
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
Example $2$: Mercury Thermometer Steps for Problem Solving
A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury?
Solution
Solution to Example 2.9.2
Steps for Problem Solving Unit Conversion
Identify the "given" information and what the problem is asking you to "find."
Given: 0.750 g
Find: mL
List other known quantities. 13.6 g/mL (density of mercury)
Prepare a concept map.
Calculate.
$0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} = 0.055147 ... \; \rm{mL} \approx 0.0551\; \rm{mL} \nonumber$
We have limited the final answer to three significant figures.
Exercise $2$
What is the volume of 100.0 g of air if its density is 1.3 g/L?
Answer
$77 \, L$
Summary
• Density is defined as the mass of an object divided by its volume.
• Density can be used as a conversion factor between mass and volume.
2.E: Measurement and Problem Solving (Exercises)
2.1: Measuring Global Temperatures
2.2: Scientific Notation: Writing Large and Small Numbers
2.3: Significant Figures: Writing Numbers to Reflect Precision
1. Define significant figures. Why are they important?
2. Define the different types of zeros found in a number and explain whether or not they are significant.
3. How many significant figures are in each number?
1. 140
2. 0.009830
3. 15,050
4. 221,560,000
5. 5.67 × 103
6. 2.9600 × 10−5
4. How many significant figures are in each number?
1. 1.05
2. 9,500
3. 0.0004505
4. 0.00045050
5. 7.210 × 106
6. 5.00 × 10−6
5. Round each number to three significant figures.
1. 34,705
2. 34,750
3. 34,570
2.4: Significant Figures in Calculations
2.5: The Basic Units of Measurement
2.6: Problem Solving and Unit Conversions
2.7: Solving Multi-step Conversion Problems
2.8: Units Raised to a Power
2.9: Density
2.10: Numerical Problem-Solving Strategies and the Solution Map | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.09%3A_Density.txt |
• 3.1: In Your Room
Matter is any substance that has mass and takes up space. Matter includes atoms and anything made of atoms, but not other energy phenomena or waves such as light or sound. While this simple definition is easily applied, the way people view matter is often broken down into two characteristic length scales: the macroscopic and the microscopic.
• 3.2: What is Matter?
Matter is anything that has mass and volume (takes up space). For most common objects that we deal with every day, it is fairly simple to demonstrate that they have mass and take up space. You might be able to imagine, however, the difficulty for people several hundred years ago to demonstrate that air has mass and volume. Air (and all other gases) are invisible to the eye, have very small masses compared to equal amounts of solids and liquids, and are quite easy to compress (change volume).
• 3.3: Classifying Matter According to Its State—Solid, Liquid, and Gas
Three states of matter exist—solid, liquid, and gas. Solids have a definite shape and volume. Liquids have a definite volume, but take the shape of the container. Gases have no definite shape or volume.
• 3.4: Classifying Matter According to Its Composition
One useful way of organizing our understanding of matter is to think of a hierarchy that extends down from the most general and complex, to the simplest and most fundamental. Matter can be classified into two broad categories: pure substances and mixtures. A pure substance is a form of matter that has a consistent composition and properties that are constant throughout the sample. A material composed of two or more substances is a mixture.
• 3.5: Differences in Matter- Physical and Chemical Properties
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, melting points, and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change.
• 3.6: Changes in Matter - Physical and Chemical Changes
Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are either classified as physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes.
• 3.7: Conservation of Mass - There is No New Matter
The law of conservation of mass states that matter can not be created or destroyed in a chemical reaction. So the mass of the product equals the mass of the reactant. The reactant is the chemical interaction of two or more elements to make a new substance, and the product is the substance that is formed as the result of a chemical reaction. Matter and its corresponding mass may not be able to be created or destroyed, but can change forms to other substances like liquids, gases, and solids.
• 3.8: Energy
When we speak of using energy, we are really referring to transferring energy from one place to another. Although energy is used in many kinds of different situations, all of these uses rely on energy being transferred in one of two ways—energy can be transferred as heat or as work.
• 3.9: Energy and Chemical and Physical Change
Phase changes involve changes in energy. All chemical reactions involve changes in energy. This may be a change in heat, electricity, light, or other forms of energy. Reactions that absorb energy are endothermic. Reactions that release energy are exothermic.
• 3.10: Temperature - Random Motion of Molecules and Atoms
Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K).
• 3.11: Temperature Changes - Heat Capacity
The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius.
• 3.12: Energy and Heat Capacity Calculations
Heat is a familiar manifestation of transferring energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand.
• 3.E: Matter and Energy (Exercises)
03: Matter and Energy
Matter is any substance that has mass and takes up space. Matter includes atoms and anything made up of atoms, but not other energy phenomena or waves such as light or sound. While this simple definition is easily applied, the way people view matter is often broken down into two characteristic length scales: the macroscopic and the microscopic.
The macroscopic scale is the length scale on which objects or phenomena are large enough to be visible almost practically with the naked eye, without magnifying optical instruments. Everything that one can see, touch, and handle in the dorm room of Figure \(1\) is within the macroscopic scale. To describe each of these objects, only a few macroscopic properties are required. However, each of these items can be decomposed into smaller microscopic scale properties.
The microscopic scale is the scale of objects and events smaller than those that can easily be seen by the naked eye, requiring a lens or microscope to see them clearly. All of the everyday objects that we can bump into, touch, or squeeze are ultimately composed of atoms. This ordinary atomic matter is in turn made up of interacting subatomic particles—usually a nucleus of protons and neutrons, and a cloud of orbiting electrons. Because of this, a large number of variables are needed to describe such a system which complicates the characterization.
Matter vs. Mass
Matter should not be confused with mass, as the two are not the same in modern physics. Matter is a physical substance of which systems may be composed, while mass is not a substance, but rather a quantitative property of matter and other substances or systems.
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.01%3A_In_Your_Room.txt |
Learning Objectives
• Define matter and explain how it is composed of building blocks known as "atoms".
We are all familiar with matter. The definition of Matter is anything that has mass and volume (takes up space). For most common objects that we deal with every day, it is fairly simple to demonstrate that they have mass and take up space. You might be able to imagine, however, the difficulty for people several hundred years ago to demonstrate that air had mass and volume. Air (and all other gases) are invisible to the eye, have very small masses compared to equal amounts of solids and liquids, and are quite easy to compress (change volume). Without sensitive equipment, it would have been difficult to convince people that gases are matter. Today, we can measure the mass of a small balloon when it is deflated and then blow it up, tie it off, and measure its mass again to detect the additional mass due to the air inside. The mass of air, under room conditions, that occupies a one quart jar is approximately 0.0002 pounds. This small amount of mass would have been difficult to measure in times before balances were designed to accurately measure very small masses. Later, scientists were able to compress gases into such a small volume that the gases turned into liquids, which made it clear that gases are matter.
Even though the universe consists of "things" as wildly different as ants and galaxies, the matter that makes up all of these "things" is composed of a very limited number of building blocks. These building blocks are known as atoms, and so far, scientists have discovered or created a grand total of 118 different types of atoms. Scientists have given a name to each different type of atom. A substance that is composed of only one type of atom is called an element. At this point, what should amaze you is that all forms of matter in our universe are made with only 118 different building blocks. In some ways, it's sort of like cooking a gourmet, five-course meal using only three ingredients! How is it possible? To answer that question, you have to understand the ways in which different elements are put together to form matter.
The most important method that nature uses to organize atoms into matter is the formation of molecules. Molecules are groups of two or more atoms that have been bonded together. There are millions of different ways to bond atoms together, which means that there are millions of different possible molecules. Each of these molecules has its own set of chemical properties, and it's these properties with which chemists are most concerned. You will learn a lot more about atoms and molecules, including how they were discovered, in a later part of the textbook.
Summary
All matter has mass and occupies space. All physical objects are made of matter. Matter itself is composed of tiny building blocks known as "atoms". There are only 118 different types of atoms known to man. Frequently, atoms are bonded together to form "molecules".
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.02%3A_What_is_Matter.txt |
Learning Objectives
• To describe the solid, liquid and gas phases.
Water can take many forms. At low temperatures (below $0^\text{o} \text{C}$), it is a solid. When at "normal" temperatures (between $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$), it is a liquid. While at temperatures above $100^\text{o} \text{C}$, water is a gas (steam). The state that water is in depends upon the temperature. Each state has its own unique set of physical properties. Matter typically exists in one of three states: solid, liquid, or gas.
The state that a given substance exhibits is also a physical property. Some substances exist as gases at room temperature (oxygen and carbon dioxide), while others, like water and mercury metal, exist as liquids. Most metals exist as solids at room temperature. All substances can exist in any of these three states. Figure $2$ shows the differences among solids, liquids, and gases at the molecular level. A solid has definite volume and shape, a liquid has a definite volume but no definite shape, and a gas has neither a definite volume nor shape.
Plasma: A Fourth State of Matter
Technically speaking, a fourth state of matter called plasma exists, but it does not naturally occur on earth, so we will omit it from our study here.
A plasma globe operating in a darkened room. (CC BY-SA 3.0; Chocolateoak).
Solids
In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle.
Solids are defined by the following characteristics:
• Definite shape (rigid)
• Definite volume
• Particles vibrate around fixed axes
If we were to cool liquid mercury to its freezing point of $-39^\text{o} \text{C}$, and under the right pressure conditions, we would notice all of the liquid particles would go into the solid state. Mercury can be solidified when its temperature is brought to its freezing point. However, when returned to room temperature conditions, mercury does not exist in solid state for long, and returns back to its more common liquid form.
Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure $3$. Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid.
Liquids
If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
Liquids have the following characteristics:
• No definite shape (takes the shape of its container).
• Has definite volume.
• Particles are free to move over each other, but are still attracted to each other.
A familiar liquid is mercury metal. Mercury is an anomaly. It is the only metal we know of that is liquid at room temperature. Mercury also has an ability to stick to itself (surface tension)—a property that all liquids exhibit. Mercury has a relatively high surface tension, which makes it very unique. Here you see mercury in its common liquid form.
Video $1$: Mercury boiling to become a gas.
If we heat liquid mercury to its boiling point of $357^\text{o} \text{C}$ under the right pressure conditions, we would notice all particles in the liquid state go into the gas state.
Gases
If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. This describes the gas state, which we will consider in more detail elsewhere. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either. The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Gases have the following characteristics:
• No definite shape (takes the shape of its container)
• No definite volume
• Particles move in random motion with little or no attraction to each other
• Highly compressible
Table $1$: Characteristics of the Three States of Matter
Characteristics Solids Liquids Gases
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions
Example $1$
What state or states of matter does each statement, describe?
1. This state has a definite volume, but no definite shape.
2. This state has no definite volume.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes the liquid state.
2. This statement describes the gas state.
3. This statement describes the liquid state.
Exercise $1$
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to each other.
2. This state has individual particles far apart from each other in space.
3. This state has a definite shape.
Answer a:
solid
Answer b:
gas
Answer c:
solid
Summary
• Three states of matter exist—solid, liquid, and gas.
• Solids have a definite shape and volume.
• Liquids have a definite volume, but take the shape of the container.
• Gases have no definite shape or volume. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.03%3A_Classifying_Matter_According_to_Its_StateSolid_Liquid_and_Gas.txt |
Learning Objectives
• Explain the difference between a pure substance and a mixture.
• Explain the difference between an element and a compound.
• Explain the difference between a homogeneous mixture and a heterogeneous mixture.
One useful way of organizing our understanding of matter is to think of a hierarchy that extends down from the most general and complex to the simplest and most fundamental (Figure \(1\)). Matter can be classified into two broad categories: pure substances and mixtures. A pure substance is a form of matter that has a constant composition (meaning that it is the same everywhere) and properties that are constant throughout the sample (meaning that there is only one set of properties such as melting point, color, boiling point, etc. throughout the matter). A material composed of two or more substances is a mixture. Elements and compounds are both examples of pure substances. A substance that cannot be broken down into chemically simpler components is an element. Aluminum, which is used in soda cans, is an element. A substance that can be broken down into chemically simpler components (because it has more than one element) is a compound. For example, water is a compound composed of the elements hydrogen and oxygen. Today, there are about 118 elements in the known universe. In contrast, scientists have identified tens of millions of different compounds to date.
Ordinary table salt is called sodium chloride. It is considered a substance because it has a uniform and definite composition. All samples of sodium chloride are chemically identical. Water is also a pure substance. Salt easily dissolves in water, but salt water cannot be classified as a substance because its composition can vary. You may dissolve a small amount of salt or a large amount into a given amount of water. A mixture is a physical blend of two or more components, each of which retains its own identity and properties in the mixture. Only the form of the salt is changed when it is dissolved into water. It retains its composition and properties.
A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture. The salt water described above is homogeneous because the dissolved salt is evenly distributed throughout the entire salt water sample. Often it is easy to confuse a homogeneous mixture with a pure substance because they are both uniform. The difference is that the composition of the substance is always the same. The amount of salt in the salt water can vary from one sample to another. All solutions are considered homogeneous because the dissolved material is present in the same amount throughout the solution.
A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. Vegetable soup is a heterogeneous mixture. Any given spoonful of soup will contain varying amounts of the different vegetables and other components of the soup.
Phase
A phase is any part of a sample that has a uniform composition and properties. By definition, a pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers. Each of the layers is called a phase.
Example \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. filtered tea
2. freshly squeezed orange juice
3. a compact disc
4. aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms
5. selenium
Given: a chemical substance
Asked for: its classification
Strategy:
1. Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.
2. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Solution
1. A) Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration.
B) Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. A) Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure.
B) Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. A) A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence, a compact disc is not chemically pure.
B) The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. A) Aluminum oxide is a single, chemically pure compound.
5. A) Selenium is one of the known elements.
Exercise \(1\)
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. white wine
2. mercury
3. ranch-style salad dressing
4. table sugar (sucrose)
Answer a:
homogeneous mixture (solution)
Answer b:
element
Answer c:
heterogeneous mixture
Answer d:
compound
Example \(2\)
How would a chemist categorize each example of matter?
1. saltwater
2. soil
3. water
4. oxygen
Solution
1. Saltwater acts as if it were a single substance even though it contains two substances—salt and water. Saltwater is a homogeneous mixture, or a solution.
2. Soil is composed of small pieces of a variety of materials, so it is a heterogeneous mixture.
3. Water is a substance. More specifically, because water is composed of hydrogen and oxygen, it is a compound.
4. Oxygen, a substance, is an element.
Exercise \(2\)
How would a chemist categorize each example of matter?
1. coffee
2. hydrogen
3. an egg
Answer a:
a homogeneous mixture (solution), assuming it is filtered coffee
Answer b:
element
Answer c:
heterogeneous mixture
Summary
Matter can be classified into two broad categories: pure substances and mixtures. A pure substance is a form of matter that has a constant composition and properties that are constant throughout the sample. Mixtures are physical combinations of two or more elements and/or compounds. Mixtures can be classified as homogeneous or heterogeneous. Elements and compounds are both examples of pure substances. Compounds are substances that are made up of more than one type of atom. Elements are the simplest substances made up of only one type of atom.
Vocabulary
• Element: a substance that is made up of only one type of atom.
• Compound:a substance that is made up of more than one type of atom bonded together.
• Mixture: a combination of two or more elements or compounds which have not reacted to bond together; each part in the mixture retains its own properties. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.04%3A_Classifying_Matter_According_to_Its_Composition.txt |
Learning Objectives
To separate physical from chemical properties.
All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).
Physical Property
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Silver is a shiny metal that conducts electricity very well. It can be molded into thin sheets, a property called malleability. Salt is dull and brittle and conducts electricity when it has been dissolved into water, which it does quite easily. Physical properties of matter include color, hardness, malleability, solubility, electrical conductivity, density, melting point, and boiling point.
For the elements, color does not vary much from one element to the next. The vast majority of elements are colorless, silver, or gray. Some elements do have distinctive colors: sulfur and chlorine are yellow, copper is (of course) copper-colored, and elemental bromine is red. However, density can be a very useful parameter for identifying an element. Of the materials that exist as solids at room temperature, iodine has a very low density compared to zinc, chromium, and tin. Gold has a very high density, as does platinum. Pure water, for example, has a density of 0.998 g/cm3 at 25°C. The average densities of some common substances are in Table \(1\). Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.”
Table \(1\): Densities of Common Substances
Substance Density at 25°C (g/cm3)
blood 1.035
body fat 0.918
whole milk 1.030
corn oil 0.922
mayonnaise 0.910
honey 1.420
Hardness helps determine how an element (especially a metal) might be used. Many elements are fairly soft (silver and gold, for example) while others (such as titanium, tungsten, and chromium) are much harder. Carbon is an interesting example of hardness. In graphite, (the "lead" found in pencils) the carbon is very soft, while the carbon in a diamond is roughly seven times as hard.
Melting and boiling points are somewhat unique identifiers, especially of compounds. In addition to giving some idea as to the identity of the compound, important information can be obtained about the purity of the material.
Chemical Properties
Chemical properties of matter describe its potential to undergo some chemical change or reaction by virtue of its composition. The elements, electrons, and bonds that are present give the matter potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually, after studying chemistry for some time, you should be able to look at the formula of a compound and state some chemical property. For example, hydrogen has the potential to ignite and explode given the right conditions—this is a chemical property. Metals in general have the chemical property of reacting with an acid. Zinc reacts with hydrochloric acid to produce hydrogen gas—this is a chemical property.
A chemical property of iron is its capability of combining with oxygen to form iron oxide, the chemical name of rust (Figure \(2\)). The more general term for rusting and other similar processes is corrosion. Other terms that are commonly used in descriptions of chemical changes are burn, rot, explode, decompose, and ferment. Chemical properties are very useful in identifying substances. However, unlike physical properties, chemical properties can only be observed as the substance is in the process of being changed into a different substance.
Table \(2\): Contrasting Physical and Chemical Properties
Physical Properties Chemical Properties
Gallium metal melts at 30 oC. Iron metal rusts.
Mercury is a very dense liquid. A green banana turns yellow when it ripens.
Gold is shiny. A dry piece of paper burns.
Example \(1\)
Which of the following is a chemical property of iron?
1. Iron corrodes in moist air.
2. Density = 7.874 g/cm3
3. Iron is soft when pure.
4. Iron melts at 1808 K.
Solution
"Iron corrodes in moist air" is the only chemical property of iron from the list.
Exercise \(\PageIndex{1A}\)
Which of the following is a physical property of matter?
1. corrosiveness
2. pH (acidity)
3. density
4. flammability
Answer
c
Exercise \(\PageIndex{1B}\)
Which of the following is a chemical property?
1. flammability
2. melting point
3. boiling point
4. density
Answer
a
Summary
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change. To identify a chemical property, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.05%3A_Differences_in_Matter-_Physical_and_Chemical_Properties.txt |
Learning Objectives
• Label a change as chemical or physical.
• List evidence that can indicate a chemical change occurred.
Change is happening all around us all of the time. Just as chemists have classified elements and compounds, they have also classified types of changes. Changes are classified as either physical or chemical changes. Chemists learn a lot about the nature of matter by studying the changes that matter can undergo. Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes.
Physical Change
Physical changes are changes in which no bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc.) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include:
• Changes of state (changes from a solid to a liquid or a gas and vice versa).
• Separation of a mixture.
• Physical deformation (cutting, denting, stretching).
• Making solutions (special kinds of mixtures).
As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change. A physical change is a change to a sample of matter in which some properties of the material change, but the identity of the matter does not. When liquid water is heated, it changes to water vapor. However, even though the physical properties have changed, the molecules are exactly the same as before. We still have each water molecule containing two hydrogen atoms and one oxygen atom covalently bonded. When you have a jar containing a mixture of pennies and nickels and you sort the mixture so that you have one pile of pennies and another pile of nickels, you have not altered the identity of the pennies or the nickels—you've merely separated them into two groups. This would be an example of a physical change. Similarly, if you have a piece of paper, you don't change it into something other than a piece of paper by ripping it up. What was paper before you started tearing is still paper when you are done. Again, this is an example of a physical change.
Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolving is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind.
Chemical Change
Chemical changes occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes.
One good example of a chemical change is burning a candle. The act of burning paper actually results in the formation of new chemicals (carbon dioxide and water) from the burning of the wax. Another example of a chemical change is what occurs when natural gas is burned in your furnace. This time, on the left there is a molecule of methane, $\ce{CH_4}$, and two molecules of oxygen, $\ce{O_2}$; on the right are two molecules of water, $\ce{H_2O}$, and one molecule of carbon dioxide, $\ce{CO_2}$. In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change.
We can't actually see molecules breaking and forming bonds, although that's what defines chemical changes. We have to make other observations to indicate that a chemical change has happened. Some of the evidence for chemical change will involve the energy changes that occur in chemical changes, but some evidence involves the fact that new substances with different properties are formed in a chemical change.
Observations that help to indicate chemical change include:
• Temperature changes (either the temperature increases or decreases).
• Light given off.
• Unexpected color changes (a substance with a different color is made, rather than just mixing the original colors together).
• Bubbles are formed (but the substance is not boiling—you made a substance that is a gas at the temperature of the beginning materials, instead of a liquid).
• Different smell or taste (do not taste your chemistry experiments, though!).
• A solid forms if two clear liquids are mixed (look for floaties—technically called a precipitate).
Example $1$
Label each of the following changes as a physical or chemical change. Give evidence to support your answer.
1. Boiling water.
2. A nail rusting.
3. A green solution and colorless solution are mixed. The resulting mixture is a solution with a pale green color.
4. Two colorless solutions are mixed. The resulting mixture has a yellow precipitate.
Solution
1. Physical: boiling and melting are physical changes. When water boils, no bonds are broken or formed. The change could be written: $\ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right)$
2. Chemical: The dark grey nail changes color to form an orange flaky substance (the rust); this must be a chemical change. Color changes indicate chemical change. The following reaction occurs: $\ce{Fe} + \ce{O_2} \rightarrow \ce{Fe_2O_3}$
3. Physical: because none of the properties changed, this is a physical change. The green mixture is still green and the colorless solution is still colorless. They have just been spread together. No color change occurred or other evidence of chemical change.
4. Chemical: the formation of a precipitate and the color change from colorless to yellow indicate a chemical change.
Exercise $1$
Label each of the following changes as a physical or chemical change.
1. A mirror is broken.
2. An iron nail corroded in moist air
3. Copper metal is melted.
4. A catalytic converter changes nitrogen dioxide to nitrogen gas and oxygen gas.
Answer a:
physical change
Answer b:
chemical change
Answer c:
physical change
Answer d:
chemical change
Separating Mixtures Through Physical Changes
Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.
Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States.
Another example for using physical properties to separate mixtures is filtration (Figure $4$). Filtration is any mechanical, physical or biological operation that separates solids from fluids (liquids or gases) by adding a medium through which only the fluid can pass. The fluid that passes through is called the filtrate. There are many different methods of filtration; all aim to attain the separation of substances. Separation is achieved by some form of interaction between the substance or objects to be removed and the filter. The substance that is to pass through the filter must be a fluid, i.e. a liquid or gas. Methods of filtration vary depending on the location of the targeted material, i.e. whether it is dissolved in the fluid phase or suspended as a solid.
Summary
• Chemists make a distinction between two different types of changes that they study—physical changes and chemical changes.
• Physical changes are changes that do not alter the identity of a substance.
• Chemical changes are changes that occur when one substance is turned into another substance.
• Chemical changes are frequently harder to reverse than physical changes. Observations that indicate a chemical change has occurred include color change, temperature change, light given off, formation of bubbles, formation of a precipitate, etc.
Contributions & Attributions
• Boundless (www.boundless.com) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.06%3A_Changes_in_Matter_-_Physical_and_Chemical_Changes.txt |
It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before. Look at Figure $1$ below. It shows that when wood burns, it combines with oxygen and changes not only to ashes, but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose you had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also suppose you had been able to measure the oxygen used by the fire and the gases produced by the fire. What would you find? The total mass of matter after the fire would be the same as the total mass of matter before the fire.
Law of Conservation of Mass
The law of conservation of mass was created in 1789 by a French chemist, Antoine Lavoisier. The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases equals the original mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant. A reactant is the chemical reaction of two or more elements to make a new substance, and a product is the substance that is formed as the result of a chemical reaction (Video $1$). Matter and its corresponding mass may not be able to be created or destroyed, but can change forms to other substances like liquids, gases, and solids.
If you witness a 300 kg tree burn to the ground, there are only ashes left after the burn, and all of them together weigh 10 kg. It may make you wonder where the other 290 kg went. The missing 290 kg was released into the atmosphere as smoke, so the only thing left that you can see is the 10 kg of ash. If you know the law of conservation of mass, then you know that the other 290 kg has to go somewhere, because it has to equal the mass of the tree before it burnt down.
Example $1$
If heating 10.0 grams of calcium carbonate (CaCO3) produces 4.4 g of carbon dioxide (CO2) and 5.6 g of calcium oxide (CaO), show that these observations are in agreement with the law of conservation of mass.
Solution
\begin{align*} \text{Mass of the reactants} &= \text{Mass of the products} \[4pt] 10.0\, \text{g of } \ce{CaCO3} &= 4.4 \,\text{g of }\ce{CO2} + 5.6\, \text{g of } \ce{ CaO} \[4pt] 10.0\,\text{g of reactant} &= 10.0\, \text{g of products} \end{align*} \nonumber
Because the mass of the reactant is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
Exercise $1$
Potassium hydroxide ($\ce{KOH}$) readily reacts with carbon dioxide ($\ce{CO2}$) to produce potassium carbonate ($\ce{K2CO3}$) and water ($\ce{H2O}$). How many grams of potassium carbonate are produced if 224.4 g of $\ce{KOH}$ reacts with 88.0 g of $\ce{CO2}$? The reaction also produces 36.0 g of water.
Answer
276.4 g of potassium carbonate
The Law is also applicable to both chemical and physical changes. For example, if you have an ice cube that melts into a liquid and you heat that liquid up, it becomes a gas. It will appear to have disappeared, but is still there.
Summary
• Burning and other changes in matter do not destroy matter.
• The mass of matter is always the same before and after the changes occur.
• The law of conservation of mass states that matter cannot be created or destroyed. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.07%3A_Conservation_of_Mass_-_There_is_No_New_Matter.txt |
Learning Objectives
• Define heat and work.
• Distinguish between kinetic energy and potential energy.
• State the law of conservation of matter and energy.
Just like matter, energy is a term that we are all familiar with and use on a daily basis. Before you go on a long hike, you eat an energy bar; every month, the energy bill is paid; on TV, politicians argue about the energy crisis. But what is energy? If you stop to think about it, energy is very complicated. When you plug a lamp into an electric socket, you see energy in the form of light, but when you plug a heating pad into that same socket, you only feel warmth. Without energy, we couldn't turn on lights, we couldn't brush our teeth, we couldn't make our lunch, and we couldn't travel to school. In fact, without energy, we couldn't even wake up because our bodies require energy to function. We use energy for every single thing that we do, whether we are awake or asleep.
Ability to Do Work or Produce Heat
When we speak of using energy, we are really referring to transferring energy from one place to another. When you use energy to throw a ball, you transfer energy from your body to the ball, and this causes the ball to fly through the air. When you use energy to warm your house, you transfer energy from the furnace to the air in your home, and this causes the temperature in your house to rise. Although energy is used in many kinds of different situations, all of these uses rely on energy being transferred in one of two ways. Energy can be transferred as heat or as work.
When scientists speak of heat, they are referring to energy that is transferred from an object with a higher temperature to an object with a lower temperature, as a result of the temperature difference. Heat will "flow" from the hot object to the cold object until both end up at the same temperature. When you cook with a metal pot, you witness energy being transferred in the form of heat. Initially, only the stove element is hot—the pot and the food inside the pot are cold. As a result, heat moves from the hot stove element to the cold pot. After a while, enough heat is transferred from the stove to the pot, raising the temperature of the pot and all of its contents (Figure $1$).
Heat is only one way in which energy can be transferred. Energy can also be transferred as work. The scientific definition of work is force (any push or pull) applied over a distance. When you push an object and cause it to move, you do work, and you transfer some of your energy to the object. At this point, it's important to warn you of a common misconception. Sometimes we think that the amount of work done can be measured by the amount of effort put in. This may be true in everyday life, but it is not true in science. By definition, scientific work requires that force be applied over a distance. It does not matter how hard you push or how hard you pull. If you have not moved the object, you haven't done any work.
So far, we've talked about the two ways in which energy can be transferred from one place, or object, to another. Energy can be transferred as heat, and energy can be transferred as work. But the question still remains—what IS energy?
Kinetic Energy
Machines use energy, our bodies use energy, energy comes from the sun, energy comes from volcanoes, energy causes forest fires, and energy helps us to grow food. With all of these seemingly different types of energy, it's hard to believe that there are really only two different forms of energy: kinetic energy and potential energy. Kinetic energy is energy associated with motion. When an object is moving, it has kinetic energy. When the object stops moving, it has no kinetic energy. While all moving objects have kinetic energy, not all moving objects have the same amount of kinetic energy. The amount of kinetic energy possessed by an object is determined by its mass and its speed. The heavier an object is and the faster it is moving, the more kinetic energy it has.
Kinetic energy is very common, and it's easy to spot examples of it in the world around you. Sometimes we even try to capture kinetic energy and use it to power things like our home appliances. If you are from California, you might have driven through the Tehachapi Pass near Mojave or the Montezuma Hills in Solano County and seen the windmills lining the slopes of the mountains (Figure $2$). These are two of the larger wind farms in North America. As wind rushes along the hills, the kinetic energy of the moving air particles turns the windmills, trapping the wind's kinetic energy so that people can use it in their houses and offices.
Potential Energy
Potential energy is stored energy. It is energy that remains available until we choose to use it. Think of a battery in a flashlight. If left on, the flashlight battery will run out of energy within a couple of hours, and the flashlight will die. If, however, you only use the flashlight when you need it, and turn it off when you don’t, the battery will last for days or even months. The battery contains a certain amount of energy, and it will power the flashlight for a certain amount of time, but because the battery stores potential energy, you can choose to use the energy all at once, or you can save it and only use a small amount at a time.
Any stored energy is potential energy. There are a lot of different ways in which energy can be stored, and this can make potential energy very difficult to recognize. In general, an object has potential energy because of its position relative to another object. For example, when a rock is held above the earth, it has potential energy because of its position relative to the ground. This is potential energy because the energy is stored for as long as the rock is held in the air. Once the rock is dropped, though, the stored energy is released as kinetic energy as the rock falls.
Chemical Energy
There are other common examples of potential energy. A ball at the top of a hill stores potential energy until it is allowed to roll to the bottom. When two magnets are held next to one another, they store potential energy too. For some examples of potential energy, though, it's harder to see how "position" is involved. In chemistry, we are often interested in what is called chemical potential energy. Chemical potential energy is energy stored in the atoms, molecules, and chemical bonds that make up matter. How does this depend on position?
As you learned earlier, the world, and all of the chemicals in it are made up of atoms and molecules. These store potential energy that is dependent on their positions relative to one another. Of course, you can't see atoms and molecules. Nevertheless, scientists do know a lot about the ways in which atoms and molecules interact, and this allows them to figure out how much potential energy is stored in a specific quantity (like a cup or a gallon) of a particular chemical. Different chemicals have different amounts of potential energy because they are made up of different atoms, and those atoms have different positions relative to one another.
Since different chemicals have different amounts of potential energy, scientists will sometimes say that potential energy depends not only on position, but also on composition. Composition affects potential energy because it determines which molecules and atoms end up next to one another. For example, the total potential energy in a cup of pure water is different than the total potential energy in a cup of apple juice, because the cup of water and the cup of apple juice are composed of different amounts of different chemicals.
At this point, you may wonder just how useful chemical potential energy is. If you want to release the potential energy stored in an object held above the ground, you just drop it. But how do you get potential energy out of chemicals? It's actually not difficult. Use the fact that different chemicals have different amounts of potential energy. If you start with chemicals that have a lot of potential energy and allow them to react and form chemicals with less potential energy, all the extra energy that was in the chemicals at the beginning, but not at the end, is released.
Units of Energy
Energy is measured in one of two common units: the calorie and the joule. The joule $\left( \text{J} \right)$ is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie $\left( \text{cal} \right)$ is the quantity of heat required to raise the temperature of 1 gram of water by $1^\text{o} \text{C}$. For example, raising the temperature of $100 \: \text{g}$ of water from $20^\text{o} \text{C}$ to $22^\text{o} \text{C}$ would require $100 \times 2 = 200 \: \text{cal}$.
Calories contained within food are actually kilocalories $\left( \text{kcal} \right)$. In other words, if a certain snack contains 85 food calories, it actually contains $85 \: \text{kcal}$ or $85,000 \: \text{cal}$. In order to make the distinction, the dietary calorie is written with a capital C.
$1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories} \nonumber$
To say that the snack "contains" 85 Calories means that $85 \: \text{kcal}$ of energy are released when that snack is processed by your body.
Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below.
$1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J} \nonumber$
We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested.
$400 \: \text{Cal} = 400 \: \text{kcal} \times \dfrac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ} \nonumber$
Summary
• Any time we use energy, we transfer energy from one object to another.
• Energy can be transferred in one of two ways: as heat, or as work.
• Heat is the term given to energy that is transferred from a hot object to a cooler object due to the difference in their temperatures.
• Work is the term given to energy that is transferred as a result of a force applied over a distance.
• Energy comes in two fundamentally different forms: kinetic energy and potential energy.
• Kinetic energy is the energy of motion.
• Potential energy is stored energy that depends on the position of an object relative to another object.
• Chemical potential energy is a special type of potential energy that depends on the positions of different atoms and molecules relative to one another.
• Chemical potential energy can also be thought of according to its dependence on chemical composition.
• Energy can be converted from one form to another. The total amount of mass and energy in the universe is conserved.
• Wikibooks | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.08%3A_Energy.txt |
Learning Objectives
• Define endothermic and exothermic reactions.
• Describe how heat is transferred in endothermic and exothermic reactions.
• Determine whether a reaction is endothermic or exothermic through observations, temperature changes, or an energy diagram.
So far, we have talked about how energy exists as either kinetic energy or potential energy and how energy can be transferred as either heat or work. While it's important to understand the difference between kinetic energy and potential energy and the difference between heat and work, the truth is, energy is constantly changing. Kinetic energy is constantly being turned into potential energy, and potential energy is constantly being turned into kinetic energy. Likewise, energy that is transferred as work might later end up transferred as heat, while energy that is transferred as heat might later end up being used to do work.
Even though energy can change form, it must still follow one fundamental law: Energy cannot be created or destroyed, it can only be changed from one form to another. This law is known as the Law of Conservation of Energy. In a lot of ways, energy is like money. You can exchange quarters for dollar bills and dollar bills for quarters, but no matter how often you convert between the two, you will not end up with any more or any less money than you started with. Similarly, you can transfer (or spend) money using cash, or transfer money using a credit card; but you still spend the same amount of money, and the store still makes the same amount of money.
A campfire is an example of basic thermochemistry. The reaction is initiated by the application of heat from a match. The reaction converting wood to carbon dioxide and water (among other things) continues, releasing heat energy in the process. This heat energy can then be used to cook food, roast marshmallows, or just keep warm when it's cold outside.
An image of a campfire with colored flames, made by the burning of a garden hose in a copper pipe. (CC SA-BY 3.0; Jared)
Exothermic and Endothermic Processes
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe: the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings are everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings are the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter \(q\). The sign of \(q\) for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of \(q\) for an exothermic process is negative because the system is losing heat.
During phase changes, energy changes are usually involved. For example, when solid dry ice vaporizes (physical change), carbon dioxide molecules absorb energy. When liquid water becomes ice, energy is released. Remember that all chemical reactions involve a change in the bonds of the reactants. The bonds in the reactants are broken and the bonds of the products are formed. Chemical bonds have potential energy or "stored energy". Because we are changing the bonding, this means we are also changing how much of this "stored energy" there is in a reaction.
Energy changes are frequently shown by drawing an energy diagram. Energy diagrams show the stored/hidden energy of the reactants and products as well as the activation energy. If, on an energy diagram, the products have more stored energy than the reactants started with, the reaction is endothermic. You had to give the reaction energy. If, on the energy diagram, the products have less stored energy than the reactants started with, the reaction is exothermic.
Example \(1\)
Label each of the following processes as endothermic or exothermic.
1. water boiling
2. gasoline burning
3. ice forming on a pond
Solution
1. Endothermic—you must put a pan of water on the stove and give it heat in order to get water to boil. Because you are adding heat/energy, the reaction is endothermic.
2. Exothermic—when you burn something, it feels hot to you because it is giving off heat into the surroundings.
3. Exothermic—think of ice forming in your freezer instead. You put water into the freezer, which takes heat out of the water, to get it to freeze. Because heat is being pulled out of the water, it is exothermic. Heat is leaving.
Exercise \(1\)
Label each of the following processes as endothermic or exothermic.
1. water vapor condensing
2. gold melting
Answer (a)
exothermic
Answer (b)
endothermic
Summary
Phase changes involve changes in energy. All chemical reactions involve changes in energy. This may be a change in heat, electricity, light, or other forms of energy. Reactions that absorb energy are endothermic. Reactions that release energy are exothermic. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.09%3A_Energy_and_Chemical_and_Physical_Change.txt |
Learning Objectives
• Identify the different between temperature and heat.
• Recognize the different scales used to measure temperature
The concept of temperature may seem familiar to you, but many people confuse temperature with heat. Temperature is a measure of how hot or cold an object is relative to another object (its thermal energy content), whereas heat is the flow of thermal energy between objects with different temperatures. Temperature is a measure of the average kinetic energy of the particles in matter. In everyday usage, temperature indicates a measure of how hot or cold an object is. Temperature is an important parameter in chemistry. When a substance changes from solid to liquid, it is because there was in increase in the temperature of the material. Chemical reactions usually proceed faster if the temperature is increased. Many unstable materials (such as enzymes) will be viable longer at lower temperatures.
Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K). Thermometers measure temperature by using materials that expand or contract when heated or cooled. Mercury or alcohol thermometers, for example, have a reservoir of liquid that expands when heated and contracts when cooled, so the liquid column lengthens or shortens as the temperature of the liquid changes.
The Fahrenheit Scale
The first thermometers were glass and contained alcohol, which expanded and contracted as the temperature changed. The German scientist, Daniel Gabriel Fahrenheit used mercury in the tube, an idea put forth by Ismael Boulliau. The Fahrenheit scale was first developed in 1724 and tinkered with for some time after that. The main problem with this scale is the arbitrary definitions of temperature. The freezing point of water was defined as $32^\text{o} \text{F}$ and the boiling point as $212^\text{o} \text{F}$. The Fahrenheit scale is typically not used for scientific purposes.
The Celsius Scale
The Celsius scale of the metric system is named after Swedish astronomer Anders Celsius (1701-1744). The Celsius scale sets the freezing point and boiling point of water at $0^\text{o} \text{C}$ and $100^\text{o} \text{C}$ respectively. The distance between those two points is divided into 100 equal intervals, each of which is one degree. Another term sometimes used for the Celsius scale is "centigrade" because there are 100 degrees between the freezing and boiling points of water on this scale. However, the preferred term is "Celsius".
The Kelvin Scale
The Kelvin temperature scale is named after Scottish physicist and mathematician Lord Kelvin (1824-1907). It is based on molecular motion, with the temperature of $0 \: \text{K}$, also known as absolute zero, being the point where all molecular motion ceases. The freezing point of water on the Kelvin scale is $273.15 \: \text{K}$, while the boiling point is $373.15 \: \text{K}$. Notice that there is no "degree" used in the temperature designation. Unlike the Fahrenheit and Celsius scales where temperatures are referred to as "degrees $\text{F}$" or "degrees $\text{C}$", we simply designate temperatures in the Kelvin scale as kelvins.
Converting Between Scales
The Kelvin is the same size as the Celsius degree, so measurements are easily converted from one to the other. The freezing point of water is 0°C = 273.15 K; the boiling point of water is 100°C = 373.15 K. The Kelvin and Celsius scales are related as follows:
$T \,\text{(in °C)} + 273.15 = T \, \text{(in K)} \tag{3.10.1} \label{3.10.1}$
$T \; \text{ (in K)} − 273.15 = T \; \text{(in °C)} \tag{3.10.2} \label{3.10.2}$
Degrees on the Fahrenheit scale, however, are based on an English tradition of using 12 divisions, just as 1 ft = 12 in. The relationship between degrees Fahrenheit and degrees Celsius is as follows: where the coefficient for degrees Fahrenheit is exact. (Some calculators have a function that allows you to convert directly between °F and °C.) There is only one temperature for which the numerical value is the same on both the Fahrenheit and Celsius scales: −40°C = −40°F. The relationship between the scales is as follows:
$°C = \dfrac{(°F-32)}{1.8} \tag{3.10.3} \label{3.10.3}$
$°F = 1.8 \times (°C)+32 \tag{3.10.4} \label{3.10.4}$
Example $1$: Temperature Conversions
A student is ill with a temperature of 103.5°F. What is her temperature in °C and K?
Solution
Converting from Fahrenheit to Celsius requires the use of Equation \ref{3.10.3}:
\begin{align} °C &= \dfrac{(103.5°F - 32)}{1.8} \ &= 39.7 \,°C \end{align} \nonumber
Converting from Celsius to Kelvin requires the use of Equation \ref{3.10.1}:
\begin{align} K &= 39.7 \,°C + 273.15 \ &= 312.9\,K \end{align} \nonumber
Exercise $1$
Convert each temperature to °C and °F.
1. the temperature of the surface of the sun (5800 K)
2. the boiling point of gold (3080 K)
3. the boiling point of liquid nitrogen (77.36 K)
Answer (a)
5527 K, 9980 °F
Answer (b)
2807 K, 5084 °F
Answer (c)
-195.79 K, -320.42 °F
Summary
Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K).
3.11: Temperature Changes - Heat Capacity
If a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. The heat capacity of an object depends on both its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the wading pool.
Heat Capacity and Specific Heat
Different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water in the same sun will not become nearly as hot. We would say that water has a high heat capacity (the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$). Water is very resistant to changes in temperature, while metals in general are not. The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. The symbol for specific heat is $c_p$, with the $p$ subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree $\left( \text{J/g}^\text{o} \text{C} \right)$ or calories per gram per degree $\left( \text{cal/g}^\text{o} \text{C} \right)$ (Table $1$). This text will use $\text{J/g}^\text{o} \text{C}$ for specific heat.
$\text{specific heat}= \dfrac{\text{heat}}{\text{mass} \times \text{cal/g}^\text{o} \text{C}} \nonumber$
Notice that water has a very high specific heat compared to most other substances.
Table $1$: Specific Heat Capacities
Substance Specific Heat Capacity
at 25oC in J/g oC
Substance Specific Heat Capacity
at 25oC in J/g oC
$\ce{H2}$ gas 14.267 steam @ 100oC 2.010
$\ce{He}$ gas 5.300 vegetable oil 2.000
$\ce{H2O(l)}$ 4.184 sodium 1.23
lithium
3.56
air 1.020
ethyl alcohol
2.460
magnesium 1.020
ethylene glycol
2.200
aluminum 0.900
ice @ 0oC
2.010
concrete 0.880
steam @ 100oC
2.010
glass 0.840
Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days.
Summary
• Heat capacity is the amount of heat required to raise the temperature of an object by $1^\text{o} \text{C}$).
• The specific heat of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by $1^\text{o} \text{C}$. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.10%3A_Temperature_-_Random_Motion_of_Molecules_and_Atoms.txt |
Learning Objectives
• To relate heat transfer to temperature change.
Heat is a familiar manifestation of transferring energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer.
The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat $\left( q \right)$ to specific heat $\left( c_p \right)$, mass $\left( m \right)$, and temperature change $\left( \Delta T \right)$ is shown below.
$q = c_p \times m \times \Delta T \nonumber$
The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by $\Delta T = T_f - T_i$, where $T_f$ is the final temperature and $T_i$ is the initial temperature.
Every substance has a characteristic specific heat, which is reported in units of cal/g•°C or cal/g•K, depending on the units used to express ΔT. The specific heat of a substance is the amount of energy that must be transferred to or from 1 g of that substance to change its temperature by 1°. Table $1$ lists the specific heats for various materials.
Table $1$: Specific Heats of Some Common Substances
Substance Specific Heat $\left( \text{J/g}^\text{o} \text{C} \right)$
Water (l) 4.18
Water (s) 2.06
Water (g) 1.87
Ammonia (g) 2.09
Ethanol (l) 2.44
Aluminum (s) 0.897
Carbon, graphite (s) 0.709
Copper (s) 0.385
Gold (s) 0.129
Iron (s) 0.449
Lead (s) 0.129
Mercury (l) 0.140
Silver (s) 0.233
The direction of heat flow is not shown in heat = mcΔT. If energy goes into an object, the total energy of the object increases, and the values of heat ΔT are positive. If energy is coming out of an object, the total energy of the object decreases, and the values of heat and ΔT are negative.
Example $1$
A $15.0 \: \text{g}$ piece of cadmium metal absorbs $134 \: \text{J}$ of heat while rising from $24.0^\text{o} \text{C}$ to $62.7^\text{o} \text{C}$. Calculate the specific heat of cadmium.
Known
• Heat $= q = 134 \: \text{J}$
• Mass $= m = 15.0 \: \text{g}$
• $\Delta T = 62.7^\text{o} \text{C} - 24.0^\text{o} \text{C} = 38.7^\text{o} \text{C}$
Unknown
• $c_p$ of cadmium $= ? \: \text{J/g}^\text{o} \text{C}$
The specific heat equation can be rearranged to solve for the specific heat.
Step 2: Solve.
$c_p = \dfrac{q}{m \times \Delta T} = \dfrac{134 \: \text{J}}{15.0 \: \text{g} \times 38.7^\text{o} \text{C}} = 0.231 \: \text{J/g}^\text{o} \text{C} \nonumber$
Step 3: Think about your result.
The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.
Since most specific heats are known (Table $1$), they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a $60.0 \: \text{g}$ of water at $23.52^\text{o} \text{C}$ was cooled by the removal of $813 \: \text{J}$ of heat. The change in temperature can be calculated using the specific heat equation:
$\Delta T = \dfrac{q}{c_p \times m} = \dfrac{813 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 60.0 \: \text{g}} = 3.24^\text{o} \text{C} \nonumber$
Since the water was being cooled, the temperature decreases. The final temperature is:
$T_f = 23.52^\text{o} \text{C} - 3.24^\text{o} \text{C} = 20.28^\text{o} \text{C} \nonumber$
Example $2$
What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?
Solution
We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, ΔT is as follows:
ΔT = TfinalTinitial = 73.3°C − 25.0°C = 48.3°C
The mass is given as 150.0 g, and Table 7.3 gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into heat = mcΔT and solve for amount of heat:
$\mathrm{heat=(150.0\: g)\left(0.108\: \dfrac{cal} {g\cdot {^\circ C}}\right)(48.3^\circ C) = 782\: cal} \nonumber$
Note how the gram and °C units cancel algebraically, leaving only the calorie unit, which is a unit of heat. Because the temperature of the iron increases, energy (as heat) must be flowing into the metal.
Exercise $1$
What quantity of heat is transferred when a 295.5 g block of aluminum metal is cooled from 128.0°C to 22.5°C? What is the direction of heat flow?
Answer
Heat leaves the aluminum block.
Example $2$
A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal? Can you identify the metal from the data in Table $1$?
Solution
The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of ΔT is as follows:
ΔT = TfinalTinitial = 22.0°C − 97.5°C = −75.5°C
If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mcΔT and solve for c:
−71.7 cal = (10.3 g)(c)(−75.5°C)
$c \,\mathrm{=\dfrac{-71.7\: cal}{(10.3\: g)(-75.5^\circ C)}}$
c = 0.0923 cal/g•°C
This value for specific heat is very close to that given for copper in Table 7.3.
Exercise $2$
A 10.7 g crystal of sodium chloride (NaCl) has an initial temperature of 37.0°C. What is the final temperature of the crystal if 147 cal of heat were supplied to it?
Answer
Summary
Specific heat calculations are illustrated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.12%3A_Energy_and_Heat_Capacity_Calculations.txt |
3.1: In Your Room
3.2: What Is Matter?
1. What is matter?
2. What does weight mean?
3. In this chapter, we'll learn about atoms, which are the building blocks of all matter in the universe. As of 2011, scientists only know of 118 different types of atoms. How do you think it's possible to generate so many different forms of matter using only 118 types of building blocks?
4. Which do you think has more matter, a cup of water or a cup of mercury? Explain.
5. Decide whether each of the following statements is true or false.
1. Mass and weight are two words for the same concept.
2. Molecules are bonded together to form atoms.
3. Alchemists couldn't make gold out of common metals because gold is an element.
4. The symbol for Gold in the periodic table is Gd.
6. Would you have more mass on the moon or on Earth?
7. Would you have more weight on the moon or on Earth? The force of gravity is stronger on the Earth than it is on the moon.
8. Match the following terms with their meaning.
Terms Definitions
(a) Mass a. a measure of the total quantity of matter in an object
(b) Volume b. a measure of how strongly gravity pulls on an object
(c) Weight c. a measure of the space occupied by an object
9. For the following statements, circle all of the options that apply:
• Mass depends on…
(a) the total quantity of matter
(b) the temperature
(c) the location
(d) the force of gravity
• Volume depends on…
(a) the total quantity of matter
(b) the temperature
(c) the object's shape (independent of size)
(d) the object's size (independent of shape)
• Weight depends on…
(a) the total quantity of matter
(b) the temperature
(c) the location
(d) the force of gravity
3.3: Classifying Matter According to Its State: Solid, Liquid, and Gas
3.4 Classifying Matter According to Its Composition
3.5: Differences in Matter: Physical and Chemical Properties
3.6: Changes in Matter: Physical and Chemical Changes
3.7: Conservation of Mass: There is No New Matter
3.8: Energy
1. Classify each of the following as energy primarily transferred as heat, or energy primarily transferred as work:
1. The energy transferred from your body to a shopping cart as you push the shopping cart down the aisle.
2. The energy transferred from a wave to your board when you go surfing.
3. The energy transferred from the flames to your hotdog when you cook your hotdog over a campfire.
2. Decide whether each of the following statements is true or false:
1. When heat is transferred to an object, the object cools down.
2. Any time you raise the temperature of an object, you have done work.
3. Any time you move an object by applying force, you have done work.
4. Any time you apply force to an object, you have done work.
3. Rank the following scenarios in order of increasing work:
1. You apply 100 N of force to a boulder and successfully move it by 2 m.
2. You apply 100 N of force to a boulder and successfully move it by 1 m.
3. You apply 200 N of force to a boulder and successfully move it by 2 m.
4. You apply 200 N of force to a boulder but cannot move the boulder.
4. In science, a vacuum is defined as space that contains absolutely no matter (no molecules, no atoms, etc.) Can energy be transferred as heat through a vacuum? Why or why not?
5. Classify each of the following energies as kinetic energy or potential energy:
1. The energy in a chocolate bar.
2. The energy of rushing water used to turn a turbine or a water wheel.
3. The energy of a skater gliding on the ice.
4. The energy in a stretched rubber band.
6. Decide which of the following objects has more kinetic energy:
1. A 200 lb. man running at 6 mph or a 200 lb. man running at 3 mph.
2. A 200 lb. man running at 7 mph or a 150 lb. man running at 7 mph.
3. A 400 lb. man running at 5 mph or a 150 lb. man running at 3 mph.
7. A car and a truck are traveling along the highway at the same speed.
1. If the car weighs 1500 kg and the truck weighs 2500 kg, which has more kinetic energy, the car or the truck?
2. Both the car and the truck convert the potential energy stored in gasoline into the kinetic energy of motion. Which do you think uses more gas to travel the same distance, the car or the truck?
8. You mix two chemicals in a beaker and notice that as the chemicals react, the beaker becomes noticeably colder. Which chemicals have more chemical potential energy, those present at the start of the reaction or those present at the end of the reaction?
3.9: Energy and Chemical and Physical Change
3.10: Temperature: Random Motion of Molecules and Atoms
3.11: Temperature Changes: Heat Capacity
3.12: Energy and Heat Capacity Calculations
1. A pot of water is set on a hot burner of a stove. What is the direction of heat flow?
2. Some uncooked macaroni is added to a pot of boiling water. What is the direction of heat flow?
3. How much energy in calories is required to heat 150 g of H2O from 0°C to 100°C?
4. How much energy in calories is required to heat 125 g of Fe from 25°C to 150°C?
5. If 250 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum?
6. If 195 cal of heat were added to 33.2 g of Hg at 56.2°C, what is the final temperature of the mercury?
7. A sample of copper absorbs 145 cal of energy, and its temperature rises from 37.8°C to 41.7°C. What is the mass of the copper?
8. A large, single crystal of sodium chloride absorbs 98.0 cal of heat. If its temperature rises from 22.0°C to 29.7°C, what is the mass of the NaCl crystal?
9. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the largest temperature change?
10. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the smallest temperature change?
11. Determine the heat capacity of a substance if 23.6 g of the substance gives off 199 cal of heat when its temperature changes from 37.9°C to 20.9°C.
12. What is the heat capacity of gold if a 250 g sample needs 133 cal of energy to increase its temperature from 23.0°C to 40.1°C?
Answers
1. Heat flows into the pot of water.
1. 15,000 cal
1. 49.0°C
1. 404 g
1. Mercury would experience the largest temperature change.
1. 0.496 cal/g•°C | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/03%3A_Matter_and_Energy/3.E%3A_Matter_and_Energy_%28Exercises%29.txt |
• 4.1: Cutting Aluminum until you get Atoms
Take some aluminum foil. Cut it in half. Now you have two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil.
• 4.2: Indivisible - The Atomic Theory
You learned earlier how all matter in the universe is made out of tiny building blocks called atoms. All modern scientists accept the concept of the atom, but when the concept of the atom was first proposed about 2,500 years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of the existence of things that are too small to see. We will spend some time considering the evidence (observations) that convince scientists of the existence of atoms.
• 4.3: The Nuclear Atom
While Dalton's Atomic Theory held up well, J. J. Thomson demonstrate that his theory was not the entire story. He suggested that the small, negatively charged particles making up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we know them as electrons. Thanks to his clever experiments and careful reasoning, J. J. Thomson is credited with the discovery of the electron.
• 4.4: The Properties of Protons, Neutrons, and Electrons
Electrons are extremely small. The mass of an electron is only about 1/2000 the mass of a proton or neutron, so electrons contribute virtually nothing to the total mass of an atom. Electrons have an electric charge of −1, which is equal but opposite to the charge of a proton, which is +1. All atoms have the same number of electrons as protons, so the positive and negative charges "cancel out", making atoms electrically neutral.
• 4.5: Elements- Defined by Their Number of Protons
Scientists distinguish between different elements by counting the number of protons in the nucleus. Since an atom of one element can be distinguished from an atom of another element by the number of protons in its nucleus, scientists are always interested in this number, and how this number differs between different elements. The number of protons in an atom is called its atomic number (Z). This number is very important because it is unique for atoms of a given element.
• 4.6: Looking for Patterns - The Periodic Table
Certain elemental properties become apparent in a survey of the periodic table as a whole. Every element can be classified as either a metal, a nonmetal, or a metalloid (or semi metal). A metal is a substance that is shiny, typically (but not always) silvery in color, and an excellent conductor of electricity and heat. Metals are also malleable (they can be beaten into thin sheets) and ductile (they can be drawn into thin wires).
• 4.7: Ions - Losing and Gaining Electrons
Atom may lose valence electrons to obtain a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result. Some atoms have nearly eight electrons in their valence shell, and can gain additional valence electrons until they have an octet. When these atoms gain electrons, they acquire a negative charge.
• 4.8: Isotopes - When the Number of Neutrons Varies
All atoms of the same element have the same number of protons, but some may have different numbers of neutrons. For example, all carbon atoms have six protons, and most have six neutrons as well. But some carbon atoms have seven or eight neutrons instead of the usual six. Atoms of the same element that differ in their numbers of neutrons are called isotopes. Many isotopes occur naturally.
• 4.9: Atomic Mass - The Average Mass of an Element’s Atoms
In chemistry, we very rarely deal with only one isotope of an element. We use a mixture of the isotopes of an element in chemical reactions and other aspects of chemistry, because all of the isotopes of an element react in the same manner. That means that we rarely need to worry about the mass of a specific isotope, but instead need to know the average mass of the atoms of an element.
04: Atoms and Elements
Take some aluminum foil. Cut it in half. Now there are two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can this exercise be taken, at least in theory? Can one continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century BC by Democritus and other ancient Greek philosophers (Figure \(1\)).
Most elements in their pure form exist as individual atoms. For example, a macroscopic chunk of iron metal is composed, microscopically, of individual atoms. Some elements, however, exist as groups of atoms called molecules. Several important elements exist as two-atom combinations and are called diatomic molecules. In representing a diatomic molecule, we use the symbol of the element and include the subscript 2 to indicate that two atoms of that element are joined together. The elements that exist as diatomic molecules are hydrogen (H2), oxygen (O2), nitrogen (N2), fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.01%3A_Cutting_Aluminum_until_you_get_Atoms.txt |
Learning Objectives
• Give a short history of the concept of the atom.
• Describe the contributions of Democritus and Dalton to atomic theory.
• Summarize Dalton's atomic theory and explain its historical development.
You learned earlier that all matter in the universe is made out of tiny building blocks called atoms. All modern scientists accept the concept of the atom, but when the concept of the atom was first proposed about 2,500 years ago, ancient philosophers laughed at the idea. It has always been difficult to convince people of the existence of things that are too small to see. We will spend some time considering the evidence (observations) that convince scientists of the existence of atoms.
Democritus and the Greek Philosophers
About 2,500 years ago, early Greek philosophers believed the entire universe was a single, huge, entity. In other words, "everything was one." They believed that all objects, all matter, and all substances were connected as a single, big, unchangeable "thing." One of the first people to propose "atoms" was a man known as Democritus. As an alternative to the beliefs of the Greek philosophers, he suggested that atomos, or atomon—tiny, indivisible, solid objects—make up all matter in the universe.
Democritus then reasoned that changes occur when the many atomos in an object were reconnected or recombined in different ways. Democritus even extended this theory, suggesting that there were different varieties of atomos with different shapes, sizes, and masses. He thought, however, that shape, size, and mass were the only properties differentiating the different types of atomos. According to Democritus, other characteristics, like color and taste, did not reflect properties of the atomos themselves, but rather, resulted from the different ways in which the atomos were combined and connected to one another.
The early Greek philosophers tried to understand the nature of the world through reason and logic, but not through experiment and observation. As a result, they had some very interesting ideas, but they felt no need to justify their ideas based on life experiences. In a lot of ways, you can think of the Greek philosophers as being "all thought and no action." It's truly amazing how much they achieved using their minds, but because they never performed any experiments, they missed or rejected a lot of discoveries that they could have made otherwise. Greek philosophers dismissed Democritus' theory entirely. Sadly, it took over two millennia before the theory of atomos (or "atoms," as they are known today) was fully appreciated.
Greeks: "All Thought and No Action"
Greek philosophers were "all thought and no action" and did not feel the need to test their theories with reality. In contrast, Dalton's efforts were based on experimentation and testing ideas against reality.
While it must be assumed that many more scientists, philosophers, and others studied composition of matter after Democritus, a major leap forward in our understanding of the composition of matter took place in the 1800's with the work of the British scientists John Dalton. He started teaching school at age twelve, and was primarily known as a teacher. In his twenties, he moved to the growing city of Manchester, where he was able to pursue some scientific studies. His work in several areas of science brought him a number of honors. When he died, over 40,000 people in Manchester marched at his funeral.
The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure \(1\)), is a fundamental concept that states that all elements are composed of atoms. Previously, we defined an atom as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of your little finger (about 1 cm).
Dalton studied the weights of various elements and compounds. He noticed that matter always combined in fixed ratios based on weight, or volume in the case of gases. Chemical compounds always contain the same proportion of elements by mass, regardless of amount, which provided further support for Proust's law of definite proportions. Dalton also observed that there could be more than one combination of two elements.
From his experiments and observations, as well as the work from peers of his time, Dalton proposed a new theory of the atom. This later became known as Dalton's atomic theory. The general tenets of this theory were as follows:
• All matter is composed of extremely small particles called atoms.
• Atoms of a given element are identical in size, mass, and other properties. Atoms of different elements differ in size, mass, and other properties.
• Atoms cannot be subdivided, created, or destroyed.
• Atoms of different elements can combine in simple whole number ratios to form chemical compounds.
• In chemical reactions, atoms are combined, separated, or rearranged.
Dalton's atomic theory has been largely accepted by the scientific community, with the exception of three changes. We know now that (1) an atom can be further subdivided, (2) all atoms of an element are not identical in mass, and (3) using nuclear fission and fusion techniques, we can create or destroy atoms by changing them into other atoms.
The evidence for atoms is so great that few doubt their existence. In fact, individual atoms are now routinely observed with state-of-the art technologies. Moreover, they can even be used for making pretty images; or as IBM research demonstrates in Video \(1\), control of individual atoms can be use used create animations.
A Boy and His Atom is a 2012 stop-motion animated short film released by IBM Research. The movie tells the story of a boy and a wayward atom who meet and become friends. It depicts a boy playing with an atom that takes various forms. It was made by moving carbon monoxide molecules viewed with a scanning tunneling microscope, a device that magnifies them 100 million times. These molecules were moved to create images, which were then saved as individual frames to make the film.
Summary
• 2,500 years ago, Democritus suggested that all matter in the universe was made up of tiny, indivisible, solid objects he called "atomos." However, other Greek philosophers disliked Democritus' "atomos" theory because they felt it was illogical.
• Dalton's Atomic Theory is the first scientific theory to relate chemical changes to the structure, properties, and behavior of the atom. The general tenets of this theory are:
• All matter is composed of extremely small particles called atoms.
• Atoms of a given element are identical in size, mass, and other properties. Atoms of different elements differ in size, mass, and other properties.
• Atoms cannot be subdivided, created, or destroyed.
• Atoms of different elements can combine in simple whole number ratios to form chemical compounds.
• In chemical reactions, atoms are combined, separated, or rearranged. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.02%3A_Indivisible_-_The_Atomic_Theory.txt |
Learning Objectives
• Explain the observations that led to Thomson's discovery of the electron.
• Describe Thomson's "plum pudding" model of the atom and the evidence for it.
• Draw a diagram of Thomson's "plum pudding" model of the atom and explain why it has this name.
• Describe Rutherford's gold foil experiment and explain how this experiment altered the "plum pudding" model.
• Draw a diagram of the Rutherford model of the atom and label the nucleus and the electron cloud.
Dalton's Atomic Theory held up well to a lot of the different chemical experiments that scientists performed to test it. In fact, for almost 100 years, it seemed as if Dalton's Atomic Theory was the whole truth. However, in 1897, a scientist named J. J. Thomson conducted some research that suggested that Dalton's Atomic Theory was not the entire story. He suggested that the small, negatively charged particles making up the cathode ray were actually pieces of atoms. He called these pieces "corpuscles," although today we know them as electrons. Thanks to his clever experiments and careful reasoning, J. J. Thomson is credited with the discovery of the electron.
Electrons and Plums
The electron was discovered by J. J. Thomson in 1897. The existence of protons was also known, as was the fact that atoms were neutral in charge. Since the intact atom had no net charge and the electron and proton had opposite charges, the next step after the discovery of subatomic particles was to figure out how these particles were arranged in the atom. This was a difficult task because of the incredibly small size of the atom. Therefore, scientists set out to design a model of what they believed the atom could look like. The goal of each atomic model was to accurately represent all of the experimental evidence about atoms in the simplest way possible.
Following the discovery of the electron, J.J. Thomson developed what became known as the "plum pudding" model in 1904. Plum pudding is an English dessert similar to a blueberry muffin. In Thomson's plum pudding model of the atom, the electrons were embedded in a uniform sphere of positive charge like blueberries stuck into a muffin. The positive matter was thought to be jelly-like or similar to a thick soup. The electrons were somewhat mobile. As they got closer to the outer portion of the atom, the positive charge in the region was greater than the neighboring negative charges, and the electron would be pulled back more toward the center region of the atom.
However, this model of the atom soon gave way to a new model developed by New Zealander Ernest Rutherford (1871-1937) about five years later. Thomson did still receive many honors during his lifetime, including being awarded the Nobel Prize in Physics in 1906 and a knighthood in 1908.
Atoms and Gold
In 1911, Rutherford and coworkers Hans Geiger and Ernest Marsden initiated a series of groundbreaking experiments that would completely change the accepted model of the atom. They bombarded very thin sheets of gold foil with fast moving alpha particles. Alpha particles, a type of natural radioactive particle, are positively charged particles with a mass about four times that of a hydrogen atom.
According to the accepted atomic model, in which an atom's mass and charge are uniformly distributed throughout the atom, the scientists expected that all of the alpha particles would pass through the gold foil with only a slight deflection or none at all. Surprisingly, while most of the alpha particles were indeed not deflected, a very small percentage (about 1 in 8000 particles) bounced off the gold foil at very large angles. Some were even redirected back toward the source. No prior knowledge had prepared them for this discovery. In a famous quote, Rutherford exclaimed that it was "as if you had fired a 15-inch [artillery] shell at a piece of tissue and it came back and hit you."
Rutherford needed to come up with an entirely new model of the atom in order to explain his results. Because the vast majority of the alpha particles had passed through the gold, he reasoned that most of the atom was empty space. In contrast, the particles that were highly deflected must have experienced a tremendously powerful force within the atom. He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus. The nucleus is the tiny, dense, central core of the atom and is composed of protons and neutrons.
Rutherford's atomic model became known as the nuclear model. In the nuclear atom, the protons and neutrons, which comprise nearly all of the mass of the atom, are located in the nucleus at the center of the atom. The electrons are distributed around the nucleus and occupy most of the volume of the atom. It is worth emphasizing just how small the nucleus is compared to the rest of the atom. If we could blow up an atom to be the size of a large professional football stadium, the nucleus would be about the size of a marble.
Rutherford's model proved to be an important step towards a full understanding of the atom. However, it did not completely address the nature of the electrons and the way in which they occupy the vast space around the nucleus. It was not until some years later that a full understanding of the electron was achieved. This proved to be the key to understanding the chemical properties of elements.
Atomic Nucleus
The nucleus (plural, nuclei) is a positively charged region at the center of the atom. It consists of two types of subatomic particles packed tightly together. The particles are protons, which have a positive electric charge, and neutrons, which are neutral in electric charge. Outside of the nucleus, an atom is mostly empty space, with orbiting negative particles called electrons whizzing through it. The figure below shows these parts of the atom.
The nucleus of the atom is extremely small. Its radius is only about 1/100,000 of the total radius of the atom. Electrons have virtually no mass, but protons and neutrons have a lot of mass for their size. As a result, the nucleus has virtually all the mass of an atom. Given its great mass and tiny size, the nucleus is very dense. If an object the size of a penny had the same density as the nucleus of an atom, its mass would be greater than 30 million tons!
Holding it all Together
Particles with opposite electric charges attract each other. This explains why negative electrons orbit the positive nucleus. Particles with the same electric charge repel each other. This means that the positive protons in the nucleus push apart from one another. So why doesn't the nucleus fly apart? An even stronger force—called the strong nuclear force—holds protons and neutrons together in the nucleus.
Summary
• Atoms are the ultimate building blocks of all matter.
• The modern atomic theory establishes the concepts of atoms and how they compose matter.
• Bombardment of gold foil with alpha particles showed that some particles were deflected.
• The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.03%3A_The_Nuclear_Atom.txt |
Learning Objectives
• Describe the locations, charges, and masses of the three main subatomic particles.
• Determine the number of protons and electrons in an atom.
• Define atomic mass unit (amu).
Dalton's Atomic Theory explained a lot about matter, chemicals, and chemical reactions. Nevertheless, it was not entirely accurate, because contrary to what Dalton believed, atoms can, in fact, be broken apart into smaller subunits or subatomic particles. We have been talking about the electron in great detail, but there are two other particles of interest to us: protons and neutrons. We already learned that J. J. Thomson discovered a negatively charged particle, called the electron. Rutherford proposed that these electrons orbit a positive nucleus. In subsequent experiments, he found that there is a smaller positively charged particle in the nucleus, called a proton. There is also a third subatomic particle, known as a neutron.
Electrons
Electrons are one of three main types of particles that make up atoms. Unlike protons and neutrons, which consist of smaller, simpler particles, electrons are fundamental particles that do not consist of smaller particles. They are a type of fundamental particle called leptons. All leptons have an electric charge of $-1$ or $0$. Electrons are extremely small. The mass of an electron is only about 1/2000 the mass of a proton or neutron, so electrons contribute virtually nothing to the total mass of an atom. Electrons have an electric charge of $-1$, which is equal but opposite to the charge of a proton, which is $+1$. All atoms have the same number of electrons as protons, so the positive and negative charges "cancel out", making atoms electrically neutral.
Unlike protons and neutrons, which are located inside the nucleus at the center of the atom, electrons are found outside the nucleus. Because opposite electric charges attract one another, negative electrons are attracted to the positive nucleus. This force of attraction keeps electrons constantly moving through the otherwise empty space around the nucleus. The figure below is a common way to represent the structure of an atom. It shows the electron as a particle orbiting the nucleus, similar to the way that planets orbit the sun. However, this is an incorrect perspective, as quantum mechanics demonstrates that electrons are more complicated.
Protons
A proton is one of three main particles that make up the atom. Protons are found in the nucleus of the atom. This is a tiny, dense region at the center of the atom. Protons have a positive electrical charge of one $\left( +1 \right)$ and a mass of 1 atomic mass unit $\left( \text{amu} \right)$, which is about $1.67 \times 10^{-27}$ kilograms. Together with neutrons, they make up virtually all of the mass of an atom.
Neutrons
Atoms of all elements—except for most atoms of hydrogen—have neutrons in their nucleus. Unlike protons and electrons, which are electrically charged, neutrons have no charge—they are electrically neutral. That's why the neutrons in the diagram above are labeled $n^0$. The zero stands for "zero charge". The mass of a neutron is slightly greater than the mass of a proton, which is 1 atomic mass unit $\left( \text{amu} \right)$. (An atomic mass unit equals about $1.67 \times 10^{-27}$ kilograms.) A neutron also has about the same diameter as a proton, or $1.7 \times 10^{-15}$ meters.
As you might have already guessed from its name, the neutron is neutral. In other words, it has no charge whatsoever and is therefore neither attracted to nor repelled from other objects. Neutrons are in every atom (with one exception), and they are bound together with other neutrons and protons in the atomic nucleus.
Before we move on, we must discuss how the different types of subatomic particles interact with each other. When it comes to neutrons, the answer is obvious. Since neutrons are neither attracted to nor repelled from objects, they don't really interact with protons or electrons (beyond being bound into the nucleus with the protons).
Even though electrons, protons, and neutrons are all types of subatomic particles, they are not all the same size. When you compare the masses of electrons, protons, and neutrons, what you find is that electrons have an extremely small mass, compared to either protons or neutrons. On the other hand, the masses of protons and neutrons are fairly similar, although technically, the mass of a neutron is slightly larger than the mass of a proton. Because protons and neutrons are so much more massive than electrons, almost all of the mass of any atom comes from the nucleus, which contains all of the neutrons and protons.
Table $1$: Properties of Subatomic Particles
Particle Symbol Mass (amu) Relative Mass (proton = 1) Relative Charge Location
proton p+ 1 1 +1 inside the nucleus
electron e 5.45 × 10−4 0.00055 −1 outside the nucleus
neutron n0 1 1 0 inside the nucleus
Table $1$ gives the properties and locations of electrons, protons, and neutrons. The third column shows the masses of the three subatomic particles in "atomic mass units." An atomic mass unit ($\text{amu}$) is defined as one-twelfth of the mass of a carbon-12 atom. Atomic mass units ($\text{amu}$) are useful, because, as you can see, the mass of a proton and the mass of a neutron are almost exactly $1$ in this unit system.
Negative and positive charges of equal magnitude cancel each other out. This means that the negative charge on an electron perfectly balances the positive charge on the proton. In other words, a neutral atom must have exactly one electron for every proton. If a neutral atom has 1 proton, it must have 1 electron. If a neutral atom has 2 protons, it must have 2 electrons. If a neutral atom has 10 protons, it must have 10 electrons. You get the idea. In order to be neutral, an atom must have the same number of electrons and protons.
Summary
• Electrons are a type of subatomic particle with a negative charge.
• Protons are a type of subatomic particle with a positive charge. Protons are bound together in an atom's nucleus as a result of the strong nuclear force.
• Neutrons are a type of subatomic particle with no charge (they are neutral). Like protons, neutrons are bound into the atom's nucleus as a result of the strong nuclear force.
• Protons and neutrons have approximately the same mass, but they are both much more massive than electrons (approximately 2,000 times as massive as an electron).
• The positive charge on a proton is equal in magnitude to the negative charge on an electron. As a result, a neutral atom must have an equal number of protons and electrons.
• The atomic mass unit (amu) is a unit of mass equal to one-twelfth the mass of a carbon-12 atom | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.04%3A_The_Properties_of_Protons_Neutrons_and_Electrons.txt |
Learning Objectives
• Define atomic number.
• Define mass number.
• Determine the number of protons, neutrons, and electrons in an atom.
It's important to be able to distinguish atoms of one element from atoms of another element. Elements are pure substances that make up all other matter, so each one is given a unique name. The names of elements are also represented by unique one- or two-letter symbols, such as $\ce{H}$ for hydrogen, $\ce{C}$ for carbon, or $\ce{He}$ for helium. However, it would more powerful if these names could be used to identify the numbers of protons and neutrons in the atoms. That's where atomic number and mass number are useful.
Atomic Number
Scientists distinguish between different elements by counting the number of protons in the nucleus (Table $1$). If an atom has only one proton, we know that it's a hydrogen atom. An atom with two protons is always a helium atom. If scientists count four protons in an atom, they know it's a beryllium atom. An atom with three protons is a lithium atom, an atom with five protons is a boron atom, an atom with six protons is a carbon atom . . . the list goes on.
Since an atom of one element can be distinguished from an atom of another element by the number of protons in its nucleus, scientists are always interested in this number, and how this number differs between different elements. The number of protons in an atom is called its atomic number ($Z$). This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms. For example, all helium atoms have two protons, and no other elements have atoms with two protons.
Name Protons Neutrons Electrons Atomic Number (Z) Mass Number (A)
Table $1$: Atoms of the First Six Elements
Hydrogen 1 0 1 1 1
Helium 2 2 2 2 4
Lithium 3 4 3 3 7
Beryllium 4 5 4 4 9
Boron 5 6 5 5 11
Carbon 6 6 6 6 12
Of course, since neutral atoms have to have one electron for every proton, an element's atomic number also tells you how many electrons are in a neutral atom of that element. For example, hydrogen has an atomic number of 1. This means that an atom of hydrogen has one proton, and, if it's neutral, one electron as well. Gold, on the other hand, has an atomic number of 79, which means that an atom of gold has 79 protons, and, if it's neutral, 79 electrons as well.
Neutral Atoms
Atoms are neutral in electrical charge because they have the same number of negative electrons as positive protons (Table $1$). Therefore, the atomic number of an atom also tells you how many electrons the atom has. This, in turn, determines many of the atom's chemical properties.
Mass Number
The mass number ($A$) of an atom is the total number of protons and neutrons in its nucleus. The mass of the atom is a unit called the atomic mass unit $\left( \text{amu} \right)$. One atomic mass unit is the mass of a proton, or about $1.67 \times 10^{-27}$ kilograms, which is an extremely small mass. A neutron has just a tiny bit more mass than a proton, but its mass is often assumed to be one atomic mass unit as well. Because electrons have virtually no mass, just about all the mass of an atom is in its protons and neutrons. Therefore, the total number of protons and neutrons in an atom determines its mass in atomic mass units (Table $1$).
Consider helium again. Most helium atoms have two neutrons in addition to two protons. Therefore the mass of most helium atoms is 4 atomic mass units ($2 \: \text{amu}$ for the protons + $2 \: \text{amu}$ for the neutrons). However, some helium atoms have more or less than two neutrons. Atoms with the same number of protons but different numbers of neutrons are called isotopes. Because the number of neutrons can vary for a given element, the mass numbers of different atoms of an element may also vary. For example, some helium atoms have three neutrons instead of two (these are called isotopes and are discussed in detail later on).
Why do you think that the "mass number" includes protons and neutrons, but not electrons? You know that most of the mass of an atom is concentrated in its nucleus. The mass of an atom depends on the number of protons and neutrons. You have already learned that the mass of an electron is very, very small compared to the mass of either a proton or a neutron (like the mass of a penny compared to the mass of a bowling ball). Counting the number of protons and neutrons tells scientists about the total mass of an atom.
$\text{mass number} \: A = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right) \nonumber$
An atom's mass number is very easy to calculate, provided that you know the number of protons and neutrons in an atom.
Example 4.5.1
What is the mass number of an atom of helium that contains 2 neutrons?
Solution
$\left( \text{number of protons} \right) = 2$ (Remember that an atom of helium always has 2 protons.)
$\left( \text{number of neutrons} \right) = 2$
$\text{mass number} = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right)$
$\text{mass number} = 2 + 2 = 4$
A chemical symbol is a one- or two-letter designation of an element. Some examples of chemical symbols are $\ce{O}$ for oxygen, $\ce{Zn}$ for zinc, and $\ce{Fe}$ for iron. The first letter of a symbol is always capitalized. If the symbol contains two letters, the second letter is lower case. The majority of elements have symbols that are based on their English names. However, some of the elements that have been known since ancient times have maintained symbols that are based on their Latin names, as shown in Table $2$.
Chemical Symbol Name Latin Name
Table $2$: Symbols and Latin Names for Elements
$\ce{Na}$ Sodium Natrium
$\ce{K}$ Potassium Kalium
$\ce{Fe}$ Iron Ferrum
$\ce{Cu}$ Copper Cuprum
$\ce{Ag}$ Silver Argentum
$\ce{Sn}$ Tin Stannum
$\ce{Sb}$ Antimony Stibium
$\ce{Au}$ Gold Aurum
$\ce{Pb}$ Lead Plumbum
Summary
• Elements are pure substances that make up all matter, so each one is given a unique name.
• The names of elements are also represented by unique one- or two-letter symbols.
• Each element has a unique number of protons. An element's atomic number is equal to the number of protons in the nuclei of any of its atoms.
• The mass number of an atom is the sum of the protons and neutrons in the atom.
• Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons in their atomic nuclei. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.05%3A_Elements-_Defined_by_Their_Number_of_Protons.txt |
Learning Objectives
• Explain how elements are organized into the periodic table.
• Describe how some characteristics of elements relate to their positions on the periodic table.
In the 19th century, many previously unknown elements were discovered, and scientists noted that certain sets of elements had similar chemical properties. For example, chlorine, bromine, and iodine react with other elements (such as sodium) to make similar compounds. Likewise, lithium, sodium, and potassium react with other elements (such as oxygen) to make similar compounds. Why is this so?
In 1864, Julius Lothar Meyer, a German chemist, organized the elements by atomic mass and grouped them according to their chemical properties. Later that decade, Dmitri Mendeleev, a Russian chemist, organized all the known elements according to similar properties. He left gaps in his table for what he thought were undiscovered elements, and he made some bold predictions regarding the properties of those undiscovered elements. When elements were later discovered whose properties closely matched Mendeleev’s predictions, his version of the table gained favor in the scientific community. Because certain properties of the elements repeat on a regular basis throughout the table (that is, they are periodic), it became known as the periodic table.
Mendeleev had to list some elements out of the order of their atomic masses to group them with other elements that had similar properties.
The periodic table is one of the cornerstones of chemistry because it organizes all of the known elements on the basis of their chemical properties. A modern version is shown in Figure \(1\). Most periodic tables provide additional data (such as atomic mass) in a box that contains each element’s symbol. The elements are listed in order of atomic number.
Features of the Periodic Table
Elements that have similar chemical properties are grouped in columns called groups (or families). As well as being numbered, some of these groups have names—for example, alkali metals (the first column of elements), alkaline earth metals (the second column of elements), halogens (the next-to-last column of elements), and noble gases (the last column of elements).
The word halogen comes from the Greek for “salt maker” because these elements combine with other elements to form a group of compounds called salts.
To Your Health: Radon
Radon is an invisible, odorless noble gas that is slowly released from the ground, particularly from rocks and soils whose uranium content is high. Because it is a noble gas, radon is not chemically reactive. Unfortunately, it is radioactive, and increased exposure to it has been correlated with an increased lung cancer risk.
Because radon comes from the ground, we cannot avoid it entirely. Moreover, because it is denser than air, radon tends to accumulate in basements, which if improperly ventilated can be hazardous to a building’s inhabitants. Fortunately, specialized ventilation minimizes the amount of radon that might collect. Special fan-and-vent systems are available that draw air from below the basement floor, before it can enter the living space, and vent it above the roof of a house.
After smoking, radon is thought to be the second-biggest preventable cause of lung cancer in the United States. The American Cancer Society estimates that 10% of all lung cancers are related to radon exposure. There is uncertainty regarding what levels of exposure cause cancer, as well as what the exact causal agent might be (either radon or one of its breakdown products, many of which are also radioactive and, unlike radon, not gases). The US Environmental Protection Agency recommends testing every floor below the third floor for radon levels to guard against long-term health effects.
Each row of elements on the periodic table is called a period. Periods have different lengths; the first period has only 2 elements (hydrogen and helium), while the second and third periods have 8 elements each. The fourth and fifth periods have 18 elements each, and later periods are so long that a segment from each is removed and placed beneath the main body of the table.
Certain elemental properties become apparent in a survey of the periodic table as a whole. Every element can be classified as either a metal, a nonmetal, or a metalloid (or semi metal), as shown in Figure \(2\). A metal is a substance that is shiny, typically (but not always) silvery in color, and an excellent conductor of electricity and heat. Metals are also malleable (they can be beaten into thin sheets) and ductile (they can be drawn into thin wires). A nonmetal is typically dull and a poor conductor of electricity and heat. Solid nonmetals are also very brittle. As shown in Figure \(2\), metals occupy the left three-fourths of the periodic table, while nonmetals (except for hydrogen) are clustered in the upper right-hand corner of the periodic table. The elements with properties intermediate between those of metals and nonmetals are called metalloids (or semi-metals). Elements adjacent to the bold line in the right-hand portion of the periodic table have semimetal properties.
Example \(1\)
Based on its position in the periodic table, classify each element below as metal, a nonmetal, or a metalloid.
1. Se
2. Mg
3. Ge
Solution
1. In Figure \(1\), selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal.
2. Magnesium lies to the left of the diagonal line marking the boundary between metals and nonmetals, so it should be a metal.
3. Germanium lies within the diagonal line marking the boundary between metals and nonmetals, so it should be a metalloid.
Exercise \(1\)
Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a metalloid?
Answer
Indium is a metal.
Another way to categorize the elements of the periodic table is shown in Figure \(3\). The first two columns on the left and the last six columns on the right are called the main group elements. The ten-column block between these columns contains the transition metals. The two rows beneath the main body of the periodic table contain the inner transition metals. The elements in these two rows are also referred to as, respectively, the lanthanide metals and the actinide metals.
Descriptive Names
As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins.
Group 1: The Alkali Metals
The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal.
The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively.
Group 2: The Alkaline Earth Metals
The alkaline earth metals are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals.
Group 17: The Halogens
The halogens are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all of the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt).
Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature.
Group 18: The Noble Gases
The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights.
Example \(2\): Groups
Provide the family or group name of each element.
1. Li
2. Ar
3. Cl
Solution
1. Lithium is an alkali metal (Group 1)
2. Argon is a noble gas (Group 18)
3. Chlorine is a halogen (Group 17)
Exercise \(2\): Groups
Provide the family or group name of each element.
1. F
2. Ca
3. Kr
Answer a:
Fluorine is a halogen (Group 17).
Answer b:
Calcium is a alkaline earth metal (Group 2).
Answer c:
Krypton is a noble gas (Group 18).
Example \(3\): Classification of Elements
Classify each element as metal, non metal, transition metal or inner transition metal.
1. Li
2. Ar
3. Am
4. Fe
Solution
1. Lithium is a metal.
2. Argon is a non metal.
3. Americium is an inner transition metal.
4. Iron is a transition metal.
Exercise \(3\): Classification of Elements
Classify each element as metal, non metal, transition metal or inner transition metal.
1. F
2. U
3. Cu
Answer a:
Fluorine is a nonmetal.
Answer b:
Uranium is a metal (and a inner transition metal too).
Answer c:
Copper is a metal (and a transition metal too).
Summary
The periodic table is an arrangement of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right.
The elements can be broadly divided into metals, nonmetals, and semi metals. Semi metals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semi metals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable). Solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, and the inner transition metals (the lanthanides, and the actinides). | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.06%3A_Looking_for_Patterns_-_The_Periodic_Table.txt |
Learning Objectives
• Define the two types of ions.
Most atoms do not have eight electrons in their valence electron shell. Some atoms have only a few electrons in their outer shell, while some atoms lack only one or two electrons to have an octet. In cases where an atom has three or fewer valence electrons, the atom may lose those valence electrons quite easily until what remains is a lower shell that contains an octet. Atoms that lose electrons acquire a positive charge as a result because they are left with fewer negatively charged electrons to balance the positive charges of the protons in the nucleus. Positively charged ions are called cations. Most metals become cations when they make ionic compounds.
Cations
A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron.
$\ce{Na \rightarrow Na^{+} + e^{-}} \nonumber$
The cation produced in this way, Na+, is called the sodium ion to distinguish it from the element. The outermost shell of the sodium ion is the second electron shell, which has eight electrons in it. The octet rule has been satisfied. Figure $1$ is a graphical depiction of this process.
Anions
Some atoms have nearly eight electrons in their valence shell and can gain additional valence electrons until they have an octet. When these atoms gain electrons, they acquire a negative charge because they now possess more electrons than protons. Negatively charged ions are called anions. Most nonmetals become anions when they make ionic compounds.
A neutral chlorine atom has seven electrons in its outermost shell. Only one more electron is needed to achieve an octet in chlorine’s valence shell. (In table salt, this electron comes from the sodium atom.)
$\ce{e^{-} +Cl -> Cl^{-}} \nonumber$
In this case, the ion has the same outermost shell as the original atom, but now that shell has eight electrons in it. Once again, the octet rule has been satisfied. The resulting anion, Cl, is called the chloride ion; note the slight change in the suffix (-ide instead of -ine) to create the name of this anion. Figure $2$ is a graphical depiction of this process.
The names for positive and negative ions are pronounced CAT-eye-ons and ANN-eye-ons, respectively.
In many cases, elements that belong to the same group (vertical column) on the periodic table form ions with the same charge because they have the same number of valence electrons. Thus, the periodic table becomes a tool for remembering the charges on many ions. For example, all ions made from alkali metals, the first column on the periodic table, have a 1+ charge. Ions made from alkaline earth metals, the second group on the periodic table, have a 2+ charge. On the other side of the periodic table, the next-to-last column, the halogens, form ions having a 1− charge. Figure $3$ shows how the charge on many ions can be predicted by the location of an element on the periodic table. Note the convention of first writing the number and then the sign on a ion with multiple charges. The barium cation is written Ba2+, not Ba+2. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.07%3A_Ions_-_Losing_and_Gaining_Electrons.txt |
Learning Objectives
• Explain what isotopes are and how an isotope affects an element's atomic mass.
• Determine the number of protons, electrons, and neutrons of an element with a given mass number.
All atoms of the same element have the same number of protons, but some may have different numbers of neutrons. For example, all carbon atoms have six protons, and most have six neutrons as well. But some carbon atoms have seven or eight neutrons instead of the usual six. Atoms of the same element that differ in their numbers of neutrons are called isotopes. Many isotopes occur naturally. Usually one or two isotopes of an element are the most stable and common. Different isotopes of an element generally have the same physical and chemical properties because they have the same numbers of protons and electrons.
An Example: Hydrogen Isotopes
Hydrogen is an example of an element that has isotopes. Three isotopes of hydrogen are modeled in Figure $1$. Most hydrogen atoms have just one proton, one electron, and lack a neutron. These atoms are just called hydrogen. Some hydrogen atoms have one neutron as well. These atoms are the isotope named deuterium. Other hydrogen atoms have two neutrons. These atoms are the isotope named tritium.
For most elements other than hydrogen, isotopes are named for their mass number. For example, carbon atoms with the usual 6 neutrons have a mass number of 12 (6 protons + 6 neutrons = 12), so they are called carbon-12. Carbon atoms with 7 neutrons have an atomic mass of 13 (6 protons + 7 neutrons = 13). These atoms are the isotope called carbon-13.
Example $1$: Lithium Isotopes
1. What is the atomic number and the mass number of an isotope of lithium containing 3 neutrons?
2. What is the atomic number and the mass number of an isotope of lithium containing 4 neutrons?
Solution
A lithium atom contains 3 protons in its nucleus irrespective of the number of neutrons or electrons.
a.
\begin{align}\text{atomic number} = \left( \text{number of protons} \right) &= 3 \nonumber \ \left( \text{number of neutrons} \right) &= 3 \nonumber\end{align} \nonumber
\begin{align} \text{mass number} & = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right) \nonumber\ \text{mass number} & = 3 + 3 \nonumber\ &= 6 \nonumber \end{align}\nonumber
b.
\begin{align}\text{atomic number} = \left( \text{number of protons} \right) &= 3 \nonumber\ \left( \text{number of neutrons} \right) & = 4\nonumber\end{align}\nonumber
\begin{align}\text{mass number} & = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right)\nonumber \ \text{mass number} & = 3 + 4\nonumber \ &= 7 \nonumber \end{align}\nonumber
Notice that because the lithium atom always has 3 protons, the atomic number for lithium is always 3. The mass number, however, is 6 in the isotope with 3 neutrons, and 7 in the isotope with 4 neutrons. In nature, only certain isotopes exist. For instance, lithium exists as an isotope with 3 neutrons, and as an isotope with 4 neutrons, but it doesn't exist as an isotope with 2 neutrons or as an isotope with 5 neutrons.
Stability of Isotopes
Atoms need a certain ratio of neutrons to protons to have a stable nucleus. Having too many or too few neutrons relative to protons results in an unstable, or radioactive, nucleus that will sooner or later break down to a more stable form. This process is called radioactive decay. Many isotopes have radioactive nuclei, and these isotopes are referred to as radioisotopes. When they decay, they release particles that may be harmful. This is why radioactive isotopes are dangerous and why working with them requires special suits for protection. The isotope of carbon known as carbon-14 is an example of a radioisotope. In contrast, the carbon isotopes called carbon-12 and carbon-13 are stable.
This whole discussion of isotopes brings us back to Dalton's Atomic Theory. According to Dalton, atoms of a given element are identical. But if atoms of a given element can have different numbers of neutrons, then they can have different masses as well! How did Dalton miss this? It turns out that elements found in nature exist as constant uniform mixtures of their naturally occurring isotopes. In other words, a piece of lithium always contains both types of naturally occurring lithium (the type with 3 neutrons and the type with 4 neutrons). Moreover, it always contains the two in the same relative amounts (or "relative abundance"). In a chunk of lithium, $93\%$ will always be lithium with 4 neutrons, while the remaining $7\%$ will always be lithium with 3 neutrons.
Dalton always experimented with large chunks of an element—chunks that contained all of the naturally occurring isotopes of that element. As a result, when he performed his measurements, he was actually observing the averaged properties of all the different isotopes in the sample. For most of our purposes in chemistry, we will do the same thing and deal with the average mass of the atoms. Luckily, aside from having different masses, most other properties of different isotopes are similar.
There are two main ways in which scientists frequently show the mass number of an atom they are interested in. It is important to note that the mass number is not given on the periodic table. These two ways include writing a nuclear symbol or by giving the name of the element with the mass number written.
To write a nuclear symbol, the mass number is placed at the upper left (superscript) of the chemical symbol and the atomic number is placed at the lower left (subscript) of the symbol. The complete nuclear symbol for helium-4 is drawn below:
The following nuclear symbols are for a nickel nucleus with 31 neutrons and a uranium nucleus with 146 neutrons.
$\ce{^{59}_{28}Ni} \nonumber$
$\ce{ ^{238}_{92}U} \nonumber$
In the nickel nucleus represented above, the atomic number 28 indicates that the nucleus contains 28 protons, and therefore, it must contain 31 neutrons in order to have a mass number of 59. The uranium nucleus has 92 protons, as all uranium nuclei do; and this particular uranium nucleus has 146 neutrons.
Another way of representing isotopes is by adding a hyphen and the mass number to the chemical name or symbol. Thus the two nuclei would be Nickel-59 or Ni-59 and Uranium-238 or U-238, where 59 and 238 are the mass numbers of the two atoms, respectively. Note that the mass numbers (not the number of neutrons) are given to the side of the name.
Example $2$: Potassium-40
How many protons, electrons, and neutrons are in an atom of $^{40}_{19}\ce{K}$?
Solution
$\text{atomic number} = \left( \text{number of protons} \right) = 19 \nonumber$
For all atoms with no charge, the number of electrons is equal to the number of protons.
$\text{number of electrons} = 19 \nonumber$
The mass number, 40, is the sum of the protons and the neutrons.
To find the number of neutrons, subtract the number of protons from the mass number.
$\text{number of neutrons} = 40 - 19 = 21. \nonumber$
Example $3$: Zinc-65
How many protons, electrons, and neutrons are in an atom of zinc-65?
Solution
$\text{number of protons} = 30 \nonumber$
For all atoms with no charge, the number of electrons is equal to the number of protons.
$\text{number of electrons} = 30 \nonumber$
The mass number, 65, is the sum of the protons and the neutrons.
To find the number of neutrons, subtract the number of protons from the mass number.
$\text{number of neutrons} = 65 - 30 = 35 \nonumber$
Exercise $3$
How many protons, electrons, and neutrons are in each atom?
1. $^{60}_{27}\ce{Co}$
2. Na-24
3. $^{45}_{20}\ce{Ca}$
4. Sr-90
Answer a:
27 protons, 27 electrons, 33 neutrons
Answer b:
11 protons, 11 electrons, 13 neutrons
Answer c:
20 protons, 20 electrons, 25 neutrons
Answer d:
38 protons, 38 electrons, 52 neutrons
Summary
• The number of protons is always the same in atoms of the same element.
• The number of neutrons can be different, even in atoms of the same element.
• Atoms of the same element that contain the same number of protons, but different numbers of neutrons, are known as isotopes.
• Isotopes of any given element all contain the same number of protons, so they have the same atomic number (for example, the atomic number of helium is always 2).
• Isotopes of a given element contain different numbers of neutrons, therefore, different isotopes have different mass numbers. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.08%3A_Isotopes_-_When_the_Number_of_Neutrons_Varies.txt |
Learning Objectives
• Explain what is meant by the atomic mass of an element.
• Calculate the atomic mass of an element from the masses and relative percentages of the isotopes of the element.
In chemistry we very rarely deal with only one isotope of an element. We use a mixture of the isotopes of an element in chemical reactions and other aspects of chemistry, because all of the isotopes of an element react in the same manner. That means that we rarely need to worry about the mass of a specific isotope, but instead we need to know the average mass of the atoms of an element. Using the masses of the different isotopes and how abundant each isotope is, we can find the average mass of the atoms of an element. The atomic mass of an element is the weighted average mass of the atoms in a naturally occurring sample of the element. Atomic mass is typically reported in atomic mass units.
Calculating Atomic Mass
You can calculate the atomic mass (or average mass) of an element provided you know the relative abundance (the fraction of an element that is a given isotope), the element's naturally occurring isotopes, and the masses of those different isotopes. We can calculate this by the following equation:
$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber$
Look carefully to see how this equation is used in the following examples.
Example $1$: Boron Isotopes
Boron has two naturally occurring isotopes. In a sample of boron, $20\%$ of the atoms are $\ce{B}$-10, which is an isotope of boron with 5 neutrons and mass of $10 \: \text{amu}$. The other $80\%$ of the atoms are $\ce{B}$-11, which is an isotope of boron with 6 neutrons and a mass of $11 \: \text{amu}$. What is the atomic mass of boron?
Solution
Boron has two isotopes. We will use the equation:
$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber$
• Isotope 1: $\%_1 = 0.20$ (Write all percentages as decimals), $\text{mass}_1 = 10$
• Isotope 2: $\%_2 = 0.80$, $\text{mass}_2 = 11$
Substitute these into the equation, and we get:
$\text{Atomic mass} = \left( 0.20 \right) \left( 10 \right) + \left( 0.80 \right) \left( 11 \right)\nonumber$
$\text{Atomic mass} = 10.8 \: \text{amu}\nonumber$
The mass of an average boron atom, and thus boron's atomic mass, is $10.8 \: \text{amu}$.
Example $2$: Neon Isotopes
Neon has three naturally occurring isotopes. In a sample of neon, $90.92\%$ of the atoms are $\ce{Ne}$-20, which is an isotope of neon with 10 neutrons and a mass of $19.99 \: \text{amu}$. Another $0.3\%$ of the atoms are $\ce{Ne}$-21, which is an isotope of neon with 11 neutrons and a mass of $20.99 \: \text{amu}$. The final $8.85\%$ of the atoms are $\ce{Ne}$-22, which is an isotope of neon with 12 neutrons and a mass of $21.99 \: \text{amu}$. What is the atomic mass of neon?
Solution
Neon has three isotopes. We will use the equation:
$\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots\nonumber$
• Isotope 1: $\%_1 = 0.9092$ (write all percentages as decimals), $\text{mass}_1 = 19.99$
• Isotope 2: $\%_2 = 0.003$, $\text{mass}_2 = 20.99$
• Isotope 3: $\%_3 = 0.0885$, $\text{mass}_3 = 21.99$
Substitute these into the equation, and we get:
$\text{Atomic mass} = \left( 0.9092 \right) \left( 19.99 \right) + \left( 0.003 \right) \left( 20.99 \right) + \left( 0.0885 \right) \left( 21.99 \right)\nonumber$
$\text{Atomic mass} = 20.17 \: \text{amu}\nonumber$
The mass of an average neon atom is $20.17 \: \text{amu}$
The periodic table gives the atomic mass of each element. The atomic mass is a number that usually appears below the element's symbol in each square. Notice that the atomic mass of boron (symbol $\ce{B}$) is 10.8, which is what we calculated in Example $1$, and the atomic mass of neon (symbol $\ce{Ne}$) is 20.8, which is what we calculated in Example $2$. Take time to notice that not all periodic tables have the atomic number above the element's symbol and the mass number below it. If you are ever confused, remember that the atomic number should always be the smaller of the two and will be a whole number, while the atomic mass should always be the larger of the two and will be a decimal number.
Exercise $1$
Chlorine has two naturally occurring isotopes. In a sample of chlorine, $75.77\%$ of the atoms are $\ce{Cl}$-35, with a mass of $34.97 \: \text{amu}$. Another $24.23\%$ of the atoms are $\ce{Cl}$-37, with a mass of $36.97 \: \text{amu}$. What is the atomic mass of chlorine?
Answer
35.45 amu
Summary
• An element's atomic mass is the weighted average of the masses of the isotopes of an element
• An element's atomic mass can be calculated provided the relative abundance of the element's naturally occurring isotopes and the masses of those isotopes are known.
• The periodic table is a convenient way to summarize information about the different elements. In addition to the element's symbol, most periodic tables will also contain the element's atomic number and the element's atomic mass. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/04%3A_Atoms_and_Elements/4.09%3A_Atomic_Mass_-_The_Average_Mass_of_an_Elements_Atoms.txt |
There are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multi-atom combinations are called molecules. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance. In some respects, a molecule is similar to an atom. A molecule, however, is composed of more than one atom.
• 5.1: Sugar and Salt
Both salt and sugar have radically different properties (both physical and chemical) than the constituent elements that make up these compounds. This difference in properties, of constituent elements and compounds, is a central feature of chemical reactions.
• 5.2: Compounds Display Constant Composition
A compound is a substance that contains two or more elements chemically combined in a fixed proportion. The elements carbon and hydrogen combine to form many different compounds. One of the simplest is called methane, in which there are always four times as many hydrogen particles as carbon particles. Methane is a pure substance because it always has the same composition. However, it is not an element because it can be broken down into simpler substances—carbon and hydrogen.
• 5.3: Chemical Formulas - How to Represent Compounds
A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements. A molecular formula is a chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of the compound. An empirical formula is a formula that shows the elements in a compound in their lowest whole-number ratio.
• 5.4: A Molecular View of Elements and Compounds
Most elements exist with individual atoms as their basic unit. It is assumed that there is only one atom in a formula if there is no numerical subscript on the right side of an element’s symbol. There are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multi-atom combinations are called molecules. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance.
• 5.5: Writing Formulas for Ionic Compounds
Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio.
• 5.6: Nomenclature- Naming Compounds
The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning which chemical compound the name refers to—each chemical name should refer to a single substance. A less important aim is to ensure that each substance has a single name, although a limited number of alternative names is acceptable in some cases. Preferably, the name also conveys some information about the structure or chemistry of a compound.
• 5.7: Naming Ionic Compounds
Ionic compounds are named by stating the cation first, followed by the anion. Positive and negative charges must balance. Some anions have multiple forms and are named accordingly with the use of roman numerals in parentheses. Ternary compounds are composed of three or more elements.
• 5.8: Naming Molecular Compounds
Molecular compounds are inorganic compounds that take the form of discrete molecules. Examples include such familiar substances as water and carbon dioxide. These compounds are very different from ionic compounds like sodium chloride. Ionic compounds are formed when metal atoms lose one or more of their electrons to nonmetal atoms. The resulting cations and anions are electrostatically attracted to each other.
• 5.9: Naming Acids
An acid can be defined in several ways. The most straightforward definition is: an acid is a molecular compound that contains one or more hydrogen atoms and produces hydrogen ions when dissolved in water.
• 5.10: Nomenclature Summary
Brief overview of chemical nomenclature.
• 5.11: Formula Mass - The Mass of a Molecule or Formula Unit
Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas.
05: Molecules and Compounds
Sodium chloride, also known as table salt, is an ionic compound with the chemical formula $\ce{NaCl}$, representing a 1:1 ratio of sodium and chloride ions. It is commonly used as a condiment and food preservative. Salt can be created by adding two very reactive elements together: sodium ($\ce{Na (s)}$ metal and chlorine ($\ce{Cl2 (g)}$ gas.
$\ce{2Na (s) + Cl2(g) \rightarrow 2NaCl (s)} \label{eq1}$
The element sodium (Figure $\PageIndex{1a}$) is a very reactive metal; given the opportunity, it will react with the sweat on your hands and form sodium hydroxide, which is a very corrosive substance. The element chlorine (Figure $\PageIndex{1b}$) is a pale yellow, corrosive gas that should not be inhaled due to its poisonous nature. Bring these two hazardous substances together, however, and they react to make the ionic compound sodium chloride (Figure $\PageIndex{1c}$), known simply as salt.
Salt is necessary for life. $\ce{Na^{+}}$ ions are one of the main ions in the human body and are necessary to regulate the fluid balance in the body. $\ce{Cl^{−}}$ ions are necessary for proper nerve function and respiration. Both of these ions are supplied by salt. The taste of salt is one of the fundamental tastes; salt is probably the most ancient flavoring known, and one of the few rocks we eat. Clearly when the elemental sodium and chlorine combine (Equation \ref{eq1}), the resulting salt product has radically different properties (both physical and chemical). This reaction is spectacular to observe (Video $1$).
Video $1$: Making Table Salt using Sodium Metal and Chlorine gas
Another compound is sugar, which is the generic name for sweet, soluble carbohydrates, many of which are used in food. Sugar has the chemical formulate $\ce{C12H22O11}$ and is constructed from different elements than salt: carbon, hydrogen and oxygen. While sugar qualitatively resembles table salt (often confused in the kitchen), the two have distinctly different physical and chemical properties. There are various types of sugar derived from different sources. While sugar is made with carbon, hydrogen, and oxygen, it is considerably harder to synthesize from its constituent elements than table salt is (Equation \ref{eq1}). However, the thermal decomposition is considerably easier and can be represented as a dehydration of sucrose to pure carbon and water vapor in Equation \ref{eq2}, and demonstrated in Video $2$.
$\ce{C12H22O11 (s) + heat → 12C (s) + 11H2O (g)} \label{eq2}$
Video $2$: A science experiment in the kitchen shows what happens to sugar molecules when they are heated. The experiment did not disappoint!
As with salt, sugar has radically different properties (both physical and chemical) than its constituent elements. This difference in properties, of constituent elements and compounds, is a central feature of chemical reactions.
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.01%3A_Sugar_and_Salt.txt |
When building a house, the starting point is a blueprint of what the house will look like. The plan states how many windows and what kind, how many doors and what style, how many rooms and what type (bedroom, kitchen, other). The blueprint shows how the different pieces will go together to make the house. As long as the blueprint is followed and exactly the same items are used, the house will be identical to its blueprint.
Compounds
A compound is a substance that contains two or more elements chemically combined in a fixed proportion. The elements carbon and hydrogen combine to form many different compounds. One of the simplest is called methane, in which there are always four times as many hydrogen particles as carbon particles. Methane is a pure substance because it always has the same composition. However, it is not an element because it can be broken down into simpler substances—carbon and hydrogen.
Recall that the components of a mixture can be separated from one another by physical means. This is not true for a compound. Table salt is a compound consisting of equal parts of the elements sodium and chlorine. Salt cannot be separated into its two elements by filtering, distillation, or any other physical process. Salt and other compounds can only be decomposed into their elements by a chemical process. A chemical change is a change that produces matter with a different composition. Many compounds can be decomposed into their elements by heating. When sugar is heated, it decomposes into carbon and water. Water is still a compound, but one which cannot be broken down into hydrogen and oxygen by heating. Instead, the passage of an electrical current through water will produce hydrogen and oxygen gases.
The properties of compounds are generally very different than the properties of the elements from which the compound is formed. Sodium is an extremely reactive soft metal that cannot be exposed to air or water. Chlorine is a deadly gas. The compound sodium chloride is a white solid which is essential for all living things (see below).
Summary
• A compound is a substance that contains two or more elements chemically combined in a fixed proportion.
• A chemical change is a change that produces matter with a different composition.
5.03: Chemical Formulas - How to Represent Compounds
Learning Objectives
• Determine the number of different atoms in a formula.
• Define chemical formula, molecular formula, and empirical formula.
A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements. Water is composed of hydrogen and oxygen in a 2:1 ratio. The chemical formula for water is \(\ce{H_2O}\). Sulfuric acid is one of the most widely produced chemicals in the United States and is composed of the elements hydrogen, sulfur, and oxygen. The chemical formula for sulfuric acid is \(\ce{H_2SO_4}\).
Certain groups of atoms are bonded together to form what is called a polyatomic ion that acts as a single unit. Polyatomic ions are discussed in more detail in Section 5.5. Polyatomic ions are enclosed in parenthesis followed by a subscript if more than one of the same ion exist in a chemical formula. The formula \(\ce{Ca3(PO4)2}\) represents a compound with the following:
3 Ca atoms + 2 PO43- ions
To count the total number of atoms for formulas with polyatomic ions enclosed in parenthesis, use the subscript as a multiplier for each atom or number of atoms.
Ca3(PO4)2
3 Ca + 2 x1 P + 2 x 4 O = 3 Ca atoms + 2 P atoms + 8 O atoms
Molecular Formula
A molecular formula is a chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of the compound. Ammonia is a compound of nitrogen and hydrogen as shown below:
Note from the example that there are some standard rules to follow in writing molecular formulas. The arrangements of the elements depend on the particular structure, which is not of concern at this point. The number of atoms of each kind is indicated by a subscript following the atom. If there is only one atom, no number is written. If there is more than one atom of a specific kind, the number is written as a subscript following the atom. We would not write \(\ce{N_3H}\) for ammonia, because that would mean that there are three nitrogen atoms and one hydrogen atom in the molecule, which is incorrect.
Empirical Formula
An empirical formula is a formula that shows the elements in a compound in their lowest whole-number ratio. Glucose is an important simple sugar that cells use as their primary source of energy. Its molecular formula is \(\ce{C_6H_{12}O_6}\). Since each of the subscripts is divisible by 6, the empirical formula for glucose is \(\ce{CH_2O}\). When chemists analyze an unknown compound, often the first step is to determine its empirical formula.
• molecular formula: \(\ce{C_6H_{12}O_6}\)
• empirical formula: \(\ce{CH_2O}\)
There are a great many compounds whose molecular and empirical formulas are the same. If the molecular formula cannot be simplified into a smaller whole-number ratio, as in the case of \(\ce{H_2O}\) or \(\ce{P_2O_5}\), then the empirical formula is also the molecular formula.
Summary
• A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements.
• If only one atom of a specific type is present, no subscript is used.
• For atoms that have two or more of a specific type of atom present, a subscript is written after the symbol for that atom.
• Polyatomic ions in chemical formulas are enclosed in parentheses followed by a subscript if more than one of the same type of polyatomic ion exist.
• Molecular formulas do not indicate how the atoms are arranged in the molecule.
• The empirical formula tells the lowest whole-number ratio of elements in a compound. The empirical formula does not show the actual number of atoms. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.02%3A_Compounds_Display_Constant_Composition.txt |
Learning Objectives
• Classify substances as atomic elements, molecular elements, molecular compounds, or ionic compounds.
Atomic Elements
Most elements exist with individual atoms as their basic unit. It is assumed that there is only one atom in a formula if there is no numerical subscript on the right side of an element’s symbol.
Molecular Elements
There are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multi-atom combinations are called molecules. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance. In some respects, a molecule is similar to an atom. A molecule, however, is composed of more than one atom.
Table \(1\): Elements That Exist as Diatomic Molecules
Hydrogen, H Oxygen Nitrogen Fluorine Chlorine Bromine Iodine
Some elements exist naturally as molecules. For example, hydrogen and oxygen exist as two-atom molecules. Other elements also exist naturally as diatomic molecules—a molecule with only two atoms (Table \(1\)). As with any molecule, these elements are labeled with a molecular formula, a formal listing of what and how many atoms are in a molecule. (Sometimes only the word formula is used, and its meaning is inferred from the context.) For example, the molecular formula for elemental hydrogen is H2, with H being the symbol for hydrogen and the subscript 2 implying that there are two atoms of this element in the molecule. Other diatomic elements have similar formulas: O2, N2, and so forth. Other elements exist as molecules—for example, sulfur normally exists as an eight-atom molecule, S8, while phosphorus exists as a four-atom molecule, P4 (Figure \(1\)).
Figure \(1\) shows two examples of how molecules will be represented in this text. An atom is represented by a small ball or sphere, which generally indicates where the nucleus is in the molecule. A cylindrical line connecting the balls represents the connection between the atoms that make this collection of atoms a molecule. This connection is called a chemical bond.
Ionic Compounds
The elements in the periodic table are divided into specific groupings; the metals, the non-metals, the semi-metals, and so on. These groupings are largely based on physical properties and on the tendency of the various elements to bond with other elements by forming either an ionic or a covalent bond. As a general rule of thumb, compounds that involve a metal binding with either a non-metal or a semi-metal will display ionic bonding. Thus, the compound formed from sodium and chlorine will be ionic (a metal and a non-metal). The basic unit of ionic compounds is the formula unit.
Molecular Compounds
Compounds that are composed of only non-metals or semi-metals with non-metals will display covalent bonding and will be classified as molecular compounds. Nitrogen monoxide (NO) will be a covalently bound molecule (two non-metals) and silicon dioxide (SiO2) will also be a covalently bound molecule (a semi-metal and a non-metal). The basic unit of molecular compounds is the molecule.
Example \(1\)
Provide the classification (i.e. atomic element, molecular element, molecular compound, or ionic compound) of each substance.
1. Fe
2. PCl3
3. LiBr
4. P4
5. oxygen gas
Solution
1. Fe (iron) is an element that is represented with no subscript, so it is an atomic element.
2. PCl3 is made up of two nonmetals, so it is a molecular compound.
3. LiBr is made up of lithium, a metal, and bromine, a nonmetal, so it is an ionic compound.
4. P4 is a substance that is made up of four atoms of the same element, so it is a molecular element.
5. The formula for oxygen gas is O2 so it is a molecular element.
Exercise \(1\)
Provide the classification (i.e. atomic element, molecular element, molecular compound, or ionic compound) of each substance.
1. I2
2. He
3. H2O
4. Al
5. CuCl
Answer a:
molecular element
Answer b:
atomic element
Answer c:
molecular compound
Answer d:
atomic element
Answer e:
ionic compound | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.04%3A_A_Molecular_View_of_Elements_and_Compounds.txt |
Learning Objectives
• Write the correct formula for an ionic compound.
• Recognize polyatomic ions in chemical formulas.
Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula.
If you know the name of a binary ionic compound, you can write its chemical formula. Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charges to cancel each other out.
Example \(1\): Aluminum Nitride and Lithium Oxide
Write the formulas for aluminum nitride and lithium oxide.
Solution
Solution to Example 5.5.1
Write the formula for aluminum nitride Write the formula for lithium oxide
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\)
2. Use a multiplier to make the total charge of the cations and anions equal to each other.
total charge of cations = total charge of anions
1(3+) = 1(3-)
+3 = -3
total charge of cations = total charge of anions
2(1+) = 1(2-)
+2 = -2
3. Use the multipliers as subscript for each ion. \(\ce{Al_1N_1}\) \(\ce{Li_2O_1}\)
4. Write the final formula. Leave out all charges and all subscripts that are 1. \(\ce{AlN}\) \(\ce{Li_2O}\)
An alternative way to writing a correct formula for an ionic compound is to use the crisscross method. In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped.
Example \(2\): The Crisscross Method for Lead (IV) Oxide
Write the formula for lead (IV) oxide.
Solution
Solution to Example 5.5.2
Crisscross Method Write the formula for lead (IV) oxide
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. \(\ce{Pb_2O_4}\)
4. Write the final formula. Leave out all subscripts that are 1. \(\ce{PbO_2}\)
Exercise \(2\)
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the calcium ion and the oxygen ion
2. the 2+ copper ion and the sulfur ion
3. the 1+ copper ion and the sulfur ion
Answer a:
CaO
Answer b:
CuS
Answer c:
Cu2S
Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions.
Example \(3\): Sulfur Compound
Write the formula for sodium combined with sulfur.
Solution
Solution to Example 5.5.3
Crisscross Method Write the formula for sodium combined with sulfur
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. This step is not necessary.
4. Write the final formula. Leave out all subscripts that are 1. \(\ce{Na_2S}\)
Exercise \(3\)
Write the formula for each ionic compound.
1. sodium bromide
2. lithium chloride
3. magnesium oxide
Answer a:
NaBr
Answer b:
LiCl
Answer c:
MgO
Polyatomic Ions
Some ions consist of groups of atoms bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. Polyatomic ions have characteristic formulas, names, and charges that should be memorized. For example, NO3 is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall 1− charge. Table \(1\) lists the most common polyatomic ions.
Table \(1\): Some Polyatomic Ions
Name Formula
ammonium ion NH4+
acetate ion C2H3O2 (also written CH3CO2)
carbonate ion CO32
chromate ion CrO42
dichromate ion Cr2O72
hydrogen carbonate ion (bicarbonate ion) HCO3
cyanide ion CN
hydroxide ion OH
nitrate ion NO3
nitrite ion NO2
permanganate ion MnO4
phosphate ion PO43
hydrogen phosphate ion HPO42
dihydrogen phosphate ion H2PO4
sulfate ion SO42
hydrogen sulfate ion (bisulfate ion) HSO4
sulfite ion SO32
The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. An example is Ba(NO3)2.
Writing Formulas for Ionic Compounds Containing Polyatomic Ions
Writing a formula for ionic compounds containing polyatomic ions also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion.
Example \(4\): Calcium Nitrate
Write the formula for calcium nitrate.
Solution
Solution to Example 5.5.4
Crisscross Method Write the formula for calcium nitrate
1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Ca^{2+}} \: \: \: \: \: \ce{NO_3^-}\)
2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation.
3. Reduce to the lowest ratio. \(\ce{Ca_1(NO_3)_2}\)
4. Write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses. \(\ce{Ca(NO_3)_2}\)
Example \(5\)
Write the chemical formula for an ionic compound composed of the potassium ion and the sulfate ion.
Solution
Solution to Example 5.5.5
Explanation Answer
Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is \(\ce{K2SO4}\). \(\ce{K_2SO_4}\)
Exercise \(5\)
Write the chemical formula for an ionic compound composed of each pair of ions.
1. the magnesium ion and the carbonate ion
2. the aluminum ion and the acetate ion
Answer a:
\(\ce{MgCO3}\)
Answer b:
\(\ce{Al(CH3COO)3}\)
Recognizing Ionic Compounds
There are two ways to recognize ionic compounds.
Method 1
Compounds between metal and nonmetal elements are usually ionic. For example, \(\ce{CaBr2}\) contains a metallic element (calcium, a group 2 [or 2A] metal) and a nonmetallic element (bromine, a group 17 [or 7A] nonmetal). Therefore, it is most likely an ionic compound (in fact, it is ionic). In contrast, the compound \(\ce{NO2}\) contains two elements that are both nonmetals (nitrogen, from group 15 [or 5A], and oxygen, from group 16 [or 6A]. It is not an ionic compound; it belongs to the category of covalent compounds discussed elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion.
Method 2
Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula \(\ce{Ba(NO3)2}\), you may recognize the “\(\ce{NO3}\)” part as the nitrate ion, \(\ce{NO3^{-}}\). (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, \(\ce{Ba}\), is actually the \(\ce{Ba^{2+}}\) ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic.
Example \(6\)
Identify each compound as ionic or not ionic.
1. \(\ce{Na2O}\)
2. \(\ce{PCl3}\)
3. \(\ce{NH4Cl}\)
4. \(\ce{OF2}\)
Solution
Solution to Example 5.5.6
Explanation Answer
a. Sodium is a metal, and oxygen is a nonmetal. Therefore, \(\ce{Na2O}\) is expected to be ionic via method 1. \(\ce{Na2O}\), ionic
b. Both phosphorus and chlorine are nonmetals. Therefore, \(\ce{PCl3}\) is not ionic via method 1 \(\ce{PCl3}\), not ionic
c. The \(\ce{NH4}\) in the formula represents the ammonium ion, \(\ce{NH4^{+}}\), which indicates that this compound is ionic via method 2 \(\ce{NH4Cl}\), ionic
d. Both oxygen and fluorine are nonmetals. Therefore, \(\ce{OF2}\) is not ionic via method 1 \(\ce{OF2}\), not ionic
Exercise \(6\)
Identify each compound as ionic or not ionic.
1. \(\ce{N2O}\)
2. \(\ce{FeCl3}\)
3. \(\ce{(NH4)3PO4}\)
4. \(\ce{SOCl2}\)
Answer a:
not ionic
Answer b:
ionic
Answer c:
ionic
Answer d:
not ionic
Summary
Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio.
5.06: Nomenclature- Naming Compounds
Nomenclature is the process of naming chemical compounds so that they can be easily identified as separate chemicals. The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning which chemical compound the name refers to—each chemical name should refer to a single substance. A less important aim is to ensure that each substance has a single name, although a limited number of alternative names is acceptable in some cases. Preferably, the name also conveys some information about the structure or chemistry of a compound. A common name will often suffice to identify a chemical compound in a particular set of circumstances. To be more generally applicable, the name should indicate at least the chemical formula. To be more specific still, the three-dimensional arrangement of the atoms may need to be specified.
Contributions & Attributions
• Wikipedia (CC-BY-SA-3.0) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.05%3A_Writing_Formulas_for_Ionic_Compounds.txt |
Learning Objectives
• To use the rules for naming ionic compounds.
After learning a few more details about the names of individual ions, you will be one step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds.
Naming Ions
The name of a monatomic cation is simply the name of the element followed by the word ion. Thus, Na+ is the sodium ion, Al3+ is the aluminum ion, Ca2+ is the calcium ion, and so forth.
We have seen that some elements lose different numbers of electrons, producing ions of different charges (Figure 3.3). Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe2+ from Fe3+. The same issue arises for other ions with more than one possible charge.
There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system, an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion. Thus, Fe2+ is called the iron(II) ion, while Fe3+ is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na+ ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion.
Table \(1\): The Modern and Common System of Cation Names
Element Stem Charge Modern Name Common Name
iron ferr- 2+ iron(II) ion ferrous ion
3+ iron(III) ion ferric ion
copper cupr- 1+ copper(I) ion cuprous ion
2+ copper(II) ion cupric ion
tin stann- 2+ tin(II) ion stannous ion
4+ tin(IV) ion stannic ion
lead plumb- 2+ lead(II) ion plumbous ion
4+ lead(IV) ion plumbic ion
chromium chrom- 2+ chromium(II) ion chromous ion
3+ chromium(III) ion chromic ion
gold aur- 1+ gold(I) ion aurous ion
3+ gold(III) ion auric ion
The second system, called the common system, is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (-ic and -ous) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. Table \(1\) lists the elements that use the common system, along with their respective cation names.
Table \(2\): Some Monatomic Anions
Ion Name
F fluoride ion
Cl chloride ion
Br bromide ion
I iodide ion
O2− oxide ion
S2− sulfide ion
P3− phosphide ion
N3− nitride ion
The name of a monatomic anion consists of the stem of the element name, the suffix -ide, and then the word ion. Thus, as we have already seen, Cl is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O2− is the oxide ion, Se2 is the selenide ion, and so forth. Table \(2\) lists the names of some common monatomic ions. The polyatomic ions have their own characteristic names, as discussed earlier.
Example \(1\)
Name each ion.
1. Ca2+
2. S2−
3. SO32
4. NH4+
5. Cu+
Solution
1. the calcium ion
2. the sulfide ion
3. the sulfite ion
4. the ammonium ion
5. the copper(I) ion or the cuprous ion
Exercise \(1\)
Name each ion.
1. Fe2+
2. Fe3+
3. SO42
4. Ba2+
5. HCO3
Answer a:
iron(II) ion
Answer b:
iron(III) ion
Answer c:
sulfate ion
Answer d:
barium ion
Answer e:
hydrogen carbonate ion or bicarbonate ion
Example \(2\)
Write the formula for each ion.
1. the bromide ion
2. the phosphate ion
3. the cupric ion
4. the magnesium ion
1. Br
2. PO43
3. Cu2+
4. Mg2+
Exercise \(2\)
Write the formula for each ion.
1. the fluoride ion
2. the carbonate ion
3. the stannous ion
4. the potassium ion
Answer a:
F-
Answer b:
CO32-
Answer c:
Sn 2+
Answer d:
K+
Naming Binary Ionic Compounds with a Metal that Forms Only One Type of Cation
A binary ionic compound is a compound composed of a monatomic metal cation and a monatomic nonmetal anion. The metal cation is named first, followed by the nonmetal anion as illustrated in Figure \(1\) for the compound BaCl2. The word ion is dropped from both parts.
Subscripts in the formula do not affect the name.
Example \(3\): Naming Ionic Compounds
Name each ionic compound.
1. CaCl2
2. AlF3
3. KCl
Solution
1. Using the names of the ions, this ionic compound is named calcium chloride.
2. The name of this ionic compound is aluminum fluoride.
3. The name of this ionic compound is potassium chloride
Exercise \(3\)
Name each ionic compound.
1. AgI
2. MgO
3. Ca3P2
Answer a:
silver iodide
Answer b:
magnesium oxide
Answer c:
calcium phosphide
Naming Binary Ionic Compounds with a Metal That Forms More Than One Type of Cation
If you are given a formula for an ionic compound whose cation can have more than one possible charge, you must first determine the charge on the cation before identifying its correct name. For example, consider FeCl2 and FeCl3 . In the first compound, the iron ion has a 2+ charge because there are two Cl ions in the formula (1− charge on each chloride ion). In the second compound, the iron ion has a 3+ charge, as indicated by the three Cl ions in the formula. These are two different compounds that need two different names. By the Stock system, the names are iron(II) chloride and iron(III) chloride (Figure \(2\)).
Table \(3\): Naming the \(FeCl_2\) and \(FeCl_3\) Compounds in the Modern/Stock System.
Name of cation (metal) + (Roman Numeral in parenthesis) + Base name of anion (nonmetal) and -ide
If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively (Figure \(3\)) .
Table \(4\): Naming the \(FeCl_2\) and \(FeCl_3\) Compounds in the Old/Common System.
"Old" base name of cation (metal) and -ic or -ous + Base name of anion (nonmetal) and -ide
-ous (for ions with lower charge)
-ic (for ions with higher charge)
Example \(4\):
Name each ionic compound.
1. Co2O3
2. FeCl2
Solution
Solutions to Example 5.7.4
Explanation Answer
a
We know that cobalt can have more than one possible charge; we just need to determine what it is.
• Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−.
• This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+.
• Therefore, the proper name for this ionic compound is cobalt(III) oxide.
cobalt(III) oxide
b
Iron can also have more than one possible charge.
• Chloride always has a 1− charge, so with two chloride ions, we have a total negative charge of 2−.
• This means that the one iron ion must have a 2+ charge.
• Therefore, the proper name for this ionic compound is iron(II) chloride.
iron(II) chloride
Exercise \(4\)
Name each ionic compound.
1. AuCl3
2. PbO2
3. CuO
Answer a:
gold(III) chloride
Answer b:
lead(IV) oxide
Answer c:
copper(II) oxide
Naming Ionic Compounds with Polyatomic Ions
The process of naming ionic compounds with polyatomic ions is the same as naming binary ionic compounds. The cation is named first, followed by the anion. One example is the ammonium sulfate compound in Figure \(6\).
Example \(5\): Naming Ionic Compounds
Write the proper name for each ionic compound.
1. (NH4)2S
2. AlPO4,
3. Fe3(PO4)2
Solution
Solutions to Example 5.7.5
Explanation Answer
a. The ammonium ion has a 1+ charge and the sulfide ion has a 2− charge.
Two ammonium ions need to balance the charge on a single sulfide ion.
The compound’s name is ammonium sulfide.
ammonium sulfide
b. The ions have the same magnitude of charge, one of each (ion) is needed to balance the charges.
The name of the compound is aluminum phosphate.
aluminum phosphate
c. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6.
To get 6+, three iron(II) ions are needed, and to get 6−, two phosphate ions are needed .
The compound’s name is iron(II) phosphate.
iron(II) phosphate
Exercise \(\PageIndex{5A}\)
Write the proper name for each ionic compound.
1. (NH4)3PO4
2. Co(NO2)3
Answer a:
ammonium phosphate
Answer b:
cobalt(III) nitrite
Figure \(1\) is a synopsis of how to name simple ionic compounds.
Exercise \(\PageIndex{5B}\)
Name each ionic compound.
1. ZnBr2
2. Al2O3
3. (NH4)3PO4
4. AuF3
5. AgF
Answer a:
zinc bromide
Answer b:
aluminum oxide
Answer c:
ammonium phosphate
Answer d:
gold(III) fluoride or auric fluoride
Answer e:
silver fluoride
Summary
• Ionic compounds are named by stating the cation first, followed by the anion.
• Positive and negative charges must balance.
• Some anions have multiple forms and are named accordingly with the use of roman numerals in parentheses.
• Ternary compounds are composed of three or more elements. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.07%3A_Naming_Ionic_Compounds.txt |
Learning Objectives
• Determine the name of a simple molecular compound from its chemical formula.
Molecular Compounds
Molecular compounds are inorganic compounds that take the form of discrete molecules. Examples include such familiar substances as water $\left( \ce{H_2O} \right)$ and carbon dioxide $\left( \ce{CO_2} \right)$. These compounds are very different from ionic compounds like sodium chloride $\left( \ce{NaCl} \right)$. Ionic compounds are formed when metal atoms lose one or more of their electrons to nonmetal atoms. The resulting cations and anions are electrostatically attracted to each other.
So what holds the atoms of a molecule together? Rather than forming ions, the atoms of a molecule share their electrons in such a way that a bond forms between a pair of atoms. In a carbon dioxide molecule, there are two of these bonds, each occurring between the carbon atom and one of the two oxygen atoms.
Larger molecules can have many, many bonds that serve to keep the molecule together. In a large sample of a given molecular compound, all of the individual molecules are identical.
Naming Binary Molecular Compounds
Recall that a molecular formula shows the number of atoms of each element that a molecule contains. A molecule of water contains two hydrogen atoms and one oxygen atom, so its formula is $\ce{H_2O}$. A molecule of octane, which is a component of gasoline, contains 8 atoms of carbon and 18 atoms of hydrogen. The molecular formula of octane is $\ce{C_8H_{18}}$.
Naming binary (two-element) molecular compounds is similar to naming simple ionic compounds. The first element in the formula is simply listed using the name of the element. The second element is named by taking the stem of the element name and adding the suffix -ide. A system of numerical prefixes is used to specify the number of atoms in a molecule. Table $1$ lists these numerical prefixes.
Table $1$: Numerical Prefixes for Naming Binary Covalent Compounds
Number of Atoms in Compound Prefix on the Name of the Element
1 mono-*
2 di-
3 tri-
4 tetra-
5 penta-
6 hexa-
7 hepta-
8 octa-
9 nona-
10 deca-
*This prefix is not used for the first element’s name.
Note
• Generally, the less electronegative element is written first in the formula, though there are a few exceptions. Carbon is always first in a formula and hydrogen is after nitrogen in a formula such as $\ce{NH_3}$. The order of common nonmetals in binary compound formulas is $\ce{C}$, $\ce{P}$, $\ce{N}$, $\ce{H}$, $\ce{S}$, $\ce{I}$, $\ce{Br}$, $\ce{Cl}$, $\ce{O}$, $\ce{F}$.
• The a or o at the end of a prefix is usually dropped from the name when the name of the element begins with a vowel. As an example, four oxygen atoms, is tetroxide instead of tetraoxide.
• The prefix is "mono" is not added to the first element’s name if there is only one atom of the first element in a molecule.
Some examples of molecular compounds are listed in Table $2$.
Table $2$
Formula Name
$\ce{NO}$ nitrogen monoxide
$\ce{N_2O}$ dinitrogen monoxide
$\ce{S_2Cl_2}$ disulfur dichloride
$\ce{Cl_2O_7}$ dichlorine heptoxide
Notice that the mono- prefix is not used with the nitrogen in the first compound, but is used with the oxygen in both of the first two examples. The $\ce{S_2Cl_2}$ emphasizes that the formulas for molecular compounds are not reduced to their lowest ratios. The o of the mono- and the a of hepta- are dropped from the name when paired with oxide.
Exercise $1$
Write the name for each compound.
1. CF4
2. SeCl2
3. SO3
Answer a:
carbon tetrafluoride
Answer b:
selenium dichloride
Answer c:
sulfur trioxide
Simple molecular compounds with common names
For some simple covalent compounds, we use common names rather than systematic names. We have already encountered these compounds, but we list them here explicitly:
• H2O: water
• NH3: ammonia
• CH4: methane
• H2O2: hydrogen peroxide
Methane is the simplest organic compound. Organic compounds are compounds with carbon atoms and are named by a separate nomenclature system.
Some Compounds Have Both Covalent and Ionic Bonds
If you recall the introduction of polyatomic ions, you will remember that the bonds that hold the polyatomic ions together are covalent bonds. Once the polyatomic ion is constructed with covalent bonds, it reacts with other substances as an ion. The bond between a polyatomic ion and another ion will be ionic. An example of this type of situation is in the compound sodium nitrate. Sodium nitrate is composed of a sodium ion and a nitrate ion. The nitrate ion is held together by covalent bonds and the nitrate ion is attached to the sodium ion by an ionic bond.
Summary
• A molecular compound is usually composed of two or more nonmetal elements.
• Molecular compounds are named with the first element first and then the second element by using the stem of the element name plus the suffix -ide. Numerical prefixes are used to specify the number of atoms in a molecule. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.08%3A_Naming_Molecular_Compounds.txt |
A spot test for gold has been in use for decades. The sample is first treated with nitric acid. Other metals may react or dissolve in this acid, but gold will not. Then the sample is added to a mixture of nitric acid and hydrochloric acid. Gold will only dissolve in this mixture. The term "acid test" arose from the California gold rush in the late 1840's when this combination was used to test for the presence of real gold. It has since come to mean, "tested and approved" in a number of fields.
Acids
An acid can be defined in several ways. The most straightforward definition is that an acid is a molecular compound that contains one or more hydrogen atoms and produces hydrogen ions $\left( \ce{H^+} \right)$ when dissolved in water.
This is a different type of compound than the others we have seen so far. Acids are molecular, which means that in their pure state they are individual molecules and do not adopt the extended three-dimensional structures of ionic compounds like $\ce{NaCl}$. However, when these molecules are dissolved in water, the chemical bond between the hydrogen atom and the rest of the molecule breaks, leaving a positively-charged hydrogen ion and an anion. This can be symbolized in a chemical equation:
$\ce{HCl} \rightarrow \ce{H^+} + \ce{Cl^-} \nonumber$
Since acids produce $\ce{H^+}$ cations upon dissolving in water, the $\ce{H}$ of an acid is written first in the formula of an inorganic acid. The remainder of the acid (other than the $\ce{H}$) is the anion after the acid dissolves. Organic acids are also an important class of compounds, but will not be discussed here.
Naming Acids
Since all acids contain hydrogen, the name of an acid is based on the anion that goes with it. These anions can either be monatomic or polyatomic.
Naming Binary acids (in aqueous form)
A binary acid is an acid that consists of hydrogen and one other element. The most common binary acids contain a halogen. The acid name begins with the prefix hydro-. followed by the base name of the anion, followed by the suffix -ic.
Naming Oxyacids
An oxyacid is an acid that consists of hydrogen, oxygen, and a third element. The third element is usually a nonmetal.
a. Oxyanions with -ite ending.
The name of the acid is the root of the anion followed by the suffix -ous. There is no prefix.
b. Oxyanions with -ate ending.
The name of the acid is the root of the anion followed by the suffix -ic. There is no prefix.
Note
The base name for sulfur containing oxyacid is sulfur- instead of just sulf-. The same is true for a phosphorus containing oxyacid. The base name is phosphor- instead of simply phosph-.
Writing Formulas for Acids
Like other compounds that we have studied, acids are electrically neutral. Therefore, the charge of the anion part of the formula must be exactly balanced out by the $\ce{H^+}$ ions. Another way to think about writing the correct formula is to utilize the crisscross method, shown below for sulfuric acid.
Formula: H2SO4
Figure $2$: Crisscross approach to writing formula for sulfuric acid.
Summary
• Acids are molecular compounds that release hydrogen ions.
• A binary acid consists of hydrogen and one other element.
• Oxyacids contain hydrogen, oxygen, and one other element.
• The name of the acid is based on the anion attached to the hydrogen. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.09%3A_Naming_Acids.txt |
Learning Objectives
• To determine the formula mass of an ionic or molecular compound.
A necessary skill for future chapters is the ability to determine the mass of the formula of an ionic compound. This quantity is called the formula mass. The formula mass is obtained by adding the masses of each individual atom in the formula of the compound. Because a proper formula is electrically neutral (with no net electrons gained or lost), the ions can be considered atoms for the purpose of calculating the formula mass.
Let us start by calculating the formula mass of sodium chloride (NaCl). This formula mass is the sum of the atomic masses of one sodium atom and one chlorine atom, which we find from the periodic table; here, we use the masses to two decimal places:
Na: 22.99 amu
Cl: +35.34 amu
Total: 58.44 amu
To two decimal places, the formula mass of NaCl is 58.44 amu.
When an ionic compound has more than one anion or cation, you must remember to use the proper multiple of the atomic mass for the element in question. For the formula mass of calcium fluoride (CaF2), we must multiply the mass of the fluorine atom by 2 to account for the two fluorine atoms in the chemical formula:
Ca: 1 x 40.08 = 40.08 amu
F: 2 x 19.00 = +38.00 amu
Total = 78.08 amu
The formula mass of CaF2 is 78.08 amu.
For ionic compounds with polyatomic ions, the sum must include the number and mass of each atom in the formula for the polyatomic ion. For example, potassium nitrate (KNO3) has one potassium atom, one nitrogen atom, and three oxygen atoms:
K: 1 x 39.10 = 39.10 amu
N: 1 x 14.00 = +14.00 amu
O: 3 x 16.00 = +48.00 amu
Total = 101.10 amu
The formula mass of KNO3 is 101.10 amu.
Potassium nitrate is a key ingredient in gunpowder and has been used clinically as a diuretic.
When a formula contains more than one polyatomic unit in the chemical formula, as in Ca(NO3)2, do not forget to multiply the atomic mass of every atom inside of the parentheses by the subscript outside of the parentheses. This is necessary because the subscript refers to the entire polyatomic ion. Thus, for Ca(NO3)2, the subscript 2 implies two complete nitrate ions, so we must sum the masses of two (1 × 2) nitrogen atoms and six (3 × 2) oxygen atoms, along with the mass of a single calcium atom:
Ca: 1 x 40.08 = 40.08 amu
N: 2 x 14.00 = +28.00 amu
O: 6 x 16.00 = +96.00 amu
Total = 164.08 amu
The key to calculating the formula mass of an ionic compound is to correctly count each atom in the formula and multiply the atomic masses of its atoms accordingly.
Example $1$
Use the atomic masses (rounded to two decimal places) to determine the formula mass for each ionic compound.
1. FeCl3
2. (NH4)3PO4
Solution
a.
Fe: 1 x 55.85 = 55.85 amu
Cl: 1 x 35.45 = +106.35 amu
________________________
Total = 162.20 amu
The formula mass of FeCl3 is 162.2 amu.
b. When we distribute the subscript 3 through the parentheses containing the formula for the ammonium ion, we see that we have 3 nitrogen atoms and 12 hydrogen atoms. Thus, we set up the sum as follows:
N: 3 x 14.00 = 42.00 amu
H: 12 x 1.00 = +12.00 amu
P: 1 x 30.97 = +30.97 amu
O: 4 x 16.00 = +64.00 amu
Total = 148.97 amu
The formula mass for (NH4)3PO4 is 149.0 amu.
Exercise $1$
Use the atomic masses (rounded to two decimal places) to determine the formula mass for each ionic compound.
1. TiO2
2. AgBr
3. Au(NO3)3
4. Fe3(PO4)2
Answer
1. 79.87 amu
2. 187.77 amu
3. 383.0 amu
To Your Health: Hydrates
Some ionic compounds have water ($\ce{H2O}$) incorporated within their formula unit. These compounds, called hydrates, have a characteristic number of water units associated with each formula unit of the compound. Hydrates are solids, not liquids or solutions, despite the water they contain.
To write the chemical formula of a hydrate, write the number of water units per formula unit of compound after its chemical formula. The two chemical formulas are separated by a vertically centered dot. The hydrate of copper(II) sulfate has five water units associated with each formula unit, so it is written as $\ce{CuSO4 \cdot 5H2O}$. The name of this compound is copper(II) sulfate pentahydrate, with the penta- prefix indicating the presence of five water units per formula unit of copper(II) sulfate.
Hydrates have various uses in the health industry. Calcium sulfate hemihydrate ($\ce{CaSO4 \cdot 1/2 H2O}$), known as plaster of Paris, is used to make casts for broken bones. Epsom salt ($\ce{MgSO4 \cdot 7H2O}$) is used as a bathing salt and a laxative. Aluminum chloride hexahydrate is an active ingredient in antiperspirants. Table $1$ lists some useful hydrates.
Table $1$: Names and Formulas of Some Widely Used Hydrates
Formula Name Uses
AlCl3•6H2O aluminum chloride hexahydrate antiperspirant
CaSO4•½H2O calcium sulfate hemihydrate (plaster of Paris) casts (for broken bones and castings)
CaSO4•2H2O calcium sulfate dihydrate (gypsum) drywall component
CoCl2•6H2O cobalt(II) chloride hexahydrate drying agent, humidity indicator
CuSO4•5H2O copper(II) sulfate pentahydrate fungicide, algicide, herbicide
MgSO4•7H2O magnesium sulfate heptahydrate (Epsom salts) laxative, bathing salt
Na2CO3•10H2O sodium carbonate decahydrate (washing soda) laundry additive/cleaner
Key Takeaway
• Formula masses of ionic compounds can be determined from the masses of the atoms in their formulas. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/05%3A_Molecules_and_Compounds/5.11%3A_Formula_Mass_-_The_Mass_of_a_Molecule_or_Formula_Unit.txt |
Chapter 6 is concerned with the amounts of substances which participate in chemical reactions, the quantities of heat given off or absorbed when reactions occur, and the volumes of solutions which react exactly with one another. These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of the density calculations, and of the calculations involving molar mass and the Avogadro constant.
• 6.1: Prelude to Chemical Composition - How Much Sodium?
Why is knowledge of composition important? Everything in nature is either chemically or physically combined with other substances. To find the amount of a material in a sample, you need to know what fraction of the sample it is.
• 6.2: Counting Nails by the Pound
The size of molecule is so small that it is physically difficult, if not impossible, to directly count out molecules. However, we can count them indirectly by using a common trick of "counting by weighing".
• 6.3: Counting Atoms by the Gram
In chemistry, it is impossible to deal with a single atom or molecule because we can't see them or count them or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect this number is going to be very large and you are right. The number of particles in this group is Avagadro's number and the name of this group is the mole.
• 6.4: Counting Molecules by the Gram
The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. Calculations for formula mass and molecular mass are described. Calculations involving conversions between moles of a material and the mass of that material are described. Calculations are illustrated for conversions between mass and number of particles.
• 6.5: Chemical Formulas as Conversion Factors
Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. In any given formula the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor.
• 6.6: Mass Percent Composition of Compounds
Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound.
• 6.7: Mass Percent Composition from a Chemical Formula
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. This is divided by the molar mass of the compound and multiplied by 100%.
• 6.8: Calculating Empirical Formulas for Compounds
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. A process is described for the calculation of the empirical formula for a compound based on the percent composition of that compound.
• 6.9: Calculating Molecular Formulas for Compounds
A procedure is described that allows the calculation of the exact molecular formula for a compound.
06: Chemical Composition
Why is knowledge of composition important? Everything in nature is either chemically or physically combined with other substances. To find the amount of a material in a sample, you need to know what fraction of the sample it is. Some simple applications of composition are: the amount of sodium in sodium chloride for a diet, the amount of iron in iron ore for steel production, the amount of hydrogen in water for hydrogen fuel, and the amount of chlorine in freon to estimate ozone depletion. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.01%3A_Prelude_to_Chemical_Composition_-_How_Much_Sodium.txt |
Counting by Weighing and Avogadro's number
The size of molecule is so small that it is physically difficult, if not impossible, to directly count out molecules (Figure $1$). However, we can count them indirectly by using a common trick of "counting by weighing".
Consider the example of counting nails in a big box at a hardware store. You need to estimate the number of nails in a box. The weight of an empty box is $213 \,g$ and the weight of the box plus a bunch of big nails is $1340\, g$. Assume that we know that the weight of one big nail is 0.450 g. Hopefully it's not necessary to tear open the package and count the nails. We agree that
$\text{mass of big nails} = 1340\, g - 213\, g = 1227 \,g \nonumber$
Therefore
$\text{Number of big nails in box} = \dfrac{1227\, g}{0.450\, g/ \text{big nail}} = 2,726.6\, \text{big nails} = 2,730 \,\text{big nails}. \label{eq2}$
You have just counted the number of big nails in the box by weighing them (rather than by counting them individually).
Now consider if the box of nails weighed the same, but the box were filled with small nails with an individual mass of $0.23\, g/\text{small nail}$ instead? You would do the same math, but use a different denominator in Equation \ref{eq2}:
$\text{Number of small nails in box} = \dfrac{1227\, g}{0.230\, g/\text{small nail}} = 5,334.7\, \text{small nails} = 5,335 \, \text{small nails}. \label{eq3}$
The individual mass is the conversion factor used in the calculation and changes, based on the nature of the nail (big or small). Let's ask a different question: how many dozens of nails are there in the same box of small nails described above?
If we know the information from Equation \ref{eq3}, we can just use the conversion of how many nails are in a dozen:
$\dfrac{5,335 \,\text{small nails} }{12\, \text{small nails/dozen}} = 444.6 \,\text{dozen small nails} \label{eq4}$
If we want to get this value from weighing, we use the "dozen mass" instead of individual mass:
$12 \times 0.23 g = 2.76\, g/\text{dozen small nails}. \label{eq5}$
So following Equation \ref{eq3}, we get:
$\text{Number of dozens of small nails} = \dfrac{1227\, g}{2.76\, g/\text{dozen small nails}} = 444.6 \,\text{dozen small nails} \label{eq6}$
and this is the same result as Equation \ref{eq4}. These calculations demonstrate the difference between individual mass (i.e., per individual) and collective mass (e.g., per dozen or per gross). The collective mass of most importance to chemistry is molar mass (i.e., mass per mole or mass per $6.022 \times 10^{23}$).
Avogadro's Number
Avogadro's number is an accident of nature. It is the number of particles that delivers a mole of a substance. Avogadro's number = $6.022 \times 10^{23}$. The reason why the value is an accident of nature is that the mole is tied to the gram mass unit. The gram is a convenient mass unit because it matches human sizes. If we were a thousand times greater in size (like Paul Bunyan) we would find it handy to use kilogram amounts. This means the kilogram mole would be convenient. The number of particles handled in a kilogram mole is 1000 times greater. The kilo Avogadro number for the count of particles in a kilomole is $6.022 \times 10^{26}$.
If humans were tiny creatures (like Lilliputians) only 1/1000 our present size, milligrams would be more convenient. This means the milligram mole would be more useful. The number of particles handled in a milligram mole (millimole) would be 1/1000 times smaller. The milli Avogadro number for the count of particles in a millimole is $6.022 \times 10^{20}$.
What do you think would happen to Avogadro's number if the American system was used and amounts were measured in pound moles? Remember 1 pound = 454 grams. Avogadro's number would be larger by a factor of 454. A pound mole of hydrogen would weigh 1 pound, which is 454 grams. A gram mole of hydrogen weighs 1 gram and contains $6.022 \times 10^{23}$ H atoms.
Molar Mass for Elements
You are able to read the periodic table and determine the average atomic mass for an element like carbon. The average mass is 12.01 amu. This mass is a ridiculously tiny number of grams. It is too small to handle normally. The molar mass of carbon is defined as the mass in grams that is numerically equal to the average atomic weight. This means
$1 g/ mole carbon = 12.01 \,g \,carbon \nonumber$
this is commonly written
$1\, mol\, carbon = 12.01\, grams\, carbon. \nonumber$
This is the mass of carbon that contains $6.022 \times 10^{23}$ carbon atoms.
• Avogadro's number is $6.022 \times 10^{23}$ particles.
This same process gives us the molar mass of any element. For example:
• $1\, mol\, neon = 20.18\, g\, neon\, Ne$
• $1 \,mol\, sodium = 22.99\, g \,sodium\, Na$
Molar Mass for Compounds
Example $1$: Molar Mass of Water
The formulas for compounds are familiar to you. You know the formula for water is $\ce{H2O}$. It should be reasonable that the weight of a formula unit can be calculated by adding up the weights for the atoms in the formula.
Solution
The formula weight for water
weight from hydrogen + weight from oxygen
The formula weight for water
2 H atoms x 1.008 amu + 1 O atom x 16.00 amu = 18.016 amu
The molar mass for water
18.016 grams water or 18 grams to the nearest gram
Example $2$: Molar Mass of Methane
The formula for methane, the major component in natural gas, is $\ce{CH4}$.
Solution
The formula weight for methane
weight from hydrogen + weight from carbon
The formula weight for methane
4 H atoms x 1.008 amu + 1 C atom x 12.01 amu = 16.04 amu
The molar mass for methane
16.04 grams per mole of methane
Example $3$: Molar Mass of Ethyl Chloride
What is its molar mass for ethyl chloride $\ce{CH3CH2Cl}$?
Solution
The formula weight
weight from hydrogen + weight from carbon + weight from chlorine
The formula weight
5 H atoms x 1.008 amu + 2 C atom x 12.01 amu + 35.5 amu = 64.5 amu
The molar mass for ethyl chloride
64.5 grams per mole of ethyl chloride
Example $4$: Molar Mass of Sulfur Dioxide
What is the molar mass for sulfur dioxide, $\ce{SO2 (g)}$, a gas used in bleaching and disinfection processes?
Solution
Look up the atomic weight for each of the elements in the formula.
• 1 sulfur atom = 32.07 amu
• 1 oxygen atom = 16.00 amu
Count the atoms of each element in the formula unit.
• one sulfur atom
• two oxygen atoms
The formula weight
weight from sulfur + weight from oxygen
The formula weight
1 sulfur atom x (32. 07 amu ) + 2 oxygen atoms x (16.00 amu)
The formula weight
$\ce{SO2}$ = 32. 07 amu + 32.00 amu = 64.07 amu = 64 amu $\ce{SO2}$
The molar mass for $\ce{SO2}$ is
64.07 grams of $\ce{SO2}$; 1 mol $\ce{SO2}$ = 64 grams per mole of $\ce{SO2}$
Exercise $1$
What is the formula weight and molar mass for alum, $\ce{KAl(SO4)2 • 12 H2O}$ ?
Answer
1. Check the periodic table for the atomic masses for each atom in the formula.
2. Count the number of each type of atom in the formula.
3. Multiply the number of atoms by the atomic mass for each element.
4. Add up the masses for all of the elements.
Table $1$: Masses of each element in alum, $\ce{KAl(SO4)2 • 12 H2O}$
element average atomic mass number of atoms in formula rounded to nearest one unit for simplicity
potassium k 39.1 amu 1 39. amu
aluminum 26.98 amu 1 27. amu
sulfur 32.07 amu 2 64. amu
oxygen 16.00 amu 8 + 12 = 20 320. amu
hydrogen 1.008 amu 2 x 12 = 24 24. amu
Molar mass is 474 grams (add up the amu of each element to find the total of 474 amu). This is a mass in grams that is numerically (474) the same as the formula weight.
1 mole alum $\ce{KAl(SO4)2• 12 H2O}$ = 474 grams alum $\ce{KAl(SO4)2• 12 H2O}$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.02%3A_Counting_Nails_by_the_Pound.txt |
Learning Objectives
• Use Avogadro's number to convert to moles and vice versa given the number of particles of an element.
• Use the molar mass to convert to grams and vice versa given the number of moles of an element.
When objects are very small, it is often inconvenient, inefficient, or even impossible to deal with the objects one at a time. For these reasons, we often deal with very small objects in groups, and have even invented names for various numbers of objects. The most common of these is "dozen" which refers to 12 objects. We frequently buy objects in groups of 12, like doughnuts or pencils. Even smaller objects such as straight pins or staples are usually sold in boxes of 144, or a dozen dozen. A group of 144 is called a "gross".
This problem of dealing with things that are too small to operate with as single items also occurs in chemistry. Atoms and molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with.
Avogadro's Number and Mole
In chemistry, it is impossible to deal with a single atom or molecule because we can't see them, count them, or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect this number is going to be very large, and you are right. The number of particles in this group is $6.02 \times 10^{23}$ particles and the name of this group is the mole (the abbreviation for mole is $\text{mol}$). One mole of any object is $6.02 \times 10^{23}$ of those objects. There is a particular reason that this number was chosen and this reason will become clear as we proceed.
When chemists are carrying out chemical reactions, it is important that the relationship between the numbers of particles of each reactant is known. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose.
The mole (symbol: mol) is the base unit of amount of substance ("number of substance") in the International System of Units or System International (SI), defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons. The current definition was adopted in November 2018, revising its old definition based on the number of atoms in 12 grams of carbon-12 (12C) (the isotope of carbon with relative atomic mass 12 Daltons, by definition). For most purposes, 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 1023 atoms, 1 mole of eggs contains 6.022 × 1023 eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.
It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023.
Converting Between Number of Atoms to Moles and Vice Versa
We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles by using the relationship
$\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms}. \nonumber$
Example $1$: Moles of Carbon
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Solution
Solutions to Example 6.3.1
Steps for Problem Solving The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is $4.72 \times 10^{24}$ atoms of carbon?
Identify the "given" information and what the problem is asking you to "find." Given: $4.72 \times 10^{24}$ C atoms
Find: mol C
List other known quantities. $1\, mol = 6.022 \times 10^{23}$ C atoms
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $4.72 \times 10^{24} \: \cancel{\text{C} \: \ce{atoms}} \times \dfrac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \cancel{\text{C} \: \ce{atoms}}} = 7.84 \: \text{mol} \: \ce{C} \nonumber$
Think about your result. The given number of carbon atoms was greater than Avogadro's number, so the number of moles of $\ce{C}$ atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures.
Molar Mass
Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of the element lithium is 6.94g, the molar mass of zinc is 65.38g, and the molar mass of gold is 196.97g. Each of these quantities contains $6.022 \times 10^{23}$ atoms of that particular element. The units for molar mass are grams per mole or g/mol. $1.00 \: \text{mol}$ of carbon-12 atoms has a mass of $12.0 \: \text{g}$ and contains $6.022 \times 10^{23}$ atoms. 1.00 mole of any element has a mass numerically equal to its atomic mass in grams and contains $6.022 \times 10^{23}$ particles. The mass, in grams, of 1 mole of particles of a substance is now called the molar mass (mass of 1.00 mole).
Converting Grams to Moles of an Element and Vice Versa
We can also convert back and forth between grams of an element and moles. The conversion factor for this is the molar mass of the substance. The molar mass is the ratio giving the number of grams for each one mole of the substance. This ratio is easily found by referring to the atomic mass of the element using the periodic table. This ratio has units of grams per mole or $\text{g/mol}$.
Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure 6.4.1 illustrates what conversion factor is needed and two examples are given below.
Example $2$: Chromium
Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.
Solution
Solutions to Example 6.3.2
Steps for Problem Solving Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.
Identify the "given" information and what the problem is asking you to "find." Given: 0.560 mol Cr
Find: g Cr
List other known quantities. 1 mol Cr = 52.00g Cr
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $0.560 \: \cancel{\text{mol} \: \ce{Cr}} \times \dfrac{52.00 \: \text{g} \: \ce{Cr}}{1 \: \cancel{\text{mol} \: \ce{Cr}}} = 29.1 \: \text{g} \: \ce{Cr} \nonumber$
Think about your result. Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the $0.560 \: \text{mol}$
Example $3$: Silicon
How many moles are in 107.6g of Si?
Solution
Solutions to Example 6.3.3
Steps for Problem Solving How many moles are in 107.6g of Si.
Identify the "given" information and what the problem is asking you to "find." Given: 107.6g Si
Find: mol Si
List other known quantities. 1 mol Si = 28.09g Si
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $107.6 \: \cancel{\text{g} \: \ce{Si}} \times \dfrac{1 \: \text{mol} \: \ce{Si}}{28.09 \: \cancel{\text{g} \: \ce{Si}}} = 3.83 \: \text{mol} \: \ce{Si} \nonumber$
Think about your result. Since 1 mol of Si is 28.09g, 107.6 should be about 4 moles.
Exercise $1$
1. How many moles are present in 100.0 g of Al?
2. What is the mass of 0.552 mol of Ag metal?
Answer a:
3.706 mol Al
Answer b:
59.5 g Ag
Summary
• A mole is defined as exactly 6.02214076×1023 particles, e.g., atoms, molecules, ions or electrons.
• There are $6.02214076 \times 10^{23}$ particles in 1.00 mole. This number is called Avogadro's number.
• The molar mass of an element can be found by referring to the atomic mass on a periodic table with units of g/mol.
• Using dimensional analysis, it is possible to convert between grams, moles, and the number of atoms or molecules.
Further Reading/Supplemental Links
• learner.org/resources/series61.html - The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge. The website has one video that relates to this lesson called The Mole.
• Using Avogadro's law, the mass of a substance can be related to the number of particles contained in that mass. The Mole: (www.learner.org/vod/vod_window.html?pid=803)
• Vision Learning tutorial: The Mole http://visionlearning.com/library/mo...p?mid-53&1=&c3=
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.03%3A_Counting_Atoms_by_the_Gram.txt |
Learning Objectives
• Define molecular mass and formula mass.
• Perform conversions between mass and moles of a compound.
• Perform conversions between mass and number of particles.
Molecular and Formula Masses
The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example $1$.
Example $1$: Ethanol
Calculate the molecular mass of ethanol, whose condensed structural formula is $\ce{CH_3CH_2OH}$. Among its many uses, ethanol is a fuel for internal combustion engines
Solution
Solutions to Example 6.4.1
Steps for Problem Solving Calculate the molecular mass of ethanol, whose condensed structural formula is $\ce{CH_3CH_2OH}$
Identify the "given"information and what the problem is asking you to "find." Given: Ethanol molecule (CH3CH2OH)
Find: molecular mass
Determine the number of atoms of each element in the molecule.
The molecular formula of ethanol may be written in three different ways:
• CH3CH2OH (which illustrates the presence of an ethyl group
• CH3CH2, and an −OH group)
• C2H5OH, and C2H6O;
All show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.
Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
1 C atom = 12.011 amu
1 H atom = 1.0079 amu
1 O atom = 15.9994 amu
Add the masses together to obtain the molecular mass.
2C: (2 atoms)(12.011amu/atom) = 24.022 amu
6H: (6 atoms)(1.0079amu/atom) = 6.0474amu
+1O: (1 atoms)(15.9994amu/atom) =15.9994amu
C2H6O : molecular mass of ethanol = 46.069amu
Exercise $1$: Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, which has a condensed structural formula of $\ce{CCl3F}$. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer
137.37 amu
Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Example $2$: Calcium Phosphate
Calculate the formula mass of $\ce{Ca3(PO4)2}$, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Solution
Solutions to Example 6.4.2
Steps for Problem Solving Calculate the formula mass of $\ce{Ca3(PO4)2}$, commonly called calcium phosphate.
Identify the "given" information and what the problem is asking you to "find." Given: Calcium phosphate [Ca3(PO4)2] formula unit
Find: formula mass
Determine the number of atoms of each element in the molecule.
• The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43 ions.
• The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.
Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
1 Ca atom = 40.078 amu
1 P atom = 30.973761 amu
1 O atom = 15.9994 amu
Add together the masses to give the formula mass.
3Ca: (3 atoms) (40.078 amu/atom)=120.234amu
2P: (2 atoms) (30.973761amu/atom)=61.947522amu
+ 8O: (8 atoms)(15.9994amu/atom)=127.9952amu
Formula mass of Ca3(PO4)2=310.177amu
Exercise $2$: Silicon Nitride
Calculate the formula mass of $\ce{Si3N4}$, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.
Answer
140.29 amu
Molar Mass
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:
Table $1$: Molar Mass of Select Substances
Substance (formula) Basic Unit Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) atom 12.011 (atomic mass) 12.011
ethanol (C2H5OH) molecule 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] formula unit 310.177 (formula mass) 310.177
Converting Between Grams and Moles of a Compound
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride $\left( \ce{CaCl_2} \right)$. Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of $\ce{CaCl_2}$ that you should measure as shown in Example $3$.
Example $3$: Calcium Chloride
Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
Solution
Solutions to Example 6.4.3
Steps for Problem Solving Calculate the mass of 3.00 moles of calcium chloride
Identify the "given" information and what the problem is asking you to "find." Given: 3.00 moles of $\ce{CaCl2}$
Find: g $\ce{CaCl2}$
List other known quantities. 1 mol $\ce{CaCl2}$ = 110.98 g $\ce{CaCl2}$
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}$
Think about your result.
Exercise $3$: Calcium Oxide
What is the mass of $7.50 \: \text{mol}$ of (calcium oxide) $\ce{CaO}$?
Answer
420.60 g
Example $4$: Water
How many moles are present in 108 grams of water?
Solution
Solutions to Example 6.4.4
Steps for Problem Solving How many moles are present in 108 grams of water?
Identify the "given" information and what the problem is asking you to "find." Given: 108 g $\ce{H2O}$
Find: mol $\ce{H2O}$
List other known quantities. $1 \: \text{mol} \: \ce{H_2O} = 18.02 \: \text{g}$ H2O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}$
Think about your result.
Exercise $4$: Nitrogen Gas
What is the mass of $7.50 \: \text{mol}$ of Nitrogen gas $\ce{N2}$?
Answer
210 g
Conversions Between Mass and Number of Particles
In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to moles. To convert from mass to number of particles or vice-versa, it will first require a conversion to moles as shown in Figure $1$ and Example $5$.
Example $5$: Chlorine
How many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Solution
Solutions to Example 6.4.5
Steps for Problem Solving How many molecules is $20.0 \: \text{g}$ of chlorine gas, $\ce{Cl_2}$?
Identify the "given" information and what the problem is asking you to "find." Given: 20.0 g $\ce{Cl2}$
Find: # $\ce{Cl2}$ molecules
List other known quantities.
• 1 mol $\ce{Cl2}$ = 70.90 g $\ce{Cl2}$,
• 1mol $\ce{Cl2}$ = $6.022 \times 10^{23}$ $\ce{Cl2}$ molecules
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $20.0 \: \cancel{\text{g} \: \ce{Cl_2}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{Cl_2}}}{70.90 \: \cancel{\text{g} \: \ce{Cl_2}}} \times \dfrac{6.02 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}}{1 \: \cancel{\text{mol} \: \ce{Cl_2}}} \[4pt] = 1.70 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}$
Think about your result. Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro's number.
Exercise $5$: Calcium Chloride
How many formula units are in 25.0 g of $\ce{CaCl2}$?
Answer
1.36 x 1023 $\ce{CaCl2}$ formula units
Summary
• Calculations for formula mass and molecular mass are described.
• Calculations involving conversions between moles of a material and the mass of that material are described.
• Calculations are illustrated for conversions between mass and number of particles. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.04%3A_Counting_Molecules_by_the_Gram.txt |
Learning Objectives
• Use chemical formulas as conversion factors.
Figure $1$ shows that we need 2 hydrogen atoms and 1 oxygen atom to make one water molecule. If we want to make two water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make five molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom.
Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of water (H2O) we can construct the relationships given in (Table $1$).
Table $1$: Molecular Relationships for Water
1 Molecule of $H_2O$ Has 1 Mol of $H_2O$ Has Molecular Relationships
2 H atoms 2 mol of H atoms $\mathrm{\dfrac{2\: mol\: H\: atoms}{1\: mol\: H_2O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{2\: mol\: H\: atoms}}$
1 O atom 1 mol of O atoms $\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: H_2O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{1\: mol\: O\: atoms}}$
The Mole is big
A mole represents a very large number! The number 602,214,129,000,000,000,000,000 looks about twice as long as a trillion, which means it’s about a trillion trillion.
(CC BY-SA NC; what if? [what-if.xkcd.com]).
A trillion trillion kilograms is how much a planet weighs. If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times.
Table $2$: Molecular and Mass Relationships for Ethanol
1 Molecule of $C_2H_6O$ Has 1 Mol of $C_2H_6O$ Has Molecular and Mass Relationships
2 C atoms 2 mol of C atoms $\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}$
6 H atoms 6 mol of H atoms $\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}$
1 O atom 1 mol of O atoms $\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}$
2 (12.01 amu) C
24.02 amu C
2 (12.01 g) C
24.02 g C
$\mathrm{\dfrac{24.02\: g\: C\: }{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{24.02\: g\: C\: }}$
6 (1.008 amu) H
6.048 amu H
6 (1.008 g) H
6.048 g H
$\mathrm{\dfrac{6.048\: g\: H\: }{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6.048\: g\: H\: }}$
1 (16.00 amu) O
16.00 amu O
1 (16.00 g) O
16.00 g O
$\mathrm{\dfrac{16.00\: g\: O\: }{1\: mol\: C_2H_6O\: molecules}}$ or $\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{16.00\: g\: O\: }}$
The following example illustrates how we can use the relationships in Table $2$ as conversion factors.
Example $1$: Ethanol
If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have?
Solution
Solutions to Example 6.5.1
Steps for Problem Solving If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have?
Identify the "given" information and what the problem is asking you to "find."
Given: 2.5 mol C2H6O
Find: mol C atoms
List other known quantities. 1 mol C2H6O = 2 mol C
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
Note how the unit mol C2H6O molecules cancels algebraically.
$\mathrm{2.5\: \cancel{mol\: C_2H_6O\: molecules}\times\dfrac{2\: mol\: C\: atoms}{1\: \cancel{mol\: C_2H_6O\: molecules}}=5.0\: mol\: C\: atoms}$
Think about your result. There are twice as many C atoms in one C2H6O molecule, so the final amount should be double.
Exercise $1$
If a sample contains 6.75 mol of Na2SO4, how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have?
Answer
13.5 mol Na atoms, 6.75 mol S atoms, and 27.0 mol O atoms
The fact that 1 mol equals 6.022 × 1023 items can also be used as a conversion factor.
Example $2$: Oxygen Mass
Determine the mass of Oxygen in 75.0g of C2H6O.
Solution
Solutions to Example 6.5.2
Steps for Problem Solving Determine the mass of Oxygen in 75.0g of C2H6O
Identify the "given" information and what the problem is asking you to "find."
Given: 75.0g C2H6O
Find: g O
List other known quantities.
1 mol O = 16.0g O
1 mol C2H6O = 1 mol O
1 mol C2H6O = 46.07g C2H6O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $\require{cancel}\mathrm{75.0\: \cancel{g\: C_2H_6O}\times\dfrac{1\: \cancel{mol\: C_2H_6O}}{46.07\:\cancel{g\: C_2H_6O}}\times\dfrac{1\: \cancel{mol\:O}}{1\: \cancel{mol\:C_2H_6O}}\times\dfrac{16.00\: g\: O}{1\: \cancel{mol\:O}}=26.0\: g\: O}$
Think about your result.
Exercise $2$
1. How many molecules are present in 16.02 mol of C4H10? How many C atoms are in 16.02 mol?
2. How many moles of each type of atom are in 2.58 mol of Na2SO4?
Answer a:
9.647 x 1024 C4H10 molecules and 3.859 x 1025 C atoms
Answer b:
5.16 mol Na atoms, 2.58 mol S atoms, and 10.3 mol O atoms
Summary
In any given formula, the ratio of the number of moles of molecules (or formula units) to the number of moles of atoms can be used as a conversion factor. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.05%3A_Chemical_Formulas_as_Conversion_Factors.txt |
Learning Objectives
• Determine percent composition of each element in a compound based on mass.
Packaged foods that you eat typically have nutritional information provided on the label. The label on a jar of peanut butter reveals that one serving size is considered to be $32 \: \text{g}$. The label also gives the masses of various types of compounds that are present in each serving. One serving contains $7 \: \text{g}$ of protein, $15 \: \text{g}$ of fat, and $3 \: \text{g}$ of sugar. By calculating the fraction of protein, fat, or sugar in one serving size of peanut butter and converting to percent values, we can determine the composition of peanut butter on a percent by mass basis.
Percent Composition
Chemists often need to know what elements are present in a compound and in what percentage. The percent composition is the percent by mass of each element in a compound. It is calculated in a similar way to that of the composition of the peanut butter.
$\% \: \text{by mass} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100\% \nonumber$
The sample problem below shows the calculation of the percent composition of a compound based on mass data.
Example $1$: Percent Composition from Mass Data
A certain newly synthesized compound is known to contain the elements zinc and oxygen. When a $20.00 \: \text{g}$ sample of the compound is decomposed, $16.07 \: \text{g}$ of zinc remains. Determine the percent composition of the compound.
Solution
Solutions to Example 6.6.1
Steps for Problem Solving When a $20.00 \: \text{g}$ sample of the zinc-and-oxygen compound is decomposed, $16.07 \: \text{g}$ of zinc remains. Determine the percent composition of the compound.
Identify the "given" information and what the problem is asking you to "find."
Given : Mass of compound = 20.00 g
Mass of Zn = 16.07 g
Find: % Composition (% Zn and %O)
List other known quantities.
Subtract to find the mass of oxygen in the compound. Divide each element's mass by the mass of the compound to find the percent by mass.
Mass of oxygen = 20.00 g - 16.07 g = 3.93 g O
Cancel units and calculate.
$\% \: \ce{Zn} = \dfrac{16.07 \: \text{g} \: \ce{Zn}}{20.00 \: \text{g}} \times 100\% = 80.35\% \: \ce{Zn} \nonumber$
$\% \: \ce{O} = \dfrac{3.93 \: \text{g} \: \ce{O}}{20.00 \: \text{g}} \times 100\% = 19.65\% \: \ce{O} \nonumber$
Calculate the percent by mass of each element by dividing the mass of that element by the mass of the compound and multiplying by $100\%$.
Think about your result. The calculations make sense because the sum of the two percentages adds up to $100\%$. By mass, the compound is mostly zinc.
Exercise $1$
Sulfuric acid, H2SO4 is a very useful chemical in industrial processes. If 196.0 g of sulfuric acid contained 64.0g oxygen and 4.0 g of hydrogen, what is the percent composition of the compound?
Answer
2.04% H, 32.65% S, and 65.3% O
Summary
• Processes are described for calculating the percent composition of a compound based on mass.
6.07: Mass Percent Composition from a Chemical Formula
Learning Objectives
• Determine the percent composition of each element in a compound from the chemical formula.
The percent composition of a compound can also be determined from the formula of the compound. The subscripts in the formula are first used to calculate the mass of each element in one mole of the compound. This is divided by the molar mass of the compound and multiplied by $100\%$.
$\% \: \text{by mass} = \dfrac{\text{mass of element in} \: 1 \: \text{mol}}{\text{molar mass of compound}} \times 100\% \nonumber$
The percent composition of a given compound is always the same, given that the compound is pure.
Example $1$
Dichlorine heptoxide $\left( \ce{Cl_2O_7} \right)$ is a highly reactive compound used in some organic synthesis reactions. Calculate the percent composition of dichlorine heptoxide.
Solution
Solutions to Example 6.7.1
Steps for Problem Solving Calculate the percent composition of dichlorine heptoxide $\left( \ce{Cl_2O_7} \right)$.
Identify the "given" information and what the problem is asking you to "find."
Given : Cl2O7
Find: % Composition (% Cl and %O)
List other known quantities.
Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g
Mass of O in 1 mol Cl2O7 , 7 O: 7 x 16.00 g = 112.00 g
Molar mass of Cl2O7 = 182.90 g/mol
Cancel units and calculate.
$\% \ce{Cl} = \dfrac{70.90 \: \text{g} \: \ce{Cl}}{182.90 \: \text{g}} \times 100\% = 38.76\% \: \ce{Cl} \nonumber$
$\% \: \ce{O} = \dfrac{112.00 \: \text{g} \: \ce{O}}{182.90 \: \text{g}} \times 100\% = 61.24\% \: \ce{O} \nonumber$
Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by $100\%$.
Think about your result. The percentages add up to $100\%$.
Percent composition can also be used to determine the mass of a certain element that is contained in any mass of a compound. In the previous sample problem, it was found that the percent composition of dichlorine heptoxide is $38.76\% \: \ce{Cl}$ and $61.24\% \: \ce{O}$. Suppose that you needed to know the masses of chlorine and oxygen present in a $12.50 \: \text{g}$ sample of dichlorine heptoxide. You can set up a conversion factor based on the percent by mass of each element.
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \dfrac{38.76 \: \text{g} \: \ce{Cl}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 4.845 \: \text{g} \: \ce{Cl} \nonumber$
$12.50 \: \text{g} \: \ce{Cl_2O_7} \times \dfrac{61.24 \: \text{g} \: \ce{O}}{100 \: \text{g} \: \ce{Cl_2O_7}} = 7.655 \: \text{g} \: \ce{O} \nonumber$
The sum of the two masses is $12.50 \: \text{g}$, the mass of the sample size.
Exercise $1$
Barium fluoride is a transparent crystal that can be found in nature as the mineral frankdicksonite. Determine the percent composition of barium fluoride.
Answer a:
78.32% Ba and 21.67% F
Summary
• Processes are described for calculating the percent composition of a compound based on the chemical formula. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.06%3A_Mass_Percent_Composition_of_Compounds.txt |
Learning Objectives
• Define empirical formula.
• Determine empirical formula from percent composition of a compound.
In the early days of chemistry, there were few tools for the detailed study of compounds. Much of the information regarding the composition of compounds came from the elemental analysis of inorganic materials. The "new" field of organic chemistry (the study of carbon compounds) faced the challenge of not being able to characterize a compound completely. The relative amounts of elements could be determined, but so many of these materials had carbon, hydrogen, oxygen, and possibly nitrogen in simple ratios. We did not know exactly how many of these atoms were actually in a specific molecule.
Determining Empirical Formulas
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen. We can also work backwards from molar ratios because if we know the molar amounts of each element in a compound, we can determine the empirical formula.
In a procedure called elemental analysis, an unknown compound can be analyzed in the laboratory in order to determine the percentages of each element contained within it. These percentages can be transformed into the mole ratio of the elements, which leads to the empirical formula.
Empirical Formula: In Steps
Steps to determine empirical formula:
1. Assume a $100 \: \text{g}$ sample of the compound so that the given percentages can be directly converted into grams.
2. Use each element's molar mass to convert the grams of each element to moles.
3. In order to find a whole-number ratio, divide the moles of each element by whichever of the moles from step 2 is the smallest.
4. If all the moles at this point are whole numbers (or very close), the empirical formula can be written with the moles as the subscript of each element.
5. In some cases, one or more of the moles calculated in step 3 will not be whole numbers. Multiply each of the moles by the smallest whole number that will convert each into a whole number. Write the empirical formula.
Example $1$
A compound of iron and oxygen is analyzed and found to contain $69.94\%$ iron and $30.06\%$ oxygen. Find the empirical formula of the compound.
Solution
Solutions to Example 6.8.1
Steps for Problem Solving Find the empirical formula of a compound of $69.94\%$ iron and $30.06\%$ oxygen.
Identify the "given" information and what the problem is asking you to "find."
Given:
$\%$ of $\ce{Fe} = 69.94\%$
$\%$ of $\ce{O} = 30.06\%$
Find: Empirical formula $= \ce{Fe}_?\ce{O}_?$
Calculate
a. Assume a $100 \: \text{g}$ sample, convert the same % values to grams.
$69.94 \: \text{g} \: \ce{Fe} \nonumber$
$30.06 \: \text{g} \: \ce{O} \nonumber$
b. Convert to moles.
$69.94 \: \text{g} \: \ce{Fe} \times \dfrac{1 \: \text{mol} \: \ce{Fe}}{55.85 \: \text{g} \: \ce{Fe}} = 1.252 \: \text{mol} \: \ce{Fe} \nonumber$
$30.06 \: \text{g} \: \ce{O} \times \dfrac{1 \: \text{mol} \: \ce{O}}{16.00 \: \text{g} \: \ce{O}} = 1.879 \: \text{mol} \: \ce{O} \nonumber$
c. Divide both moles by the smallest of the results.
$\mathrm{Fe:\:\dfrac{1.252\:mol}{1.252}}$
$\mathrm{O:\:\dfrac{1.879\:mol}{1.252}}$
The "non-whole number" empirical formula of the compound is $\ce{Fe_1O}_{1.5}$
Multiply each of the moles by the smallest whole number that will convert each into a whole number.
Fe:O = 2 (1:1.5) = 2:3
Since the moles of $\ce{O}$ is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number.
Write the empirical formula. The empirical formula of the compound is $\ce{Fe_2O_3}$.
Think about your result. The subscripts are whole numbers and represent the mole ratio of the elements in the compound. The compound is the ionic compound iron (III) oxide.
Exercise $1$
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Answer
HgCl2
Summary
• A process is described for the calculation of the empirical formula of a compound, based on the percent composition of that compound. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.08%3A_Calculating_Empirical_Formulas_for_Compounds.txt |
Learning Objectives
• Understand the difference between empirical formulas and molecular formulas.
• Determine molecular formula from percent composition and molar mass of a compound.
Below, we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people can distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way to tell glucose and sucrose apart is to determine the molar masses—this approach allows you to easily tell which compound is which.
Molecular Formulas
Molecular formulas give the kind and number of atoms of each element present in the molecular compound. In many cases, the molecular formula is the same as the empirical formula. The chemical formula will always be some integer multiple ($n$) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).
$\text{ Molecular Formula} = n (\text{Empirical formula}) \nonumber$
therefore
$n = \dfrac{\text{Molecular Formula}}{\text{Empirical Formula}} \nonumber$
The integer multiple, n, can also be obtained by dividing the molar mass, $MM$, of the compound by the empirical formula mass, $EFM$ (the molar mass represented by the empirical formula).
$n = \dfrac{MM ( molar mass)}{EFM (empirical formula molar mass)} \nonumber$
Table $1$ shows the comparison between the empirical and molecular formula of methane, acetic acid, and glucose, and the different values of n. The molecular formula of methane is $\ce{CH_4}$ and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is $\ce{C_2H_4O_2}$. Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is $\ce{C_6H_{12}O_6}$. The structures of both molecules are shown in Figure $2$. They are very different compounds, yet both have the same empirical formula of $\ce{CH_2O}$.
Table $1$: Molecular Formula and Empirical Formula of Various Compounds.
Name of Compound Molecular Formula Empirical Formula n
Methane $\ce{CH_4}$ $\ce{CH_4}$ 1
Acetic acid $\ce{C_2H_4O_2}$ $\ce{CH_2O}$ 2
Glucose $\ce{C_6H_{12}O_6}$ $\ce{CH_2O}$ 6
Empirical formulas can be determined from the percent composition of a compound as discussed in section 6.8. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:
1. Calculate the empirical formula molar mass (EFM).
2. Divide the molar mass of the compound by the empirical formula molar mass. The result should be a whole number or very close to a whole number.
3. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.
Example $1$
The empirical formula of a compound of boron and hydrogen is $\ce{BH_3}$. Its molar mass is $27.7 \: \text{g/mol}$. Determine the molecular formula of the compound.
Solution
Solutions to Example 6.9.1
Steps for Problem Solving Determine the molecular formula of $\ce{BH_3}$.
Identify the "given" information and what the problem is asking you to "find."
Given:
Empirical formula $= \ce{BH_3}$
Molar mass $= 27.7 \: \text{g/mol}$
Find: Molecular formula $= ?$
Calculate the empirical formula mass (EFM). $\text{Empirical formula molar mass (EFM)} = 13.84 \: \text{g/mol} \nonumber$
Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.
$\dfrac{\text{molar mass}}{\text{EFM}} = \dfrac{27.7 g/mol}{13.84 g/mol} = 2 \nonumber$
Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula. $\ce{BH_3} \times 2 = \ce{B_2H_6} \nonumber$
Write the molecular formula. The molecular formula of the compound is $\ce{B_2H_6}$.
Think about your result. The molar mass of the molecular formula matches the molar mass of the compound.
Exercise $1$
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What are the empirical and chemical formulas for ascorbic acid?
Answer Empirical Formula
C3H4O3
Answer Molecular Formula
C6H8O6
Summary
• A procedure is described that allows the calculation of the exact molecular formula for a compound. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/06%3A_Chemical_Composition/6.09%3A_Calculating_Molecular_Formulas_for_Compounds.txt |
• 7.2: Evidence of a Chemical Reaction
In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.
• 7.3: The Chemical Equation
A chemical reaction is the process in which one or more substances are changed into one or more new substances. Chemical reactions are represented by chemical equations. Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right.
• 7.4: How to Write Balanced Chemical Equations
In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products—they are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.
• 7.5: Aqueous Solutions and Solubility - Compounds Dissolved in Water
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes.
• 7.6: Precipitation Reactions
A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions.
• 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations
Precipitation is a process in which a solute separates from a supersaturated solution. In a chemical laboratory, precipitation usually refers to a solid crystallizing from a liquid solution, but in weather reports it applies to liquid or solid water separating from supersaturated air.
• 7.8: Acid–Base and Gas Evolution Reactions
A gas evolution reaction is a chemical process that produces a gas, such as oxygen or carbon dioxide.
• 7.9: Oxidation–Reduction Reactions
An oxidation-reduction reaction is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation is the full or partial loss of electrons or the gain of oxygen. Reduction is the full or partial gain of electrons or the loss of oxygen. A redox reaction is another term for an oxidation-reduction reaction.
• 7.10: Classifying Chemical Reactions
Chemical reactions are classified into types to help scientists analyze them, and also to help scientists predict what the products of the reaction will be. The five major types of chemical reactions are synthesis, decomposition, single replacement, double replacement, and combustion.
• 7.11: The Activity Series- Predicting Spontaneous Redox Reactions
Metals and halogens are ranked according to their ability to displace other metals or halogens below them in the series. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series.
07: Chemical Reactions
Learning Objectives
• Identify the evidence for chemical reactions.
In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.
To identify a chemical reaction, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include: reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure \(1\)).
Video\(1\): Evidence of a Chemical Reaction
Example \(1\): Evidence of a Chemical Reaction
Which of the following is a chemical reaction?
1. Freezing liquid mercury.
2. Adding yellow to blue to make green.
3. Cutting a piece of paper into two pieces.
4. Dropping a sliced orange into a vat of sodium dydroxide.
5. Filling a balloon with natural air.
Solution
A, B, C, & E involve only physical changes. A sliced orange has acid (citric acid) that can react with sodium hydroxide, so the answer is D.
Exercise \(1\)
Which of the following is a chemical reaction?
1. Painting a wall blue.
2. A bicycle rusting.
3. Ice cream melting.
4. Scratching a key across a desk.
5. Making a sand castle.
Answer
B
Example \(2\): Evidence of a Chemical Reaction
Which of the following is not a chemical reaction?
1. Shattering glass with a baseball.
2. Corroding metal.
3. Fireworks exploding.
4. Lighting a match.
5. Baking a cake.
Solution
Shattering glass with a baseball results in glass broken into many pieces but no chemical change happens, so the answer is A.
Exercise \(2\)
Which of the following is NOT a chemical reaction?
1. Frying an egg.
2. Slicing carrots.
3. A Macbook falling out of a window.
4. Creating ATP in the human body.
5. Dropping a fizzy tablet into a glass of water.
Answer
B and C
Summary
Chemical reactions can be identified via a wide range of different observable factors including change in color, energy change (temperature change or light produced), gas production, formation of precipitate and change in properties. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.02%3A_Evidence_of_a_Chemical_Reaction.txt |
Learning Objectives
• Identify the reactants and products in any chemical reaction.
• Convert word equations into chemical equations.
• Use the common symbols, $\left( s \right)$, $\left( l \right)$, $\left( g \right)$, $\left( aq \right)$, and $\rightarrow$ appropriately when writing a chemical reaction.
In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.
Reactants and Products
To describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants and the substances present at the end are called products.
Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are just that—what is produced—or the result of what happens to the reactants when we put them together in the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter, sugar, vanilla, baking soda, salt, egg, and chocolate chips. What would be the products? Cookies! The reaction vessel would be our mixing bowl.
$\underbrace{\text{Flour} + \text{Butter} + \text{Sugar} + \text{Vanilla} + \text{Baking Soda} + \text{Eggs} + \text{Chocolate Chips}}_{\text{Ingredients = Reactants}} \rightarrow \underbrace{\text{Cookies}}_{\text{Product}} \nonumber$
Writing Chemical Equations
When sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur dioxide and oxygen, $\ce{SO_2} + \ce{O_2}$, are reactants and sulfur trioxide, $\ce{SO_3}$, is the product.
$\underbrace{\ce{2 SO2(g) + O2(g) }}_{\text{Reactants}} \rightarrow \underbrace{\ce{2SO3(g)}}_{\text{Products}} \nonumber$
In chemical reactions, the reactants are found before the symbol "$\rightarrow$" and the products are found after the symbol "$\rightarrow$". The general equation for a reaction is:
$\text{Reactants } \rightarrow \text{Products} \nonumber$
There are a few special symbols that we need to know in order to "talk" in chemical shorthand. In the table below is the summary of the major symbols used in chemical equations. Table $1$ shows a listing of symbols used in chemical equations.
Table $1$: Symbols Used in Chemical Equations
Symbol Description Symbol Description
$+$ used to separate multiple reactants or products $\left( s \right)$ reactant or product in the solid state
$\rightarrow$ yield sign; separates reactants from products $\left( l \right)$ reactant or product in the liquid state
$\rightleftharpoons$ replaces the yield sign for reversible reactions that reach equilibrium $\left( g \right)$ reactant or product in the gas state
$\overset{\ce{Pt}}{\rightarrow}$ formula written above the arrow is used as a catalyst in the reaction $\left( aq \right)$ reactant or product in an aqueous solution (dissolved in water)
$\overset{\Delta}{\rightarrow}$ triangle indicates that the reaction is being heated
Chemists have a choice of methods for describing a chemical reaction.
1. They could draw a picture of the chemical reaction.
2. They could write a word equation for the chemical reaction:
"Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor."
3. They could write the equation in chemical shorthand.
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right) \nonumber$
In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products, while symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations.
We could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand we could write:
$\ce{Ca(NO_3)_2} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Ca(OH)_2} \left( s \right) + 2 \ce{NaNO_3} \left( aq \right) \nonumber$
How much easier is that to read? Let's try it in reverse. Look at the following reaction in shorthand and write the word equation for the reaction:
$\ce{Cu} \left( s \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Cu(NO_3)_2} \left( aq \right) + \ce{Ag} \left( s \right) \nonumber$
The word equation for this reaction might read something like "solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper (II) nitrate with solid silver."
To turn word equations into symbolic equations, we need to follow the given steps:
1. Identify the reactants and products. This will help you know which symbols go on each side of the arrow and where the $+$ signs go.
2. Write the correct formulas for all compounds. You will need to use the rules you learned in Chapter 5 (including making all ionic compounds charge balanced).
3. Write the correct formulas for all elements. Usually this is given straight off of the periodic table. However, there are seven elements that are considered diatomic, meaning that they are always found in pairs in nature. They include those elements listed in the table.
Table $1$: Diatomic Elements
Element Name Hydrogen Nitrogen Oxygen Fluorine Chlorine Bromine Iodine
Formula $H_2$ $N_2$ $O_2$ $F_2$ $Cl_2$ $Br_2$ $I_2$
Example $1$
Transfer the following symbolic equations into word equations or word equations into symbolic equations.
1. $\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$
2. Gaseous propane, $\ce{C_3H_8}$, burns in oxygen gas to produce gaseous carbon dioxide and liquid water.
3. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide.
Solution
a. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water.
b. Reactants: propane ($\ce{C_3H_8}$) and oxygen ($\ce{O_2}$)
Product: carbon dioxide ($\ce{CO_2}$) and water ($\ce{H_2O}$)
$\ce{C_3H_8} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber$
c. Reactants: hydrogen fluoride and potassium carbonate
Products: potassium fluoride, water, and carbon dioxide
$\ce{HF} \left( g \right) + \ce{K_2CO_3} \left( aq \right) \rightarrow \ce{KF} \left( aq \right) + \ce{H_2O} \left( l \right) + \ce{CO_2} \left( g \right) \nonumber$
Exercise $1$
Transfer the following symbolic equations into word equations or word equations into symbolic equations.
1. Hydrogen gas reacts with nitrogen gas to produce gaseous ammonia.
2. $\ce{HCl} \left( aq \right) + \ce{LiOH} \left( aq \right) \rightarrow \ce{LiCl} \left( aq \right) + \ce{H_2O} \left( l \right)$
3. Copper metal is heated with oxygen gas to produce solid copper(II) oxide.
Answer a
$H_2 (g) + N_2 (g) \rightarrow NH_3 (g)$
Answer b
An aqueous solution of hydrochloric acid reacts with an aqueous solution of lithium hydroxide to produce an aqueous solution of lithium chloride and liquid water.
Answer c
$Cu (s) + O_2 (g) \rightarrow CuO (s)$
Summary
• A chemical reaction is the process by which one or more substances are changed into one or more new substances.
• Chemical reactions are represented by chemical equations.
• Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.03%3A_The_Chemical_Equation.txt |
Learning Objectives
• Explain the roles of subscripts and coefficients in chemical equations.
• Balance a chemical equation when given the unbalanced equation.
• Explain the role of the Law of Conservation of Mass in a chemical reaction.
Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products—they are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.
Coefficients and Subscripts
There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products; and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced.
The subscripts are part of the formulas and once the formulas for the reactants and products are determined, the subscripts may not be changed. The coefficients indicate the number of each substance involved in the reaction and may be changed in order to balance the equation. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper (II) nitrate and two atoms of solid silver.
Balancing a Chemical Equation
Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $1$.
Original molecule H2O: if the coefficient 2 is added in front, that makes 2 water molecules; but if the subscript 2 is added to make H2O2, that's hydrogen peroxide.
The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.
Steps in Balancing a Chemical Equation
1. Identify the most complex substance.
2. Beginning with that substance, choose an element(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides.
3. Balance polyatomic ions (if present on both sides of the chemical equation) as a unit.
4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
Example $1$: Combustion of Heptane
Balance the chemical equation for the combustion of Heptane ($\ce{C_7H_{16}}$).
$\ce{C_7H_{16} (l) + O_2 (g) → CO_2 (g) + H_2O (g) } \nonumber$
Solution
Solutions to Example 7.4.1
Steps Example
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $C_7H_{16}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients.
a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:
$\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber$
• 7 carbon atoms on both reactant and product sides
b. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side:
$\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber$
• 16 hydrogen atoms on both reactant and product sides
3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
4. Balance the remaining atoms.
The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:
$\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber$
• 22 oxygen atoms on both reactant and product sides
5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.
Example $2$: Combustion of Isooctane
Combustion of Isooctane ($\ce{C_8H_{18}}$)
$\ce{C8H18 (l) + O2 (g) -> CO_2 (g) + H_2O(g)} \nonumber$
Solution
The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water.
Solutions to Example 7.4.2
Steps Example
1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $\ce{C8H18}$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.
2. Adjust the coefficients.
a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:
$\ce{C8H18 (l) + O2 (g) -> }\underline{8} \ce{ CO2 (g) + H2O(g)}\nonumber$
• 8 carbon atoms on both reactant and product sides
b. 18 hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:
$\ce{C8H18 (l) + O2 (g) -> 8CO2 (g) + }\underline{9} \ce{ H2O(g)} \nonumber$
• 18 hydrogen atoms on both reactant and product sides
3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.
4. Balance the remaining atoms.
The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient ($\dfrac{25}{2}$) to balance the oxygen atoms:
$\ce{C8H18 (l) + } \underline{ \dfrac{25}{2} } \ce{O2 (g)→ 8CO2 (g) + 9H2O(g) }\nonumber$
• 25 oxygen atoms on both reactant and product sides
The equation is now balanced, but we usually write equations with whole number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2:
$\underline{2} \ce{C8H18(l) + } \underline{25} \ce{O2(g) ->} \underline{16} \ce{CO2(g) + } \underline{18} \ce{H2O(g)} \nonumber$
5. Check your work.
The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.
Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.
Example $3$: Precipitation of Lead (II) Chloride
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.
Solution
Solutions to Example 7.4.3
Steps Example
1. Identify the most complex substance.
The most complex substance is lead (II) chloride.
$\ce{Pb(NO3)2(aq) + NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$
2. Adjust the coefficients.
There are twice as many chloride ions in the reactants as there are in the products. Place a 2 in front of the NaCl in order to balance the chloride ions.
$\ce{Pb(NO3)2(aq) + }\underline{ 2} \ce{NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$
• 1 Pb atom on both reactant and product sides
• 2 Na atoms on reactant side, 1 Na atom on product side
• 2 Cl atoms on both reactant and product sides
3. Balance polyatomic ions as a unit.
The nitrate ions are still unbalanced. Place a 2 in front of the NaNO3. The result is:
$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → } \underline {2} \ce{NaNO3(aq) + PbCl2(s)} \nonumber$
• 1 Pb atom on both reactant and product sides
• 2 Na atoms on both reactant and product sides
• 2 Cl atoms on both reactant and product sides
• 2 NO3- atoms on both reactant and product sides
4. Balance the remaining atoms. There is no need to balance the remaining atoms because they are already balanced.
5. Check your work.
$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)} \nonumber$
• 1 Pb atom on both reactant and product sides
• 2 Na atoms on both reactant and product sides
• 2 Cl atoms on both reactant and product sides
• 2 NO3- atoms on both reactant and product sides
Exercise $1$
Is each chemical equation balanced?
1. $\ce{2Hg(ℓ)+ O_2(g) \rightarrow Hg_2O_2(s)}$
2. $\ce{C_2H_4(g) + 2O_2(g)→ 2CO_2(g) + 2H_2O(g)}$
3. $\ce{Mg(NO_3)_2(s) + 2Li (s) \rightarrow Mg(s)+ 2LiNO_3(s)}$
Answer a
yes
Answer b
no
Answer c
yes
Exercise $2$
Balance the following chemical equations.
1. $\ce{N2 (g) + O2 (g) → NO2 (g) }$
2. $\ce{Pb(NO3)2(aq) + FeCl3(aq) → Fe(NO3)3(aq) + PbCl2(s)}$
3. $\ce{C6H14(l) + O2(g)→ CO2(g) + H2O(g)}$
Answer a
N2 (g) + 2O2 (g) → 2NO2 (g)
Answer b
3Pb(NO3)2(aq) + 2FeCl3(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Answer c
2C6H14(l) + 19O2(g)→ 12CO2(g) + 14H2O(g)
Summary
• To be useful, chemical equations must always be balanced. Balanced chemical equations have the same number and type of each atom on both sides of the equation.
• The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.
Vocabulary
• Chemical reaction - The process in which one or more substances are changed into one or more new substances.
• Reactants - The starting materials in a reaction.
• Products - Materials present at the end of a reaction.
• Balanced chemical equation - A chemical equation in which the number of each type of atom is equal on the two sides of the equation.
• Subscripts - Part of the chemical formulas of the reactants and products that indicate the number of atoms of the preceding element.
• Coefficient - A small whole number that appears in front of a formula in a balanced chemical equation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.04%3A_How_to_Write_Balanced_Chemical_Equations.txt |
Learning Objectives
• Define and give examples of electrolytes.
When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.
Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure $1$).
Ionic Electrolytes
Water and other polar molecules are attracted to ions, as shown in Figure $2$. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes.
Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K+ and Cl ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure $2$ shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system, as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat.
In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust).
Solubility Rules
Some combinations of aqueous reactants result in the formation of a solid precipitate as a product. However, some combinations will not produce such a product. If solutions of sodium nitrate and ammonium chloride are mixed, no reaction occurs. One could write a molecular equation showing a double-replacement reaction, but both products, sodium chloride and ammonium nitrate, are soluble and would remain in the solution as ions. Every ion is a spectator ion and there is no net ionic equation at all. It is useful to be able to predict when a precipitate will occur in a reaction. To do so, you can use a set of guidelines called the solubility rules (Tables $1$ and $2$).
Table $1$: Solubility Rules for Soluble Substances
Soluble in Water Important Exceptions (Insoluble)
All Group IA and NH4+ salts none
All nitrates, chlorates, perchlorates and acetates none
All sulfates CaSO4, BaSO4, SrSO4, PbSO4
All chlorides, bromides, and iodides AgX, Hg2X2, PbX2 (X= Cl, Br, or I)
Table $2$: Solubility Rules for Sparingly Soluble Substances
Sparingly Soluble in Water Important Exceptions (Soluble)
All carbonates and phosphates Group IA and NH4+ salts
All hydroxides Group IA and NH4+ salts; Ba2+, Sr2+, Ca2+ sparingly soluble
All sulfides Group IA, IIA and NH4+ salts; MgS, CaS, BaS sparingly soluble
All oxalates Group IA and NH4+ salts
Special note: The following electrolytes are of only moderate solubility in water: CH3COOAg, Ag2SO4, KClO4. They will precipitate only if rather concentrated solutions are used.
As an example on how to use the solubility rules, predict if a precipitate will form when solutions of cesium bromide and lead (II) nitrate are mixed.
$\ce{Cs^+} \left( aq \right) + \ce{Br^-} \left( aq \right) + \ce{Pb^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \rightarrow ? \nonumber$
The potential precipitates from a double-replacement reaction are cesium nitrate and lead (II) bromide. According to the solubility rules table, cesium nitrate is soluble because all compounds containing the nitrate ion, as well as all compounds containing the alkali metal ions, are soluble. Most compounds containing the bromide ion are soluble, but lead (II) is an exception. Therefore, the cesium and nitrate ions are spectator ions and the lead (II) bromide is a precipitate. The balanced net ionic reaction is:
$\ce{Pb^{2+}} \left( aq \right) + 2 \ce{Br^-} \left( aq \right) \rightarrow \ce{PbBr_2} \left( s \right) \nonumber$
Example $1$: Solubility
Classify each compound as soluble or insoluble
1. Zn(NO3)2
2. PbBr2
3. Sr3(PO4)2
Solution
1. All nitrates are soluble in water, so Zn(NO3)2 is soluble.
2. All bromides are soluble in water, except those combined with Pb2+, so PbBr2 is insoluble.
3. All phosphates are insoluble, so Sr3(PO4)2 is insoluble.
Exercise $1$: Solubility
Classify each compound as soluble or insoluble.
1. Mg(OH)2
2. KBr
3. Pb(NO3)2
Answer a
insoluble
Answer b
soluble
Answer c
soluble
Summary
Substances that dissolve in water to yield ions are called electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water. Solubility rules allow prediction of what products will be insoluble in water. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.05%3A_Aqueous_Solutions_and_Solubility_-_Compounds_Dissolved_in_Water.txt |
Learning Objectives
• To identify a precipitation reaction and predict solubility.
A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. When a colorless solution of silver nitrate is mixed with a yellow-orange solution of potassium dichromate, a reddish precipitate of silver dichromate is produced.
$\ce{ AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)} \label{4.2.1}$
This unbalanced equation has the general form of an exchange reaction:
$\ce{ AC + BD \rightarrow } \underset{insoluble}{\ce{AD}} + \ce{BC} \label{4.2.2}$
Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Precipitation reactions are used to isolate metals that have been extracted from their ores, and to recover precious metals for recycling.
Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of aqueous $NaCl$ solution is mixed with 500 mL of aqueous $KBr$ solution, the final solution has a volume of 1.00 L and contains $\ce{Na^{+}(aq)}$, $\ce{Cl^{−}(aq)}$, $\ce{K^{+}(aq)}$, and $\ce{Br^{−}(aq)}$. As you will see in (Figure $1$), none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other.
Predicting Precipitation Reactions
A precipitation reaction occurs when a solid precipitate forms after mixing two strong electrolyte solutions. As stated previously, if none of the species in the solution reacts then no net reaction occurred.
Predict what will happen when aqueous solutions of barium chloride and lithium sulfate are mixed.
Change the partners of the anions and cations on the reactant side to form new compounds (products):
Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains $\ce{Ba^{2+}}$, $\ce{Cl^{−}}$, $\ce{Li^{+}}$, and $\ce{SO4^{2−}}$ ions. The only possible exchange reaction is to form $\ce{LiCl}$ and $\ce{BaSO4}$.
Correct the formulas of the products based on the charges of the ions.
No need to correct the formula as both compounds already have their charges balanced.
$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4 + LiCl} \nonumber$
Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate.
$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4(s) + LiCl(aq)} \nonumber$
Table 7.5.1 from the previous section shows that $\ce{LiCl}$ is soluble in water, but $\ce{BaSO4}$ is not soluble in water.
Balance the equation:
$\ce{BaCl_2(aq) + Li_2SO_4(aq) \rightarrow BaSO_4(s) + 2LiCl(aq)} \nonumber$
Although soluble barium salts are toxic, $\ce{BaSO4}$ is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine $\ce{BaSO4}$ particles in water.
Example $1$
Predict what will happen if aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
Solution
Solutions to Example 7.6.1
Steps Example
Change the partners of the anions and cations on the reactant side to form new compounds (products).
Correct the formulas of the products based on the charges of the ions. $\ce{RbOH(aq) + CoCl2(aq) \rightarrow RbCl + Co(OH)_2}$
Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate. $\ce{RbOH(aq) + CoCl_2(aq) \rightarrow RbCl(aq) + Co(OH)_2(s)}$
Balance the equation.
Coefficients already balanced.
$\ce{RbOH(aq) + CoCl_2(aq) \rightarrow RbCl(aq) + Co(OH)_2(s)}$
Example $2$
Predict what will happen if aqueous solutions of strontium bromide and aluminum nitrate are mixed.
Solution
Solutions for Example 7.6.2
Steps Example
Change the partners of the anions and cations on the reactant side to form new compounds (products).
Correct the formulas of the products based on the charges of the ions. $\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow Sr(NO_3)_2 + AlBr_3}$
Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate.
$\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow Sr(NO_3)_2(aq) + AlBr_3(aq)}$
According to Table 7.5.1 from the previous section, both $\ce{AlBr3}$ (rule 4) and $\ce{Sr(NO3)2}$ (rule 2) are soluble.
If all possible products are soluble, then no net reaction will occur.
$\ce{SrBr_2(aq) + Al(NO_3)_3(aq) \rightarrow}$
NO REACTION
Exercise $2$
Using the information in Table 7.5.1 from the previous section, predict what will happen in each case involving strong electrolytes.
1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.
Answer a
Fe(OH)2 precipitate is formed.
Answer b
Hg3(PO4)2 precipitate is formed.
Answer c
No Reaction.
Answer d
CaCO3 is precipitate formed.
Summary
In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when two electrolyte solutions are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.06%3A_Precipitation_Reactions.txt |
A typical precipitation reaction occurs when an aqueous solution of barium chloride is mixed with one containing sodium sulfate. The complete chemical equation can be written to describe what happens, and such an equation is useful in making chemical calculations.
$\underbrace{\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}}_{\text{Complete Chemical Equation}}\label{1}$
However, Equation $\ref{1}$ does not really represent the microscopic particles (that is, the ions) present in the solution. Below is the complete ionic equation:
$\underbrace{\ce{Ba^{2+}(aq) + \overbrace{2Cl^{-}(aq)}^{spectator} + \overbrace{2Na^{+}(aq)}^{spectator} + SO4^{2-}(aq) -> BaSO4(s) + \overbrace{2Na^{+}(aq)}^{spectator} + \overbrace{Cl^{-}(aq)}}^{\text{spectator}}}_{\text{Complete Ionic Equation}}\label{2}$
Equation $\ref{2}$ is rather cumbersome and includes so many different ions that it may be confusing. In any case, we are often interested in the independent behavior of ions, not the specific compound from which they came. A precipitate of $\ce{BaSO4(s)}$ will form when any solution containing $\ce{Ba^{2+}(aq)}$ is mixed with any solution containing $\ce{SO4^{2–}(aq)}$ (provided concentrations are not extremely small). This happens independently of the $\ce{Cl^{–}(aq)}$ and $\ce{Na^+(aq)}$ ions in Equation $\ref{2}$. These ions are called spectator ions because they do not participate in the reaction. When we want to emphasize the independent behavior of ions, a net ionic equation is written, omitting the spectator ions. For precipitation of $\ce{BaSO_4}$ the net ionic equation is
$\underbrace{\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}}_{\text{Net Ionic Equation}} \label{3}$
Example $1$
1. When a solution of $\ce{AgNO3}$ is added to a solution of $\ce{CaCl2}$, insoluble $\ce{AgCl}$ precipitates. Write three equations (complete chemical equation, complete ionic equation, and net ionic equation) that describe this process.
2. Write the balanced net ionic equation to describe any reaction that occurs when the solutions of $\ce{Na2SO4}$ and $\ce{NH4I}$ are mixed.
Solution
Solutions to Example 7.7.1
Equation Type Example $\PageIndex{1a}$ Example $\PageIndex{1b}$
Complete Chemical Equation
$\ce{2AgNO3(aq) + CaCl2(aq) ->} \ \ce{2AgCl(s) + Ca(NO3)2(aq)}$
The proper states and formulas of all products are written and the chemical equation is balanced.
$\ce{Na2SO4(aq) + 2NH4I(aq) ->} \ \ce{2NaI(aq) + (NH4)2SO4(aq)}$
Both products are aqueous so there is no net ionic equation that can be written.
Complete Ionic Equation
$\ce{2Ag^+(aq) + 2NO3^{-}(aq) + Ca^{2+}(aq) + 2Cl^{-}(aq) -> } \ \ce{2AgCl(s) + Ca^{2+}(aq) + 2NO3^{-}(aq)}$
AgCl is a solid so it does not break up into ions in solution.
Net Ionic Equation
$\ce{Ag^+(aq) + Cl^{-} (aq) -> AgCl(s)}$
All spectator ions are removed.
The chemical equation is written using the lowest common coefficients.
$\ce{NaI}$ and $\ce{(NH4)2SO4}$ are both soluble.
No net ionic equation.
There is no reaction.
The occurrence or nonoccurrence of precipitates can be used to detect the presence or absence of various species in solution. A $\ce{BaCl2}$ solution, for instance, is often used as a test for the presence of $\ce{SO4^{2–}(aq)}$ ions. There are several insoluble salts of $\ce{Ba}$, but they all dissolve in dilute acid except for $\ce{BaSO4}$. Thus, if $\ce{BaCl2}$ solution is added to an unknown solution which has previously been acidified, the occurrence of a white precipitate is proof of the presence of the $\ce{SO4^{2–}}$ ion.
$\ce{AgNO3}$ solutions are often used in a similar way to test for halide ions. If $\ce{AgNO3}$ solution is added to an acidified unknown solution, a white precipitate indicates the presence of $\ce{Cl^{–}}$ ions, a cream-colored precipitate indicates the presence of $\ce{Br^{–}}$ ions, and a yellow precipitate indicates the presence of $\ce{I^{–}}$ ions (Figure $1$). Further tests can then be made to see whether perhaps a mixture of these ions is present. When $\ce{AgNO_3}$ is added to tap water, a white precipitate is almost always formed. The $\ce{Cl^{–}}$ ions in tap water usually come from the $\ce{Cl2}$ which is added to municipal water supplies to kill microorganisms.
Exercise $1$
Write balanced net ionic equations to describe any reaction that occurs when the following solutions are mixed.
1. $\ce{K2CO3 + SrCl2}$
2. $\ce{FeSO4 + Ba(NO3)2 }$
Answer a
$\ce{Sr^{2+}(aq) + CO3^{2-} (aq) -> SrCO3 (s)} \nonumber$
Answer b
$\ce{Ba^{2+}(aq) + SO4^{2-} (aq) -> Ba(SO4) (s)} \nonumber$
Precipitates are also used for quantitative analysis of solutions, that is, to determine the amount of solute or the mass of solute in a given solution. For this purpose it is often convenient to use the first of the three types of equations described above. Then the rules of stoichiometry may be applied. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.07%3A_Writing_Chemical_Equations_for_Reactions_in_Solution-_Molecular_Complete_Ionic_and_Net_Ionic_Equations.txt |
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