Chapter 25 Organic and Biological Chemistry

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Chapter 25Organic and Biological Chemistry
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten
John D. Bookstaver St. Charles Community
College St. Peters, MO ? 2006, Prentice Hall, Inc.
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Organic Chemistry

• The chemistry of carbon compounds.
• Carbon has the ability to form long chains.
• Without this property, large biomolecules such as
  proteins, lipids, carbohydrates, and nucleic
  acids could not form.

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Structure of Carbon Compounds

• There are three hybridization states and
  geometries found in organic compounds
• sp3 Tetrahedral
• sp2 Trigonal planar
• sp Linear

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Hydrocarbons

• Four basic types
• Alkanes
• Alkenes
• Alkynes
• Aromatic hydrocarbons

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Alkanes

• Only single bonds.
• Saturated hydrocarbons.
• Saturated with hydrogens.

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Formulas

• Lewis structures of alkanes look like this.
• Also called structural formulas.
• Often not convenient, though

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Formulas

• so more often condensed formulas are used.

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Properties of Alkanes

• Only van der Waals force London force.
• Boiling point increases with length of chain.

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Structure of Alkanes

• Carbons in alkanes sp3 hybrids.
• Tetrahedral geometry.
• 109.5 bond angles.

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Structure of Alkanes

• Only ?-bonds in alkanes
• Free rotation about CC bonds.

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Isomers

• Have same molecular formulas, but atoms are
  bonded in different order.

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Organic Nomenclature

• Three parts to a compound name
• Base Tells how many carbons are in the longest
  continuous chain.

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Organic Nomenclature

• Three parts to a compound name
• Base Tells how many carbons are in the longest
  continuous chain.
• Suffix Tells what type of compound it is.

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Organic Nomenclature

• Three parts to a compound name
• Base Tells how many carbons are in the longest
  continuous chain.
• Suffix Tells what type of compound it is.
• Prefix Tells what groups are attached to chain.

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To Name a Compound

• Find the longest chain in the molecule.
• Number the chain from the end nearest the first
  substituent encountered.
• List the substituents as a prefix along with the
  number(s) of the carbon(s) to which they are
  attached.

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To Name a Compound

• If there is more than one type of substituent in
  the molecule, list them alphabetically.

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SAMPLE EXERCISE 25.1 Naming Alkanes
Solution Analyze We are given the structural
formula of an alkane and asked to give its
name. Plan Because the hydrocarbon is an alkane,
its name ends in -ane. The name of the parent
hydrocarbon is based on the longest continuous
chain of carbon atoms, as summarized in Table
25.1. Branches are alkyl groups, named after the
number of C atoms in the branch and located by
counting C atoms along the longest continuous
chain.
The parent compound is thus heptane. There are
two CH3 (methyl) groups that branch off the main
chain. Hence, this compound is a dimethylheptane.
To specify the location of the two methyl groups,
we must number the C atoms from the end that
gives the lowest number possible to the carbons
bearing side chains. This means that we should
start numbering with the upper left carbon. There
is a methyl group on carbon 3, and one on carbon
4. The compound is thus 3,4-dimethylheptane.
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SAMPLE EXERCISE 25.1 continued
Answer 2,4-dimethylpentane
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SAMPLE EXERCISE 25.2 Writing Condensed Structural
Formulas
Write the condensed structural formula for
3-ethyl-2-methylpentane.
Solution Analyze We are given the systematic
name for a hydrocarbon and asked to write its
structural formula. Plan Because the compounds
name ends in -ane, it is an alkane, meaning that
all the carboncarbon bonds are single bonds. The
parent hydrocarbon is pentane, indicating five C
atoms (Table 25.1). There are two alkyl groups
specified, an ethyl group (two carbon atoms,
C2H5) and a methyl group (one carbon atom, CH3).
Counting from left to right along the five-carbon
chain, the ethyl group will be attached to the
third C atom and the methyl group will be
attached to the second C atom.
Solve We begin by writing a string of five C
atoms attached to each other by single bonds.
These represent the backbone of the parent
pentane chain CCCCC
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SAMPLE EXERCISE 25.2 continued
The formula can be written even more concisely in
the following style CH3CH(CH3) CH(C2H5)
CH2CH3 In this formula the branching alkyl
groups are indicated in parentheses.
PRACTICE EXERCISE Write the condensed structural
formula for 2,3-dimethylhexane.
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Cycloalkanes

• Carbon can also form ringed structures.
• Five- and six-membered rings are most stable.
• Can take on conformation in which angles are very
  close to tetrahedral angle.
• Smaller rings are quite strained.

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Reactions of Alkanes

• Rather unreactive due to presence of only CC and
  CH ?-bonds.
• Therefore, great nonpolar solvents.

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Alkenes

• Contain at least one carboncarbon double bond.
• Unsaturated.
• Have fewer than maximum number of hydrogens.

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Structure of Alkenes

• Unlike alkanes, alkenes cannot rotate freely
  about the double bond.
• Side-to-side overlap makes this impossible
  without breaking ?-bond.

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Structure of Alkenes

• This creates geometric isomers, which differ
  from each other in the spatial arrangement of
  groups about the double bond.

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Properties of Alkenes

• Structure also affects physical properties of
  alkenes.

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Nomenclature of Alkenes

• Chain numbered so double bond gets smallest
  possible number.
• cis- alkenes have carbons in chain on same side
  of molecule.
• trans- alkenes have carbons in chain on opposite
  side of molecule.

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Reactions of Alkenes

• Addition Reactions
• Two atoms (e.g., bromine) add across the double
  bond.
• One ?-bond and one ?-bond are replaced by two
  ?-bonds therefore, ?H is negative.

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Mechanism of Addition Reactions

• Two-step mechanism
• First step is slow, rate-determining step.
• Second step is fast.

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Mechanism of Addition Reactions

• In first step, ?-bond breaks and new CH bond
  and cation form.

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Mechanism of Addition Reactions

• In second step, new bond forms between negative
  bromide ion and positive carbon.

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SAMPLE EXERCISE 25.5 Identifying the Product of a
Hydrogenation Reaction
Write the structural formula for the product of
the hydrogenation of 3-methyl-1-pentene.
Solution Analyze We are asked to predict the
compound formed when a particular alkene
undergoes hydrogenation (reaction with H2). Plan
To determine the structural formula of the
reaction product, we must first write the
structural formula or Lewis structure of the
reactant. In the hydrogenation of the alkene, H2
adds to the double bond, producing an alkane.
(That is, each carbon atom of the double bond
forms a bond to an H atom, and the double bond is
converted to a single bond.)
Comment The longest chain in the product alkane
has five carbon atoms its name is therefore
3-methylpentane.
PRACTICE EXERCISE Addition of HCl to an alkene
forms 2-chloropropane. What is the alkene?
Answer propene
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Alkynes

• Contain at least one carboncarbon triple bond.
• Carbons in triple bond sp-hybridized and have
  linear geometry.
• Also unsaturated.

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Nomenclature of Alkynes
4-methyl-2-pentyne

• Analogous to naming of alkenes.
• Suffix is -yne rather than ene.

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Reactions of Alkynes

• Undergo many of the same reactions alkenes do.
• As with alkenes, impetus for reaction is
  replacement of ?-bonds with ?-bonds.

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SAMPLE EXERCISE 25.3 Drawing Isomers
Draw all the structural and geometric isomers of
pentene, C5H10, that have an unbranched
hydrocarbon chain.
Solution Analyze We are asked to draw all the
isomers (both structural and geometric) for an
alkene with a five-carbon chain. Plan Because
the compound is named pentene and not pentadiene
or pentatriene, we know that the five-carbon
chain contains only one carboncarbon double
bond. Thus, we can begin by first placing the
double bond in various locations along the chain,
remembering that the chain can be numbered from
either end. After finding the different distinct
locations for the double bond, we can consider
whether the molecule can have cis and trans
isomers.
Solve There can be a double bond after either
the first carbon (1-pentene) or second carbon
(2-pentene). These are the only two possibilities
because the chain can be numbered from either
end. (Thus, what we might erroneously call
4-pentene is actually 1-pentene, as seen by
numbering the carbon chain from the other end.)
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SAMPLE EXERCISE 25.3 continued
Because the first C atom in 1-pentene is bonded
to two H atoms, there are no cis-trans isomers.
On the other hand, there are cis and trans
isomers for 2-pentene. Thus, the three isomers
for pentene are
(You should convince yourself that cis- or
trans-3-pentene is identical to cis- or
trans-2-pentene, respectively.)
PRACTICE EXERCISE How many straight-chain isomers
are there of hexene, C6H12?
Answer five (1-hexene, cis-2-hexene,
trans-2-hexene, cis-3-hexene, trans-3-hexene)
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SAMPLE EXERCISE 25.4 Naming Unsaturated
Hydrocarbons
Name the following compounds
Solution Analyze We are given the structural
formulas for two compounds, the first an alkene
and the second an alkyne, and asked to name the
compounds. Plan In each case the name is based
on the number of carbon atoms in the longest
continuous carbon chain that contains the
multiple bond. In the case of the alkene, care
must be taken to indicate whether cis-trans
isomerism is possible and, if so, which isomer is
given.
Solve (a) The longest continuous chain of
carbons that contains the double bond is seven in
length. The parent compound is therefore heptene.
Because the double bond begins at carbon 2
(numbering from the end closest to the double
bond), the parent hydrocarbon chain is named
2-heptene. A methyl group is found at carbon atom
4. Thus, the compound is 4-methyl-2-heptene. The
geometrical configuration at the double bond is
cis (that is, the alkyl groups are bonded to the
double bond on the same side). Thus, the full
name is 4-methyl-cis-2-heptene.
(b) The longest continuous chain of carbon atoms
containing the triple bond is six, so this
compound is a derivative of hexyne. The triple
bond comes after the first carbon (numbering from
the right), making it a derivative of 1-hexyne.
The branch from the hexyne chain contains three
carbon atoms, making it a propyl group. Because
it is located on the third carbon atom of the
hexyne chain, the molecule is 3-propyl-1-hexyne.
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SAMPLE EXERCISE 25.4 continued
PRACTICE EXERCISE Draw the condensed structural
formula for 4-methyl-2-pentyne.
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Aromatic Hydrocarbons

• Cyclic hydrocarbons.
• p-Orbital on each atom.
• Molecule is planar.
• Odd number of electron pairs in ?-system.

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Aromatic Nomenclature

• Many aromatic hydrocarbons are known by their
  common names.

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Reactions of Aromatic Compounds

• Unlike in alkenes and alkynes, ?-electrons do not
  sit between two atoms.
• Electrons are delocalized this stabilizes
  aromatic compounds.

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Reactions of Aromatic Compounds

• Due to stabilization, aromatic compounds do not
  undergo addition reactions they undergo
  substitution.
• Hydrogen is replaced by substituent.

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Structure of Aromatic Compounds

• Two substituents on a benzene ring could have
  three possible relationships
• ortho- On adjacent carbons.
• meta- One carbon between them.
• para- On opposite sides of ring.

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Reactions of Aromatic Compounds
Halogenation
Friedel-Crafts Reaction

• Reactions of aromatic compounds often require a
  catalyst.

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Functional Groups

• Term used to refer to parts of organic molecules
  where reactions tend to occur.

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Alcohols

• Contain one or more hydroxyl groups, OH

• Named from parent hydrocarbon suffix changed to
  -ol and number designates carbon to which
  hydroxyl is attached.

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Alcohols

• Much more acidic than hydrocarbons.
• pKa 15 for most alcohols.
• Aromatic alcohols have pKa 10.

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Ethers

• Tend to be quite unreactive.
• Therefore, they are good polar solvents.

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Carbonyl Compounds

• Contain CO double bond.
• Include many classes of compounds.

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Aldehydes

• At least one hydrogen attached to carbonyl
  carbon.

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Ketones

• Two carbons bonded to carbonyl carbon.

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Carboxylic Acids

• Have hydroxyl group bonded to carbonyl group.
• Tart tasting.
• Carboxylic acids are weak acids.

CH3COOH
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Carboxylic Acids
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Esters

• Products of reaction between carboxylic acids and
  alcohols.
• Found in many fruits and perfumes.

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SAMPLE EXERCISE 25.6 Naming Esters and Predicting
Hydrolysis Products
In a basic aqueous solution, esters react with
hydroxide ion to form the salt of the carboxylic
acid and the alcohol from which the ester is
constituted. Name each of the following esters,
and indicate the products of their reaction with
aqueous base.
Solution Analyze We are given two esters and
asked to name them and to predict the products
formed when they undergo hydrolysis (split into
an alcohol and carboxylate ion) in basic solution.
Plan Esters are formed by the condensation
reaction between an alcohol and a carboxylic
acid. To name an ester, we must analyze its
structure and determine the identities of the
alcohol and acid from which it is formed. We can
identify the alcohol by adding an OH to the alkyl
group attached to the O atom of the carboxyl
(COO) group. We can identify the acid by adding
an H group to the O atom of the carboxyl group.
Weve learned that the first part of an ester
name indicates the alcohol portion and the second
indicates the acid portion. The name conforms to
how the ester undergoes hydrolysis in base,
reacting with base to form an alcohol and a
carboxylate anion.
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SAMPLE EXERCISE 25.6 continued
The products are benzoate ion and ethanol.
The products are butyrate ion and phenol.
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SAMPLE EXERCISE 25.6 continued
PRACTICE EXERCISE Write the structural formula
for the ester formed from propyl alcohol and
propionic acid.
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Amides

• Formed by reaction of carboxylic acids with
  amines.

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Amines

• Organic bases.
• Generally have strong, unpleasant odors.

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Chirality

• Carbons with four different groups attached to
  them are handed, or chiral.
• Optical isomers or stereoisomers
• If one stereoisomer is right-handed, its
  enantiomer is left-handed.

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Chirality
S-ibuprofen

• Many pharmaceuticals are chiral.
• Often only one enantiomer is clinically active.

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SAMPLE EXERCISE 25.8 Identifying Chiral Centers
How many chiral carbon atoms are there in the
open-chain form of glucose (Figure 25.28)?
Solution Analyze We are given the structure of
glucose and asked to determine the number of
chiral carbons in the molecule. Plan A chiral
carbon has four different groups attached
(Section 25.7). We need to identify those carbon
atoms in glucose.
Carbon atoms 1 and 6 have only three different
substituents on them. Thus, there are four chiral
carbon atoms in the glucose molecule.
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SAMPLE EXERCISE 25.8 continued
PRACTICE EXERCISE How many chiral carbon atoms
are there in the open-chain form of fructose
(Figure 25.28)?
Answer three
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Amino Acids and Proteins

• Proteins are polymers of ?-amino acids.
• A condensation reaction between the amine end of
  one amino acid and the acid end of another
  produces a peptide bond.

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Amino Acids and Proteins

• Hydrogen bonding in peptide chains causes coils
  and helices in the chain.
• Kinking and folding of the coiled chain gives
  proteins a characteristic shape.

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Amino Acids and Proteins

• Most enzymes are proteins.
• The shape of the active site complements the
  shape of the substrate on which the enzyme
  acts?hence, the lock- and-key model.

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SAMPLE EXERCISE 25.7 Drawing the Structural
Formula of a Tripeptide
Draw the full structural formula for
alanylglycylserine.
Solution Analyze We are given the name of a
substance with peptide bonds and asked to write
its full structural formula. Plan The name of
this substance suggests that three amino
acidsalanine, glycine, and serinehave been
linked together, forming a tripeptide. Note that
the ending -yl has been added to each amino acid
except for the last one, serine. By convention,
the first-named amino acid (alanine, in this
case) has a free amino group and the last-named
one (serine) has a free carboxyl group. Thus, we
can construct the structural formula of the
tripeptide from its amino acid building blocks
(Figure 25.23).
We can abbreviate this tripeptide as Ala-Gly-Ser.
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SAMPLE EXERCISE 25.7 continued
Answer serylaspartic acid Ser-Asp
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Carbohydrates

• Simple sugars are polyhydroxy aldehydes or
  ketones.

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Carbohydrates

• In solution they form cyclic structures.
• These can form chains of sugars that form
  structural molecules such as starch and cellulose.

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Nucleic Acids

• Two of the building blocks of RNA and DNA are
  sugars (ribose or deoxyribose) and cyclic bases
  (adenine, guanine, cytosine, and thymine or
  uracil).

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Nucleic Acids

• These combine with a phosphate to form a
  nucleotide.

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Nucleic Acids

• Nucleotides combine to form the familiar
  double-helix form of the nucleic acids.

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SAMPLE INTEGRATIVE EXERCISE Putting Concepts
Together
It is formed in the body from carbohydrate
metabolism. In the muscle it is reduced to lactic
acid in the course of exertion. The
acid-dissociation constant for pyruvic acid is
3.2 ? 103. (a) Why does pyruvic acid have a
higher acid-dissociation constant than acetic
acid? (b) Would you expect pyruvic acid to exist
primarily as the neutral acid or as dissociated
ions in muscle tissue, assuming a pH of about 7.4
and an acid concentration of 2 ? 104 M? (c)
What would you predict for the solubility
properties of pyruvic acid? Explain. (d) What is
the hybridization of each carbon atom in pyruvic
acid? (e) Assuming H atoms as the reducing agent,
write a balanced chemical equation for the
reduction of pyruvic acid to lactic acid (Figure
25.17). (Although H atoms dont exist as such in
biochemical systems, biochemical reducing agents
deliver hydrogen for such reductions.)
Solution (a) The acid ionization constant for
pyruvic acid should be somewhat greater than that
of acetic acid because the carbonyl function on
the ?-carbon atom exerts an electron-withdrawing
effect on the carboxylic acid group. In the
COH bond system the electrons are shifted
from hydrogen, facilitating loss of the hydrogen
as a proton. (Section 16.10)
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SAMPLE INTEGRATIVE EXERCISE continued
Solving for x, we obtain x3.2 ? 103 4.0 ?
108 6.4 ? 107.
The second term in the brackets is negligible
compared to the first, so x Pv 6.4 ?
107/3.2 ? 103 2 ? 104 M. This is the initial
concentration of acid, which means that
essentially all the acid has dissociated. We
might have expected this result because the acid
is quite dilute and the acid-dissociation
constant is fairly high.
(c) Pyruvic acid should be quite soluble in
water because it has polar functional groups and
a small hydrocarbon component. It is miscible
with water, ethanol, and diethyl ether.
(d) The methyl group carbon has sp3
hybridization. The carbon carrying the carbonyl
group has sp2 hybridization because of the double
bond to oxygen. Similarly, the carboxylic acid
carbon is sp2 hybridized.
Essentially, the ketonic functional group has
been reduced to an alcohol.