UNIT 11:
Organic Chemistry-Structure
First, we need to review a few structural characteristics of the carbon atom. These are ideas which were part of your general chemistry courses, but it will help if we briefly restate them.
- Carbon is tetracovalent.
That means that a carbon atom typically makes four bonds to other atoms
and that these bonds are covalent--formed by sharing an electron pair
between the two atoms joined by the bond. Such arrangements provide eight
valence electrons for a carbon atom, so that it's
electronic configuration is like that of the very stable noble gas neon.
Similarly, hydrogen forms one
covalent bond, oxygen two, and nitrogen three.
- Carbon can form multiple covalent bonds. That is, a single carbon atom can form a double (to C, O or N) or triple (to C or N) bond to another atom. A double bond would involve two electron pairs between the bonded atoms and a triple bond would involve three electron pairs.
- Bonds between carbon and atoms other than carbon or hydrogen are polar. That is, in a bond between carbon and oxygen or nitrogen the electrons are closer to the more electronegative element (oxygen or nitrogen) than to the carbon, so the carbon has a slightly positive charge. (Fluorine is the most electronegative element, and the elements close to fluorine in the periodic table are also quite electronegative.)
- Bonds between one carbon atom and another and between a carbon and a hydrogen are non polar. That is, the electron pair forming the bond is quite evenly shared by the atoms.
- We can predict the geometry of the bonds around an atom by using the idea that electron pairs and groups of electron pairs (such as in double or triple bonds) repel each other (Valence Shell Electron Pair Repulsion--VSEPR--Theory).
1. What is the
difference between a structural formula and a molecular formula?
Answer:
Structural Formulas
are diagrams that show which atoms are bonded to which other atoms in a
molecule of a compound. There are two
different types of structural formulas:
(1)
Complete
Lewis Structures, and
(2)
Condensed
Structural Formulas
Lewis Structures
symbolize a bonding pair of electrons as a pair of dots or as a dash between
two bonding atoms. Lone pairs of nonbonding electrons are shown as pairs of
dots. Condensed Structural Formulas are written without showing all of the
individual bonds. Each central atom is
shown with the atoms bonded to it.
Molecular formulas
give the number of atoms of each element in one molecule of a compound.
Example:
1-butanol
Molecular Formula =
C4H10O
Condensed
Structural Formula =
CH3(CH2)3OH
H H H H
| | |
|
Lewis Structure = H─C
─ C ─ C ─ C ─ O ─ H
| | |
|
H H H H
2. What
is the difference between the condensed structural, the expanded structural and
the line structure/ line-angle drawing of a compound?
Answer:
Expanded structural formulas show all the covalent bonds, while condensed structural formulas show no covalent bonds or only selected bonds.
Let's look at an example case, one with several carbon atoms. A molecular formula is not enough to specify the structure of an organic compound which includes more than three carbons. We need information about which atoms, particularly carbons, nitrogens and oxygens, are connected to each other. This is often represented in a condensed formula like the one for butanal, also called butyraldehyde:
CH3CH2CH2CHO
To take a more detailed look at the structure of this compound, we need to expand its representation. We begin by ignoring the hydrogens and focusing on the carbons and the oxygen to arrive at a skeleton. A trial skeleton might be (where the unused valences are again shown as lines that connect to only one atom):
If we compare this skeleton to the condensed formula above, the left end carbon seems to be associated with three hydrogens in the condensed formula and there are three unused valences on the left end carbon as shown in the skeleton, so that matches well. Similarly, the two middle carbons have two hydrogens each, and there are two unused valences on each of those carbons in the skeleton. If we match all these up we arrive at:
The right end carbon and the oxygen are possibly troublesome. We have only one hydrogen left and there are three unused valences to deal with. Since oxygen has two valences and can form a double bond, we can try a double bond between carbon and oxygen and see if it helps (it does). That gives us this skeleton:
Adding the final hydrogen, we arrive at this expanded structure:
It is often inconvenient to show all the carbons and hydrogens in detail, especially since the parts of a molecule which are made up of only carbon and hydrogens joined by single bonds usually do not play a significant role in the reactions of that molecule. (These portions of the molecule are known as "R-groups.") We can represent these atoms and how they are connected by using an abbreviated structural type known as a bond-line structure (also called a line-angle formula, skeletal structure, or a stick figure ). Such a structure for the compound, butanal, that we just worked with is shown below:
In a bond-line structure (also called a line-angle formula, skeletal structure, or a stick figure ), carbons are shown by the "empty" ends of lines and by junctions (corners) between lines. Letters at the end of a line represent atoms of the designated heteroatom (Heteroatoms are atoms other than carbon or hydrogen. Hydrogen is often shown where necessary for clarity.) Double or triple bonds are represented by two or three parallel lines joining the same two atoms. Hydrogens are added as needed to fill up the remaining unused valences.
We can "flesh out" the skeleton above by first filling in the carbons at the corners and "empty" ends:
Then we count how many bond each carbon has showing, and add enough hydrogens to each carbon to bring its number of bonds up to four. (For other atoms like oxygen or nitrogen which can make more than one bond, hydrogens are added as needed to arrive at an appropriate number of bonds.) For example, the left end carbon has one bond showing, so we need to add three hydrogens:
Continuing in the same way with the middle carbons, each of which has two bonds showing, we arrive at the final expanded structure:
Notice that we didn't need to do anything to the right end carbon (the "carbonyl" carbon) or the oxygen, since these atoms already have the appropriate number of bonds showing.
Sometimes we'll represent a molecule by showing its R-groups in a condensed fashion and its reactive parts (functional groups) in an expanded fashion:
3.
What are structural isomers?
Answer:
In general Isomers are different compounds with
the same molecular formula. But as we
know from above, the molecular formula is not sufficient to specify the
structure of the compound. Different compounds
with different chemical properties can have the same molecular formula. Structural isomers (also called constitutional isomers) are compounds
with the same molecular formula that differ in their bonding sequence; that is,
their atoms are connected differently.
Example: n-butane and
isobutane
( Molecular formula C4H10 ) are structural
isomers.
n-butane (normal butane) has a condensed structural formula CH3CH2CH2CH3
isobutane has a condensed structural formula CH(CH3)3
The expanded structural formula for n-butane is CH3─CH2─CH2─CH3
CH3
|
The expanded structural formula for isobutane is CH3─CH─CH3
4.
What are the four types of tetrahedral carbon?
I have been
unable to find a reference to answer this rather vague question. Is he thinking of sp3 hybridization? Diamond versus graphite?
Who can tell? This is a very vague question.
My guess is probably wrong in that
it is likely too complex and not what the instructor intended. See my wild guess below. I recommend you look
to your textbook for the answer to this one.
1. acyclic
hydrocarbons, saturated and unsaturated, branched and straight-chain
2. monocyclic
hydrocarbons, substituted and
unsubstituted and aromatics
3. fused
polycyclic hydrocarbons
4. bridged hydrocarbons, bicyclic
systems, polycyclic systems, and hydrocarbon bridges
5.
What are functional groups?
Answer:
Functional groups
are the reactive nonalkane part of an organic molecule. Most nonalkane organic
compounds are characterized and classified by the functional groups they
contain. A molecule that contains a
functional group may be represented as the functional group with alkyl groups attached. An alkyl
group is the alkane portion of a molecule with one hydrogen atom removed to
allow bonding to the functional group.
Often the alkyl groups bonded to a functional group are often designated
using the symbol R to refer to the substituent.
The R group is merely an alkyl group.
Using this notation, we presume that the exact nature of the alkyl group
is not important.
6.
What are the names and structures of the most common functional groups?
Answer:
Other than combustion, Alkanes undergo very few reactions and are therefore not included as functional groups by most authors. When a molecule contains an alkane portion and a nonalkane portion, we often ignore the alkane portion because it is relatively unreactive. On the other hand, Alkenes contain one or more carbon-carbon double bond(s). Because a carbon-carbon double bond is relatively reactive, it is considered to be a functional group and thus Alkenes are often included in the list of functional groups. Some authors also include Alkynes and Aromatics as functional groups due to their reactivity. The major functional groups are named and shown below:
Class Structure Functional
Group
Alkenes R─CH=CH─R* carbon-carbon double bond
Alkynes R─C≡C─R* carbon-carbon triple bond
Alkyl Halide R─X X = F, Cl, Br, or I
Aromatics 
R benzene
ring
Alcohols R─OH hydroxyl group
Phenols OH hydroxyl
group on Aromatic ring
Thiols R─SH sulfhydril group
Ethers R─O─R* oxygen between two alkyl groups
║
Ketones R─C─R* carbonyl group
║
Aldehydes R─C─H carbonyl group
O
║
Carboxylic Acid R─C─OH carboxyl group
O
║
Esters R─C─O─R* carboalkoxy group
║
Amides R─C─NH2 carboxamide group
Amines R─NH2 amino group
Nitriles R─C≡N cyano group
Nitroalkanes R─NO2 nitro group
Is there any
difference between vanillin made synthetically and vanillin extracted from the
vanilla bean?
Answer: No, there is no difference.
Historically, when chemists discovered the Law of Definite Proportions at the beginning of the 19th century, it appeared that this law did not apply to the various compounds they had isolated from plant and animal sources. Carbon compounds can be so complex that the ratios of elements in them did not appear to be simple numbers. For example, ordinary table sugar has the molecular formula C12H22O11, not the kind of simple ratio seen with the oxides of copper, Cu2O or CuO, for example. Chemists imagined that organic compounds were held together by a mysterious "vital force".
The beginning of the end of the "vital force" hypothesis is generally considered to be Friedrich Wφhler's synthesis of urea in 1828. He started with lead cyanate, which is about as "dead" as any chemical can be, and ammonium hydroxide or chloride, also "dead", which generated ammonium cyanate, NH4+OCN- (empirical formula CH4N2O) (still "dead"). When he heated the ammonium cyanate, he got urea, H2NCONH2 (molecular formula also CH4N2O, but atoms arranged differently). Urea is just what the name sounds like, a major ingredient in urine, and was thought at the time to be a purely "organic" chemical. Other syntheses of "organic" compounds from "inorganic" materials soon convinced chemists that organic compounds obeyed the same laws of chemistry as other chemicals.
Although chemists gave up the "vital force" hypothesis at least 150 years ago, a shadow of it lingers on in the popular notion that "natural" organic materials are somehow safer or more healthful than synthetic chemicals. This popular notion ignores the fact that we would not know, for example, what vitamin C is if we could not find out its molecular structure, synthesize it, and show that the synthetic material is in every way identical with the vitamin C that the famous Hungarian chemist Szent-Gyorgy (pronounced "Saint-George") first isolated from Hungarian peppers. This bit of popular culture also ignores the toxicity of nicotine, strychnine, pufferfish toxin, and botulism toxin, the last of which, although quite natural and produced naturally by the bacteria Clostridium Botulinum, is the most poisonous chemical known.
List three reasons
why functional groups are important in organic chemistry
This
is another vague question that will depend on exactly what your textbook has to
say about it. Here are my ideas on the
best way to answer the question.
- Functional groups are important as a means of organizing, characterizing, and classifying many thousands of organic compounds into different families of compounds according to the nature and the structure of their functional groups.
- When organized in this way, the features and reactions of the various families of organic molecules become readily comprehensible.
- The necessity of memorizing vast numbers of seemingly unrelated facts disappears, for we find that the reactions of the simplest member of a family of organic compounds are often the same for hundreds and sometimes thousands of compounds in the same family.
UNIT 10: Acids
and Bases
Workshop Questions
1. Write the formula of the conjugate base of
the ammonium ion
Answer: NH3
Explanation: According to the Bronsted-Lowry theory of acids and bases, an acid is a
proton donor and a base is a proton acceptor. Once, an acid
has given up a proton, the remaining part can be a proton acceptor, and thus a base.
In this regard, an acid and a base are closely related to one another.
H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of
Acid-
For example:
NH3
+ H2O = NH4+ +
OH-
HAc = H+ + Ac-
Thus, NH4+
and NH3 are
a pair of conjugate acids and bases, as are HAc and Ac-.
2. Write the formula of the
conjugate acid of the carbonate ion
Answer: HCO3
Explanation: Carbonate ion is CO32
H+ + CO32 = HCO3
3. List the following acids in
order of decreasing strength: H2SO4, HF, H2S
Answer: H2SO4 HF
H2S
Explanation: Acid strength depends only on the value of Ka and not on the
concentration.
In
general, the larger the value of Ka, the stronger the acid, and vice versa
Strong
acids have large values of Ka, weak acids have small values of Ka and
very weak acids have very small values of Ka. See the Table of Ka values.
4. List the following bases in
order of decreasing strength: HSO4-, F-, HS-
Answer: HS F HSO4
Explanation: In
general, the larger the value of Ka, the weaker the base and vice versa.
Strong
bases have very small values of Ka, weak bases have small values of Ka and
very weak bases have large values of Ka. See the Table of Ka values.
4. Explain
the relationship between your answers to 3 and 4
Answer: A strong acid has a very weak conjugate
base,
a weak acid has a weak conjugate base, and
a very weak acid has a strong conjugate base.
5. Write
the net ionic equation for the following reactions. For each reaction, predict
if the formation of products will be favored at equilibrium
i. HI
and HC2O4-
Answer: HI (hydroiodic acid) is a very strong acid
( Ka = 3.2 x 109 ) and so it will
dissociate 100% in aqueous solution.
HI = H+ + I Formation of the dissociated ionic products
is favored at equilibrium
HC2O4 (hydrogen oxalate ion) is a weak base (Ka = 5.4 * 10-2) and therefore oxalic acid (H2C2O4) is a weak acid.
A weak acid is only sparingly dissociated in water and therefore HC2O4 will readily accept and hold free H+ ions forming H2C2O4
H+ + HC2O4 = H2C2O4 Formation of the oxalic acid product is favored at equilibrium
The
net ionic equation is HI + HC2O4 = H2C2O4 + I Formation of the products is favored at
equilibrium.
ii.
H3BO3 and HC2H3O2 2-
Answer: H3BO3 (Boric acid) is a weak acid (Ka = 5.8
* 10-10)and will dissociate sparingly.
I regret that I am
unable to identify the compound having the molecular formula HC2H3O2
2-
Therefore, Solution still pending upon further thought
*Will carbon dioxide be evolved
as a gas when sodium bicarbonate us added to an
aqueous solution of each compound? Explain
sulfuric acid
Ethanol C2H5OH
Ammonium chloride NH4CL
Solution still pending upon further thought
6. What
is neutralization reaction in terms of: Bronsted-Lowry
concept and Arrhenius concept? Give two examples of
each.
Answer: In
general, neutralization consists of
Acid + Base =
Salt + Water
Solution still pending upon further thought
7. Without using a calculator, fill in the blanks in
the following table:
|
pH |
pOH |
H30+ |
|
Acidic, Basic or Neutral |
|
3 |
11 |
1*10-3 |
1*10-11 |
Acidic |
|
6 |
8 |
1*10-6 |
1*10-8 |
Acidic |
|
10 |
4 |
1*10-10 |
1*10-4 |
Basic |
|
9 |
5 |
1*10-9 |
10-5 |
Basic |
Explanation:
pH = -
log [H3O+] pOH
= - log [
Detailed Discussion of pH and pOH
Adding an acid to water increases the H3O+ ion
concentration and decreases the
[H3O+][
The table below lists pairs of H3O+ and
Pairs of Equilibrium
Concentrations of H3O+ and
|
Concentration (mol/L) |
||||
|
[H3O+] |
|
[ |
|
|
|
1 |
|
1 x 10-14 |
|
|
|
1 x 10-1 |
|
1 x 10-13 |
|
|
|
1 x 10-2 |
|
1 x 10-12 |
|
|
|
1 x 10-3 |
|
1 x 10-11 |
Acidic Solution |
|
|
1 x 10-4 |
|
1 x 10-10 |
|
|
|
1 x 10-5 |
|
1 x 10-9 |
|
|
|
1 x 10-6 |
|
1 x 10-8 |
|
|
|
1 x 10-7 |
|
1 x 10-7 |
|
Neutral Solution |
|
1 x 10-8 |
|
1 x 10-6 |
|
|
|
1 x 10-9 |
|
1 x 10-5 |
|
|
|
1 x 10-10 |
|
1 x 10-4 |
|
|
|
1 x 10-11 |
|
1 x 10-3 |
Basic Solution |
|
|
1 x 10-12 |
|
1 x 10-2 |
|
|
|
1 x 10-13 |
|
1 x 10-1 |
|
|
|
1 x 10-14 |
|
1 |
|
|
Data from this table are plotted in the figure below over a narrow range of
concentrations between 1 x 10-7 M and 1 x 10-6 M.
The point at which the concentrations of the H3O+ and
It is impossible to construct a graph that includes all the data from the
table given above. In 1909, the Danish biochemist S. P. L. Sorenson proposed
using logarithmic mathematics to condense the range of H3O+
and
log (10-7)
= -7
Since the concentrations of the H3O+ and
pH = - log [H3O+]
Similarly, pOH is the negative of the logarithm of the
pOH = - log [
pH + pOH = 14
The equation above can be used to convert from pH to pOH, or vice versa, for
any aqueous solution at 25?C, regardless of how much
acid or base has been added to the solution. By converting the H3O+
and
8. Use
a calculator to fill in the blanks in the following table:
|
pH |
pOH |
H30+ |
|
Acidic, Basic or Neutral |
|
8.12 |
5.88 |
7.6 x10-9 |
1.3 x10-6 |
Basic |
|
10.56 |
3.44 |
2.8 x10-11 |
3.6 x10-4 |
Basic |
|
8.47 |
5.53 |
3.4 x 10-9 |
3.0 x10-6 |
Basic |
|
7.08 |
6.92 |
8.3 x 10-8 |
1.2 x10-7 |
Almost Neutral |
10. What are acid-base indicators? Why are
they used? Give two examples of acid-base
indicators.
Answer:
What are acid-base indicators?
An acid-base
indicator is a weak acid or a weak base. The undissociated
form of the indicator is a different color than the iogenic
form of the indicator. An Indicator does not change color from pure acid to
pure alkaline at specific hydrogen ion concentration, but rather, color change
occurs over a range of hydrogen ion concentrations. This range is termed the
color change interval. It is expressed as a pH range.
Why are they used?
Acid-base indicators
are useful in titrations and for determination of the pH of a solution. Weak
acids should be titrated in the presence of indicators which change color under
slightly alkaline conditions. Weak bases should be titrated in the presence of
indicators which change color under slightly acidic conditions.
Some common acid-base indicators
Tried-and-true
indicators include: thymol blue, tropeolin
OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol
blue, phenol red, neutral red, phenolphthalein, thymolphthalein,
alizarin yellow, tropeolin O, nitramine,
and trinitrobenzoic acid. Data in this table are for
sodium salts of thymol blue, bromphenol
blue, tetrabromphenol blue, bromcresol
green, methyl red, bromthymol blue, phenol red, and
cresol red.
11. Will
the solution formed by dissolving each of the following salts in water be
basic,
neutral or
acidic? Write the equation to show how each ion of each salt affects the
pH of the
solution.
i.Na3PO4
Solution still pending upon further thought
ii.NH4Cl
Solution still pending upon further thought
iii.CH3COONa
Solution still pending upon further thought
12. Write
the equation for the reaction of hydrochloric acid with the following:
i. Sodium carbonate
HCl + Na2CO3 = H+ + Cl + 2Na+ + CO32
All ions
remain solvated and dissociated in water and no precipitate forms
ii. Ammonia
HCl + NH3
= NH4+ + Cl
iii. Sodium hydroxide
HCl + NaOH =
H+ + Cl + Na+ +
13. A
buffer was prepared from equal molar quantities of acetic acid, HC2H3O2
and sodium acetate, NaC2H3O2. Explain how the buffer keeps the pH of the solution
at about the same value (a) after the addition of a few drops of a dilute HCl solution and (b) after the addition of a few drops of a
dilute NaOH solution. Use reaction equations to
describe the chemical changes that occur.
Solution still pending upon further thought
14. How
would your answer to question 12 be different if the buffer was prepared from
equal molar quantities of sodium bicarbonate and sodium
carbonate?
Solution still pending upon further thought
15. Determine
the equivalent mass of each of the following:
Answer:
For acid-base reactions, equivalent weights or equivalent masses are based on the fact that
one H+(aq) ion reacts with one
that supplies one mole of H+(aq) ions, and one equivalent of a base is the amount of the base
which supplies one mole of
The
equivalent weight or equivalent mass is therefore given by the general
equation,
equivalent mass =
molar mass/a
where "a" for an acid is the number of
moles of H+(aq)
supplied by one mole of acid in the reaction
taking place, and for a base "a" is the
number of moles of
reaction taking place.
i.
HCl
molar mass = 36.4
grams per mole
a = 1
equivalent mass = 36.4 grams per mole
ii.
H2SO4
molar mass = 98 grams per mole
a = 2
equivalent mass = 49 grams per mole
iii. KOH
molar mass = 56
grams per mole
a = 1
equivalent mass = 56 grams per mole
iv.
Ba(OH)2
molar mass = 171
grams per mole
a = 2
equivalent mass = 85.5 grams per mole
v. Al(OH)3
molar mass = 78
grams per mole
a = 3
equivalent mass = 26 grams per mole
16. How
is the formula VacidNacid =
VbaseNbase affected
when the acid used in the
titration is
HCl instead of H2SO4? How is the formula affected when the base is Mg(OH)2 instead of NaOH?
This
equation is derived from the idea that equivalents of acid will equal
equivalents of base when a system is at the endpoint of a titration.
If the
units of the terms on the left side of the equal sign and the units of the
terms on the right side of the equal sign are analyzed,
the
resulting units will be (equivalents acid) = (equivalents base).
In the
case of using HCl instead of H2SO4, HCl provides only half the protons per mole compared to H2SO4
So for
the same volume and normality of base, one would need HCL in twice as much
volume as H2SO4
In the case of using
Mg(OH)2 instead of NaOH, Mg(OH)2 provides twice as many
So for the same
volume and normality of acid, one would need Mg(OH)2
in half as much volume as NaOH
17.
Your group is to prepare an HCO3-
solution that approximates the normal
concentration in
blood, which is 24meq/L. Give a complete description of how you
would make 10.0L
of this solution.
Solution still pending upon further thought