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1 -- Part 3: STRUCTURE AND BONDING
The two compounds are isomers because they have the same formula.
They are constitutional isomers because their atoms are different.
The carbon atoms in butane are in a straight chain.
As the number of carbon atoms increases, the number of isomers increases as well.
Constitutional isomers may have different ways of branching their carbon chain.
The constitutional isomers all have different ways in which atoms are bonded to other atoms.
These compounds are not bonds.
The atoms are in the same order.
2-ene are not constitutional isomers because they have the same bonds.
They are stereoisomers because they differ only in how they are attached to the double bond.
The atoms of the cis isomer are oriented in space.
Stereoisomerism includes Cis and trans isomers.
Chapter 5 of the study of organic chemistry is devoted to stereochemistry.
The cis isomer has the same therapeutic effects.
Quinine and the trans isomer have similar groups on product isolated from the bark of opposite sides of the double bond.
For example, but-1-ene has two hydrogens on one end of Quinidine.
They don't give a different compound by changing their positions.
It is used to treat irregular heartbeat.
There are two identical groups on one end of the double bond.
Changing the groups doesn't give a different compound.
These compounds can't show cis-trans isomerism.
No isomers of those that do draw the cis and trans double-bonded carbons.
There is a relationship between the structures.
There is a glossary at the end of each chapter.
The index serves the purpose of these glossaries, which are more than just a dictionary to look up unfamiliar terms.
The glossary can be used to review the chapter.
If you read through the glossary carefully, you can see if you understand and remember the terms.
The page number in the glossary listing should be used to review anything that seems unfamiliar.
The cis isomer and trans isomer have similar groups on opposite sides.
The sharing of electrons in the region between the two nuclei is bonding.
One pair of electrons sharing a bond.
Two pairs of electrons are shared in a bond.
Three pairs of electrons are shared in a bond.
A charge is spread out.
We usually draw resonance forms to show how the charge can appear on each atom.
The product of the charge separation times the bond length is a measure of the bond's polarity.
Four electrons are between the two nuclei.
A region of space has a relative probability of finding an electron.
A measure of an element's magnetism.
Elements with higher electronegativities attract electrons.
Most of the time, the EPM uses red to show electron-rich regions and blue or purple to show electron-poor regions.
The intermediate colors are orange, yellow, and green.
A method for keeping track of charges, showing what charge would be on an atom in a par ticular Lewis structure.
When there are two or more unfilled orbitals of the same energy, the lowest-energy configuration places the electrons in different orbitals.
Bonding occurs when oppositely charged ion are attracted to each other.
The formation of a large, three-dimensional crystal lattice is a result of ion bonding.
The wave functions of new orbitals can be added to each other.
The original number of orbitals is the same as the number of new ones.
A structural formula with bonds.
Wherever two lines meet or a line begins or bends, carbon atoms are implied.
Hydrogen atoms are not shown unless they are on an atom that is drawn.
Four bonds are assumed for each carbon atom.
Structural information is not given by the empirical formula or the molecular formula.
The overlap of atomic orbitals on different atoms creates an orbital.
The bonding MOs are filled in most stable molecules, so they can be either bond ing or antibonding.
The energy of an electron in a bonding MO is lower than it is in an atomic orbital.
The energy of an electron is higher in an antibonding MO than it is in an atomic orbital.
Valence electrons are not used for bonding.
Noble-gas conations are formed when atoms form bonding arrangements that give them filled shells of electrons.
The second-row elements have eight electrons.
The allowed energy state for an electron bound to a nucleus is the probability function.
The study of compounds derived from living organisms.
The pi bond has two lobes, one above and one below the line joining the nucleus.
The electrons are shared equally.
Two or more valid Lewis structures can be drawn from a molecule or ion, but only in the placement of the electrons.
We can estimate the relative energies of individual resonance forms.
A cylindri cally symmetrical bond is a bond with most of its electron density centered along the line.
sigma bonds are single bonds.
atoms are oriented in space
Section 1-10 has examples.
There is a bond between two nuclei.
One pair of electrons forms a sigma bond, and the other two pairs form two pi bonds at right angles to each other.
The electrons are in the shell.
The largest possible angles for bonds and lone pairs around a central atom are about 180 degrees for two, 120 degrees for three, and 109.5 degrees for four.
The square of the wave function is determined by the electron density.
Each skill is followed by problem numbers.
There are many possible structures for a given formula.
Lewis, Condensed, and line-angle structural formulas can be drawn and interpreted.
Predict patterns of bonding involving C, H, O, N, and the halogens.
Compare the relative importance of their Problems 1-26, 29, 36, 40, 41, 42, resonance forms.
A double bond, a triple bond, and a single bond can be drawn.
Predict the hybridization and geometry of the atoms in a molecule and draw a three-dimensional representation of the molecule.
Predict the hybridization and geometry of a resonance hybrid by drawing its resonance forms.
Predict the major contributor.
Predict which compounds can be constitutional isomers and which compounds can be cis-trans isomers.
It's easy to think you understand organic chemistry when you don't.
You have not learned to combine and use the facts and ideas that you have read through the book.
You don't really understand the material during an examination.
The best way to learn organic chemistry is to use it.
You will need to read and re-read the material in the chapter, but this level of understanding is just the beginning.
Problems are provided so that you can apply the ideas to new compounds and reactions that you have never seen before.
You force yourself to use the material by working problems.
You can do well on exams if you have self-confidence.
There are a lot of problems in the chapter.
There are problems within the chapters that provide examples and drill for the material that is covered.
As you read through the chapter, make sure you understand the problems.
The back of the book contains answers to many of the in-chapter problems.
The study problems at the end of the chapter give you more experience with the material and force you to think in more depth about the ideas.
There are more difficult problems with red stars that require extra thought and possibly an extension of the material presented in the chapter.
Taking organic chemistry without working the problems is like skydiving without a parachute.
Initially, there is a sense of freedom.
Those who went unprepared will get a shock at the end.
The element that corresponds to each configuration is named.
You need to know a small part of the periodic table to do organic chemistry.
Use the following steps to make this part from memory.
A list of the elements in the first two rows of the periodic table can be made from memory.
The first two rows of the periodic table can be constructed using this list.
Sulfur, phosphorus, chlorine, bromine, and iodine are found in organic compounds.
If the compound is a mixture of covalent and ionic, state that.
Both compounds are stable.
Lewis structures should be drawn for these compounds.
All attempts to synthesise NF5 have failed.
Explain why a Lewis structure is unlikely.
Each species has a Lewis structure.
Each compound has a Lewis structure.
Draw a stable structure for each formula.
There is a rule for the number of hydrogen atoms in stable hydrocarbons.
For the following compounds, draw complete Lewis structures.
The direction of the dipole moments of the bonds can be seen from what you remember.
Predict whether the dipole moment is large or small.
Draw a Lewis structure and fill it with nonbonding electrons.
The formal charge on each atom is calculated.
Determine if the following structures are actually different compounds or simply resonance forms of the same compounds.
The delocalization of charges is shown in the resonance forms.
The major resonance form is indicated in each case.
The major and minor contributors are labeled in the following resonance forms.
Add any missing forms.
Predict which resonance form is likely to be the major contributor by drawing the important resonance forms.
Determine which ion is more stable.
You can use resonance forms to explain your answers.
The smell of dirtyock sweats comes from Compound X, which is isolated from the sheep's wool fat.
A careful analysis showed that compound X has a high amount of carbon and hydrogen.
Nitrogen and halogen were not found.
A determination of compound X's weight was made.
There are many structures that have this formula.
In 1934, Edward A. Doisy of Washington University isolated estradiol from 3000 lbs of hog ovaries.
The sample was burned in oxygen and the CO2 and H2O were generated.
Determine the formula for estradiol.
The weight of estradiol was found to be 272.
Determine the formula of estradiol.
There would be two stereoisomers if the carbon atom were flat.
The carbon atom is ahedral.
Make a model of this compound and see if there are any stereoisomers.
Most other cycloalkenes are more reactive than the three-membered ring of cyclopropene.
A Lewis structure is needed for cyclopropene.
Tell us why cyclopropene is so reactive.
Give the bond angles around each atom.
Any lone pairs of electrons are included in a three-dimensional diagram.
A Lewis structure can be drawn.
There are different types of orbitals that overlap to form bonds.
Give bond angles to each atom.
The pi bond is shown with its proper geometry.
There are six coplanar atoms in this compound.
Circle the coplanar atoms with the trans isomer.
There are four atoms in a straight line.
To draw a three-dimensional representation of this molecule, use dashed lines and wedges to circle the four atoms that are in a straight line.
The cis and trans isomers should be drawn.
The relationships between the structures are given.
The possible relationships are the same compound, cis-trans isomers, constitutional isomers, and not isomers.
There is an anti- inflammatory rub for race horses.
The C O carbon atom in acetone and the S O sulfur atom in DMSO have similar structures.
Draw Lewis structures for acetone and predict the hybridizations.
There are features that cause compounds to be polar and engage in hydrogen bonding.
General trends in physical properties include boiling points and solubilities.
Use curved arrows to show the flow of electrons in Lewis acid-base reactions.
There are two water-soluble organic acids in the fruit.
Ascorbic acid is needed for growth and repair of collagen, a structural protein in connective tissue, because it plays a central role in metabolism.
Humans cannot make their own ascorbic acid.
The functional groups determine the properties of their parent compounds, and they serve as basic sites within the molecule.
Understanding the basic properties of organic compounds is important.
In an acidic environment, most organic compounds can react with both acids and bases.
You are surrounded by acids and bases, from the food you eat to the products you use to clean up afterwards.
In this chapter, we look at more about the distribution of electrons in the molecule and how they affect the properties.
The compounds react as acids and bases.
The transfer of protons is one of the many reactions that involve the acid-base reactions of organic compounds.
Many acid- base reactions don't involve protons at all, but depend on changes in electron distribution.
We will show how the functional groups in each class determine the electron distribution in the molecule and their reactivity.
We will show how the concepts of relative acidity and basicity affect organic reactions in this course.
The concept of polar covalent bonds between atoms with different electronegativities was reviewed in Section 1-6.
We are going to combine the concept with the geometry to study the chemistry of the molecule.
Bond polarities can range from non polar to totally ionic.
There is an ionic bond between the methylammonium ion and the chloride ion.
Dipole moments are expressed in units of the debye.
The bond has a partial negative charge on oxygen and a partial positive charge on carbon.
The carbon atom has about an eighth of a positive charge, and the oxygen atom has about an eighth of a negative charge.
The C " O double bond has a long and short bond.
It's a good indicator of a molecule's direction.
Bond dipole moments can be estimated by comparing various compounds, but they can't be measured directly.
The magnitude and direction of each individual bond dipole moment are reflected in this sum.
For example, carbon dioxide has two polar C " O bonds.
CO2's dipole moment is zero, but we might expect it to have a larger one.
This result is explained by the symmetry of the carbon dioxide molecule.
The structures of both carbon dioxide and formaldehyde are shown here.
The directions of the bond dipole moments are shown in the maps, with red at the negative ends and blue at the positive ends.
The bond dipole moments are oriented in opposite directions in carbon dioxide.
The figure shows some examples of dipole moments.
It is necessary to have dipole moments in directions.
The bonds have a smaller dipole moment than CH3Cl.
The dipole moments of bonds and molecules are contributed by lone pairs of electrons.
The nucleus has a partial positive charge balanced by the negative charge of the lone pair.
Notice how the lone pairs contribute to the large dipole moments.
The red areas in the maps show high negative potential in the regions of the lone pairs.
The sum of the individual bonds is referred to as a molecular dipole moment.
The Lewis structure can be drawn.
The compound may have a large, small, or zero dipole moment.
The N in NH + 4 is a positive charge.
The four polar bonds have a symmetrical arrangement.
O3 must be bent.
There are partial negative charges on the outer oxygens and partial positive charges on the central oxygen.
The net dipole is small.
There are two isomers of 1,2-dichloroethene.
One has a dipole moment of 2.4 D while the other has zero.
Explain why one has zero dipole moment by drawing the two isomers.
They attract or repel each other when they approach each other.
This interaction can be described simply in the case of atoms or simple molecules.
The forces are attractive until they come close to each other.
The small attractive force becomes a large repulsive force when it happens.
These attractive and repulsive forces are more difficult to predict with complicated organic molecules.
We can show how the forces affect the physical properties of organic compounds.
Solids and liquids have attractions between them.
The molecule are in contact with each other in thecondensed phases.
The effects of these forces can be seen in the melting points, boiling points, and solubilities of organic compounds.
The strength of intermolecular forces is indicated by the boiling points.
Molecules have permanent dipole moments because of their polar bonds.
There is a positive and a negative end to each molecule.
The positive end of one dipole is close to the negative end of another.
When two negative ends or two positive ends approach each other, they repel, but they may turn and orient themselves in the more stable positive-to-negative arrangement.
The net force is attractive and the polar molecule is mostly oriented in the positive-to-negative arrangement.
The attraction must be overcome when the liquid is vaporized, resulting in larger heats of vaporization and higher boiling points for strongly polar compounds.
The interaction is attractive if their approach is positive and negative.
The boiling point of Carbon tetrachloride is higher than that of chloroform.
There must be some kind of force other than the dipole-dipole forces.
The London force is caused by temporary dipole moments in a molecule.
The electrons are not always evenly distributed.
A small temporary dipole moment occurs when one molecule approaches another molecule in which the electrons are slightly displaced from a symmetrical arrangement.
The attractive dipole-dipole interaction results from the slightly displaced electrons in the approaching molecule.
The net force of these temporary dipoles is attractive because they last only a fraction of a second.
The force is proportional to the surface area of the two molecules.
The intermolecular London dispersion attractions between carbon tetrachloride and chloroform are stronger than those between hydrogen and chlorine.
The effects of London forces can be seen in the boiling points of hydro crevices.
A lizard's toe has a lot of carbons.
If we compare the boiling points of several isomers, the isomers with larger of tiny hairs, with each hair split into surface areas, have higher boiling hundreds of tiny tips.
Van 5H12 isomers attract the boiling points of three C tip to the surface.
pentane has the highest boiling point.
As the amount increases, the molecule becomes more spherical and the weak intermolecular attractions decrease.
The smallest surface area supports the gecko's weight.
If hydrogen is bonded to oxygen, nitrogen, or fluorine, a hydrogen atom can participate.
H hydrogens to be hydrogen bonds.
The hydrogen atom has a partial positive charge because of the strong H bonds.
The nonbonding electrons on oxygen or nitrogen atoms have a strong affinity for this hydrogen.
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