Organic Chemistry

Organic ChemistryTutorial

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IUPACIUPAC, The International Union of Pure and Applied Chemistry, is a worldwide organization which was established in1918. It’s main purpose is to aid in chemical sciences as well as their applications with humankind. Organic nomenclature, being a large focus of IUPAC, was established in 1892 when chemists created a list of rules called the Geneva rules. This group of chemists ultimately became IUPAC and is who we should thank for organic nomenclature.

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IUPAC Nomenclature RulesIUPAC has 4 fundamental steps when naming organic compounds. The following steps should be considered:

(1) Identify and name the parent chain.

The Parent Chain is usually the single largest continuous chain of carbon atoms within an organic molecule. Its name is dependent on the number of carbon atoms. Its name should be determined by the following prefixes.

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Parent Chain Prefixes

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Carbon Atoms

Parent Prefix

1 Meth

2 Eth

3 Prop

4 But

5 Pent

6 Hex

7 Hept

8 Oct

9 Non

10 Dec

Carbon Atoms

Parent Prefix

11 Undec

12 Dodec

13 Tridec

14 Tetradec

15 Pentadec

20 Eicos

30 Triacont

40 Tetracont

50 Pentacont

100 Hect

IUPAC Nomenclature Rules(2) Identify and name the substituent’s.

The Substituent’s are the chain like projections from the parent chain. The majority of these chains follow the parent chain prefixes. They contain theSuffix “yl” added to the prefix.

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You can see that the longest possible chain is 9 without the incorporation of the

2 carbon chain. This is an example of a substituent.

IUPAC Nomenclature Rules(3) Number the parent chain and assign a locant to each substituent.

A Locant is the carbon number of the substituent attached to the parent chain. It is important that you number a parent chain such that the sum of all locant’s is the least possible. This standard is used to prevent the same molecule being named twice simply by false numbering.

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IUPAC Nomenclature Rules(4) Assemble the substituent's in alphabetical order.

You should only consider the alphabetical order of the parent chain prefix, not additional sub-prefixes (di, tri, etc.) which are attached to the beginning of a prefix for a substituent if there exist more than one on a parent chain of the same type.

Know that you have the 4 fundamental IUPAC steps…Let’s start naming!

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AlkanesAlkanes are organic compounds that contain fully saturated parent chains. That is, they are the hydrocarbons that contain the most hydrogen atoms possible (No Double or Triple Bonds). These molecules contain Sigma bonds between all C-C Bonds. When naming alkanes, the suffix “ane” is attached to the parent chain prefix.

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CycloalkanesCycloalkanes are the alkanes which exhibit a ring structure. These molecules will contain the “cyclo” suffix added to the parent chain prefix if the parent chain being considered is the ring structure. The “cyclo” suffix can also be conjoint with the parent chain suffix if considered an Alykl Group (substituent's that are of alkane essence). For the second case, the parent chain will be named as normal alkanes.

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(1) QUESTION:

Name the following Organic Molecule:

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NEXT QUESTION ANSWER HELP

(1) ANSWER:

STEP 1: Identify and name the parent.

Since there are 6 carbons there, the parent name should be hexane. And since there are no substituent's, there is not need to follow the next 3 steps.

ANSWER: Hexane

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(2) QUESTION:

Name the following Organic Molecule in two ways:

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(2) ANSWER:

Consider the Cycloalkane as the Parent


The parent is the 6 carbon atom cycloalkane. Which should be named cyclohexane.

STEP 2: Identify and name the substituent's.

The only substituent's is an 5 carbon atom alkyl group. Which should be name pentyl.

Since the other two steps aren’t required for this molecule we achieve a name of ANSWER: Pentyl Cyclohexane

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NEXT QUESTION ANSWER HINT

(2) ANSWER:

Consider the Alkane as the Parent


The parent is the 5 carbon atom alkane. Which should be named pentane.


The only substituent's is the 6 carbon cycloalkane. Which should be name cyclohexyl.

Since the other two steps aren’t required for this molecule we achieve a name of ANSWER: 1 - Cyclohexyl Pentane

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(3) QUESTION:


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(3) ANSWER:


The parent chain is the 7 carbon chainWhich should be named HEPTANE.


The two substituent's that exist are ETHYL and the CYCLOHEXYL

STEP 3: Number the parent chain and assign a locant to each. It is clear by the diagrams below that the best option is 1 for cyclohexyl and 4 for

Ethyl since the sum is less than 4 and 7.

STEP 4: Alphabetize

ANSWER: 4-ETHYL-1-CYCLOHEXYLHEPTANE

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12

34

56

7 76

54

32

1

(4) QUESTION:

Alkanes are typically straight chain molecules due to the alignmentof sigma bonds.

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(4) ANSWER:

Alkanes aren’t typically straight chain molecules. Straight chainmolecules are due to the overlapping of pi bonds and are usuallyfound in alkynes.

ANSWER: FALSE

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Complex SubstituentsThe following structures have special names associated with their type of bonding:

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Isopropyl Sec-Butyl Tert-Butyl Isobutyl Phenyl

3-Carbon 4-Carbon Benzene

Alkyl HalidesThese substituents are those in the halogen column of the periodic table:

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Fluoro Chloro Bromo Iodo

AlkenesAlkenes are organic compounds that contain at least one unsaturated carbon. That is, they are the hydrocarbons that contain at least one double bond. These molecules contain Sigma bonds between all C-C Bonds and a Pi bond between all C=C. When naming alkenes, the suffix “ene” is attached to the parent chain prefix as well as the appropriate location of the double bond.

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AlkynesAlkynes are organic compounds that contain at least one triple bond. These molecules contain Sigma bonds between all C-C Bonds and a 2 Pi bond between all C≡C. When naming alkynes, the suffix “yne” is attached to the parent chain prefix as well as the appropriate location of the triple bond.

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(5) QUESTION:


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(5) ANSWER:

STEP 1: Parent chain is 10 Carbons long which must include the triplebond. We will include the double bond as a side chain for ease ofnaming and since the triple bond has that higher authority.

STEP 2: 4 substituents (3 alkyl halides: 2 chlorines and 1 fluorine, 1double bond ethyl side chain)

STEP 3: Chloro at 7 and 9 (must count from the carbon closest to the triple bond) , Fluoro at 6, allyl (the name for the ethyldouble bond) at 8.

STEP 4: Alphabetize (allyl, dichloro, fluoro)

ANSWER: 8-allyl-7,9-dichloro-6-fluoro-2-decyne

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(6) QUESTION:

Count the number of carbon’s, Hydrogens. Name this molecule.

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(6) ANSWER:

STEP 1: Parent chain is 9 Carbons long which must include the double bond.

STEP 2: 4 substituents (1 alkyl halide, 1 Tert Butyl,1 alkyl, 1 cycloalkyl)

STEP 3: iodo at 7, tert Butyl at 6, methyl at 3,Cyclohexyl at 2.

STEP 4: Alphabetize (tert Butyl, cyclohexyl, Iodo, methyl)

ANSWER: 6-tert Butyl-2 cyclohexyl-7-iodo-3 methyl-1,3-nondiene (20 CARBONS, 33 HYDROGENS)

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(7) QUESTION:

Name this molecule:

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(7) ANSWER:

STEP 1: Parent chain is 7 Carbons long.

STEP 2: 6 substituents (4 phenol, 2 alkyl halides)

STEP 3: Phenyl at 1,2,6,7 and chloro at 3,5.

STEP 4: Alphabetize (chloro, phenyl)

ANSWER: 3,5-dichloro-1,2,6,7-tetraphenyl-heptane

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Alcohols and PhenolsAlcohols are organic compounds that contain an OH, hydroxyl group, and are named according to an ending in “ol”.

Phenols are alcohols comprised of a hydroxyl group attached directly to a phenyl ring.

When numbering the parent chain, alcohol should receive lowest numbering despite presence of alkyl substituent's or pi bonds.

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EthersEthers are compounds that contain an oxygen atom bonded between two R groups, where each R group can be an alkyl, aryl (aromatic compounds like benzene), or vinyl (double bonded side chains).

Two methods of naming: 1) Name the 2 substituents alphabetically followed by “Ether”2) Place “oxy” between a side chain and a parent chain

respectively.

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EpoxidesEpoxides are cyclic ethers that contain an oxygen within a cycloalkane. A special type is Oxirane, which is the triangular epoxide that is more reactive than other ethers due to a significant ring strain.When naming as parents use the following Parent Chains:

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Oxirane Oxetane Oxolane Oxane

When naming as side chain, merge epoxy with the position of the “overlapping carbons”.

(8) QUESTION:

Name this molecule in two ways:

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(8) ANSWER:

OPTION 1

Sides chains are butyl and pentyl

Thus: Butyl Pentyl Ether

OPTION 2

Take Pentane as the Parent chain and Butyl as the side chains

Thus: Butoxypentane

ANSWER: Butyl Pentyl Ether Butoxypentane

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(9) QUESTION:

Name all the following molecules. Using your knowledge fromprevious chemistry courses, order the following molecules in orderof increasing boiling point.

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(9) ANSWER:

Compound 1: 1-Butanol

Compound 2: 1,4-Butandiol

Compound 3: butane

Compound 4: methoxyethane

ANSWER: Butane < Methoxyethane < 1-Butanol < 1,4-Butandiol

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2 hydrogen bond interactions per molecule

1 hydrogen bond interaction per molecule

No hydrogen bonds no major electronegative change

Slight electronegative change between the oxygen atom and carbon atoms, creates dipole on the oxygen.

(10) QUESTION:

Name all the following molecule:

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(10) ANSWER:

STEP 1: Parent chain is 12 carbons long withA triple bond, thus dodecyne.

STEP 2: There are 5 substituents (2 isopropyl,1 epoxy, 1 phenyl, 1 methyl)

STEP 3: isopropyl is at the 6 and 9 position, The epoxy at 7 and 8, phenyl at 10 and methyl at 2.

STEP 4: Alphabetize (epoxy, isopropyl, methyl, phenyl)

ANSWER: Butane < Methoxyethane < 1-Butanol < 1,4-Butandiol

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Thiols and SulfidesSince Sulphur is under oxygen in the periodic table, it is often the case the oxygen contain organic molecules will have a sulphur derivative. Alcohols, oxygen bound to hydrogen and carbon, is referred to as a THIOL when replaced with sulphur. Add the term “thiol” to the parent chain for SH containing molecules. If other functional groups are present, use “mercapto” as the substituent name. Ethers, oxygen bound to two carbons on either side, is referred to as a SULFIDE when replaced with sulphur. Replace “sulfide” with ether or replace “ylthio” with “oxy” for parent chains and substituents respectively.

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Thiols and SulfidesHere are some examples of these molecules:

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1,4-Pentanedithiol EthylthioethaneDiethyl Sulfide Diethyl Sulfoxide

Sulfides that undergo oxidation reactions are called Sulfoxide (for one oxygen) and Sulfone (for two oxygen)

Diethyl Sulfone 4,6-dimercapto-3-octyne

Aldehydes and KetonesA CARBONYL group is an oxygen group double bonded to a Carbon. Within an organic molecule this can happen at two locations, First, the very end or very beginning of the molecule. This type of molecule is referred to as an ALDEHYDE. Attach the suffix “al” to the parent chain when naming. Use “carbaldehyde” when an aldehyde group is attached on the first carbon after a cyclic compound. Secondly, anywhere except the very end and the very beginning. This type of molecule is referred to as a KETONE. Attach the suffix “one” to the parent chain when naming. You may also use the method similar to ether, where you will place the parent name of each chain and followed by “ketone”.

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Aldehydes and KetonesHere are some examples of these molecules:

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Methyl propyl ketone Pentanal Cyclopentanecarbaldehyde2-pentanone

MethanalFormaldehyde

Formaldeyhde is a popular preservative as it has the ability to prevent the growth of common bacteria and fungi. It is the simplest form of aldehyde and is a significant consideration to human health as it may be carcinogenic.

(11) QUESTION:

Name the molecule:

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(11) ANSWER:


STEP 2: substituents ( 1 phenyl, 2 carbonyl groups, 1 thiol)

STEP 3: Phenyl at 3, Carbonyl groups at 2, 6, Thiol at 4.

STEP 4: Alphabetize (mercapto, phenyl)

ANSWER: 4-mercapto-3-phenyl-2,6-heptandial

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(12) QUESTION:

Name all the following molecules. Using your knowledge fromprevious chemistry courses, order the following molecules in orderof increasing boiling point.

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(12) ANSWER:

Ethanethiol 2-mercaptoethanol

1,2-Ethanedithiol Ethanedial

2-mercaptoethanal

ANSWER: Ethanethiol < Ethanedial < 2-mercaptoethanal < 1,2-Ethanedithiol < 2-mercaptoethanol

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(13) QUESTION:

The following molecule is the commonly known analgesic (Drug that reduces pain) and antipyretic (Drug that reduces fever), Acetaminophen (Tylenol™). Name all functional groups within theMolecule.

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(13) ANSWER:

ANSWER: hydroxyl, ketone, amine, benzene

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Amine

Ketone

Benzene

Hydroxyl

Carboxylic AcidsCARBOXYLIC ACIDS are those molecules containing a carbon bonded to a hydroxyl and bonded to a carbonyl group simultaneously. When naming these molecules, ensure the addition of “oic” to the parent name followed by “acid”.

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The Carboxylic Acid Functional Group

Amines and AmidesAMINES are the derivatives of Ammonias in which the hydrogen atoms (protons) have been replaced with alkyl groups (carbon) or aryl groups (benzene). As a parent chain the addition of the suffix “amine” is used. As a substituent, “amino is used. AMIDES are derivatives of carboxylic acids where the hydroxyl group has been substituted for a nitrogen group. Amides will often include structure that branch from the nitrogen group. The use of the suffix “amide” is used for as the parent chain.

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The Amine Functional Group The Amide

Functional Group

EstersESTERS are those molecules where a carbonyl group and a ether groups are situated on the same carbon. Naming esters involves using the suffix “oate”. You will have two side chains, and thus will name them as follows. The side chain whose derivative is a carboxylic acid will be considered the parent side chain, and the side chain whose derivative is an alcohol will be a substituent.

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The Carboxylic Acid Derivative

The Ester Function Group

The Alcohol Derivative

Butyl PropanoateH₂O

(14) QUESTION:

Name the molecule:

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(14) ANSWER:


STEP 2: substituents ( 1 amino)

STEP 3: Amino at 2

ANSWER: 2-Aminopropanoic acid Alanine

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(15) QUESTION:

Viagra™ the popular erectile dysfunction drug contains many interesting functional groups. Name all functional groups and state the number of tertiary amine groups.

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(15) ANSWER:

ANSWER: 3 Tertiary Amines, Sulfone, Amine, Amide, Benzene (alkane and alkene)

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(16) QUESTION:

Name the following Molecule:

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(16) ANSWER:

ANSWER: (3 amino-5-hydroxy-4-mercapto-2-oxo)-propyl propanoate

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OthersHere are some other names you must be familiar with:

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PEROXIDE

ARENE

NITRILE

DO YOU REMEMBER

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Periodic Table TrendsEFFECTIVE NUCLEAR CHARGE

Is the average nuclear charge felt by an individual electron in an atom, taking into consideration the “shielding” effect of inner-shell electrons. Since negatively charged electrons are attracted to the positively charge protons in the nucleus, and at the same time, repelled by other electrons in the atom, we use effective nuclear charge to determine the positive charge on a specific electrons in a specific orbital. A general approach to calculating nuclear charge is using this equation where Z is the total number of protons in an atom and S is the number of electrons in the inner-orbitals of the orbital at interest.

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Although this is typically not a trend discussed in class, it is very important in

understanding other periodic trends.

EFFECTIVE NUCLEAR CHARGE

Periodic Table Trends

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Na 11+

-

-

--

-

-

-

-

-

-

- Electron of interest

Cl 17+

-

-

--

-

-

-

-

-

-

-

-

-

-

--

-

Electron of interest

And so we see that sodium has a less effective nuclear

charge than chlorine.

Periodic Table TrendsATOMIC RADII

Is the measure of the size of an element from it’s nucleus to its outer most cloud of electrons. Typically, we see as we go from left to right on the periodic table, Atomic Radii DECREASES since the effective nuclear charge increases and electrons in the outer most orbital are more effectively attracted to the positive nucleus. We also see that if we go from top to bottom of the periodic table Atomic Radii INCREASES since electrons are filling new outer orbitals.

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Periodic Table TrendsIONIZATION ENERGYThe amount of energy required to remove an electron from a gaseous atom (i.e. an atom that is all by itself, not hooked up to others is in a solid or liquid). Typically, the ionization energy INCREASES from left to right on the periodic table since the effective nuclear charge increases, meaning that electrons are more effectively attracted to the nucleus and this will require more energy to be extracted. The ionization energy DECREASES from top to bottom of the periodic table since electrons in the distance from the nucleus to the outermost shell is increased and thus the effective nuclear charge on an electron is weakened due to the distance of attraction.

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Periodic Table TrendsELECTRONEGATIVITY

Is the power of an atom to attract electrons to Itself. Typically, as we go from left to right onthe periodic table we see thatelectronegativity INCREASES because effectivenuclear charge increases and thus an atom’sability to effectively attract electrons is higher.as we go from top to the bottom of theperiodic table we see that electronegativityDECREASES since effective nuclear charge isweakened due to the distance from the nucleus to the further most orbital.

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(17) QUESTION:

, a radioactive isotope, contains two extra neutrons. What effect would this have on the atomic radii of carbon? Explain.

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(17) ANSWER:

The addition of neutrons within an atom will have NO (by our definition of nuclear charge) effect on the atomic radii. Since Atomic Radii is the distance from the nucleus to the outer most cloud of electrons which is determined by the effective nuclear charge and its ability to effectively attract electrons toward the nucleus, we notice that the addition of neutrons will not affect our nuclear charge and thus will exhibit an identical Atomic Radii.

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(18) QUESTION:

What is true about :

i. has a greater atomic radii than ii. has a greater electronegativity than iii. has a smaller ionization energy than

a) ib) iic) iiid) i & iie) ii & iii

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(18) ANSWER:

i) This answer is TRUE. More electrons in a orbital with the same effective nuclear charge between atoms implies that the electrons are not as tightly bound as they are when more are present.

ii) This answer is FALSE. Ionic oxygen has already received two electrons and is stable. The addition of any more electrons would imply the placement in a new orbital which would result in a electron that has a 0 effective nuclear charge since all inner most electrons pair up with a proton.

iii) This answer is FALSE. Ionic oxygen is stable and thus the amount of energy to remove an electron is more than if oxygen wasn’t stable.

ANSWER: A

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(19) QUESTION:

Arrange the following Atoms in increasing Atomic Radii, if the inequality is reversed, does this match the definition of increasing electronegativity?(Phosphorus, Neon, chlorine, Arsenic)

a) Chlorine < Neon < Phosphorus < Arsenicb) Arsenic < Phosphorus < Chlorine < Neonc) Neon < Chlorine < Phosphorus < Arsenicd) Arsenic < Phosphorus < Neon < Chlorine e) Phosphorus < Neon < Chlorine < Arsenic

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(19) ANSWER:

Following the Trend on the periodic table, we notice the appropriate increasing order of atomic radii of these elements is Neon < Chlorine < Phosphorus < Arsenic.

If this inequality is switched, we have Neon > Chlorine > Phosphorus > Arsenic, which is false in terms of increasing electronegativity as noble gases have a electronegativity of 0.

ANSWER: C , False

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Bond TypeThere are three different types of bonds we will consider:

NONPOLAR COVALENT: The sharing of electrons equally, typically found in identical atoms or atoms with slight electronegativity difference.

POLAR COVALENT: The sharing of electrons unequally due to the effective nuclear charge on one atom having a higher effective pull of electrons than that of the other atom.

IONIC: The taking of electrons from an atom whose effective nuclear charges are on both sides of the “spectrum”, results in atoms being stable and similar to that of their respected noble gas.

Determining bond types by electronegativity difference:

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.4 1.7

Determining Lewis Structure

When determining Lewis Structure of a molecule, we utilize the valence electrons number as the electrons available to form bonds and lone pairs within a molecule. We will also use the octet rule which allows states that all atoms must have 8 electrons around it. Lets determine the Structure of the following Molecule () :

Valence ElectronsC = 2x4 = 8O = 2x6 = 12H = 1x4 = 4 -> 28 available electrons to form bonds and lone pairs

*** Keep in mind of the most common forms of each individual atom…Oxygen “typically” has two lone pairs and two bonds, Carbon “typically” has 4 bonds. Hydrogen “typically” has 1 bond.

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Formal ChargeFormal Charge, is defined as the charge associated with an atom that does not exhibit the appropriate number of valence electrons. In other terms, atoms situated in a molecule all have a standard valence electron value according to their position in the periodic table. If for some reason the number of electrons they are contributing within a molecule is different then their valence electron value, we assign a Formal Charge according to the addition or loss of electrons contributing within a molecule for the particular atom.

In general, we can define Formal Charge as follows:

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Oxygen has 6 valence electrons, in this molecule, oxygen has 3 pairs of lone

pairs (6 valence electrons) + 1 electron contributing in a

sigma bond. Thus, we assign a Formal Charge of -1

Chlorine has 7 valence electrons, in this molecule, chlorine has 2 lone pairs (4

valence electrons) and 2 electrons contributing in

sigma bonds. Thus, we assign a Formal Charge of +1.

(20) QUESTION:

Determine the Formal Charge on Oxygen in the following Molecules:

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(20) ANSWER:

ANSWER:0 , +1(if considered a standard molecule), -1, +1

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Oxygen has 6 valence electrons, in this

molecule, oxygen has 2 pairs of lone pairs (4 valence electrons)

+ 2 electron contributing in a

sigma bond. Thus, we assign a Formal

Charge of 0

The Carbon Atom in this molecule breaks this octet rule, under standard conditions, this molecule does not exist. Though, if we were to assign a

Formal Charge, it would be +1

Oxygen has 6 valence electrons, in this

molecule, oxygen has 3 pairs of lone pairs (6 valence electrons)

+ 1 electron contributing in a

sigma bond. Thus, we assign a Formal

charge of -1.

Oxygen has 6 valence electrons, in this molecule, oxygen has 1 pairs of lone pairs (2 valence electrons) + 3 electrons

contributing in sigma bonds. Thus, we assign a Formal charge of +1.

(21) QUESTION:

Determine the Lewis Structure of the following Molecular Formulas:

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(21) ANSWER:

Valence Electrons C = 4 = 4H = 2x1 = 2O = 2x6 = 12

= 18e⁻

C typically has 4 Bonds, H typically has 2 Bonds, O typically has 2 bonds and 2 lone pairs

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Solving these problems usually requires multiple figures. But we can use them to determine the appropriate figure. Lets move one of the hydrogen atoms to the oxygen and make the other oxygen a double bond.

Satisfies all Requirements

(21) ANSWER:

Valence Electrons C = 4x4 = 16H = 3x1 = 3O = 2x6 = 12

= 1e⁻= 32e⁻

Electrons in outer most orbitalC = 8x4 = 32H = 3x2 = 6O = 2x8 = 16

= 54e⁻

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Since there exists an additional electron, we know that the general rules for atoms cannot be applied, thus we have a new rule that determines the amount of bonds in the molecule:

We can deduce the number of bonds by (54-32)/2 = 11. Since this results in bonds with other molecules, lone pairs would be included in the valence electrons, which we have subtracted.

(21) ANSWER:

ANSWER

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This molecules is missing the negative charge. Lets moving the hydrogen on the oxygen, creating the negative charge and place it on the last carbon.

This molecule breaks the octet rule on carbon 4. Let remove the double bond and make a triple bond between carbon 2 and 3. This molecule satisfies all requirements!

Though notice the negative charge could exist on the double bonded oxygen. Will discuss soon.

(22) QUESTION:

Epinephrine, the hormone used to aid with anaphylaxis, exhibitsmultiple bond types. State the bonds types present within the molecule and give an example of their location.

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(22) ANSWER:

ANSWER: Multiple Answers Possible, Note, no ionic bonds present!

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Polar Covalent Bonds

Non-Polar Covalent Bonds

VSEPRValence Shell Electron Pair Repulsion (VSEPR) Theories main purpose is to show that “electron pairs” (i.e. bonds and lone pairs) will exist within a molecule such that there location in space is to minimize repulsion with other atoms and electron pairs. This theory has given us the following observed shapes within molecule and these structure have been proven via other methods (i.e. X-ray Crystallography).

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TETRAHEDRAL all atoms

spaced 109.5° (ex. Methane)

TRIGONAL PLANAR all atoms spaced 120°

(ex. Boron Trifluoride)

LINEARall atoms spaced 180°

(ex. Beryllium Difluoride)

TRIGONAL PYRIMIDAL

atoms spaced 107°(ex. Ammonia)

BENTatoms spaced

104.5°(ex. Water)

Atomic Orbitals

Some Principles About Placement of Electrons

AUFBAU PRINCIPLE: lowest-energy filled first.

PAULI EXCLUSION PRINCIPLE: each orbital can have a maximum of two electrons that have opposite spins.

HUND’S RULE: when dealing with degenerate orbitals, such as p orbitals, one electron is placed in each degenerate orbital first.

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= the probability of where electrons exist in space, determined from , which is the solved wave equation (wave function) that takes into account the wave property of electrons.

HybridizationIf we look back at VSEPR, it tells us the positioning of atoms relative to each other, (i.e. Methane = 109°). If we look at Orbitals we would expect methane to have angles of 90°…but this is not observed??? If we consider the electron configuration of Carbon in methane we find:

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__ __ __2P__2S__1S

ENER

GY

This does not coincide with what we see. So, let’s say during formation, carbon undergoes a

“HYBRIDIZATON” of 2S and 2P orbitals, we will call them sp³ hybridized orbitals as the are the average of

1 s orbital and 3 p orbitals.

HYBRIDIZE

__ __ __ __SP³

__1S

ENER

GY

An average of 3 p orbitals and 1 s orbital, an sp³ orbital

HybridizationMETHANE

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__ __ __2P__2S__1S

ENER

GY HYBRIDIZE

__ __ __ __SP³

__1S

ENER

GYH

HHH

HybridizationBORON TRIFLUORIDE

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__ __ __2P__2S__1S

ENER

GY HYBRIDIZE

__2P

__ __ __SP²

__1S

ENER

GYF

FF

HybridizationBERYLLIUM DIHYDRIDE

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__ __ __2P__2S__1S

ENER

GY HYBRIDIZE

__ __2P

__ __SP

__1S

ENER

GY

HH

HybridizationSIGMA BONDS: are covalent bonds formed from the overlap of atomic orbitals

PI BONDS: are covalent bonds formed from the overlap of p atomic orbitals.

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__2P

__ __ __SP²

__1S

ENER

GY

The Hybridized electron configuration of Carbon in ethylene (ethene). Notice the 2P orbital, it requires one more electron, which it receives from the second carbon

(23) QUESTION

Tetrahydrocannabinol, is the principal psychoactive constituent of the cannabis plant. It’s chemical structure is very interesting! What is the number of pi bondsand sigma bonds?

a)b)c)d) e)

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(23) ANSWER:

Count…All bonds (including theOH bond and double bonds) is considered a Sigma bond.All Pi bonds are double bonds Or Triple bonds (Note,Though this molecule does notContain a triple bond, tripleBonds account for two piBonds).

ANSWER: A

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(24) QUESTION

When a double bond is formed between two atoms, one of the bonds is a sigma bond and the other is a pi bond. The pi bond is created by the overlap of...

a) spb) sp²c) sp³d) P orbitalse) S orbitals

RESOURCE: http://chemistry.boisestate.edu/people/richardbanks/inorganic/mc111/vol10/mcquestions111j.htm

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http://chemistry.boisestate.edu/people/richardbanks/inorganic/mc111/vol10/mcquestions111j.htm


(24) ANSWER:

ANSWER: B

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Ethene, an example of a double bond, is formed from the overlap of two sp² orbitals.

(25) QUESTION

What is the electron-pair geometry of the central oxygen atom of ozone (O3)?

a) Linearb) Trigonal planarc) Bentd) Tetrahedrale) Trigonal Pyramidal

RESOURCE: http://chemistry.boisestate.edu/people/richardbanks/inorganic/mc111/vol10/mcquestions111j.htm

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(25) ANSWER:

Central Atom is Trigonal Planar

Lone Pair

ANSWER: B

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ResonanceSuppose we had a divided chamber of gas where one chamber the gas had a higher pressure than the other:

If the slit in the chamber were to be removed, What would happen?

Suppose we were studying the orbital diagram of the following molecule:

Notice the hybridization of the positively charge carbon, The P orbital is empty and all SP² orbitals have been filled by 2 bonds between 2 hydrogen atoms and a bond between a carbon atom.

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__ __ __2P__2S__1S

ENER

GY HYBRIDIZE

__2P

__ __ __SP²

__1S

ENER

GY

The hybridizing of a positive carbon.

ResonanceThus, the orbital diagram (if we only draw pi bond) would look like:

If we consider the electron cloud over Carbon 1 to Carbon 2 as the chamber of high pressure and the electron cloud from Carbon 2 to Carbon 3 as the chamber of low pressure, what should we expect?

Some sort of combination between the first structure and this structure:

This is what we will refer to as RESONANCE, and each of these structuresWill be called Resonating Structures.

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C C C

Note, the empty P orbital

C C C

ResonanceThe problem is with our method of drawing structures, Lewis Diagrams, but since these are so practical and so well known, we must deal with learning resonance via these structures though they should be thought an average , not a equilibrium, among electron pi clouds. Though, we still must abide by standard rules including the Octet Rule. So, we will draw all Resonating Structures as Lewis Diagrams but think of them as an average of all structures combined:

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Rules for Resonance

• Only electrons can move between resonance structures.• The position of the atoms does not change• The only electrons that can participate in resonance are non-

bonding electrons (lone pairs) and pi electrons (double or triple bonds)

• Sigma bonds cannot participate in resonance (cannot break sigma bonds)

• Cannot EXCEED the octet rule for elements in rows 1 and 2 (8 electrons max)

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(26) QUESTION

Determine the Resonating Structures of the following molecule

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(26) ANSWER:

ANSWER:

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C C C C C C CC

-

(27) QUESTION

Determine the Resonating Structures of the following molecule:

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(27) ANSWER:

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(28) QUESTION

Determine the Resonating Structures of the following molecule:

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(28) ANSWER:

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(29) QUESTION

Determine the Major Contributor of this molecule:

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(29) ANSWER:

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(30) QUESTION

Considering all of the major resonance structures for the molecule below; the number of major resonance contributors in which a negative charge resonates onto an oxygen atom is?

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(30) ANSWER:

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Intermolecular Forces

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