Organic chemistry
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Organic chemistry

Organic Chemistry

Tutorial


Notes on presentation

  • This presentation is completely interactive

  • In order for this presentation to work you MUST follow the indicated tabs on each slide

  • Answer the question on a separate piece of paper and compare with the slides

  • This presentations purpose is to try and find what you haven’t understood or what you haven’t attempted

  • Follow it properly and you will succeed!

Notes on Presentation

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Iupac

  • IUPAC, 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.

IUPAC

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Iupac nomenclature rules

  • IUPAC 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.

IUPAC Nomenclature Rules

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Parent chain prefixes

Parent Chain Prefixes

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Iupac nomenclature rules1

  • (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 the

  • Suffix “yl” added to the prefix.

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

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Iupac nomenclature rules2

  • (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.

IUPAC Nomenclature Rules

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Iupac nomenclature rules3

  • (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!

IUPAC Nomenclature Rules

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Alkanes

  • Alkanes 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.

Alkanes

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Cycloalkanes

  • Cycloalkanes 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.

Cycloalkanes

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Question section

  • QUESTION:

  • Name the following Organic Molecule:

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ANSWER

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Question section1

  • 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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Question section2

  • (2) QUESTION:

  • Name the following Organic Molecule in two ways:

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Question section3

  • (2) ANSWER:

  • Consider the Cycloalkane as the Parent

  • STEP 1: Identify and name 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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Question section4

  • (2) ANSWER:

  • Consider the Alkane as the Parent

  • STEP 1: Identify and name the parent.

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

  • STEP 2: Identify and name the substituent's.

  • 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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Question section5

  • (3) QUESTION:

  • Name the following Organic Molecule:

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Question section6

  • (3) ANSWER:

  • STEP 1: Identify and name the parent.

  • The parent chain is the 7 carbon chain

  • Which should be named HEPTANE.

  • STEP 2: Identify and name the substituent's.

  • 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

Question Section

2

4

6

6

4

2

1

3

5

7

7

5

3

1

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Question section7

  • QUESTION:

  • Alkanes are typically straight chain molecules due to the alignment

  • of sigma bonds.

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Question section8

  • (4) ANSWER:

  • Alkanes aren’t typically straight chain molecules. Straight chain

  • molecules are due to the overlapping of pi bonds and are usually

  • found in alkynes.

  • ANSWER: FALSE

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Complex substituents

  • The following structures have special names associated with their type of bonding:

Complex Substituents

Phenyl

Isopropyl

Sec-Butyl

Tert-Butyl

Isobutyl

3-Carbon

4-Carbon

Benzene

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Alkyl halides

  • These substituents are those in the halogen column of the periodic table:

Alkyl Halides

Fluoro

Chloro

Bromo

Iodo

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Alkenes

  • Alkenes 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.

Alkenes

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Alkynes

  • Alkynes 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.

Alkynes

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Question section9

  • (5) QUESTION:

  • Name the following Organic Molecule:

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Question section10

  • (5) ANSWER:

  • STEP 1: Parent chain is 10 Carbons long which must include the triple

  • bond. We will include the double bond as a side chain for ease of

  • naming and since the triple bond has that higher authority.

  • STEP 2: 4 substituents (3 alkyl halides: 2 chlorines and 1 fluorine, 1

  • double 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 ethyl

  • double bond) at 8.

  • STEP 4: Alphabetize (allyl, dichloro, fluoro)

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

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Question section11

  • (6) QUESTION:

  • Count the number of carbon’s,

  • Hydrogens.

  • Name this molecule.

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Question section12

  • (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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Question section13

  • (7) QUESTION:

  • Name this molecule:

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Question section14

  • (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 phenols

  • Alcohols 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.

Alcohols and Phenols

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Ethers

  • Ethers 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:

  • Name the 2 substituents alphabetically followed by “Ether”

  • Place “oxy” between a side chain and a parent chain respectively.

Ethers

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Epoxides

  • Epoxides 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:

Epoxides

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

Oxane

Oxirane

Oxetane

Oxolane

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Question section15

  • (8) QUESTION:

  • Name this molecule in two ways:

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Question section16

  • (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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Question section17

  • (9) QUESTION:

  • Name all the following molecules. Using your knowledge from

  • previous chemistry courses, order the following molecules in order

  • of increasing boiling point.

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Question section18

  • (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

1 hydrogen bond interaction per molecule

Question Section

2 hydrogen bond interactions per molecule

No hydrogen bonds no major electronegative change

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

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ANSWER

HELP


Question section19

  • (10) QUESTION:

  • Name all the following molecule:

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Question section20

  • (10) ANSWER:

  • STEP 1: Parent chain is 12 carbons long with

  • A 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 sulfides

  • Since 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.

Thiols and Sulfides

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Thiols and sulfides1

  • Here are some examples of these molecules:

Thiols and Sulfides

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

Ethylthioethane

1,4-Pentanedithiol

Diethyl Sulfoxide

Diethyl Sulfide

Diethyl Sulfone

4,6-dimercapto-3-octyne

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Aldehydes and ketones

  • A 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”.

Aldehydes and Ketones

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Aldehydes and ketones1

  • Here are some examples of these molecules:

Aldehydes and Ketones

Pentanal

Methyl propyl ketone

Cyclopentanecarbaldehyde

2-pentanone

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.

Methanal

Formaldehyde

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Question section21

  • (11) QUESTION:

  • Name the molecule:

Question Section

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Question section22

  • (11) ANSWER:

  • STEP 1: Parent chain is 7 Carbons long.

  • 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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Question section23

  • (12) QUESTION:

  • Name all the following molecules. Using your knowledge from

  • previous chemistry courses, order the following molecules in order

  • of increasing boiling point.

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Question section24

  • (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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Question section25

  • (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 the

  • Molecule.

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Question section26

  • (13) ANSWER:

  • ANSWER: hydroxyl, ketone, amine, benzene

Amine

Question Section

Ketone

Hydroxyl

Benzene

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Carboxylic acids

  • CARBOXYLIC 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”.

Carboxylic Acids

The Carboxylic Acid Functional Group

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Amines and amides

  • AMINES 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, “aminois 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.

Amines and Amides

The Amine Functional Group

The Amide Functional Group

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Esters

  • ESTERS 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.

Esters

The Alcohol Derivative

The Ester Function Group

The Carboxylic Acid Derivative

Butyl Propanoate

H₂O

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Question section27

  • (14) QUESTION:

  • Name the molecule:

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Question section28

  • (14) ANSWER:

  • STEP 1: Parent chain is 3 Carbons long.

  • STEP 2: substituents ( 1 amino)

  • STEP 3: Amino at 2

  • ANSWER: 2-Aminopropanoic acid Alanine

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Question section29

  • (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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Question section30

  • (15) ANSWER:

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

Question Section

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Question section31

  • (16) QUESTION:

  • Name the following Molecule:

Question Section

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Question section32

  • (16) ANSWER:

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

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Others

  • Here are some other names you must be familiar with:

Others

PEROXIDE

NITRILE

ARENE

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Do you remember

DO YOU REMEMBER

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Periodic table trends

  • EFFECTIVE 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.

Although this is typically not a trend discussed in class, it is very important in understanding other periodic trends.

Periodic Table Trends

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Periodic table trends1

  • EFFECTIVE NUCLEAR CHARGE

Electron of interest

Periodic Table Trends

-

Electron of interest

-

-

-

-

-

-

-

-

-

-

-

And so we see that sodium has a less effective nuclear charge than chlorine.

-

Cl 17+

Na 11+

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-

-

-

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-

-

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-

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-

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Periodic table trends2

  • ATOMIC 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.

Periodic Table Trends

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Periodic table trends3

  • IONIZATION ENERGY

  • The 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.

Periodic Table Trends

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Periodic table trends4

  • ELECTRONEGATIVITY

  • Is the power of an atom to attract electrons to

  • Itself. Typically, as we go from left to right on

  • the periodic table we see that

  • electronegativity INCREASES because effective

  • nuclear charge increases and thus an atom’s

  • ability to effectively attract electrons is higher.

  • as we go from top to the bottom of the

  • periodic table we see that electronegativity

  • DECREASES since effective nuclear charge is

  • weakened due to the distance from the nucleus to the further most orbital.

Periodic Table Trends

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Question section33

  • (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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Question section34

  • (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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Question section35

  • (18) QUESTION:

  • What is true about :

  • has a greater atomic radii than

  • has a greater electronegativity than

  • has a smaller ionization energy than

  • a) i

  • b) ii

  • c) iii

  • d) i & ii

  • e) ii & iii

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Question section36

  • (18) ANSWER:

  • 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.

  • 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.

  • 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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Question section37

  • (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 < Arsenic

  • b) Arsenic < Phosphorus < Chlorine < Neon

  • c) Neon < Chlorine < Phosphorus < Arsenic

  • d) Arsenic < Phosphorus < Neon < Chlorine

  • e) Phosphorus < Neon < Chlorine < Arsenic

Question Section

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Question section38

  • (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

Question Section

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Bond type

  • There 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:

Bond Type

1.7

.4

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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 Electrons

  • C = 2x4 = 8

  • O = 2x6 = 12

  • H = 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.

Determining Lewis Structure

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Formal charge

  • Formal 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:

Formal Charge

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.

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Question section39

  • (20) QUESTION:

  • Determine the Formal Charge on Oxygen in the following Molecules:

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Question section40

  • (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.

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Question section41

  • (21) QUESTION:

  • Determine the Lewis Structure of the following Molecular Formulas:

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Question section42

  • (21) ANSWER:

  • Valence Electrons

  • C = 4 = 4

  • H = 2x1 = 2

  • O = 2x6= 12

  • = 18e⁻

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

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.

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Satisfies all Requirements

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Question section43

  • (21) ANSWER:

  • Valence Electrons

  • C = 4x4= 16

  • H = 3x1 = 3

  • O = 2x6= 12

  • = 1e⁻

  • = 32e⁻

  • Electrons in outer most orbital

  • C = 8x4= 32

  • H = 3x2= 6

  • O = 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.

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Question section44

  • (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.

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Question section45

  • (22) QUESTION:

  • Epinephrine, the hormone used to aid with anaphylaxis, exhibits

  • multiple bond types.State the bonds types present within the

  • molecule and give an example of their location.

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Question section46

  • (22) ANSWER:

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

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Polar Covalent Bonds

Non-Polar Covalent Bonds

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Vsepr

  • Valence 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).

VSEPR

LINEAR

all atoms spaced 180°

(ex. Beryllium Difluoride)

TETRAHEDRAL all atoms spaced 109.5° (ex. Methane)

TRIGONAL PYRIMIDAL

atoms spaced 107°

(ex. Ammonia)

BENT

atoms spaced 104.5°

(ex. Water)

TRIGONAL PLANAR all atoms spaced 120°

(ex. Boron Trifluoride)

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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.

= 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.

Atomic Orbitals

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Hybridization

  • If 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:

Hybridization

__ __ __ __

SP³

__

1S

__ __ __

2P

__

2S

__

1S

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

ENERGY

ENERGY

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

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Hybridization1

  • METHANE

Hybridization

__ __ __

2P

__

2S

__

1S

__ __ __ __

SP³

__

1S

ENERGY

HYBRIDIZE

ENERGY

H

H

H

H

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Hybridization2

  • BORON TRIFLUORIDE

__

2P

__ __ __

SP²

__

1S

Hybridization

__ __ __

2P

__

2S

__

1S

ENERGY

HYBRIDIZE

ENERGY

F

F

F

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Hybridization3

  • BERYLLIUM DIHYDRIDE

__ __

2P

__ __

SP

__

1S

Hybridization

__ __ __

2P

__

2S

__

1S

ENERGY

HYBRIDIZE

ENERGY

H

H

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Hybridization4

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

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

Hybridization

__

2P

__ __ __

SP²

__

1S

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

ENERGY

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Question section47

  • (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 bonds

  • and sigma bonds?

  • a)

  • b)

  • c)

  • d)

  • e)

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Question section48

  • (23) ANSWER:

  • Count…All bonds (including the

  • OH bond and double bonds)

  • is considered a Sigma bond.

  • All Pi bonds are double bonds

  • Or Triple bonds (Note,

  • Though this molecule does not

  • Contain a triple bond, triple

  • Bonds account for two pi

  • Bonds).

  • ANSWER: A

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Question section49

  • (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...

  • sp

  • sp²

  • sp³

  • P orbitals

  • S orbitals

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

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Question section50

  • (24) ANSWER:

  • ANSWER: B

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

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Question section51

  • (25) QUESTION

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

  • Linear

  • Trigonal planar

  • Bent

  • Tetrahedral

  • Trigonal Pyramidal

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

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Question section52

  • (25) ANSWER:

  • Central Atom is Trigonal Planar

  • Lone Pair

  • ANSWER: B

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Resonance

  • Suppose 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.

Resonance

__

2P

__ __ __

SP²

__

1S

__ __ __

2P

__

2S

__

1S

The hybridizing of a positive carbon.

ENERGY

HYBRIDIZE

ENERGY

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Resonance1

  • Thus, 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 structures

  • Will be called Resonating Structures.

Resonance

Note, the empty P orbital

C

C

C

C

C

C

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Resonance2

  • The 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:

Resonance

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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)

Rules for Resonance

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Question section53

  • (26) QUESTION

  • Determine the Resonating Structures of the following molecule

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Question section54

  • (26) ANSWER:

  • ANSWER:

-

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C

C

C

C

C

C

C

C

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Question section55

  • (27) QUESTION

  • Determine the Resonating Structures of the following molecule:

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Question section56

  • (27) ANSWER:

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Question section57

  • (28) QUESTION

  • Determine the Resonating Structures of the following molecule:

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Question section58

  • (28) ANSWER:

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Question section59

  • (29) QUESTION

  • Determine the Major Contributor of this molecule:

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Question section60

  • (29) ANSWER:

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Question section61

  • (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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Question section62

  • (30) ANSWER:

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Intermolecular forces

Intermolecular Forces

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