sbu che 321 exam 1 f13 review slides

Exam 1 ReviewCHE 321Fall 2013

Don’t Forget!

• You are allowed to bring a 5”x8” cheat sheet to the exam.• Your card must be completely

handwritten.• Make it ASAP.• You may also bring a model kit.

Your presenters this evening are…

•Chapter 1: Timothy Shub•Chapter 2: April Slamowitz

S

Chapter 1 Review

Timothy Shub

IHD (Index of hydrogen deficiency)Index of unsaturation

Slide created by Olivia Cheng

C7H12 =

(2C +2) – (H + X – N)

2IHD =

C = # of carbonsH = # of hydrogensX = # of halogens (e.g. Cl, I, Br)N + # of nitrogens

(2*7+2) – (12 + 0 – 0)

2

(16) – (12)

2= = 2

1 index can be either a double bond or ring2 indices can be 2 double bonds, 2 rings, a double bond and a ring, or a triple bond

Possible Isomers

Counts as 2 double bonds

In fond memory of my orgo review..

Periodic Table Trends

Electronegativity

Valence Electrons

The Valence shell is the outermost shell

# Of Valence Electrons (electrons in the valence shell) at ground state can be determined by the group number.

Valence Electrons are responsible for creating bonds with other atoms to form other compounds.

Transition elements have varying valence electrons

Group # - Valence Electrons

Group #’s Valence Electrons

How many carbons are in this structure?A. 3

B. 4

C. 5

D. 6

E. 7

How many hydrogens are in this structure?A. 10

B. 11

C. 12

D. 13

E. 14

Formal Charge

How do we calculate formal charge? SIMPLE!

Valence e- - bonded e- - unshared e-

There is 1 bonded e- per bond, so in a double bond there are 2 bonded e-

The unshared e- are simply the lone pairs or the “dots”

What charge does the oxygen on the right obtain?

A) -2 C) Same charge of 0 E) +2

B) -1 D) +1

What charge does the double bonded oxygen obtain?

Calculating Formal Charge

Oxygen (all the way on right)- 6 – 2 – 4 = 0

Oxygen (double bonded oxygen)-6 – 2 – 4 = 0

Nitrogen–5 – 3 – 2 = 0 ___________________________

Oxygen (all the way on right)- 6 – 3 – 2 = +1

Oxygen (formerly double bonded)- 6 – 1 – 6 = -1

Nitrogen-5 – 3 – 2 = 0

(Valence e- ) – (bonded e- ) – ( lone e- ) = Formal Charge

____________________________________

Resonance

Electrons have the ability to move and so lone pairs can become a multiple bond (double bond or triple bond) or a multiple bond can become a lone pair!

Electrons can only move to adjacent bonds or atoms

Many atoms cannot exceed the octet (like carbon) but they can be deficient in their octet

What about that Nitrogen?

The nitrogen is positively charged which is OKAY!

However, that carbon has too many electrons, so many in fact it literally has no more space to put them all. It exceeded octet.

That fact makes this resonance structure impossible.

-

So what makes a resonance structure good?

1. Octet Rule/ more bonding

2. Formal Charge (Minimize formal charge separation)

3. Atom/charge compatibility

For example: A negative charge is not “compatible” with a carbon.

Geometrical Shapes

Much simpler to determine than most people think. .

CHECK WIKIPEDIA!

http://en.wikipedia.org/wiki/Molecular_geometry

Geometrical Shape Examples

A) Linear

B) Bent

C) Trigonal Pyramidal

D) Tetrahedral

E) Trigonal Planar

*Note – Triple Bonds (Alkynes) are ALWAYS linear.

How many carbons are in this structure?A. 3

B. 4

C. 5

D. 6

E. 7

Hybridization

Nothing to be scared about!

General Rule of Thumb: Bond Pairs (not counting pi bonds) + Lone Pairs = “ # ”

Match “#” with the total number of s and p orbitals.

s(1)p(1) (1+1) = 2 s(1)p2 (1+ 2) = 3s(1)p3 (1 + 3) = 4

REMEMBER! For this system only count the sigma bonds when counting bond pairs! NOT Pi BONDS!

Let’s play a game called.. DETERMINE THE HYBRIDIZATION!

A. sp

B. sp2

C. sp3

D. s

E. p

A little table to help you get it down..

Hybridization state of central atom

Shape of molecule or ion

Examples

sp Linear BeH2, CO2

sp2 Trigonal planar BF3, CH3+

sp3 Tetrahedral CH4, NH4+

~sp3 (w/1 Lone pair) Trigonal Pyramidal NH3, CH3-

~sp3 (w/2Lone pairs) Angular/bent H2O

How do we determine how much S character and P

character

Simply count how many orbitals of s or p are present then divide by the total amount of orbitals!

If something is sp3. It has 1 s orbital and 3 p orbitals. To determine how much s character it has simply take 1 and divide it by 4 (1+3) then multiply by 100 to find the %!

Likewise, if something is sp2 and you want to know how much p character it has, simply take 2 (since 2 p orbitals are present) and divide it by 3 (2+1).

More s character means MORE STABLE!

Examplezzzz (We love examplez…)

s1p3 1+3 = 4 ¼ = 25% s character

s1p2 1+2 = 3 1/3 = 33% s character

s1p1 1+1 = 2 ½ = 50% s character

What is a Constitutional Isomer?

Different compounds with the same molecular formula!

Note that these are not resonance structures!

Another problem for good measure..

Remember: • Constitutional Isomers = Diff compounds, same molecular

formula• Contributing structures (resonance)= Different electronic

configurations of the same molecule!

And that’s about it. . .

GOOD LUCK ON YOUR EXAM!

and now I welcome my esteemed colleague APRIL SLAMOWITZ!!!!!

To help you guys with chapter 2 ^_^

Chapter Two ReviewExam 1

Functional groups and intermolecular forces

APRIL SLAMOWITZ

APRIL.SLAMOWITZ@STONYBROOK.EDU

OFFICE HOURS THURSDAY 11:30-12:30PM CLC

Functional Groups Organic Lingo: “common and specific arrangements of atoms that impart predictable reactivity and properties to a molecule”

Average Human Lingo: “combinations of atoms that occur together on a molecule which influence the way a molecule reacts and its properties”

Exam 1, 2012 Q1

Hydrocarbons: functional groups that contain only Hydrogen and Carbon

Alkanes: contain only SINGLE bonds between carbons Natural gases (octane, pentane)Symbolized by R

Alkenes: contain DOUBLE bonds between carbonsIndustry products such as ethene and propeneSymbolized by R

Alkynes: contain TRIPLE bonds between carbons Symbolized by R

Aromatic compounds: contain a special type of benzene ringsBonds are in resonance, therefore bond length is between a single and double bond (1.5 bonds) Symbolized by Ar (for aromatic)

Solved Problem 2.1

Alkyl Halides: A hydrocarbon with a halide replacing a hydrogen

Carbon attached to a halide (Br, Cl, I, F)

Represented by R-X

Classified further by naming the carbon atom that the halide is attached to (primary, secondary, tertiary)

Review Problem 2.10

Alcohols: a hydrocarbon attached to an –OH group (hydroxyl)

Represented by R-OH

Classified further by naming the carbon atom that the hydroxyl is attached to (primary, secondary, tertiary)

Review Problem 2.12

Ethers: can be thought of as derivatives of water in which both hydrogen groups are replaced by alkyl groups

Represented by R-O-R or R-O-R’Two of the same side chains or two different side chains

Amines: can be considered organic derivatives of ammonia

Represented by R-NH2, R2-NH

Classified further by naming the carbon atom that the amino group is attached to (primary, secondary, tertiary)

Trigonal pyramidal shape Lone pair of electrons on the nitrogen atom and three substituentssp3 hybridized

Review Problem 2.16

Review Problem 2.17

Aldehydes: a carbon chain that is attached to a double bonded oxygen and hydrogen at the end of the molecule

Represented by RCHO

Contains a carbonyl group (carbon with a double bonded oxygen)

Workshop One Part Two

Ketones: a carbon attached to a double bonded oxygen and another chain of carbons

Represented by RCOR or RCOR’

Contains a carbonyl group

Workshop One Part One

Review Problem 2.18

Carboxylic Acids: a carbon that is connected to a carbonyl and a hydroxyl

Represented by R-COOH or R-CO2H

Contains a carbon attached to a double bonded oxygen (carbonyl) and an –OH group (hydroxyl)

Carbonyl + Hyrdroxyl = CARBOXYLic acid

Review Problem 2.20

Esters: a carbon that is attached to a carbonyl which is attached to an oxygen which is attached to more carbons

Represented by R-CO2R or R-CO2R’

Contain a carbon attached to a double bonded oxygen and another oxygen which is connected to another chain of carbons

Review Problem 2.23

Amides: a carbon attached to a carbonyl group and an amino group

Represented by RCONH2 (in which H can be replaced by R)

Contains a carbon that is attached to a double bonded O (carbonyl) and an NH2 (in which H can be replaced by R) group (amino group)

Further classified by the amount of R groups attached to the nitrogen atom Unsubtituted, N-substituted, N,N-substituted

Workshop One Part Two

Review Problem 2.24

Nitriles: a carbon tripled bonded to a nitrogen

Represented by R-CN

Contains a carbon that has a triple bond to a nitrogen atom which contains one lone pair

Both carbon and nitrogen are sp hybridized

Exam One 2010

In Summary…

Now go home and do this. (It helps, I promise…)

For Your Entertainment… http://www.youtube.com/watch?v=mAjrnZ-znkY

Intermolecular Forces (IMF)Forces that are found between different molecules (not within the same molecule) that determine its physical properties and molecular structureDetermines melting point (MP), boiling point (BP), solubility, color, reactivity, etc…

IMF (ion-ion )

Ion-Ion Forces: Intra-molecular force between two or more ions of opposite charge

Organic Lingo: “strong electrostatic lattice forces that act between the positive and negative ions in the orderly crystalline structure”

Average Human Lingo: “strong forces that hold together positive and negative ions in a very structured way”

Forces WITHIN a molecule (not between two molecules)

Usually very strong forces

Ionic Bonds: an intra-molecular bond between a metal and a nonmetal

Because of large electronegativity differences, electrons are taken from one atomThis creates a positive ion and a negative ionThese ions are held together by ion-ion forces(EN is the “ability of an atom to attract electrons that’s it is sharing”, a numerical value of how much an

atom wants an electron)

Covalent Bonds: an intra-molecular within a molecule that are held together by the sharing of electrons

Forces WITHIN a molecule (not between two molecules)

Electrons are shared between two or more molecules Can be shared equally to form NONPOLAR covalent bonds

Due to no difference in electronegativity

Can be shared unequally to form POLAR covalent bonds Due to electronegativity differences (EN is the “ability of an atom to attract electrons that’s it is sharing”, a numerical value of how

much an atom wants an electron)

Hydrochloric acid has a covalent bond. The electrons are shared

unequally between the two atoms, creating a slightly positive and

slightly negative end.

Covalent Bonds: more examples (Polar and Nonpolar)

Solved Problem 2.4

Intermolecular Forces (IMF)Forces that hold two or more molecules of the same or different species together

Not as strong as those between ions

Influence physical properties of a compound (boiling point, melting point, solubility, etc)

Ion-dipole forces

van der waals forcesDipole-dipole forcesHydrogen bondsDispersion forces

van der Waals Forces: dipole-dipole

forces Interactions between molecules that have permanent dipoles due to differences in electronegativity (non-uniform electron distribution)Not fully ionic

van der Waals Forces: hydrogen bonds Very strong dipole-dipole forces between hydrogen and small, highly electronegative atom (F, O, N)Remember these three with “E.T FON home”Stronger than dipole-dipole, but weaker than covalent bondsBDE = 4 to 38kJ/mole

Responsible for base-pairing of DNA nucleotides (Adenine and Thymine, Cytosine and Guanine)

The reason that water is a liquid at room temperatureIf water did not have hydrogen bonds, water would boil at -80C and there would be no life probably

What is life?!

van der Waals Forces: dispersion forces

IMF between two nonpolar molecules NP molecules exhibit no positive or negative ends that could attract each other So how do two

molecules that are nonpolar interact with one another?

Because electrons are always moving, there is a small instant of time where one side of the molecule is more negatively or positively charged than the otherA small, temporary dipole moment will occur and this will cause an attraction between two molecules

Factors that determine the magnitude of dispersion forcesThe bigger the molecule, the further the electrons are from the nucleus, the more ability they have to

move and create more and larger dipole momentsThe smaller the molecule, the closer electrons are to the nucleus, the less ability they have to move and

create less and weaker dipole momentsThe larger the surface area, the larger the dipole moments (more attraction) because there is more

available space for IMF interactions

Ion-dipole Forces The interaction with an ion and a permanent dipole

Usually result in solvation of the ion within a solution

For Your Entertainment… http://www.youtube.com/watch?v=_M9khs87xQ8&list=PLD5F5F0334A6B06A2

Boiling Point (BP)Organic Lingo: “the temperature at which vapor pressure equals atmospheric pressure”

Average Human Lingo: “the temperature where a liquid gets all bubbly and turns into a gas”

To go from the liquid to the gaseous phase, molecules must separate from one another and become more randomTherefore, IMF play a crucial role in determining the temperature at which a substance boils

Generally, the weaker the IMF, the easier it is to separate the molecules from one another, the lower the BPDispersion forces < dipole-dipole < hydrogen bonds < ion-dipole < covalent < ion-ion forces

Generally, the more symmetrical the molecule, the higher the BP

Review Problem 2.25

Solved Problem 2.6

2010 Exam One

SolubilityThe dissolution of a solid into a liquid

Ions are always more soluble in water because they are able to form ion-dipole bonds with water!

The solid must have some shared properties with the liquid it is dissolving in “like dissolves like”

Hyrdophobic: water fearing, will not dissolve in water, usually nonpolarHyrdophillic: water loving, will dissolve in water, usually polar *Water is polar!*

Exam One 2010

Random, Important Information You NEED to Know

pH < pKa then the BASE is favored and pH > pKa then the ACID is favored

Something is most stable when : it has the most amount of bonds, the least amount of charge separation, the most EN contains the – charge (usually)

Acidity and BasicityBAAD : Bases Accept, Acids Donate (Protons, that is..)If the hydrogen on a molecule is likely to leave (to be donated), that means it is an acidic hydrogen

This also means it is least stable, most reactive! Usually if it is creating a possible resonance structure, it is the one that will get removed most easily (because resonance is a stabilizing factor) In D2O reactions, just replace D with the most acidic hydrogen(s)

Lone pairs on molecules MAKE A DIFFERENCE!They will change the shape of the molecules! They count when determining hybridization!They are sometimes reactive, attacking positive centers!

Curved arrows are your friendsFish are friends. Push electrons toward electronegative atoms and positive charges!

Exam One 2011

Exam One 2011

Exam One 2011

Exam One 2011

Exam One 2012

Exam One 2012

That’s it! Go home!

Good luck!

Get some sleep!

Don’t go partying on Tuesday night!

Come to office hours!

Study hard!