O Chem Test 2

chirality

describes molecules without a line of symmetry (left and right hand)

Enantiomers

molecules that are not identical to their mirror images. Usually have chirality centers. Always come in a pair of two molecules. Mirror images of chiral molecules

A few examples of molecules that cannot show chirality

CH₂, CH₃, C double bonded to C, and C triple bonded to C

Achiral molecules

Have a plane of symmetry

Chirality Centers

The point in a molecule where 4 different atoms or groups attach to one carbon.

Formula to find number of stereoisomers using # of chirality centers

2^x (where x = # of chirality centers in molecule)

Specific rotation formula

dextrorotory molecules

rotate light clockwise (+) (right)

levorotatory molecules

rotates light counter clockwise (-) (left)

Racemic mixture

1:1 mixture of enantiomers, unable to rotate the plane-polarized light (due to cancelling specific rotations. One molecule is dectrorotatory, while the other is equally levorotatory)

Nomenclature of chiral molecules

1. rank atoms 1-4 that connect to chirality center. Largest atomic # = highest ranking

2. If two atoms have the same atomic number, look at the second atom(s), then the third, and so on.

3. Multiple bonded atoms are equivalent to the same number of single bonded atoms (1 double bond = 2 single bonds)

R configuration

(when highest ranking is pointing away) Clockwise (right)

S configuration

(When highest ranking is pointing away) counter-clockwise. Left

Diastereomers

Require at least 2 chirality centers. First, find R and S of the chiral carbons, then compare to the other molecule. If they differ from each other (except when they are totally opposites), they’re diastereomers. For example, if the first molecule is RSR, and the second is SSS, they’d be steroisomers because they differ in 2 chiral centers

Meso compound

An archiral compound (has a plane of symmetry) with chirality centers. Always will have that plane of symmetry.

Resolution

the process of seperaing enantiomers. Normally done by adding a chiral resolving agent which can react reversibly to make two diastereomers. The distereomers then can be seperated by recrystallization.

Addition reactions

2 molecules combine; no atoms are left over.

Elimination Reactions

One molecule splits into 2 products. Often involves the formation of a small molecule such as water or HBr

substitution reactions

parts from two molecules exchange. Most common organic reaction

Rearrangement reaction

A molecule undergoes changes in the way its atoms are connected. Just one molecule changing.

Symetrical bond breaking

HOMOLYTIC. Each product gets an electron. Half arrows are used.

Unsymmetrical bond breaking

HETEROLYTIC. One product gets both electrons. Full arrow is used.

Symmetrical bond forming

one electron comes from each reactant. Half arrow is used.

Unsymmetrical bond forming

both electrons come from one reactant. This is usually seen with ions. Full arrow is used.

Carbon in functional groups

Usually positively charged. Only exception is when it’s bonded to metals. Otherwise, C will be an electrophile.

Carbonyl vs alkene stability

Carbonyl groups (C = O) are much more stable than alkenes (C = C)

Polarizability

the tendency for atoms to undergo polarization. Larger atoms are typically more polarizable.

Charge density

Atoms can have more or less space to distribute their charge. Larger atoms have way more space to spread it out, leading to a lower density.

Nucleophile

Loves nucleus (which is positively charged). These species are either negatively chargerd or have excess electrons. Technically a lewis base, since it donated LP electrons. If no negative charge, look for LP electrons or pi bonds)

Eletrophile

loves electrons. These species are electron-poor, and may have open orbitals. Technically a lewis acid, since it accepts LP electrons. Look for empty orbitals and metals.

Radical reactions

Highly reactive because it contains a species with an odd number of electrons (usually 7) in its valence shell. No charge on radicals.

Chlorination of Alkane (Methane) reaction steps

1. Initiation: Cl₂ splits, creating two Cl radicals

2. Propagation: One radical generates many more radicals in a self-sustaining cycle.

3. Termination: Combination of two radicals to form a stable product. May lead to a very low % of unused product.

K_eq>1

more material present as products

K_eq<1

more material present as reactants

K_eq formula

Gibbs Free energy

the max amount of work that may be performed by a thermodynamic close system at constant temperature and pressure

If K_eq>1 ____

gibbs free energy < 0. Energy is released to the surroundings (exothermic) Forward reaction is favored

If K_eq<1, ____

Gibbs free energy > 0. Energy is absorbed from surroundings (endergonic). Reverse reaction is favored

Bond dissociation energy

Amount of energy needed to break a bond in gas phase to produce 2 radicals

Transition state

absolute highest point (peaks) on an energy diagram

Activation energy E_a

Energy needed to go from reactants to transition state

High E_a favors fast or slow reaction?

slowwwww

Small E_a favors fast or slow reaction?

faster reaction

How much energy is needed to react two radicals together

0

negative delta G°

energy is RELEASED

positive delta G°

energy is ABSORBED

saturated compound

single bonds and no rings. Just C and H

Degree of unsaturation (DU)

Summer number of multiple bonds or rings (triple bond gives 2 DU)

Example of how to use alkane formula (C_nH_2n+2) to find DU

ex) you have C₄H₆. Compared to C₄H₁₀, theres 4 deficient Hydrogens. Divide by 2, to get a DU of 2.

DU of halogens

halogens substitute one hydrogen. They act just like a hydrogen. Add halogens and hydrogens, and compare to alkane formula to find DU.

DU of oxygen/sulfur

Totally ignore them. Cross them out. After they’re gone, use the alkane formula to find DU.

DU of nitrogen

subtract H by amount of N. ex) C₅H₉N —> C₅H₈. Then use alkane formula to find that this has a DU of 2.

Naming alkenes

1. find longest parent chain that contains double bond

2. number the C in the chain, giving the double bond the lowest number (if there is a substitution, still prioritize the double bond, but if the double bond is in the middle, give any substituent a lower number)

3. Write full name. Arrange alphabetically. Indicate double bond position by writing the number of the first carbon atom of the double bond and place that number in front of the parent name. Rings have cyclo. (ex. 1,5-dimethylcyclopentene. 2-ethyl-1-pentene)

Cis-trans isomerism in alkenes

must have that double bond, and cannot have two of the same atom or molecule on the same side.

What to do in tri or tetra substituted alkenes?

Compare the top and bottom of each side. (Higher vs lower) From there, you can figure out if they are lined up or not. (Cis or trans)

Zame zide Z

If the starting alkene is unsymmetrical…

The reaction prefers to go towards the higher carbon (the one with more hydrogens). Hydrogen connects to the lower carbon, halogen gets the higher one.

Regioselectivity

“selective region” The preference of forming/breaking bonds in only one direction over any other direction

Markovnikov’s orientation

halogen attached to the higher carbon, hydrogen attaches to the lower one.

more hydrogens connected to a carbocation results in more or less stability?

less. The carbocation prefers to be tertiary if possible.

Carbocation rearrangements

if a carbocation has an immediate neighbor with a higher carbon, rearrangement will happen. The carbocation will move to the 3° carbon over a 2° carbon. This results in more stability.