exam 3 (chapters 5 & 6)

Question 1

conformational vs configurational isomers

Answer 1

conformational: the different positions that a molecule can twist into

configurational: R/S, matter of left or right handed

depends on whether or not the compounds can be interconverted by the rotation about single bonds

Question 2

enantiomers vs diastereomers

Answer 2

both are non-superimposable

enantiomers: mirror images, “twins”
- can only have 1 other enantiomers (come in pairs)
- all chiral centers have opposite configurations

diastereomers: not mirror images, “cousins”
- can be many of them
- opposite configruration at some chiral centers but not all

Question 3

how to determine if a molecule is identical or not

Answer 3

identical means that its superimposable

• superimposable = no chiral centers
• if the molecule has at least 1 chiral center, then it can be non-superimposable
• superimposable molecules are achiral
• b/c then its mirror image will also be the exact same

Question 4

stereocenters

Answer 4

single atom that is bonded to 4 different groups not specifically just 4 different atoms

Question 5

identifying chiral centers (stereocenters) in ring structures

Answer 5

ring with a group = stereocenter

• if ring doesn’t have a group attached on it somewhere, it’s not a stereocenter because either direction you travel around the ring, it’s the same
• if carbon in ring has substituents that make each “route” unique as you go clockwise and counterclockwise, it can be a chiral center
• or if the ring has symmetry in going around the ring both ways then it’s also not a chiral center

Question 6

how to determine the configuration of a stereocenter

Answer 6

• each group around the carbon is given priority based on atomic number of the element directly attached
• hydrogen always gets lowest priority
• if 2 same elements: list the atoms of the atom attached to stereocenter (basically looking for next point of difference)
• priority group going (decreasing) clockwise = R
• priority group order going counter clockwise = S

BUT MAKE SURE H IS THE ONE ON THE DASH OTHERWISE NEED TO FLIP IT AND SWITCH THE CONFIGURATION

Question 7

rotation of plane-polarized light in enantiomers + diastereomers

Answer 7

configuration does not have to do with direction of polarized light!!

• enantiomers rotate plane-polarized light in opposite directions so if one is rotating clockwise (+) then the other will rotate counter clockwise (-)
• if a sample has equal amounts of both enantiomers (racemic mixture), the rotations cancel each other out, and the mixture is optically inactive
• diastereomers are not predictable, they can either both rotate + or both - or opposite

Question 8

nonmenclature for R and S for multiple chiral centers

Answer 8

(3R, 4S) -

goes in the beginning for multiple (specify location of stereocenter), otherwise just the configuration in the beginning

Question 9

properties of enantiomers

Answer 9

the same physidcal (melting and boiling points) and chemical properties

• only difference is the rotation of plane polarized light

Question 10

difference in configuration between enantiomers

Answer 10

ALL chiral centers have different configurations

ex. (3R, 4S) and (3S, 4R)

Question 11

properties of diastereomers

Answer 11

have different physical properties (like melting points or solubility) and different chemical behaviors

Question 12

difference in configuration between diastereomers

Answer 12

have different configurations at some but not all stereocenters

Question 13

racemic mixture

Answer 13

enantiomers are mixed together in equal concentrations

• 1 R and 1 S enantiomer
• because of this, each rotates light in the opposite direction so overall rotation cancels out = racemic mixtures do not rotate plane-polarized light

Question 14

how to tell if a molecule is chiral

Answer 14

has chiral centers and no internal symmetry

• if it has symmetry its a meso compound

Question 15

aryls

Answer 15

aromatic rings

(ally’s are the ones that look like an A)

Question 16

E/Z naming system

Answer 16

1. look at the groups on both sides of the double bond
2. between the groups on each side, label the higher priority group
3. if the higher priority groups are both on the same side, its Z and opposite sides is E

E = trans, Z = cis

and put this E/Z in the beginning of the name and -ene at the end (for the double bond)

• cis/trans can only be used when the 2 groups are the same across the double bond, but this is not the case with E/Z naming

Question 17

2 ways to draw enantiomers

Answer 17

simpler way: invert ALL stereo centers (convert dashes to wedges and wedges to dashes)
- this puts the mirror behind the compound
- if all of them are not inverted, then they’re not enantiomers - they’re diastereomers

other way: create a mirror image but placing a mirror and flipping the molecule
- this puts the mirror on the side of the compound

• first method doesn’t work all the time, esp when the dashes and wedges are implied and not drawn

Question 18

E/Z isomers are what type of stereoisomers

Answer 18

diastereomers because they are stereoisomers that are not mirror images of each other

Question 19

optically inactive molecules

Answer 19

do not rotate plane- polarized light

• no chiral centers or plane of symmetry present
• meso compounds and racemic mixtures

Question 20

meso compound

Answer 20

HAS chiral centers but overall chiral because of an internal plane of symmetry

• makes them optically inactive
• also be careful about rotating a carbon because if you can rotate a carbon and gain symmetry, then it is also meso
• that one lonely guy that is a mirror image of itself (b/c drawing the mirror image will just give you itself)
• if it didn’t even have chiral centers, then it will be superimposable and just itself
• also every R is inverted to S and every S is inverted to R but have plane of symmetry

Question 21

constitutional isomers

Answer 21

same molecular formula but different connectivity

Question 22

how to calculate the maximum number of optically active stereoisomers (enantiomers + diastereomers) that a single compound can have

Answer 22

max number of optically active isomers = 2^n

where n = # of chiral centers

but meso compounds are considered inactive b/c their plane of symmetry throws them off, even though they have many chiral centers

if enantiomers then arranged as multiple pairs of enantiomers so 2^2 = 4 arranged as 2 pairs of enantiomers

Question 23

cis vs trans compounds dipole moments + IMFs

Answer 23

cis: dipole moment so stronger IMFs (higher boiling point), but lower melting points (b/c cant form crystals and squash together)

trans: no dipole moment so weaker IMFs but higher melting points (can form crystals bc can squash together)

Question 24

what region will the nucleophile tend to attack?

Answer 24

the most positive one

Question 25

what type of compound is more likely to undergo an Sn1 reaction?

Answer 25

a **tertiary** (central carbon bonded to 3 other carbons) **substrate** rather than a secondary or primary one since the carbonation must be formed spontaneously - also when phenyl group is attached to the tertiary carbon, also increases stability (but phenyl group directly attached to leaving group leads to no reaction)

Question 26

formula for calculating enantiomeric excess

Answer 26

**enantiomeric excess (ee)**: used to indicate the extent to which a sample contains more of one enantiomer than the racemic mixture **ee % = (observed rotation/rotation of pure enantiomer) x 100** also: **ee% = ([major enantiomer] - [minor enantiomer] / [total enantiomer]) x 100** answer means that the mixture has ____% MORE of one enantiomer than the other, its not the EXACT value (need to calculate exact value if it's given)

Question 27

cis and trans alkene isomers contain what type of stereogenic centers

Answer 27

trigonal

Question 28

stereogenic center vs chiral center

Answer 28

**stereogenic center**: border term that refers to any atom in a molecule at which exchanging 2 groups produces a different stereoisomer - *ex. cis-trans isomers* (since swapping substituents around leads to different stereoisomers) - hybridization can be *sp2 or sp3* - can be trigonal or tetrahedral - results in stereoisomers - may or not be symmetrical **chiral centers**: type of stereogenic center - sp3 hybridized - all bonds are sigma bonds (single) so tetrahedral shape - asymmetrical

Question 29

quick ways to find multiple chiral centers

Answer 29

**look for dashes and wedges**: they're commonly used to indicate 3D arrangements around a chiral center, is often a clue that the carbon is chiral **ignore symmetrical carbons**: carbons in symmetrical parts of the molecule are often not chiral centers **skip carbons with double or triple bonds**: can't be chiral bc not bonded to 4 different substituents **look for alkyl substituents, hydrogen, OH, and carbonyls (C=O)**: combination of thee groups on a carbon often indicates a chiral center **count chiral centers based on functional group positioning**: single rule is that cyclic compounds with substituents at different positions often have chiral centers where the ring structures restricts rotation

Question 30

what does α represent

Answer 30

observed optical rotation

Question 31

what does 0% and 100% ee (enantiomeric excess) represent

Answer 31

**0%**: indicates racemic mixture, where both enantiomers are present in equal amounts **100%**: represents a sample containing only one enantiomer, considered "optically pure"

Question 32

how to answer this question: A mixture of the 2-butanol enantiomers showed a specific rotation of +6.76. An **enantiomerically pure sample** of (S)-(+)-2-butanol shows a specific rotation of +13.52. What is the actual stereoisomeric composition of the mixture?

Answer 32

**1. calculate % ee**: (+6.76/+13.53) x 100 = 50% so enantiomeric excess of the (S)-(+)-2-butanol is 50% 2. of the total mixture, 50% consists of the racemic form, which contains equal number of the 2 enantiomers - so **half of this 50%**, or 25% is the (-) enantiomer and the other 25% is the (+) enantiomer 3. in total, 50 + 25 is the (+) enantiomer (*because the other 50% of the mixture is also the (+) enantiomer) so answer: 75% of the mixture is the (+) enantiomer and 25% is the (-) enantiomer

Question 33

reasoning behind this question: the reaction of CH₃CH₂CH₂C(=O)CH₂CH(CH₃)CH₃ with H2/Ni forms: (R)- and (S)-6-methyl-3-heptanol

Answer 33

- hydrogenation will reduce C=O to C-OH - it has a chiral center because different groups attached so possible configurations are either (R) or (S) - since reaction doesn't favor one enantiomer over the other, it will produce a **racemic mixture** so can be both (R) and (S) - can't determine whether (R) or (S) specifically because H is non-selective of stereo and **can either attack from the front or back (which determines R or S)**

Question 34

how to draw multiple stereoisomers for compounds containing multiple chiral centers

Answer 34

- make the mirror image - swap one of the dash/wedge group - swap another one and so on

Question 35

how to draw Newman projections

Answer 35

the circle represents the back carbon, the 3 lines represent the front carbon **looking from the left**: atom on the wedge is on the RIGHT side of Newman projection - dash: on the left **looking from the right side**: atom on the wedge is on the LEFT side of the Newman projection - dash: on the right

Question 36

absolute vs. relative configuration

Answer 36

absolute configuration - the R/S relative configuration - cis/trans stuff, spatial arrangement of molecules relative to another chiral molecule without knowing the exact R and S, which group on dash and which on wedge stuff

Question 37

whats allowed in rotation of fisher projections

Answer 37

- allowed to rotate **all groups 180º** (this means that the bottom goes to the top (moves 2 places each) - NOT allowed to rotate **all groups 90º** b/c this would put the vertical groups on a horizontal and changes the absolute configuration - allowed to rotate **3 groups keeping 1 same** b/c this is like a double swap which is allowed

Question 38

how to draw the enantiomer of a fisher projection

Answer 38

simply exchange the right and left horizontal bonds

Question 39

drawing Fischer projections - guess and check method

Answer 39

- randomly guess and put the groups on the Fischer template - find the configuration (R/S) of each chiral carbon on both the 3D molecule and the template - **whichever line up b/w 3D & template = keep** - **whatever don't line up = swap the horizontal groups**

Question 40

fischer projections - what everything represents

Answer 40

- horizontal lines represent **wedges** = coming out towards you - vertical lines represent **dashes** = going away from you - the intersection points represent the carbons

Question 41

how to go from Newman projections to Fischer projections

Answer 41

- first make the conformation **eclipsed**: - keep the front carbon pointing downward - groups go through a **180º rotation** so groups go across (*bottom right goes eclipsed behind top left*) and so on - then simply add them onto the fischer template in the order that you see them - the group thats pointing up at the top of the fischer projection is really the one pointing straight down in the Newman projection

Question 42

compounds that are cis/trans but have a plane of symmetry

Answer 42

NOT meso even though they have symmetry - can be diastereomers or enantiomers depending on the configuration changes

Question 43

how does being cis/trans + having plane of symmetry + having more than 2 chiral centers affect the amount of stereosiomers possible?

Answer 43

it **reduces them** because if the compound is drawn cis, because of the plane of symmetry, the compound becomes **meso** (only 1 stereoisomer) when it's drawn trans, there can be R and S possibilities *ex. 1,3-dimethylcyclohexane* has only 3 stereoisomers even though it has 2 chirality centers - **usually the meso compound only occurs with cis**

Question 44

inversion/retention of configuration based on whether or not the bonds to the chirality center are broken

Answer 44

chirality center bonds broken + backside attack = inversion of configuration chirality center bonds not broken + front side attack = retention of configuration (absolute configuration changed, but relative configuration retained)

Question 45

SN1 reactions + 2 key points

Answer 45

*SN1 = substitution nucleophilic unimolecular* (first order reaction) - reactions occurs in **2 steps**: **step 1**: leaving group breaks off carbon forming a **carbocation** (positively charged carbon atom - this is the slow step and the **rate determining step** **step 2**: nucleophile attacks carbocation, replacing the leaving group **key points**: - rate of reaction depends ONLY on the concentration of the SUBSTRATE (molecule that has the leaving group) - the more stable the carbocation, the faster the reaction (*tertiary carbocations are more stable than secondary and secondary are more stable than primary*) ALSO ONLY WORKS FOR SP3 HYBRIDIZED REACTIONS

Question 46

carbon-halogen bond

Answer 46

generally polar with the **partial positive** on the carbon and **partial neg**. on the halogen

Question 47

SN2 reactions + 3 key points

Answer 47

*SN2 = substitution nucleophilic bimolecular* (second order reaction) - reaction occurs in **1 step**: nucleophile attacks carbon that is attached to the leaving group from the opposite side, pushing leaving group out while bonding to the carbon - attack is simultaneous with the departure of the leaving group **key points**: - rate of reactions depends on the concentration of **both substrate and nucleophile** - **primary carbons** are more likely to undergo SN2 because there is less steric hinderance (crowding around carbon) while **tertiary carbons are too crowded** and dont undergo SN2 easily - reaction involves **backside attack** leading to **inversion of configuration** (dashes and wedges switch, absolute configuration may or not change) at the chiral center (if there is one)

Question 48

nucleophile, leaving group, and substrate

Answer 48

**substrate**: contains leaving group and electrophile **nucleophile**: attacks the substrate (usually has a neg. charge or abundance of electrons) **leaving group**: molecule that departs (BYE)

Question 49

good leaving groups + how to identify good leaving group

Answer 49

**good leaving group**: after breaking off, can stabilize the electrons it takes with it - large atoms = better leaving groups b/c neg. charge is spread out over large area are usually **weak bases** (b/c weak bases can hold onto electrons after leaving, less likely to attract back the protons after they leave) **common good leaving groups**: - halides (Cl-, Br-, I-) and sulfonates (S=O2) - *good leaving groups are usually conjugate bases of strong acids* - some leaving groups also **depart as neutral molecules** (ex. H2O) - this means it must have had a formal charge when it was bonded to the substrate

Question 50

stability of carbocations

Answer 50

**tertiary carbocations** are the most stable b/c the positive charge is spread out over the 3 carbon atoms (**more methyl groups = more stable carbocation**) - primary carbocations are very unstable and do not form easily

Question 51

physical properties of alkyl halides (mostly to do with solubility)

Answer 51

- low solubilities in water - miscible with each other and relatively non polar solvents - used as solvents for non polar and moderately polar compounds

Question 52

what type of hybridization only do SN reactions apply to?

Answer 52

sp3

Question 53

how to tell if racemic mixture is formed?

Answer 53

SN2 reactions dont produce racemic mixture, only 100% one enantiomer with inversion of configuration SN1 reactions usually result in racemic mixture and this means that product is **overall chiral** b/c effect of R and S cancel each other out - they also produce carbocation intermediate (flat, positively charged carbon) that allows the nucleophile to attack from either side

Question 54

protic solvent

Answer 54

protic means it has an H attached to an electronegative atom - CH3 groups are aprotic

Question 55

which types of compounds are prone to nucleophilic substitutions (Sn) & Elimination reactions and which ones are not?

Answer 55

alkyl halides with sp3 hybridization - akenyl halides (double bonds), aryl (phenyl) halides, and alkynyl halides (triple bonded) do not generally undergo Sn or E reactions

Question 56

what is the charge of the product that results from SN reactions

Answer 56

**neutral** the product has to be neutral so make sure has no charge (+ or -) in the example, attacking H2O became OH when bonded to keep the charge neutral

Question 57

heterolytic bond cleavage

Answer 57

when a bond breaks and **one atom gets both of the electrons** and the other gets none - creates ions where the one that takes all is negatively charged (anion) and the one with none is positively charged one (cation) *happens in SN reactions*

Question 58

nucleophilic substitution by neutral nucleophile results in what

Answer 58

a positively charged product

Question 59

poor leaving groups are _____

Answer 59

**strong bases** like OH-, NH2- , OR- unstable on their own and tend to hold onto electrons tightly, making them less likely to leave the molecule F- is also poor among halides b/c small and holds electrons tightly

Question 60

what is the transition state in a free energy diagram for SN2 reaction + what does higher products than reactants mean

Answer 60

the highest point of the curve at the top where the carbocation is being formed **higher products than reactants**: means reactants are more stable so reaction wants to go towards the reactants (reverse direction)

Question 61

free energy of activation (∆G+) vs. free energy change for the reaction (∆G)

Answer 61

**free energy of activation (∆G+)**: difference in energy between reactants and the transition state **free energy change for the reaction (∆G)**: difference in energy between the reactants and the products

Question 62

a 10ºC increase in temperature causes what to reaction

Answer 62

causes the reaction rate to double for many reactions taking place near room temperature

Question 63

when do chirality centers in different molecules have the same relative configuration

Answer 63

when the chirality centers share 3 groups in common and if these groups with the central carbon can be superimposed in a pyramidal arrangement

Question 64

the slow step in SN1 reactions

Answer 64

is the rate determining step - results in formation of the carbocation

Question 65

how many electrons are present in the p orbital of a carbocation

Answer 65

**none** - it can only accept electrons

Question 66

factors that stabilize carbocations

Answer 66

**resonance** (in aromatic groups, only for benzylic groups NOT phenyls) **hyperconjugation**: presence of neighboring C-H or C-C bonds can interact with empty p-orbital and help spread out the positive charge, making it more stable (*helping hand*) - shown with that straight cross looking thing where the bonds move and then the H just becomes a floating H+

Question 67

based on what do the amount of hyperconjugation structures increase?

Answer 67

more the # of carbons attached to the central carbon, more the amount of hyperconjugation structures

Question 68

solvolysis (hydrolysis + methanolysis)

Answer 68

**solvolysis**: substitution reaction where solvent breaks down the compound, acting as the nucleophile **hydrolysis**: water is the solvent and nucleophile, results in formation of alcohol group (-OH) **methanolysis**: methanol (CH3OH) is the solvent and nucleophile, results in formation of ether group (-OCH3) *1 H pops off from each* - many Sn1 reactions would be called solvolysis reactions

Question 69

the more stable the carbonation formed in Sn1 reactions, the __________ the reaction

Answer 69

faster

Question 70

the 7 strong acids

Answer 70

- HCl - HNO3 - H2SO4 - HBr - HI - HClO4 - HClO3

Question 72

which nucleophiles are better in protic/aprotic solvents?

Answer 72

**protic solvents**: stronger nucleophiles are larger ions like I- **aprotic solvents**: smaller ions like F- are better nucleophiles because protons surround and support smaller ions better and their electrons are not as accessible