Orgo Exam Dump

chapter 1: 

- sp3 is single, sp2 is double and sp is a triple bond

- molecular geometry:

- tetrahderal: ex would be CH4 assuming that there are 4 bonds sp3 attached and no lone pairs with a bond angle of 109.5

- trig pyramidal: ex would be NH3, there are three bonds attached and one lone pair however the bond angle is not 109.5 but rather 107 due to the extra push the lone pair provides, still sp3

- bent: ex would be H2O, there are four things attached however the oxygen atom has two bonds and two lone pairs, the bond angle is 105, smaller than NH3 as now there are two lone pairs instead of one that push the bonds towards each other

- trigonal planar: ex would be BF3, sp2 with a bond angle of 120, the reason that the hybridization is different is because boron has a steric number 3 so nothing else must occupy it

- lineaer: ex would be BeH2, sp bond angle 180, Be has a steric number of 2

- IMFs are the attractive forces between individual molecules and they are electrostatic forces

- dipole dipole (between atoms of diff EN), H bonding (between H and highly EN atom such as F, O, N), and instantaneous dipole dipole (LDF, has to do with the movement of electron density that slightly shifts around each compound)

- in terms of alcohols, the more OH present the stronger the IMFs therefore the higher the BP, smaller alcohols will have lower BP than larger alcohols due to IMFs interacting 

- solubility: like dissolves like, in the context of alcohols then a hydrocabon chain longer than 8 carbons will be insoluble in water 

- alcohols like soap have a hydrophic and hydrophillic center; hydrophic center is the hydrocarbon chain as that is nonpolar while the hydrophillic center is where the OH group(s) are/is located as hydroxyl groups are polar

- EN increases as you move across a period but decreases as you go down a group

Chapter 2:

- functional groups

- ether: one O carbons both side

- ester: two O, one double bonded to a carbon with another O attached to the same carbon

- ketone: double bonded O carbons both side

- aldehyde: double bonded O with H on same carbon (Ester but with H instead of C)

- carboxylic acid: ester group but with H

- anhydride: ester group but extra double bonded O

- amide: ketone but with N instead of C on free end

- amine: N surrounding by R

- resonance occurs due to deplocalized electrons; molecules and ions are stabilized by the delocalization of electrons 

- resonance patterns:

- allylic lone pair: lone pair adjacent to double bond, requires two arrows 

- allylic carbo: posiive charge adjacent to double bond, requires one arrow that switches their places

- lone pair adjacent to carbo: in the name, requires one arrow where lone pair is moved to form bond between that atom and the carbo thereby moving the positive charge on the atom with the lone pair

- pi bond between atoms of diff EN: in the same, one curved arrow 

- conjugated pi bond enclosed in ring: in name, imagine a bezene ring and how those bonds can move around

- rules for deciding importance of resonance structures by importance:

- most sig res structure is one where the greatest number of octets are filled

- if that is the same, the structure with the least formal chargese is the more sig one

- if those two are met, structure with the negative charge on the more EN atom will be sig

- if those three are met, then the structures are equally as sig

- resonance hybrid is basically a combo of all sig resonance structures; wherever a bond is in both stuctures a dashed line must be drawn on top of the bond to show that either connection is present in the hybrid, whereever a charge is present negative or positive, partials should be drawn indicating that and if the resonance structures include the same atom with the same charge multiple times then that partial should be drawn bigger to indicate its presence 

- delocalized lone pairs can move around while localized (cannot participate in resonance) cannot ; if an atome has lone pairs and a double bond the lone pairs are localized and cannot move

Chapter 3:

- 3.2: just know how electrons move

- 3.4: if the conjugate is stable (weak) original acid is strong, if the conjugate is unstable (strong) the original acid is weak when comparing anions

- ARIO is used to determine stability: A= atom, R= resonance,I= inductoin, O= orbital ; important thing to note when comparing atoms in the same period EN is dominant as you go across the table EN increases the more EN atom can better stabilize the negative charge, when comparing atoms in the same group polarzaiblity is dominant as the more polarizable atom can better stabilize the negative charge

- 3.8: if the alcohol is bulkier then when deprotonated, compared to an alcohol that is less bulky, will be less acidic since the solvant cannot stabilize the charge due to steric hindrance. 

Chapter 4:

- 4.6: staggered confor is the lowest energy while eclipsed is highest in energy; staggered confors that are the same energy are degenerate ; the difference between the energies of stag and eclipsed is torisional strain 

- 4.8: anti: big groups are farthest apart ; gauche: big groups are next to each other and is a form of steric interaction

- 4.10: most stable conformation of cyclohexane is the chair conf as it has no torisional strain

- 4.13: wedge means up while dash means down, either one could be axial or eq depending on the carbon ; ring flips do not change the configuration of a compound ; relatvie positions on the ring can be drawn anywhere so long as it meets the requirement ; more stable conformation is the one where the larger sub is on the eq position 

Chapter 5:

- 5.2: when a mirror is identical to the og it is superimposible ; objects that are not superimposable on their mirrors are chiral ; chiral centers are carbons that bear four different groups ; enantiomers are nonsuperimposable mirrors of each other (they will have opposite configs RvsS)

- 5.3: counterclockwise is S clockwise is R for assigning config

- 5.5: diastereomers are nonsuperimpossable nonmirrors of each other (one of the configs has changed not the other) ; cis and trans isomers are diastereomers ; 2-methylcyclohexan-ol can be written with both groups on wedges/dashes or one group on a wedge and the other on a dash  the compounds with same wedge or same dash are enantiomers while the other ones are enantiomers to each other (the ones with alternating wedge and dash) but comparing them to each other they are diastereomers

- 5.6: if compounds have (plane of symmetry) reflectional symmetry then they are not chiral but this does not apply if they have rotational symmetry (axis of symmetry) ; compounds that have chiral centers but also have a plane of symmetry are meso and are not chiral compound (compounds can have chiral centers but not be chiral, they can also not have chiral centers and still be chiral like in the case of the allenes)

- 5.7: fischer projections are ass and look weird, in a fischer projectino if two compounds have subs on opposite sides they are enantiomers but if not all the subs are on diff sides then they are diastereomers; assiging config is easy with these projections since horizontal lines are wedges and vertical lines are dashes, basically if you want to draw a fischer projection take the molecule and rotate it so that its entire backbone is facing you

Chapter 6:

 - arrows

Chapter 7: 

- 7.10: 

- tosylates: sulfonate ions are good leaving groups since they are the conjugate bases of very strong acids; in terms of alcohols since they are bad LG the compound can be treated with TsCl/pyr to switch out the group but in doing so the configuration of the alcohol (given that there was one) does not change since the C-O bond is not broken in the process, once the tosylate has been attached it can undergo SN2 with a SN or elim with a SB depending on what its treated with; since tosylates are good LG then the same table will apply to them in regards to which rxn will take place with primary, secondary or tertiary substrates 

- alcohols (more throughly covered in chapter 12) are bad LG, treating them with an acid such as HBr can undergo SN chemistry depending on the type of alcohol it is (primary will udnergo SN2, secondary (depending on if RAR is possible will) undergo SN2 or SN1 and tertiary will  undergo SN1 (SN2 chemistry will not occur with a tertiary substrate) ; if treated with H2SO4 they will undergo E1 giving the zaitsev major product producing an alkene , because its unimolecular that means that RAR can occur if possible but if that leads to a primary carbo then the rxn proceeds through E2, alcohols can only be protonated under strongly acidic conditions and cannot occur with strong bases at the same time

- for E2 reactions, there is an anti coplanar req that must be met stating that both the LG and the beta hydrogen must be anti to each other (one is dash other is wedge vice versa), if this req is not met then E2 cannot occur at that location 

Chapter 8:

- 8.6: acid catalyzed hydration is markov addition of water across the alkene, has carbo intermediate, no specific stereochem, reagents are H3O+ for protonation and H2O for NA then PT again

- 8.7: oxymerc: markov addition of water across alkene but no carbo since mercurinium ion so no RAR, no specific stereochem. addition of mercuric cation occurs first which is why stereochem is not of concern. reganets are Hg(OAc)2 for oxymercuration and NaBH4 for demercuration

- 8.8: hydrobaration oxi: anti markov addition of water across alkene, no carbo due to concerted process, proceeds through syn addition since addition of borane and H occur at the same time, reagents are BH3 THF for addition of borane and H2O2 NaOH for oxidation

Chapter 12: LO

- name alcohol: follows same IUPAC rules except now alcohol takes priority, replace suffix with -ol remove "e", if there are two OH groups add "e" back and is now diol at the end

- draw alcohol given name: follows same princple as 1

- classify alcohols by type: 1 one alkly group, 2 two alkyl groups, 3 three alkyl groups attached to the carbon attached to the OH

- identify CB and CA of alcohols: look at it

- predict trends in alcohol acidity based on structure: ARIO 

- recall how to synthesize an alcohol using substituion chemistry: the OH must be turned into a good LG either through tosylate or through protonation using an acid (HBr for SN1/SN2 and H2SO4 for E1) but if using TsCl/pyr then since its a good LG any SN2 or SN1 rxn can be used with it

- predict the product when a carbonyl group reacts with H2/metal catalyst: carbonyl reduced to an alcohol although this requires extreme conditions high tmp and pressure so it is rarely used

- predict product when hydride reagent is reacted with a ketone, an aldehyde, formaldehyde or an ester: 

- ketone+aldehyde: NaBH4 and LiAlH4 can be used to reduce a ketone to an alcohol following NA and then PT, the nucleophilic hydride will attack the carbon in the double bond and the solvent (usually MeOH for NaBH4 and H2O or H3O+ for LiAlH4) will protonate the O ; the product of these reactions produces a racemic mixture as the carbonly group is planar

- formaldehyde: as it is just a C=O bond, it will also undergo NA then PT to give methanol but it is not a racemic mixture as it does not produce any chiral centers

- ester: ester reactions with LiAlH4/H3O+ will reduce the ester all the way to an alcohol through NA, loss of LG, NA, PT, the leaving group is the methanol attached ; NaBH4 will not react with an ester as it is not strong enough

- explain why LiAlH4 and H2O must be sequential: it is very strong and very reactive and will have a violent reaction with protic solvents therefore the workup must be in another flask 

-  describe the preparation of a grignard reagent from an alkyl/vinyl/aryl halide: Mg will simply come in between the halogen and the carbon 

- predict product when grignard reagent followed by H2O ir reacted with a ketone, aldehyde, formaldehyde or carboxylic acid: 

- ketone + aldehyde: grignard reagents are strong, will proceed through NA and then PT, the nuc is the negatively charged carbon that will attack the carbon in the double bond then in the work up with H2O or H3O+ PT will occur to neutrlize it ; formation of a secondary alcohol will produce enantiomers but tertiary will only give you one specific product (same with primary) 

- formaldehyde: will folow the same mechanism basically, just adding a hydrocabon extention 

- carboxylic acid: will not react as it as an acidic proton and the grignard will react as a base instead ; grignard reagents cannot be used with acidic protons since it is a really strong base

- predict the product when OH reacts with strong oxidizing agents such as NaCr2O7/H2SO4

- primary alcohols can produce aldehydes and then carboxylic acids: when reacted with sodium dichromate it will produce a carboxylic acid fully and the aldehyde middle cannot be isolated since the reagent is so reactive 

- secondary can only produce aldehydes: sodium dichromate will only oxidieze the alcohol to a aldehyde

- tertiary cannot be oxidized

- predict the product when OH reacts with an anhydrous oxidizing agents such as PCC

- primary alcohols can produce aldehydes as PCC cannot oxidize it further into a carboxylic acid 

- secondary can only produce aldehydes

- tertiary cannot be oxidized

- predict the product when an alcohol is reacted with HX: basically the subsitution; HBr can proceed through normal SN2 or SN1 but with HCl need the ZnCl2 catalyst 

- synthesize a tosylate from an alcohol: react wiht TsCl/pyr

- predict product when OH reacted with TsCl/pyr followed by nuc: depends on nuc and substrate type (either 1, 2 or 3) since if its a bad nuc and tertiary then SN1 (but primary and secondary will not react) SN2 with primary and secondary since the tosylate is now a good LG but tertiary is SN1

- predict product when OH reacted with TsCl/pyr followed by base: E2 or E1 depending on base and substrate type (E2 needs regiochemical req of anti beta hydrogen) reaction since the tosylate is now a good LG

- SN/SB PRIMARY IS SN2 MAJOR E2 MINOR, SECONDARY IS E2 MAJOR SN2 MINOR, TERTIARY IS E2 ONLY

- SN/WB PRIMARY AND SECONDARY (DEPENDS ON LG) SN2 TERTIARY IS SN1

- WN/SB PRIMARY SECONDARY AND TERTIARY IS E2

- WB/WN NO RXN WITH PRIMARY SECONDARY BUT TERTIARY WILL GIVE EQUAL AMOUNTS SN1 E1 PRODUCTS

- predict the product when OH reacted with PX3 and SOCl2: sub reaction SN2 for both will simply replace OH with the halogen specified