⋆.˚˖࿔ ࣪ orgo 1 ; exam 2 info!

:-)

☆ how do you identify a stereocenter?

☐ a stereocenter is an atom where swapping 2 groups gives a stereoisomer

☐ in orgo 1, the most common stereocenter is an sp³ carbon bonded to 4 different groups

☐ if 2 attached groups are identical, that carbon is not a stereocenter

☆ what is an asymmetric atom?

☐ an asymmetric atom is usually a tetrahedral atom bonded to 4 different groups

☐ an asymmetric carbon is a common source of chirality

☐ not every chiral molecule has only carbon stereocenters, but carbon is the main case here

☆ how do you assign R/S configuration with CIP rules?

☐ rank the 4 attached atoms by atomic number

☐ if there is a tie, move outward to the first point of difference

☐ point the lowest-priority group away from you

☐ clockwise 1→2→3 = R

☐ counterclockwise 1→2→3 = S

☆ how are multiple bonds treated in CIP ranking?

☐ a double bond is treated as if the atom is bonded to duplicate atoms

☐ a triple bond is treated as if the atom is bonded to triplicate atoms

☐ this matters when breaking ties in priority

☆ how do you classify 2 isomers?

☐ constitutional isomers = same formula, different connectivity

☐ enantiomers = non-superimposable mirror images

☐ diastereomers = stereoisomers that are not mirror images

☐ geometric isomers are a subtype of diastereomers, usually cis/trans or E/Z pairs

☆ what are enantiomers?

☐ enantiomers are mirror-image stereoisomers

☐ they have opposite configuration at every stereocenter

☐ they have identical physical properties in achiral environments except optical rotation

☆ what are diastereomers?

☐ diastereomers are stereoisomers that are not mirror images

☐ they differ at one or more stereocenters, but not all

☐ cis/trans isomers are commonly diastereomers

☆ how do you draw an enantiomer from a given structure?

☐ reverse the configuration at every stereocenter

☐ equivalently, swap wedge and dash at every stereocenter

☐ the result should be the mirror image of the original molecule

☆ how do you draw a diastereomer from a given structure?

☐ change the configuration at one stereocenter but not all stereocenters

☐ if only 1 stereocenter exists, a diastereomer does not exist

☐ the new molecule must keep the same connectivity

☆ how do you draw a constitutional isomer from a given structure?

☐ keep the same molecular formula

☐ change which atoms are connected to which

☐ a constitutional isomer differs in connectivity, not just 3D arrangement

☆ how do you draw a correct line-and-wedge structure from a given configuration?

☐ draw the tetrahedral center with 2 bonds in the plane, 1 wedge, and 1 dash

☐ place groups so the assigned priorities match the requested R or S configuration

☐ always verify by re-checking the 1→2→3 order after drawing

☆ what is a meso compound?

☐ a meso compound has stereocenters but is overall achiral

☐ it contains an internal symmetry element

☐ a meso compound is superimposable on its mirror image

☆ how do you recognize whether a compound has a meso stereoisomer?

☐ look for 2 or more stereocenters

☐ check whether a configuration pattern can create an internal mirror plane

☐ if internal symmetry makes the molecule achiral, that stereoisomer is meso

☆ what symmetry elements can make a molecule achiral?

☐ mirror plane

☐ center of inversion

☐ improper rotation axis in more advanced cases

☐ in this course, the key one is usually a mirror plane

☆ why is a meso compound optically inactive?

☐ its stereocenters’ effects cancel because of internal symmetry

☐ the molecule is achiral overall

☐ it does not rotate plane-polarized light as a pure compound

☆ what is specific rotation?

☐ specific rotation is a normalized measure of optical rotation

☐ it relates observed rotation to sample concentration and path length

☐ enantiomers have equal-magnitude but opposite-sign specific rotations

☆ how are observed rotation and specific rotation related?

☐ observed rotation = specific rotation × path length × concentration

☐ solve for the missing value if the other 3 are known

☐ opposite enantiomers give opposite signs

☆ what is enantiomeric excess (ee)?

☐ ee = |% major enantiomer − % minor enantiomer|

☐ ee also equals observed rotation ÷ pure-enantiomer rotation × 100%

☐ higher ee means the mixture is farther from racemic

☆ how do you find enantiomeric ratio (er) from ee?

☐ major % = (100 + ee) / 2

☐ minor % = (100 − ee) / 2

☐ er = major : minor

☆ what is a racemic mixture?

☐ a racemic mixture contains 50:50 enantiomers

☐ a racemic mixture has ee = 0

☐ its net optical rotation is 0

☆ how are conformations related stereochemically?

☐ different conformations can be identical, enantiomeric, or diastereomeric

☐ rapid bond rotation often connects identical conformations of the same compound

☐ some conformations become enantiomeric only when the whole shape is chiral

☆ what is inversion at a central atom?

☐ inversion flips the arrangement around a pyramidal center

☐ trigonal pyramidal atoms like amines can invert through a planar transition state

☐ fast inversion can prevent isolation of separate stereoisomers

☆ when can central-atom inversion prevent isolable enantiomers?

☐ if inversion is fast, one form rapidly converts into the other

☐ ordinary amines usually invert too fast to isolate separate enantiomers

☐ heavier atoms like phosphorus or sulfur may invert slowly enough to give isolable stereoisomers

☆ what are atropisomers?

☐ atropisomers are stereoisomers caused by restricted rotation

☐ rotation is too slow to average the structures

☐ they can sometimes be isolated as separate enantiomers

☆ how can enantiomers be separated?

☐ convert them into diastereomers using a chiral resolving agent

☐ separate the diastereomers because they have different physical properties

☐ then regenerate the separated enantiomers

☆ what is resolution?

☐ resolution is the separation of enantiomers from a racemic mixture

☐ one classic method uses diastereomeric salt formation

☐ selective crystallization can separate the diastereomeric salts

☆ why can cyclopropane and cyclobutane be strained?

☐ their bond angles are forced away from ideal tetrahedral angles

☐ they also have torsional strain from eclipsing interactions

☐ this makes them less stable than larger rings

☆ why is cyclopentane less strained than cyclopropane or cyclobutane?

☐ its bond angles are closer to ideal

☐ it can pucker to reduce eclipsing

☐ it still has some strain, but less than smaller rings

☆ why is cyclohexane especially stable?

☐ the chair conformation has near-ideal bond angles

☐ adjacent bonds are staggered

☐ angle strain and torsional strain are both minimized

☆ how do you convert a chair into a planar line-and-wedge structure?

☐ identify whether each substituent is up or down

☐ convert up/down into wedge/dash on the flat hexagon

☐ do not confuse up/down with axial/equatorial

☆ how do you convert a planar line-and-wedge cyclohexane into a chair?

☐ read each substituent as up or down from the planar drawing

☐ place each substituent onto a chair position with the same up/down relationship

☐ then determine whether each one is axial or equatorial in that chair

☐ drawing the ring-flipped chair gives the alternate conformation of the same compound

☆ what happens during cyclohexane chair interconversion?

☐ axial becomes equatorial

☐ equatorial becomes axial

☐ up remains up

☐ down remains down

☆ what is the stereochemical relationship between 2 chair forms of the same monosubstituted cyclohexane?

☐ they are identical conformations of the same compound

☐ ring flip does not change connectivity or up/down identity

☐ up substituents stay up

☐ down substituents stay down

☐ every axial substituent becomes equatorial

☐ every equatorial substituent becomes axial

☐ it only changes conformation

☆ when can 2 chair drawings be identical, enantiomers, or diastereomers?

☐ identical if one becomes the other by ring flip or simple rotation

☐ enantiomers if they are mirror images and non-superimposable

☐ diastereomers if they differ stereochemically but are not mirror images

☆ what are axial and equatorial positions?

☐ axial bonds point straight up or down

☐ equatorial bonds point outward around the ring

☐ every carbon in a chair has 1 axial and 1 equatorial bond

☆ what are cis and trans relationships in cyclohexane?

☐ cis = both substituents on the same side of the ring

☐ trans = substituents on opposite sides of the ring

☐ cis/trans depends on up/down, not axial/equatorial

☆ what are 1,3-diaxial interactions?

☐ an axial substituent interacts sterically with axial groups on C3 and C5

☐ these are 1,3-diaxial interactions

☐ they make axial conformers less stable

☆ why is equatorial usually more stable than axial in monosubstituted cyclohexanes?

☐ equatorial avoids most 1,3-diaxial strain

☐ axial places the group in steric conflict with the ring

☐ larger groups show a bigger equatorial preference

☆ how do you compare chair conformers of disubstituted cyclohexanes?

☐ place substituents on both possible chairs

☐ compare how many groups are axial versus equatorial

☐ the chair with more and larger equatorial groups is more stable

☆ what do you do with given free-energy differences for monosubstituted cyclohexanes?

☐ use each substituent’s axial penalty as an estimate

☐ add axial penalties together in a disubstituted chair

☐ the lower total energy chair is more stable

☆ what is a chair’s stability order among cyclohexane conformations?

☐ chair is lowest in energy

☐ twist-boat is higher

☐ boat is higher still

☐ half-chair is highest

☆ why is boat less stable than chair?

☐ boat has eclipsing interactions

☐ boat has flagpole steric interactions

☐ chair avoids both of these problems much better

☆ what are the key disubstituted cyclohexane stability patterns?

☐ trans-1,2 can be diequatorial

☐ cis-1,3 can be diequatorial

☐ trans-1,4 can be diequatorial

☐ diequatorial conformers are usually strongly favored

☆ what are the key disubstituted cyclohexane patterns that cannot be diequatorial?

☐ cis-1,2 cannot be diequatorial

☐ trans-1,3 cannot be diequatorial

☐ cis-1,4 cannot be diequatorial

☐ these usually have one axial and one equatorial group in each chair

☆ how do you use chair, Newman, and wedge-dash drawings together?

☐ wedge-dash shows up/down stereochemistry

☐ chair shows axial/equatorial and conformational stability

☐ Newman helps visualize bond rotation and steric relationships

☐ all 3 represent the same molecule in different ways

☆ what are spiro, fused, and bridged bicyclic compounds?

☐ spiro compounds share 1 atom between 2 rings

☐ fused compounds share 2 adjacent atoms and the bond between them

☐ bridged compounds share 2 nonadjacent bridgehead atoms connected by bridges

☆ how are simple bicyclic compounds named?

☐ use bicyclo[x.y.z]alkane for bridged/fused bicyclic systems

☐ x, y, z count carbons in the 3 bridges between bridgeheads

☐ write the bridge lengths from largest to smallest

☆ what are bridgehead carbons?

☐ bridgehead carbons are the 2 shared carbons that connect the bridges

☐ they are used to define bicyclo names

☐ they are also central in Bredt’s rule

☆ how do cis- and trans-fused bicyclic six-membered rings differ?

☐ fused bicyclic compounds can differ by whether the ring-junction substituent relationships are cis or trans

☐ cis-fused means the ring-junction substituent directions are on the same side

☐ trans-fused means they are on opposite sides

☐ trans-fused systems are often more rigid and locked

☐ cis- and trans-fused systems are stereoisomers

☆ how do you convert a chair conformation into a planar line-and-wedge structure?

☐ identify whether each substituent is up or down on the chair

☐ keep the carbon connectivity unchanged

☐ translate up/down orientation into wedges and dashes on the planar ring drawing

☐ axial versus equatorial is lost in the planar drawing

☐ up/down stereochemistry must be preserved

☐ the planar drawing represents configuration, not a specific chair conformation

☆ how do you tell whether two chair drawings are the same compound, enantiomers, or diastereomers?

☐ first compare connectivity

☐ then compare the up/down orientation of every substituent

☐ if only axial/equatorial changes while up/down stays the same, they are the same compound

☐ if all stereocenters are inverted, the pair may be enantiomers

☐ if some but not all stereocenters are inverted, the pair are diastereomers

☆ how can energy differences from monosubstituted cyclohexanes help estimate disubstituted-chair stability?

☐ treat each axial substituent as adding an energy penalty

☐ use known axial-versus-equatorial preferences from monosubstituted cyclohexanes

☐ add the penalties for all axial substituents in a chair

☐ the chair with the smaller total axial penalty is predicted to be more stable

☐ this is an estimate based on conformational analysis

image

☆ why are cycloalkanes of ring size C3-C6 different in stability?

☐ their stabilities depend on ring strain

☐ ring strain includes angle strain and torsional strain

☐ cyclopropane has severe angle strain and torsional strain

☐ cyclobutane has significant strain

☐ cyclopentane reduces strain compared with smaller rings

☐ cyclohexane can adopt a chair conformation with very low strain

☆ what causes ring strain in small cycloalkanes?

☐ angle strain comes from bond angles forced away from the ideal tetrahedral angle

☐ torsional strain comes from eclipsing interactions

☐ small rings cannot easily achieve both ideal bond angles and full staggering

☐ this raises their energy

☆ what is a functional group?

☐ a functional group is a specific arrangement of atoms that gives a molecule characteristic properties and reactions

☐ the carbon framework supports the functional group

☐ identifying the functional group is a key first step in classification and naming

☆ what are the halogen-containing functional groups?

☐ alkyl halide or haloalkane = carbon bonded to F, Cl, Br, or I

☐ common halogen substituent names are fluoro-, chloro-, bromo-, and iodo-

☐ haloalkanes are classified as functionalized hydrocarbons

☆ what are the oxygen-containing functional groups?

☐ alcohol = R–OH

☐ ether = R–O–R

☐ epoxide = three-membered cyclic ether

☆ how are alcohols classified?

☐ primary alcohol = the carbon bearing OH is attached to 1 other carbon

☐ secondary alcohol = the carbon bearing OH is attached to 2 other carbons

☐ tertiary alcohol = the carbon bearing OH is attached to 3 other carbons

☐ classification depends on the carbon attached to OH

☆ how are alkyl halides classified?

☐ primary alkyl halide = the carbon bearing the halogen is attached to 1 other carbon

☐ secondary alkyl halide = the carbon bearing the halogen is attached to 2 other carbons

☐ tertiary alkyl halide = the carbon bearing the halogen is attached to 3 other carbons

☐ classification depends on the carbon bonded to the halogen

☆ what are alpha and beta carbons?

☐ the alpha carbon is the carbon directly attached to the functional group

☐ the beta carbon is the next carbon away

☐ alpha and beta labels help describe positions near a functional group

☆ what is a carbonyl group?

☐ a carbonyl group is a C=O unit

☐ it appears in several important functional groups

☐ recognizing the carbonyl helps identify aldehydes, ketones, carboxylic acids, and esters

☆ what are aldehydes and ketones?

☐ aldehyde = terminal carbonyl, R–CHO

☐ ketone = internal carbonyl, R–C(=O)–R

☐ both contain a carbonyl group

☐ they differ by what is attached to the carbonyl carbon

☆ what is a carboxylic acid?

☐ a carboxylic acid has the pattern R–COOH

☐ it contains a carbonyl and an –OH on the same carbon

☐ the carboxyl group is a distinct functional group

☆ what is an ester?

☐ an ester has the pattern R–C(=O)–O–R

☐ it contains a carbonyl adjacent to an oxygen bonded to another carbon

☐ esters differ from acids and ethers in connectivity

☆ what are the nitrogen-containing functional groups?

☐ amine = nitrogen bonded to carbon and/or hydrogen without an adjacent carbonyl

☐ amide = nitrogen directly attached to a carbonyl carbon

☐ the carbonyl next to nitrogen identifies an amide

☆ what is the basic naming priority rule for functional groups?

☐ the suffix functional group has naming priority

☐ the parent name is modified to reflect the highest-priority functional group

☐ other groups may be named as substituent prefixes

☐ identifying the suffix group is essential before naming the molecule

☆ why do some IUPAC names drop a vowel before the suffix?

☐ a vowel may be dropped to avoid awkward double-vowel combinations

☐ this often happens when joining the parent ending to the suffix

☐ the final name is written in the standard IUPAC form

☆ what is the difference between a hydrogen-bond donor and a hydrogen-bond acceptor?

☐ a hydrogen-bond donor provides the hydrogen atom in a hydrogen bond

☐ the donor must have an H directly bonded to an electronegative atom, usually O or N

☐ a hydrogen-bond acceptor provides a lone pair to interact with that hydrogen

☐ donor examples:

☐ O–H groups

☐ N–H groups

☐ acceptor examples:

☐ oxygen atoms with lone pairs

☐ nitrogen atoms with lone pairs

☐ some atoms or groups can act as both donor and acceptor

☐ an –OH oxygen can accept, while its H can donate

☐ an amine nitrogen (bonded to carbon group) can often accept, and if it has N–H, it can also donate

☐ hydrogen bonding affects solubility, boiling point, and intermolecular interactions

☆ how does a halogen add to an alkene?

☐ X₂ means Cl₂ or Br₂
☐ the alkene π bond acts as a nucleophile toward the halogen
☐ the reaction forms a vicinal dihalide
☐ the two halogens add across the double bond

☐ the reaction does not proceed through a free carbocation
☐ a bridged halonium ion intermediate forms first
☐ the halonium ion is three-membered and positively charged

☆ what is the mechanism of halogenation of an alkene?

☐ step 1: the alkene attacks X₂
☐ the X–X bond breaks as one halogen bonds to both alkene carbons
☐ a halide ion and a halonium ion form

☐ step 2: X⁻ attacks one carbon of the halonium ion from the back side
☐ the ring opens as the C–X bond to the bridging halogen breaks at one carbon
☐ the product is an anti addition product

☐ curved arrows must begin at the π bond and at the C–X bond of the halonium ion
☐ charge must be conserved through both steps

☆ why is halogen addition anti?

☐ the bridged halonium ion blocks attack from the same face
☐ the nucleophile must open the ring from the opposite face
☐ this gives anti addition overall

☐ anti means the two newly added groups end up on opposite faces
☐ anti addition is a stereochemical outcome of the mechanism
☐ it is not just a naming convention

☆ how is halogen addition regioselective when the alkene is unsymmetrical?

☐ the halonium ion is unsymmetrical
☐ one carbon bears more positive character than the other
☐ nucleophilic attack occurs preferentially at the more substituted carbon

☐ the more substituted carbon better stabilizes developing positive charge
☐ this resembles opening of an unsymmetrical bridged cation
☐ the product orientation follows attack at the more electrophilic carbon

☆ what happens when a halogen adds to a cyclic alkene?

☐ halogenation still proceeds through a halonium ion
☐ backside attack gives anti addition
☐ in a ring, anti addition often gives a trans relationship between the new substituents

☐ the product may be a pair of enantiomers if new stereocenters form
☐ do not assume only one drawing represents the full stereochemical outcome
☐ ring geometry helps reveal whether the product is cis or trans

☆ what happens when an alkene reacts with X₂ in water?

☐ water competes with X⁻ as the nucleophile
☐ the reaction forms a halohydrin
☐ one carbon receives X and the other receives OH

☐ a halonium ion still forms first
☐ water opens the halonium ion from the back side
☐ the addition is anti overall

☐ the OH group comes from water
☐ deprotonation gives the neutral halohydrin product

☆ how is halohydrin formation regioselective?

☐ water attacks the more substituted carbon of the halonium ion
☐ that carbon has greater positive character
☐ therefore OH ends up on the more substituted carbon

☐ X ends up on the less substituted carbon
☐ the real reason is nucleophilic attack at the more electrophilic carbon
☐ the overall addition remains anti

☆ what is the mechanism of halohydrin formation?

☐ step 1: the alkene reacts with X₂ to form a halonium ion

☐ step 2: H₂O attacks the more substituted carbon from the back side
☐ the ring opens to give a protonated alcohol

☐ step 3: another water molecule removes H⁺
☐ the neutral halohydrin forms
☐ the overall addition is anti

☐ this uses the same halonium logic as ordinary halogenation
☐ the main difference is the nucleophile that opens the ring

☆ how is halohydrin formation regioselective?

☐ water attacks the more substituted carbon of the halonium ion
☐ that carbon has greater positive character
☐ therefore OH ends up on the more substituted carbon

☐ X ends up on the less substituted carbon
☐ the real reason is nucleophilic attack at the more electrophilic carbon
☐ the overall addition remains anti

☆ what is the mechanism of halohydrin formation?

☐ step 1: the alkene reacts with X₂ to form a halonium ion

☐ step 2: H₂O attacks the more substituted carbon from the back side
☐ the ring opens to give a protonated alcohol

☐ step 3: another water molecule removes H⁺
☐ the neutral halohydrin forms
☐ the overall addition is anti

☐ this uses the same halonium logic as ordinary halogenation
☐ the main difference is the nucleophile that opens the ring

☆ how do you predict halogenation versus halohydrin formation?

☐ first identify the alkene
☐ then determine whether the reagent is X₂ alone or X₂ in water

☐ X₂ alone gives a vicinal dihalide
☐ X₂/H₂O gives a halohydrin
☐ both reactions proceed by anti addition

☐ for an unsymmetrical alkene, X₂/H₂O places OH on the more substituted carbon
☐ the halogen ends up on the less substituted carbon

☆ what does oxymercuration–demercuration do to an alkene?

☐ it hydrates an alkene to give an alcohol
☐ the net result is addition of H and OH across C=C
☐ OH appears on the more substituted carbon

☐ the reaction gives Markovnikov hydration
☐ it avoids carbocation rearrangements
☐ common reagents are Hg(OAc)₂/H₂O followed by NaBH₄

☆ why does oxymercuration avoid rearrangements?

☐ the reaction does not form a free carbocation
☐ a bridged mercurinium ion intermediate forms instead
☐ this prevents hydride or alkyl shifts before nucleophilic attack

☐ water attacks the more substituted carbon of the bridged intermediate
☐ demercuration later replaces the Hg-containing group with H
☐ the product is a rearrangement-free alcohol

☆ what is the mechanism of oxymercuration–demercuration?

☐ step 1: the alkene reacts with Hg(OAc)₂ to form a bridged mercurinium ion

☐ step 2: water attacks the more substituted carbon

☐ step 3: deprotonation forms an organomercury alcohol

☐ step 4: NaBH₄ replaces the Hg-containing group with H
☐ the net result is hydration of the alkene
☐ OH ends up on the more substituted carbon

☆ what does hydroboration–oxidation do to an alkene?

☐ it hydrates an alkene to give an alcohol
☐ the net result is addition of H and OH across C=C
☐ OH appears on the less substituted carbon

☐ the reaction gives anti-Markovnikov hydration
☐ it proceeds without a carbocation intermediate
☐ common reagents are BH₃ followed by H₂O₂, OH⁻

☆ why is hydroboration anti-Markovnikov?

☐ boron bonds to the less substituted carbon in the hydroboration step
☐ hydrogen bonds to the more substituted carbon at the same time
☐ oxidation later replaces B with OH at the same carbon

☐ the regiochemistry is set in the first step
☐ steric effects favor boron delivery to the less hindered carbon
☐ the final alcohol retains that orientation

☆ what stereochemistry does hydroboration give?

☐ hydroboration is a concerted syn addition
☐ H and B add to the same face of the alkene
☐ oxidation preserves the stereochemical relationship

☐ the overall result is syn addition of H and OH
☐ syn means both added groups come from the same face
☐ stereochemistry is especially visible in cyclic or substituted systems

☆ what is the mechanism logic of hydroboration–oxidation?

☐ step 1: BH₃ adds across the alkene in a concerted step
☐ no carbocation intermediate forms
☐ B attaches to the less substituted carbon and H to the more substituted carbon

☐ step 2: oxidation replaces C–B with C–OH
☐ the OH appears where B was originally attached
☐ the net product is an anti-Markovnikov alcohol

☆ how do the three alkene hydration methods compare?

☐ acid-catalyzed hydration gives Markovnikov alcohols
☐ acid-catalyzed hydration can rearrange

☐ oxymercuration–demercuration gives Markovnikov alcohols
☐ oxymercuration avoids rearrangements

☐ hydroboration–oxidation gives anti-Markovnikov alcohols
☐ hydroboration is syn and rearrangement-free

☆ what happens when a peroxyacid reacts with an alkene?

☐ the alkene is converted into an epoxide
☐ an epoxide is a three-membered cyclic ether
☐ the reaction is called epoxidation

☐ the oxygen is delivered in a single concerted step
☐ the alkene π bond is replaced by two C–O σ bonds
☐ common reagents include peroxyacids such as mCPBA

☆ what is the stereochemistry of epoxidation?

☐ epoxidation is a concerted reaction
☐ both C–O bonds form in the same step
☐ the relative stereochemistry of the alkene is retained in the epoxide

☐ a cis alkene gives a cis-substituted epoxide framework
☐ a trans alkene gives a trans-substituted epoxide framework
☐ no carbocation intermediate forms

☆ how are epoxides opened under basic conditions?

☐ a strong nucleophile attacks an epoxide carbon by backside attack
☐ the ring opens in an SN2-like step
☐ attack occurs at the less substituted carbon in an unsymmetrical epoxide

☐ the C–O bond breaks at the attacked carbon
☐ oxygen remains attached to the other carbon as an alkoxide
☐ protonation later gives an alcohol

☐ the opening gives an anti relationship between the nucleophile and OH
☐ basic opening is controlled mainly by sterics

☆ how are epoxides opened under acidic conditions?

☐ the epoxide oxygen is protonated first
☐ protonation makes the ring more electrophilic
☐ a nucleophile then attacks and opens the ring

☐ in unsymmetrical epoxides, attack usually occurs at the more substituted carbon
☐ the protonated epoxide has more positive character there
☐ backside attack still gives anti opening overall

☆ how do basic and acidic epoxide opening compare?

☐ both involve backside attack
☐ both give anti opening
☐ both relieve ring strain

☐ under basic conditions, the nucleophile attacks the less substituted carbon
☐ steric effects dominate under basic conditions

☐ under acidic conditions, the epoxide is protonated first
☐ the nucleophile usually attacks the more substituted carbon
☐ positive-charge development affects regioselectivity

☆ what does ozonolysis of an alkene accomplish?

☐ ozonolysis cleaves the C=C bond
☐ each alkene carbon becomes a carbonyl carbon
☐ the double bond is divided into two fragments

☐ product identity depends on the substituents attached to each alkene carbon
☐ products may be aldehydes, ketones, or carboxylic acids depending on workup
☐ always track each alkene carbon into its own carbonyl product

☆ how do you predict ozonolysis products from an alkene?

☐ locate the C=C bond
☐ break the bond between the two alkene carbons
☐ give each alkene carbon a double bond to oxygen

☐ an alkene carbon bearing H may become an aldehyde under reductive workup
☐ an alkene carbon bearing two carbon groups becomes a ketone
☐ cyclic alkenes open into acyclic dicarbonyl products

☆ what is the difference between reductive and oxidative ozonolysis workup?

☐ both begin by cleaving the alkene with ozone
☐ the difference appears in the workup step

☐ reductive workup preserves aldehydes
☐ oxidative workup converts aldehydes into carboxylic acids
☐ ketones remain ketones under either workup

☐ identifying whether an alkene carbon originally had H is crucial

☆ how do you work backward from ozonolysis products to an alkene?

☐ identify the carbonyl carbons in the products
☐ remove the oxygens conceptually
☐ connect the carbonyl carbons with a double bond

☐ if both carbonyls are in one product, the original alkene may have been cyclic
☐ if two separate molecules form, the original alkene may have been acyclic
☐ carbon counting helps confirm the reconstruction

☆ what is a carbene?

☐ a carbene is a neutral carbon species with six valence electrons
☐ the carbene carbon has only two σ bonds
☐ it is electron-deficient and highly reactive

☐ carbenes can add to alkenes
☐ this addition forms cyclopropanes
☐ methylene, :CH₂, is the simplest carbene example

☆ how does carbene addition to an alkene affect stereochemistry?

☐ carbene addition forms a cyclopropane
☐ both new C–C bonds form in one event
☐ the reaction is stereospecific

☐ a cis alkene gives a cis-substituted cyclopropane
☐ a trans alkene gives a trans-substituted cyclopropane
☐ the original alkene geometry is retained

☆ what does a Simmons–Smith reaction do?

☐ the Simmons–Smith reaction converts an alkene into a cyclopropane
☐ it delivers a CH₂ unit across the double bond
☐ the reaction is stereospecific

☐ alkene geometry is retained
☐ a cis alkene gives a cis cyclopropane
☐ a trans alkene gives a trans cyclopropane

☆ how do common alkene reactions compare by regiochemistry and stereochemistry?

☐ HX addition usually gives a Markovnikov product through a carbocation
☐ acid hydration gives a Markovnikov alcohol and can rearrange
☐ oxymercuration gives a Markovnikov alcohol without rearrangement
☐ hydroboration gives an anti-Markovnikov alcohol by syn addition

☐ X₂ gives an anti vicinal dihalide
☐ X₂/H₂O gives an anti halohydrin with OH on the more substituted carbon
☐ epoxidation retains the alkene’s relative stereochemistry
☐ catalytic hydrogenation gives syn addition of H₂