Topic 10 - Organic Chemistry Flashcards

Question

condensed structural formula

Answer 1

- often omits bonds where they can be assumed - groups atoms together - contains the minimal info required to describe the molecule unambiguously

Answer 2

- shows relative positions of atoms and groups around carbon in 3D - bonds sticking forwards are shown as a wedge - bonds sticking backwards are shown as hashed lines - bonds along the plane of the paper are shown as solid lines

Answer 3

1. Identify the longest straight chain of carbon atoms 2. Identify the functional group 3. Identify subchains and substituent groups

Answer 4

- side chains or functional groups in addition to the one used as the suffix - they are given as the prefix

Answer 5

- molecules with the same molecular formula but with differing arrangements of atoms - each isomer is a distinct compound with distinct physical and chemical characteristics

Answer 6

attached to: - the functional group - 2+ H atoms

Answer 7

attached to: - functional group - 1 H atom - 2 alkyl groups

Answer 8

attached to: - functional group - 3 alkyl groups - 0 H atoms

Answer 9

- class of compounds derived from benzene (C6H6) | - they form a special branch of organic compounds known as aromatics

Answer 10

- 1:1 ratio of C:H indicates high degree of unsaturation (greater than alkenes/alkynes) - but unlike other unsaturated molecules, benzene has no structural isomers - benzene is also unwilling to undergo addition reactions

Answer 11

- cyclic structure - a framework of single bonds attaches each C to the one on either side and to a H atom - each of the 6 Cs are sp2 hybridized - forms 3 sigma bonds with angles of 120 degrees - planar shaped - stable arrangement

Answer 12

- there's 1 unhybridized p e- on each carbon atom - their dumbbell shapes are perpendicular to the ring - instead of forming discrete alternating pi bonds, they effectively overlap in both directions - thus spreading themselves out evenly and forming a delocalized pi e- cloud - electron density is concentrated above & below the plane of the ring - this lowers the internal energy of the molecule

Answer 13

- all C-C bond lengths in benzene are equal and intermediate in length - because each bond contains a share of 3 e-s between the bonded atoms

Answer 14

- it has been experimentally proven that benzene is more stable than the Kekule structure predicts - this is bc delocalization minimizes the repulsion between electrons - this gives benzene a more stable structure by reducing its internal energy by (experimental value - theoretical value) and thus its resonance energy by extension

Answer 15

- AKA stabilization energy | - energy required to overcome the stability of the delocalized ring

Answer 16

- benzene is reluctant to undergo addition reactions - more likely to undergo substitution reactions - bc addition reactions are energetically not favored - they would disrupt the entire cloud of delocalized electrons - resonance energy would be required - without the delocalized ring of e-s, the product would be less stable as well - substitution reactions are preferred as they preserve the stable ring structure

Answer 17

- only one isomer exists of each compound - because benzene is a symmetrical molecule with no alternating single/double bonds - so all adjacent positions in the ring are equal

Answer 18

- a framework consisting of carbon and hydrogen only (this is known as the hydrocarbon skeleton) - functional group (differs between homologous series)

Answer 19

more branching = lower boiling point

Answer 20

- saturated hydrocarbons with strong C-C and C-H bonds - thus require high activation energy to break those bonds - so alkanes are generally stable under most conditions and can be stored/transported/compressed safely - as C-H and C-C bonds are non-polar, they aren't susceptible to attack by common reactants - so alkanes are generally very unreactive

Answer 21

- alkanes release a significant amount of energy when broken - so they are widely used as fuels - alkane combustion reactions are highly exothermic because of the large amounts of energy released when forming CO2 and H2O - the products are fully oxidized, so alkane undergoes complete combustion

Answer 22

- incomplete combustion - CO and H2O produced instead - in extreme oxygen limitation, just C and H2O will be produced

Answer 23

- under complete or incomplete combustion depending on oxygen availability - large amounts of energy are released generally - as C:H ratio increases with unsaturation, so does the smokiness of the flame due to unburned hydrocarbon

Answer 24

- CO2 and H2O are greenhouse gases - thus they contribute to global warming and climate change - CO is a toxin as it combines irreversibly with blood hemoglobin, preventing it from carrying oxygen - unburned carbon is released into the air as particulates - they have a direct effect on human health - they also catalyze the formation of smog in polluted air - they are also the source of global dimming

Answer 25

- main reaction undergone by alkanes - occurs when another reactant (halogen) takes the place of a hydrogen atom in the alkane - light-dependent as energy from UV rays is needed to break covalent bonds in the halogen molecule - energy splits the halogen molecule into free radicals - the radicals start a chain reaction in which a mixture of products (including the halogenoalkane) is formed

Answer 26

- the chain reaction of the substitution reaction | - it occurs as a sequence of steps

Answer 27

- initiation - propagation - termination

Answer 28

- AKA photochemical homolytic fission - the bond between the 2 halogen atoms in the diatomic halogen molecule is broken - photochemical refers to the light dependency of the rxn - homolytic refers to the the fact that the 2 products have an equal assignment of bond e-s upon splitting

Answer 29

- AKA chain reaction | - series of reactions that use and produce free radicals

Answer 30

- remove free radicals from the rxn mixture | - by causing them to react together and pair up their e-s

Answer 31

- bromine water changes from brown to colorless - for alkanes: occurs in UV light only - for alkenes: will occur rapidly at room temp

Answer 32

C(n)H(2n+2)

Answer 33

- unsaturated hydrocarbons - double bond is made up of 1 sigma and 1 pi bond - the C atoms are sp2 hybridized - trigonal planar shape

Answer 34

- more reactive than alkanes - as its double bond is the site of reactivity of the molecule - the pi bond is broken relatively easily - so the double bond can be readily broken for a reaction - 2 new bonding positions are created on the C atoms - this enables alkenes to undergo addition reactions - they can form a range of differing saturated products

Answer 35

hydrogen + alkene --> alkane - AKA hydrogenation - catalyzed by nickel - ideal temp: 150°C

Answer 36

- used in margarine industry - converts unsaturated hydrocarbon chains into saturated compounds with higher melting points - so that margarine is solid at room temp

Answer 37

trans fats are produced by partial hydrogenation

Answer 38

halogen + alkene --> dihalogenoalkane - occur quickly at room temp - color change observed (color to colorless) - alkene's double bond is broken and halogen atoms attach to each of the 2 C atoms that were in a double bond

Answer 39

hydrogen halide + alkene --> halogenoalkane - occur rapidly at room temp - reactivity in order: HI > HBr > HCl - this is because the strength of the halide bond decreases down the group - so as HI has the weakest bond it reacts most readily

Answer 40

water + alkene --> alcohol - AKA hydration - ideal condition when heated with steam - conc. H2SO4 as catalyst - involves an intermediate/transition state - in which the double bonds are broken, then H+ and HSO4- ions bond to the C atoms that were in a double bond - final state occurs with hydrolysis and replacement of HSO4- with OH-, and H2SO4 then reforms

Answer 41

- used to manufacture ethanol on a large scale | - as ethanol is a very important solvent

Answer 42

- long chains of alkenes - forms when alkenes are joined together in addition reactions - this occurs bc alkenes can readily break their double bonds to react

Answer 43

- used to symbolize polymer structures - it's a single unit with repeating bonds at each end - put in a bracket with n in subscript e.g. ethene in polymer form is called polythene (polyethene) propene in polymer form is polypropylene (polypropene)

Answer 44

C(n)H(2n+1)

Answer 45

as the -OH group is polar, it increases alcohols' solubility in water relative to alkanes

Answer 46

- like hydrocarbons, alcohol combustion produces significant amounts of energy along with CO2 and H2O - the amount of energy released per mole increases down the homologous series

Answer 47

like hydrocarbons, the combustion of alcohol produces CO and H2O instead of CO2 and H2O

Answer 48

- combustion completely oxidizes alcohol molecules - but alcohols can also react with oxidizing agents that selectively oxidize the O atom in the -OH group - this keeps the carbon skeleton intact - allows alcohols to be oxidized into other organic compounds - the most common oxidant used as acidified potassium dichromate (VI) [K2Cr2O7], and the bright orange solution is reduced to just green Cr (III) over the course of the rxn - oxidants are often represented as + [O] in diagrams

Answer 49

- primary alcohols are oxidized to form aldehydes, then further oxidized to form carboxylic acids - the second oxidation (aldehydes to carboxylic acids) needs to be heated under reflux

Answer 50

- bacteria will slowly oxidize the ethanol to ethanoic acid | - i.e. alcohol to carboxylic acid

Answer 51

- the oxidation of primary alcohols can be stopped on the first step of oxidation (alcohols to aldehydes) - distillation can be utilized to remove the aldehydes from the reaction mixture - this is possible because aldehydes have lower boiling points than alcohols and carboxylic acids

Answer 52

- simply leave the aldehyde alone with the oxidant for an extended period of time - preferably heated under reflux

Answer 53

- secondary alcohols are oxidized to ketone | - preferably heated under reflux

Answer 54

- tertiary alcohols don't readily oxidize - their carbon structure needs to be broken and this requires a lot more energy than the oxidation of other alcohols - so unlike the other 2 oxidations, oxidizing tertiary alcohols with K2Cr2O7 won't result in a color change as there is no reaction

Answer 55

carboxylic acid + alcohol ester + water - reversible reaction - type of condensation reaction - catalysed by concentrated H2SO4

Answer 56

- as the ester has the lowest boiling point, they can be removed via distillation - the presence of esters can be identified by their distinct fruity, sweet smell - as they don't have -OH groups (unlike the reactants), they can't form hydrogen bonds and will remain as an insoluble layer on the surface of water

Answer 57

C(n)H(2n+1)X

Answer 58

- basically like alkanes except 1 halogen atom is substituted for 1 of the hydrogen atoms - as halogenoalkanes are saturated molecules, they undergo substitution reactions

Answer 59

- halogenoalkanes contain a polar bond - this makes them more reactive than alkanes - the halogen atom is more electronegative than the C atom - so they exert a stronger pull on the shared e-s in the C-H bond - thus halogen exerts a partial -tive charge while the C atom exerts a partial +tive charge - the C atom is then said to be e- deficient - this e- deficient C atom defines much of a halogenoalkane's reactivity

Answer 60

- reactions in which substitution of the halogen (From halogenoalkane) occurs - due to a nucleophile being attracted to the e- deficient C atom

Answer 61

- stable because of the delocalized ring of e-s - that delocalized ring of e-s represents an area of e- density and is the site of reactivity - prefers substitution reactions - where a H atom is replaced by an incoming group - preferred because the arene ring is conserved - doesn't favor addition reactions - as addition reactions would lead to the loss of the stable arene ring - this is because the products would have higher energy than the reactants

Answer 62

- electrophiles are attracted to the e- rich benzene ring | - results in the substitution of one of the H atoms with an ion from the electrophilic molecule

Answer 63

- polar C-Halogen bond results in e- deficient C - C is therefore attacked by a nucleophilic ion (e.g. OH-) - the C-Halogen bond breaks and the halogen is the leaving group (leaves in the form of a halide) - as the halide takes both e-s from the bond, this is heterolytic fission

Answer 64

when a covalent bond breaks and one product takes both e-s in the bond

Answer 65

the group that gets displaced in a reaction

Answer 66

- it can - but it lacks a negative charge - so it's a relatively weak nucleophile - the rxn occurs more slowly

Answer 67

- stands for substitution nucleophilic bimolecular - favored by polar, aprotic solvents - 1-step concerted reaction with a transition state - stereospecific - has inversion effect (like umbrella being blown inside out) on the final product reactants -> transition state -> products

Answer 68

when the rxn mechanism of a rxn is dependent on both reactants e.g. for halogenoalkanes rate = k [halogenoalkane] [nucleophile]

Answer 69

- when the 3D arrangement of the reactants determines the 3D arrangement of the products - occurs because in transition state, the bond formation occurs before fission - so the stereochemistry of the C attacked isn't lost

Answer 70

- solvents unable to form hydrogen bonds - due to not containing -OH or -NH groups - they still may have strong dipoles though - in a reaction, they will solvate a metal ion over a nucleophile - thus increasing the rate of rxn

Answer 71

- SN2 - as H atoms are small, the C atom is open to attack by nucleophiles - an unstable transition state forms in which the C atom is weakly bonded to both the halogen and the nucleophile - the C-Halogen bond then breaks heterolytically - this releases the halogen and forms an alcohol product

Answer 72

- the nucleophile attacks on the opposite side from the leaving group - thus the arrangement of atoms become inverted when the halogenoalkane undergoes fission - like an umbrella blowing inside out)

Answer 73

rate = k [halogenoalkane] - substitution nucleophilic unimolecular - 2 steps - the rate-determining step is determined only by the halogenoalkane concentration - favored by polar, protic solvents - non-stereospecific reactants --> carbocation intermediate --> products

Answer 74

- when large groups bonded to a specific atom makes it difficult for nucleophiles to attack that atom - due to their size

Answer 75

- occurs when the C atom breaks its C-Halogen bond (heterolytic fission) - and the C atom is left with a temporary +tive charge - this phenomenon is called the carbocation intermediate - carbocation intermediate molecules are unstable as they don't fulfil the octet rule

Answer 76

- when a group has an electron-donating effect | - this helps stabilize an unstable structure

Answer 77

- SN1 - the tertiary C atom is bonded to 3 alkyl groups and 1 halogen - the 3 alkyl groups provide steric hindrance 1st step (slow): - the C atom breaks its C-Halogen bond heterolytically - carbocation intermediate forms - the 3 alkyl groups have a positive inductive effect - stabilizes the carbocation to persist long enough for the 2nd step to occur - as the carbocation intermediate has a planar shape, the nucleophile can attack from any position ``` 2nd step (fast): - the C atom bonds with the nucleophile ```

Answer 78

e. g. water - contains an -OH or -NH group - can form hydrogen bonds - thus are effective in stabilizing the carbocation intermediate

Answer 79

can be SN1, can be SN2

Answer 80

- type of SN - polarity of C-Halogen bond - strength of C-Halogen bond - choice of solvent

Answer 81

- SN1 faster than SN2 | - overall speeds: tertiary > secondary > primary

Answer 82

- less polar bonds would result in C being less e- deficient - thus the more polar the bonds, the more vulnerable C is to a nucleophilic attack - overall speeds: fluoroalkane > ... > iodoalkane

Answer 83

- less reactive halogen = weaker bond | - overall speeds: iodoalkane > ... > fluoroalkane

Answer 84

- the strength dominates the polarity | - so overall speeds: iodoalkane > ... > fluoroalkane

Answer 85

- SN1 favored by polar protic solvents | - SN2 favored by polar aprotic solvents

Answer 86

- the halide may appear - if AgNO3 is added, it will react with the halide - forms a silver halide precipitate with a distinct color

Answer 87

- Cs are sp2 hybridized - they form a planar triangular shape - bond angle 120 - open structure, makes it easy for incoming groups to attack - 1 central sigma bond and 2 pi bonds above & below - as pi bonds are areas of e- density, they attract electrophiles - pi bonds are much weaker than sigma bonds so they break a lot more easily during addition reactions

Answer 88

- reactions in which an electrophile molecule attacks 2 double-bonded C atoms - breaks one of the bonds so the = bond becomes single bond - the molecule will split to attach to each C atom

Answer 89

alkene + halogen --> dihalogenoalkane e. g. ethene + bromine --> dibromoethene - ethene gas is bubbled through bromine step 1 (slow): - although bromine is non-polar, a temporary dipole can be induced by electron repulsion as bromine approaches the pi bond (which is e- dense) - the bromine closer to the pi bond gains a partial +tive charge and splits heterolytically - the +tive bromine acts as an electrophile - results in unstable carbocation intermediate that doesn't follow octet rule ``` step 2 (fast): - the carbocation reacts with the -tive bromide ion ```

Answer 90

- rxn between ethene + bromine in the presence of Cl- - no dichloro compounds formed - thus confirming that the initial attack was by the electrophile Br+

Answer 91

alkene + hydrogen halide --> halogenoalkane - ethene gas is bubbled through hydrogen bromide - same mechanism as alkene + halogen

Answer 92

alkene + hydrogen halide --> halogenoalkane - there are multiple potential stereoisomers of each halogenoalkane depending on the stereoisomer of alkene and on which C atom is attacked by the electrophile - as all alkenes after ethene have stereoisomers - each stereoisomer reactant has 2 potential stereoisomer products (although only 1 will be produced) - if the electrophile attacks a primary C, in carbocation state there's a positive inductive effect from the 1 alkyl group - positive inductive effect doubles for secondary C and triples for tertiary C, etc - the more stable the carbocation, the more likely it is that the reaction will complete with a halogenoalkane

Answer 93

the more electropositive part of the reacting species will always bond to the C atom that has the least bonds with other C atoms

Answer 94

- despite high unsaturation, benzene doesn't behave like alkenes - as its stable aromatic ring will be altered in addition reactions, substitution is preferred - but it's still attracted to electrophiles - so benzene undergoes electrophilic substitution

Answer 95

- mostly specific to arenes - this reaction requires high activation energy - benzene's delocalized cloud of pi electrons seeks electrophiles and forms a new bond as a H atom is lost 1st step (slow): - e- pair from benzene is attracted to electrophile - symmetry of delocalized pi e-s is disrupted - unstable carbocation intermediate forms - both the entering group and leaving group are temporarily bonded 2nd step (fast): - leaving group leaves - 2 e-s from the C-H bond move to regenerate the symmetry of the aromatic ring

Answer 96

mixture of concentrated nitric acid and sulfuric acid

Answer 97

- substitution of H by -NO group - catalyst: conc H2SO4 - ideal temp: 50°C - nitronium is used as electrophile - generated by using a nitrating mixture at 50°C - sulfuric acid is stronger so it protonates nitric acid - the nitrating mixture then loses H2O to form nitronium ions - NO2+ is a strong electrophile and reacts with pi e-s to form carbocation intermediate - the loss of a proton (H+) leads to reformation of the arene ring - final product: C6H5NO2 + H2O

Answer 98

compounds containing C=O groups

Answer 99

oxidation of: primary alcohol --> aldehyde --> carboxylic acid secondary alcohol --> ketone these oxidation reactions can be reversed using: - sodium borohydride (NaBH4) in aqueous/alcoholic soln - lithium aluminium hydride (LiAlH4) in anhydrous conditions - both reagents produce a hydride ion (H-) to act as a nucleophile - NaBH4 is safer but not strong enough to reduce carboxylic acids - LiAlH4 can reduce carboxylic acids -

Answer 100

- nitrobenzene can be reduced to form phenylamine - 2-stage reduction process 1. - C6H5NO2 reacts w a mixture of tin (Sn) and conc HCl - heated under reflux (boiling water bath) - produces C6H5NH3+ (phenylammonium ions) which is protonated because of the acidic conditions 2. - reacted w NaOH to remove H+ - forms C6H5NH (phenylamine)

Answer 101

- series of discrete steps used to convert compounds from one form to another - by organizing reactions in a sequence so that the product of one reaction is the reactant of the next reaction - its basis lies in the interconversions of functional groups

Answer 102

- technique for working out synthetic routes | - works backwards from the target molecule through precursors all the way to the starting material

Answer 103

compounds with the same molecular formula but different arrangement of atoms

Answer 104

- structural isomerism | - stereoisomerism

Answer 105

atoms and functional groups are attached in different ways

Answer 106

- similar attachments | - different spatial (3D) arrangements

Answer 107

- configurational isomerism | - conformational isomerism

Answer 108

can only be interconverted by breaking covalent bonds

Answer 109

- can be interconverted by free rotation about sigma bonds - usually spontaneously interconvert via rotation so cannot be isolated separately - but some conformers are more stable than others (these are favored)

Answer 110

- cis-trans or E-Z isomerism | - optical isomerism

Answer 111

- exists where rotation is restricted around atoms | - so that they become fixed in space relative to each other

Answer 112

isomers with the same groups on the same side of the double bonds/ring (i.e. reference plane)

Answer 113

isomers with the same groups on opposite sides of the double bonds/ring (i.e. reference plane)

Answer 114

the plane of the ring in a cyclic compound

Answer 115

DOUBLE BONDS - if the double bond consists of 1 sigma and 1 pi bond (the pi bond must be formed by sideways overlap of p orbitals) - thus free rotation is not possible as the p orbitals would be pushed out of position, breaking the pi bond CYCLIC MOLECULES - most cycloalkanes exhibit aromaticity and the ring of C atoms restricts rotation - bond angles are strained from the tetrahedral angles (e.g. cyclobutane has bond angles of 90°) - when the molecule contains 2+ different groups attached to the double bonds of the ring, they can be arranged to give 2 different isomers (cis or trans) - substituted groups don't have to be on adjacent carbon atoms (it's their position relative to the reference plane that matters)

Answer 116

used where cis-trans can't explain the structure e.g. if all groups attached to the double bond are different, or if the C atoms are bonded to 2+ different substituents

Answer 117

- out of all the atoms bonded to C-atoms of the = bond, higher atomic number = higher priority - longer carbon chains have higher priority e. g. C3H7 > C2H5 > CH3 > H & Br > Cl > F - E-isomers: higher priority groups on opposite sides of the = bond - Z-isomers: higher priority groups on the same side of the = bond

Answer 118

AKA stereocentre OR asymmetric compounds - when a C atom is attached to 4 different atoms/groups - the 4 attached atoms/groups have bond angles of 109.5° - the compound has no plane of symmetry so is non-superimposable

Answer 119

- a chiral molecule has 2 alternative 3D configurations (both mirror images of each other) - they are said to be optical isomers - most molecules have multiple chiral atoms so they'll have more than just 2 optical isomers

Answer 120

- think of it like hands - hands are mirror images but non-superimposable - as the fingers and thumbs don't line up - similarly, isomers of chiral molecules are mirror images yet non-superimposable

Answer 121

- enantiomers | - diastereomers

Answer 122

- optical isomers - mirror images - they have opposite configurations at each chiral centre

Answer 123

AKA racemate - mixture containing equal amounts of 2 enantiomers (must be mirror images of each other) - optically inactive

Answer 124

- molecules that have opposite configurations at 1+ (but not all) chiral centres - so not mirror images of each other - common sugars are diastereomers of each other (e.g. glucose, galactose)

Answer 125

- optically active | - reacts with other chiral molecules

Answer 126

- light that has been passed through a polarizer - so that only light oscillating along a specific plane passes through - while all other planes are blocked out

Answer 127

- optical isomers will show a difference in a specific interaction with light - when plane-polarized light is passed through optical isomers, the plane of polarization is rotated - different enantiomers of the same compound (& at same concs) will rotate plane-polarized light in equal amounts but in opposite directions

Answer 128

- polarimeter - measures direction of rotation & how much the light has been rotated PROCEDURE - light is plane-polarized - then passed through a solution - then passed through a second polarizer (the analyser) - the analyser is rotated until the light can pass through it (thus identifying the extent and direction of rotation)

Answer 129

- rotation to the right is denoted by + | - vice versa for left (-)

Answer 130

- if a racemic mixture reacts with an enantiomer of another chiral compound, the 2 enantiomers of the racemic mixture will react to produce diff products - the products have distinct chemical and physical properties and thus can be easily separated - this is one way to separate a racemic mixture into its enantiomer constituents

Answer 131

- biological systems are chiral environments - when ingesting drugs, one enantiomer may produce a helpful product while the other may produce a life-endangering product

Answer 132

- if a racemic mixture reacts with an enantiomer of another chiral compound, the 2 enantiomers of the racemic mixture will react to produce diff products - the products have distinct chemical and physical properties and thus can be easily separated