Organic Chemistry

Question 1

Nucleophile

Answer 1

an electronegative functional group which has an electron pair that can be used to form a covalent bond

Question 2

Electrophile

Answer 2

an electron deficient functional group which accepts an electron pair to form a covalent bond
must have an empty orbital

Question 3

Aromaticity and Hunkel’s Rule

Answer 3

to be aromatic a compound must be planar and cyclic and have 4n+2 delocalised pi electrons, where n is an integer

Question 4

Hybridization

Answer 4

the mixing of atomic orbitals to make molecular orbitals

Question 5

sp3 hybridization

Answer 5

4 identical hybrid orbitals
tetrahedral
can have lone pair not involved in bonding available to hydrogen bond

Question 6

sp2 hybridization

Answer 6

3 identical hybrid orbitals
trigonal planar
p orbital not involved in bonding is at right angles to the plane and can for m a double bond
partial double bonds and resonance require this

Question 7

Tautomers

Answer 7

structural isomers differing in position of a hydrogen and a double bond
eg. keto to enol or imine to enamine

Question 8

Hydroxyl Group

Answer 8

• sp3 hybridisation
• 2 lone pairs
• electronegative
• H bonding

Question 9

Ether

Answer 9

• sp3 hybridization
• lone pairs but poor nucleophile
• unreactive

Question 10

Carboxylic Acid

Answer 10

• sp2 hybridisation of oxygen promotes delocalisation of lone pair
• forms carboxylate ion which acts as a base

Question 11

Ester

Answer 11

• sp3 hybridization of oxygen promotes delocalisation of lone pair
• reduces electrophilicity of carbonyl carbon and the nucleophilicity of the alpha carbon
• less reactive than carbonyl compounds
• less acidic because the ester O pushes its electrons into the system, competing with the carbonyl electrons
• partial double bond character and planar

Question 12

Carbonyl compounds (ketons/aldehydes)

Answer 12

• sp2 hybridization of oxygen promotes delocalisation of lone pair
• reactive electrophilic carbon
• acidic alpha carbon that acts as a nucleophile when deprotonated

Question 13

Enolate ion

Answer 13

• H removal from alpha carbon by a base from a carbonyl compound
• stabilised by delocalisation of charge over the system
• removal changes hybridisation to sp2 so that pi orbital overlap can occur to form a double bond

Question 14

Aldehydes

Answer 14

More reactive than ketones

• less steric hindrance as it lacks the R group
• inductive effect in which the alpha carbon shares some of its electron cloud with the carbonyl carbon, partially quenching the partial positive charge

Question 15

Amine

Answer 15

• sp3 hybridization
• one lone pair
• electronegative
• can be a base and a nucleophile only if deprotonated
• when protonated can donate a proton (acid)

Question 16

Amide

Answer 16

• sp3 hybridization of N promotes delocalisation of lone pair and reduces electrophilicity of carbonyl carbon, ie. reactivity reduced
• N is not basic or a nucleophile
• planar molecule

Question 17

Imine

Answer 17

• similar to carbonyl compounds
• N protonated at physiological pH so it is strongly polarised
• carbon highly electrophilic

Question 18

Schiff Base

Answer 18

• protonation of imine causes N to pull more electrons towards it because of its positive charge
• p orbital sharing to form double bond between N and C

Question 19

Thiol

Answer 19

• sulfur equivalent of hydroxyl
• more nucleophilic because its lone pairs are in level 3 and highly polarisable
• orbitals are too big for effective overlap with hydrogen so no hydrogen bonding
• not basic
• good leaving group

Question 20

Thioester

Answer 20

• sp3 hybridization as it couldn’t effectively overlap with 2p orbitals if sp2 hybridized
• no electron sharing with carbonyl
• similar reactivity to aldehydes
• no resonance
• alpha carbon is nucleophilic as S doesn’t interfere with enolate anion formation
• good leaving group as it is highly polarisable

Question 21

Nucleophilic Substitution (Sn1)

Answer 21

• two step mechanism
• nucleophilic attack from either side of the plane to form racemic mixture
• carbocation intermediate formed as X removes electrons

Question 22

Nucleophilic Substitution (Sn2)

Answer 22

• one step mechanism (concerted)
• forms transition state where both groups are attached via stretched p orbitals
• TS is sp2-like in the same plane as its substituents
• inversion of stereochemistry

Question 23

Elimination

Answer 23

• loss of a leaving group from a tetrahedral carbon together with a hydrogen from the adjacent carbon to form a C-C double bond
• need a base
• base pulls off hydrogen and leaving group leaves
• pi bond formed from p orbital overlap
• p orbitals form as groups leave
• change to sp2 hybridization in both carbons

Question 24

E2

Answer 24

• single step mechanism
• B attacks hydrogen
• electron flow back into C-C bond
• X pulls off electrons and leaves
• formation of sp2 like TS

Question 25

E1

Answer 25

- two step mechanism - X leaves first and pulls off electrons - carbocation intermediate with empty p orbital forms - base attacks hydrogen and electrons flow into C-C bond

Question 26

Electrophilic Addition to C-C double bond

Answer 26

- pi cloud of the double bond acts as the nucleophile - protons are electrophiles - addition of a proton (from water) to the double bond gives carbocation intermediate that reacts with nucleophile (water) to give product

Question 27

Nucleophilic Addition

Answer 27

- carbonyl compounds without a leaving group - nucleophilic attack at electrophilic carbonyl carbon (sp2 hybridized) - as the bond forms between the carbonyl carbon and the nucleophile, the carbon and oxygen of the carbonyl group begin to change their hybridisation to sp3 - this provides the orbitals needed for bonding to the nucleophile and for the pi electrons to finish as a lone pair on the oxygen - base pulls hydrogen off hydroxyl group - attack on carbon and flow of electrons onto oxygen - tetrahedral intermediate that is protonated to give the product - protonation via attack on acid and breakage of acid hydrogen bond

Question 28

Imine formation

Answer 28

- nucleophilic addition | - addition of amine to aldehyde (or ketone) followed by loss of oxygen as water

Question 29

Hemiacetal Formation

Answer 29

- alcohol and aldehyde - nucleophilic addition mechanism - this then reacts with a second alcohol to form an acetal via a glycosidic linkage * see notes for mechanism*

Question 30

Schiff Base (Imine) Formation

Answer 30

- amine + aldehyde eg. lysine side chain and PLP - nucleophilic attack of N of amine on the carbonyl carbon - electron flow back onto oxygen - NH electron flow into N-C bond to make it a double bond and hydroxyl is protonated and leaves as water (is pushed off) - unstable tetrehedral intermediate

Question 31

Nucleophilic Substitution

Answer 31

- carbonyl compounds with leaving groups - mechanism the same as addition up to the formation of the tetrahedral intermediate - base deprotonates water and nucleophilic attack on carbonyl carbon - electrons flow back onto oxygen - oxonium ion intermediate - because there is a leaving group, electrons can flow back to reform the carbonyl pi cloud and the leaving group departs - sp2 collapsed product

Question 32

Aldol Reactions

Answer 32

- nucleophilic attack of an enolate anion on a second carbonyl - alpha carbon of second carbonyl is the nucleophile

Question 33

Aldol Addition

Answer 33

- electrophile doesn't have a leaving group - carbanion attack on carbonyl carbon - electrons flow onto oxygen, which is protonated by an acid - product: B hydroxy carbonyl

Question 34

Aldo Substitution

Answer 34

- electrophile has leaving group - carbanion attack on carbonyl carbon - electron flow onto oxygen - formation of negative oxygen intermediate - electrons flow back into C-O bond and leaving group leaves - product: B keto carbonyl compound