3-4: ORGANIC CHEMISTRY MECHANISMS

Question 1

SN1

Answer 1

• has intermediate
• needs v.good leaving group
• Nu not involved in first step
• Nu attacks carbocation from either side of plane
• in chemical reaction, makes racemic mixture because it does not go w/stereochemical integrity like sn2
• in enzyme-catalysed reaction results in stereochemical purity of the product (one/other isomer depends on enzyme)
• 1st order - 1 reagent in RDS

Question 2

Sn2

Answer 2

• Nu attacks electrophilic s+ C whilst leaving group starts to leave so an orbital is available for Nu to attack C
• tetrahedron has inverted itself so sn2 reaction kinetic goes w/inversion of stereochemistry
• transition state C is sp2 like
• concerted mechanism

Question 3

E1

Answer 3

• makes C=C double bond
• involves bases not nucleophiles but many FG’s can be both
• same reactant as SN
• RDS involves leaving group leaving (same as step 1. in SN1)
• makes sp2 carbocation
• has empty p bond waiting for e pair to be shared into it
• as bond H-C starts to break, starts to hybridise sp2 so starts to generate empty p-orbital on that C as well as carbocation
• double bond is going to be formed by sideways overlap of p orbitals in the carbons
• stepwise/sequential reaction

Question 4

E2

Answer 4

• 2nd order; both base and molecule w/leaving group are involved in RDS
• both C are becoming sp2-like in TS; progresses to form sp2 hybridisation
• pair of e in sp3 hybrid orbital (H-C) of original molecule are going to move into the pi-cloud to be sharing w/empty p-orbital thats formed on C w/LG because LG is taking the e pair
• concerted reaction
• BH+ and X-

Question 5

electrophilic addition to C=C double bond (product = alcohol)

Answer 5

• e in double bond are behaving as nucleophile and they’re attacking one of the H in the water molecule to form a carbocation intermediate (transiently you have OH-)
• once carbocation is formed, another H2O comes in and acts as nucleophile and attacks C+
• transiently the water is now an OH- that will be protonated, but that H+ will be rapidly removed in an enzyme reaction by a base in the active site
• product is alcohol

Question 6

nucleophilic addition at carbonyl compounds

e.g. addition of alcohol to aldehyde to form hemiacetal

Answer 6

• reaction takes place in active site of an enzyme side-chain
• basic side chain is positioned near alcohol and a H-bonded acid positioned near the carbonyl acceptor
• base removes proton from alcohol and at the same time nucleophilic pair of e from H-O bond directly attack carbonyl
• C can react (orbitals overlap) because the e flow out onto the O
• product is hemiacetal

Question 7

carbonyl compounds

Answer 7

• carbonyl systems is sp2 planar
• Cc is electrophilic because O is pulling e away from it
• acidic H can be removed by base (easily because of formation of enolate anion where -ve charge is delocalised over carbonyl; stability of enolate ion depends on nature of X)
• after removal of that proton, Ca becomes nucleophilic
• Nu can come from either above/below plane but will have consequences on stereochemistry of product

Question 8

hemiacetal to acetal

Answer 8

• hemiacetal can react w/another alcohol or second molecule of same alcohol to form acetal
• OH group leaves as water because O of ether part of hemiacetal is able to use its LP to make an oxonium ion (a carbonation intermediate where O is +ve)
• pair of e on second alcohol attacks electrophilic C of oxonium ion and e flow back into O resulting in product, acetal

Question 9

carbonyl addition:

amine + aldehyde = schiff base

Answer 9

• e.g. lyside side chain + pyridoxal phosphate (PLP)
• lysine cannot be a nucleophile if its protonated, so base has to remove H+ first
• need a second base to catalyse reaction
• LP of e come from amine
• LP attacks carbonyl, e flow out and form a tetrahedral intermediate after protonation; its v. unstable due to stability of imine
• product forms when LP on N flows into N-C bond (can do that because OH is a good leaving group as water)

Question 10

nucleophilic substitution w/ thioester

Answer 10

• reaction is hydrolysis of thioester to make carb.acid + thiol
• form sp3 tetrahedral intermediate
• after intermediate is formed, e flow back and regenerate the C=O
• carboxylic acids are not protonates at physiological pH so have to indicate when protons are taken off

Question 11

aldol condensation

Answer 11

• involves water in different way than usual condensation
• have to have enolate anion so first make sure base electrophilic attacks Ca of aldehyde/ketone
• have 2 carbonyl compounds: enolate and aldehyde
• one behaves as nucleophile in carbonyl reaction using its Ca (deprotonated enolate anion w/delocalisation of e over carbonyl system)
• the other is the electrophile
• addition and substitution: if electrophile does not have leaving group then its addition

Question 12

aldol addition

Answer 12

• Nu attack on carbonyl in enzyme thats H-bonded to acid that will be the proton in reaction
• product is ß-hydroxy-carbonyl irrespective of what starting materials have been chosen
• reaction is reversible but reverse reaction requires another enzyme
• catabolism (breaks C-C), insert OH at ß position of carbonyl, or it oxidised an alcohol to make a carbonyl which is in a ß-position to OH so creates arrangement allowing retro-aldol to take place

Question 13

aldol substitution

Answer 13

• nucleophilic substitution using 2 carbonyls
• X can be anything in any carbonyl compound
• using Ca as nucleophile
• makes tetrahedral intermediate; important to not protonate because once its formed the e flow back to regenerate the carbonyl w/thiol leaving group
• product is ß-keto-carbonyl compound
• reaction is also reversible; needs different enzyme; C-C is readily broken