s&r topic 9

Question 1

in the C=O bond, give the hybridisation for each atom

Answer 1

C

• sp2 hybridised
• 3 sp2 orbitals, one used for σ(C-O)
• one p-orbital left for π-system

O

• sp2 hybridsed
• one sp2 orbital used for σ(C-O), left over 2 are n.b. and contain two lone pairs
• one p-orbital for π-system

Question 2

in the C=O bond, describe the pi-system and its polarisation

Answer 2

p-orbital overlap between C and O lead to a pi-bond

difference in electronegativity between C and O = polarisation of bond = pi-bond is skewed towards O

π* orbital is skewed towards C

Question 3

in the C=O bond, where is the LUMO

Answer 3

LUMO is the π* orbital

which is electrophilic at carbon, nucleophile likely to attack the carbon

Question 4

nucleophilic attack on C=O

• where does the nucleophile attack?
• which two orbitals are interacting?
• describe transfer of electrons

Answer 4

• nucleophile attacks the LUMO of the carbonyl group, the π* orbital that is empty and skewed towards carbon
• the HOMO of the nucleophile interacts with LUMO of carbonyl
• electrons are being transferred from HOMO to LUMO

Question 5

nucleophilic attack on C=O

• how does the nucleophilic attack change the bonding and hybridisation of the carbonyl atoms?

Answer 5

• nucleophilic attack leads to a tetrahedral intermediate
• electrons filling π* orbital causes pi-bond to break and sigma bond is formed between the carbon and the nucleophile
• carbon is now sp3 hybridised
• pair of electrons that were once in the pi-bond are now a negative charge on the oxygen

Question 6

nucleophilic attack on C=O

what are two outcomes for the tetrahedral intermediate

Answer 6

• nucleophilic addition: the O- can gain a proton and give an alcohol product
• nucleophilic substitution: the O- can collapse and form a carbonyl group again and leaving group is removed

Question 7

define leaving group

Answer 7

a molecule fragment that leaves a molecule with a pair of electrons

Question 8

nucleophilic substitution on C=O

what determines the stability of the product when the C=O reforms?

Answer 8

the stability of the product depends on the stability of the leaving group anion

most stable anion = best leaving group

more stable anion = more electronegative

Question 9

how is leaving group ability determined

how does reactivity change with leaving group ability

Answer 9

pKa can show leaving group ability

if X- is the leaving group, the lower the pKa value of the acid HX, the better the leaving group

stronger acids make better conjugate base leaving groups

as you increase pKa (worse leaving groups) reactivity decreases

Question 10

describe what happens to the MOs of the carbonyl group when an amide (NH2 group) is present

Answer 10

in an amide, the nitrogen lone pair is in a p-orbital and will overlap with the empty π* orbital of the carbonyl, this is delocalisation

this creates a new bonding and anti-bonding MO

• bonding MO is lower in energy and stabilised
• anti-bonding MO becomes too high in energy, making it less ready to react with nuclephiles

Question 11

what are 3 conditions for nucleophilic substitution of C=O

Answer 11

need a good enough nucleophile

need an electrophilic enough carbonyl

need a good leaving group