Organic Chemistry

Question 1

Induction

Answer 1

Carbocation draws electron density from its substituents

Therefore the more substituents, the more stable the carbocation

Question 2

The Zaitsev Rule

Answer 2

More substituted alkene will have a more stable carbocation intermediate and will therefore be he major product

Question 3

Nucleophiles and Nucleophile Trends

Answer 3

• electron rich atom which is capable of donating an electron
• typically have lone electron pair to donate
• usually negative or neutral
• lewis base
• the more polarisable an atom (ability to unevenly distribute electron density), the stronger the nucleophile
• the larger the atom, the more polarisable because the electrons are further from the nucleus
• more negative = better nucleophile than corresponding neutral molecule (OH- > H2O)
• less electronegative = better nucleophile
• an atom with have strong polar bonds when it is electronegative

Question 4

Examples of Strong Nucleophiles

Answer 4

Br-, I-
CH3S-, RS-
OH-, CH3O-, RO-

Question 5

Electrophile

Answer 5

• electron deficient atom which is capable of accepting an electron pair
• lewis acid
• often has empty p-orbital

Question 6

Sn2 Mechanism

Answer 6

rate = k [substrate] [nucleophile]

• substitution (s), nucleophilic (n) and biomolecular (2)
• concerted (one step)
• nucleophile attacks with simultaneous loss of leaving group
• transition state cannot be isolated

Question 7

Sn1 Mechanism

Answer 7

rate = k [substrate], there is only one rate determining step

Rate depends on structure of substrate:
methyl < primary < secondary < tertiary
Tertiary substrate will react quickest (most reactive

• substitution (s), nucleophilic (n) and unimolecular (1)
• leaving group leaves first, then nucleophile attacks substrate
• transition state can be isolated
• reaction energy diagram has two humps
• tertiary carbocations are more stable and will have a smaller Ea
• accompanied by competing E1 –> mixture of products
• more substituted alkene is the major product (Zaitsev Rule)

Question 8

Substrate Structure and Substitution Reactions

Answer 8

tertiary –> Sn1

secondary –> Sn1 or Sn2

primary –> Sn2

Question 9

E2 Mechanism

Answer 9

rate = k [substrate] [base]

• elimination (e) and bimolecular (2)
• concerted (one step)
• simultaneous bond breaking and bond making
• base removes proton from beta-carbon
• at the same time, leaving group departs and a double bond is formed
• tertiary substrates undergo E2 readily (reagent acts as a base not as a nucleophile)
• favoured with a strong base

Question 10

E1 Mechanism

Answer 10

rate = k [substrate]

• elimination (e) and unimolecular (1)
• first, loss of leaving group
• next, base transfers proton to form double bond
• intermediate can be isolates
• accompanied by competing Sn1 –> mixture of products
• tertiary favours E1 with weak bases

Question 11

Beta/Alpha carbons

Answer 11

Alpha: first carbon attached to functional group

Beta: second carbon attached to functional group

Question 12

Strong Acids

Answer 12

HCl
HI
HBr
HClO4
HNO3
H2SO4

Question 13

Strong Bases

Answer 13

KOH
NaOH
Ba(OH)2
Ca(OH)2
LiOH
C2H5O-

Question 14

Primary haloalkanes

Answer 14

do not form stable carbocations so only undergo Sn2/E2

Question 15

Tertiary haloalkanes

Answer 15

• undergo E1 with weak bases
• undergo E2 with strong bases
• form stable carbocations so undergo Sn1

Question 16

Secondary haloalkanes

Answer 16

• undergo Sn2 with strong nucleophiles
• undergo E1 with weak bases
• undergo E2 with strong bases

Question 17

[O]

Answer 17

KMnO4
K2Cr2O7 / H+
CrO3 / H+