Chap 13: Halogen Deratives (RX)

Question 1

What are nucleophiles?

Answer 1

Nucleophiles are electron-rich species that form a bond by donating a lone pair of electrons to an electron-deficient species/
they are also Lewis base (lone pair donor)

Question 2

What are the reactions of RX compounds?

Answer 2

(i) Reaction w NaOH (aq)
N.S.
NaOH (aq), heat under reflux
(converts R-X to R-OH)

(ii) Reaction w ethanolic NaCN
N.S.
Ethanolic NaCN, heat under reflux
(step-up rxn, increases carbon chain by one C atom)
(converts R-X to R-CN)
R-CN can undergo acidic hydrolysis (H2SO4/HCl (aq), heat under reflux) to form R-COOH
R-CN can undergo alkaline hydrolysis (NaOH (aq), heat under reflux) to form R-COO-
R-CN can be reduced to form R-CH2NH2 (LiAlH4 in dry ether/H2 w Ni, heat or H2 w Pd/Pt catalyst)

(iii) Reaction w ethanolic ammonia
N.S.
excess NH3 (alc), heat in a sealed tube
(converts R-X to R-NH2)

Question 3

Why does the SN1 mechanism produce optically inactive product?

Answer 3

tertiary 3 RX predominantly undergoes Sn1 mechanism since it has the most electron-donating -CH3 groups which stabilises the carbocation.

the nucleophile can attack the trigonal planar carbonation centre from either the top or the bottom of the plane with equal probability, producing equal amounts of each enantiomer (racemic mixture)

the product mixture is optically inactive as each enantiomer rotate plane-polarised light in the opposite direction by the same magnitude hence the rotating powers of the enantiomers cancel out.

pdt formed is optically inactive

Question 4

Why does SN2 mechanism produce products that are inverted?

Answer 4

primary 1 RX predominantly undergo SN2 mechanism since it is the least bulky and would provide least steric hindrance to incoming nucleophiles

the single step reaction involves nucleophile attacking from the direction directly opposite the group that leaves (backside attack), this results in an inversion of stereochemical configuration during the rxn

pdt formed is optically active

Question 5

What are the conditions for elimination rxn?

Answer 5

Elimination
ethanolic NaOH OR NaOH (alc), heat under reflux
(eliminates H and X atoms to form alkene)
by Saytzeff’s rule, more highly substituted double bond will be formed.

Question 6

Why the C-Cl bond in chlorobenzene cannot be broken?

Answer 6

(i) the p-orbital of Cl in chlorobenzene overlaps with the pi e- cloud of the benzene ring, resulting in delocalisation of the lone pair of e- on the Cl atom into the benzene ring. this introduces a partial double bond btw C and Cl and strengthening the C-Cl bond

(ii) the rear side of the C-X bonf is sterically hindered by the bulky benzene ring

(iii) the high e- density of the aromatic ring repels the approaching nucleophile

Question 7

Describe the distinguishing tests for RX

Answer 7

method 1:
(1) Addition of NaOH (aq) and heat
RX + NaOH (aq) –> ROH + NaX
(2) Addition of dilute HNO3 (aq)
HNO3 + NaOH –> NaNO3 + H2O
(3) Addition of AgNO3 (aq)
X- + Ag+ –> AgX (s)
(4) Addition of aq/conc. NH3
(it may be difficult to see colour of ppt hence additional solubility test is conducted)

R-I (immediate yellow ppt, AgI)
R-Br (immediate cream ppt, AgBr)
R-Cl (immediate white ppt, AgCl)

method 2:
ethanolic AgNO3, warm
CH3CH2I (iodoethane): yellow ppt, AgI formed immediately
CH3CH2Br (bromoethane): cream ppt, AgBr formed aft awhile
CH3CH2Cl (chloroethane): white ppt, AgCl formed after a long time

Down the group, atomic radius of halogens increases. hence bond length increases.
Orbital overlap becomes less effective down the group and bond strength decreases
Down the group, the weaker C-X bond is broken more easily
hence rate of hydrolysis: (slowest) RCl < RBr < RI (fastest)