Download F7 AL Chemistry (Tutorial 31)

C. Y. Yeung (AL Chemistry)
F7 AL Chemistry (Tutorial 31)
~ Suggested Solution ~
1.
(a) “acyclic compound” = organic molecule without benzene ring
“fruity smell” = ester
“not produce derivative with 2,4-dinitrophenylhydrazine” = not carbonyl compound
“not produce derivative with propanoyl chloride” = not an alcohol / phenol / amine
Four possible structures for compound X (ester):
O
O
CH3CH2–C–O–CH3
H–C–O–CH2CH2CH3
O
O
CH3–C–O–CH2CH3
H–C–O–CHCH3
CH3
(b) “reduction with LiAlH4” = reduction of ester gives two alcohols
“gives only one product” = the two alcohols are the same = “symmetrical” ester
“reacts with ethanoic anhydride” = the alcohol reacts with anhydride to form ester
O
structures of X:
CH3CH2–C–O–CH3
structures of Y:
2.
CH3CH2OH
Compound A is a substituted amide.
Compound B is a ketone with a substituted amino group.
Treating compounds A and B with 2,4-dinitrophenylhydrazine separately,
only compound B would give a bright colored crystal.
[It is useful test for carbonyl compounds!]
O
CH3CH2CCH2NHCH3
+
H2N–NH
NO2
O2N
CH3CH2
CH3NHCH2
C
N–NH
NO2
O2N
1/3
3.
C. Y. Yeung (AL Chemistry)
Treating the compounds A and B with iodine in aqueous sodium hydroxide (I2 / NaOH),
only compound B will give a yellow precipitate (iodoform).
[Iodoform test!]
O
C
4.
O
I2 / NaOH
C
CH3
O- Na+
+
CHI3
“does not react with chlorine in darkness” = saturated hydrocarbon / do not bear C=C
“react with Cl2 under UV, only two monochlorinated products” = alkane with 3 carbons
“treated with warm NaOH” = alkaline hydrolysis of RX (SN reaction), to form ROH
“followed by KMnO4, Q gives carboxylic acid” = Q gives a primary ROH after hydrolysis
“R gives a ketone” = R gives a secondary ROH after hydrolysis
P:
Propane
CH3CH2CH3
Q:
1-chloropropane
CH3CH2CH2Cl
S:
Propanoic acid
CH3CH2COOH
Cl
R:
2-chloropropane
CH3CHCH3
O
T:
Propanone
CH3–C–CH3
5.
O
C
O
heat
Condensation
C
(i) excess NH3
amide formation
(ii) acid
O
Hydrolysis of
Nitrile
O
COOH
C
COOH
COOH
Br2, NaOH
dil. H2SO4, heat
NH2
Hofmann
degradation
CN
NH2
COOH
COOH
CuCN
Replacement of
Diazonium group
+
NN ClCOOH
HNO2, 50C
Formation of
Diazonium salt
2/3
C. Y. Yeung (AL Chemistry)
6.
Let the mass of compound A be 100g,
C
H
O
64.9
13.5
21.6
64.9 / 12
13.5/1
21.6/16
= 5.41
=13.5
=1.35
5.41/1.35
13.5/1.35
1.35/1.35
=4
=10
=1
Mass
No. of moles
Molar ratio

The molecular formula of compound A (and isomers B, C) is C4H10O.
Compound A molecule contains one oxygen atom, it may be an alcohol, an aldehyde,
a ketone or an ether.
“A does not react with K2Cr2O7/H+” = A cannot be oxidized = A is a ketone.
“C forming an acid” = C is a 10 ROH or an aldehyde.
“B is oxidized by K2Cr2O7/H+, but not forming an acid” = B is a 20 ROH.
“isomers B and C contain the same functional group” = both B and C are alcohols.
O
A:
B.
CH3CH2–C–CH2CH3
OH
CH3CH2CHCH3
C.
CH3CH2CH2CH2OH
7.
(a)
1. O3, CH3CCl3, <200C
O
O
ozonolysis
HCCH2CH2CH2CH
2. Zn / H2O
mild reduction of
carbonyl group
(b)
O
NaBH4
HC
(c)
O
O
COCH3
H2O
O
1. I2 / NaOH
CCH3
2. H3O+
iodoform reaction
HOCH2
O
COH
COCH3
O
PCl3
CCl
formation of acyl chloride
/ halogenation of carboxylic acid
3/3