H2 Chemistry 2010/11: Tutorial 18 Suggested Solutions – Nitro Nitro Cpds Discussion Questions Question 1 (a)
O2N
NH2
(b)
CH2NH2
(c)
NH CH3
(d)
NHCOCH3
(e)
NH2
Expected order of increasing basic strength: (d) < (a) < (e) < (b) < (c). The basic strength of each given compound depends on the availability of the lone pair of electrons on the nitrogen atom to form a dative covalent bond with a proton. The greater this availability, the greater the basic strength of the compound. compound . (d) is an amide and is the least basic (effectively neutral) because its lone pair of electrons on the nitrogen atom not only partially delocalises into the benzene ring but also interacts with the pi-electron cloud of the C=O bond. As such, this lone pair is is the least available for co-ordination to a proton. proton . The aromatic amines, (a) and (e), are less basic than the aliphatic amines, (b) and (c), because of the partial delocalisation of the lone pair of electrons on the nitrogen atom into the benzene ring . (a) is less basic than (e) because the electron-withdrawing –NO2 group increases the extent of this delocalisation and hence further reduces the availability of the nitrogen lone pair for co-ordination to a proton. proton. The aliphatic amines, (b) and (c), are stronger bases because the alkyl groups bonded to the nitrogen atom are electron-releasing and this increases the electron density at the nitrogen atom, making the lone pair of electrons on the nitrogen atom much more readily available for co-ordination to a proton . (c) is a stronger base than (b) since there are two electron-releasing alkyl groups bonded to the nitrogen atom in (c) compared to only one in (b).
Question 2 (a) CH3CH2Cl
CH3CH2CH2NH2 alcoholic KCN
CH3 CH2
CH3 CH2
C
Cl + -CN
N + 4[H]
CN + Cl
CH3CH2 reflux with heat LiAlH4/ dry ether
H CH3 CH2
C
H N H
H
(b) CH2=CH2
CH3 CONH CH2CH2 NHCO CH3
Synthesis of amine Br 2/CCl4 H2 C
CH2 + Br 2
Br H2 C
CH2
Br Br H2C
CH2
NH2
excess NH3 in alchohol + 2NH3
heat in sealed tube
Br
H2C
CH2
NH2
Synthesis of acyl chloride
-1-
+ 2 HBr
CH2 + H2O
H2 C
CH2 + 2[O]
K2Cr 2O7(aq)/ H2SO4(aq)
CH2
O H3 C
heat
O H3 C
H3 C
2. Add H2O, warm
OH H3 C
OH
1. cold conc. H 2SO4
+ H2O
C OH
O
pyridine + SOCl2
C
H3C
OH
+ SO2 + HCl
C Cl
Synthesis of amide NH2
O
CH2 + 2 H3 C
H2 C
H3 C
C
C
Cl
NH2
(c) CH3CONH2
H
H
N
CH2 CH2 N
C
CH3 + 2 HCl
O
O
CH3CONHCH2CH3
Synthesis of amine O H3C
C
LiAlH4/ dry ether NH2 + 4[H]
H3C
C NH2 + H2O H2
Synthesis of acyl chloride O H3 C
C
+ H + + H 2O
NH2
O H3 C
C
O
H2SO4 (aq)
H3 C
heat
OH + NH 4+
C
O
PCl5
H3 C
OH + PCl5
C
Cl + POCl3 + HCl
Synthesis of amide O
O H3C
C
Cl
HC
+
H2 N
H3C
C CH3 H2
COOH
+ HCl
C N H
C CH3 H2
(CH3CH2)2NH
(d) 3 Synthesis of acyl chloride
O H3 C
H3 C
COOH + PCl5
C
Cl + POCl3 + HCl
Synthesis of amine O
O H3 C
C
O H3 C
C
H3 C
Cl + NH3
+ 4[H]
NH2
C
+ HCl
NH2
LiAlH4/dry ether heat
H H3 C
C
NH2
+ H2O
H
Synthesis of amide O
O H3 C
C
Cl
+
H2 NCH2 CH3
H3 C
C
N
CH2 CH3 + HCl
H
O H3 C
C
LiAlH4 /dry ether N
CH2 CH3 + 4[H]
N
CH2 CH3
H
H
(e)
CH3 CH2
CHO
Chain extension required.
CH2CH2NH2
-2-
LiAlH4 CHO + 2[H]
CH2 OH
dry ether
PCl PCL55
CH2 Cl + POCl3 + HCl
CH2OH + PCl5
alc. KCN CH2 CN + KCl
CH2 Cl + KCN
reflux
LiAlH4 CH2 CN + 4[H]
CH2 CH2 NH2
dry ether
Br Br
(f)
NH2 Br
Conc. H2SO4/ Conc HNO3 + HNO3
NO2 + H2O
55 oC
Sn/conc. HCl
NO2 + 6[H]
NH2
reflux
Synthesis of arylamine, MUST use Sn reagent with aryl nitro cpd
+ 2H2O
Br NH2 + 3HBr
Br
NH2 + 3Br 2(aq)
Br
(g)
NH2
CHCH (g) 3CH 2CH2Br 3CH2CH2Br
CH3CH2
C
COOH
H Br + OH-
CH3 CH2 CH2
OH + [O]
O
O CH3 CH2 C
heat with immediate distillation
OH
HCN/ trace KCN + HCN
CH3 CH2
C
Cl
PCl5 CN + PCl5
CH3 CH2
Cl C H
C
CN + POCl3 + HCl
H
H
CH3CH2
CN
H
OH C
+ H2O H
H
CH3CH2
OH + Br
CH3CH2 CH2
reflux K2Cr 2O7(aq)/ H2SO4(aq)
CH3 CH2 CH2
CH3 CH2 C
aq NaOH
NH2
ethanol CN
+ NH3(excess)
heat in sealed tube
CH3CH2
C H
-3-
CN + HCl
NH2 CH3 CH2
C
NH2
1. HCl(aq) reflux CH3 CH2
+ 2H2O
CN
C
2. Neutralise
H
COOH + NH3
H + NH3
CH3 CH2
COO-
C H
Note product is normally zwitterionic
Question 3 (a) phenylamine and (phenylmethyl)amine Test: Add aqueous bromine to each of the two samples. Observation: For phenylamine, there would be decolourisation of orange aqueous bromine and the formation of a white ppt.
NH2
NH2
Br
Br
+ 3Br 2(aq)
+ 3 HBr Br (white ppt.)
Note: 3 moles of Br 2 required!
Observation: For (phenylmethyl)amine, there would be no decolurisation of bromine nor formation of white ppt.
CH2NH2
no electrophilic substitution reaction no decolourisation of bromine no formation of white ppt.
Br 2(aq)
(b)
+
CH3CH2CH2NH3 Cl and CH3CONH2
Reagents and conditions: At first, add NaOH(aq) and warm gently, test for any gas evolved with moist red litmus paper Observations: Compound which gives a vapour turning moist red litmus paper blue is ammonium salt. +
-
CH3CH2CH2NH3 Cl + NaOH(aq)
CH3CH2CH2NH2 + NaCl(aq) + H2O(l)
Follow up. Heat remaining unknown with NaOH(aq) in boiling water bath . Test for any gas evolved with moist red litmus paper Observations: Compound which gives a gas turning moist red litmus paper blue is amide Amide is hydrolysed on heating, and yields ammonia gas. CH3CONH2 + OH
-
CH3COO
-
+ NH3(g)
(c) (d)
H2N
COCH3
and
CH3CONH
Test: Add 2,4-dinitrophenylhydrazine to each of the two samples. Observations: An orange ppt will be seen for
No orange ppt will be seen for
H2N
COCH3
CH3CONH
COCH C O3
H
H3C
CH3
CH3 H2N
Alternative test can be iodoform test.
+
H2N
N H
NO2
H2N C
H NO2
-4-
N
N
C
H + H2ONO2
H N N NO 2
+
O2N
H2O
NO2
Question 4 (a) A
B
C CH3
H +
HOCH2
N
CO2CH2CH3
H3N
D
+
HO
C
CH CH2 N R H
CO2CH2CH3
(b) A
B
+
+
N
CO2H
H3N
D
CH3
H
NH3
+
HO2C
C
+
+
(c) A
H
CO2CH2CH3
S CH +
N
C CH
CH3 CH3 CO2H
H
CH3CH2OH
+ RCOOH
B
CH3 C N
CH
CO2H
CH3CH2OH
No reaction.
H
OH
+
C O
No reaction.
CH3 C O CH2
O H CH3 C O O
O O C
CH3
CH CH2 N C R
CH3
O
Question 5 Observation Deduction A reflux with aq. NaOH gives Alkaline hydrolysis. salt B and cpd C A is a amide. B and C are salt of carboxylic acid and amine respectively. C: (M r = 93) with aqueous bromine gives D: C6H4Br 3N Electrophilic substitution C has strongly activating group in the benzene ring. Bromination without catalyst. C is phenylamine. NH2
M r =
93
Hence D is Br Br
NH2 Br
Since formula for A: C8H9NO and C has 6 carbons, B therefore has 2 carbons B is a sodium salt of ethanoate. O CH3 C O-Na+ And hence A is phenylethanamide O N
C
CH3
H
-5-
E: isomer of A C8H9NO Alkaline hydrolysis of primary amide in E to yield carboxylate Reflux E with NaOH(aq) and NH3. NH3 evolved, + salt F F: refluxed with alkaline Oxidation of aryl side chains to COOH. KMnO4, followed by F is therefore disubstituted. F is therefore : acidification yields HOOC
COOH
COO-Na+ COO Na+
H3 C Hence E is
O C NH2
H3 C G: isomer of A C8H9NO G refluxed with aq NaOH obtains alkaline vapour H and salt I:C7H5O2Na
Hydrolysis of substituted amide (G) to yield amine (H) and carboxylate (I) I has 7 carbons, so H has only 1 carbon (G) has 8 carbons total)
H C NH2 H is methanamine 3 I is a 7 carbon carboxylate, with a benzene ring, so it must be sodium benzoate. COO Na+ COO-Na+
H reacts with CH3COCl to give J: C3H7NO
Reaction of amine H with acyl halide to give amide J O H3 C
J is
N
C
CH3
H
Therefore G is H C N CH3 O
-6-
Optional Question COOH 1. Sn/c.HCl
H+/MnO4
CH3
heat
NO2
COOH
2. NaOH
NO2
NH2
F
E
reduction of NO2 in F to NH2 in E
oxidation of alkyl sidechain of a benzene ring
B:
B is
Na2CO3
C9H9NO2
-gives zwitterions =>contains COOH and NH2 groups
CO2 gas evolved => B has -COOH group
Br 2/H2O
C:
C9H7Br 4NO2 - from change in molecular formula 2 H atoms in B replaced by 2 Br atoms, 2 more Br atoms added => electrophilic substitution on phenylamine, with aq Br 2 no catalyst required. disubstitution, confirms that 2 position on phyneylamine occupied by side chain; on phenylamine occupied by side chain; substitution at 4 and 6 positions. substitution at 4 and 6 position
- has which is also in E index of unsaturation is 6. since the benzene ring and COOH account for 5 units of unsaturation a C=C may be in B
- 2 Br atoms added; electrophilic addition on C=C in sidechain
HBrNO CC 98 8Br 3NO 9H 3 3 D similar - electrophilic substituition substitution ofof2 2HHatoms atoms ininBB to -electrophilic togive give D similar to formation formation of to ofCC - from change in molecular formula addition of another Br and OH took place. Electrophilic addition to C=C in B D:
H+/ MnO4 heat
E +
also a 2 substituted phenylamine compound
2 CO2
-B has alkyl side chain and NH2 group on a 1,2-disubstituted benzene ring similar to E - Oxidative cleavage of C=C in B took place to give COO group in E -B has a 3 carbon side chain (total 9 carbon atoms) with a C=C and a COOH group
Br Br H C C COOH H
B
Br
C H
C
C H
OH Br COOH
Br
C H
D
NH2
NH2
C H
COOH
NH2
Br
Br
Oxidation of B to E H C C COOH H
+ 5[O]
COOH
H+/ MnO4
NH2
+ 2CO2 + H2O NH2
heat E
B
Acid-carbonate neutralisation of B H C C COOH H + 1/2Na2CO3 NH2
H + C C COO Na H + 1/2CO2 + 1/2 H2O NH2
B
-7-
Electrophilic substitution and addition of H C C COOH H + 3Br 2(aq)
B to
give C
Br Br
Br
C H
NH2
C H
COOH + 2HBr
NH2 Br C
B
Electrophilic substitution and addition of B to give D H C C COOH H + 3Br (aq) + H O 2
OH Br Br
C H
2
NH2
NH2
Br
B
-8-
D
C H
COOH + 3HBr
