Download Carboxylic Acid Derivatives and Nitrogen Cpds

JC2 H2 CHEMISTRY (9647)
CHEMISTRY REVISION PROGRAMME WORKSHEET (Ans)
Name:
Class:
TOPICS: Carboxylic Acids and Derivatives, Nitrogen Compounds
1. Summary of reactions of Carboxylic Acids and Derivatives
No.
Reactants
Carboxylic Acid
1
(e.g. CH3COOH)
Carboxylic Acid
2
Type of reaction / Remarks
Reagents & Conditions
Products
Reactive metals
Carboxylate salt
-
(e.g. Na),
(e.g. CH3COO )
r.t.p.
+ ½ H2(g)
Carboxylate salt
Aqueous NaOH, r.t.p.
(e.g. CH3COO )
3
3
Carbonate or hydrogen
carbonate
(e.g. Na2CO3, NaHCO3),
r.t.p.
Carboxylate salt
-
(e.g. CH COO )
+ CO2 + H2O
Carboxylic Acid
4
5
(e.g. CH3COOH)
Carboxylic Acid
(e.g. CH3COOH)
Neutralisation
-
(e.g. CH3COOH)
Carboxylic Acid
(e.g. CH3COOH)
Acid-metal displacement
½ mol of H2 formed for each
–OH group present
Primary Alcohol
LiAlH4 in dry ether, r.t.p.
1. Anhydrous PCl3, heat
or
(e.g. CH3CH 2OH)
Acyl Chloride
2. Anhydrous PCl5, r.t.p.
or
(e.g. CH3COCl)
Neutralisation
Only carboxylic acids are
acidic enough to react with
carbonates
and
hydrogen
carbonates.
(used as a chemical test –
effervescence of CO2 that
forms a white ppt with
Ca(OH)2. Heating is required
to produce more CO2 to
make white ppt more visible )
Reduction
Not reduced by the other
reagents that can reduce
carbonyl compounds
Nucleophilic Substitution
(PCl5 used as a chemical test
– white fumes of HCl formed.
Alcohols also give similar
observation)
3. Anhydrous SOCl2, heat
Acyl Chloride
6
(e.g. CH3COCl)
Carboxylic Acid
H2O, r.t.p.
(e.g. CH3COOH)
+ HCl
1
Nucleophilic Substitution
(used as a chemical test –
white fumes of HCl formed)
Acyl chlorides produce very
acidic solutions in water due
to the HCl formed.
No.
7
Reactants
Reagents & Conditions
Products
Carboxylic Acid
Alcohol (e.g. CH3OH)
conc. H2SO4 catalyst, heat
Ester
(e.g. CH3COOH)
Type of reaction / Remarks
e.g. CH3COOCH3
Alcohol or phenol
8
Acyl Chloride
Ester
(e.g. ROH), r.t.p.
e.g. CH3COOR
(e.g. CH3COCl)
Nucleophilic Substitution
Phenols do NOT react with
carboxylic acids
Nucleophilic Substitution
When a phenol is used,
aqueous NaOH is often used
to produce the phenoxide
ion, which is a stronger
nucleophile than the phenol
(since the electron density on
the O atom is increased).
Choice of alcohol or phenol
depends on the type of ester
to be produced.
Ammonia or amines
9
Acyl Chloride
(e.g. CH3COCl)
Amide
(e.g. RNH2),
(e.g. CH3CONHR)
r.t.p.
Nucleophilic Substitution
When synthesizing amides in
the lab, acyl chlorides are
used, NOT carboxylic acids
because the yield will be poor
due to the neutralization
reaction between the acid
and the basic reagent.
Choice of ammonia or amine
depends on the type of amide
to be produced.
Carboxylic Acid
10
Ester
(e.g. CH3COOH)
e.g. CH3COOR
+ Alcohol or
phenol (depends
on the ester given)
Aqueous HCl (or H2SO4),
heat
Carboxylic Acid
Amide
11
(e.g. CH3COOH)
e.g. CH3CONH2
+ NH4
+
Carboxylate Salt
-
12
(e.g. CH3COO )
Ester
+ Alcohol or
phenoxide salt
(depends on the
ester given)
e.g. CH3COOR
Aqueous NaOH, heat
Carboxylate Salt
13
Acidic Hydrolysis
For acidic hydrolysis of 2° and
3° amides, an amine salt is
+
formed instead of NH4 .
Amide
-
(e.g. CH3COO )
e.g. CH3CONH2
+ Ammonia
2
Basic Hydrolysis
The NH3 produced from basic
hydrolysis of a 1° amide can
be used as a chemical test –
gas evolved turns moist red
litmus paper blue.
For basic hydrolysis of 2° and
3° amides, an amine is formed
instead of NH3.
2. Summary of reactions of Nitrogen Compounds
No.
Reactants
Reagents & Conditions
Products
Amine
e.g. CH3NH2
Aqueous HCl (or H2SO4),
r.t.p.
Amine salt
1
e.g. CH3NH3
Type of reaction / Remarks
Neutralisation
Actually
other
acids
e.g.
carboxylic acids can also
neutralize amines but are not
often used (e.g. you don’t want
to have the carboxylate salt
being formed also otherwise
it’s difficult to obtain the amine
salt)
+
Nucleophilic substitution
2
Amide
Amine
Acyl chloride
e.g. CH3NH2
(e.g. CH3CH2COCl), r.t.p.
e.g.
CH3NHCOCH2CH3
Carboxylic acids NOT used for
laboratory synthesis of amides
due to neutralization reaction
that can take place instead
Electrophilic substitution
NH2
NH2
X
X
Aqueous X2, r.t.p.
3
(because
of
the
strong
electron-donating –NH2 group
which activates the ring)
(aqueous
(X = Cl or Br)
Br2
used
as
a
chemical test – decolourisation
X
of reddish-brown Br2 solution
and white ppt formed)
Reduction
NO 2
(i) Sn in conc. HCl, heat
4
(ii) aqueous NaOH
Amino Acids
5
NH2
e.g.
H2NCH2COOH
Same reagents and conditions needed and same
type of products formed as for the individual
amines and carboxylic acids
3
After step (i), the salt is formed
instead of the amine. Hence in
step (ii), excess NaOH is
added
to
liberate
the
phenylamine
Neutralisation
Structured Questions
Time allocation: 15 mins
1
Trichloroethanal (chloral), CCl3CHO, was much used up to the middle of this
century as a drug and a stimulant.
(a)
(i)
When it is warmed with aqueous sodium hydroxide, one organic product
is trichloromethane (chloroform). Construct an equation for this reaction
and name the other organic product.
CCl3CHO + NaOH → CHCl3 + HCO2-Na+ [1]
Sodium methanoate [1]
This reaction is linked to the iodoform test
for carbonyl compounds
e.g. for CH3CHO, the intermediate is
CI3CHO, which gives the final products
CHI3 and HCO2-Na+
(ii)
Hence, or otherwise, suggest how chloral can be converted into ethyl
methanoate, stating any other required reagents and conditions used in
the conversion.
[4]
Cl
Cl
O
C
C
Cl
O
NaOH (aq)
H
+
C
H
heat
O-Na+
H2SO4 (aq)
r.t.p.
[½]
[½]
O
O
CH3CH2OH
H
C
H
O
CH2 CH 3
CHCl 3
C
conc H2SO4, heat
[½]
OH
[½]
(b)
Chloral is easily oxidized to trichloroethanoic acid.
(i)
Draw a displayed formula of trichloroethanoic acid
Cl
Cl C
O
C
O
Cl
H
4
Remember to show
ALL bonds!
[1]
(ii)
Explain why this acid is a much stronger acid than ethanoic acid
CH3CO2CCl3CO2-
CH3CO2H
CCl3CO2H
H+
H+
+
+
In trichloroethanoic acid, there are 3 electron-withdrawing chloro
groups [1] that disperses the negative charge on the carboxylate
anion hence stabilising the carboxylate anion relative to the acid [1]
[3]
(c)
Arrange the following in order of increasing acid strength and explain your
reasoning.
O
OH
C
OH
OH
CH3
[3]
Acid strength:
O
OH
OH
OH
C
CH 3
<
<
[1]
Benzoic acid
O
OH
O
C
O
C
+
H+
The carboxylate anion is resonance stabilized [½] by the delocalization
of the negative charge over the C atom and both oxygen atoms [½] in the
carboxylate anion, hence stabilizing the carboxylate anion relative to acid. As
such, benzoic acid has a greater tendency to ionize compared to phenol, thus is
the strongest acid.
5
CH3
2-methyl phenol
OH
OCH 3
CH3
+
H+
The electron-donating alkyl group decreases the intensity of the negative
charge on the phenoxide anion, hence destabilizing the phenoxide anion
relative to acid. As such, benzoic acid has a greater tendency to ionize compared
to phenol, thus 2-methyl phenol is the weakest acid.
(Modified N96/III/3a, b(i) and (ii))
[Total: 10]
Time allocation: 15 mins
2
The limescale that collects in kettles in hard water areas is mostly calcium
carbonate. It can be removed fairly harmlessly by a warm solution of vinegar,
which contains ethanoic acid. The limescale dissolves with fizzing and a solution
of calcium ethanoate remains.
(a)
Write a balanced equation for the reaction between ethanoic acid and calcium
carbonate.
[1]
2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O [1]
(b)
When the solution produced in (a) is evaporated, and the resulting solid calcium
ethanoate heated strongly in a test-tube, an organic compound G is formed which
condenses into a colourless liquid. The residue in the tube consists of calcium
carbonate.
Compound G is neutral and water-soluble. G does not react with sodium metal
nor with Fehling’s solution but it does react with alkaline aqueous iodine. G has a
Mr of about 57.9. Suggest a structural formula for G. Justify your answer by
reference to these properties of G.
[5]
G does NOT undergo acid-metal displacement with Na
⇒ G is NOT an alcohol. [1]
G does NOT undergo oxidation with Fehling’s solution
⇒ G is NOT an aliphatic aldehyde [1]
G undergoes oxidation with alkaline aqueous iodine
⇒ G has the H3C C
group in the compound [1]
O
Based on the above properties and the given Mr, G is likely to be CH3COCH3, [1]
propanone, which would be neutral and is water-soluble since it can form
hydrogen bonds with water. [1]
6
(c)
Construct a balanced equation for the formation of G by the action of heat on
calcium ethanoate.
[1]
(CH3CO2)2Ca → CH3COCH3 + CaCO3 [1]
(d)
Suggest a simple one-step test you could carry out to confirm the identity of the
functional group present in G. You should give the reagent and the observation
you would make.
[2]
Add 2,4-DNPH to G and heat. [1] An orange ppt [1] should be observed.
(e)
Suggest the structural formula of the organic product you might expect when
calcium propanoate, (CH3CH2CO2)2Ca, is heated strongly.
[1]
CH3CH2COCH2CH3 [1]
(Similar to example in (c), since (CH3CO2)2Ca gives CH3COCH3)
(Modified from J2000/I/8)
Time allocation: 15 mins
3
Compound H, C8H11NO, is weakly basic. It dissolves in dilute aqueous
hydrochloric acid to give a solution from which a crystalline solid J can be isolated.
H decolourises aqueous bromine with the formation of a white precipitate K.
No orange crystals are observed when 2,4-dinitrophenyihydrazine is added to H.
However when H is heated with alkaline aqueous iodine and then followed by
careful acidification, some yellow crystals are produced together with L, C7H7NO2.
L can be produced from 4-nitromethylbenzene in a 2-step reaction via an
intermediate M. 4-nitromethylbenzene is first heated with tin in the presence of
concentrated hydrochloric acid to form M, and then followed by adding hot
acidified potassium manganate(Vll) to produce L.
Deduce the structures of the compounds H to M. Explain your reasoning and
write balanced equations for the reactions involved where appropriate.
[10]
[Total: 10]
H undergoes neutralization with dil. aq. HCl to give a crystal J [½]
⇒ H may contain an amine group [½]
⇒ J is a salt
H undergoes electrophilic substitution with aq. bromine to give white ppt, K [½]
⇒ H may be a phenylamine [½]
⇒ K may be a bromosubstituted phenylamine [½]
7
H does not undergo condensation with 2,4-DNPH
⇒ H does not contain a carbonyl group
[½]
H undergoes oxidation with alkaline aq. iodine and acidification to form a yellow
crystals of CHI3 and L. [½]
⇒ H may contains the CH3CH(OH)- group
[½]
⇒ L is a carboxylate salt [½]
4-nitromethylbenzene undergoes reduction with tin and conc. HCl to form M [½]
⇒ M could be
CH3
NH2
[½]
M undergoes oxidation with hot potassium manganate to form L [½]
⇒ L could be
COOH
NH2
[½]
H could be
CH3
HO
C
NH2
H
[½]
8
J could be
CH3
HO
C
H
NH3+Cl-
[½]
K could be
CH3
HO
C
H
Br
Br
NH2
[½]
Equations
CH3
HO
C
CH3
H
HO
+
C
HCl
NH2
NH3+Cl-
CH3
HO
C
H
[1]
CH3
H
HO
+
C
H
2Br2
+ 2HBr
Br
NH2
Br
NH2
9
[1]
Remember to show
ALL bonds!
CH3
CH3
+
4[H]
+
NO2
NH2
2H2O
[1]
CH3
HO
C
COO-Na+
H
+ 3I2 + 4NaOH
+ CHI3 + 3NaI + 3H2O
NH2
NH2
[1]
[Total: 12, max = 10]
Time allocation:
4(a)
9 mins
Draw a displayed formula of aminoethanoic acid (glycine)
H
H
N
H
(b)
C
H
[1]
O
C
O
H [1]
Penicillin is widely used to kill bacteria. The general structure of a penicillin
molecule is given below.
R = an aryl group
(i)
Mark with an asterisk * on the above structure any carbon atoms which
are chiral centres.
[2]
10
*
*
For a complete answer, you need to show understanding
that you recognize the amide groups in penicillin are neutral
(as compared to amines which are basic)
*
[3 = 2 marks; 2 = 1 mark; 1 or 0 = 0 mark]
(ii)
Bacteria mistake penicillin for protein molecules and bond on to the amino
acid sequence present in penicillin. This sequence is also present in the
skeleton of the aminoethanoic acid molecule you have drawn in (a).
Draw a ring around the part of the penicillin molecule which bacteria
mistake for an amino acid.
[1]
|
Circle at the bottom right portion of the molecule – N – CH – CO2H
(c)
[1]
Would you expect penicillin to be acidic, neutral or basic? Explain your answer as
fully as you can.
[2]
(N95/III/5)
Acidic [1] because of the acidic –COOH group present and the amide groups
are neutral. [1]
Time allocation:
5(a)
14 mins
How would you expect the melting point and the solubility in water, of an
unionised covalent form of glycine (aminoethanoic acid) to compare with the
actual properties of the zwitterionic form?
[2]
Melting point for the unionised form < zwitterionic form because of the
weaker hydrogen bonding between the amino acid molecules compared to the
stronger electrostatic forces of attraction between the zwitterions. [1]
Solubility for the unionised form < zwitterionic form because the zwitterions
can form ion-dipole interactions with water which releases energy for the
detachment of the ions from the crystal lattice for hydration. [1]
11
(b)
The following scheme of reactions illustrates the synthesis of a dipeptide, E.
(i)
What type of reaction is step I?
Nucleophilic substitution [1]
(ii)
Suggest a reagent for step II.
PCl5, r.t.p. [1]
(iii)
During which step is the peptide bond formed? Step III [1]
(iv)
What are the products of hydrolysis of E, and how may the reaction be
carried out?
Dilute H2SO4, heat [1]
+
+
→ Products: H3N–CH(CH3)COOH [1] + H3NCH2CO2H [1]
OR
Dilute NaOH, heat [1]
→ Products: H2N–CH(CH3)COO- [1] + H2NCH2CO2- [1]
(v)
Suggest the type of reaction in step IV.
Reduction [1]
This is an unusual reaction which involves the addition of H2 (reduction)
with the cleavage of C-N bond occurring simultaneously.
[7]
(Modified N96/I/10)
12