New Way Chemistry for Hong Kong A-level
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(Book 3B, p. 110 – p.120)
Part 9
Chemistry of Organic Compounds
50.
(i)
carbonyl group (aldehydes and ketones)
(ii)
aldehydes
(iii)
alcohols or carboxylic acids
(a)
(b)
(i) & (ii)
The displayed formula of Q is:
Q gave an orange precipitate with 2,4-dinitrophenylhydrazine reagent. This means that it
contains a carbonyl group. However, it is not an aldehyde as it gives a negative result on
Fehling’s test. Moreover, it does not contain alcoholic —OH because it does not react with
phosphorus pentachloride.
(c)
(i)
The reddish brown Br2(aq) is decolourised and a white precipitate is formed.
(ii)
The displayed formula of the resulting organic product is:
(iii)
Electrophilic substitution reaction
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51.
(a)
Stage 1: KCN in ethanol, heat
Stage 2: LiAlH4 in dry ether
(b)
Propylamine is more basic than ammonia. This is because the basicity depends on the ability of the
lone pair electrons on nitrogen to form a dative bond with a proton. The propyl group is an
electron-donating group which makes the lone pair electrons on nitrogen more available.
52.
(a)
(i)
HCN + trace amount of NaOH; 10°C – 20°C
Product:
(ii)
LiAlH4 in dry ether; room temperature [or NaBH4 in aqueous solution]
Product:
(iii)
Fehling’s solution [or Tollen’s reagent or hot acidfied KMnO4 (or acidified K2Cr2O7)]
Product:
(iv)
Cold, alkaline KMnO4 [or hot alkaline KMnO4 or ozone]
Product:
(b)
To distinguish cinnamaldehyde from compound A:
Test: Add bromine in 1,1,1-trichloroethane at room temperature (in the dark).
Cinnamaldehyde decolourizes the reddish brown bromine but compound A does not.
To distinguish cinnamaldehyde from compound B:
Test: Add 2,4-dinitrophenylhydrazine.
Cinnamaldehyde gives an orange precipitate but compound B does not.
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53.
(a)
(i)
(ii)
There are twwo types of reactions taking place. They are electrophilic substitution (nitration
of phenol) and acid-base reaction with the primary amine.
(b)
The structural formula of G is:
Step
Reagent
Conditions
I
Excess concentrated NH3 solution
Heat in sealed tube
II
KCN in ethanol
Heat under reflux
III
LiAlH4 in dry ether
Room temperature
(Note:Other possible reagents for step III include using H2/Ni at 150°C – 200°C, sodium in ethanol
or tin with concentrated HCl.)
54.
(a)
Structures B and C are cis-trans isomers.
(b)
A reacts with steam to give butan-2-ol, CH3CH2CH(OH)CH3, which is chiral.
(c)
(i)
A diol is formed when D is subject to very mild oxidation.
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(ii)
Strong oxidation of D results in the cleavage of the C = C double bond.
(iii)
D undergoes addition polymerization.
55.
(a)
Reagents: concentrated HNO3 and concentrated H2SO4
Conditions: reflux at 55°C
(b)
(i)
Step I: Acidified KMnO4; heat
Step II: Ethanol and trace amount of concentrated H2SO4 as catalyst; heat
Step III: Tin and concentrated HCl; heat
(ii)
Molar mass of 4-nitromethylbenzene (C7H7NO2)
= (12.0 × 7 + 1.0 × 7 + 14.0 + 16.0 × 2) g mol-1= 137 g mol-1
The molar mass of benzocaine (C9H11NO2)
= (12.0 × 9 + 1.0 × 11 + 14.0 + 16.0 × 2) g mol-1 = 165 g mol-1
As 1 mole of nitromethylbenzene produces 1 mole of benzocaine,
 1.0

∴ mass of benzocaine produced = 
165  g
 137

= 1.20 g
56.
(a)
(i)
E:
CH2 = CHCH2OH
F:
G:
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(ii)
(1)
Add sodium metal to E, F and G separately. E gives effervescence of H2(g) indicating
that it is an alcohol, while F and G do not.
CH2 = CHCH2OH + Na  CH2 = CHCH2O─Na+ +
(2)
1
H2
2
Add 2,4-dinitrophenylhydrazine to E, F and G separately and warm. F and G give
orange precipitates showing that they are carbonyl compounds, while E does not.
(3)
Add Fehling’s solution to E, F and G separately and warm. G gives a red precipitate
of Cu2O showing that it is an aldehyde, while E and F do not.
CH3CH2CHO + 2Cu2+ + 5OH─  CH3CH2CO2─ + Cu2O + 3H2O
(b)
(i)
The ketone groups are reduced to secondary alcohols.
(ii)
Methyl ketones are oxidized to give CHI3 and acid salts.
57.
(a)
Step I: NaCN and H2SO4 ; 10°C – 20°C
Step II: dilute H2SO4; heat
The structural formula of intermediate J is:
(b)
(i)
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(ii)
(iii)
(iv)
(v)
58.
(a)
(b)
CH2 = CHCl + H2  CH3CH2Cl
Conditions: Ni as catalyst, at temperature above 150°C
(c)
(i)
(ii)
Electrophilic addition
(iii)
Hydrolysis of A and followed by elimination of water
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(d)
(i)
Optical isomerism
(ii)
59.
(a)
Step I: Elimination (or dehydrohalogenation)
Reagent: NaOH in ethanol
(b)
It is because each isomer has H atom on one C atom and Cl atom on adjacent C atom. Both
isomers would undergo elimination of HCl to give CH2 = CHCl.
(c)
(i)
Initiation:
Propagation:
CH3CH3 + Cl   CH3CH2  + HCl
CH3CH2  + Cl2  CH3CH2Cl + Cl 
CH3CH2Cl + Cl   CH3CHCl  + HCl
CH3CHCl  + Cl2  CH3CHCl2 + Cl 
Termination:
Cl  + Cl   Cl2
CH3CH2  + Cl   CH3CH2Cl
CH3CH2  + CH3CH2   CH3(CH2)2CH3
CH3CHCl  + Cl   CH3CHCl2
(ii)
One possible by-product is CH3CH2Cl and it can be separated from CH3CHCl2 by fractional
distillation.
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60.
(d)
Step II: addition polymerization
(a)
(i)
Step I: KCN in ethanolic solution; heat
Step II: H2O/H+; heat
(ii)
Step I: nucleophilic substitution
Step III: reduction
(b)
(i)
Test: Add I2(aq) and NaOH(aq) to both isomers separately and heat.
CH3CH(OH)CH3 gives a yellow precipitate of CHI3 whereas CH3CH2CH2OH does not.
(ii)
Test: Add 2,4-dinitrophenylhydrazine to both isomers separately.
CH3COCH3 gives an orange precipitate whereas CH2 = CHCH2OH does not.
Or
Test: Add I2(aq) and NaOH(aq) to both isomers separately and heat.
CH3COCH3 gives a yellow precipitate of CHI3 whereas CH2 = CHCH2OH does not.
Or
Test: Add bromine in 1,1,1-trichloroethane at room temperature (in the dark).
The bromine solution is decolourized with CH2 = CHCH2OH but not with
CH3COCH3.
Or
Test: Add acidified KMnO4 (or acidified K2Cr2O7) and heat.
The purple KMnO4 is decolourized with CH2 = CHCH2OH but not with CH3COCH3.
(Or The orange K2Cr2O7 turns green with CH2 = CHCH2OH but not with
CH3COCH3.)
Or
Test: Add sodium metal at room temperature.
Bubbles of H2 gas are given out with CH2 = CHCH2OH but not with CH3COCH3.
(iii)
Test: Add Tollen’s reagent to both isomers separately and heat.
A silver mirror is observed with CH3CH2CH2CH2CHO but not with CH3CH3COCH2CH3.
Or
Test: Add Fehling’s solution and heat.
CH3CH2CH2CH2CHO gives a red precipitate of Cu2O whereas CH3CH3COCH2CH3
does not.
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Or
Test: Add acidified KMnO4 (or acidified K2Cr2O7) and heat.
The purple KMnO4 is decolourized with CH3CH2CH2CH2CHO but not with
CH3CH3COCH2CH3. (Or The orange K2Cr2O7 turns green with
CH3CH2CH2CH2CHO but not with CH3CH3COCH2CH3.)
(iv)
Test: Add I2(aq) and NaOH(aq) to both isomers separately and heat.
CH3COCH2OH gives a yellow precipitate of CHI3 whereas CH3CH2CO2H does not.
Or
Test: Add acidified KMnO4 (or acidified K2Cr2O7) and heat.
The purple KMnO4 is decolourized with CH3COCH2OH but not with CH3CH2CO2H.
(Or The orange K2Cr2O7 turns green with CH3COCH2OH but not with
CH3CH2CO2H.)
Or
Test: Add 2,4-dinitrophenylhydrazine to both isomers separately.
CH3COCH2OH gives an orange precipitate whereas CH3CH2CO2H does not.
Or
Test: Add sodium carbonate.
Bubbles of CO2 gas are given out with CH3CH2CO2H bu tnot with CH3COCH2OH.
61.
(a)
C6H5OH + NaOH  C6H5O─Na+ + H2O
C6H5CO2H + NaOH  C6H5CO2─Na+ + H2O
2C6H5CO2H + Na2CO3  2C6H5CO2─Na+ + CO2 + H2O
(b)
These observations indicate that the relative acidities of the three compounds are in the order:
C6H5CO2H > C6H5OH > C6H5CH2OH
C6H5CH2OH is the least acidic because the negative charge of the anion C6H5CH2O─ is not
delocalized causing the anion C6H5CH2O─ is the least stable.
C6H5OH is more acidic than C6H5CH2OH because the electron-withdrawing benzene ring weakens
the O—H bond. Therefore, C6H5OH dissociates readily to give H+ ion. Moreover, the anion
C6H5O─ is stabilized by delocalizing the negative charge over the ring.
C6H5CO2H is even more acidic than C6H5OH because it has two electron-withdrawing groups. This
further weaken the O—H bond and C6H5CO2H dissociates more readily to give H+ ions.
Furthermore, the anion is further stabilized by delocalizing the negative charge over two oxygen
atoms.
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62.
(a)
Ethanal reacts with hydrogen cyanide in the presence of trace amount of sodium cyanide as
catalyst to give cyanohydrins.
The type of reaction undergone is nucleophilic addition.
Step 1: generation of nucleophile, CN─
HCN is a weak acid and a poor source of CN─, therefore a trace amount of NaCN is added to speed
up the reaction.
NaCN  Na+ + CN─
(HCN shows no nucleophilic properties because the nitrogen is electronegative and pulls its lone
pair electrons in.)
Step 2: addition of nucleophile, CN─, to the electron-deficient carbon centre of the carbonyl group.
(SLOW).
Step 3: formation of the cyanohydrin product by taking up a proton with the resulting negatively
charged oxygen. (FAST)
CN─ ion is regenerated in the reaction to replace the one which consumed in the initial attack; i.e.
NaCN acts as a catalyst.
(b)
Optical isomerism
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(c)
(i)
Test: Add Na2CO3(aq) separately to each compound at room temperature.
Effervescence of CO2 is observed with 3-oxobutanoic acid but not with propanone.
2CH3COCH2CO2H + Na2CO3  2CH3COCH2CO2─Na+ + CO2 + H2O
Or
Test: Add sodium or PCl5 (or SOCl2).
3-oxobutanoic acid gives effervescence of H2 with sodium or fumes of HCl with
PCl5 or SOCl2, but propanone does not.
(ii)
Test: Add 2,4-dinitrophenylhydrazine separately to each compound.
3-oxobutanoic acid reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate, as it
contains a ketone group. 3-Hydroxybutanoic acid does not give such precipitate.
Or
Test: Add acidified K2Cr2O7 separately to each compound and reflux.
3-Hydroxybutanoic acid is oxidized to a ketone and the orange acidified K2Cr2O7 turns
green.
3-oxobutanoic acid cannot be oxidized, therefore, the solution remain orange.
63.
Boiling point:
•
The high boiling points of propanoic acid and butan-1-ol can be explained by the presence of
intermolecular hydrogen bonding. Propanoic acid has even a higher boiling point than butan-1-ol,
because hydrogen bonding is more extensive in propanoic acid.
In the case of 1-aminobutane, intermolecular hydrogen bonding is also present but it is not as
extensive as in propanoic acid and butan-1-ol. Therefore, the boiling point of 1-aminobutane is
slightly lower.
Pentane has a very low boiling point because its molecules are held together by weak
intermolecular van der Waals’ forces which are easily broken.
Density:
•
The increase in intermolecular forces of attraction cause these compounds to be more closely
packed together and this is responsible for the slight increase in their density.
Solubility in water:
•
Propanoic acid and 1-aminobutane are soluble in water (a polar solvent) as they are polar
molecules. They can also form hydrogen bonds with water molecules.
The solubility of butan-1-ol is reduced by the presence of non-polar C4H9 group. Therefore,
butan-1-ol is only slightly soluble in water.
Pentane is a non-polar molecule. Thus, it is insoluble in polar solvents like water.
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64.
(a)
(b)
(i)
Alcoholic KOH; reflux
(ii)
Aqueous KOH; reflux
(i)
But-2-ene
(ii)
Geometrical isomerism
cis
trans
They are geometrical isomers because they have different arrangements of their atoms in
space due to restricted rotation about the C = C bond.
(c)
(i)
2-Methylbutanoic acid
(ii)
The structural formula of D is:
(iii)
Optical isomerism
(iv)
A to D: Alcoholic KCN; reflux
D to E: Aqueous HCl; reflux
65.
(d)
Nucleophilic substitution
(a)
CH3CH2CH2CH2OH is a primary alcohol. It is oxidized to butanal (an aldehyde) and then to
butanoic acid (a carboxylic acid) by acidified K2Cr2O7 on warming.
CH3CH2CH2CH2OH + [O]  CH3CH2CH2CHO + H2O
CH3CH2CH2CHO + [O]  CH3CH2CH2COOH
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CH3CH(OH)CH2CH3 is a secondary alcohol. It is oxidized to butan-2-one (a ketone) by acidified
K2Cr2O7 on warming. However, further oxidation to a carboxylic acid is not allowed.
(b)
CH3CH2CH2CH2OH dehydrates to give but-1-ene by heating with concentrated H2SO4 at 180°C.
CH3CH(OH)CH2CH3 dehydrates to give but-2-ene (as major product) and but-1-ene (as minor
product).
(c)
CH3CH = CHCH3 reacts with HBr at room temperature to give 2-bromobutane (one product only).
CH3CH2CH = CH2 reacts with HBr to give 2-bromobutane (as major product) and 1-bromobutane
(as minor product).
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(d)
(
) is hydrolyzed by refluxing with aqueous NaOH to give phenylmethanol. The
.
reaction is a kind of nucleophilic substitution.
(
) is not hydrolyzed because the C — Cl bond is very strong due to the interaction of
the lone pair electrons on Cl with the  electron cloud of the benzene ring.
66.
Methyl propanoate
(a)
Ethyl ethanoate
Propyl methanoate
Methylethyl methanoate
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(b)
Take methyl propanoate as an example. The reaction between methyl propanoate and aqueous
sodium hydroxide can be shown by the following equation:
Sodium
propanoate
Methanol
This type of reaction is used to make soap (saponification of esters of fatty acids).
(c)
The distillates from X and Y gave a yellow precipitate when treated with iodine and aqueous alkali.
This shows that they are alcohols containing the
The distillates of X and Y are ethanol (
(
group.
) from ethyl ethanoate and propan-2-ol
) from methylethyl methanoate.
Besides, the aqueous residue from Y gave a red precipitate when warmed with Fehling’s solution.
This shows that the aqueous residue from Y contains aldehydes. As ethanol can be oxidized to an
aldehyde and propan-2-ol cannot (propan-2-ol can only be oxidized to a ketone), we can conclude
that Y is methylethyl methanoate and X is ethyl ethanoate.
(d)
This isomer must contain a carboxy group since it liberates carbon dioxide gas (colourless gas)
when treated with aqueous sodium carbonate. The two possible structures for this isomer are:
CH3CH2CH2COOH
or
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(e)
67.
(a)
(b)
In cyclohexene, the π electrons are localized on the two carbon atoms only. This provides an
electron-rich centre for electrophiles to attack and electrophilic additions are susceptible to occur.
Thus, cyclohexene reacts with bromine to give a colourless product, 1,2-dibromocyclohexane.
In benzene, the π electrons are not localized on specific carbon atoms. They are delocalized above
and below the carbon skeleton. The delocalization of π electrons gives extra stability to the
benzene molecule. Though benzene is a highly unsaturated molecule, it does not undergo
electrophilic addition because the addition of atoms or groups of atoms into the benzene ring
would destroy the stable structure provided by the delocalization of π electrons. As a result,
bromine does not react with benzene and the solution remains brown.
68.
(a)
Test: Add sodium metal.
Effervescence of H2(g) after the addition of sodium metal shows that the ester contain an –OH
group.
(b)
Aqueous sodium hydroxide solution followed by addition of concentrated HCl
(c)
Test: Add sodium carbonate.
Effervescence of CO2(g) after the addition of sodium carbonate shows that the –CO2H group is
present in the citric acid.
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69.
(a)
Reagent
Reaction I
Reaction II
Reaction III
phosphorus tribromide
acidified potassium
concentrated sulphuric
(PBr3)
dichromate
acid (H2SO4)
(K2Cr2O7/H+)
Condition
(b)
heat
heat
heating at 180°C
For 2-methylpropan-2-ol as the starting material, the products of reactions I, II and III are shown in
the following table:
Product
70.
(a)
Reaction I
Reaction II
Reaction III
2-bromo-2-methylpropane
No reaction
2-methylpropene
(i)
(ii)
(b)
(i)
(ii)
Nucleophilic addition reaction
(iii)
Due to the difference in electronegativity between carbon and oxygen, the carbonyl carbon
bears a partial positive charge and the carbonyl oxygen bears a partial negative charge. As
the carbonyl carbon carries a partial positive charge, the carbonyl group is susceptible to
nucleophilic attack. When the nucleophile attacks the carbonyl group, it uses its lone pair
electrons to form a bond with the carbonyl carbon atom. The carbonyl carbon atom can
accept this electron pair because one pair of bonding electrons of the carbon-oxygen bond
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can shift out to the carbonyl oxygen atom. This electron-rich oxygen then transfers its
electron pair to an electropositive agent.
(iv)
Ethene contains a C = C double bond. As there is no difference in electronegativity between
two carbon atoms, the electrons in the double bond are evenly distributed over the C = C
double bond. Besides, the C = C double bond is an electron-rich centre. Therefore, there is
no electron-deficient centre for the attack of the nucleophile CN-. Thus, ethene would not
react with HCN in a similar way.
71.
(a)
(i)
Monomer 1: ClOC – (CH2)4 – COCl
Monomer 2: H2N – (CH2)6 – NH2
(ii)
(b)
(i)
(ii)
(c)
Condensation polymer
Use as non-stick coating on frying pan
The polymer in (a) is a polyamide which can be hydrolyzed by water, acids or alkalis into its
constituent monomers. However, the polymer in (b) is very unreactive and has a high boiling point.
Therefore, the polymer in (a) is easier to break down and hence causes less environmental hazards.
72.
(a)
No. Because there is no chiral carbon centre in the molecule.
(b)
Yes. Because the groups (or atoms) on the sides of both double bonds are different. Thus
geometrical isomerism would occur.
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(c)
(i)
(ii)
(iii)
(d)
No. Tollen’s reagent (i.e. ammoniacal solution of silver nitrate) reacts with aldehydes only.
-ionone is a ketone which cannot react with Tollen’s reagent.
(e)
(i)
Mix 1 cm3 of iodine solution with a few drops of the unknown compound. Then add aqueous
NaOH solution drop by drop until the brown colour of the solution almost disappears. Warm
the tube in a beaker of hot water if necessary.
(ii)
The formation of a yellow precipitate is the indication of a positive result of the test.
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(iii)
73.
Yes. It is because  -ionone contains the methyl ketone (
) group.
(a)
(b)
(i)
It is because the spatial arrangements of the atoms or groups of atoms are different due to
restricted rotation about the C = C double bond.
(ii)
The plane of plane polarized monochromatic light rotates.
(c)
(d)
(i)
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(ii)
Add 3 cm3 of the product formed in (i) to a test tube. Then add 2 cm3 of
2,4-dinitrophenylhydrazine to the test tube. An orange precipitate is formed. The formation
of the orange precipitate shows that the product formed in (i) contains a carbonyl group.
Add Fehling’s solution (an alkaline solution of copper(II) tartrate) to a test tube. The
solution is deep blue. Then add a few drops of the product formed in (i) to the test tube. The
solution remains deep blue showing that there is no reaction. Therefore the product formed
in (i) is not an aldehyde. So the product should be a ketone.
74.
(a)
(i)
(ii)
In CH3COCl, the acyl carbon is highly electron-deficient. It is because both oxygen and
chlorine are electron-withdrawing groups. They draw the electron in the bonds towards
themselves making the carbon atom highly electron-deficient. Therefore, CH3COCl is
susceptible to nucleophilic attack.
However, in the case of CH3CH2Cl, the degree of electron deficiency of the carbon adjacent
to the chlorine atom is much lower than that in the acyl carbon of CH3COCl. Thus, the
reaction between water and CH3CH2Cl is extremely slow.
(b)
(i)
(ii)
It is because the esterification between ethanoic acid and ethanol is a reversible reaction
while the ester formation between ethanoyl chloride and ethanol is an irreversible reaction.
Moreover, ethanoyl chloride is much more reactive than ethanoic acid. Thus, the reaction
with ethanoyl chloride gives a better yield of the ester.
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75.
(a)
(i)
UV light provides energy for chlorine molecules to break homolytically, forming reactive
free chlorine radicals.
(ii)
The reaction between ethane and chlorine proceeds by a free radical chain mechanism. The
chlorine radical formed attacks an ethane molecule, forming CH3CH2• radical which attacks
a chlorine molecule to form chloroethane. The chloroethane formed may be attacked by
another chlorine radical to form CH3CHCl• radical. The CH3CHCl• radical may further
attack a chlorine molecule to form dichloroethane. The chain reaction continues to occur
until hexachloroethane is reached. The formation of polychlorinated ethane depends on the
relative amounts of chlorine and ethane. If chlorine is in excess, the major product is
hexachloroethane. If ethane is in excess, chloroethane becomes the major product. If both
were in similar amounts, more than one chlorine-containing organic product would be
formed.
(iii)
Butane. It is formed in the chain-terminating step when two CH3CH2• radicals are joined
together.
(b)
(i)
(ii)
76.
One of the products formed in the reaction is HCl gas which is highly corrosive.
(a)
Free radicals are electrically neutral atoms or groups of atoms possessing an unpaired electron.
(b)
Substitution reactions are the reactions with an atom or a group of atoms of the reactant molecule
being replaced by another atom or group of atoms.
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(c)
The chlorine radical attacks one of the hydrogen atoms in the ethane molecule. Then, one of the
bonding electrons in the C – H bond forms a bond with the chlorine radical forming hydrogen
chloride. Another electron of the C – H bond then shifts to the carbon atom giving a carbon free
radical.
(d)
77.
(a)
CH3CH(Br)COOH + OH–  CH3CH(OH)COOH + Br–
(b)
2-Bromopropanoic acid contains a chiral centre which bonded to four different atoms or groups of
atoms. Therefore, the 2-bromopropanoic acid molecule and its mirror image are not
superimposable.
If the plane of polarized light is rotated when the solution of 2-bromopropanoic acid is placed in
the polarizer, we can conclude that 2-bromopropanoic acid shows optical isomerism.
(c)
As the reaction between 2-bromopropanoic acid and sodium hydroxide solution proceeds by the
SN1 mechanism, this would lead to the formation of the carbocation which is planar. When it reacts
with a nucleophile, the nucleophile may either attack from the frontside or the backside. This
results in the formation of racemic mixture (i.e. the mixture of containing equal amounts of both
enantiomers). Thus, the optical activity of the product is lost.
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78.
(a)
(b)
Chloromethane has a tetrahedral shape. According to the electron pair repulsion theory, the
electron pairs in the outermost shell of the central atom in a molecule should stay as far apart as
possible to minimize the electrostatic repulsion between electron pairs in the valence shell. In
chloromethane, carbon is the central atom of the molecule and it carries four bond pairs of
electrons. The four electron pairs in three dimensions are separated at a maximum with bond
angles of 109.5. The shape of the molecule is therefore tetrahedral.
(c)
As chloromethane is a polar molecule, the molecules are held together by dipole-dipole
interactions. But in the case of methane, its molecules are non-polar. Therefore, methane molecules
are only held together by instantaneous dipole-induced dipole interactions. As dipole-dipole
interactions are stronger than instantaneous dipole-induced dipole interactions, more energy is
required to separate chloromethane molecules in the process of boiling. As a result, chloromethane
has a higher boiling point than methane.
(d)
In methanol, there are extensive intermolecular hydrogen bonds between methanol molecules.
However, chloromethane molecules are only held together by dipole-dipole interactions. As
hydrogen bonds are much stronger than dipole-dipole interactions, more energy is required to
separate methanol molecules in the process of boiling. Therefore, methanol has a higher boiling
point than chloromethane.
79.
(a)
(i)
Bromine in 1,1,1-trichloroethane
(ii)
1,2-Dibromoethane
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(b)
(c)
(i)
(ii)
H–CC–H
(i)
A chiral molecule is one that is not superimposable on its mirror image.
(d)
(e)
(ii)
80.
(a)
(b)
Propyl ethanoate
(c)
As catalyst
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81.
(a)
(b)
(c)
Ketones
(d)
Firstly, the test below is done in order to show that X is a carbonyl compound.
1.
Add 1 – 2 drops of X into a test tube.
2.
Add 2 cm3 of 2,4-dinitrophenylhydrazine to the test tube.
As an orange precipitate is formed in the test, this shows that X is a carbonyl compound.
In order to show that X is a ketone, Tollen’s test can be carried out. The procedure is shown as
follows:
1.
Add 1 cm3 of 0.05 M AgNO3 with 3 – 4 drops of NaOH into a very clean test tube.
2.
Add NH3 solution drop by drop until the Ag2O precipitate is nearly dissolved.
3.
Add 1 – 2 drops of X and shake the tube gently. Warm the tube in a hot water bath.
As a silver mirror is not appeared in the test tube, this shows that X is a ketone.
(e)
and
It is because X does not contain
The test is a characteristic test for the above two groups.
82.
(a)
(i)
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(ii)
(iii)
(b)
83.
(a)
(b)
(i)
Electrophilic addition
(ii)
Chloroethane
(i)
CH3CH2CH2CN
(ii)
CH3CH2CH2NH2 ; (CH3CH2CH2)2NH ; (CH3CH2CH2)3N
The procedure of the test is shown below:
1.
Add 5 drops of 2-chlorobutane and 5 drops of butan-2-ol into each of two test tubes.
2.
Add 2 cm3 of iodine solution into each test tube.
3.
For each test tube, add NaOH solution drop by drop until the brown colour of the solution
almost disappears.
The test tube containing 2-chlorobutane remains unchanged. But in the test tube containing
butan-2-ol, a pale yellow precipitate (CHI3) is formed.
(c)
(i)
2-Methylpropan-2-ol
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(ii)
The procedure of the test is given below:
1.
Add 5 drops of butan-2-ol and 5 drops of 2-methylpropan-2-ol into each of two test
tubes.
2.
Add K2Cr2O7/H+ into each test tube and warm the test tube in hot water.
The test tube containing butan-2-ol will change from orange to green. The test tube
containing 2-methylpropan-2-ol will remain orange.
(iii)
Butan-2-ol is a secondary alcohol which can be oxidized by acidified potassium
dichromate(VI) to form butanone while the orange dichromate(VI) ion is reduced to green
chromium(III) ion. As 2-methylpropan-2-ol is a t-pertiary alcohol, it is resistant to oxidation
because any oxidation would immediately involve the breakage of the high energy C – C
bond in the molecule.
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