Download STD. XII Sci. Perfect Chemistry - II, NCERT Questions and Answers

08 Target Publications Pvt. Ltd.
d and f‐Block Elements
Chapter 08: d and f - Block Elements
Q.1. How is the variability of oxidation states of the transition elements different from that of the
non-transition elements? Illustrate with examples.
(NCERT)
Ans: i.
The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in
such a way that their oxidation states differ from each other by unity.
eg. Fe2+ and Fe3+, Cu+ and Cu2+, etc.
ii.
In case of non-transition elements, the oxidation states differ by units of two.
eg. Pb2+ and Pb4+, Sn2+ and Sn4+, etc.
iii. In transition elements, the higher oxidation states are more stable for heavier elements in a group.
eg. In group 6, Mo (VI) and W (VI) are more stable than Cr (VI).
iv. In p-block elements, the lower oxidation states are more stable for heavier members due to inert pair
effect.
eg. In group 14, Pb (II) is more stable than Pb (IV).
Q.2. Describe the oxidizing action of potassium dichromate and write the ionic equations for its
reaction with: i.
iodide
ii.
iron (II) solution
iii. H2S.
(NCERT)
Ans: Potassium dichromate, K2Cr2O7 is a strong oxidizing agent and is used as a primary standard in volumetric
analysis involving oxidation of iodides, ferrous ions and S2 ions, etc.
i.
It oxidizes potassium iodide to iodine.
Cr2 O 72 14H   6I  
 2Cr 3  3I 2  7H 2 O
ii.
It oxidizes iron (II) salt to iron (III) salt.
Cr2 O72 14H   6Fe 2  
 2Cr 3  6Fe3  7H 2 O
iii.
It oxidizes H2S to S.
Cr2O72  14H   3S2 
 3S  2Cr 3  7H 2O
Q.3. Decide which of the following atomic numbers are the atomic numbers of the inner transition
elements: 29, 59, 74, 95, 102, 104.
(NCERT)
Ans: Inner transition elements have atomic numbers from 58 to 71 and 90 to 103. Hence, the atomic numbers 59,
95 and 102 are the atomic numbers of the inner transition elements.
Q.4. Write the electronic configurations of the elements with the atomic numbers 61, 91 and 101.
(NCERT)
Ans: The electronic configurations are given in the following table.
Element
Promethium
Protactinium
Mendelevium
Symbol
Pm
Pa
Md
Atomic Number
61
91
101
Electronic Configuration
[Xe] 4f5 5d0 6s2
[Rn] 5f2 6d1 7s2
[Rn] 5f13 6d0 7s2
Q.5. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this
statement by giving some examples from the oxidation state of these elements.
(NCERT)
Ans: Lanthanoids show limited number of oxidation state, viz., +2, +3 and +4 (out of which +3 is most
common). This is because of large energy gap between 4f and 5d-subshells. The dominant oxidation state of
actinoids is also +3 but they show a number of other oxidation states also. eg. uranium (Z = 92) and
plutonium (Z = 94), show +3, +4, +5 and +6, neptunium (Z = 93) shows +3, +4, +5, +6 and +7 etc. This is
due to small energy difference between 5f, 6d and 7s-subshells of the actinoids.
1
Std. XII Sci.: Perfect Chemistry ‐ II Q.6. Compare the chemistry of actinoids with that of the lanthanoids with special reference to :
(NCERT)
i.
iii.
electronic configuration
atomic size and ionic sizes
ii.
iv.
oxidation state
chemical reactivity
Ans:
ii.
Characteristics
Electronic
configuration
Oxidation state
iii.
Atomic and ionic sizes
iv.
Chemical reactivity
i.
2
Lanthanoids
It may be represented by
[Xe]4f014 5d0 or 1 6s2
Show +3 oxidation state only, except
in few cases where it is +2 or +4.
They never show more than +4
oxidation state.
The ionic radii of M3+ ions in
lanthanoids series show a regular
decrease in size of ions with increase
in atomic number. This decrease is
known as lanthanoid contraction.
These are less reactive metals and
form oxides, sulphides, nitrides,
hydroxides and halides etc. These
also form H2 with acids. They show
a lesser tendency for complex
formation.
Actinoids
It may be represented by
[Rn]5f014 6d0 or 1 7s2
Show higher oxidation states such
as +4, +5, +6, +7 also in addition to
+3 oxidation state.
There is a greater and gradual
decrease in the size of atoms or M3+
ions across the series. This greater
decrease is known as actinoid
contraction.
These are highly reactive metals
especially in finely divided state.
They form a mixture of oxide and
hydride by action of boiling water.
They combine with non-metals even
at moderate temperature. They
show a greater tendency for
complex formation.
09 Target Publications Pvt. Ltd.
Coordination Compounds
Chapter 09: Coordination Compounds
Q.1. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4
solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain
why?
(NCERT)
Ans: i.
1 : 1 molar mixture of FeSO4 and (NH4)2SO4 forms a double salt, FeSO4.(NH4)2SO4.6H2O (Mohr’s
salt) which exists only in solid state. In aqueous solution, it dissociates into ions Fe2+, NH4+ and
SO42. Hence, it gives the test of Fe2+ ion.
ii.
1 : 4 molar mixture of CuSO4 and ammonia forms a coordination compound, [Cu(NH3)4]SO4. In
aqueous solution, it retains its identity and does not dissociate into ions. Hence, it does not give the
test of Cu2+ ion.
Q.2. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution
of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed
through this solution?
(NCERT)
Ans: i.
When excess of aqueous KCN is added to an aqueous solution of copper sulphate, the coordination
entity formed is [Cu(CN)4]2.
[Cu(H2O)4]2+ + 4CN  [Cu(CN)4]2 + 4H2O
ii.
[Cu(CN)4]2 is a coordination compound and hence, retains its identity in the aqueous solution. It does
not dissociate to give free Cu2+ ions. Hence, no precipitate of copper sulphide is obtained when H2S(g)
is passed through this solution.
1
10 Halogen Derivatives of Alkanes and Arenes Chapter 10: Halogen Derivatives of Alkanes
and Arenes Q.1. Write the structures of the following compounds:
i.
2-Chloro-3-methylpentane
ii.
iii. 1,4-Dibromobut-2-ene
iv.
v.
2-(2-Chlorophenyl)-1-iodooctane
vi.
4-tert-Butyl-3-iodoheptane
1-Chloro-4-ethylcyclohexane
2-Bromobutane
CH3
Ans: i.
1
CH3
5
4
3
2
(NCERT)
ii.
CH3 – CH  CH – CH2 – CH3
H3C – C – CH3
7
6
5
4
3
2
1
H3C – CH2 – CH2 – C  CH – CH2 – CH3
Cl
H
I
Cl
iii.
1
2
3
Br
1
v.
4
iv.
H2C – CH = CH – CH2
C2H5
Br
2
I  CH2  CH  (CH2)5CH3
1
Cl
vi.
1
2
3
4
CH3  CH  CH2  CH3
Br
2
Q.2. Write the isomers of compound having formula C4H9Br.
Ans: The isomers of compound having formula C4H9Br are as follows:
i.
H3C  CH2  CH2  CH2  Br
ii.
1-Bromobutane
(NCERT)
H3C  CH2  CH  CH3
Br
2-Bromobutane
CH3
iii.
H3C  CH  CH2Br
iv.
CH3
H3C  C  Br
CH3
1-Bromo-2-methylpropane
2-Bromo-2-methylpropane
Q.3. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical
chlorination yields:
i.
a single monochloride
ii.
three isomeric monochlorides
iii. four isomeric monochlorides
(NCERT)
Ans: i.
In neopentane (C5H12), all the H-atoms are equivalent and hence, on photochemical chlorination
yields only one type of monohalogen derivative.
1
Std. XII Sci.: Perfect Chemistry ‐ II CH3
CH3  C  CH3
CH3
Neopentane
(2,2-Dimethylpropane)
ii.
In n-Pentane(C5H12), replacement of a,b or c type of H-atom leads to formation of three different
types of monohalogen derivatives which are isomers of each other.
a
b
c
b
a
CH3  CH2  CH2  CH2  CH3
n-Pentane
(Pentane)
iii.
In isopentane (C5H12), replacement of a, b, c or d type of H-atoms leads to formation of four different
types of monohalogen derivatives which are isomers of each other.
a
b
c
d
CH3  CH  CH2  CH3
a
CH3
Isopentane
(2Methylbutane)
Q.4. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with
sodium ethoxide in ethanol and identify the major alkene:
i.
1-Bromo-1-methylcyclohexane
ii.
2-Chloro-2-methylbutane
iii.
3-Bromo-2,2,3-trimethylpentane
Ans: i.
(NCERT)
In 1-Bromo-1-methylcyclohexane, the -hydrogen atoms on either side of the Br atoms are
equivalent, therefore, only one type of product i.e., 1-methylcyclohexene is formed.
HH
H
Br
CH3
H
CH3

C2 H5 ONa /C2 H5 OH
 HBr
1-Methylcyclohexene
1-Bromo-1-methylcyclohexane
ii.
2-Chloro-2-methylbutane has two different sets of equivalent -hydrogen atoms and hence, in
principle can give two alkanes (I and II). But according to Saytzeff’s rule, more highly substituted
alkene (II), being more stable, is the major product.
  CH3
CH3
C
CH3
 C2 H5 ONa / C2 H5 OH
CH2CH3 
CH2 = C  CH2CH3 + CH3  C = CHCH3
 HCl
2-Methylbut-1-ene (I)
(Minor product)
Cl
2-Chloro-2-methylbutane
iii.
2
CH3
2-Methylbut-2-ene (II)
(Major product)
3-Bromo-2,2,3-trimethylpentane has two different sets of -hydrogen atoms and hence, in principle,
can give two alkenes (I and II). But according to Saytzeff's rule, more highly substituted alkene (II),
being more stable, is the major product.
Chapter 10: Halogen Derivatives of Alkanes
and Arenes 1
H3C
4
C2H5ONa/C2H5OH
3
CH2
2
 CH3 – C – C  CH2 – CH3
 HBr
 H3C
H3C CH3
5 4 1 2 3 CH3 – C – C  CH2 – CH3

H3C Br
3-Bromo-2,2,3-trimethylpentane
2-Ethyl-3,3-dimethylbut-1-ene (I)
(Minor product)
H3C CH3
5
C2H5ONa/C2H5OH
4
3
2
1  CH3 – C – C = CH – CH3
 HBr
H3C
3,4,4-Trimethylpent-2-ene (II)
(Major product)
Q.5. Identify A, B, C and R in following:
i.
ii.
Dry ether
H 2O
Br + Mg 
 A 
B
Dry ether
D2O
R  Br + Mg 
 C 
 CH3CHCH3
D
Ans: i.
Dry ether
Br + Mg 

H 2O
MgBr 

Cyclohexyl
magnesium
bromide
(A)
Cyclohexyl
bromide
ii.
(NCERT)
+ Mg(OH)Br
Cyclohexane
(B)
D2 O
Dry ether
 R  Mg  Br 
 CH3CHCH3
R  Br + Mg 
Alkyl
magnesium
bromide
Alkyl
bromide
D
Since, D gets attached to same C-atom on which MgX was present.
Therefore, R  CH3  CH  CH3
Thus, structure of ‘C’ is CH3  CH  CH3
MgBr
Isopropyl magnesium bromide
Q.6. Write the structure of the major organic product in each of the following reactions:
i.
Dry acetone
CH3CH2CH2Cl + NaI 

Heat
ii
Ethanol
(CH3)3CBr + KOH 

Heat
iii.
Water
CH3CH(Br)CH2CH3 + NaOH 
iv.
aq.ethanol
CH3CH2Br + KCN 

v.
C6H5ONa + C2H5Cl 

vi.
CH3CH2CH2OH + SOCl2 

vii.
Peroxide
CH3CH2CH = CH2 + HBr 

viii. CH3CH = C(CH3)2 + HBr 

(NCERT)
Ans: i.
CH3CH2CH2Cl
1-Chloropropane
+ NaI
Dry acetone, heat


Finkelstein reaction
CH3CH2CH2I + NaCl
1-Iodopropane
3
Std. XII Sci.: Perfect Chemistry ‐ II CH3
ii.
(CH3)3CBr
Ethanol, 

CH3  C = CH2 +
Dehydrohalogenation
+ KOH
Water

Hydrolysis
+ NaOH
CH3CHCH2CH3
CH3  CH  CH2  CH3
Br
OH
2-Bromobutane
iv.
Butan-2-ol
CH3CH2Br
+ KCN
C6H5ONa+
+ C2H5Cl
aq.ethanol


1-Bromoethane
v.
CH3CH2CN
+
KBr
C6H5OCH2CH3
+
NaCl
Propane nitrile
Williamson 's


synthesis
Phenetole
Sodium phenoxide
vi.
CH3CH2CH2OH
Reflux

Δ
+ SOCl2
Propan-1-ol
vii.
H2O
2-Methylpropene
2-Bromo-2-methylpropane
iii.
KBr +
CH3CH2CH = CH2
+ HBr
CH3CH2CH2Cl +
HCl + SO2
1-Chloropropane
Peroxide

Anti-Markownikoff's rule
CH3CH2CH2CH2Br
1-Bromobutane
(Major product)
But-1-ene
CH3
viii. CH3CH = C  CH3
Markownikoff's rule


+ HBr
CH3CH2  C  CH3
CH3
Br
2-Methylbut-2-ene
2-Bromo-2-methylbutane
Q.7. What happens when:
i.
n-butyl chloride is treated with alcoholic KOH?
ii.
bromobenzene is treated with Mg in the presence of dry ether?
iii.
ethyl chloride is treated with (aq) KOH?
iv.
methyl bromide is treated with sodium in the presence of dry ether?
v.
Ans: i.
methyl chloride is treated with KCN?

 CH3CH2CH = CH2 + KCl + H2O
CH3CH2CH2CH2Cl + KOH(alc.) 
But-1-ene
n-Butyl chloride
ii.
C6H5Br
+
Mg 

Phenyl magnesium
bromide
Bromobenzene
iii.
CH3CH2Cl
C6H5MgBr
Dry ether

+ KOH(aq) 
 CH3CH2OH + KCl + H2O
Ethyl chloride
iv.
2CH3Br
Methyl
bromide
v.
CH3Cl
Methyl
chloride
4
Ethanol
Dry ether
+ 2Na 
Wurtz  CH3CH3 + 2NaBr
reaction
+ KCN
(alc.)
Ethane
 CH3CN + KCl
Methyl
cyanide
(NCERT)
Chapter 10: Halogen Derivatives of Alkanes
and Arenes Q.8. Arrange the compounds of each set in order of reactivity towards S N2 displacement:
i.
ii.
iii.
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.
1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane.
1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane,
1-Bromo-3-methylbutane.
(NCERT)
CH3
Ans: i.
CH3  C  CH2  CH3
CH3  CH2  CH2  CH2 CH2  Br
CH3  CH  CH2  CH2  CH3
1-Bromopentane (1)
Br
Br
2-Bromo-2-methylbutane (3)
2-Bromopentane (2)
Since, due to steric reasons, the order of reactivity in SN2 reactions follows the order : 1 > 2 >3,
therefore, order of reactivity of the given alkyl bromide is
1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane.
CH3
ii.
4
3
CH3
1
2
CH3  CH  CH2  CH2  Br
1-Bromo-3-methylbutane (1)
1
3
2
1
4
CH3  C  CH2  CH3
2
CH3
3
4
CH3  CH  CH  CH3
Br
Br
2-Bromo-2-methylbutane (3)
2-Bromo-3-methylbutane (2)
Since, due to steric reasons, the order of reactivity of alkyl halides in SN2 reaction follows the order:
1 > 2 > 3, therefore, the order of reactivity of the given alkyl bromides is
1-Bromo-3-methylbutane (1) > 2-Bromo-3-methylbutane (2) > 2-Bromo-2-methylbutane (3).
CH3
iii.


CH3  C  CH2  Br
CH3  CH2  CH2  CH2  Br
1-Bromobutane
CH3
(1 With no branching)
1-Bromo-2,2-dimethylpropane
(1 with two -methyl groups)
CH3
 CH3

 1-Bromo-2-methylbutane
(1 with one -methyl group)


CH3  CH CH2 CH2  Br
CH3  CH2  CH  CH2  Br
1-Bromo-3-methylbutane
(1 with one methyl group at -position)
Since, in case of 1 alkyl halides, steric hindrance increases in the order: n-alkyl halides, alkyl halide
with a substituent at position other than the -position, one substituent at the -position, two
substituents at the -position, therefore, the reactivity decreases in the same order. Thus, the reactivity
of the given alkyl bromides decreases in the order:
1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane.
Q.9. Predict the order of reactivity of the following compounds in S N1 and S N2 reactions.
i.
ii.
Ans: i.
The four isomeric bromobutanes.
C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br
(NCERT)
CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (S N1 )
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (S N 2 )
5
Std. XII Sci.: Perfect Chemistry ‐ II Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more
stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of
(CH3)2CH – group. Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1
reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertiary bromide. Hence,
ii.
the above order is followed in SN1 . The reactivity in SN 2 reactions follows the reverse order as the
steric hindrance around the electrophilic carbon increases in that order.
C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (S N1 )
C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br < C6H5CH(CH3)Br < C6H5CH2Br (S N2 )
Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more
stable than obtained from C6H5CH(CH3)Br because it is stabilised by two phenyl groups due to
resonance. Therefore, the former bromide is more reactive than the latter in SN1 reactions. A phenyl
group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than
C6H5CH(CH3)Br in SN2 reaction.
Q.10. Which compound in each of the following pairs will react faster in S N2 reaction with OH?
i.
CH3Br or CH3I
ii.
(CH3)CCl or CH3Cl
(NCERT)


Ans: i.
Since I ion is a better leaving group than Br ion, therefore, CH3I reacts faster than CH3Br in SN2
reaction with OH ion.
ii.
On steric grounds, 1 alkyl halides are more reactive than tert-alkyl halides in SN 2 reactions.
Therefore, CH3Cl will react at a faster rate than (CH3)3CCl in a SN2 reaction with OH ion.
Q.11.Write the structures of the following organic halogen compounds:
i.
p-Bromochlorobenzene
ii.
Perfluorobenzene
Ans: i.
ii.
F
Cl
F
F
F
Br
(NCERT)
F
F
Perfluorobenzene
p-Bromochlorobenzene
Q.12. Draw the structures of major monohalo products in each of the following reactions:
CH2OH
CH2CH3
heat
i.
ii.
+ HCl 

Br2 ,heat or



UV light
HO
O2N
CH2OH
Ans: i.
+
HO
heat
HCl 

4(Hydroxymethyl)phenol
4
CH2Cl
1
HO
4-Chloromethylphenol
Only alcoholic but not phenolic OH groups are replaced by Cl on heating with HCl.
Br
ii.
CH2CH3
O2N
4-Ethylnitrobenzene
6
Br2 ,heat or


UV light
4
CH  CH3
1
O2N
4-(1-Bromoethyl)nitrobenzene
(NCERT)
11 Alcohols, Phenols and Ethers
Chapter 11: Alcohols, Phenols and Ethers
Q.1. Write structures of the compounds whose IUPAC names are as follows:
i.
Cyclohexylmethanol
ii.
3-Cyclohexylpentan-3-ol
iii. Cyclopent-3-en-1-ol
iv. 3-(Chloromethyl)pentan-1-ol
Ans:
CH2  OH
i.
ii.
(NCERT)
CH3  CH2  C  CH2  CH3
OH
Cyclohexylmethanol
3-Cyclohexylpentan-3-ol
OH
Cl
iii.
iv.
HO  CH2  CH2  CH  CH2
CH2  CH3
3-(Chloromethyl)pentan-1-ol
Cyclopent-3-en-1-ol
Q.2. Name the following compounds according to IUPAC system:
CH2OH
CH2OH
i.
CH3  CH2  CH  CH  CH  CH3
CH2Cl
ii.
CH3  CH  CH2  CH – CH – CH3
CH3
CH3
OH
OH
iii.
iv.
H2C = CH – CH – CH2  CH2  CH3
Br
v.
OH
(NCERT)
CH3 – C = C – CH2OH
CH3 Br
Ans:
CH2OH
i.
CH3  CH2  CH  CH  CH  CH3
CH2Cl
CH3  CH  CH2  CH – CH – CH3
CH3
CH3
3-Chloromethyl-2-isopropylpentan-1-ol
iii.
CH2OH
ii.
OH
2,5-Dimethylhexane-1,3-diol
iv.
OH
H2C = CH – CH – CH2  CH2  CH3
OH
Hex-1-en-3-ol
Br
3-Bromocyclohexanol
1
Std. XII Sci.: Perfect Chemistry ‐ II v.
CH3 – C = C – CH2OH
CH3 Br
2-Bromo-3-methylbut-2-en-1-ol
Q.3. Write the IUPAC names of the following compounds:
CH3
i.
CH3 OH
iii.
ii.
CH3  CH  CH  C  CH3
H3C  CH  CH2  CH  CH  CH2  CH3
OH
CH3
CH3  CH  CH  CH3
iv.
HO  CH2  CH  CH2  OH
OH
OH OH
Ans:
C2H5
OH
(NCERT)
CH3
i.
ii.
CH3  CH  CH  C  CH3
CH3 OH
OH
CH3
2,2,4-Trimethylpentan-3-ol
iii.
OH
C2H5
5-Ethylheptane-2,4-diol
iv.
CH3  CH  CH  CH3
OH
H3C  CH  CH2  CH  CH  CH2  CH3
HO  CH2  CH  CH2  OH
OH
OH
Propane-1,2,3-triol
Butane-2,3-diol
Q.4. Draw the structures of all isomeric alcohols of molecular formula C5H12O, give their IUPAC names
and classify them as primary, secondary and tertiary alcohols.
(NCERT)
Ans: Eight isomers are possible with molecular formula C5H12O. These are:
OH
i.
CH3  CH2  CH2  CH2  CH2  OH
ii.
Pentan-1-ol (1 alcohol)
CH3  CH2  CH2  CH  CH3
Pentan-2-ol (2 alcohol)
CH3
OH
iii.
iv.
CH3  CH2  CH  CH2  CH3
Pentan-3-ol (2 alcohol)
2-Methylbutan-1-ol (1 alcohol)
CH3
CH3
v.
CH3  CH2  CH  CH2 – OH
CH3  CH  CH2 – CH2  OH
vi.
3-Methylbutan-1-ol (1 alcohol)
CH3  C  CH2 – CH3
OH
2-Methylbutan-2-ol (3 alcohol)
CH3
vii.
CH3 OH
CH3  C  CH2 – OH
viii. CH3  CH  CH  CH3
3-Methylbutan-2-ol (2 alcohol)
CH3
2,2-Dimethylpropan-1-ol (1 alcohol)
2
Chapter 11: Alcohols, Phenols and Ethers
Q.5. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on
methanal?
i. CH3  CH  CH2OH
ii.
(NCERT)
CH2OH
CH3
In the preparation of isobutyl alcohol (CH3  CH – CH2OH) from methanal, CH2  OH is the part which
Ans: i.
CH3
comes from methanal and the remaining part comes from the Grignard reagent.
O
Dry
H  C  H + CH3  CH  Mg  Br 

ether
CH3
Formaldehyde
(Methanal)
ii.
H
H
C
O  Mg  Br
CH  CH3
HOH
CH3  CH  CH2  OH+ Mg(OH)Br

 
H
CH3
CH3
Isopropyl magnesium
bromide
Isobutyl alcohol
(2-Methylpropan-1-ol)
Addition compound
In the preparation of cyclohexylmethanol from methanal, CH2  OH is the part which comes from
methanal and the remaining part comes from Grignard reagent.
O
Mg  Br
HCH+
Formaldehyde
(Methanal)


Dry
ether
H
H
Cyclohexyl
magnesium
bromide
C
O  Mg  Br
CH2OH


HOH
H
+ Mg(OH)Br
Cyclohexyl
methanol
Addition compound
Q.6. Starting from appropriate alkyl halide and aldehyde/ketone, suggest synthesis of following alcohols.
Wherever possible suggest more than one combinations.
CH3
|
i.
CH3  CH  CH2  OH
ii.
CH3  C CH2  OH
|
|
CH3
CH3
iii.
Ans: i.
CH3  CH  OH
|
CH2  CH2  CH3
iv.
C2H5
|
C2H5  C  CH3
|
OH
Isobutyl alcohol can be prepared as follows:
a.
Isobutyl bromide, on heating with dilute KOH, gives isobutyl alcohol.
Boil
CH3  CH  CH2  Br + KOH 
 CH3  CH  CH2  OH + KBr
|
(aq)
|
CH3
CH3
Isobutyl bromide
Isobutyl alcohol
OR
CH3  CH  CH2  Br + AgOH  CH3  CH  CH2  OH + AgBr
moist silver oxide

CH3
CH3
Isobutyl bromide
Isobutyl alcohol
3
Std. XII Sci.: Perfect Chemistry ‐ II b.
Isobutyl alcohol is prepared by hydrogenation of isobutyraldehyde using nickel as a catalyst at
413 K.
Raney.Ni
CH3  CH  CHO + H2 
 CH3  CH  CH2OH
413 K
CH3
CH3
Isobutyraldehyde
c.
Isobutyl alcohol
Isobutyl alcohol is prepared by reduction of isobutyraldehyde using sodium amalgam in water.
Na  Hg /H 2 O
CH3  CH  CHO + 2[H] 
 CH3  CH  CH2OH
|
|
CH3
CH3
Isobutyraldehyde
ii.
Isobutyl alcohol
Neopentyl alcohol can be prepared as follows:
a.
Neopentyl chloride on heating with dilute KOH gives neopentyl alcohol.
CH3
CH3
|
|

CH3  C  CH2  Cl + KOH(aq.) 
 CH3  C  CH2  OH + KCl
|
|
CH3
CH3
Neopentyl chloride
Neopentyl alcohol
OR
CH3
CH3
|
|
moist silver oxide
CH3  C  CH2  Cl + AgOH 
CH

C
 CH2  OH + AgCl
3

|
|
CH3
CH3
Neopentyl chloride
b.
Neopentyl alcohol
Neopentyl alcohol is prepared by hydrogenation of 2,2-dimethylpropanal using nickel as a
catalyst at 413 K.
CH3
CH3
CH3  C  CHO
+
H2
Raney Ni


413 K
CH3
CH3
2,2-Dimethylpropanal
c.
CH3  C  CH2OH
Neopentyl alcohol
(2,2-Dimethylpropan-1-ol)
Neopentyl alcohol is prepared by reduction of 2,2-dimethylpropanal using sodium amalgam in water.
CH3
CH3
Na  Hg/ H 2 O
 CH3  C  CH2OH
CH3  C  CHO + 2[H] 
CH3
CH3
2,2-Dimethylpropanal
iii.
Neopentyl alcohol
(2,2-Dimethylpropan-1-ol)
Pentan-2-ol can be prepared as follows:
a.
2-Bromopentane on heating with aqueous potassium hydroxide gives pentan-2-ol.

CH3  CH  CH2  CH2  CH3 + KOH(aq.) 
 CH3  CH  CH2  CH2  CH3 + KBr
Br
OH
2-Bromopentane
4
Pentan-2-ol
Chapter 11: Alcohols, Phenols and Ethers
OR
moist silver oxide
CH3  CH  CH2  CH2  CH3 + AgOH 
CH3  CH  CH2  CH2  CH3 + AgBr

OH
Br
2-Bromopentane
Pentan-2-ol
O
b.
Raney Ni
CH3  C – CH2 – CH2 – CH3 + H2 
 CH3 – CH – CH2 – CH2 – CH3
413 K
OH
Pentan-2-one
Pentan-2-ol
OR
O
Na-Hg / H 2 O
CH3  C – CH2 – CH2 – CH3 + 2[H] 

 CH3 – CH – CH2 – CH2 – CH3
OH
Pentan-2-one
c.
O
||
CH3 – C – H
+
Pentan-2-ol
dry ether
CH3 – CH2 – CH2 – Mg – Br 


n-Propyl magnesium
bromide
Ethanal
O – Mg – Br
|
CH3 – C – H
|
CH2 – CH2 – CH3
OH
|
CH3 – C – H
+
|
CH2 – CH2 – CH3
Complex
Mg(OH)Br
HOH
H+
Pentan-2-ol
OR
O
O – Mg – I
||
|
dry ether
CH3 – CH2 – CH2 – C – H + CH3 – Mg – I 

 CH3 – CH2 – CH2 – C – H
|
Methyl magnesium
Butanal
CH3
iodide
Complex
OH
|
CH3 – CH2 – CH2 – C – H +
|
CH3
iv.
HOH
H+
Pentan-2-ol
3-Methylpentan-3-ol can be prepared as follows:
a.
3-Bromo-3-methylpentane on heating with aqueous KOH gives 3-Methylpentan-3-ol.
C2H5
|
C2H5  C  CH3
|
Br
+
KOH(aq.)
3-Bromo-3-methylpentane
Mg(OH)I
C2H5
|


 C2H5  C  CH3
|
OH
+ KBr
3-Methylpentan-3-ol
5
Std. XII Sci.: Perfect Chemistry ‐ II OR
C2H5
|
C2H5  C  CH3
+
AgOH
C2H5
|
C2H5  C  CH3


moist silver oxide
OH
Br
3-Bromo-3-methylpentane
b.
+ AgBr
3-Methylpentan-3-ol
Diethyl ketone, on treatment with methyl magnesium iodide, forms complex, which on acid
hydrolysis gives 3-Methylpentan-3-ol.
OMgI
O
||
dry ether
C2H5  C  C2H5 + CH3MgI 

 C2H5  C  C2H5
Methyl
magnesium
iodide
Diethyl ketone
OH
HOH

 C2H5  C  C2H5 + Mg(OH)I
H
CH3
CH3
3-Methylpentan-3-ol
Complex
OR
O
O  Mg – Br
dryether
 C2H5  C  CH3
C2H5  C  CH3 + C2H5  Mg  Br 
Butan-2-one
Ethyl
magnesium
bromide
C2H5
Complex
OH
C2H5  C  CH3 + Mg(OH)Br
HOH
H+
C2H5
3-Methylpentan-3-ol
Q.7. Give structures of the products you would expect when each of the following alcohol reacts with:
i.
HCl – ZnCl2
ii.
HBr
iii. SOCl2.
a.
Butan-1-ol.
b.
2-Methylbutan-2-ol.
(NCERT)
Ans: i.
HCl – ZnCl2:
a.
Butan-1-ol being a primary alcohol does not react with Lucas reagent (HCl – ZnCl2) at room
temperature. However, cloudiness appears only upon heating.
concHCl

CH3CH2CH2CH2OH 
anhydrous ZnCl2 , Heat
Butan-1-ol
b.
1-Chlorobutane
But 2-methylbutan-2-ol (3) reacts at room temperature giving cloudiness immediately.
CH3
CH3
|
|
concHCl
CH3 – C – CH2 – CH3 
 CH3 – C – CH2 – CH3
Room temperature
|
|
OH
Cl
2-Methylbutan-2-ol
ii.
2-Chloro-2-methylbutane
(Cloudiness)
HBr:
Both alcohols react with HBr to give corresponding alkyl bromides.

a.
CH3CH2CH2CH2OH + HBr

 CH3CH2CH2CH2Br
Butan-1-ol
6
CH3CH2CH2CH2Cl
1-Bromobutane
Chapter 11: Alcohols, Phenols and Ethers
CH3
|
CH3 – C – CH2 – CH3
|
OH
b.
+
CH3
|
CH3 – C – CH2 – CH3
|
Br



HBr
2-Methylbutan-2-ol
iii.
2-Bromo-2-methylbutane
SOCl2:
Both alcohols react to give corresponding alkyl chlorides.
Pyridine
a.
CH3CH2CH2CH2OH + SOCl2 
CH3CH2CH2CH2Cl + SO2 + HCl
reflux
Butan-1-ol
1-Chlorobutane
CH3
|
CH3 – C – CH2 – CH3 + SOCl2
|
OH
b.
Pyridine

reflux
2-Methylbutan-2-ol
CH3
|
CH3 – C – CH2 – CH3 + SO2 + HCl
|
Cl
2-Chloro-2-methylbutane
Q.8. Write IUPAC names of the following compounds:
i.
CH3
OH
ii.
CH3
iii.
CH3
CH3
iv.
OH
CH3
OH
v.
OH
OH
CH3
C2H5
(NCERT)
C2H5
Ans: i.
iv.
2-Methylphenol
2,6-Dimethylphenol
ii.
v.
4-Methylphenol
2,3-Diethylphenol
iii.
2,5-Dimethylphenol
Q.9. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
(NCERT)
Ans: Three isomers of monohydric phenols with molecular formula C7H8O are possible. They are as follows:
OH
OH
OH
CH3
CH3
2-Methylphenol
CH3
3-Methylphenol
4-Methylphenol
Q.10. Write chemical reaction for the preparation of phenol from chlorobenzene.


Ans:
OH
Cl
ONa
dil.HCl,  NaCl


or H 2 O  CO2 ,

+ 2NaOH 
 NaCl,  H 2 O
613K,300atm
Chlorobenzene
(NCERT)
 NaHCO3
Sodium
phenoxide
Phenol
7
Std. XII Sci.: Perfect Chemistry ‐ II Q.11. Give the equations of reactions for the preparation of phenol from cumene.
Ans:
CH3
CH3
H3C  C  O  O H
H3C  C  H
+
Cumene or
(Isopropyl benzene
or 2-Phenylpropane)
O2
(Air)


Cobalt naphthenate
423K
(alkaline medium)
(NCERT)
OH


dil.H 2SO 4

Cumene hydroperoxide
O
+ CH3  C  CH3
Phenol
Acetone
Q.12. Suppose you are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of
phenol using these reagents.
(NCERT)
conc.H 2SO 4 / 
H 2SO 4
fused NaOH



Ans: C6H6 

C6H5SO3H
C6H5ONa

 C6H5OH
 Na 2SO3
(Sulphonation )
Hydrolysis
Benzene
Benzene sulphonic acid
Sodium phenoxide
Q.13. Give the IUPAC names of the following ethers:
i.
C2H5OCH2  CH  CH3
ii.
|
CH3
iii.
O2N  C6H4  OCH3 (p)
iv.
v.
H3C
vi.
CH3
Phenol
CH3  O  CH2CH2Cl
CH3  CH2  CH2  OCH3
OC2H5
OC2H5
vii.
CH3  O  CH2  CH  CH3
|
CH3
ix.
CH3  CH2  O  CH  CH2 CH3
|
CH3
1-Ethoxy-2-methylpropane
4-Nitroanisole
4-Ethoxy-1,1-dimethylcyclohexane
1-Methoxy-2-methylpropane
2-Ethoxybutane
Ans: i.
iii.
v.
vii.
ix.
viii. C6H5  O  C7H15 (n )
(NCERT)
ii.
iv.
vi.
viii.
2-Chloro-1-methoxyethane
1-Methoxypropane
Ethoxybenzene
1-Phenoxyheptane
Q.14. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and
3-methylpentan-2-ol.
(NCERT)
HBr
Ans:
CH3CH2OH 
 CH3CH2Br

Ethanol
Bromoethane
Na
2CH3CH2  CH  CH  OH 
 2CH3CH2  CH  CH  O Na+ + H2
|
|
|
|
CH3 CH3
CH3 CH3
3-Methylpentan-2-ol
8
Sodium 3-methylpentan-2-oxide
Chapter 11: Alcohols, Phenols and Ethers
CH3CH2  CH  CH  O Na+ + CH3CH2Br 
 CH3CH2  CH  CH  OCH2CH3 + NaBr

|
|
|
|
CH3 CH3
CH3 CH3
Sodium 3-methylpentan-2-oxide
2-Ethoxy-3-methylpentane
Bromoethane
Q.15. How is 1-propoxypropane synthesized from propan-1-ol? Write mechanism of this reaction.
(NCERT)
P,Br2
Ans: CH3CH2CH2OH 
 CH3CH2CH2Br + H3PO3
Propan-1-ol
1-Bromopropane
1
H 2
2
Na
CH3CH2CH2OH 
 CH3CH2CH2ONa +
Sodium
n-propoxide
Propan-1-ol

CH3CH2CH2ONa+
Sodium
n-propoxide
Dry ether
+ CH3CH2CH2  Br 
 CH3CH2CH2OCH2CH2CH3 + NaBr
1-Propoxypropane
It can also be prepared by dehydration of propan-1-ol with conc. H2SO4 at 413K.

H
CH3CH2CH2OH 
 CH3CH2CH2OCH2CH2CH3
413K
Propan-1-ol
1-Propoxypropane
Q.16.Write the names of reagents and equations for the preparation of the following ethers by
Williamson’s synthesis:
i.
1-Propoxypropane
ii.
Ethoxybenzene
iii. 2-Methoxy-2-methylpropane
iv. 1-Methoxyethane.
(NCERT)
1-Propoxypropane:
Ans: i.
Heat
CH3CH2CH2ONa+ + CH3CH2CH2  Br 
 CH3CH2CH2  O  CH2CH2CH3 + NaBr
Sodium propoxide
ii.
1-Bromopropane
1-Propoxypropane
Ethoxybenzene:
O Na+
OCH2CH3
+
Sodium phenoxide
iii.
Heat
CH3CH2  Br 

Bromoethane
Ethoxybenzene
CH3
|
Heat
+ CH3  Br 
 CH3  C  OCH3 + NaBr
Bromomethane
|
CH3
Sodium 2-methyl-2-propoxide
2-Methoxy-2-methylpropane
1-Methoxyethane:
CH3CH2ONa+
Sodium ethoxide
NaBr
2-Methoxy-2-methylpropane:
CH3
|
CH3  C  O Na+
|
CH3
iv.
+
Heat
+ CH3  Br 
 CH3CH2  O  CH3 + NaBr
Bromomethane
1-Methoxyethane
9
Std. XII Sci.: Perfect Chemistry ‐ II Q.17. Predict the product of the following reactions:
i.
CH3  CH2  CH2  O  CH3 + HBr 
OC2H5
ii.
+ HBr 
OC2H5
conc.H 2 SO 4


conc.HNO 3
iii.
iv.
Ans: i.
HI
(CH3)3C  OC2H5 

(NCERT)
Both the alkyl groups attached to the oxygen atom are primary, therefore attack of Br – ion occurs on
the smaller methyl group leading to the formation of Propan-1-ol and Bromomethane.
373K
CH3  CH2  CH2  O  CH3 + HBr 
 CH3  CH2  CH2  OH +
n-Propyl methyl ether
ii.
Propan-1-ol
Due to resonance, C6H5-O bond has some double bond character and hence, is stronger than OC2H5
bond. Therefore, the cleavage of the weaker OC2H5 bond occurs to yield Phenol and Bromoethane.
OC2H5
OH
373K
+ HBr 

+ CH3CH2  Br
Ethoxybenzene
iii.
Phenol
Bromoethane
Due to the strong +R effect of the OC2H5 group, it is activating as well as o, p-directing . Therefore
nitration of ethoxybenzene will give a mixture of 2-Ethoxynitrobenzene. and 4-Ethoxynitrobenzene.
2 OC2H5
4 OC2H5
OC2H5
conc.H 2SO 4


 conc.HNO3
Ethoxybenzene
iv.
CH3  Br
Bromomethane
+
1
O2N
1
4-Ethoxynitrobenzene
(Major product)
NO2
2-Ethoxynitrobenzene
(Minor product)
Since tert-Butyl carbocation is much more stable than ethyl carbocation, reaction follows S N1
mechanism leading to the formation of tert-butyl iodide and ethanol as shown below:
HI  H+ + I
Step I:
Formation of carbocation:
CH3
CH3
CH3
+

S
H
N
CH3  C  O  CH2CH3 
CH3  C  O  CH2CH3 
 CH3  C+ + CH3CH2OH
Slow
CH3
CH3 H
tert-Butyl ethyl ether
+
CH3
|
CH3  C +
|
CH3
Fast


10
CH3
|
CH3  C  I
|
CH3
tert-Butyl iodide
tert-Butyl
carbocation
CH3
tert-Butyl
carbocation
Step II:
Formation of tert-Butyl iodide:
I
1
Ethanol
12 Aldehydes, Ketones and Carboxylic Acids Chapter 12: Aldehydes, Ketones and
Carboxylic Acids Q.1. Write the structures of products of the following reactions:
O
C
+ C2H5
i.
Anhyd AlCl
3
Cl 

CS 2
ii.
(C6H5CH2)2Cd + 2CH3COCl 
CH3
iii.
2
Hg , H 2SO4

H3C  C  C  H 
iv.
(NCERT)
i. CrO 2Cl 2

 
ii.H 3O
NO2
Ans: i.
Benzene reacts with C2H5COCl, to yield propiophenone (Friedel Craft’s acylation).
O
O
C
C
Anhyd AlCl3
Cl 

CS2
+ H5C2
Benzene
ii.
Propanoyl
chloride
Propiophenone
Dibenzyl cadmium reacts with acetyl chloride to give Benzyl methyl ketone.
O
O
2CH3  C  Cl
Acetyl chloride
iii.
C2H5 + HCl
+ (C6H5CH2)2Cd  2CH3 C  CH2C6H5 + CdCl2
Benzyl methyl ketone
Dibenzyl cadmium
Propyne, in presence of hot dil. H2SO4 and HgSO4, forms acetone.
O
dil.H 2SO 4  HgSO 4
Tautomerism
H3C  C  CH + H2O 
H3C  C = CH2 

 CH3  C  CH3
333K
Propyne
OH
Acetone
Prop-1-en-2-ol
iv.
Oxidation of methyl group by chromyl chloride gives a chromium complex. Acid hydrolysis of the
chromium complex gives corresponding aldehyde.
CH3
CH(OCrOHCl2)2

CrO 2 Cl2


CS2
NO2
4-Nitrotoluene
CHO
H 3O


NO2
NO2
Chromium complex
4-Nitrobenzaldehyde
1
Std. XII Sci.: Perfect Chemistry ‐ II Q.2. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition
reactions:
i.
Ethanal, Propanal, Propanone, Butanone
ii.
Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
(Hint: Consider steric effect and electronic effect.)
(NCERT)
Ans: i.
Ethanal, Propanal, Propanone, Butanone:
a.
Due to inductive (+I) effect aldehydes have more electrophilic carbonyl carbon than ketones.
b.
Hence aldehydes are more susceptible (to the attack of nucleophile) than ketones.
c.
Hence the reactivity of propanal and ethanal is higher than that of butanone and propanone.
d.
As the steric hindrance increases, reactivity decreases because the attack of the nucleophile to
the carbonyl carbon becomes more difficult.
e.
Hence the reactivity of propanal is lower than that of ethanal. Also the reactivity of butanone is
lesser than that of propanone.
f.
The increasing order of reactivity in nucleophilic addition reactions is,
Butanone < Propanone < Propanal < Ethanal.
ii.
Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone:
a.
The effect of methyl group at para position of aromatic aldehydes is not significant. Hence the
reactivity of benzaldehyde and p-tolualdehyde is comparable.
b.
The nitro group at para position is an electron withdrawing group and makes the carbonyl
carbon more electrophilic. Hence p-nitrobenzaldehyde is more reactive than benzaldehyde.
c.
Aldehydes have more electrophilic carbonyl carbon than ketones. Hence aldehydes are more
subceptible (to the attack of nucleophile) than ketones.
d.
hence benzaldehyde is more reactive than acetophenone.
e.
The increasing order of reactivity in nucleophilic addition reactions is,
Acetophenone < p-Tolualdehyde , Benzaldehyde < p-Nitrobenzaldehyde.
Q.3. Predict the products of the following reactions:
i.
O
H
+ HO  NH2 
O2N
O
+ H2N  NH
ii.
NO2 

O
iii.
iv.
Ans: i.

H
R  CH = CH  CHO + H2N  C  NH  NH2 
O
C

H
CH3 + CH3CH2NH2 
(NCERT)
Cyclopentanone reacts with hydroxyl amine to form oxime.
O
N  OH

H
+ HO  NH2 
Cyclopentanone
ii.
2
Cyclopentanone
oxime
Cyclohexanone reacts with 2,4-dinitro phenyl hydrazine to form 2,4-dinitro phenyl hydrazone.
O2N
O2N
O
NO2
N  NH
+ H2N  NH
NO2 

+ H2O
Cyclohexanone
Hydroxyl
amine
+ H2O
2,4-Dinitro phenyl hydrazine
2,4-Dinitro phenyl hydrazone
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids iii.
,  unsaturated aldehyde reacts with semicarbazide (H2NCONHNH2) to form semicarbazone.
O
O
H
R  CH = CH  CHO + H2N  C  NH  NH2  R  CH = CH  CH = N  C  NH  NH2 + H2O
, Unsaturated aldehyde
iv.
Semicarbazide
Semicarbazone
Acetophenone reacts with ethyl amine to form an imine
N  CH2  CH3
O
C
C

H
CH3 + CH3CH2NH2 
Acetophenone
Ethylamine
CH3
Substituted imine
Q.4. Draw structures of the following derivatives:
i.
The 2,4-dinitrophenylhydrazone of benzaldehyde
ii.
Cyclopropanone oxime
iii.
Acetaldehydedimethylacetal
iv.
The semicarbazone of cyclobutanone
v.
The ethylene ketal of hexan-3-one
vi.
The methyl hemiacetal of formaldehyde
Ans: i.
2,4-dinitrophenylhydrazone of benzaldehyde
CH = N  NH
iii.
ii.
Cyclopropanone oxime
N  OH
NO2
NO2
Acetaldehydedimethylacetal
iv.
The semicarbazone of cyclobutanone
N  NH  C  NH2
O CH3
CH3  CH
v.
(NCERT)
O
O CH3
The ethylene ketal of hexan-3-one
vi.
H3C  CH2  C  CH2  CH2  CH3
O
O
H2C
The methyl hemiacetal of formaldehyde
H
HO  C  O  CH3
CH2
H
Q.5. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents:
ii.
Tollen’s reagent
i.
PhMgBr and then H3O+
iii. Semicarbazide and weak acid
iv. Excess ethanol and acid
v.
Zinc amalgam and concentrated hydrochloric acid
(NCERT)
Ans: i.
The reaction of cyclohexanecarbaldehyde with PhMgBr followed by acid hydrolysis gives an alcohol.
OH
CHO
1. PhMgBr

 
2. H3O
Cyclohexane
Carbaldehyde
C
Ph
H
1-Cyclohexyl-1-phenylmethanol
3
Std. XII Sci.: Perfect Chemistry ‐ II ii.
Oxidation of cyclohexane carbaldehyde with Tollen’s reagent gives cyclohexane carboxylate ion.
O
CHO
C
O
Tollen's reagent


Cyclohexane
carbaldehyde
iii.
Cyclohexane
carboxylate ion
The reaction of cyclohexane carbaldehyde with semicarbazide and weak acid gives a semicarbazone.
O
CHO
O
CH = NNH  C  NH2
H
+ H2NNH  C  NH2 
Semicarbazide
Cyclohexane
carbaldehyde
iv.
The reaction of cyclohexane carbaldehyde with excess ethanol and acid gives an acetal.
H
CHO
C OC2H5
excess ethanol


acid
OC2H5
Cyclohexane
carbaldehyde
v.
Cyclohexane carbaldehyde
semicarbazone
Cyclohexane carbaldehyde
diethyl acetal
Zinc amalgam and concentrated hydrochloric acid reduces CHO group to CH3 group
(Clemmensen’s reduction).
CHO
CH3
Zinc
amalgam


conc.HCl
1-Methyl
cyclohexane
Cyclohexane
carbaldehyde
Q.6. Give simple chemical tests to distinguish between the following pairs of compounds:
i.
Acetophenone and Benzophenone
ii.
Phenol and Benzoic acid
iii. Benzoic acid and ethyl benzoate
iv. Benzaldehyde and acetophenone
v.
Ethanal and propanal
(NCERT)
Ans: i.
Acetophenone and Benzophenone: Acetophenone being a methyl ketone when treated with NaOI
(I2 / NaOH) gives yellow precipitate of iodoform whereas benzophenone does not.
PhCOCH3 + 3NaOI  PhCOONa + CHI3 + 2NaOH
Sodium
benzoate
Acetophenone
Iodoform
(yellow precipitate)
PhCOPh + 3NaOI  No yellow precipitate
Benzophenone
ii.
Phenol and Benzoic acid: Phenol is a weak acid it does not react with weak base NaHCO3 whereas
benzoic acid is a strong acid. It reacts with NaHCO3 to form a sodium salt alongwith evolution of
CO2.
OH
+
Phenol
4
NaHCO3(aq)  No reaction
Sodium
bicarbonate
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids O
O
C  OH
C  ONa 

NaHCO3(aq) 
+
Sodium
bicarbonate
Benzoic acid
iii.
+ H2O + CO2
Sodium
benzoate
Benzoic acid and ethyl benzoate: Benzoic acid is a carboxylic acid and reacts with NaHCO3 to form
a sodium salt alongwith evolution of CO2. Ethylbenzoate is an ester. It does not react with NaHCO3.
O
O

C  ONa 
C  OH
+
NaHCO3(aq) 
+ H2O + CO2
Sodium
bicarbonate
Benzoic acid
Sodium
benzoate
O
COCH2CH3
+
Sodium
bicarbonate
Ethyl benzoate
iv.
NaHCO3(aq)  No reaction
Benzaldehyde and acetophenone: Benzaldehyde being an aldehyde reduces Tollen’s reagent to
shining silver mirror whereas acetophenone being a ketone does not.
PhCHO
+
Benzaldehyde
PhCOCH3

2[Ag(NH3)2]+ + 3OH 
 PhCOO + 2Ag + 4NH3 + 2H2O
Benzoate
ion
Tollen’s
reagent
Silver
metal
Tollen 's reagent

 No silver mirror
Acetophenone
v.
Ethanal and propanal: Ethanal contains CH3CO group. Hence on treatment with NaOI (I2 / NaOH)
gives yellow precipitate of iodoform whereas propanal does not.
CH3CHO + 3NaOI  HCOONa+ + CHI3 + 2NaOH
Ethanal
Sodium
formate
Iodoform
(yellow
precipitate)
CH3CH2CHO + 3NaOI  No yellow precipitate
Propanal
Q.7. i.
ii.
Ans:
i.
Write structural formulae and names of four possible aldol condensation products from
propanal and butanal.
In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
(NCERT)
CH2CH3
i. dil NaOH
CH3CH2CHO + CH3CH2CH2CHO 
 CH3CH2C = C  CHO

Propanal
(Electrophile)
Butanal
(Nucleophile)
ii.  H 2 O
H
2-Ethylpent-2-enal
CH3
ii.
i. dil.NaOH
CH3CH2CHO + CH3CH2CH2CHO 
 CH3CH2CH2C = C  CHO

Propanal
(Nucleophile)
Butanal
(Electrophile)
ii.  H 2 O
H
2-Methylhex-2-enal
5
Std. XII Sci.: Perfect Chemistry ‐ II CH3
iii.
i. dil NaOH
CH3CH2CHO + CH3CH2CHO 
 CH3CH2C = C  CHO

Propanal
(Electrophile)
Propanal
(Nucleophile)
ii.  H 2 O
H
2-Methylpent-2-enal
CH2CH3
iv.
i. dil NaOH
 CH3CH2CH2C = C  CHO
CH3CH2CH2CHO + CH3CH2CH2CHO 

Butanal
(Electrophile)
Butanal
(Nucleophile)
ii.  H 2 O
H
2-Ethylhex-2-enal
Q.8. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction
and which neither? Write the structures of the expected products of aldol condensation and
Cannizzaro reaction:
i.
Methanal
ii.
2-Methylpentanal
iii. Benzaldehyde
iv. Benzophenone
v.
Cyclohexanone
vi. 1-Phenylpropanone
vii. Phenylacetaldehyde
viii. Butan-1-ol
ix. 2,2-Dimethylbutanal (NCERT)
Ans:
Compound
Reaction
Product
i.
HCHO
Cannizzaro
H3C  OH + HCOONa+
Methanal
Methanol
ii.
CH3
Sodium formate
Aldol condensation
CH3
CH3CH2CH2  C
CH3CH2CH2CHCHO
2-Methylpentanal
HC
CHO
CH3
C
CH2
CH2
CH3
2,4-Dimethyl-2-propylhept-3-enal
iii.
CH2OH
Cannizzaro
CHO
+
Benzyl
alcohol
Benzaldehyde
COONa+
Sodium
benzoate
iv.
PhCOPh

Neither
Benzophenone
v.
Aldol condensation
O
Cyclohexanone
6
O
vi.
Ph  CH2COCH3
Aldol condensation
1-Phenyl propanone
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids CH3
CO CH3
Ph  C = C  CH2  Ph
4-Methyl-3,5-diphenylpent-3-en-2-one
vii.
Ph  CH2CHO
Aldol condensation
Phenyl acetaldehyde
CHO
PhCH2C = C  Ph
2,4-Diphenylbut-2-enal
viii. CH3CH2CH2CH2OH

Neither
Butan-1-ol
CH3
ix.
Cannizzaro reaction
CH3
CH3CH2C  CH2OH
CH3  CH2  C  CHO
CH3
CH3
2,2-Dimethylbutanol
2,2-Dimethylbutanal
+
CH3
CH3CH2C  COOH
CH3
2,2-Dimethylbutanoic acid
Q.9. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative reduces Tollen’s
reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic
acid. Identify the compound.
(NCERT)
Ans: i.
The organic compound (A) is 2-Ethylbenzaldehyde
CHO
CH2CH3
ii.
iii.
iv.
Its molecular formula is C9H10O.
Since it is an aldehyde, it can form 2,4-DNP derivative and can reduce Tollens’ reagent. It can also
undergo Cannizzaro reaction, as it does not contain H atom.
On vigorous oxidation, both CHO group and the ethyl side chain are oxidized to COOH groups to
give 1,2-benzene dicarboxylic acid HOOC  C6H4  COOH.
Q.10. What is meant by the following terms? Give an example of the reaction in each case.
i.
Cyanohydrin
ii.
Acetal
iii. Semicarbazone
iv. Aldol
v.
Hemiacetal
vi. Oxime
vii. Ketal
viii. Imine
ix. 2,4-DNP-derivative
x.
Schiff’s base
(NCERT)
Ans: i.
Cyanohydrin: It is a compound in which cyano and hydroxyl groups are present on same carbon
atom.
OH
HCN
CH3CHO 
 CH3  C  CN
Ethanal
H
Acetaldehyde
cyanohydrin
7
Std. XII Sci.: Perfect Chemistry ‐ II ii.
Acetal: It contains two alkoxy groups, one alkyl group and one H atom on the same carbon atom.
CH2 CH2
CH3CHO +
Ethanal
CH2  OH
CH2  OH
O
O
H 3O 

C
H3C
H
Ethylene glycol
iii.
Acetal
Semicarbazone: It is the condensation product of an aldehyde or ketone with semicarbazide.
O
NNHCNH2
CH3COCH3 + H2NNHCONH2
Acetone

Semicarbazide
C
+ H2O
CH3
H3C
Acetone semicarbazone
iv.
Aldol: It is  hydroxy aldehyde or ketone obtained by the condensation of two aldehyde or ketone
molecules in presence of a dilute alkali.
OH
dil. NaOH
CH3CHO + CH3CHO 
 CH3  CH  CH2  CHO
Ethanal
v.
3-Hydroxy butanal (aldol)
Ethanal
Hemiacetal: It contains one hydroxyl and one alkoxy group on the same carbon atom.
OH
CH3CHO
Acetaldehyde
Dry HCl gas
+ CH3OH 
CH3  C  OCH3
Methanol
H
Hemiacetal
vi.
Oxime: It is the condensation product of an aldehyde or ketone with hydroxyl amine.
H
CH3CHO + H2NOH  CH3  C = NOH
Ethanal
vii.
Hydroxyl
amine
Oxime
Ketal: It contains two alkoxy groups and two alkyl groups on the same carbon atom.
CH2 CH2
CH3COCH3 +
Acetone
CH2  OH
CH2  OH


C
H3C
Ethylene glycol
viii. Imine: It contains
O
O
H 2O
CH3
Ketal
C = N  group.
It is the condensation product of aldehydes and ketones with ammonia derivatives.
NH
CH3COCH3 + NH3  CH3  C  CH3
Acetone
8
Ammonia
Imine
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids ix.
2,4-DNP derivative:
Also known as 2,4-dinitro phenyl hydrazone, it is condensation product of an aldehyde or ketone with
2,4-dinitro phenyl hydrazine (2,4-DNP).
CH3
NH2
N=C
CH3
NH
NH
NO2
NO2
 H 3O
CH3COCH3 +


Acetone
NO2
NO2
2,4-dinitrophenyl
hydrazine
x.
2,4-DNP derivative
Schiff’s base: It is azomethine obtained by the condensation of aldehydes and ketones with primary
amines.
H
CH3CHO + C2H5NH2 
CH3CH = NCH2CH3 + H2O

Ethanal
Ethanamine
Schiff’s base
Q.11. Give the IUPAC names of the following compounds:
i.
PhCH2CH2COOH
ii.
(CH3)2C = CHCOOH
iii.
iv.
NO
CH
2
3
COOH
COOH
Ans: i.
PhCH2CH2COOH
ii.
NO2
O2N
(CH3)2C = CHCOOH
3-Phenylpropanoic acid
iii.
CH3
(NCERT)
3-Methylbut-2-enoic acid
NO2
iv.
COOH
COOH
NO2
O2N
2-Methylcyclopentane
carboxylic acid
2,4,6-Trinitrobenzoic acid
Q.12. Show how each of the following compounds can be converted to benzoic acid.
i.
Ethyl benzene
ii.
Acetophenone
iii. Bromobenzene
iv. Phenylethene (Styrene)
(NCERT)
Oxidation of ethyl benzene with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid.
Ans: i.
CH2CH3

KMnO 4



KOH
Ethyl benzene
ii.
H 3O


Potassium
benzoate
Benzoic acid
Oxidation of acetophenone with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid.
COOH
COOK
COCH3

KMnO 4



KOH
Acetophenone
COOH
COOK
H 3O


Benzoic acid
Potassium
benzoate
9
Std. XII Sci.: Perfect Chemistry ‐ II iii.
Bromobenzene reacts with magnesium to form Grignard reagent which attacks carbon dioxide to
form an intermediate which on acid hydrolysis gives benzoic acid.
COOH
O
MgBr
Br
OMgBr
Mg
OCO
HOH





dil.HCl
Phenyl magnesium
bromide
Bromo benzene
iv.
Benzoic acid
Oxidation of phenyl ethene with alkaline KMnO4 followed by acid hydrolysis gives benzoic acid.
COOK
COOH
CH = CH2

KMnO4



KOH
Phenyl ethene
H 3O


Potassium
benzoate
Benzoic acid
Q.13.An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to
give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C)
on dehydration gives but-1-ene. Write equations for the reactions involved.
(NCERT)
Ans: The organic compound A is butyl butanoate
dil H 2SO4
 CH3CH2CH2COOH + CH3CH2CH2CH2OH
CH3CH2CH2COOCH2CH2CH2CH3 
hydrolysis
Butanoic acid
(B)
Butyl butanoate[C
8H16O2]
(A)
Butanol
(C)
[O] Oxidation
CH3CH2CH2CH2OH 
 CH3CH2CH2COOH
Chromic acid
(H 2 CrO4 )
Butanol
(C)
Butanoic acid
(B)
Dehydration
CH3CH2CH2CH2OH 
CH3CH2CH = CH2
Butanol
(C)
But-1-ene
Q.14.Complete each synthesis by giving missing starting material, reagent or products.
CH2CH3
i.
ii.
COOH
SOCl 2


KMnO4
heat


KOH, heat
COOH
O
iii.
v.
H 2 NCONHNH 2
C6H5CHO 

O

 Ag(NH ) 
3 2



iv.
vi.
C6H5CHO
+
CH3CH2CHO
ix.
xi.
10
dil.NaOH


(i ) O
viii.
NaCN/ HCl


(i ) NaBH 4
CH3COCH2COOC2H5 
 
(ii ) H
CH2 

x.
CrO
3
OH 

3

 2
(ii ) Zn  H 2O
CHO
COOH
CHO
vii.
C


CHO
(NCERT)
O
Ans: i.
CH2CH3
KMnO4


KOH, heat
Ethyl
benzene
COOK
Potassium
benzoate
COOH
ii.
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids COCl
SOCl2


heat
COOH
COCl
Phthalic acid
Phthaloyl chloride
O
iii.
H 2 NCONHNH 2
C6H5CHO 
C6H5CH = NNHC  NH2
Benzaldehyde semicarbazone
Benzaldehyde
O
C
C6 H5COCl


AlCl
iv.
3
Benzene
v.
Benzophenone
O
Ag( NH ) 

3 2



O
OH
COO
CHO
OH
vi.
CHO
C
CN
H
COOH
NaCN/ HCl


COOH
1-Formylbenzoic acid
vii.
Cyanohydrin
C6H5CHO
i.dil.NaOH

 C6H5CH = C  CHO
+
ii.
CH3CH2CHO
CH3
OH
i. NaBH 4
viii. CH3COCH2COOC2H5 
 CH3CCH2COOC2H5

ii. H
H
ix.
CrO3
OH 

Cyclohexanol
x.
xi.
O
Cyclohexanone
i. BH3
CH2 

ii. H 2 O 2 /OH
iii.PCC
i. O3

2
ii. Zn  H 2 O
CHO
O
11
Std. XII Sci.: Perfect Chemistry ‐ II Q.15.Give plausible explanation for each of the following:
i.
Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
ii.
There are two NH2 groups in semicarbazide. However, only one is involved in the formation of
semicarbazones.
iii. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an
acid catalyst, the water or the ester should be removed as soon as it is formed.
(NCERT)

Ans: i.
In cyclohexanone, the carbonyl carbon is not hindered. Hence, the nucleophile CN ion can easily
attack the carbonyl carbon. However in 2,2,6-trimethyl cyclohexanone, the carbonyl carbon is
sterically hindered due to presence of methyl groups. Hence, the nucleophile CN ion cannot easily
attack the carbonyl carbon.
ii.
Out of two NH2 groups in semicarbazide, one is a part of amide functional group. In this, the lone
pair on nitrogen atom is involved in the resonance with carbonyl group. Hence, this NH2 group
cannot act as a nucleophile. Hence, it is not involved in the formation of semicarbazones.
iii. The formation of esters from a carboxylic acid and alcohol in the presence of an acid catalyst is a
reversible reaction. If water or ester is not removed as soon as it is formed, the ester will hydrolyse to
give starting materials carboxylic acid and alcohol.
Q.16.How will you bring about the following conversions in not more than two steps?
i.
Propanone to Propene
ii.
Benzoic acid to Benzaldehyde
iii. Ethanol to 3-Hydroxybutanal
iv. Benzene to m-Nitroacetophenone
v.
Benzaldehyde to Benzophenone
vi. Bromobenzene to 1-Phenylethanol
vii. Benzaldehyde to 3-Phenylpropan-1-ol
viii. Benzaldehyde to -Hydroxyphenylacetic acid
(NCERT)
Ans: i.
Propanone to Propene:
OH
NaBH 4 ,CH3OH
Conc.H 2SO4
CH3COCH3 
CH3  CH  CH3 
 CH3CH = CH2
reduction
443K
Propanone
ii.
Benzoic acid to Benzaldehyde:
COOH
CHO
COCl
SOCl2

 SO2 ,  HCl
H 2 , Pd / BaSO 4


Benzoyl
chloride
Benzoic acid
iii.
Propene
Propan-2-ol
Benzaldehyde
Ethanol to 3-Hydroxybutanal:
OH
Cu,573K
dil.NaOH
CH3CH2OH 
 CH3CHO 

 CH3  CH  CH2CHO
Aldol
Ethanol
iv.
condensation
Ethanal
Benzene to m-Nitroacetophenone:
COCH3
COCH3
(CH3CO)2O


Anhyd.AlCl3
Benzene
v.
3-Hydroxybutanal
Conc.HNO3


Conc.H 2SO 4
Acetophenone
NO2
m-Nitroacetophenone
Benzaldehyde to Benzophenone:
K 2 Cr2 O7
CaCO3
Distil
C6H5CHO 
 C6H5COOH 
 (C6H5COO)2Ca 
 C6H5COC6H5

Benzaldehyde
12
H
Benzoic acid
Calcium benzoate
Benzophenone
Chapter 12: Aldehydes, Ketones and
Carboxylic Acids vi.
Bromobenzene to 1-Phenylethanol:
Br
MgBr
Mg



dry ether
Bromobenzene
vii.
OH
CH  CH3
i. CH3CHO

 
ii. H3O
Phenyl magnesium
bromide
1-Phenyl ethanol
Benzaldehyde to 3-Phenylpropan-1-ol:
CHO
CH = CHCHO
+
CH3CHO
dil NaOH, 



cross aldol
H 2 , Ni


Catalytic
hydrogenation
condensation
Benzaldehyde
Acetaldehyde
CH2CH2CH2OH
3-Phenyl prop-2-enal
3-Phenylpropan-1-ol
viii. Benzaldehyde to -Hydroxyphenylacetic acid:
CHO
OH
OH
CH  CN
CHCOOH

H , H2O


Hydrolysis
NaCN, HCl


pH 9 10
Benzaldehyde
Benzaldehyde
cyanohydrin
-Hydroxy phenyl
acetic acid
13