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CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 10 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Alcohols, Phenols And Ethers
Quiz 10
Dehydration of 2-butanol yields
Report Question
0%
1-butene
0%
2-butene
0%
2-butyne
0%
both (A) and (B)
Which of the following alcohol does not give a stable compound on dehydration?
Report Question
0%
Ethyl alcohol
0%
Methyl alcohol
0%
n-propyl alcohol
0%
n-butyl alcohol
Explanation
Alcohols undergo dehydration (removal of water) to form an alkene.
To form alkene, we need at least two carbon atoms. But, methanol
(
C
H
3
O
H
)
(
C
H
3
O
H
)
has only one carbon atom. So, it does not give a stable compound on dehydration.
Hence, option
B
B
is correct.
Which of the following would undergo dehydration most readily?
Report Question
0%
1-phenyl-1butanol
0%
2-phenyl-2-butanol
0%
1-phenyl-2-butanol
0%
2-phenyl-1-butanol
Explanation
Dehydration of alcohols involves the formation of a carbocation intermediate.
Higher the stability of carbocation, higher is the ease of dehydration.
The order of stability of carbocation is
T
e
r
t
i
a
r
y
>
S
e
c
o
n
d
a
r
y
>
P
r
i
m
a
r
y
T
e
r
t
i
a
r
y
>
S
e
c
o
n
d
a
r
y
>
P
r
i
m
a
r
y
On dehydration, tertiary alcohol forms tertiary carbocation, secondary alcohol forms secondary alcohol, tertiary alcohol forms tertiary carbocation.
From the figure, we can see that only 2-phenyl-2-butanol is a tertiary phenol. So, it would undergo dehydration most readily.
Hence, option
B
B
is correct.
Which compound is soluble in water
Report Question
0%
C
S
2
C
S
2
0%
C
2
H
5
O
H
C
2
H
5
O
H
0%
C
C
l
4
C
C
l
4
0%
C
H
C
l
3
C
H
C
l
3
Explanation
C
2
H
5
O
H
C
2
H
5
O
H
is soluble in water due to H-bonding.
Which of the following reagents convert the propene to 1-propanol
Report Question
0%
H
2
O
,
H
2
S
O
4
H
2
O
,
H
2
S
O
4
0%
Aqueous KOH
0%
M
h
S
O
4
,
N
a
B
H
4
/
H
2
O
M
h
S
O
4
,
N
a
B
H
4
/
H
2
O
0%
B
2
H
6
,
H
2
O
2
,
O
H
−
B
2
H
6
,
H
2
O
2
,
O
H
−
Explanation
C
H
3
−
C
H
=
C
H
2
+
a
q
.
K
O
H
P
r
o
p
e
n
e
−
1
→
C
H
3
−
C
H
2
−
C
H
2
O
H
P
r
o
p
a
n
o
l
−
1
The order of melting point of ortho, para, meta-nitrophenol is:
Report Question
0%
o
>
m
>
p
0%
p
>
m
>
o
0%
m
>
p
>
o
0%
p
>
o
>
m
Explanation
For
orthonitrophenol
, hydrogen bonding would be intramolecular, i.e. confined to the same molecule. For
meta
and
para nitrophenols
, hydrogen bonding would become intermolecular, a force between molecules.
Thus, o-nitrophenol has higher boiling and melting point than m-nitrophenol and p-nitrophenol.
Among all options,
A
denotes o-nitrophenol has the highest melting point.
Hence, option
A
is correct.
The product
(
B
)
is:
Report Question
0%
0%
0%
0%
All
Explanation
Only the acidic
A
r
O
H
is converted first to its conjugate base then to
M
e
ether.
So, the product formed in the above reaction is
p
−
(
M
e
O
)
−
C
6
H
4
−
C
H
2
O
H
Hence, Option "B" is the correct answer.
The major product of (B) is:
Report Question
0%
0%
0%
0%
Explanation
Complete reaction can be illustrated as shown above.
Hence, Option "A" is the correct answer.
Which compound is soluble in
H
2
O
Report Question
0%
H
C
H
O
0%
C
H
3
C
H
O
0%
C
H
3
C
O
C
H
3
0%
All
Explanation
Reason: H-bonding
Heating ethanol at 443 K with excess concentrated sulphuric acid results in the dehydration of ethanol to give ethane
Report Question
0%
True
0%
False
Explanation
C
2
H
5
O
H
443
K
→
C
2
H
4
+
H
2
O
The keto form of phenol contains
Report Question
0%
3
π
,12
δ
4 non-bonding electrons
0%
3
π
,9
δ
4 non-bonding electrons
0%
3
π
,9
δ
2 non-bonding electrons
0%
3
π
,8
δ
4 non-bonding electrons
Identify
Z
in the following series:
C
2
H
5
O
H
P
B
r
3
→
X
a
l
c
.
K
O
H
→
(
i
)
H
2
S
O
4
→
(
i
i
)
H
2
O
+
heat
Z
Report Question
0%
C
H
2
=
C
H
2
0%
C
H
3
−
C
H
2
O
H
0%
C
H
3
−
C
H
2
−
O
−
C
H
2
−
C
H
3
0%
None
Explanation
Hence, Option "B" is the correct answer.
IUPAC name of the compound is:
C
H
3
−
C
|
H
C
H
2
C
H
3
−
C
H
2
−
C
H
O
H
−
C
H
3
is
Report Question
0%
4
-Methyl-
3
-hexanol
0%
Heptanol
0%
4
-Methyl-
2
-hexanol
0%
None of these
Explanation
Longest carbon chain has 6 atoms. We number from right hand side as alcohol group at C-2 gets least number in this case and suffix used for alcohol is "-ol". Methyl group is present at C-4 position.
Thus, IUPAC name is 4-Methyl-2-hexanol.
Hence , option C is correct .
Which of the following compound has wrong IUPAC name ?
Report Question
0%
C
H
3
−
C
H
2
−
C
H
2
−
C
O
O
−
C
H
2
C
H
2
ethyl butanoate
0%
C
H
3
−
C
|
H
C
H
3
−
C
H
2
−
C
H
O
3
-methylbutanal
0%
C
H
3
−
C
|
H
O
H
−
C
|
H
C
H
3
−
C
H
3
2
-methyl-
3
-butanol
0%
C
H
3
−
C
|
H
C
H
3
−
C
∥
O
−
C
H
2
−
C
H
3
2
-methyl-
3
-pentanone
Explanation
In option C as we see -OH group is attached . As from IUPAC nomenclature it will be named first .
Longest carbon chain is of 4 . hence name will be butanol . (As OH is present )
-OH is present at 2nd Carbon . Hence name will be butan-2-ol .
Now , as we see ,
−
C
H
3
is also attached at 3rd Carbon .
Hence , IUPAC name is 3-methyl-2-butanol .
Hence , option C is correct .
Statement-1: In the fermentation process of molasses, along with yeast,
(
N
H
4
)
2
S
O
4
and
(
N
H
4
)
3
P
O
4
are added.
Statement-2: Both
(
N
H
4
)
3
P
O
4
and
(
N
H
4
)
2
S
O
4
act as food and help in the growth of yeast.
Report Question
0%
Statement-1 is True, statement-2 is True; statement-2 is a correct explanation for statement-1
0%
Statement-1 is True, statement-2 is True; statement-2 is NOT a correct explanation for statement-1
0%
Statement-1 is True, statement-2 is False
0%
Statement-1 is False, statement-2 is True
Explanation
In fermentation process of molasses along with yeast
(
N
H
4
)
2
S
O
4
and
(
N
H
4
)
3
P
O
4
are added. Due to this the element such as
N
,
H
,
P
,
O
increases in processing of yeast.
In yeast fermentation
(
N
H
4
)
3
P
O
4
and
(
N
H
4
)
2
S
O
4
both act as a growth of yeast to fulfill the nutrition and quantity of elements
(
N
,
H
,
P
,
O
)
.
Teritary alcohol is obtained as major product in
Report Question
0%
(
C
H
3
)
2
C
H
−
C
H
=
C
H
2
o
x
y
m
e
r
c
u
t
i
o
n
/
d
e
m
e
r
c
u
t
i
o
n
→
0%
(
C
H
3
)
3
C
−
C
H
=
C
H
2
d
i
l
H
2
S
O
4
→
0%
(
C
H
3
)
2
C
H
−
C
H
=
C
H
2
H
y
d
r
o
b
o
r
a
t
i
o
n
,
O
x
i
d
a
t
i
o
n
→
0%
All of the above
Explanation
Tertiary alcohol is obtained when an alkene reacts with dilute sulphuric acid.
Which reagent can be used to convert ethanol to diethyl ether?
Report Question
0%
A
l
2
O
3
at
250
0
C
0%
conc.
H
2
S
O
4
at
170
0
C
0%
conc.
H
2
S
O
4
at
130
0
−
140
0
C
0%
A
l
2
O
3
at
350
0
C
Explanation
In this case, the ethanol is treated by concentrated sulphuric acid
(
H
2
S
O
4
)
to form diethyl ether. The acid in an aqueous medium produces hydronium
H
3
O
+
ion.
H
+
protonate the oxygen atom of ethanol and the ethanol molecule gets a positively charged oxygen.
After that, the oxygen atom of unprotonated ethanol abstracts a
H
2
O
molecule from ethanol and produces diethyl ether.
C
H
3
C
H
2
O
H
+
H
2
S
O
4
140
o
C
→
C
H
3
C
H
2
O
H
+
2
+
H
2
O
C
H
3
C
H
2
O
H
+
2
+
C
H
3
C
H
2
O
H
→
C
H
3
C
H
2
O
C
H
2
C
H
3
+
H
2
O
+
H
+
So, the correct option is C
Hydrogen chloride and
S
O
2
are the side products in the reaction of ethanol with thionyl chloride. Which of the following is main product in this reaction?
Report Question
0%
C
2
H
5
−
O
−
C
2
H
5
0%
C
2
H
6
0%
C
H
3
C
l
0%
C
2
H
5
C
l
Explanation
Chlorination of ethanol is done by using thionyl chloride. The complete reaction is:
C
2
H
5
O
H
+
S
O
C
l
2
→
C
2
H
5
C
l
+
H
C
l
+
S
O
2
Here, the major product formed is ethyl chloride.
So, the correct option is 'D'.
Which of the following substance can be used as a raw material for obtaining alcohol?
Report Question
0%
Potatoes
0%
Molasses
0%
Maize
0%
All of the above
Explanation
All of them contain starch in certain quantities which can be fermented to give alcohol.
Option D is correct.
Which of the following catalyst is used in the conversion of water gas and hydrogen into methyl alcohol?
Report Question
0%
M
n
O
0%
Raney
N
i
0%
F
e
0%
Z
n
O
−
C
r
2
O
3
Explanation
Water gas is an equimolar mixture of
C
O
and
H
2
.
A mixture of zinc oxide and chromium oxide
(
Z
n
O
−
C
r
2
O
3
)
converts water gas into methanol
(
C
H
3
O
H
)
.
The temperature of the reaction is 573 K.
C
O
+
H
2
Z
n
O
−
C
r
2
O
3
→
C
H
3
O
H
Option D is correct.
Oxymercuration-demercuration reaction
(
H
g
(
O
A
c
)
2
/
T
H
F
−
H
2
O
)
followed by
(
N
a
B
H
4
,
C
2
H
5
O
H
)
on
C
H
3
C
H
=
C
H
2
produces
:
Report Question
0%
C
H
3
C
H
2
C
H
2
C
O
0%
C
H
3
C
H
(
O
H
)
C
H
3
0%
C
H
3
C
H
(
O
H
)
C
H
2
O
H
0%
C
H
3
C
O
C
H
3
Explanation
In oxymercuration - demercuration, more stable carbocation is formed and water attacks on more substituted carbon to form alcohol.
So, the correct option is 'B'.
Hydroboration-oxidation of
C
H
3
C
H
=
C
H
2
produces:
Report Question
0%
C
H
3
C
H
2
C
H
2
O
H
0%
C
H
3
C
H
(
O
H
)
C
H
3
0%
C
H
3
C
H
(
O
H
)
C
H
2
O
H
0%
C
H
3
C
O
C
H
3
Explanation
A Hydroboration-oxidation reaction is a two-step organic reaction that converts an alkene into neutral alcohol by the net addition of water across the double bond. It follows Anti-Markownikoff's rule.
An industrial method of preparation of methanol is:
Report Question
0%
catalytic reduction of carbon monoxide in presence of
Z
n
O
−
C
r
2
O
3
0%
by reacting methane with steam at
900
0
C
with a nickel catalyst
0%
by reducing formaldehyde with lithium aluminium hydride
0%
by reacting formaldehyde with aqueous sodium hydroxide solution
Explanation
The industrial method of preparation of methanol includes catalytic reduction of carbon monoxide. For this purpose, water gas is used.
Water gas is an equimolar mixture of
C
O
and
H
2
.
A mixture of zinc oxide and chromium oxide
(
Z
n
O
−
C
r
2
O
3
)
converts water gas in methanol
(
C
H
3
O
H
)
.
The temperature of the reaction is 573 K.
C
O
+
2
H
2
Z
n
O
−
C
r
2
O
3
→
C
H
3
O
H
Option A is correct.
In which one of the following reagents
O
−
H
bond fission in alcohol can not take place?
Report Question
0%
Na
0%
C
H
3
C
O
O
H
0%
C
H
3
M
g
B
r
0%
P
C
l
5
Explanation
Only
P
C
l
5
gives a substitution reaction with alcohol replacing the OH group with Cl atom.
Which of the following reactions are correctly represented?
Report Question
0%
0%
0%
0%
Explanation
The negative part of the unsymmetrical reagent is added to the carbon atom bearing less number of hydrogen atoms.
Thus the addition occurs according to the Markownikoff's rule.
The correct reactions are the reactions of options A and D.
Which of the following pairs of reagents will not form ether?
Report Question
0%
C
2
H
5
B
r
+
C
2
H
5
O
N
a
0%
C
2
H
5
B
r
+
C
H
3
O
N
a
0%
C
H
3
B
r
+
C
2
H
5
O
N
a
0%
C
2
H
5
B
r
+
H
C
O
O
N
a
Explanation
If the reaction in D takes place, it will give esters and not ethers.
H
C
O
O
N
a
+
C
2
H
5
B
r
→
H
C
O
O
C
2
H
5
+
N
a
B
r
Hence, D is the answer.
Compounds formed from P and Q are, respectively:
Report Question
0%
Optically active S and optically active pair (T, U)
0%
Optically inactive S and optically inactive pair (T, U)
0%
Optically active pair (T, U) and optically active S
0%
Optically inactive pair (T, U) and optically inactive S
Explanation
Therefore, the answer is B.
(
C
O
+
H
2
)
+
H
2
673
k
,
300
a
t
m
,
/
C
r
2
O
3
−
Z
n
O
→
The above shown reaction may by used for manufacture of:
Report Question
0%
H
C
H
O
0%
C
H
3
C
O
O
H
0%
H
C
O
O
H
0%
C
H
3
O
H
Explanation
Methanol is prepared on an industrial scale from a mixture of CO and H2. The mixture is subjected to 200atm pressure and passed over ZnO and Cr2O3 catalyst at 400 to 450 degrees celsius.
2- phenyl propene on acidic hydration gives :
Report Question
0%
2-phenyl-2-propanol
0%
2-phenyl-1-propanol
0%
3-phenyl-1-propanol
0%
1-phenyl-2-propanol
Explanation
Hence, option
A
is correct.
C
H
3
M
g
X
C
H
3
C
O
O
C
2
H
5
→
A
N
a
→
B
C
2
H
5
B
r
→
C
Here, C is
:
Report Question
0%
C
2
H
5
O
C
2
H
5
0%
C
H
3
C
O
O
C
2
H
5
0%
C
H
3
C
O
O
C
H
3
0%
(
C
H
3
)
3
C
O
C
2
H
5
Explanation
In first step
C
H
3
M
g
−
X
+
C
H
3
C
O
O
C
2
H
5
d
r
y
e
t
h
e
r
→
(
C
H
3
)
3
C
−
O
H
+
C
2
H
5
O
M
g
X
Here the tertiary butyl alchol formed which then reacts with
N
a
to give
(
C
H
3
)
3
C
O
−
N
a
+
0.5
H
2
So when there is primary alkyl halide reacts with alkoxide then Williamson's ether synthesis takes place rather than elimination.
Williamson's ether synthesis goes through the
S
N
2
mechanism.
Option D is correct.
How many structures of F are possible?
Report Question
0%
2
0%
5
0%
6
0%
3
Explanation
3
isomers of compound F are possible. They are shown above.
Protonation in acidic conditions of the hydroxyl group of secondary alcohols converts the hydroxyl group into a leaving group, which departs as water.
The carbocation intermediate can rearrange and it will then be deprotonated in a
E
1
elimination mechanism. This is the reverse of acid-catalyzed hydration of alkenes.
Hence, the correct option is
D
An organic compound (A) containing C, H and O has a pleasant odour with boiling point
78
0
. On boiling (A) with concentrated
H
2
S
O
4
, colourless gas is evolved which decolorizes bromine water and alkaline
K
M
n
O
4
. The organic compound (A) is
:
Report Question
0%
C
2
H
5
C
l
0%
C
2
H
5
C
O
O
C
H
3
0%
C
2
H
5
O
H
0%
C
2
H
6
Explanation
The organic compound is ethanol,
C
2
H
5
O
H
. It has a pleasant odour. Its boiling point is
78
0
C
. On treating it with conc.
H
2
S
O
4
, the colourless gas formed is
C
2
H
4
that decolourises bromine water and alkaline
K
M
n
O
4
.
The structure of compound
I
is:
Report Question
0%
0%
0%
0%
Explanation
Hence, option
A
is correct.
Ortho-nitrophenol is less soluble in water than para and meta-nitrophenols because:
Report Question
0%
o- nitrophenol is more volatile in steam than those of m- and p- isomers.
0%
o- nitrophenol shows intramolecular
H
−
bonding.
0%
o-nitrophenol shows intermolecular
H
−
bonding.
0%
none of these.
Explanation
Intramolecular hydrogen bonding decreases water solubility of o-nitrophenol, while para and meta-nitrophenols show intermolecular hydrogen bonding which increases their solubility in water. Hence, ortho-nitrophenol is less soluble in water than para and meta-nitrophenols.
Hence, the correct option is 'B'.
Identify the structure of A:
Report Question
0%
C
H
3
−
C
H
2
−
C
H
2
−
C
H
|
O
H
−
C
H
3
0%
(
C
H
3
)
2
C
H
C
H
O
H
C
H
3
0%
(
C
H
3
)
3
C
−
C
H
2
−
O
H
0%
C
H
3
−
C
H
2
−
C
H
|
O
H
−
C
H
2
−
C
H
3
Explanation
The structure of A in the following reaction is option B.
On oxymercuration-demercuration, the given compound produces the major product:
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0%
0%
0%
0%
Explanation
The oxymercuration reaction is an electrophilic addition organic reaction that transforms an alkene into a neutral alcohol. In oxymercuration, the alkene reacts with mercuric acetate (AcOHgOAc) in aqueous solution to yield the addition of an acetoxymercuri (HgOAc) group and a hydroxy (OH) group across the double bond. Carbocations are not formed in this process and thus rearrangements are not observed. The reaction follows Markovnikov's rule (the hydroxy group will always be added to the more substituted carbon) and it is an anti addition (the two groups will be trans to each other).
The main product of the above reaction is:
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0%
0%
0%
Which of the following will be the correct product (P) for the given reaction?
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0%
0%
0%
Explanation
Dehydration of alcohol involves the formation of carbocation which further undergoes rearrangement as the tertiary carbocation is more stable than secondary carbocation, which then accepts the electron from
π
bond of cyclohexene and forms bicyclic compound.
Option C is correct.
The addition of mercuric acetate in the presence of water is called oxymercuration. The adduct obtained gives alcohol on reduction with
N
a
B
H
4
in alkaline medium. This is known as demercuration. Oxymercuration-demercuration allows Markownikoff's addition of H, OH without rearrangement. The net result is the addition of
H
2
O
.
Based on the above information, find product A in the given reaction.
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Identify the correct option(s) related with the physical properties of the unknown compounds [X] and [Y] formed as major and minor product respectively in the given reaction.
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Water solubility of [Y] is greater than [X]
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Boiling point of [Y] is greater than [X]
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Melting point of [Y] is greater than [X]
0%
[X] is more volatile than [Y]
P
r
o
p
a
n
−
1
−
o
l
can be prepared from propene by :
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H
2
O
/
H
2
S
O
4
0%
C
o
n
c
.
H
2
S
O
4
0%
B
2
H
6
followed by
H
2
O
2
/
O
H
0%
C
H
3
C
O
2
H
/
H
2
S
O
4
Which of the following reagents should be used to prepare tert-butyl ethyl ether?
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t
e
r
t
-butyl bromide and sodium ethoxide
0%
t
e
r
t
-butyl alcohol and ethyl bromide
0%
t
e
r
t
-butyl alcohol and ethanol
0%
potassium
t
e
r
t
-butoxide and ethyl bromide
Explanation
t
e
r
t
-butyl ethyl ether can be prepared by the reaction of potassium
t
e
r
t
-butoxide with ethyl bromide.
The process is Williamson synthesis.
This reaction follows
S
N
2
mechanism.
Option D is correct.
Ethanol
P
B
r
3
→
X
a
l
c
.
K
O
H
→
Y
(
i
)
H
2
S
O
4
r
o
o
m
t
e
m
p
e
r
a
t
u
r
e
→
(
i
i
)
H
2
O
,
H
e
a
t
Z
;
The product
Z
in the above reaction is:
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0%
C
H
3
C
H
2
−
O
H
0%
C
H
2
=
C
H
2
0%
C
H
3
C
H
2
−
O
−
C
H
2
C
H
3
0%
C
H
3
C
H
2
−
O
−
S
O
3
H
Explanation
When ethanol reacts with
P
B
r
3
it undergoes substitution reaction to form ethyl bromide as a product.
This ethyl bromide when reacts with
a
l
c
.
K
O
H
, it undergoes
β
-elimination reaction to form ethene.
This ethene on reaction with sulphuric acid undergoes sulphonation followed by hydration to form ethanol as a product.
A
(
C
7
H
8
O
)
+
N
a
/
Δ
⟶
No gaseous product
A
+
B
r
2
/
F
e
B
r
3
⟶
C
7
H
7
O
B
r
(Two isomers)
A
can be:
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0%
Explanation
When compound A is heated with sodium no gaseous product is obtained.
When phenol is heated with sodium, hydrogen gas is evolved.
Thus presence of phenolic OH group is ruled out. Thus options A, B and C are ruled out.
Option D (anisole) is the correct answer.
Anisole on bromination with bromine in presence of ferric bromide gives two isomers ortho and para bromo derivatives of anisole.
C
(
C
H
3
)
2
=
C
(
C
H
3
)
2
C
l
2
→
H
2
O
A
O
H
−
→
B
The compound
B
is:
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0%
0%
(
C
H
3
)
2
C
|
−
O
H
C
|
C
l
(
C
H
3
)
2
0%
(
C
H
3
)
2
C
|
−
O
H
C
|
O
H
(
C
H
3
)
2
0%
none of the above
Explanation
The given reaction is base catalyzed cyclization of vicinal chlorohydrins.
Alkene reacts with chlorine in presence of water to form vicinal chlorohydrin.
This is followed by elimination of
H
C
l
in presence of base to form epoxide.
The compound
A
and
B
are :
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0%
0%
None of the above
Explanation
Alkene is oxidized with cold alkaline potassium permanganate solution to form diol
A
.
Of the two hydroxyl groups, the secondary hydroxyl group is oxidized with chromium trioxide in presence of acetic acid to form hydroxy ketone
B
. Under these, conditions, tertiary hydroxyl group is not oxidized as it requires drastic conditions.
Which of the following are possible products in the above reaction?
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0%
0%
0%
Give the expected major product of the above reaction.
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0%
0%
0%
Which method is useful for the synthesis of ether?
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0%
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Which of the followings is/are true regarding phenol?
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It turns pink when exposed to air
0%
It is less reactive than anisole in aromatic electrophilic substitution reaction
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It is more reactive than aniline in an aromatic electrophilic substitution reaction
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Acetylation of phenol decreases its reactivity in aromatic electrophilic substitution reaction
Explanation
When phenol is exposed to air, a mixture of quinones is formed. This results in pink colour of phenol.
Acetylation of phenol results in formation of phenyl acetate. The acetate group reduces the electron density of aromatic nucleus through resonance and hence, decreases the reactivity towards electrophilic aromatic substitution reaction. Same is done by the oxygen atom in anisole.
For the similar reason, aniline is more reactive than phenol as nitrogen increases the electron density.
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Practice Class 12 Medical Chemistry Quiz Questions and Answers
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