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CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 9 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Alcohols,Phenols And Ethers
Quiz 9
In the following sequence of reactions
C
H
3
C
H
2
C
H
2
B
r
K
O
H
(
a
l
c
)
⟶
(
A
)
H
B
r
⟶
(
B
)
K
O
H
(
a
q
.
)
⟶
(
C
)
.
The product (C) is
Report Question
0%
Propan - 2 - ol
0%
Propan - l - ol
0%
Propyne
0%
Propene
One of the following that cannot undergo dehydro - halogenation is
Report Question
0%
Iso - propyl bromide
0%
Ethanol
0%
Ethyl bromide
0%
None of these
Explanation
Ethanol doesn't have any halogen group. Thus ethanol cannot undergo dehydrohalogenation.
The boiling point of alcohol are _________ than corresponding thiols.
Report Question
0%
More
0%
Same
0%
Either of these
0%
Less
Explanation
In alcohol, hydrogen bonding is present between oxygen and a hydrogen atom.
Sulfur is less electronegative than oxygen and hence, forms a weaker hydrogen bond than in alcohol.
Stronger the hydrogen bonding, higher is the boiling point.
Hence, the boiling point of alcohol is more than the corresponding thiol.
Hence, option
A
is correct.
Which of the following is most soluble in water?
Report Question
0%
n-butyl alcohol
0%
Isobutyl alcohol
0%
Tertiary butyl alcohol
0%
Secondary butyl alcohol
Explanation
The solubility of alcohols and phenols in water due to their ability to form hydrogen bonds with water molecules.
The solubility decreases with an increase in size of alkyl group which is a hydrophobic group, which makes the alcohol less hydrophilic.
In the case of isomers, the order of solubility in water is
1
0
>
2
0
>
3
0
due to decrease in polar character.
Hence, among all, n-butyl alcohol is the most soluble in water.
The answer is option
A
.
Dehydration of ethanol gives
Report Question
0%
Acetic acid
0%
Ethane
0%
Ethylene
0%
Acetylene
Explanation
C
H
3
−
C
H
2
−
O
H
C
o
n
c
.
H
2
S
O
4
→
170
0
C
H
2
=
C
H
2
+
H
2
O
Option C is correct.
Which of the following is not characteristic of alcohols
Report Question
0%
They are lighter than water
0%
Their boiling points rise fairly uniformly with increasing molecular weight
0%
Lower members are insoluble in water and organic solvents but solubility regularly increases with molecular weight
0%
Lower members have pleasant smell and burning taste, while higher members are odourless and tasteless
Explanation
Lower members are soluble in water and solubility decreases with increasing molecular mass because the length of the hydrophobic chain increases.
Option C is correct.
In the following series of chemical reactions, identify Z
C
3
H
7
O
H
C
o
n
c
.
H
2
S
O
4
→
160
−
180
0
C
X
B
r
2
→
Y
A
l
c
.
K
O
H
→
E
x
c
e
s
s
o
f
Z
Report Question
0%
C
H
3
−
C
H
|
N
H
2
−
C
H
2
|
N
H
2
0%
C
H
3
−
C
H
|
O
H
−
C
H
2
|
O
H
0%
C
H
3
−
C
|
O
H
=
C
H
2
0%
C
H
3
C
≡
C
H
Explanation
C
H
3
C
H
2
C
H
2
O
H
c
o
n
c
.
H
2
S
O
4
→
160
−
18060
C
C
H
3
C
H
=
C
H
2
B
r
2
→
C
H
3
−
C
H
|
B
r
−
C
H
2
|
B
r
A
l
c
.
K
O
H
→
C
H
3
−
C
≡
C
H
P
r
o
p
y
n
e
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
)
has only one carbon atom. So, it does not give a stable compound on dehydration.
Hence, option
B
is correct.
Primary alcohols on dehydration give
Report Question
0%
Alkenes
0%
Alkanes
0%
Both (a) and (b)
0%
None of these
Explanation
R
−
C
H
2
−
C
H
2
−
O
H
1
0
a
l
c
o
h
o
l
C
o
n
c
.
H
2
S
O
4
→
170
0
C
R
−
C
H
=
C
H
2
+
H
2
O
A
l
k
e
n
e
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
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
is correct.
Which compound is soluble in water
Report Question
0%
C
S
2
0%
C
2
H
5
O
H
0%
C
C
l
4
0%
C
H
C
l
3
Explanation
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
0%
Aqueous KOH
0%
M
h
S
O
4
,
N
a
B
H
4
/
H
2
O
0%
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 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.
Aqueous sulphuric acid reacts with 2-methyl-1-butene togive predominantly
Report Question
0%
Isobutyl hydrogen sulphate
0%
2-methyl-2-butanol
0%
2-methyl-1-butanol
0%
Secondary butyl hydrogen sulphate
Explanation
C
H
2
=
C
H
3
|
C
−
C
H
2
2
−
m
e
t
h
y
l
−
1
−
b
u
t
a
n
e
−
C
H
3
+
H
2
O
H
2
S
O
4
→
M
a
r
k
o
w
n
i
k
o
f
f
s
r
u
l
e
C
H
3
−
C
H
3
|
C
|
O
H
2
−
M
e
t
h
y
l
1
−
2
−
b
u
t
a
n
o
l
−
C
H
2
−
C
H
3
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
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'.
The main product of the above reaction is:
Report Question
0%
0%
0%
0%
On oxymercuration-demercuration, the given compound produces the major product:
Report Question
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).
Which reagent converts propene to propan-1-ol?
Report Question
0%
H
2
O
,
H
2
S
O
4
0%
B
2
H
6
,
H
2
O
2
,
O
H
−
0%
H
g
(
O
A
c
)
2
,
N
a
B
H
4
/
H
2
O
0%
Aq.
K
O
H
Explanation
Diborane (
B
2
H
6
) along with
H
2
O
2
when reacts with propene forms propan-1-ol.
In this reaction, the Anti-Markownikoff's addition of the diborane group takes place which on reaction with hydrogen peroxide forms propan-1-ol.
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.
Report Question
0%
Water solubility of [Y] is greater than [X]
0%
Boiling point of [Y] is greater than [X]
0%
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 :
Report Question
0%
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 one of the following compounds is most
acidic?
Report Question
0%
0%
0%
0%
Explanation
In case of ortho nitrophenol, the nitro group attached to a conjugated system shows a strong
−
M
and
−
I
effect and this decreases the electron density more at ortho position, i.e why o-nitrophenol is more acidic in nature.
Among the following four compounds
the acidity order is:
(a) Phenol
(b) methyl phenol
(c) meta nitrophenol
(d)
para-nitrophenol
Report Question
0%
c
>
d
>
a
>
b
0%
a
>
d
>
c
>
b
0%
b
>
a
>
c
>
d
0%
d
>
c
>
a
>
b
Explanation
Para-nitrophenol(d) is more acidic because base abstract proton from OH group of phenol to form phenoxide ion is stabilized by nitro group present at Para position through resonance effect. Stability increase acidity also increases.
Meta-nitrophenol(c) is less acidic as compared to (d), because base abstract proton to form phenoxide is stabilized by electron-withdrawing inductive effect and this effect is less effective as compared to Resonance effect.
Phenol(a) is less acidic as compared (c)&(d), because No electron withdrawing group present on Phenol and resonance & inductive effect is absent here.
Methyl phenol shows the electrondonating Inductive effect that is donating group present on phenol as result destabilize phenol/ phenoxide, So it is least acidic
Above discussion concluded that acidity order is as follows d>c>a>b
Option -D
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:
Report Question
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.
Identify the correct sequence of given steps for the conversion of calcium carbide to methyl alcohol.
(a) Reaction with aqueous KOH
(b) Hydrolysis
(c) Reaction with soda lime
(d) Reaction with
H
g
S
O
4
/
H
2
S
O
4
(e)
Potassium dichromate
,
K
2
C
r
2
O
7
(f) Reaction with
P
C
l
5
Report Question
0%
b d e c f a
0%
b d c a f e
0%
b f d e a c
0%
d b c e f a
Explanation
The steps involved can be written as:
i) Hydrolysis of calcium carbide to give acetylene as:
C
a
C
2
+
2
H
2
O
→
C
2
H
2
+
C
a
(
O
H
)
2
ii) Hydration of acetylene in the presence of
H
g
S
O
4
a
n
d
H
2
S
O
4
as:
H
C
≡
C
H
H
g
2
+
,
H
+
→
C
H
3
C
H
O
iii) Acetaldehyde is converted to methyl alcohol as:
C
H
3
C
H
O
K
2
C
r
2
O
7
→
C
H
3
C
O
O
H
N
a
O
H
→
C
H
3
C
O
O
N
a
P
C
l
5
→
C
H
3
C
l
K
O
H
→
C
H
3
O
H
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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