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CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 7 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Hydrocarbons
Quiz 7
Match the column I with Column II and mark the appropriate choice.
Column I
Column II
(A)
n
−
B
u
t
a
n
e
→
2
−
M
e
t
h
y
l
p
r
o
p
a
n
e
(i)
Free radical substitution
(B)
C
H
4
+
C
l
2
h
v
→
C
H
3
C
l
(ii)
Wurtz reaction
(C)
R
C
O
O
N
a
+
S
o
d
a
l
i
m
e
→
R
H
(iii)
Isomerisation
(D)
R
X
+
N
a
e
t
h
e
r
→
R
−
R
(iv)
Decarboxylation
Report Question
0%
(
A
)
→
(
i
i
i
)
,
(
B
)
→
(
i
)
,
(
C
)
→
(
i
v
)
,
(
D
)
→
(
i
i
)
0%
(
A
)
→
(
i
i
)
,
(
B
)
→
(
i
v
)
,
(
C
)
→
(
i
)
,
(
D
)
→
(
i
i
i
)
0%
(
A
)
→
(
i
)
,
(
B
)
→
(
i
i
)
,
(
C
)
→
(
i
v
)
,
(
D
)
→
(
i
i
i
)
0%
(
A
)
→
(
i
v
)
,
(
B
)
→
(
i
)
,
(
C
)
→
(
i
i
i
)
,
(
D
)
→
(
i
i
)
Explanation
(A)
n
−
B
u
t
a
n
e
⟶
2
−
M
e
t
h
y
l
p
r
o
p
a
n
e
Isomerisation is a reaction where one organic compound converts to another compound with same molecular mass but different structures.
n-Butane and 2-Methylpropane, both are isomers and the reaction is isomerisation
→
(iii) Isomerisation.
(B)
C
H
4
+
C
l
2
h
ν
→
C
H
3
C
l
The above reaction is Free radical substitution which takes place in presence of sunlight
→
(i) Free radical substitution.
(C)
R
C
O
O
−
N
a
+
+
N
a
O
H
+
C
a
O
S
o
d
a
l
i
m
e
⟶
R
H
+
N
a
2
C
O
3
The above reaction is known as
Decarboxylation
used to prepare hydrocarbons. Sodium salt of carboxylic acid releases
C
O
2
to form alkanes and
N
a
2
C
O
3
→
(iv) Decarboxylation.
(D)
R
X
+
N
a
e
t
h
e
r
→
R
−
R
The above reaction is termed as Wurtz reaction which unites two alkyl halides to form one alkane in presence of sodium and ether
→
(ii) Wurtz reaction.
Chlorination of alkanes is a photochemical process. It is initiated by the process of:
Report Question
0%
heterolysis
0%
homolysis
0%
pyrolysis
0%
hydrolysis
Explanation
Photochemical chlorination of alkane takes place by free radical mechanism which are possible by homolysis of
C
l
−
C
l
bond by sunlight.
C
l
2
h
ν
→
C
l
∙
+
C
l
∙
C
H
3
−
C
H
3
+
C
l
∙
⟶
∙
C
H
2
C
H
3
+
H
C
l
C
l
−
C
l
+
∙
C
H
2
C
H
3
⟶
C
l
−
C
H
2
C
H
3
+
C
l
∙
The cycloalkane formed when
1
,
4
−
d
i
c
h
l
o
r
o
p
e
n
t
a
n
e
is heated with sodium in presence of dry ether is ________.
Report Question
0%
methylcyclopentane
0%
cyclopentane
0%
methylcyclobutane
0%
cyclobutane
Which one is incorrect name?
Report Question
0%
Propyne
0%
But-2-yne
0%
Pent-3-yne
0%
But-1 -yne
Explanation
The correct answer is option C, because As the structural formula of Pent-3-yne is
C
H
3
−
C
H
2
−
C
≡
C
−
C
H
3
, According to the IUPAC nomenclature the numbering of the carbon atom is done from that side of the carbon chain from which the triple bond is nearest as in this structure the numbering can also be done from right side. Therefore, the correct IUPAC name of the compound is
P
e
n
t
−
2
−
y
n
e
.
Which one of the following IUPAC name is not correctly matched?
Report Question
0%
0%
0%
0%
Explanation
The parent chain consists of 6 carbon atoms in a cyclic structure.
There is an isopropyl group present at carbon 1.
Hence, the name of the structure is Isopropyl cyclohexane.
Therefore, option C is not correctly matched.
Ethane is formed during the formation of chloromethane by chlorination of methane because:
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0%
higher members of the hydrocarbons are generally formed during reactions
0%
two methyl free radicals may combine during chlorination to give ethane
0%
two chloromethane molecules react to form ethane
0%
chlorine free radical reacts with methane to give ethane.
Explanation
Chlorination of methane is a free radical reaction. The free radicals formed during the reaction are
C
l
∙
and
C
H
∙
3
.
During the process if two methyl radicals collapse they form ethane. So, ethane is also formed as a byproduct.
2
C
l
∙
→
C
l
2
H
3
C
∙
+
C
l
∙
⟶
C
H
3
C
l
H
3
C
∙
+
H
3
C
∙
⟶
C
H
3
−
C
H
3
Finely divided platinum and palladium commonly known as platinum and palladium black, may be prepared by reducing their,soluble salts with:
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0%
H
2
O
0%
C
2
H
5
O
H
0%
H
C
H
O
0%
C
6
H
6
Cis-dicholoroethylene can be converted to trans-dicholorethylene by irradiations. During the interconversion.
Report Question
0%
Both the
π
and
σ
bonds are broken and reformed
0%
Only the
π
bond is broken and reformed
0%
The
C
−
C
l
and
C
−
H
bonds are broken,
H
and
C
l
atoms interchanges their positions and new bonds are formed
0%
Both the
π
and
σ
bonds rotate along with the groups
Explanation
Cis-dicholoroethylene can be converted to trans-dicholorethylene by irradiation. During the inter-conversion, only the
π
bond is broken and reformed. Rotation about
C
−
C
single bond converts cis isomer into trans isomer and vice versa.
Which of the following has the lowest boiling point?
Report Question
0%
2-Methylbutane
0%
2-Methylpropane
0%
2,2-Dimethylpropane
0%
n-pentane
Explanation
Boiling point increases with increase in molecular mass. For the compounds with the same molecular mass, boiling point decreases with an increase in branching.
The molecular mass of 2-Methylbutane:
72
g
m
o
l
−
1
The molecular mass of 2-Methylpropane:
58
g
m
o
l
−
1
The molecular mass of 2,2-Dimethylpropane:
72
g
m
o
l
−
1
The molecular mass of 2-Methylbutane:
72
g
m
o
l
−
1
2-Methylpropane has the lowest molecular mass among all of the given compounds.
Thus, 2-Methylpropane has the lowest boiling point among the given options.
During halogenation of alkanes, the halogens and alkane show a specific trend. Which of the following statements is not correct?
Report Question
0%
The reactivity of halogens is in the order
F
2
>
C
l
2
>
B
r
2
>
I
2
.
0%
For a given halogen the reactivity of hydrocarbon is in the order of
3
o
>
2
o
>
1
o
.
0%
Bromine is less reactive than chlorine towards a particular alkane.
0%
On chlorination monosubstituted product is formed while on bromination disubstituted products are formed.
Explanation
Number of monohalogen products of alkane depends upon the number of different types of replaceable hydrogen atoms in the alkane, not on the halogen. So chlorine and bromine will give the same number of substitutions.
So (D) is incorrect.
The heat of combustion of carbon to
C
O
2
is -393.5 kJ/mol. The heat released upon formation of 35.2 g of
C
O
2
from catbon and oxygen gas will be :
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0%
+315 kJ/mole
0%
-31.5 kJ/mole
0%
-315 kJ/mole
0%
+31.5 kJ/mole
Explanation
Heat of combustion for
1
mole of
C
O
2
is
−
393.5
K
J
Number of moles in
35.2
g
of
C
O
2
=
35.2
44
=
0.8
mole
Thus heat of combustion for
0.8
mole of
C
O
2
is
−
393.5
×
0.8
=
−
315
K
J
What is the carbon-carbon bond length in benzene?
Report Question
0%
1.20
A
˚
and
1.31 \mathring { A }
0%
1.39 \mathring { A }
0%
1.39 \mathring { A }
and
1.20 \mathring { A }
0%
1.20 \mathring { A }
Explanation
All
C-C
and
C=C
bond lengths are the same in benzene. Hence,
C-C
bond length in benzene is 1.39
\mathring { A }
.
Nitration and chlorination of benzene are:
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0%
nucleophilic and electrophilic substitution respectively
0%
electrophilic and nucleophilic substitution respectively
0%
electrophilic substitution in both the reactions
0%
nucleophilic substitution in both the reactions.
Explanation
Since nitration and halogenation, both are carried out by the help of electrophile
{N{O}_{2}}^{+} \; \& \; {Cl}^{+}
.
Hence, both reactions are electrophilic substitution reactions.
Although benzene is highly unsaturated it does not undergo addition reactions. The explanation of this can be suggested as:
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0%
\pi
-electrons of benzene ring are delocalised
0%
since
\pi
-electrons are present inside the ring, addition cannot take place
0%
cyclic structures do not show addition reactions
0%
benzene is not a reactive compound.
Explanation
(A)
\pi
-electrons of
the benzene
ring are delocalised.
There are delocalised
\pi
-electrons above and below the plane of the ring. The presence of the delocalised
\pi
-electrons makes benzene particularly stable. Benzene resists addition reactions because that would involve breaking the delocalisation and losing that stability.
What happens when calcium carbide is treated with water?
Report Question
0%
Ethane is formed.
0%
Methane and ethane are formed.
0%
Ethyne is formed.
0%
Ethene and ethyne are formed.
Explanation
If we mix calcium carbide with water ethyne is released due to endothermic reaction between them. This is the industrial method of producing ethyne.
Ca^{2+}{[{C}\equiv C]}^{2-} + 2{H}_{2}O \longrightarrow \underset{Ethyne}{CH \equiv CH} + Ca{(OH)}_{2}
Fill in the blanks with appropriate words. Benzene has a planar structure. All carbon atoms in benzene are
I
hybridised. The ring structure of benzene was proposed by
II
. It shows
III
substitution reactions. It reacts with
IV
in presence of aluminium chloride to form acetophenone.
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0%
I = sp_2 \,\,\, II = Kekule \,\,\, III= electrophilic \,\,\, IV = acetyl\, chloride
0%
I = sp \,\,\, II = Dewar \,\,\, III= nucleophilic \,\,\, IV =chloromethane
0%
I = sp_3 \,\,\, II = Ladenberg \,\,\, III= electrophilic \,\,\, IV = chloroethane
0%
I = sp_2 \,\,\, II = Baeyer \,\,\, III= nucleophilic \,\,\, IV = methyl\, bromide
Explanation
Benzene has a planar structure. All carbon atoms in benzene are
s{p}^{2}
hybridised being bonded to two other carbon atoms and one hydrogen atom. The ring structure of benzene was proposed by Kekule. It shows electrophillic substitution reactions due to electron rich character. It reacts with acetyl chloride in presence of aluminium chloride to form acetophenone and the reaction is known as acylation of benzene.
Hence, A is correct answer
When n-butane is treated with chlorine, how many monochloro products are formed?
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+{ Cl }_{ 2 }\rightarrow { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }{ Cl }+{ CH }_{ 3 }{ CH }_{ 2 }{ - }\underset { \overset { | }{ Cl } }{ CH } -{ CH }_{ 3 }
What is the order of reactivity of hydrogen atoms attached to the carbon atom in an alkane for free radical substitution?
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0%
3^o >1^o >2^o
0%
2^o >1^o >3^o
0%
3^o >2^o >1^o
0%
1^o >2^o >3^o
Explanation
REF: free radical will have
9 \alpha H.
CH_{3}-CH_{2}-CH_{3}
free radical will have
6 \alpha H
CH_{3}-CH_{3}-H
free radical will have
3\alpha H
As no.of
\alpha -H
increase stability of free radical increase
reactivity order of H
3^{\circ} > 2^{\circ} > 1^{\circ}
Hence, te correct option is
\text{C}
One mole of 1,2-dibromopropane on treatment with
X
moles of
NaNH_2
followed by treatment with ethyl bromide gave a 2-pentyne. The value of
X
is:
Report Question
0%
one
0%
two
0%
three
0%
four
Explanation
C{H}_{3}CH(Br)C{H}_{2}(Br) \xrightarrow{2NaN{H}_{2}} C{H}_{3}-C \equiv CH \xrightarrow{NaN{H}_{2}} {C{H}_{3}-C \equiv C}^{-}{Na}^{+} \xrightarrow{{C}_{2}{H}_{5}Br} C{H}_{3}C \equiv CC{H}_{2}C{H}_{3}
Hence,
X=3
.
Mark the incorrect statement from the following.
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0%
Benzene has a planar structure
0%
Benzene is an unsaturated hydrocarbon and shows addition reactions like alkenes
0%
In benzene carbon uses two p-orbitals for hybridisation
0%
Aromatic hydrocarbons contain high percentage of carbon hence burn with sooty flame
Explanation
The statement given in the second option is incorrect.
Benzene can show both additions as well substitution reaction. It does not show an addition reaction like alkene. This is because, if it shows an addition reaction like alkene, then it will lose its aromaticity and become less stable.
The compound formed when an alcoholic solution of ethylene dibromide is heated with granulated zinc is:
Report Question
0%
ethene
0%
ethyne
0%
ethane
0%
bromoethane
Explanation
The given reaction is dehalogenation reaction. Zinc removes bromine from ethylene dibromide as zinc bromide and forms ethene.
\underset{\text{Ethylene dibromide}}{Br{H}_{2}C-C{H}_{2}Br} + Zn \xrightarrow[alcohol]{heat} {H}_{2}C=C{H}_{2} +Zn{Br}_{2}
For the given Question (1,2,3) consider the following reaction.
How many monohalo derivatives are possible (excluding stereoisomers) ?
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0%
3
0%
4
0%
5
0%
6
Explanation
Refer to image
There are total 4 products formed.
Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make
1^{\circ} ,2^{\circ} ,3^{\circ}
radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}
The relative rate of abstraction of hydrogen by Cl is
\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}
CH_3 - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}} -CH_2 - \overset{CH_3}{\overset{|}CH} - CH_3 \xrightarrow{Br_2/hv} CH_3 - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}} - CH_2 - CH_2 - \overset{CH_3}{\overset{|}{\underset{Br}{\underset{|}C}}} - CH_3
2,Bromo-2,5,5 trimethyl hexane(x%)
What is the value of x(% yield of product) ?
Report Question
0%
18 %
0%
82 %
0%
90 %
0%
60 %
Explanation
CH_3 - \!\!\!\overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}C}}}\!\!\! -CH_2 - \overset{CH_3}{\overset{|}CH} - CH_3 \xrightarrow{Br_2/hv} CH_3 -\!\!\! \overset{\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}C}}}\!\!\! - CH_2 -\!\!\! \overset{\,\,\,\,CH_3}{\overset{|}{\underset{Br}{\underset{|}C}}} - CH_3
Here, tertiary bromination occurs-
To find the percentage
(3^o\%)=\cfrac {3^o}{All}\times 100
As,
3^o\rightarrow 1\times 1600=1600
2^o\rightarrow 2\times 82=164
1^o\rightarrow 15\times 1=15
So, All is
1600+164+15=1779
\therefore \%=\cfrac {1600}{1779}\times 100=90\%
Which of the following is aromatic in nature?
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0%
0%
0%
0%
0%
None of these
Explanation
R.E.F image
Non Aromatic
R.E.F image
Antiaromatic
4\pi e^{-}s
R.E.F image
Aromatic
2\pi e^{-}s
R.E.F image
Non Aromatic
Select the chain propagation steps in the free-radical chlorination of methane.
(1)
Cl_2\, \rightarrow 2Cl^{\cdot}
(2)
Cl^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3C1 \,+ \, H^\cdot
(3)
C1^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3 ^\cdot \, + \, HCI
(4)
H^\cdot \, + \, Cl_2 \, \xrightarrow \, HCl+Cl^\cdot
(5)
CH_3 ^\cdot \, + \, Cl_2 \, \xrightarrow \, CH_3CI+Cl^\cdot
Report Question
0%
2,3,5
0%
1,3,6
0%
3,5
0%
2,3,4
Explanation
(1)
Cl_2 \longrightarrow 2Cl^\cdot
[Initiation]
(2)
Cl^\cdot+CH_4 \longrightarrow CH_3Cl+H^\cdot
(3)
Cl^\cdot + CH_4 \longrightarrow CH_3^\cdot+ HCl
[Propagation]
(4)
H^\cdot+Cl_2 \longrightarrow HCl+Cl^\cdot
(5)
CH_3^\cdot+Cl_2 \longrightarrow CH_3Cl+Cl^\cdot
[Propagation]
In the first propagation step (3), a
Cl^\cdot
combines with a hydrogen on
CH_4
giving
HCl
and
^\cdot CH_3
. In the
2^{nd}
propagation step (5)
Cl_2
combines with
^\cdot CH_3
to form
CH_3Cl
and
Cl^\cdot
.
Rank the transition states that occur during the following reaction steps in order of increasing stability
(least \xrightarrow \,most stable)
1.
H_3C \, - \, \overset{+}{OH_2} \, \xrightarrow \, CH_3^{+} \, + \, H_2O
2.
(CH_3)_3C \, - \, \overset{+}{OH_2} \xrightarrow \, - \,(CH_3)_3C \, + \, H_2O
3.
(CH_3)_2 \, CH \, - \, \overset{+}{OH_2} \xrightarrow \,(CH_3)_2CH^+ \, + \, H_2O
Report Question
0%
1 < 2 < 3
0%
2 < 3 < 1
0%
1 < 3 < 2
0%
2 < 1 < 3
Explanation
(1) (Refer to Image 1)
(2) (Refer to Image 2)
(3) (Refer to Image 3)
Now stability of carbocations: tertiary
>
secondary
>
primary. Therefore transition state corresponding to tertiary carbocation will be most stable. So stability order will be
1 < 3 <2
.
n-Butane
\xrightarrow{Cl_2/hv}
Give the total number of monochioro products (including stereoisomers), which are possible in the above reaction.
Report Question
0%
2
0%
3
0%
4
0%
5
Explanation
n-Butane
gives three products i.e.
S-2-chlorobutane, R-2-chlorobutane
and
1-chlorobutane
when
n-Butane
is treated with
Cl_2
in presence of light.
An alkane
(mol.\ wt.\ = 86)
on bromination gives only two monobromo derivatives (excluding stereoisomers). The alkane is:
Report Question
0%
CH_3-\underset{CH_3}{\underset{|}{C}}H-CH_2-CH_2-CH_3
0%
CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_2-CH_3
0%
CH_3-\underset{CH_3}{\underset{|}{C}}H-\underset{CH_3}{\underset{|}{C}}H-CH_3
0%
CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_3
Explanation
CH_3-\underset {|}{CH}-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 5
monobromoderivatives
CH_3
A
CH_3
CH_3-\underset {|}{\overset {|}{C}}-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 3
monobromoderivatives
CH_3
B
CH_3-\underset {|}{CH}-\underset {|}{CH}-CH_3 \xrightarrow [h\nu]{Br_2} 2
monobromoderivatives
CH_3
CH_3
Among them
A,B
and
C
have only molecular weight
86
. Now
A
has
5
different equivalent
H
atoms and
B
has
3
different equivalent
H
atoms, so
A
and
B
will give
5
and
3
monobromo derivatives respectively. But
C
has
2
different equivalent
H
atoms, so
C
will give
2
monobromo derivatives.
\xrightarrow[hv]{cl_2}
(x). (x) = total number of di-chloro product S-2 chloro hexane.
Report Question
0%
7
0%
6
0%
16
0%
none
Explanation
Cl
H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Cl_2} (x)
H
The possible product may be
Cl
(1)
H_2\underset {|}{C}-\overset {|}{\underset {|}C^*}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^1=2
stereoisomers
Cl
H
Cl
(2)
H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^o= 1
stereoisomer
Cl
Cl
Cl
(3)
H_3C-_*\overset {|}{\underset {|}C}-_*\overset {|}{\underset {|}C}-CH_2-CH_2-CH_3= 2^2=4
stereoisomers
H
H
Cl
Cl
(4)
H_3C-^*\overset {|}{\underset {|}C}-CH_2-\overset {|}{\underset {|}C^*}-CH_2-CH_3=2^2=4
stereoisomers
H
H
Cl
Cl
(5)
H_3C-_*\overset {|}{\underset {|}C}-CH_2-CH_2-_*\overset {|}{\underset {|}C}-CH_3=
as it is symmetrical molecule=
2^{2-1}+2^{2/2-1}=3
stereoisomers
H
H
Cl
(6)
H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_2-Cl= 2^1=2
stereoisomers
H
So, total number of di chloro product obtained=
2+1+4+4+3+2=16
product
Arrange the following compounds in decreasing order of their heats of combustion:
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0%
(iii) > (ii) > (i)
0%
(ii) > (i) > (iii)
0%
(iii) > (i) > (ii)
0%
(i) > (ii) > (iii)
Explanation
Heat of combustion is inversely proportional to stability.
Since three membered ring is most unstable due to angular strain so it is having highest heat of combustion.
So the order for heat of combustion is:
(i) > (ii) > (iii)
Among the following free radical bromination reactions, select those in which 2 halide is the major product
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P, Q, R, S
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P, R, U
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P, R, S, T
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P, Q, R, S, T
Explanation
Solution:- (B) P, R, U
Bromine is more selective.
Q, S, T
form
3°
halide as major and
P, R, U
form
2°
halide as major.
Which is not an aromatic compound?
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Pyridine
0%
Xylene
0%
Naphthalene
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Cyclohexane
Explanation
Cyclohexane is not aromatic since it does not have alternating double and single bonds between the C-atoms.
Rank the hydrogen atoms
(H_a, H_b, H_c)
in the following molecules according to their acidic strengths:
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a > b > c
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b > a > c
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b > c > a
0%
a > c > b
Explanation
Removal of
H_c
creates a negative charge on
sp^2
hybridized carbon which is highly unstable, due to which
H_c
will be least acidic.
Removal of
H_b
creates a negative charge which is stabilized by resonance due to conjugation with double bond, There is no such factor in case of
H_a
Hence the order of acid strength is
b>a>c
.
In this reaction
SbF_5
acts as:
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0%
an acid
0%
a base
0%
a nucleophile
0%
an electrophile
Explanation
As
F
is a good nucleophile, so it will attack on
SbF_5
, which is electrophilic in nature and act as electrophile.
What is the final product (B) of the sequence
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0%
0%
0%
In order to prepare 1- chloro-propane which of the following reactants can be employed?
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Propene and HCl in the presence of peroxide
0%
Propene and
Cl_2
followed by treatment with aq.KOH
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Propanol-1 and
SOCl_2
/pyridine
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Any of the above can be used
Explanation
Propanol-1 and
SOCl_{2}
with pyridine as a solvent can be used to prepare 1 - chloro - propane. The by products
SO_{2}
and
HCl
are both gaseous so they go off leaving the actual product.
(A) \, + \, Cl_2 \, \xrightarrow {hv} \, monochloro \, product
To maximise the yield of monochloro product in the above reaction?
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Cl_2
must be added in excess
0%
Reactant (A) must be added in excess
0%
Reaction must be carried out in dark
0%
Reaction must be carried out with equimolar mixture of
Cl_2
and A
Explanation
To maximize the yield of monochloro product the alkane/alkene/alkyne i.e. the hydrocarbon reactant has to be taken in excess. On the other hand, to maximize the yield of polychloro product, chlorine has to be taken in excess.
Choose the appropriate product for this reaction.
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0%
0%
0%
To carry our above conversion,
(A)
and
(B)
respectively, are :
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NaNH_{2}, Cl - CH_{2} - CH_{2} - CH_{2} - Br
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NaNH_{2}, F - CH_{2} - CH_{2} - CH_{2} - Br
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NaNH_{2}, I - CH_{2} - CH_{2} - CH_{2} - Br
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NaNH_{2}, I - CH_{2} - CH_{2} - CH_{2} - I
Explanation
Alkynes reacts with sodamides to form sodium acetylenes which are then converted to higher acetylenes by treating them with suitable alkyl halides. So the reaction goes as follows.
CH_3MgI
will give methane with:
Report Question
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CH_3CH_3
0%
CH_3-CH_2-NH_2
0%
CH_3-CO-CH_3
0%
all of these
Explanation
When the source of protons reacted to Grignard reagent it will definitely give respective alkane, therefore, the only molecule that will react with given Grignard reagent to give methane is ethaneamine.
In the reaction:
H-C\equiv CH\xrightarrow [ (i)NaNH_{ 2 }/Liq.NH_{ 3 } ]{ { (ii)CH_{ 3 }CH_{ 2 }Br } } X\xrightarrow [ (i)NaNH_{ 2 }/Liq.NH_{ 3 } ]{ { (ii)CH_{ 3 }CH_{ 2 }Br } } Y
X
and
Y
are:
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X=2
-butyne,
Y=2
-hexyne
0%
X=1
-butyne,
Y=2
-hexyne
0%
X=1
-butyne,
Y=3
-hexyne
0%
X=2
-butyne,
Y=3
-hexyne
Reaction
H-C\equiv C-H+HOCl\rightarrow
product, here product will be:
Report Question
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ClCH_2-CHO
0%
CHO-CHO
0%
CH-Cl=CHCl
0%
CHCl_2-CHCl_2
Explanation
when Acetylene reacts with hypochlorus acid the product formed is dichloroacetaldeyde.
Using anhydrous
AlCl_3
as catalyst, which one of the following reaction produces ethylbenzene
(PhEt)
?
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H_2C=CH_2+C_6H_6
0%
H_3C-CH_3+C_6H_6
0%
H_3C-CH_2OH+C_6H_6
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CH_3-CH=CH_2+C_6H_6
Explanation
R.E.F image
CH_{2} = CH_{2} \overset{AlCl_{3}/HCl}{\rightarrow} CH_{3}-C^{\oplus}H_{2}+AlCl_{4}^{\circleddash }
In the following sequence of reactions:
Toluene
\overset{KMnO_4}{\rightarrow} A\overset{SOCl_2}{\rightarrow} B\overset{H_2/Pd}{\underset{BaSO_4}{\rightarrow}}C
.
What is C?
Report Question
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Benzaldehyde
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Benzoic acid
0%
Benzoyl chloride
0%
Acetophenone
IUPAC name of the compound is:
Report Question
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butane
0%
\gamma
-Butane
0%
cyclo-Butane
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n-Butane
Explanation
Since, it is cyclic compound not acyclic straight chain, so 'cyclo' prefix is used for naming and name would be cyclo-butane.
How will you prepare ethane from methyl chloride?
Report Question
0%
Williamson's synthesis
0%
Wurtz synthesis
0%
Hydrogenation of methyl chloride
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Dehydrohalogenation of methyl chloride
Explanation
C_{2}H_{6}
will be lest peppered by ethane synthesis from
CH_{3}Cl
CH_{3}Cl +2Na\rightarrow CH_{3}^{-}Na^{+}+NaCl
CH_{3}^{-}Na^{+}+CH_{3}-Cl\rightarrow C_{2}H_{6}+NaCl
Which of the following compound, on reaction with excess of RMgX, gives 3 moles of alkane (R-H)?
Report Question
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HC \equiv C -{\underset{OH}{\underset{|}{C}}H} - COOH
0%
CH_3 - C \equiv C -{\underset{OH}{\underset{|}{C}}H} - COOH
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HC \equiv C -{\underset{OCH_3}{\underset{|}{C}}H} - COOH
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CH_2 = CH -{\underset{OH}{\underset{|}{C}}H} - COOH
Explanation
HC \equiv C - \underset{OH}{\underset{|}{C}H} - COOH \xrightarrow{3 R-MgX} \,^{\circleddash}C \equiv C - \underset{\,\,O^{\circleddash}}{\underset{|}{C}}\!\!H - COO^{\circleddash} + 3R-H
The barrier for rotation about indicated bonds will be maximum in which of these compounds?
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X
0%
Y
0%
Z
0%
Same in all
The compound
B
formed in the following sequences of reactions is:
CH_3CH_2CH_2OH\overset{PCl_3}{\rightarrow}A\overset{Alc. KOH}{\rightarrow}B
.
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propyne
0%
propene
0%
propanol
0%
propane
Explanation
CH_{3}CH_{2}CH_{2}OH\overset{PCl_{3}}{\rightarrow}\underset{propyl\ chloride}{CH_3CH_2CH_2Cl}\overset{alc.KOH}{\rightarrow}\underset{propene}{CH_3CH=CH_2}
In the first step the propyl alcohol undergo substitution reaction to form propyl chloride as a proudct which in presence of alc. KOH undergoes elimination reaction to form propene as a product.
Write chemical equations for the preparation of following alcohols in the options starting from appropriate alkenes.
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0%
0%
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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