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CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 2 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Hydrocarbons
Quiz 2
Reduction of benzene is presence of
N
i
N
i
gives:
Report Question
0%
cyclohexane
0%
heterocyclic
0%
alicyclic
0%
unsaturated Aliphatic
Explanation
Hydrogenation of benzene is also possible using a palladium catalyst or a rhodium
(
R
h
)
(
R
h
)
catalyst or
N
i
N
i
catalyst and cyclohexane forms as a product.
Alkane which is liquid at room temperature is:
Report Question
0%
C
2
H
6
C
2
H
6
0%
C
3
H
8
C
3
H
8
0%
C
4
H
10
C
4
H
10
0%
C
8
H
18
C
8
H
18
Explanation
Alkane from
C
1
−
C
4
C
1
−
C
4
are gases.
Alkane from
C
5
−
C
17
C
5
−
C
17
are liquids.
Alkane from
C
18
C
18
onwards is solids.
Hence, alkane
C
8
H
18
C
8
H
18
is liquid at room temperature and the correct option is (D).
Which of the following is the most stable cycloalkane?
Report Question
0%
Cyclopropane
0%
Cyclobutane
0%
Cyclopentane
0%
Cyclohexane
Explanation
Explanation:
Six membered rings are the most stable. As the ring size decreases the stability decreases. This stability of the six-membered ring is due to the chair conformation of cyclohexane as shown in the figure.
Step 1: Identification of ring size in each option
A) Cyclopropane: three-membered ring.
B) Cyclobutane: four-membered ring.
C) Cyclopentane: five-membered ring.
D) Cyclohexane: Six membered ring.
Step 2: So stability order will be
D
>
C
>
B
>
A
D
>
C
>
B
>
A
Final answer: Cyclohexane
Hence, option
D
D
is correct.
C
H
2
=
C
H
2
+
H
2
N
i
/
300
0
C
→
C
H
3
−
C
H
3
C
H
2
=
C
H
2
+
H
2
N
i
/
300
0
C
−
−−−−−
→
C
H
3
−
C
H
3
The above reaction is known as:
Report Question
0%
Kolbe's
0%
Wurtz
0%
Sabatier - Senderens
0%
Berthelot
Explanation
Sabatier - Senderens
R
−
C
H
=
C
H
2
+
H
2
N
i
/
Δ
→
R
−
C
H
2
−
C
H
3
R
−
C
H
=
C
H
2
+
H
2
N
i
/
Δ
−
−−
→
R
−
C
H
2
−
C
H
3
Berthelot reaction
The reaction of ammonia with phenol hypochlorite to give indophenol.
C
C
is correct
On heating sodium propionate with sodium hydroxide and quicklime, the gas evolved is :
Report Question
0%
C
2
H
2
C
2
H
2
0%
C
2
H
6
C
2
H
6
0%
C
H
4
C
H
4
0%
C
2
H
4
C
2
H
4
Explanation
C
H
3
C
H
2
C
O
O
N
a
+
N
a
O
H
C
a
O
→
C
2
H
6
+
N
a
2
C
O
3
C
H
3
C
H
2
C
O
O
N
a
+
N
a
O
H
C
a
O
−
−
→
C
2
H
6
+
N
a
2
C
O
3
On heating sodium propanoate with sodium hydroxide and quicklime, the gas evolved is ethane.
Option B is correct.
Which of the following compounds is formed, when n-hexane is heated at
500
0
C
500
0
C
and 10 - 20 atm in presence of
C
r
2
O
3
C
r
2
O
3
?
Report Question
0%
Cyclohexane
0%
Benzene
0%
2-Methyl pentane
0%
Cyclobutane
Explanation
Aromatisation: Conversion of aliphatic compounds into aromatic compound is known as aromatisation. In this process
C
r
2
O
3
C
r
2
O
3
is used as catalyst under high temperature and pressure.
C
H
3
−
C
H
2
−
C
H
2
−
C
H
2
−
C
H
2
−
C
H
3
C
r
2
O
3
a
t
10
−
20
a
t
m
→
C
6
H
6
C
H
3
−
C
H
2
−
C
H
2
−
C
H
2
−
C
H
2
−
C
H
3
C
r
2
O
3
a
t
10
−
20
a
t
m
−
−−−−−−−−−−
→
C
6
H
6
Order of rate of reaction of ethane with halogens is:
Report Question
0%
I
>
B
r
>
C
l
>
F
I
>
B
r
>
C
l
>
F
0%
C
l
>
I
>
B
r
>
F
C
l
>
I
>
B
r
>
F
0%
C
l
>
B
r
>
F
>
I
C
l
>
B
r
>
F
>
I
0%
F
>
C
l
>
B
r
>
I
F
>
C
l
>
B
r
>
I
Explanation
The reactivity of the halogens decreases in the following order:
F
2
>
C
l
2
>
B
r
2
>
I
2
F
2
>
C
l
2
>
B
r
2
>
I
2
Fluorine is so explosively reactive it is difficult to control, and iodine is generally unreactive.
Chlorination and bromination are normally exothermic.
Chlorination of ethane in presence of diffused light is an example of:
Report Question
0%
dehydration
0%
elimination reaction
0%
substitution reaction
0%
addition reaction
Explanation
Alkanes participate in substitution reactions like halogenation, nitration.
Halogenation reaction:
C
2
H
6
C
l
2
/
h
ν
→
C
2
H
5
C
l
+
H
C
l
C
2
H
6
C
l
2
/
h
ν
−
−−−
→
C
2
H
5
C
l
+
H
C
l
H
H
is substituted by
C
l
C
l
. It is a substitution reaction.
Option C is correct.
Which of the following alkane has the least boiling point?
Report Question
0%
n-butane
0%
n-pentane
0%
isopentane
0%
neopentane
Explanation
This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface area. Therefore, the intermolecular attractive forces which depend upon the surface area, also become small in magnitude on account of branching. Consequently, the boiling points of the branched-chain alkanes are less than the straight-chain isomers.
Torsional strain in eclipsed conformations of a molecule is due to:
Report Question
0%
repulsion between the eclipsed carbons
0%
repulsion between aligned electron pairs of the eclipsed bond
0%
repulsions betwen aligned ‘H’
0%
repulsions between C-H bonds of eclipsed conformation
Explanation
The eclipsed bonds are parallel to each other facing repulsions due to aligned electron pairs in the bonds.
The hybridisation of carbon atoms in ethane doesn’t
change during :
Report Question
0%
pyrolysis
0%
combustion
0%
halogenation
0%
both halogenation and combustion
Explanation
A. Pyrolysis : The pyrolysis product of ethane is ethylene. In ethylene molecule, the carbon atom is
s
p
2
s
p
2
hybridized.
B. Combustion : In combustion reaction, ethane is turned into carbon dioxide, which is also
s
p
2
s
p
2
hybridized.
C. Halogenation : The halogenation reaction is a replacement type of reaction where one hydrogen atom of ethane gets replaced by a halogen atom. So, the hybridisation state does not change during the reaction. So, the correct answer is (C).
Statement- 1: Alkanes participate in substitution reactions.
Statement- 2: Alkanes are saturated hydrocarbons.
Report Question
0%
Both statement- 1 and statement-2 are correct statement-2 is a correct explanation of statement-1.
0%
Both statement-1 and statement-2 are correct statement-2 is not a correct explanation of statement-1.
0%
Statement-1 is correct , statement-2 is incorrect.
0%
Statement-1 is incorrect , statement-2 is correct.
Explanation
statement-1: Alkanes participate in substitution reactions.
The example includes free radical halogenation reaction in which H atom of an alkane is replaced with a halogen atom.
C
H
3
C
H
3
C
l
2
→
l
i
g
h
t
C
H
3
C
H
2
C
l
C
H
3
C
H
3
C
l
2
−
−
→
l
i
g
h
t
C
H
3
C
H
2
C
l
statement-2: Alkanes are saturated hydrocarbons.
Saturated hydrocarbons are hydrocarbons that contain only single bonds between carbon atoms. They are called saturated because each carbon atom is bonded to as many hydrogen atoms as possible.
In other words, the carbon atoms are saturated with hydrogen
Thus both statement-1 and statement-2 are correct statement-2 is a correct explanation of statement-1.
The deviation of bond angle in cyclopropane:
Report Question
0%
49
0
.5
′
0%
24
0
.44
′
0%
−
5
0
.16
′
0%
−
9
0
.33
′
Explanation
Deviation of the bond angle in cyclopropane is 109.5 - 60
=
49.5
Identify A and B in the following reaction sequence:
C
2
H
6
A
→
C
2
H
5
C
l
Z
n
/
T
H
F
→
B
Report Question
0%
C
l
2
/
u
v
light and
C
4
H
10
0%
P
C
l
3
and
C
2
H
4
0%
H
C
l
and
C
2
H
6
0%
C
l
2
and
C
2
H
2
Explanation
The reactions are as follows:
Halogenation reaction:
C
2
H
6
C
l
2
/
h
v
→
C
2
H
5
C
l
+
H
C
l
Frankland's reaction:
C
2
H
5
C
l
+
Z
n
+
C
2
H
5
C
l
→
C
2
H
5
−
C
2
H
5
+
Z
n
C
l
2
Option A is correct.
When cyclohexane is poured on water, it floats
because cyclohexane is :
Report Question
0%
In boat form
0%
In chair form
0%
In crown form
0%
Less dense than water
Alkanes can be iodinated in the presence of:
Report Question
0%
H
I
0%
I
2
and P
0%
H
I
O
3
0%
P
I
3
Explanation
This is because iodination is reversible and thus needs a reducing agent to drive the reaction to forward direction according to Le Chatelier's principle.
The reaction is as follow:
C
H
4
+
I
2
↔
C
H
3
I
+
H
I
The HI formed in a product makes the reaction reversible. Thus,
H
I
O
3
is added which reduce HI to
I
2
and stop the reversible reaction.
H
I
O
3
+
5
H
I
→
3
H
2
O
+
3
I
2
Option C is correct.
In which of the following pair both angle strain and torsional strain is present?
Report Question
0%
Cyclopropane, Cyclobutane
0%
Cyclopropane, Cyclohexane
0%
Cyclobutane, Cyclohexane
0%
Cyclohexane, Cyclopentane
Explanation
3-Membered and 4-membered cycloalkane rings are the least stable containing angle strain(due to large deviation from tetrahedral angle) as well as torsional strain (due to electronic repulsions between the bonds), lying close to each other.
A single substitution of H atom in an alkane of molar mass by 72g / mole on chlorination produces only one product. The alkane is :
Report Question
0%
n- pentane
0%
2- methyl butane
0%
2,2-dimethyl propane
0%
n- butane
Explanation
In 2,2-Dimethylpropane the central carbon atom is quaternary,
meaning all four of its valencies are satisfied by Carbon atoms; which
implies that there are no Hydrogen atoms which can be substituted by
Chlorine(or halogen), and all the remaining 4 carbon atoms are
identical, so substitution on any one of those will result in the same
compound.
The most strained cycloalkane is :
Report Question
0%
cyclopropane
0%
cyclobutane
0%
cyclopentane
0%
cyclohexane
Explanation
In cyclopropane,the C-C-C bond angles are 60° whereas tetrahedral 109.5° bond angles are expected.
The intense angle strain leads to nonlinear orbital overlap of its
s
p
3
orbitals.
Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.
The H-C-H bond angle is 115° whereas 106° is expected as in the
C
H
2
groups of propane.
Option A is correct.
Which of the following statement(s) is/are not correct?
Report Question
0%
Alkanes are non polar in nature
0%
Alkanes are soluble in non-polar solvents
0%
Straight chain alkanes have lesser boiling point than corresponding branched chain isomers
0%
Alkanes exhibit alternations in melting point
Explanation
Alkanes are made of C & H only, so, they are non polar in nature and also due to non polar nature they dissolved in non polar solvent only.
Branched alkanes have lower boiling point than straight-chain alkanes.
Melting point increases with increase in mass.
Option C is correct.
The structure of 1,3-dicyclohexyl propane is:
Report Question
0%
0%
0%
0%
Explanation
1,3-dicyclohexyl propane contains 2 cyclohexyl ring attached with propane molecule at carbon no. 1 & 3. So option B is correct.
Condition for maximum yield of
C
2
H
5
C
l
is:
Report Question
0%
C
2
H
6
(
E
x
c
e
s
s
)
+
C
l
2
U
V
l
i
g
h
t
→
0%
C
2
H
6
+
C
l
2
→
R
.
T
0%
C
2
H
6
+
C
l
2
(
E
x
c
e
s
s
)
U
V
l
i
g
h
t
→
0%
C
2
H
6
+
C
l
2
→
D
a
r
k
Explanation
Halogenation reaction:
C
2
H
6
C
l
2
/
h
v
→
C
2
H
5
C
l
Bond angle in planar ring of cyclopropane is
Report Question
0%
90
0
0%
108
0
0%
60
0
0%
120
0
Explanation
Arrangement of the three carbon atoms of the cyclopropane ring is of the form of an equilateral triangle thus bonds angles are 60 degrees each.
Bromination of an alkane as compared to chlorination proceeds _____________.
Report Question
0%
at a slower rate
0%
at a faster rate
0%
with equal rate
0%
with equal rate (or) different rate depending upon the temperature
Explanation
Bromination of an alkane occurs at a slower rate than chlorination of an alkane. As we know that halogenation involves the formation of a carbon-halogen bond. Now, the ease of halogenation will certainly depend on the carbon-halogen bond strength.
Bond formation energy of
C
−
C
l
is higher than that of
C
−
B
r
bond. So, the ease of bromination is lower than the ease of chlorination.
So, the correct answer is option (A).
2.84 gms of methyl iodide was completely converted into
C
H
3
M
g
I
and the product was decomposed by the excess of ethanol. The volume of the gaseous hydrocarbon produced at NTP
will be_________.
Report Question
0%
22.4 lit.
0%
2240 lit.
0%
0.448 lit.
0%
224 lit.
Explanation
In this reaction, one mole of
Grignard reagent reacts with one mole of alcohol to produce hydrocarbon.
C
H
3
I
→
C
H
3
M
g
I
+
C
2
H
5
O
H
→
C
H
4
+
C
2
H
5
O
M
g
I
Mass of methyl iodide = 2.84 gm (given)
No. of mole of methyl iodide = 2.84/142 =0.02
So hydrocarbon produced = 0.02 mole
So the volume of produced hydrocarbon = 0.448 lit
(one mole produced 22.4 lit).
Option C is correct.
Which one of the following is reduced with
Z
n
−
H
g
and
H
C
l
to give alkane ?
Report Question
0%
Ethyl acetate
0%
Acetic acid
0%
Acetamide
0%
Butan-2-one
The reaction
C
12
H
26
→
C
6
H
12
+
C
6
H
14
represents :
Report Question
0%
Substitution
0%
synthesis
0%
Cracking
0%
Addition.
Explanation
Hydrocarbon cracking is the process of breaking a long-chain of hydrocarbons into short ones.
The compound obtained by the reaction of methanol and
C
H
3
M
g
X
is :
Report Question
0%
propanone
0%
methane
0%
ethane
0%
ethanol
Explanation
C
H
3
M
g
X
is the grignard reagent.
Grignard reagent first of all reacts with the active hydrogen if it is present on the reactant after that it further goes for the substitution reaction.
Active hydrogen is the hydrogen which is attached with high electronegative elements like oxygen, nitrogen, chlorine.
For the given reaction:
C
H
3
M
g
X
+
C
H
3
O
H
→
C
H
4
+
C
H
3
O
M
g
X
Which of the following will produce methane on hydrolysis?
Report Question
0%
B
e
C
2
0%
A
l
3
C
4
0%
A
l
4
C
3
0%
B
e
2
C
Explanation
B
e
C
2
a
n
d
A
l
3
C
4
not even exist.
These two carbides are called methanides. They usually prepare methane on hydrolysis.
H
2
C
=
C
H
2
+
H
2
N
i
/
300
0
C
→
C
2
H
6
to get back the ethene, ethane must be:
Report Question
0%
dehydrogenated in the presence of
C
u
at
300
0
C
0%
heated to
450
0
C
in the presence of
H
N
O
3
vapoar
0%
heated
170
0
C
in the presence of conc.
H
2
S
O
4
0%
heated to
750
0
C
Explanation
Alkanes formed from alkenes, can be reverted back to alkenes by catalytic cracking or thermal decomposition, when heated in the absence of air, gives a mixture of ethylene and hydrogen.
Cleavage of C-H bonds: This type of fission results in the formation of alkenes. It is catalyzed by the oxides of chromium, vanadium, molybdenum.
On halogenation, an alkane (
C
5
H
12
) gives only one mono halogenated product. The alkane is:
Report Question
0%
n- pentane
0%
2- methyl butane
0%
2,2-dimethyl propane
0%
None of these
Which of the following compounds will form a hydrocarbon on reaction with a Grignard reagent?
Report Question
0%
Ethanol
0%
Ethanal
0%
Propanone
0%
None of the above
Explanation
A Grignard reagent reacts with compounds containing active H atom, such as water, ammonia, amines, alcohols, alkynes etc., to form a hydrocarbon.
It does not form hydrocarbon with ethanal and propanone.
R
M
g
X
+
R
′
O
−
H
→
R
−
H
+
M
g
X
O
R
′
Option A is correct.
Methane can not be prepared by :
Report Question
0%
Wurtz reaction
0%
Decarboxylation
0%
Kolbes reaction
0%
Sabatier - senderens reaction
Explanation
Hint: The formula of methane is
C
H
4
. It has a single carbon atom.
Step 1: Wurtz Reaction
Wurtz reaction is-
C
H
3
C
l
+
2
N
a
+
C
H
3
C
l
→
C
2
H
6
+
2
N
a
C
l
Here, the product will have more than one carbon. So, it can not form methane.
Step 2: Decarboxylation
In a decarboxylation reaction, a carboxylic acid undergoes decarboxylation to form an alkane.
C
H
3
C
O
O
H
→
C
H
4
+
C
O
2
So, here methane can be produced.
Step 3: Kolbe reaction
In Kolbe reaction,
2
moles of the potassium salt of ester react with water to form alkane, carbon dioxide, hydrogen gas.
2
C
H
3
C
O
O
K
+
H
2
O
→
C
2
H
6
+
2
K
O
H
+
2
C
O
2
+
H
2
So, here methane can not be produced.
Step 4: Sabatier - senderens reaction
Sabatier senderens reaction is-
C
n
H
2
n
+
H
2
→
C
n
H
2
n
+
2
+
h
e
a
t
Here, an alkene gets converted into an alkane that has the same number of carbon atoms. So it can not produce an alkane with only one carbon as alkenes have two or more carbons.
Final Step: Correct options (A),(C), and (D).
2-Methyl butane on reacting with bromine in presence of sunlight gives mainly :
Report Question
0%
1-Bromo-2-methyl butane
0%
2-Bromo-2-methyl butane
0%
2-Bromo-3-methyl butane
0%
1-Bromo-3-methyl butane
Explanation
C
H
3
C
H
(
C
H
3
)
C
H
2
C
H
3
+
B
r
2
s
u
n
l
i
g
h
t
⟶
H
3
C
C
(
B
r
)
(
C
H
3
)
C
H
2
C
H
3
The reaction proceeds via the free radical mechanism.
Since the radical formed in this case is tertiary, as it is more stable than primary and secondary, 2-Bromo-2-methyl butane is formed.
Option B is correct.
The IUPAC name of given compound is :
Report Question
0%
2 - methyl propane
0%
iso butane
0%
n-butane
0%
propane
Explanation
The parent hydrocarbon contains 3 carbon atoms. Its name is propane.
A methyl group is present at second carbon atom.
Hence, the IUPAC name of the compound is 2 - methyl propane.
The
C
−
C
bond length of the following molecules is in the order:
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0%
C
H
2
>
C
2
H
4
>
C
6
H
6
<
C
2
H
2
0%
C
2
H
2
<
C
2
H
4
<
C
6
H
6
<
C
2
H
6
0%
C
2
H
6
<
C
2
H
2
<
C
6
H
6
<
C
2
H
4
0%
C
2
H
4
<
C
2
H
6
<
C
2
H
2
<
C
6
H
6
Explanation
Bond length is related to bond order, when more electrons participate in bond formation the bond is shorter. By approximation, the bond lengths of two different atoms are the sum of the two individual covalent radii.
The bond length of carbon-carbon triple bond is shorter than carbon-carbon double bond and is shorter than resonance bonding in benzene and is shorter than carbon-carbon single bond.
Therefore,
C
2
H
2
<
C
2
H
4
<
C
6
H
6
<
C
2
H
6
Difference in the molecular formulae of
D
and
C
is:
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0%
C
2
H
3
0%
C
2
H
4
0%
C
H
2
0%
C
3
H
6
Explanation
R
M
g
X
M
e
t
h
a
n
o
l
→
C
H
3
C
H
3
A
1
m
o
l
e
o
f
c
h
l
o
r
i
n
e
→
h
ν
C
H
3
C
H
2
−
C
l
B
C
H
3
C
H
2
−
C
l
B
Z
n
−
C
u
→
E
t
h
a
n
o
l
C
H
3
C
H
3
C
C
H
3
C
H
2
−
C
l
B
N
a
/
d
r
y
e
t
h
e
r
→
C
H
3
C
H
2
C
H
2
C
H
3
D
Difference in the molecular formulae of
D
and
C
is
C
2
H
4
.
IUPAC name of neo-pentane is
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0%
2-ethylpentane
0%
2,2-dimethylpentane
0%
2,2-dimethylpropane
0%
2-methylpropane
Explanation
Structure of neo-pentane is as shown in the figure and its IUPAC name is 2,2-dimethylpropane.
IUPAC name is
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0%
3, 4-dimethyl -3- propylnonane
0%
4-ethyl -4, 5-dimethyldecane
0%
6, 7-dimethyl -7-ethylnonane
0%
6, 7-dimethyl-1-7-ethyldecane
Explanation
The IUPAC name of the given compound is 4-ethyl -4, 5-di methyl decane.
The parent hydrocarbon contains 10 carbon atoms and is called decane.
An ethyl group is present on the fourth carbon atom and two methyl groups are present on fourth and fifth carbon atoms.
Which of the following has the maximum
C
−
H
bond length?
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0%
C
2
H
4
0%
C
2
H
2
0%
C
2
H
6
0%
C
6
H
6
Explanation
Ethane
C
2
H
6
has maximum
C
−
H
bond length.
Bond length decreases with the increase in s character since an s-orbital is smaller than a
p
−
orbital.
Compound
Hybridization
C
−
H
bond length (
A
o
)
Alkane
s
p
3
1.093
Alkene
s
p
2
1.087
Alkyne
s
p
1.057
The ratio of sigma and pi bonds in benzene is :
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0%
3
:
1
0%
4
:
1
0%
2
:
1
0%
1
:
1
Explanation
The molecular formula of benzene is
C
6
H
6
.
Benzene molecule contains 12
σ
and 3
π
bonds.
Hence, the ratio is
4
:
1
.
The IUPAC name of the given structure:
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0%
2,2 - dimethylbutane
0%
isohexane
0%
2,3 - dimethylbutane
0%
di-isohexane
Explanation
⇒
methyl substituent at C-2 and C-3
So, IUPAC name of the given structure is 2,3-dimethylbutane
The IUPAC name of the given compound is:
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0%
3 - ethylpentane
0%
1-ethylpentane
0%
2-ethylpentane
0%
3-methylpentane
Explanation
3-Methylpentane is a branched-chain alkane with the molecular formula
C
6
H
1
4
It is a structural isomer of hexane composed of a methyl group bonded to the third carbon atom in a pentane chain.
⇒
1 methyl substituent c-3.
so the name of given compound will be 3-methyl Pentane
The ratio of pi to sigma bonds in benzene is:
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0%
3 : 1
0%
1 : 6
0%
1 : 4
0%
1 : 2
Explanation
no of
π
Bonds (c-c)=3
no of
σ
Bonds (c-c)=6
no of
σ
bonds (c-h)=6
total
σ
bonds=6+6=12
ratio of
π
and
σ
=
3
12
=
1
:
4
Which of the following is the chain propagation step in chlorination of methane?
Report Question
0%
C
H
3
+
C
l
→
C
H
3
C
l
0%
C
H
3
+
C
l
−
C
l
→
C
H
3
−
C
l
+
C
l
0%
C
l
−
C
l
→
C
l
+
C
l
0%
C
l
+
C
l
→
C
l
2
Explanation
Chain initiation
C
l
2
h
2
→
˙
C
l
+
C
˙
l
chain propagation:
C
H
3
+
C
l
−
C
l
→
C
H
3
C
l
+
C
˙
l
C
H
3
C
l
+
C
l
−
C
l
→
C
H
2
C
l
+
C
˙
l
C
H
2
C
l
2
+
C
l
−
C
l
→
C
H
C
l
3
+
C
˙
l
C
H
C
l
3
+
C
l
−
C
l
→
C
C
l
4
+
C
˙
l
The IUPAC name of the given compound is :
Report Question
0%
3, 4, 4-trimethylheptane
0%
3, 4, 4- trimethyloctane
0%
2-butyl-2-methyl-3-ethylbutane
0%
2-ethyl-3,3-dimethylheptane
What is the IUPAC name of the given compound?
Report Question
0%
2-Methylbutane
0%
3-Methylbutane
0%
Pentane
0%
Isopentane
Explanation
Methyl substituent at 2nd carbon.
The chain is of four carbon.
IUPAC name of the given compound is 2-methyl butane.
Hence, option
A
is correct.
The
C
−
H
bond distance is the longest in:
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0%
C
2
H
6
0%
C
2
H
2
B
r
2
0%
C
2
H
4
0%
C
2
H
2
Explanation
⇒
Higher the
s
−
character means that the Bonded pair electrons are held more closely, ensuring a smaller Bond length and a stronger bond.
So,
s
p
has the least bond length and it is the strongest
s
p
3
has longest bond length and weakest.
So,
A
is the correct answer.
The IUPAC name of the given structure is :
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0%
tertiarybutane
0%
2,2 - dimethylpropane
0%
neopentane
0%
neobutane
Explanation
⇒
2 methyl groups are substituted at c-2
2-2 dimethylpropane.
The correct IUPAC name of
C
H
3
−
C
H
2
−
C
H
(
C
H
3
)
−
C
H
(
C
2
H
5
)
2
is:
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0%
4-ethyl-3-methylhexane
0%
3-ethyl-4-methylhexane
0%
4-methyl-3-ethylhexane
0%
2,4 - diethylpentane
Explanation
Solution:
List of some rules for nomenclature
First we have to find out the long chain and name it.
Then look for the side chains and functional groups.
If the chains of equal length are engaging for selection of the parent chain, the selection goes in the series to:
(i) the greatest number of side chains having by the parent chain.
(ii) the chain whose substituent have the lowest numbers.
(iii) the greatest number of carbon atoms in smaller side chain having by the chain.
(iv) the least branched side chains having by the chain.
The IUPAC name of the compound
C
H
3
−
C
H
2
−
C
H
(
C
H
3
)
−
C
H
(
C
2
H
5
)
2
is 3-ethyl-4-methylhexane.
Hence, the correct option is (B).
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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