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CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 4 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Hydrocarbons
Quiz 4
Select the structure of the major product formed when 1-methylcyclopentane reacts with $$Br_2$$ in the presence of light.
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0%
0%
0%
Explanation
(Stable free radical at tertiary carbon)
Bromine forms a free radical in the presence of sunlight, then bromine abstracts a hydrogen to form most of the stable radical.
Bromination occurs selectively at the highest substituted carbon of alkane.
Intermediate free radical is most stable due to a tertiary carbon.
So it will be the major product.
$$CH_3I + xNa + CH_3I \longrightarrow CH_yCH_y + xNaI$$
Find the value of x and y :
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$$ x = 2 , y = 1$$
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$$ x = 2 , y = 2$$
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$$ x = 3 , y = 3$$
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$$ x = 2 , y = 3$$
Which one of the following compound is not formed in the 'catalytic cracking' of octane?
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Pentane
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Butene
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Propene
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Hydrogen gas
Explanation
Catalytic cracking of octane produces lower hydro carbons like pentane, butene and propene but not hydrogen gas.
Which of these steps are part mechanism for chlorination of ethane under free radical conditions?
$$I: CH_{3}CH_{3}+Cl^{*}\rightarrow CH_{3}CH_{2}Cl+H^{*}$$
$$II: CH_{3}CH_{3}+Cl^{*}\rightarrow CH_{3}CH_{2}^{*}+HCl$$
$$III: CH_{3}CH_{2}^{*}+Cl_{2}\rightarrow CH_{3}CH_{2}Cl+Cl^{*}$$
$$IV: 2Cl^{*}\rightarrow Cl_{2}$$
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$$III $$ and $$ IV$$
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$$I,III $$ and $$ IV$$
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$$I.II,III, IV,V $$
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$$II,III,IV$$
Explanation
Chlorination of ethane occurs in three steps:
Chain initiation:$$Cl_2\rightarrow Cl^{*}+Cl^{*}$$.
Chain propagation:
$$
CH_{3}CH_{3}+Cl^{*}\rightarrow CH_{3}CH_{2}^{*}+HCl$$
$$CH_{3}CH_{2}^{*}+Cl_{2}\rightarrow CH_{3}CH_{2}Cl+Cl^{*}$$.
Chain termination:$$2Cl^{*}\rightarrow Cl_{2}$$.
$$II$$ and $$ III$$ are propagation steps while $$IV$$ is one of the termination step.
Hence options A & D are correct.
What is(are) true about free radical chlorination of a alkane ?
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Taking a large excess of alkane over chlorine minimizes the chances of dechlorination and polychlorination.
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Propagation is preferred over termination because the radicals are highly reactive.
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Major product is always formed at the carbon where the most stable free radical is formed in the intermidiate step.
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Reaction is usually carried out in aqueous medium.
Explanation
a)Taking large excess of alkane increases the probability of H-abstaction from alkane rather than product alkyl halide in the propagation step,hence minimises the chance of dichlorination
b) Progagation is preferred over termination because free radicals are highly reactive, immediately reacts rather than recombine in termination steps.
Hence options A & B are correct.
What is(are) the expected product(s) in free radical chlorination of propane?
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Hexane
0%
2-methyl pentane
0%
2,3-dimethyl butane
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2,2-dimethyl butane
Explanation
As shown in above mechanism, three types of products are possible as a result of radical additions. So option (A), (B) and (C) are all possible.
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is wrong but reason is correct.
Explanation
Bromine is much less reactive but more selective
than chlorine in radical halogenation
.
It is because the bromine atom has a significant $$E_{act}$$ in the first step of propagation.
The greater the difference between activation energies, the larger the selectivity.
S
o the reaction is much more controllable and selective.
Which of the following statement regarding free radical halogenation of alkane is not true?
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Hydrogen abstraction by halogen radical in the propagation step is exothermic in both chlorination and bromination
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Hydrogen abstraction by halogen radical in the propagation step is exothermic in cholorination but endothermic in bromination
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A single halogen radical may bring about halogenation to thousands of alkane molecule
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Continuous source of energy is required to continue the halogenation reaction
Explanation
Hydrogen abstraction by halogen radical in the propagation step is exothermic in chlorination but endothermic in bromination. Hence, the option (A) is an incorrect statement.
Which of the following is(are) true statement regarding free redical bromination of an alkane?
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It occur repidly but very less selectively
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It occur with difficulty as compared to chlorination but a highly selective reaction
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Transition state fot H abstraction in propagation step is closer to reactant than product in terms of potential energy
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Transition state fot H abstraction in propagation step is closer to product than reactant in terms of potential energy
Explanation
b) Bromination is less reactive but highly selective
d) Bromination product is less stable, comes closer to transition state in terms of potential energy.
The bromination process requires $$AlBr_{3}$$ as a catalyst. Whereas in chlorination there is no need of such catalyst, Also the bromination forms only one product unlike chlorination where the mixture of a number of product is obtained, So, bromination is very selective.
The bromination product are not as stable as chorination product as their energies are very high.
Which is the best way of preparing propene from chloroethene?
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$$CH_{2}=CHCl+CH_{3}Cl \xrightarrow[Heat]{Na/Ether}$$
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$$CH_{2}=CHCl+(CH_{3})_{2}CuLi\rightarrow$$
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$$CH_{3}Cl+(CH_{2}=CHCl)_{2}CuLi\rightarrow$$
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$$CH_{2} =CHCl+CH_{3} Li\rightarrow$$
Explanation
The reaction of option C is the best way of preparing propene from chloroethene
$$\displaystyle S_{N} 2$$ reactions occur most efficiently with $$\displaystyle CH_{3} Cl$$. Methyl chloride reacts with lithium dichloroethenyl cuprate
to form propene
$$\displaystyle CH_{3}Cl+(CH_{2}=CHCl)_{2}CuLi\rightarrow$$ $$\displaystyle CH_3-CH=CH_2$$
What is the major organic product of the following reaction?
Ethyl cyclohexane $$ + Br_{2}/hv \rightarrow$$
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0%
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0%
Explanation
In case of bromination of the most stable carbocation is formed and then bromine is added. this is major product in case of bromination
Tertiary hydrogen is removed to form most stable carbocation and then bromine is added to give major organic product above.
What is(are) true regarding free radical iodination of the alkane?
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It occur very rapidly because of very small value of bond energy of $$I_{2}$$
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Direct iodination of alkane with $$I_{2}$$ in presence of light is impractical
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Iodination of alkane can be achieved successfully using an oxidizing agent catalyst
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Presence of some $$HCI$$ active the free radical iodination
Explanation
The statements (B) and (C) are true regarding free radical iodination of the alkane.
(B) Direct iodination of alkane with iodine in presence of light is impractical.
During this reaction, $$Hl$$, a strong reducing agent is obtained as byproduct, which catalyzes the reverse reaction thereby preventing direct iodination of alkane.
(C) Iodination of alkane can be achieved successfully using an oxidizing agent catalyst.
Presence of an oxidizing agent oxidizes unwanted byproduct $$Hl$$ and enables iodination of alkane to proceed.
Which of the following will produce a pair of diastereomers on free radical chlorination giving a dichloro derivative?
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1-chloropropane
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1-chlorobutane
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2-chlorobutane
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2-chloro-2,4-dimethyl pentane
Explanation
1-chloropropane will produce a pair of diastereomers on free radical chlorination giving a dichloro derivative.
Diastereomers are non superimposable and not mirror images of each other.
Which compound will be most easily eliminated to give an alkyne?
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0%
0%
0%
Explanation
The alkenyl halide undergoes elimination reaction in the presence of a strong base to form an alkyne. For the reaction, the reactant haloalkene should have the double bond and halogen on the same carbon with an ionizable hydrogen at $$\beta$$ position.
According to this, compound $$C$$ and $$D$$ are applicable to form an alkyne via elimination. But compound $$C$$ after elimination forms strained cyclopentyne ring which is unstable thus chloride from compound $$D$$ can most easily eliminated to form an alkyne .
Even though the rings are different size, cyclopropane and cyclobutane have similar overall strain energy. This is because:
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they have similar angle compression from the optimal $$\displaystyle 109.5^{\circ}$$
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strain energy is a kinetic phenomenon
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cyclobutane has more eclipsing interactions but less angle strain
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cyclopropane has a larger overall heat of combustion than cyclobutane
Explanation
Even though the rings are different size,cyclopropane and cyclobutane have similar overall strain energy. This is because cyclobutane has more eclipsing interactions
than cyclopropane
but less angle strain.
trans-3, 6-Dimethyl oct-4-ene (A) exists in two diastereomers (I) and (II). The total number of stereoisomers for (A) is :
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6
0%
8
0%
4
0%
5
Explanation
Two asymmetric C atoms with the same terminal group (n = number of asymmetric C atom and is even) Number of O. A. I. = $$2^{n\, -\, 1}$$ = 2
Number of mcso forms = $$2^{(n\, -\, 2)/2}$$ = I
Total optical isomers = 3
One double bond = 2 disteromers
$$Total stereoisomers\, =\, 3\, \times\, 2\, =\, 6$$
An organic compound 'A’ ($$C_8H_6$$) reacts with dilute $$H_2SO_4$$ and $$HgSO_4$$ giving ‘B’ which reacts
with $$NaOH$$ and $$I_2$$ forming compounds ‘C’ and ‘D’. ‘D’ upon reaction with soda lime gives
hydrocarbon ‘E’ which on reaction with acetyl chloride in presence of anhydrous aluminum
chloride gives ‘B’.
The compound 'B' would be:
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Acetophenone
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Phenyl acetylene
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Iodoform
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Potassium benzoate
The percentage of isomers formed during monobromination of $$2, 3-dimethyl$$ butane at room temperature relative to $$1, 2, 3$$ H atoms to chlorination is $$(1.0 : 3.8 : 5.0)$$
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81%
0%
56%
0%
15%
0%
22%
Explanation
(A) is obtained from the reaction of $$12$$ equivalent $$1$$ H atoms.
(B) is obtained from the reaction of two equivalent $$3$$ H atoms.
Percentage of (A) $$ = \cfrac{12\times 100}{33}\, = 54.5%$$
Percentage of (B) $$ = \cfrac{10\times 100}{22}\, = 45.5$$
$$12\times 1 = 12.0$$
$$2\times 5 = 10.0$$
Total $$= 22.0$$
$$Trans-3,6-dimethyl oct-4-ene$$ (A) exists in two diastereomers (I) and (II). Which statement is true about (I) and (II)? If the stereochemistry about the double bond in (A) is cis-, the total number of stereoisomers for (A) is:
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2, pair of enatiomers
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3, one pair of enantiomers and one meso form (optically inactive due to the presence of centre of symmetry)
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3, one pair of enantiomers and one meso form (Optically inactive due to the presence of plane of symmetry)
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2, both optically inactive due to the presence of both centre and plane of symmetry.
Explanation
Trans-3, 6-Dimethyl oct-4-ene (A) exists in two diastereomers (I) and (II). (I) is optically active and (II) is not as it contains a center of symmetry.
If compound (a) is cis, we get two diastereomers, (III) and (IV). (III) is meso (optically inactive) due to plane of symmetry. (IV) is O.A. (two enantiomers).
Total stereoisomers when (A) is cis = 3.
A fuel with high octane number has less tendency to knock, whereas furl with high cetane number has more tendency to knock in an automobile engine.
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True
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False
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Ambiguous
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None of these
Explanation
Octane number is the percentage of isooctane in the mixture of iso-octane and n-heptane. Octane number is for gasoline (petrol) fuel. A gasoline with octane number of 80 means 80% of isooctane and 20% of n-heptane, i.e., branched - chain alkane is more, so knocking is less.
Branched-chain hydrocarbons burn more smoothly to form more stable, less reactive 3 radicals, whereas straight-chain hydrocarbons (e.g., n-heptane form less stable, more reactive 1 and 2 radicals).
In contrast, cetane number is used for diesel fuel. It is defined as the percentage of cetane in mixture of cetane and $$\alpha$$ -methyl naphthalene. A diesel fuel with cetane number 80 means 80% of cetane $$(C_{16}H_{34})$$ (a straight-chain hydrocarbon) and 20% of $$\alpha$$ - methyl naphalene (aromatic compound). So, it will not burn smoothly and produce igh knocking. But straight-chain hydrocarbones, e.g., cetane, ignite spontaneously in comparison to aromatic compounds. So, cetane number determines the ignition of fuel. Higher cetane number means fuel will ignite faster but will produce higher knocking. Lower cetane number means fuel will ignite slowly and will produce lesser knocking (eating sound).
Which of the following is not a cumulated diene?
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$$Hexa-1,2-diene$$
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$$Hexa-2,3-diene$$
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$$Penta-2,3-diene$$
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$$Penta-1,3-diene$$
Which of the following compounds would have the same vapour pressure ?
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(I) and (II)
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(II) and (III)
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(III) and (IV)
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(II) and (IV)
Explanation
(III) and (IV) are enantiomers and have the same physical and chemical properties. So they will have the same vapour pressure.
Give the IUPAC name of above given compound.
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2, 2-Dimethyl-3-propyl-4-isopropyl heptane
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4-lsopropyl-5-t-butyl octane
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4-t-Butyl-5-isopropyl octane
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2-Methyl-3-propyl-4-isopropyl heptane
Outline the reaction sequence for the conversion of ethene to ethyne (the number of steps should not be more than two).
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$$A: HOCl $$ B: KOH
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A: Bromine water B: Diethyl amine
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A: Cl$$_2,\,h\nu$$ B: Ca(OH)$$_2$$
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A: Bromine water $$B: KOH$$
Explanation
With bromine water the alkene undergoes bromination and form $$Dibromoethane$$ that on reaction with $$KOH$$ forms $$ethyne$$ by dehydrohalogenation reaction.
Which of the following has the highest boiling point?
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$$2- Methylpentane$$
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$$2, 3 - Dimethyl butane$$
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$$2, 2 - Dimethyl butane$$
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Cannot be determined
Explanation
2 - Methyl pentane has the highest boiling point because the boiling point in alkane depends upon the chains of atoms and the molecular mass of the molecules. Straight-chain molecules have higher boiling point than the branched-chain compounds.
This is because intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. The larger the surface area, the more opportunity for such attractions to exist and therefore a stronger force. Thus boiling point decreases with branching.
The boiling point order is as follows :
$$2 - Methyl pentane > 2, 3 - Dimethyl butane > 2, 2 - Dimethyl butane$$
Option A is correct.
Identify the product (A):
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0%
0%
0%
All of the above
Explanation
In this reaction, the compound follows free radical mechanism where the formation of $$2^0$$ free radical takes place. This free radical undergoes bromination reaction to form corresponding alkyl bromide which is shown option C.
An automobile engine fuel has cetane number ofWhich of the following statements is/are true ?
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Fuel contains 80% of $$\alpha$$-methyl naphthalene and 20% of $$C_{16}H_{34}$$
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Fuel contains 80% of cetane and 20% of $$\alpha$$-methyl naphthalene.
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Knocking property of the given fuel compared to the knocking property of a fuel with cetane number of 90 is high.
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Cetane number determines the quality of diesel fuel in terms of ignition properties.
Explanation
a. Wrong (by the definition of cetane number).
b. True (by the definition of cetane number).
c. Wrong (higher cetane number means higher percentage of straight-chain hydrocarbon
(
C
16
H
32
)
.
whose knocking property is high.
Therefor, cetane number of 90 hs high knocking property (makes more rattling sound).
d. True (cetane number determines the quality of diesel fuel in terms of spontaneous ignition).
Higher cetane number means that fuel ignites faster but makes more rattling sound (i.e., more knocking).
Hence options B & D are correct.
Vicinal dihalides undergo double dehydrohalogenation to give terminal alkyne. How many moles of $$NaNH_2$$ are used in the overall reaction ?
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One
0%
Two
0%
Three
0%
Four
Explanation
In the first step, one mole $$ NaNH_2$$ is the base in an elimination reaction to give the alkenyl bromide.
In the second reaction, likewise a second equivalent of $$NaNH_2$$ performs a second elimination reaction to form the alkyne.
For a terminal alkyne, any excess $$NaNH_2$$ will remove the acidic hydrogen from terminal $$C-H$$ and give the alkynyl anion. So if a terminal alkyne is formed, three equivalents of $$NaNH_2$$ will be consumed; the alkyne is protonated upon workup, usually by adding water.
Pyrolysis of ethyl acetate gives:
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$$CH_3COCH_3$$
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$$CH_2=CH_2$$
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$$CH_2=CO=O$$
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$$CH_3-CHO$$
Explanation
Pyrolysis of ethyl acetate gives ethylene $$\displaystyle (CH_2=CH_2)$$
$$\displaystyle CH_3COOC_2H_5 \xrightarrow [Pyrolysis] {Heat} CH_2=CH_2+ CH_3COOH$$
Ethyl acetate is heated in presence of liquid nitrogen and glass wool
Methane does not react with chlorine in dark.
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True
0%
False
Explanation
i. $$CH_{4}$$ does not react with $$Cl_{2}$$ in dark because the reaction proceeds via free-radical mechanism, Cl (chlorine radicals) are obtained from $$Cl_{2}$$ in the presence of light.
ii. It is in accordance with Markovnikov's rule which predicts the stability of $$2^{\circ} C^{\oplus}$$ over $$1^{\circ} C^{\oplus}$$.
What effect does branching of an alkane chain has on its melting point ?
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Increases its melting point
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Decreases its melting point
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doesn't change its melting point
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Cannot be determined
Explanation
The melting point of alkanes with even number of carbon atoms is higher than that of the odd numbered from the immediate next lower alkanes. For example,
C3
H8
is 85.9 K while that of
C6
H14
is 179 K. Since the packing is symmetrical in even numbered alkanes, they have higher melting point.
Therefore, neopentane with a spherical molecule has the highest melting point out of the three isomers due to more compact packing.
The melting point order is
(i) > (ii) > (iii)
(256.4 K > 143.3 K > 113.1 K)
The boiling point order is
(ii) > (iii) > (i)
(309 K > 300.9 K > 282.5 K)
There is no ring strain in cyclohexane, but cyclobutane has an angle strin of $$9^\circ\, {44}'.$$ If $$\Delta H_c^\circ$$ of cyclohexane per $$(CH_2)$$ group is 660 kJ $$mol^{-1}$$ and $$\Delta H_c^\circ$$ of cyclobutane is 2744 kJ $$mol^{-1}$$, what is the ring strain in kJ $$mol^{-1}$$ of cyclobutane ?
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0%
- 104
0%
104
0%
- 2084
0%
2084
Explanation
The $$\Delta H_c^\circ$$ value of cyclobutane is given. From this if we substract the $$\Delta H_c^\circ$$ value of cyclobutane calculated when there is no angle strain, we will get the value of ring strain of cyclobutane.
The $$\Delta H_c^\circ$$ value of cyclobutane when there is no ring strain can be calculated from $$\Delta H_c^\circ$$ value of cyclohexane per $$CH_2$$ group.
Ring strain = ($$\Delta H_c^\circ$$ of cyclobutane - $$4\, \times\, \Delta H_c^\circ$$ of per $$(CH_2)$$ group of cyclohexane) = 2744 - ($$4\, \times\, 660$$) = 104 kJ $$mol^{-1}$$
Which of the following statements is/are correct ?
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The product is
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0%
The product (B) is favourable because the strain of both three and four membered rigns is relieved.
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The product (C) is favourable because the strain of four-membered ring in relieved.
Explanation
The product (B) is favourable because the strain of both three and four-membered ring is relieved.
Which one of the following compounds gives methane on treatment with water?
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$${Al_4C_3}$$
0%
$${CaC_2}$$
0%
VC
0%
SiC
Explanation
$${Al_4C_3}$$
gives methane on treatment with water
$${Al_4C_3+12 H_2O\rightarrow 4Al(OH)_3+3CH_4}$$.
Hence option A is correct.
Which of the following has the highest knocking property?
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Olefins
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Straigth-chain paraffins
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Branched-chain paraffins
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Aromatic hydrocarbons
Explanation
An antiknock agent is a gasoline additive used to reduce engine knocking and increase the fuel's octane rating by raising the temperature and pressure at which auto-ignition occurs.
Since, straight chain alkane has minimum octane number. Hence, it produces maximum knocking.
Straight Chain paraffins have the highest knocking property.
Option B is correct.
Out of the five isomeric hexanes, The isomer that can give two monochlorinated compounds is:
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2,3-Dimenthyl butane
0%
2,2-Dimenthyl butane
0%
2-Dimenthyl pentane
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n-Hexane
Explanation
The isomeric hexane should have two different types of H atoms and four similar types of H atoms to gives two monochlorinated products.
In 2,3-dimethyl butane, two carbons have tertiary $$H$$ atoms are the same and at four primary carbons rest of $$H$$ are the same. So, two position are available for chlorination.
The decreasing order of the acidic character of the following is:
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$$(I) > (II) > (III) > (IV)$$
0%
$$(I) > (III) > (II) > (IV)$$
0%
$$(IV) > (III) > (II) > (I)$$
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$$(IV) > (II) > (III) > (I)$$
Explanation
The decreasing order of the acidic character of the following is (I) > (III) > (II) > (IV)
Cycloalkanes are more acidic than the corresponding alkanes.
Cyclopropane is most acidic and n butane is least acidic.
In general alkanes are inert towards acids and bases due to non polar nature of C-C and C-H bonds and their inability to donate or accept electrons.
The decreasing order of boiling points of the following compounds is
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(I) > (II) > (III) > (IV)
0%
(IV) > (III) > (II) > (I)
0%
(III) > (IV) > (I) > (II)
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(IV) > (III) > (II) < (I)
Explanation
The decreasing order of boiling points of the following compounds is (III) > (IV) > (I) > (II).
As the molecular weight of alkanes decreases, the boiling point decreases.
The boiling point of alkanes is lower than the boiling point of corresponding cycloalkanes.
The highest boiling point is expected for:
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Isooctane
0%
n-Octane
0%
2,2,3,3-Tetramethyl butane
0%
n-Butane
Explanation
Hint: Straight-chain exacerbate large surface area and more van der Waals forces of attraction which causes high boiling points.
Correct Answer: B
Explanation for correct option :
Boiling point is characterized as the specific temperature at which the vapour pressure of the fluid is equivalent to the environmental pressure.
Among the given options
n-Octane is a straight-chain exacerbate that has a very large surface area. Along these lines, there are more van der Waals forces of attraction, bringing about high boiling points.
Option B is correct.
Explanation for incorrect options:
The increase of surface area increases the ability of individual molecules to attract each other. Branching in molecules decreases the surface area thereby decreasing the attractive force between individual molecules. As a result, the boiling point decreases.
Hence, all other options are wrong.
State whether the given statement is true or false :
In benzene, carbon uses all the three $$p-orbitals$$ for hybridization.
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0%
True
0%
False
Which content(s) of middle oil separate on cooling?
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0%
Naphthalene
0%
Phenol
0%
Benzene
0%
Pyridine
Explanation
The content of middle oil separate on cooling is naphthalene.
Hence the correct option is A.
The IUPAC name of the above structure is:
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0%
3-methyl-3-isopropyl hexane
0%
3-isopropyl-3-methyl hexane
0%
3-ethyl-2,3-dimethyl hexane
0%
2,3-dimethyl-3-ethyl hexane
Which of the following hydrocarbons will undergo
substitution reactions?
(i) $$C_2 H_6$$ (ii) $$C_4 H_6$$ (iii) $$C_3 H_6$$
(iv) $$CH_4$$ (v) $$C_4 H_{10}$$ (vi) $$C_3 H_4$$
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0%
(i) and (iv) only
0%
(ii), (iii) and (vi) only
0%
(i), (iv) and (v) only
0%
(iii) and (v) only
Explanation
As we know,
alkanes undergo substitution reaction while alkenes undergo addition reactions so
(i), (iv) and (v) will undergo substitution reaction as they are alkanes.
The decreasing order of melting points of the following compounds is:
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(I) > (II) > (III) > (IV)
0%
(IV) > (III) > (II) > (I)
0%
(II) > (I) > (IV) > (III)
0%
(III) > (IV) > (I) > (II)
Explanation
For alkanes, melting point increases as molecular weight increases due to increase in Van der Waal's forces of attractions.
However, Cycloalkanes, due to their compact nature, have high boiling and melting points than the corresponding alkanes.
Hence, the correct option is $$\text{A}$$
2-Methyl butane reacting with $$Br_2$$ in sunlight mainly gives:
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1-Bromo-2-methyl butane
0%
2-Bromo-2-methyl butane
0%
2-Bromo-3-methyl butane
0%
1-Bromo-3-methyl butane
Explanation
Hence, the correct option is B
A hydrocarbon (A) was found to have vapour densityIt forms only single mono chlorosubstitution product.Suggest (A).
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0%
2,2-dimethylpropane
0%
n-pentane
0%
isobutane
0%
None of these
Explanation
$$Molecular\quad wt.=2\times vapour\quad density=2\times 36=72$$
For an alkane, compound formula will be $${ C }_{ n }{ H }_{ 2n+2 }$$
$$Molecular\quad wt.=12n+(2n+2)\times 1\quad =\quad 14n+2$$
$$\Longrightarrow 14n+2=72\\ \Longrightarrow n=5$$
Compound formula is
$${ C }_{ 5 }{ H }_{ 12 }$$
Since only one type of monochloro product is formed after monochlorination, only one type of hydrogen must be present.
2,2-dimethyl propane has only primary hydrogens.
Hence, A is the correct option.
Propene is allowed to react with B$$_2$$ D$$_6$$ and the product is treated with acetic acid. The final product obtained is:
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0%
1-deuteriopropane
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2-deuteriopropane
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1-deuteriopropene
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2-deuteriopropene
Explanation
this is an example of hydroboration reaction (followed by hydrolysis) in which diborane is isotopically labeled.
A molecule of $$HD$$ is added across the double bond of propene to form 2-deuteriopropane.
$$CH_2CHCH_3 \xrightarrow [CH_3COOH] {B_2D_6} CH_3-CHDCH_3$$
Statement-I : Neopentane forms only one monochlorinated product.
because
Statement-II : Neopentane has four identical methyl group attached to a quaternary carbon.
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Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I
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Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I
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Statement-I is True, Statement-II is False
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Statement-I is False, Statement-II is True
Explanation
Neopentane forms only one monochlorinated product because Neopentane has four identical methyl group attached to a quaternary carbon.
Statement-I: Iodination of alkanes is carried out in the presence of iodic acid.
Because
Statement-II: Iodic acid removes $$I_2$$ gas from the reaction mixture.
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Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I
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Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I
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Statement-I is True, Statement-II is False
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Statement-I is False, Statement-II is True
Explanation
Iodination of alkanes is carried out in the presence of iodic acid.
The iodination reaction is a reversible reaction.
HI is a strong reducing agent and it reduces alkyl iodide back to alkane.
$$R-H +I_2 \rightleftharpoons R-I + HI$$
To prevent this, some oxidizing agents such as mercuric oxide (HgO), iodic acid ($$HIO_3$$) or dil. nitric acid
($$HNO_3$$) are used which decompose HI acid.
$$5R-H +2I_2 + HIO_3 \rightarrow 5R-I + 3H_2O$$.
Hence the correct option is C.
The above compound undergoes elimination on heating to yield which of the following products?
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Explanation
The products obtained when the quaternary ammonium hydroxide is heated are as shown.
This is an example of Hofmann elimination.
(i) Methyl group cannot be converted into alkene.
(ii) The less substituted alkene (Hofmann product) is the main product rather than the more substituted alkene (saytzeff product).
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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