Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry
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71.
Which of the following steps is not correct in the mechanism of electrophilic substitution of benzene?
A
Generation of electrophile like $${X^ + },{R^ + },NO_2^ + ,$$ etc.
B
Attack of electrophile resulting in the formation of arenium ion in which one of the carbon is $$s{p^3}$$ hybridised.
C
Addition of proton on benzene ring to give carbocation.
D
Removal of proton from $$s{p^3}$$ carbon atom to restore aromatic character.
Answer :
Addition of proton on benzene ring to give carbocation.
Arenium ion is formed as intermediate by the attack of electrophile not by addition of proton. Proton is substituted by electrophile.
72.
Name the products of the following reactions.
(I) $${C_6}{H_6}$$ reacts with methyl chloride in presence of $$AlC{l_3}.$$
(II) $${C_6}{H_6}$$ reacts with acetyl chloride in presence of $$AlC{l_3}.$$
(III) $${C_6}{H_6}$$ reacts with fuming nitric acid in presence of cone. $${H_2}S{O_4}.$$
(IV) $${C_6}{H_6}$$ is catalytically hydrogenated.
I
II
III
IV
(a)
Chloromethane
Toluene
Nitrobenzene
$$n$$ - Hexane
(b)
Methylbenzene
Chlorobenzene
Phenylnitrite
Trimethylbenzene
(c)
Benzyl chloride
Trimethylbenzene
Trinitrotoluene
Toluene
(d)
Toluene
Acetophenone
Trinitrobenzene
Cyclohexane
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(d)
73.
Choose the correct comparison of heat of hydrogenation for the following alkenes.
A
II < IV < III < V < I
B
III < IV < I < V < II
C
V < IV < III < I < II
D
IV < V < I < III < II
Answer :
V < IV < III < I < II
Greater is the stability of alkene, lower the heat of hydrogenation.
74.
A mixture of 1-iodoethane and 1-iodopropane is treated with sodium metal and dry ether to carry out Wurtz reaction. Which of the following hydrocarbons will be formed?
75.
In the following the most stable conformation of $$n$$ - butane is
A
B
C
D
Answer :
Key Idea The conformation in which the heavier groups are present at maximum possible distances, so that the forces of repulsion get weak, is more stable. Among the given conformations of $$n$$ - butane, the conformation shown in option (B), ie. anti conformation is most stable as in it the bulkier group, i.e. $$ - C{H_3}$$ group are present at maximum possible distance and get lower energy.
76.
Ozonolysis products of 2-pentyne after decomposition of ozonide with water and subsequent oxidation are
A
ethanoic acid and propanoic acid
B
ethanoic acid and propanone
C
ethanoic acid
D
formic acid and glyoxal
Answer :
ethanoic acid and propanoic acid
77.
Which of the following will yield a mixture of 2-chlorobutene and 3-chlorobutene on treatment with $$HCl ?$$
A
B
C
D
Answer :
78.
The addition of $$HBr$$ to 1-butene gives a mixture of products (I), (II) and (III) :
$$\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - Br}\limits_{\left( {{\text{III}}} \right)} $$
The mixture consists of
A
(I) and (II) as major and (III) as minor products
B
(II) as major, (I) and (III) as minor products
C
(II) as minor, (I) and (III) as major products
D
(I) and (II) as minor and (III) as major products
Answer :
(I) and (II) as major and (III) as minor products
\[\underset{\text{1-Butene}}{\mathop{C{{H}_{3}}C{{H}_{2}}-CH=C{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix}
\text{Markovnikov}\,\text{ }\!\!'\!\!\text{ }\,\text{s} \\
\text{addition}
\end{smallmatrix}]{HBr}\] \[\underset{\begin{smallmatrix}
\text{(exists in 2 enantiomers I and II)} \\
\text{(Major product) }
\end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
Br\,\,\,\,\,
\end{smallmatrix}}{\overset{*\,\,\,\,\,}{\mathop{-CH-}}}\,C{{H}_{3}}}}\,+\] \[\underset{\text{III (Minor product)}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br}}\,\]
79.
The maximum number of isomers for an alkene with the molecular formula $${C_4}{H_8}$$ is
A
2
B
3
C
4
D
5
Answer :
4
Four isomers
80.
Chlorination of alkanes is a photochemical process. It is initiated by the process of
A
heterolysis
B
homolysis
C
pyrolysis
D
hydrolysis
Answer :
homolysis
Photochemical chlorination of alkanes is a free radical process and the reaction is initiated by homolysis of halogens to form free radicals.