Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry
Learn Hydrocarbons (Alkane, Alkene and Alkyne) MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
81.
Although benzene is highly unsaturated it does not undergo addition reactions. The explanation of this can be suggested as
A
$$\pi $$ - electrons of benzene ring are delocalised
B
since $$\pi $$ - electrons are present inside the ring, addition cannot take place
C
cyclic structures do not show addition reactions
D
benzene is not a reactive compound
Answer :
$$\pi $$ - electrons of benzene ring are delocalised
$$\pi $$ - electrons of benzene ring are delocalised throughout the molecule. This makes the molecule very stable. The stability resists breaking of double bonds for addition.
82.
Identify the reagent from the following list which can easily distinguish between 1-butyne and 2-butyne
A
bromine, $$CC{l_4}$$
B
$${H_2},$$ Lindlar catalyst
C
dilute $${H_2}S{O_4},HgS{O_4}$$
D
ammonical $$C{u_2}C{l_2}$$ solution
Answer :
ammonical $$C{u_2}C{l_2}$$ solution
Ammonical cuprous chloride gives precipitate (red) only with terminal alkynes, i.e. alkynes having acidic hydrogen atom, such as butyne-1 $$\left( {CH \equiv CC{H_2}C{H_3}} \right)$$
83.
Which of the following reactions of methane is incomplete combustion?
A
\[2C{{H}_{4}}+{{O}_{2}}\xrightarrow{Cu/523\,K/100\,atm}2C{{H}_{3}}OH\]
B
\[C{{H}_{4}}+{{O}_{2}}\xrightarrow{M{{o}_{2}}{{O}_{3}}}HCHO+{{H}_{2}}O\]
C
$$C{H_4} + {O_2} \to {C_{\left( s \right)}} + 2{H_2}{O_{\left( l \right)}}$$
D
$$C{H_4} + 2{O_2} \to C{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}}$$
Answer :
$$C{H_4} + {O_2} \to {C_{\left( s \right)}} + 2{H_2}{O_{\left( l \right)}}$$
$$C{H_4} + {O_2} \to {C_{\left( s \right)}} + 2{H_2}{O_{\left( l \right)}}$$ is incomplete combustion. In complete combustion, hydrocarbons release $$C{O_2}$$ and $${H_2}O$$ whereas in this reaction, $$C$$ and $${H_2}O$$ are released.
84.
Which of the following reactions is expected to readily give a hydrocarbon product in
good yields ?
A
\[RCOOK\xrightarrow[\text{oxidation}]{\text{Electrolytic}}\]
B
\[RCOOAg\xrightarrow{{{l}_{2}}}\]
C
\[C{{H}_{3}}C{{H}_{3}}\xrightarrow[h\nu .]{C{{l}_{2}}}\]
D
\[{{\left( C{{H}_{3}} \right)}_{2}}CCl\xrightarrow{{{C}_{2}}{{H}_{5}}OH}\]
Electrolysis of sodium or potassium salt of carboxylic acid gives good yield of hydrocarbon
\[2\,RCOOK\xrightarrow[\text{oxidation}]{\text{Electrolytic}}2\,\underset{\text{Anode}}{\mathop{RCO{{O}^{-}}}}\,+\underset{\text{Cathode}}{\mathop{2{{K}^{+}}}}\,\]
At anode $${\text{2}}\,{\text{RCO}}{{\text{O}}^ - } \to 2\,RCO\dot O + 2\,{e^ - }$$
$$2\,RCO\dot O \to R - R + 2C{O_2}$$
When $${O_3}$$ reacts with alkene, it forms ozonide, which on reaction with $$Zn$$ and acid or $${H_2}/Ni$$ gives aldehydes and/or ketones. These products helps in locating the position of a double bond as
86.
Isomers of hexane, based on their branching, can be divided
into three distinct classes as shown in the figure.
The correct order of their boiling point is
A
I > II > III
B
III > II > I
C
II > III > I
D
III > I > II
Answer :
III > II > I
Greater the extent of branching, lesser is the boiling point of the hydrocarbon, so order of b.p is III > II > I.
87.
Which of the following will exhibit aromatic character?
A
I, III
B
III, IV
C
II, IV
D
II, III
Answer :
II, IV
Since in azulene (IV) two rings share a double bond, therefore, two electrons of this double bond can be counted towards each. As a result, each ring has \[6\,\pi \] - electrons and hence azulene is aromatic. Pyrrole (II), is aromatic because it contains \[6\pi \] - electrons ( four from the two double bonds + one lone pair of electrons on the $$N$$ atom ).
88.
The two compounds $$A$$ and $$B$$ obtained from 1-butyne can be distinguished by
\[B\xleftarrow[\left( ii \right)\,{{H}_{2}}{{O}_{2}}/O{{H}^{-}}]{\left( i \right)\,B{{H}_{3}}}C{{H}_{3}}C{{H}_{2}}C\equiv CH\xrightarrow{{{H}^{+}}/H{{g}^{2+}}}A\]
A
$$NaHS{O_3}$$
B
$${\text{litmus solution}}$$
C
$${\text{iodoform test}}$$
D
$$2,4{\text{ - }}DNP$$
Answer :
$${\text{iodoform test}}$$
\[\underset{\text{(an aldehyde)}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CHO}}\,\xleftarrow[\left( ii \right)\,{{H}_{2}}{{O}_{2}}/O{{H}^{-}}]{\left( i \right)\,B{{H}_{3}}}C{{H}_{3}}C{{H}_{2}}C\equiv CH\xrightarrow{{{H}^{+}}/H{{g}^{2+}}}\underset{\text{(}A\,\,\text{ketone with}\,-COC{{H}_{3}}\,\,\text{group)}}{\mathop{C{{H}_{3}}C{{H}_{2}}COC{{H}_{3}}}}\,\]
89.
In the following sequence of reactions, the alkene affords region 3
the compound $$'B’$$
\[C{{H}_{3}}-CH=CH-C{{H}_{3}}\xrightarrow{{{O}_{3}}}A\xrightarrow[Zn]{{{H}_{2}}O}B.\]
The compound $$B$$ is
A
$$C{H_3}C{H_2}CHO$$
B
$$C{H_3}COC{H_3}$$
C
$$C{H_3}C{H_2}COC{H_3}$$
D
$$C{H_3}CHO$$
Answer :
$$C{H_3}CHO$$
Completing the sequence of given reactions,
Thus $$'B’$$ is $$C{H_3}CHO$$
Hence (D) is correct answer.
90.
Match the column I with column II and mark the appropriate choice.
Column I
Column II
a.
Alkyl halide + Sodium in presence of dry ether
1.
Sulphonation
b.
Arene + Acid halide in presence of $$AlC{l_3}$$
2.
Wurtz reaction
c.
Arene + Fuming sulphuric acid
3.
Catalytic hydrogenation
d.
Arene + Hydrogen in presence of $$Ni$$
4.
Friedel-Crafts reaction
A
a - 1, b - 3, c - 2, d - 4
B
a - 4, b - 2, c - 3, d - 1
C
a - 3, b - 1, c - 4, d - 2
D
a - 2, b - 4, c - 1, d - 3
Answer :
a - 2, b - 4, c - 1, d - 3
No explanation is given for this question. Let's discuss the answer together.