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.
181.
Reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the
order
A
tertiary > primary > secondary
B
primary > secondary > tertiary
C
Both (A) and (B)
D
tertiary > secondary > primary
Answer :
tertiary > secondary > primary
Tertiary alkanes are more reactive as they form tertiary free radical which is more stable. On the other hand, primary alkanes are less reactive because they form $${1^ \circ }$$ free radicals which are less stable.
182.
Fill in the blanks with appropriate words.
Benzene has a planar structure. All carbon atoms in benzene are $$\underline {\,\,{\text{I}}\,\,} $$ hybridised. The ring structure of benzene was proposed by $$\underline {\,\,{\text{II}}\,\,} .$$ It shows $$\underline {\,\,{\text{III}}\,\,} $$ substitution reactions. It reacts with $$\underline {\,\,{\text{IV}}\,\,} $$ in presence of aluminium chloride to form acetophenone.
I
II
III
IV
(a)
$$s{p^2}$$
Kekule
electrophilic
acetyl chloride
(b)
$$sp$$
Dewar
nucleophilic
chloromethane
(c)
$$s{p^3}$$
Ladenberg
electrophilic
chloroethane
(d)
$$s{p^2}$$
Baeyer
nucleophilic
methyl bromide
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(a)
No explanation is given for this question. Let's discuss the answer together.
183.
Given,
The enthalpy of hydrogenation of these compounds will be in the order as
A
I > II > III
B
III > II > I
C
II > III > I
D
II > I > III
Answer :
III > II > I
The enthalpy of hydrogenation of given compounds is inversely proportional to stability of
alkene.
Hence, correct order is III > II > I
184.
The highest boiling point is expected for :
A
iso-octane
B
$$n$$ - octane
C
2, 2, 3, 3 - tetramethylbutane
D
$$n$$ - butane
Answer :
$$n$$ - octane
For isomeric alkanes, th one having longest straight chain has highest b.p. because of larger surface area.
185.
1-Bromo-3-chlorocyclobutane is treated with two equivalents of $$Na,$$ in the presence of ether. Which of the following compounds will be formed?
A
B
C
D
Answer :
Since alkyl bromides are more reactive than alkyl chlorides, therefore, intramolecular Wurtz reaction occurs on the side of $$Br$$ atom.
186.
The alkene $$R - CH = C{H_2}$$ reacts readily with $${B_2}{H_6}$$ and formed the product $$B$$ which on oxidation with alkaline $${H_2}{O_2}$$ produces
A
B
C
D
Answer :
Here, $$\frac{1}{2}\,mole$$ of $${B_2}{H_6}$$ react with alkene by Markownikoff's addition and form trialkylborone called Hydroboration, $$\frac{{{H_2}{O_2}}}{{O{H^ - }}}$$ gives oxidation. So, trialkyborone oxidise in alcohols and reaction is also called Hydroboration-oxidation.
\[3R-CH=C{{H}_{2}}\xrightarrow{{{B}_{2}}{{H}_{6}}}\] \[\underset{B}{\mathop{{{\left( R-C{{H}_{2}}-C{{H}_{2}} \right)}_{3}}}}\,-B\xrightarrow{\frac{{{H}_{2}}{{O}_{2}}}{{{H}^{+}}}}\] $$3R - C{H_2} - C{H_2} - OH$$
187.
Which step is chain propagation step in the following mechanism?
\[\begin{align}
& \left( \text{i} \right)C{{l}_{2}}\xrightarrow{h\upsilon }\dot{C}l+\dot{C}l \\
& \left( \text{ii} \right)\dot{C}l+C{{H}_{4}}\to \dot{C}{{H}_{3}}+HCl \\
& \left( \text{iii} \right)\dot{C}l+\dot{C}l\to C{{l}_{2}} \\
& \left( \text{iv} \right)\dot{C}{{H}_{3}}+\dot{C}l\to C{{H}_{3}}Cl \\
\end{align}\]
A
(i)
B
(ii)
C
(iii)
D
(iv)
Answer :
(ii)
No explanation is given for this question. Let's discuss the answer together.
188.
\[R-C{{H}_{2}}-CC{{l}_{2}}-R\xrightarrow{\text{Reagent}}\] $$R - C \equiv C - R.$$ The reagent is
A
$$Na$$
B
$$HCl\,\,{\text{in}}\,\,{H_2}O$$
C
$$KOH\,\,{\text{in}}\,\,{C_2}{H_5}OH$$
D
$$Zn\,\,{\text{in alcohol}}$$
Answer :
$$KOH\,\,{\text{in}}\,\,{C_2}{H_5}OH$$
Dehydrohalogenation reaction,
In presence of ethanolic $$KOH,$$ substrate gives elimination reaction.
189.
Identify $$(A), (B)$$ and $$(C)$$ in the following sequence of reactions. \[\xrightarrow[\begin{smallmatrix}
{{H}_{2}}S{{O}_{4}} \\
\left( {{50}^{\circ }}C \right)
\end{smallmatrix}]{HN{{O}_{3}}}\left( A \right)\xrightarrow[FeB{{r}_{3}}]{B{{r}_{2}}}\left( B \right)\] \[\xrightarrow[HCl]{Sn}\left( C \right)\]