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.
51.
Which of the following alkynes can be identified and distinguished from the rest of the alkynes on reaction with ammoniacal silver nitrate?
A
$$C{H_3}C \equiv C - C{H_3}$$
B
$$C{H_3}C{H_2}C \equiv CH$$
C
$$C{H_3}C{H_2}C \equiv CC{H_3}$$
D
$$C{H_3}C \equiv CC{H_2}C{H_2}C{H_3}$$
Answer :
$$C{H_3}C{H_2}C \equiv CH$$
Only terminal alkynes react with ammoniacal silver nitrate to give white precipitate of silver acetylides.
52.
The number of optically active products obtained from the
complete ozonolysis of the given compound is :
A
0
B
1
C
2
D
4
Answer :
0
53.
Which compound would give 5 - keto - 2 - methylhexanal ipon ozonolysis ?
A
B
C
D
Answer :
When 1, 3 - dimethylcyclopentene is heated with ozone and then with zinc and acetic acid, oxidative cleavage leads to keto - aldehyde.
54.
A compound $$\left( X \right)\left( {{C_5}{H_8}} \right)$$ reacts with ammonical $$AgN{O_3}$$ to give a white precipitate, and on oxidation with hot alkaline $$KMn{O_4}$$ gives the acid, $${\left( {C{H_3}} \right)_2}CHCOOH,$$ therefore $$X$$ is –
A
$$C{H_2} = CH - CH = CH - C{H_3}$$
B
$$C{H_3} - CH = CH - C{H_2} - C{H_3}$$
C
$${\left( {C{H_3}} \right)_2}CH - C \equiv CH$$
D
$${\left( {C{H_3}} \right)_2}C = C = C{H_2}$$
Answer :
$${\left( {C{H_3}} \right)_2}CH - C \equiv CH$$
Compound $$X$$ reacts with ammonical $$AgN{O_3}$$ solution, so it must be a terminal alkyne. Formation of $${\left( {C{H_3}} \right)_2}CHCOOH$$ on oxidation of $$X$$ with hot alkaline $$KMn{O_4}$$ further confirms that $$X$$ is $${\left( {C{H_3}} \right)_2}CHC \equiv CH.$$
55.
What is the product when acetylene reacts with the formation of
hypochlorous acid ?
A
$$C{H_3}COCl$$
B
$$ClC{H_2}CHO$$
C
$$C{l_2}CHCHO$$
D
$$ClCHCOOH$$
Answer :
$$C{l_2}CHCHO$$
56.
The product$$(s)$$ obtained via oxymercuration $$\left( {HgS{O_4} + {H_2}S{O_4}} \right)$$ of 1- butyne would be
A
\[C{{H}_{3}}-C{{H}_{2}}\overset{\begin{smallmatrix}
O \\
\parallel
\end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]
B
$$C{H_3} - C{H_2} - C{H_2} - CHO$$
C
$$C{H_3} - C{H_2} - CHO + HCHO$$
D
$$C{H_3}C{H_2}COOH + HCOOH$$
Answer :
\[C{{H}_{3}}-C{{H}_{2}}\overset{\begin{smallmatrix}
O \\
\parallel
\end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]
TIPS/Formulae :
Hydration of alkynes via mercuration takes place in
accordance with Markovnikov's manner rule
\[C{H_3}C{H_2}C = CH\xrightarrow[{HgS{O_4}/{H_2}S{O_4}}]{{ + 2{H_2}O}}\] \[\left[ C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix}
| \\
\,\,\,OH
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,\,\,\,OH \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{3}} \right]\xrightarrow{-{{H}_{2}}O}\] \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix}
\parallel \\
O
\end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]
57.
Ozonolysis of 2, 3-dimethylbut-1-ene followed by reduction with zinc and water gives
A
methanal and hexanoic acid
B
methanoic acid and butanone
C
methanal and 3-methylbutan-2-one
D
butanoic acid and 2, 3-dimethylbutanoic acid
Answer :
methanal and 3-methylbutan-2-one
58.
Arrange the following in decreasing order of their boiling points
(i) $$n$$ - Butane
(ii) 2 - Methylbutane
(iii) $$n$$ - Pentane
(iv) 2, 2 - Dimethylpropane
A
(i) > (ii) > (iii) > (iv)
B
(ii) > (iii) > (iv) > (i)
C
(iv) > (iii) > (ii) > (i)
D
(iii) > (ii) > (iv) > (i)
Answer :
(iii) > (ii) > (iv) > (i)
Boiling point of alkanes increases with increase in molecular mass and for the same alkane the boiling point decreases with branching.
Thus, the decreasing order of their boiling points is :
$$\mathop {n{\text{ - Pentane }}}\limits_{\left( {{\text{iii}}} \right)} {\text{ > }}\mathop {{\text{2 - Methylbutane}}}\limits_{\left( {{\text{ii}}} \right)} {\text{ }}$$ $${\text{ > }}\mathop {{\text{2, 2 - Dimethylpropane }}}\limits_{\left( {{\text{iv}}} \right)} $$ $${\text{ > }}\mathop {n{\text{ - Butane }}}\limits_{\left( {\text{i}} \right)} $$
59.
$$HOCl$$ reacts on 3-methyl-2-pentene, the main product will be :
A
B
C
D
Answer :
The reaction follows Markownikoff rule
60.
Benzene easily shows
A
ring fission reactions since it is unstable
B
addition reactions since it is unsaturated
C
electrophilic substitution reactions due to stable ring and high $$\pi $$ electron density
D
nucleophilic substitution reactions due to stable ring and minimum electron density.
Answer :
electrophilic substitution reactions due to stable ring and high $$\pi $$ electron density
No explanation is given for this question. Let's discuss the answer together.