Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry
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261.
Which of the following compounds will not undergoes Friedel-Craft’s reaction easily?
A
Cumene
B
Xylene
C
Nitrobenzene
D
Toluene
Answer :
Nitrobenzene
Nitro group being electron withdrawing, deactivates the benzene nucleus to such an extent. that an electrophile cannot attack on benzene ring easily due to deactivation of benzene ring. Hence becomes incapable to give Friedel-Craft’s reaction. NOTE
Nitrobenzene because of its unreactivity towards Friedel-Craft's reaction is used as a solvent for this reaction.
262.
In the presence of peroxide, $$HCl$$ and $$HI$$ do not give anti-Markownikoff’s addition of alkenes because :
A
One of the steps is endothermic in $$HCl$$ and $$HI$$
B
Both $$HCl$$ and $$HI$$ are strong acids
C
$$HCl$$ is oxidizing and the $$HI$$ is reducing
D
All the steps are exothermic is $$HCl$$ and $$HI$$
Answer :
One of the steps is endothermic in $$HCl$$ and $$HI$$
Anti-Markownikoff addition is possible only in case of $$HBr$$ and not in $$HCl$$ and $$HI.$$ In $$HBr$$ both the chain initiation and propagation steps are exothermic, while in $$HCl,$$ first step is exothermic, and second step is endothermic and in $$HI,$$ no step is exothermic. Hence $$HCl$$ and $$HI$$ do not undergo anti-Markownikoff’s addition.
263.
The reaction of propene with $$HOCl$$ proceeds via the addition
of
A
$${H^ + }$$ in the first step
B
$$C{l^ + }$$ in the first step
C
$$O{H^ - }$$ in the first step
D
$$C{l^ + }$$ and $$O{H^ - }$$ in a single step
Answer :
$$C{l^ + }$$ in the first step
TIPS/Formulae :
Alkenes undergo electrophilic addition reactions.
$$HOCl$$ undergoes self - ionization
$$\eqalign{
& \left( {HOCl + HOCl \to {H_2}O + OC{l^ - } + C{l^ + }} \right) \cr
& {\text{to}}\,{\text{give}}\,{H_2}{O^ + } + OC{l^ - } + C{l^ + }. \cr} $$
So, it is the $${C{l^ + }}$$ that attacks in the first step.
264.
Identify $$A, B$$ and $$C$$ in following reactions :
A
$$A = Ni,\,B = {H_2}O{\text{ (liquid),}}\,C = {H_2}O$$
B
$$A = Zn,\,B = {H_2}O{\text{ (steam),}}\,C = LiAl{H_4}$$
C
$$A = Mg,\,B = {H_2}O\,{\text{(liquid),}}\,C = HCl$$
D
$$A = Sn,\,B = {H_2}O\,({\text{boiling),}}\,C = SnC{l_2}$$
No explanation is given for this question. Let's discuss the answer together.
265.
Identify the product, P in the following reaction :
$$C{H_3} - CH = C{H_2} + NOCl \to P$$
A
B
C
D
Answer :
Nitrosyl chloride adds on olefins according to
Markovnikof’s rule, where $$N{O^ + }$$ constitutes the positive
part of the addendum.
266.
In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the specie
A
$$X$$
B
$${X^ - }$$
C
$${X^ + }$$
D
$$\mathop X\limits^{\,\, \bullet } $$
Answer :
$${X^ + }$$
No explanation is given for this question. Let's discuss the answer together.
267.
Consider the following sequence of reactions
\[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[700K]{C{{l}_{2}}}A\xrightarrow[420K,12\,atm]{N{{a}_{2}}C{{O}_{3}}}B\]
Compound $$'B'$$ is
A
B
C
D
Answer :
268. \[+\,Cl-C{{H}_{2}}C{{H}_{2}}-C{{H}_{3}}\xrightarrow{AlC{{l}_{3}}}P\] \[\xrightarrow[\left( \text{ii} \right){{H}_{3}}{{O}^{+}}]{\left( \text{i} \right)\frac{{{O}_{2}}}{\Delta }}Q+\text{phenol}\]
The major products $$P$$ and $$Q$$ are
A
B
C
D
Answer :
It is cumene hydroperoxide rearrangement reaction.
269.
Acid catalyzed hydration of alkenes except ethene leads to the formation of
A
mixture of secondary and tertiary alcohols
B
mixture of primary and secondary alcohols
C
secondary or tertiary alcohol
D
primary alcohol
Answer :
secondary or tertiary alcohol
Addition follows Markovnikov’s rule.
270.
An inhibitor is described as,
A
a substance that slows down or stops a reaction
B
a substance which inhibits the properties of a catalyst
C
a substance formed during the reaction and does not participate in the reaction
D
a substance which prevents formation of products in a reaction being most reactive
Answer :
a substance that slows down or stops a reaction
No explanation is given for this question. Let's discuss the answer together.