Hydrocarbons Test

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Hydrocarbons Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The molecular formula of Wilkinson catalyst, used in hydrogenation of alkenes is :

A
$$Co (CO)_8$$

B
$$(Ph_3P)_3 RhCl$$

C
$$[Pt (NH_3)_2 Cl_2]$$

D
$$ K [Ag (CN)_2]$$

Solution

Wilkinson's catalyst, is the common name for Chloridotris (triphenylphosphane) rhodium(I)

Coordination complex of rhodium with the formula $$RhCl{(P{Ph}_3)}_3$$ ($$Ph$$ = phenyl).

Option B is correct.
Question 2
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Benzyl chloride $$C_6 H_5 CH_2 Cl$$ can be prepared from toluene by chlorination with:

A
$$Cl_2$$

B
$$SO_2Cl_2$$

C
$$SOCl_2$$

D
$$NaOCl$$

Solution

$$\displaystyle Benzyl \: chloride$$ $$\displaystyle C_6H_5CH_2Cl$$

can be prepared from $$\displaystyle toluene$$ by chlorination with $$\displaystyle Cl_2$$

$$\displaystyle C_6H_5CH_3+ Cl_2 \rightarrow C_6H_5CH_2Cl + HCl$$
Question 3
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Which one of the following is the most important halogen in terms of preparation of halogen derivatives of alkane?

A
Bromine

B
Chlorine

C
Iodine

D
Fluorine
Question 4
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The major product of the above reaction is:

A

B

C

D

Solution

As the base is strong. elimination reaction will take place.

$$CH_2=CHBr\xrightarrow{NaNH_2}CH\equiv CH$$
Question 5
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Which of the following can be used for the preparation of propane?

A
$$ CH_3 CH \equiv CH_2 \xrightarrow [(ii)\ AgNO_3 /NaOH ] { (i)\ B_2 H_6 } $$

B
$$ CH_3 CH_2 CH_2 Cl \xrightarrow [(ii)H_2O_2]{(i)\ Mg / ether} $$

C
$$ CH_3 CH_2 CH_2 I \xrightarrow { HI / 150^{\circ}C} $$

D
$$ CH_3 CH_2 CH_2 COONa \xrightarrow [\triangle]{ NaOH/CaO} $$

Solution

The chemical formula for propane is $$CH_3CH_2CH_3$$.

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide $$CO_2$$ from a compound.

Usually, decarboxylation refers to the reaction of carboxylic acids or their derivatives, removing a carbon atom from a carbon chain in the form of $$CO_2$$.

An alkane with one carbon atom less than the parent carbon chain will be formed in this process.

$$CH_3CH_2CH_2COONa\ \xrightarrow[\Delta]{NaOH/CaO}\ CH_3CH_2CH_3+Na_2CO_3$$

The given reaction involves the removal of a carbon dioxide molecule, therefore, it is a decarboxylation reaction. Therefore, propane is formed by the decarboxylation of sodium propanoate

Hence, option (D) is correct.
Question 6
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$$C_7H_8 \ \ {\xrightarrow{3Cl_2, heat}}\ \ A\ \ {\xrightarrow{Fe/Br_2}} \ \ \ B\ \ {\xrightarrow{Zn/HCl}} \ \ C $$.

Here the compound C is:

A
$$3$$-bromo-$$2, 4, 5, 6$$-trichloro tolucne

B
o-bromo toluene

C
p-bromo toluene

D
m-bromo toluene

Solution

In the given sequence of reactions, compound C is $$m-$$bromotoluene.

Hence, the correct option is $$\text{D}$$
Question 7
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The complete IUPAC name of the compound:

A
[R]-1-Bromo-1-phenyl ethane

B
[S]-1-Bromo-1-phenyl ethane

C
[E]-1-Bromo-1-phenyl ethane

D
[Z]-1-Bromo-1-phenyl ethane.

Solution

From $$CIP$$ rules, numbering the substituent.
The substituents are named in alphabetic order. The largest claim of carbon is ethane therefore,
$$[R]-1-Bromo-1-phenyl-ethane$$
Question 8
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Cycloalkanes are isomeric with:

A
Alkanes

B
Alkenes

C
Alkynes

D
Arenes
Question 9
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The main product of the reaction would be
$$2-butene+chloroform\xrightarrow [ ]{ NaOH } $$

A
Butanoic acid

B
2-methyl butanoic acid

C
1,1,1-trichloro-2-methyl butane

D
1,4-butane diol

Solution

$$CH_3CH = CH - CH_3+CHCl_3\longrightarrow CH_3-\overset{CCl_3}{\overset{|}{C}H}-CH_2-CH_3\xrightarrow[hydrolysis]{NaOH}$$
$$CH_3-\!\!\overset{\,\,\,C(OH)_3}{\overset{|}{C}H-}CH_2-CH_3\xrightarrow{-H_2O} \underset{2-methyl\,butanoic\,acid}{CH_3-\overset{COOH}{\overset{|}{C}H-}CH_2CH_3}$$
Answer is (b).
Question 10
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How many monochloro structural isomers are expected in free radical monochlorination of $$2-$$methylbutane?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

E
$$6$$

Solution

Number of mono chloro isomers depends upon the types of hydrogen atoms present in the substate alkane. The structure of $$2-$$methylbutane is
$$\overset {d}{C}H_{3} - \overset {c}{C}H_{2} - \overset {b}{H}\underset {\underset{a}{C}H_3}{\underset{|}{C} -} \overset {a}{C}H_{3}$$
Because of the presence of four types of hydrogen, it gives $$four + two (d$$ and $$l$$ forms) $$=$$ six monochloro products on monochlorination as
$$CH_3-CH_2\underset{CH_3}{\underset{|}{C}H}-CH_3\xrightarrow[hv]{Cl_2}\underset{ \begin{matrix}\text{d and l forms}\\\text{1-chloro-2-methyl butane}\end{matrix}}{CH_3CH_2\dot{C}H-CH_2Cl}+ \underset{2-chloro-2-methyl\,butane}{CH_3-CH_2-\overset{Cl}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}-}CH_3}$$

$$+\underset{2-chloro-3-methyl\,butane}{CH_3-\underset{Cl}{\underset{|}{\dot{C}}}H-CH(CH_3)_2}+\underset{1-chloro-3-methyl\,butane}{CH_2ClCH_2-\underset{CH_3}{\underset{|}{C}H}-CH_3}$$