Hydrocarbons Test

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Hydrocarbons Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Fill in the blanks with appropriate words. Benzene has a planar structure. All carbon atoms in benzene are I hybridised. The ring structure of benzene was proposed by II . It shows III substitution reactions. It reacts with IV in presence of aluminium chloride to form acetophenone.

A
$$ I = sp_2 \,\,\, II = Kekule \,\,\, III= electrophilic \,\,\, IV = acetyl\, chloride$$

B
$$ I = sp \,\,\, II = Dewar \,\,\, III= nucleophilic \,\,\, IV =chloromethane$$

C
$$ I = sp_3 \,\,\, II = Ladenberg \,\,\, III= electrophilic \,\,\, IV = chloroethane$$

D
$$ I = sp_2 \,\,\, II = Baeyer \,\,\, III= nucleophilic \,\,\, IV = methyl\, bromide$$

Solution

Benzene has a planar structure. All carbon atoms in benzene are $$s{p}^{2}$$ hybridised being bonded to two other carbon atoms and one hydrogen atom. The ring structure of benzene was proposed by Kekule. It shows electrophillic substitution reactions due to electron rich character. It reacts with acetyl chloride in presence of aluminium chloride to form acetophenone and the reaction is known as acylation of benzene.
Hence, A is correct answer
Question 2
1 / -0

When n-butane is treated with chlorine, how many monochloro products are formed?

A
1

B
2

C
3

D
4

Solution

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+{ Cl }_{ 2 }\rightarrow { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }{ Cl }+{ CH }_{ 3 }{ CH }_{ 2 }{ - }\underset { \overset { | }{ Cl } }{ CH } -{ CH }_{ 3 }$$
Question 3
1 / -0

Mark the incorrect statement from the following.

A
Benzene has a planar structure

B
Benzene is an unsaturated hydrocarbon and shows addition reactions like alkenes

C
In benzene carbon uses two p-orbitals for hybridisation

D
Aromatic hydrocarbons contain high percentage of carbon hence burn with sooty flame

Solution

The statement given in the second option is incorrect.
Benzene can show both additions as well substitution reaction. It does not show an addition reaction like alkene. This is because, if it shows an addition reaction like alkene, then it will lose its aromaticity and become less stable.
Question 4
1 / -0

The compound formed when an alcoholic solution of ethylene dibromide is heated with granulated zinc is:

A
ethene

B
ethyne

C
ethane

D
bromoethane

Solution

The given reaction is dehalogenation reaction. Zinc removes bromine from ethylene dibromide as zinc bromide and forms ethene.
$$\underset{\text{Ethylene dibromide}}{Br{H}_{2}C-C{H}_{2}Br} + Zn \xrightarrow[alcohol]{heat} {H}_{2}C=C{H}_{2} +Zn{Br}_{2}$$
Question 5
1 / -0

The negative part of the addendum (the molecule to be added) adds on to the carbon atom of the double bond containing the least number of hydrogen atoms. This rule is known as?

A
Saytzeffs rule

B
Peroxide rule

C
Markovnikovs rule

D
Hoffmann rule.

Solution

According to the Markovnikovs rule, the negative part of the unsymmetrical reagent adds to less hydrogenated ( more substituted) carbon atom of the double bond. So here the negative part of the addendum gets attach to that carbon which possess lesser number of hydrogen atoms.
Question 6
1 / -0

Which of the following is aromatic in nature?

A

B

C

D

E
None of these

Solution

R.E.F image
Non Aromatic
R.E.F image
Antiaromatic $$ 4\pi e^{-}s $$
R.E.F image
Aromatic $$ 2\pi e^{-}s $$
R.E.F image
Non Aromatic
Question 7
1 / -0

Which of the following reactions follows Markovnikovs rule?

A
$$C_2H_4 \, + \, HBr$$

B
$$C_3H_6 \, + \, Cl_2$$

C
$$C_3H_6 \, + \, HBr$$

D
$$CH_6 \, + \, Br_2$$

Solution

When HBr is added to an alkene in the absence of peroxides it obey Markovnikovs rule. When HBr reacts with unsymmetrical alkene in the presence of peroxides HBr adds in the opposite direction to that predicted by Markovnikov's rule.
$$ CH_3 CH = CH_2 + HBr $$
Question 8
1 / -0

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

Solution

The 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:
$$\underset{III}{n-Pentane} > \underset{II}{2-Methylbutane} > \underset{IV}{2,2-Dimethylpropane} > \underset{I}{n-Butane}$$
Hence, $$D$$ is the correct answer.
Question 9
1 / -0

The increasing order of reduction of alkyl halides with zinc and dilute $$HCl$$ is:

A
$$R-Cl < R-I < R-Br$$

B
$$R-Cl < R-Br < R-I$$

C
$$R-I < R-Br < R-Cl$$

D
$$R-Br < R-I < R-Cl$$

Solution

As the size of the halogen increases down the group, the strength of $$C-X$$ bond decreases, as a result, reactivity increases. Hence, The increasing order of reduction of alkyl halides with zinc and dil. $$HCl$$ is $$R-Cl < R-Br < R-I$$.
Hence, $$B$$ is the correct answer.
Question 10
1 / -0

Arrange the halogens $$F_2,Cl_2,Br_2,I_2,$$ in order of their increasing reactivity with alkanes.

A
$$I_2 < Br_2 < Cl_2 < F_2$$

B
$$Br_2 < Cl_2 < F_2 < I_2$$

C
$$F_2 < Cl_2 < Br_2 < I_2$$

D
$$Br_2 < I_2 < Cl_2 < F_2$$

Solution

Since reactivity decreases down the group as the electronegativity of the halogen decreases down the group. Thus, rate of reaction of alkanes with halogens is $${I}_{2}<{Br}_{2}<{Cl}_{2}<{F}_{2}$$.