Hydrocarbons Test

Result

Hydrocarbons Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The reagent used for converting Acetylene tetrabromide into ethyne is:

A
Alcoholic $$ KOH$$

B
Conc.$$H_{2}SO_{4}$$ at $$170^{0}$$ C

C
Aqueous potash

D
Zinc dust

Solution

Alc.KOH is used in preparation of acetylene from dihalides and Zn dust is used in preparation of acetylene from tetra halides.
Reaction:
Question 2
1 / -0

Which of the following is true :

A
Acetylene is more reactive than ethylene to an electrophilic attack

B
Acetylene is less reactive than ethylene to an electrophilic attack

C
Acetylene may show more reactivity or less reactivity towards electrophilic reagent.

D
Acetylene and ethylene show indentical reactivities towards an electrophilic attack

Solution

Acetylene should be more reactive than ethylene to an electrophilic attack due to dense structure but they are less reactive than ethylene.
Question 3
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Which of the following method is not used in the preparation of acetylene :

A
Dehydrohalogenation

B
Dehalogenation

C
Hydrolysis

D
Hydrogenation

Solution

Alkyne can be produced by
Kolbe electrolysis:
acetylene can be produced by electrolysis of aq. solution of sodium/potassium maleate or fumarate.
Dehydrohalogenation:
Preparation of alkyne by heating ethylene bromide with alc. KOH followed by sodamide in ammonia solution.
Synthesis:
$$2C+H_2\rightarrow C_2H_2$$ at high temp.
and by hydrolysis.
Question 4
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$${ CH }_{ 2 }={ CH }_{ 2 }\xrightarrow { HCl } A\xrightarrow [ Mg ]{ Dry\ ether } B\xrightarrow { Hydrolysis } C$$ then `C' is:

A
$${ CH }_{ 3 }-{ CH }_{ 2 }-Cl$$

B
$${ CH }_{ 3 }-{ CH }_{ 2 }-MgCl$$

C
$${ CH }_{ 4 }$$

D
$${ CH }_{ 3 }-{ CH }_{ 3 }$$

Solution

$$CH_2=CH_2+HCl \rightarrow C_2H_5Cl$$(A)
dry ether
$$CH_3CH_2Cl+Mg \rightarrow CH_3CH_2MgCl(B)$$

$$CH_3CH_2MgCl+H_2O \rightarrow CH_3CH_3$$
Question 5
1 / -0

Ethyne is not prepared by:

A
heating iodoform with $$Ag$$ powder

B
hydrolysis of calcium carbide

C
heating $$C_{2}H_{2}Cl_{4}$$ with zinc dust

D
heating iodoform with $$Na$$

Solution

In all the given reactions Acetylene or ethylene is formed except heating Idoform with Na. As iodoform doesn't even reacts with $$Na$$ so this will not produce alkyne.
Option D is correct.
Question 6
1 / -0

The number of structural isomers possible with molecular formula $$C_{5}H_{8}$$ is:

A
more than 7

B
6

C
5

D
4

Solution
Question 7
1 / -0

In benzene the overlap between carbon-carbon is of the type :

A
$$p-p$$

B
$$sp^{2}-sp^{2}$$

C
$$sp^{3}-sp^{3}$$

D
None of the above

Solution

In benzene all carbon atom are $$sp^2$$ hybridized so all bonds overlap are in between $$sp^2$$ - $$sp^2$$ hybridized carbon atom.
Question 8
1 / -0

Benzene reacts with $$CH_3Cl$$ in the presence of anhydrous $$AlCl_3$$ to form:

A
xylene

B
toluene

C
chlorobenzene

D
benzyl chloride

Solution

Benzene reacts with $$CH_{3}Cl$$ in presence of anhydrous $$AlCl_{3}$$ to give Toulene. This is fridel crafts Alkylation and an electrophillic substitution.
Question 9
1 / -0

The ratio of the number of hybrid and pure orbitals in $$C_{6}H_{6}$$ is :

A
3 : 1

B
2 : 3

C
1 : 1

D
4 : 3

Solution

In Benzene each carbon is $$sp^2$$ hybridizes. 2s and 2p orbitals of carbon participate in hybridization to form 3 $$sp^2$$ hybrid orbitals but its 1s and one 2p orbital remain unhybridized.
H has only one 1s orbital and the single orbital can not undergo hybridization, so it remains in pure form.
Thus there are 18 hybridized orbitals (3 from each carbon) and 18 pure orbitals ( 2 from each carbon and 1 from each hydrogen) in a benzene ring.
So ratio of hybridize orbitals to pure orbitals is 1:1.
Option C is correct.
Question 10
1 / -0

Structural formula of ethyne is :

A
$$HC\equiv CH$$

B
$$H_{3} C C H$$

C

D

Solution

Correct formula of Ethyne is: $$HC\equiv CH$$.