Hydrocarbons Test

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Hydrocarbons Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, how many different position isomers are possible?

A
2

B
3

C
5

D
6

Solution

the three isomers will be ortho(i.e.1,2- or 1,4-disubstituted) , meta(i.e.1,3- or 1,5-disubstituted) and para(1,4-disubstituted)
Question 2
1 / -0

The unusual behavior of benzene is explained by

A
kekule structure

B
sp² hybridization

C
delocalized π electron cloud

D
sp³ hybridization

Solution

resonance stabilisation explains the unusual stability of benzene.There is pi electron delocalisation in the ring which enhances the stability of the molecule.
Question 3
1 / -0

Aromatic compounds are charachterized by the presence of _____ π electrons in the ring where n is an integer (n = 0, 1, 2, . . .).

A
(4n + 1)

B
(4n + 2)

C
(4n - 2)

D
4n

Solution

Huckel rule which explains the unusual stability of cyclic conjugated polyenes(i.e. aromatic compounds) over other cyclic unsaturatrd compounds which do not contain (4n+2) pi electrons.
Question 4
1 / -0

Which of the following method is not a synthetic method for production of Benzene?

A
By reduction of phenol

B
Decarboxylation of aromatic acids

C
isolation from coal tar

D
Cyclic polymerization of ethyne

Solution

Since isolation of benzene from coal tar involves separation of benzene from other components through a physical operation "fractional distillation" , it is NOT a synthetic method for production of benzene.
Question 5
1 / -0

In Friedel-Crafts alkylation reaction, when benzene is treated with an alkyl halide in the presence of a catalyst, alkylbenene is formed. The catalyst used is

A
Silver

B
Palladium

C
anhydrous aluminium chloride

D
H₃PO₄

Solution

the acid chloride or anhydride reacts with anhydrous aluminium chloride to form the acylium ion which serves as an electrophile which then attacks the benzene ring to give alkylbenzene.
Question 6
1 / -0

Some of the polynuclear hydrocarbons having fused benzene ring system have the property of being

A
conductive

B
carcinogenic

C
nonflammable

D
malleable

Solution

Polynuclear hydrocarbons are carcinogenic,i.e. cancer producing
Question 7
1 / -0

Arrange the following in decreasing order of their boiling points. (A) n–butane (B) 2–methylbutane (C) n-pentane (D) 2,2–dimethylpropane

A
C > B > D > A

B
D > C > B > A

C
B > C > D > A

D
A > B > C > D

Solution

Boiling point depends on molecular mass and surface area.As the number of carbon atoms increase,boiling point increases ,hence n-butane has the minimum boiling point.Among isomeric alkanes,boiling point decreases with branching.Hence the order.
Question 8
1 / -0

Arrange the halogens F₂, Cl₂, Br₂, I₂,in order of their increasing reactivity with alkanes.

A
Br₂ < Cl₂ < F₂ < I₂

B
F₂< Cl₂< Br₂< I₂

C
Br₂< I₂< Cl₂< F₂

D
I₂< Br₂< Cl₂< F₂

Solution

The reaction follows free radical mechanism in which the first step is generation of halogen free radical which is the fastest for fluorine.Hence the order.
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–Br < R–I < R–Cl

C
R–I < R–Br < R–Cl

D
R–Cl < R–Br < R–I

Solution

As we move from I to Br to Cl,atomic size decreases,it becomes difficult to release the halogen as the bond length decreases.Hence I^- is the best leaving group among halogens and reacts the fastest.

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Hydrocarbons Test - 4

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Hydrocarbons Test - 4

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SHARING IS CARING

Question 1

2

3

5

6

Solution

Question 2

kekule structure

sp2 hybridization

delocalized π electron cloud

sp3 hybridization

Solution

Question 3

(4n + 1)

(4n + 2)

(4n - 2)

4n

Solution

Question 4

By reduction of phenol

Decarboxylation of aromatic acids

isolation from coal tar

Cyclic polymerization of ethyne

Solution

Question 5

Silver

Palladium

anhydrous aluminium chloride

H3PO4

Solution

Question 6

conductive

carcinogenic

nonflammable

malleable

Solution

Question 7

C > B > D > A

D > C > B > A

B > C > D > A

A > B > C > D

Solution

Question 8

Br2 < Cl2 < F2 < I2

F2< Cl2< Br2< I2

Br2< I2< Cl2< F2

I2< Br2< Cl2< F2

Solution

Question 9

R–Cl < R–I < R–Br

R–Br < R–I < R–Cl

R–I < R–Br < R–Cl

R–Cl < R–Br < R–I

Solution

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sp² hybridization

sp³ hybridization

H₃PO₄

Br₂ < Cl₂ < F₂ < I₂

F₂< Cl₂< Br₂< I₂

Br₂< I₂< Cl₂< F₂

I₂< Br₂< Cl₂< F₂