Hydrocarbons Test

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Hydrocarbons Test - 19

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1 - 8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Arrange the following compounds in increasing order of polarity

A
I < II < III

B
III < II < I

C
II < I < III

D
III < I < II

Solution

In case 1, the bond is broken in oxygen’s favor and it will attain its octet. Also, carbon becomes sp² hybridized, so there is a chance of polarity.
In case 2, if the bond is broken in favor of oxygen, then the ring will become anti-aromatic which is highly unstable and the bond won’t be broken in that way. If the bond is broken in favor of carbon in the ring, then although the ring becomes aromatic but oxygen will bear +ve charge which is very unstable. So, there is no chance to break the bond.
In case 3, if the double bond is broken in favor of oxygen, then oxygen will acquire a negative charge and the ring will become aromatic. So, it is a highly favorable case of double bond breaking.
Therefore, the order of polarity: - III>I>II
Question 2
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In principle, what is true regarding benzene and 1, 3, 5-cyclohexatriene?

A
Theoretically they are the same molecules

B
Both have same lengths of all of their C — C bonds

C
Both have same enthalpy of hydrogenation

D
Cyclohexatriene has three different C - C bond lengths while benzene has only one type of C — C bond length.

Solution

The correct answer is Option D.
In 1,3,5-cyclohexatriene, there are three C = C having a bond length 134 pm and three C - C having a bond length 154 pm. In benzene, all the six C - C bonds have the same bond length 139 pm.
Question 3
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Enthalpy of hydrogenation of cyclohexene is -119 kJ/mol and that of benzene is -208 kJ/mol. Based on these information, resonance energy of benzene can be calculated to be

A
-119 kJ/mol

B
-238 kJ/mol

C
-149 kJ/mol

D
-565 kJ/mol

Solution

The correct answer is Option C.
ΔH_hyd (cyclohexene) = - 119 kJ/mol (1C = C)
ΔH_exp (benzene) = -3 × 119 = -357 (3C = C)
ΔH_cal (benzene) = -208 KJ/mol
∴ resonance energy = ΔH_exp - ΔH_cal
= -357 + 208
= -149 kJ/mol
Question 4
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Select the species which is not aromatic.

A

B

C

D

Solution

Here, we can see that Nitrogen gave its lone pair to make the system aromatic. The same case happens with option b and d. But with option c, Boron is not having any lone pair to donate. So option c is correct answer
Question 5
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Consider the following bromides :

A
Both decomposes to form Br^-

B
Both decomposes to forms Br⁺

C
I decomposes to give Br^- and II gives Br⁺

D
I decomposes to give Br⁺ while II gives Br^-

Solution

In the compound I, Br will dispatch as Br⁺ so that a -ve charge appears on carbon which will give us 6π electrons. So ring will become aromatic.(4n+2 π electron is needed for aromaticity)
In compound II, Br will dispatch as Br-. So that carbon has +ve charge and all the double bond will circulate in the ring. This will maintain 4n+2 π electron and the molecule will remain as aromatic.
Question 6
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Which is not true regarding 1, 3, 5, 7-cyclooctatetraene?

A
It delcolourises purple colour of cold, dilute, alkaline KMnO₄ (Baeyer’s reagent)

B
It decolourises brown colour of Br₂ - H₂O solution

C
It is absorbed in cold, concentrated H₂SO₄ solution

D
It is an aromatic system

Solution

The correct answer is Option D.

1, 3, 5, 7-cyclooctatetraene is a system with 8 electrons but it is a nonaromatic compound as it adopts a tub like shape to escape anti-aromaticity. Hence, d option is wrong.
Question 7
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The following compound , when treated with excess of AgBF₄ gives a red precipitate leaving a highly conducting filtrate, due to

A
A dianion aromatic system is formed

B
A dication aromatic system is formed

C
A covalent ditetrafluoroborate compound is formed

D
An aromatic salt precipitate which is red coloured

Solution

The correct answer is Option B.
An aromatic dication is formed, the no of electrons become 2 thus it obeys 4n+2 rule hence it is aromatic and highly stable.
Question 8
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When potassium metal is added to 1, 3, 5, 7-cyclooctatetraene, a highly conducting salt is formed without evolution of H₂ gas because

A
reduction of cyclooctatetraene into anionic

B
reduction of cyclooctatetraene into anionic

C
reduction of cyclooctatetraene into

D
oxidation of cyclooctatetraene into

Solution

Cyclooctatetraene readily reacts with potassium metal to form the salt, which contains the dianion C₈H₈^2-. The dianion is both planar and octagonal in shape and aromatic with a Hückel electron count of 10.
Question 9
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Direction (Q. Nos. 9 - 13) This section is based on Statement i and Statement II. Select the correct answer from the codes given below.
Q.
Statement I : Benzene has very high stability than a general triene.
Statement II : Benzene is a completely conjugated system.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement I is incorrect but Statement II is correct

Solution

Benzene has very high stability than a general triene due to Aromaticity and not just completely conjugated system. Hence, Both Statement I and Statement II are correct but, Statement II is not the correct explanation of Statement I
Question 10
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Statement I: Pyrene, although aromatic, decolorises brown colour of bromine water.
Statement II: It has one pi-bond extra which is not part of the aromatic system.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement I is incorrect but Statement II is correc

Solution

Pyrenes are strong electron donor materials and has one pi-bond extra which is not the part of aromatic system Thus, decolourises brown colour of bromine water. Hence, Option A is correct.
Bromine Test: Bromine solution is brown. In this test when bromine solution is added to the unsaturated hydrocarbon the brown colour disappears if the hydrocarbon is unsaturated. Bromine forms an addition product with the unsaturated hydrocarbon. .
Question 11
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Statement I : The follow compound is optically active.
Statement II : It has a chiral carbon.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement I is incorrect but Statement II is correct

Solution

The correct answer is option C
Statement 1 is correct because Plane Of Symmetry and Centre Of Symmetry are absent in the compound .
Statement 2 is clearly incorrect as there is No 'Sp3 ' Hybridized Carbon atom .
Question 12
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Statement I : Furan is an aromatic system, has resonance energy comparable to that of benzene.
Statement II : Furan decolourises the brown colour of bromine water solution.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement I is incorrect but Statement II is correct

Solution

Furan has resonance energy comparable to benzene. But the double bonds are involved in aromaticity, so it won’t decolourise Bromine water
Bromine Test: Bromine solution is brown. In this test when bromine solution is added to the unsaturated hydrocarbon the brown colour disappears if the hydrocarbon is unsaturated. Bromine forms an addition product with the unsaturated hydrocarbon. .
Question 13
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Arrange the halogens F₂, Cl₂, Br₂, I₂ in order of their increasing reactivity with alkanes.

A
I₂ < Br₂ < Cl₂ < F₂

B
Br₂ < Cl₂ < F₂ < I₂

C
F₂< Cl₂ < Br₂ < I₂

D
Br₂ < I₂ < Cl₂ < F₂

Solution

The correct answer is Option A.
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
I₂ < Br₂ < Cl₂ 2