Chemistry Test 274

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Chemistry Test 274

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following correctly ranks the cycloalkanes in order of increasing ring strain per methylene group?

A
Cyclopropane < Cyclobutane < Cyclopentane < Cyclohexane

B
Cyclohexane < Cyclopentane < Cyclobutane < Cyclopropane

C
Cyclohexane < Cyclobutane < Cyclopentane < Cyclopropane

D
Cyclopropane < Cyclopentane < Cyclobutane < Cyclohexane

Solution

The correct answer is Option B.

The C-C-C bond angles in cyclopropane (60^o) and cyclobutane (90^o) are much different than the ideal bond angle of 109.5^o.This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5^o.
Question 2
4 / -1

Which of the following cycloalkanes exhibits the greatest molar heat of combustion per —CH₂ — group?

A
Methylcyclobutane

B
Cyclopentane

C
Cyclobutane

D
Cyclopropane

Solution

The correct answer is option D

Cyclopropane is a cycloalkane molecule with the molecular formula C₃H₆, consisting of three carbon atoms linked to each other to form a ring, with each carbon atom bearing two hydrogen atoms resulting in D₃H molecular symmetry. The small size of the ring creates substantial ring strain in the structure.
Question 3
4 / -1

The correct lUPAC name of the compound

A
4-formyl-2-oxocydohexanecarboxylic acid

B
4-carboxy-2-oxocydohexanal

C
4-carboxy-1 -formylcyclohexanone

D
2-carboxy-5-formyl-1 -oxocyclohexane

Solution

Priority of functional groups is

- COOH > - CHO >>C = O.

Hence, the lUPAC name is 4-formyl-2-oxocyclohexane carboxylicacid.
Question 4
4 / -1

How many σ and π bonds are present in HC≡C−CH=CH−CH₃?

A
9σ, 4π

B
10σ, 3π

C
6σ, 6π

D
5σ, 5π

Solution
Question 5
4 / -1

Arrange in increasing order of basicity. HC≡C−, CH₃CH=CH−, CH₃CH₂−

A
HC≡C⁻< CH₃CH = CH− < CH₃CH₂⁻

B
HC≡C⁻> CH₃CH = CH⁻> CH₃CH₂⁻

C
HC≡C⁻> CH₃CH = CH− < CH₃CH₂⁻

D
HC≡C− = CH₃CH=CH− = CH₃CH₂−

Solution
Question 6
4 / -1

What are the hybridization and shapes of the following molecules?

(i) CH₃F

(ii) HC ≡ N

A
(i) sp², trigonal planar; (ii) sp³, tetrahedral

B
(i) sp³, tetrahedral; (ii) sp, linear

C
(i) sp, linear; (ii) sp², trigonal planar

D
(i) sp², trigonal planar, (ii) sp², trigonal planar

Solution

CH₃F - sp³ hybridised carbon, tetrahedral shape

(ii) HC = N - sp hybridised carbon, linear shape.
Question 7
4 / -1

Q. Consider the following compounds.

The correct statement regarding properties of above mentioned compounds is/are

A
Both have all their C—C bonds of equal length

B
I does not decolourises brown colour of bromine water solution but II does

C
I is not absorbed in cold, dil. H₂SO₄ but II is absorbed

D
All are correct
Solution
- Both have all their C—C bonds of equal length due to conjugation.
- I does not decolorises brown colour of bromine water solution but II does as The π bonds in Cyclooctatetraene (Compound II) react as usual for olefins, rather than as aromatic ring systems.
- I is planar but II is not as it adopts a tub conformation.
- Cyclooctatetraene shows various other addition reactions including Sulfonation.
Hence, Option A, B and C are correct.
Question 8
4 / -1

What is true about the 1,3,5,7-cyclooctatetraene?

A
It is an aromatic compound

B
It is an anti-aromatic compound

C
It on reaction with K metal absorbs two electrons

D
option B and option C are correct

Solution

1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.

Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:

2 K + C8H8 → K2C8H8

The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.
Question 9
4 / -1

The compound shown below evolve hydrogen gas when refluxed with potassium metal, why?

A
Loss of H₂ introduces a new pi-bond that makes system aromatic

B
Loss of H₂ converts the given compound into an aromatic cation

C
Loss of H₂ converts the given compound into an aromatic anion

D
Deprotonation of the above compound converts it into an aromatic anion

Solution

Metals like Potassium tend to loose their own electrons and the excess electrons will complete the aromaticity of the system.Thus, Deprotonation of the above compound converts it into an aromatic anion witn 6 pi electrons.
Question 10
4 / -1

Which of the following behaves both as a nucleophile and as an electrophile?

A
CH₃C ≡ N

B
CH₃OH

C
CH₂ = CHCH₃

D
CH₃NH₂

Solution

R − C^δ = N^δ: has electrophilic C-atom and nucleophilic N-atom.

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Chemistry Test 274

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Chemistry Test 274

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Question 1

Cyclopropane < Cyclobutane < Cyclopentane < Cyclohexane

Cyclohexane < Cyclopentane < Cyclobutane < Cyclopropane

Cyclohexane < Cyclobutane < Cyclopentane < Cyclopropane

Cyclopropane < Cyclopentane < Cyclobutane < Cyclohexane

Solution

Question 2

Methylcyclobutane

Cyclopentane

Cyclobutane

Cyclopropane

Solution

Question 3

4-formyl-2-oxocydohexanecarboxylic acid

4-carboxy-2-oxocydohexanal

4-carboxy-1 -formylcyclohexanone​

2-carboxy-5-formyl-1 -oxocyclohexane

Solution

Question 4

9σ, 4π

10σ, 3π

6σ, 6π

5σ, 5π

Solution

Question 5

HC≡C− < CH3​CH = CH− < CH3​CH2​−

HC≡C− > CH3​CH = CH− > CH3​CH2​−

HC≡C− > CH3​CH = CH− < CH3​CH2​−

HC≡C− = CH3​CH=CH− = CH3​CH2​−

Solution

Question 6

(i) sp2, trigonal planar; (ii) sp3, tetrahedral

(i) sp3, tetrahedral; (ii) sp, linear

(i) sp, linear; (ii) sp2, trigonal planar

(i) sp2, trigonal planar, (ii) sp2, trigonal planar

Solution

Question 7

Both have all their C—C bonds of equal length

I does not decolourises brown colour of bromine water solution but II does

I is not absorbed in cold, dil. H2SO4 but II is absorbed

All are correct

Solution

Question 8

It is an aromatic compound

It is an anti-aromatic compound

It on reaction with K metal absorbs two electrons

option B and option C are correct

Solution

Question 9

Loss of H2 introduces a new pi-bond that makes system aromatic

Loss of H2 converts the given compound into an aromatic cation

Loss of H2 converts the given compound into an aromatic anion

Deprotonation of the above compound converts it into an aromatic anion

Solution

Question 10

CH3C ≡ N

CH3OH

CH2 = CHCH3

CH3NH2

Solution

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4-carboxy-1 -formylcyclohexanone

HC≡C⁻< CH₃CH = CH− < CH₃CH₂⁻

HC≡C⁻> CH₃CH = CH⁻> CH₃CH₂⁻

HC≡C⁻> CH₃CH = CH− < CH₃CH₂⁻

HC≡C− = CH₃CH=CH− = CH₃CH₂−

(i) sp², trigonal planar; (ii) sp³, tetrahedral

(i) sp³, tetrahedral; (ii) sp, linear

(i) sp, linear; (ii) sp², trigonal planar

(i) sp², trigonal planar, (ii) sp², trigonal planar

I is not absorbed in cold, dil. H₂SO₄ but II is absorbed

Loss of H₂ introduces a new pi-bond that makes system aromatic

Loss of H₂ converts the given compound into an aromatic cation

Loss of H₂ converts the given compound into an aromatic anion

CH₃C ≡ N

CH₃OH

CH₂ = CHCH₃

CH₃NH₂