Chemistry Test 275

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Chemistry Test 275

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Weekly Quiz Competition

Question 1
4 / -1

Butanone is a four-carbon compound with the functional group –

A
ketone

B
carboxylic acid

C
aldehyde

D
alcohol

Solution

In organic chemistry, the word butane shows the presence of four carbons.Thus, we can conclude that the functional group present in the compound is ketone.
Question 2
4 / -1

A tertiary butyl carbocation is more stable than a secondary butyl carbocation because-

A
+ R effect of -CH₃ groups

B
- R effect of -CH₃ groups

C
Hyperconjugation

D
- I effect of -CH₃ groups

Solution

Due to more hyperconjugation, tertiary butyl carbocation is more stable than secondary butyl carbocations.
Question 3
4 / -1

Match items of column I and II

A
A-(ii), B-(iv), C-(i), D-(iii)

B
A-(iii), B-(iv), C-(ii), D-(i)

C
A-(i), B-(iii), C-(ii), D-(iv)

D
A-(ii), B-(iii), C-(iv), D-(i)

Solution

Water and dichloromethane can be separated by differential extraction.

C₆H₁₂O₆ and NaCl can be separated by crystallization.
Question 4
4 / -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : Benzene is more stable than hypothetical cyclohexatriene

Reason R : The delocalised π electron cloud is attracted more strongly by nuclei of carbon atoms.

In the light of the above statements, choose the correct answer from the options given below:

A
A is true but R is false

B
Both A and R are correct and R is the correct explanation of A

C
Both A and R are correct but R is NOT the correct explanation of A

D
A is false but R is true

Solution

Benzene is more stable than hypothetical cyclohexatriene due to the delocalization of electrons in the π electron cloud of the aromatic ring. This delocalization of electrons leads to a more stable electronic configuration in benzene, as the electrons are shared more evenly between all six carbon atoms in the ring. This is in agreement with Assertion A.

Reason R is also correct, as the delocalized π electron cloud in the aromatic ring is attracted more strongly by the nuclei of the carbon atoms. This is due to the fact that the π electrons in the ring are more spread out and are not localized to a single atom, which makes them more attracted to the positively charged nuclei of the carbon atoms. This explanation supports Assertion A.
Question 5
4 / -1

A
Both Statement I and Statement II are correct.

B
Both Statement I and Statement II are incorrect.

C
Statement I is correct but Statement II is incorrect.

D
Statement I is incorrect but Statement II is correct.

Solution

Compound (A) in Statement-I and compound in Statement-II is not the mirror image of (I).
Question 6
4 / -1

Which of the following compounds react most readily with Br(g)?

A
C₂H₂

B
C₃H₆

C
C₂H₄

D
C₄H₁₀

Solution

The reactivity of a compound with Br(g) (Bromine gas) depends on the type of carbon-carbon bonds present in the compound. Compounds with double or triple bonds react more readily than those with single bonds because they have a higher electron density, which is more attractive to the electrophilic bromine.

Reactivity of Given Compounds:

C₂H₂: This compound has a triple bond between the two carbon atoms, which gives it a high electron density. However, the bond is also very strong, which makes the reaction slower.

C₃H₆: This compound has a double bond, which gives it a high electron density and makes it more reactive than compounds with single bonds. The double bond is weaker than the triple bond in C2H2, which makes the reaction faster.

C₂H₄: This compound has a double bond like C₃H₆. While it would also react readily with Br(g), the reaction would not be as fast as with C₃H₆ because C₂H₄ has fewer carbon atoms, and therefore less electron density.

C₄H₁₀: This compound only has single bonds, which makes it the least reactive of the four options.

In conclusion,C₃H₆reacts most readily with Br(g) because it has a good balance of high electron density due to its double bond and faster reaction rate due to the bond's relative weakness. This makes it more attractive to the electrophilic bromine and allows the reaction to proceed more quickly.
Question 7
4 / -1

When propene reacts with HBr in the presence of peroxide, it gives rise to

A
Allyl bromide

B
Isopropyl bromide

C
n-propyl bromide

D
3-bromopropane

Solution

Explanation of the Reaction of Propene with HBr in the Presence of Peroxide

The reaction of propene with hydrogen bromide in the presence of peroxide follows the rule of anti-Markovnikov addition. This rule states that the hydrogen (H) from HBr will add to the carbon with the most hydrogen atoms already attached, and the bromine (Br) will add to the other carbon. Peroxide promotes this anti-Markovnikov addition.

In the case of propene (CH₃-CH=CH₂), the hydrogen from HBr will add to the terminal carbon, which already has two hydrogens. The bromine will add to the middle carbon.
Question 8
4 / -1

Ethylene bromide on treatment with Zn gives

A
Alkyne

B
Alkene

C
Alkane

D
All of the above

Solution

Reaction of Ethylene Bromide with Zinc

Ethylene bromide, also known as 1,2-dibromoethane, is a halogenated hydrocarbon. When it is treated with zinc, an alkene is formed as a result. This reaction can be detailed as follows:

Why not Alkyne or Alkane?

An alkyne would require the removal of two pairs of hydrogen and bromine atoms, which does not occur in this reaction. An alkane would not have any double bonds, and the reaction with zinc specifically creates a double bond.

In conclusion, the correct answer is B: Alkene, because the reaction of ethylene bromide with zinc results in the formation of an alkene, specifically ethene. You can learn more about organic chemistry reactions on the EduRev platform.
Question 9
4 / -1

The position of double bond in alkenes can be located by :

A
Hydrogenation of oil

B
Ozonolysis

C
Photolysis

D
Hydration

Solution

Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon-carbon bonds have been replaced by a double bond to oxygen. The outcome of the reaction depends on the type of multiple bonds being oxidized.

Bromine water can be also used to identify the position of a double bond. In this reaction, red-brown colour of bromine gets turned into colourless, indicating that there is a double bond.
Question 10
4 / -1

The compound C₃H₄ has a triple bond, which is indicated by its reaction with

A
Bromine water

B
Bayers reagent

C
Fehling solution

D
Ammonical silver nitrate

Solution

CH₃−C ≡ C−H + AgNO₃ → CH₃ −C ≡ C−Ag

Propyne Ammonical Silver salt of Propyne

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Chemistry Test 275

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Chemistry Test 275

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Question 1

ketone

carboxylic acid

aldehyde

alcohol

Solution

Question 2

+ R effect of -CH3 groups

- R effect of -CH3 groups

Hyperconjugation

- I effect of -CH3 groups

Solution

Question 3

A-(ii), B-(iv), C-(i), D-(iii)

A-(iii), B-(iv), C-(ii), D-(i)

A-(i), B-(iii), C-(ii), D-(iv)

A-(ii), B-(iii), C-(iv), D-(i)

Solution

Question 4

A is true but R is false

Both A and R are correct and R is the correct explanation of A

Both A and R are correct but R is NOT the correct explanation of A

A is false but R is true

Solution

Question 5

Both Statement I and Statement II are correct.

Both Statement I and Statement II are incorrect.

Statement I is correct but Statement II is incorrect.

Statement I is incorrect but Statement II is correct.

Solution

Question 6

C2H2

C3H6

C2H4

C4H10

Solution

Question 7

Allyl bromide

Isopropyl bromide

n-propyl bromide

3-bromopropane

Solution

Question 8

Alkyne

Alkene

Alkane

All of the above

Solution

Question 9

Hydrogenation of oil

Ozonolysis

Photolysis

Hydration

Solution

Question 10

Bromine water

Bayers reagent

Fehling solution

Ammonical silver nitrate

Solution

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+ R effect of -CH₃ groups

- R effect of -CH₃ groups

- I effect of -CH₃ groups

C₂H₂

C₃H₆

C₂H₄

C₄H₁₀