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Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - NEET MCQ


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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Organic Chemistry - Nomenclature and Isomerism (September 9)

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Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 1

 Which of the following compounds will exhibit cis-trans isomerism? 

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 1

The compounds with each doubly bonded carbon attached to two different groups (like Cab=Cab, Cab=Ccd) exhibit geometrical isomerism i.e., cis and trans forms. The geometrical isomerism arises due to restricted rotation of double bond.

However, even though there is restricted rotation for triple bond, alkynes do not exhibit geometrical isomerism, since the triply bonded carbons are attached to one group each only.

Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 2

For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would be expected to have the same values?

[AIEEE 2004]

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 2

Vapor density = Molecular weight/2

As both the compounds have same molecular weights, both will have the same vapour density. Hence, gaseous density of both ethanol and dimethyl ether would be same under identical conditions of temperature and pressure. The rest of these three properties; Vapour pressure, boiling point and heat of vaporization will differ as ethanol has hydrogen bonding whereas ether does not.

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Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 3

Which of the following statements are correct?

I. A pair of positional isomers differs in the position of the same functional group.
II. A pair of. structural isomers have the same relative molar mass.
Ill. A pair of functional group isomers belongs to different homologous series. 

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 3

To determine which of the statements are correct, let's analyze each statement one by one.

Step 1: Analyze Statement I
Statement I: A pair of positional isomers differs in the position of the same functional group.

Explanation: Positional isomers are compounds that have the same molecular formula and the same functional group but differ in the position of that functional group within the molecule. For example, consider butanol:
- 1-butanol (C₄H₁₀O) has the hydroxyl (-OH) group on the first carbon.
- 2-butanol (C₄H₁₀O) has the hydroxyl (-OH) group on the second carbon.

Thus, this statement is correct.

Step 2: Analyze Statement II
Statement II: A pair of structural isomers have the same relative molar mass.

Explanation: Structural isomers have the same molecular formula but differ in the connectivity of their atoms. Since they have the same number and types of atoms, their molar masses will be the same. For example, both butanol and isobutanol (C₄H₁₀O) have the same molar mass despite differing structures.

Thus, this statement is correct.

Step 3: Analyze Statement III
Statement III: A pair of functional group isomers belongs to different homologous series.

Explanation: Functional group isomers are compounds that have the same molecular formula but different functional groups. For example, ethanol (C₂H₆O) and dimethyl ether (C₂H₆O) are functional group isomers. They belong to different functional groups but can belong to the same homologous series if they have similar structural features. Therefore, this statement is incorrect.

Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 4

How many alcohols are structural isomers with the formula: C5H11OH?

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 4

Structural isomers of alcohol C5H11OH are as follow:-

Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 5

Which of the following has three different stereoisomers?

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 5

The correct answer is Option B.
 
 
CH3−CH=CH−CH=CH−CH3
          2,4-hexadiene
 
Three different geometrical isomers of cis-cis, trans-trans and sic-trans are possible.
 

*Multiple options can be correct
Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 6

Direction (Q. Nos. 14-18) This section contains 5 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q.

Which compound below can have a non-polar steroisomers ?

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 6


However, there is no such arrangement in structure i, it won’t have any non polar structure.

Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 7

Which compound below exhibit stereolsomerism?

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 7

*Multiple options can be correct
Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 8

Consider the following free radical chlorination reaction.

The correct statement about product(s) is/are

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 8

There are 4 final products, they are as follow:-

2 products of the structure above due to presence of a chiral carbon.


1 product of structure above due to absence of Chiral Carbon.

1 product of structure above due to absence of Chiral Carbon.

Hence, Options A, B are correct.

*Multiple options can be correct
Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 9

The correct statement(s) about the compound given below is/are

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 9

The correct answer is option A & D
The given image is Self-Explanatory.
We know that following are the cases when a compound is Optically Inactive:
1. When it does not have a chiral Carbon Atom eg.: Methyl bromide
2. When it has a Plane or Axis of Symmetry eg.: Tartaric Acid (these compounds are called as Meso Compounds)
3. When it forms a Racemic Mixture eg.: Equimolar mixture of d-Lactic Acid and l-Lactic Acid.
The given compound in the question possesses an Axis of Symmetry (shown in the attached image). Hence, it is Optically Inactive.
 

*Multiple options can be correct
Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 10

Which of the following statements is/are correct regarding the below compound?

Detailed Solution for Test: Organic Chemistry - Nomenclature and Isomerism (September 9) - Question 10

Correct option is
A:
It is optically inactive due to the plane of symmetry &

B: It is optically inactive due to the center of symmetry

The presence of any kind of symmetry makes the compound optically inactive.

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