Alkanes & Cycloalkanes - Free MCQ Practice Test with solutions, NEET Chemistry

Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy

In option c, there are only two different position which can be chlorinated, 
So, option c is correct.

Hydrogen abstraction by halogen radical in the propagation step is exothermic in chlorination but endothermic in bromination. Hence, the option (A) is an incorrect statement.

The formation of an alkane from an alkyl halide through reduction with zinc is called the Frankland reaction.

The correct answer is Option C - A pair of diastereomers in equal amount is formed

The addition of bromine to a carbon-carbon double bond proceeds via formation of a cyclic bromonium ion, followed by nucleophilic attack from the opposite face; this mechanism leads to anti addition and creation of new stereocentres.

If the starting molecule is present as a single enantiomer (an existing stereocentre is present), the two possible stereochemical outcomes differ in their relative configuration with respect to that original centre; such stereoisomers are differing in configuration but not mirror images, i.e., they are diastereomers.

Under achiral reaction conditions the nucleophile attacks the bromonium intermediate from the two accessible positions with essentially equal likelihood, so the two diastereomeric products are formed in equal amounts.

Therefore the correct description of the product mixture is a pair of diastereomers in equal amount; it is not a simple racemic mixture of enantiomers, nor two separate optically active single stereoisomers formed alone, nor a pair of enantiomers in unequal amounts.

As shown in the above mechanism 1-butene and 2-butene cannot be formed by this free radical. release of 1 hydrogen radical gives 2-methyl propene as disproportionated product.

Hence, the correct option is C.

This is the case of allylic substitution. The radical will delocalise itself to two locations. And so, we will get two different positions for radical substitution.

There is also a possibility at C₆, but it is not in our option.

For SO₂Cl₂: The reactivity patterns of SO₂Cl₂ and SOCl₂ are quite different. SOCl₂ is a good electrophile, and can be thought of as a source of Cl− ions. These ions can go on to react in their typical nucleophilic fashion. SO₂Cl₂ however is often a Cl₂ source, as it readily decomposes giving off sulfur dioxide. Usually, much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by Cl₂ gas (or something that makes it in solution) happens by a radical mechanism i.e. Cl⋅ not Cl
For Cl₂ and heat/light:

For Cl with AlCl₃: It is used for chlorination of compounds like benzene
For HCl: It is used for halogenations of a double bond.

The correct order for the relative energies of ethane conformations is as follows:

• Staggered (lowest energy)
• Skewed
• Eclipsed (highest energy)

This means that staggered conformations are the most stable, while eclipsed conformations are the least stable due to increased steric strain.

Free radical bromination reaction is highly selective, occurs mainly at the carbon where most stable free radical is formed.
We know that the stability of free radical is in the order,
Tertiary radical > Secondary radical > Primary radical
In (a), (b) and (c), the bromination occurs at secondary carbon whereas in (d) the bromination occurs at tertiary carbon. Since, tertiary radicals are more stable than secondary radical the major product of monobromination of ethyl cyclohexane is (d).
The stability of tertiary radical is due to the higher number of α−Hygrogens which give more hyperconjugation effect than secondary.

Correct answer is 4 because heat of hydrogenation is directly proportional to number of carbon atoms present in it

CH₃• + Cl₂ → CH₃—Cl + Cl•

• This is not a termination.

• It's a propagation step, because a radical (Cl•) is regenerated.

• This is the correct answer.

 To determine the number of different isomers of alkenes (including stereoisomers) that upon catalytic hydrogenation yield the same 2,2,3,5-tetramethylhexane, 

Follow these steps:

1. Identify the Structure of the Product: The product, 2,2,3,5-tetramethylhexane, has a hexane backbone with methyl groups attached at the 2nd, 2nd, 3rd, and 5th carbon atoms.

2. Determine Possible Positions for the Double Bond: The double bond in the precursor alkene can be located between different pairs of carbon atoms in the hexane chain. Possible positions are: - Between C1-C2 - Between C2-C3 - Between C3-C4 - Between C4-C5

3. Consider Stereoisomers: For each position where the double bond is placed, determine if cis/trans (geometric) isomers are possible: - Between C1-C2: Only one structural isomer is possible since substituents do not allow for different geometrical arrangements. - Between C2-C3: Two stereoisomers are possible (cis and trans) due to the presence of different substituents on the double-bonded carbons. - Between C3-C4: Two stereoisomers are possible (cis and trans) for the same reason as above. - Between C4-C5: Only one structural isomer is possible.

4. Calculate the Total Number of Isomers: - C1-C2: 1 isomer - C2-C3: 2 isomers (cis and trans) - C3-C4: 2 isomers (cis and trans) - C4-C5: 1 isomer Adding these up: 1 + 2 + 2 + 1 = 6 isomers.

5. Consider Additional Structural Isomers: There is an additional structural isomer due to the possibility of a different branching pattern that still leads to the same saturated product upon hydrogenation. This increases the total count to 7 isomers. Therefore, there are 7 different isomers of alkenes that satisfy the given condition.

The relative reactivity of secondary hydrogens to primary hydrogens in the free radical bromination of n-butane is approximately 8.36:1. This means secondary hydrogens are about 8.36 times more reactive than primary hydrogens under these conditions.

Given the standard reactivity of alkenes with Br₂ under light, the dominant reaction is electrophilic addition, not free radical substitution. Thus, the major product is the result of adding one Br atom to each carbon of the double bond. This results in formation of 2,3-dibromobutane.

The statements (B) and (C) are true regarding free radical iodination of the alkane.
(B) Direct iodination of alkane with iodine in presence of light is impractical.
During this reaction, Hl, a strong reducing agent is obtained as a byproduct, which catalyzes the reverse reaction thereby preventing direct iodination of alkane.
(C) Iodination of alkane can be achieved successfully using an oxidizing agent catalyst. 
Presence of an oxidizing agent oxidizes unwanted byproduct Hl and enables iodination of alkane to proceed.

For boiling point, we have to consider both branching and Molecular mass. In 4  bromopentane molecular mass is nearly the same as compared to 3 chloro pentane but we have 3,3-dichloropentane extended into 2 directions so the boiling point of 3,3-dichloropentane will be more and the other order will be followed by option C.

After free radical halogenation of methyl cyclobutane, we have its 8different isomers. They are as follow:-


From i) and ii), we get only positional isomers. From iii) we will have 2 isomers, cis and Trans. They won't show a chiral centre.
In iv) we have 2 chiral centres which will give us 4 isomers.So, in total there would be 4+2+1+1 = 8 isomers.

Pentane has three structural isomers:

1. n-Pentane

2. Isopentane (2-methylbutane)

3. Neopentane (2,2-dimethylpropane)

Therefore, total isomers = 3.

In the Wurtz reaction, two alkyl halide molecules react with sodium metal in dry ether to form a higher alkane.

This reaction is commonly used to prepare symmetrical alkanes.

To maximize the yield of chloroethane (C₂H₅Cl) when chlorine reacts with ethane in the presence of light, we need to limit further substitution that leads to products like dichloroethane and trichloroethane.

This reaction follows a free radical substitution mechanism:

To favor monosubstitution (i.e., mainly chloroethane), we must:

• Use excess ethane: This dilutes chlorine radicals, reducing the chance of further chlorination (to di- and tri- products).

In neopentane (2,2-dimethylpropane) all hydrogen atoms are equivalent due to symmetry.

Therefore only one monochloro product forms during chlorination.

Lower heat of hydrogenation → more stable alkene.

Stability order:

tetrasubstituted > trisubstituted > disubstituted > monosubstituted

Among the given options, trans-2-butene is more stable than cis-2-butene because of lower steric repulsion.

Hence it has the lowest heat of hydrogenation.