Test: Reaction Mechanism - NEET MCQ

• Initiation step: This is where the free radical reaction begins. Free radicals are formed through homolytic bond cleavage, often using heat or light.

• Propagation step: The radicals formed in the initiation step react with stable molecules, creating more free radicals and continuing the chain reaction.

• Termination step: Free radicals combine to form stable molecules, effectively removing radicals rather than forming them.

Correct Answer: D: both A and B

The observed order of the stability of the carbocation is:

To solve this, recall the general order of carbocation stability:

Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl

This is due to:

• Hyperconjugation

• +I (inductive) effect of alkyl groups

• Resonance (not involved here directly)

Analyzing the given carbocations:

• CH₃⁺ → methyl carbocation (least stable)

• CH₃CH₂⁺ → primary (1°) carbocation

• (CH₃)₂CH⁺ → secondary (2°) carbocation

• (CH₃)₃C⁺ → tertiary (3°) carbocation (most stable)

Correct stability order:
CH₃⁺ < CH₃CH₂⁺ < (CH₃)₂CH⁺ < (CH₃)₃C⁺

Heterolytic fission, also known as heterolysis, is a type of bond fission in which a covalent bond between two chemical species is broken in an unequal manner, resulting in the bond pair of electrons being retained by one of the chemical species (while the other species does not retain any of the electrons from the bond pair). When a neutrally charged molecule undergoes heterolytic fission, one of the products will have a positive charge whereas the other product will have a negative charge.

The correct answer is option A
Reaction mechanism: A sequential account of each step, describing details of electron movement, energetics during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism.
 

The stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon due to the +I (inductive) effect and hyperconjugation, which help delocalize and stabilize the positive charge. - Option (a) is a primary carbocation, which is the least stable. - Option (b) is a primary carbocation adjacent to a tertiary carbon, but the positive charge is still on a primary carbon. - Option (c) is a secondary carbocation, more stable than primary but less than tertiary. - Option (d) is a tertiary carbocation ((CH₃)₃C⁺), where the positive charge is stabilized by three alkyl groups. Hence, it is the most stable carbocation among the given options. Therefore, option (d) is correct.

Carbocation is most reactive as it lacks electrons. So it need an electron to attain stability.

The correct answer is option B
Electron poor species are stabilized by neighboring atoms that can donate electron density. The most common way to interpret rich neighbors here is the observation that increasing the number of alkyl groups on the carbon bearing the free radical increases its stability. 
Radical stability increases in the order methyl < primary < secondary < tertiary.
 

A carbanion has a negatively charged, trivalent carbon atom that has eight electrons in its valence shell. Thus, a carbanion is not electron deficient. As a result, carbanions have pyramidal geometry. Carbocations, radicals, and carbanions can be stabilized by resonance.

Carbon atoms with its sextet are called carbocation. It is an electron deficient species.

In heterolytic bond cleavage the bond breaks unevenly and the shared pair of electrons is accommodated by one of the products, which produces one or more ions.As heterolytic bond cleavage gives ions. So the reaction which proceeds through heterolytic bond cleavage is an ionic reaction.

Hence B is the correct answer.