Match List - I with List - II.

Choose the correct answer from the options given below:

(A)-(III), (B)-(I), (C)-(II), (D)-(IV)

The correct answer is option 4. (A)-(III), (B)-(I), (C)-(II), (D)-(IV).

Lets examine each match between List - I and List - II in detail

(A) \(S_N1\) reaction(III) Racemisation:

In an \(S_N1\) reaction, the process involves the formation of a carbocation intermediate, which is a key step in the reaction mechanism. Because the carbocation is planar and can be attacked from either side, this often leads to racemization of the product. Racemization occurs because the nucleophile can attack the planar carbocation from either side, resulting in a mixture of enantiomers.

Here is a brief overview:

\(S_N1\) Reaction: This is a type of nucleophilic substitution reaction where the rate-determining step is the formation of a carbocation intermediate. The reaction generally proceeds through a two-step mechanism:

1. Formation of the Carbocation: The leaving group departs, forming a carbocation.

2. Nucleophilic Attack: The nucleophile attacks the planar carbocation, resulting in the formation of the product.

Racemization: This refers to the process where a chiral molecule converts into a racemic mixture, which contains equal amounts of both enantiomers. In the context of \(S_N1\) reactions, racemization occurs because the planar carbocation intermediate allows the nucleophile to attack from either side, leading to both possible configurations of the chiral center being formed.

So, if we are associating \(S_N1\) reactions with racemization, it's correct to say that \(S_N1\) reactions often result in racemization due to the formation of a planar carbocation intermediate.

(B) Bromination of alkenes(I) vic-dibromide:

The bromination of alkenes is a reaction where bromine (\(Br_2\)) adds across the double bond of an alkene. This reaction typically results in the formation of a vic-dibromide, which is a diatomic bromine compound with bromine atoms attached to adjacent carbons. Here is a step-by-step outline of the bromination process:

Formation of the Bromonium Ion: The alkene reacts with bromine to form a bromonium ion intermediate. In this step, the \( \pi \)-electrons of the alkene attack one of the bromine atoms, leading to the formation of a cyclic bromonium ion intermediate and the release of a bromide ion (\(Br^-\))

Nucleophilic Attack: The bromide ion (\(Br^-\)) formed in the first step then attacks the more substituted carbon of the bromonium ion intermediate, leading to the formation of the vic-dibromide product.

The result is a vic-dibromide, where each of the original alkene carbons now has a bromine atom attached, and the product is a 1,2-dibromide with the bromine atoms on adjacent carbons (vicinal positions).

So, when you associate the bromination of alkenes with vic-dibromide, you are referring to the product formed when bromine adds across the double bond of an alkene.

(C) Alkylidene halides (II) gem-dibromide:

Alkylidene halides are compounds where a carbon atom is doubly bonded to a halogen atom, with the general formula \(R-C(X)=CH_2\), where X can be a halogen like chlorine or bromine. The term "gem-dibromide" refers to a compound where two bromine atoms are attached to the same carbon atom. This is also known as a geminal dibromide. Here is how alkylidene halides relate to gem-dibromides:

Alkylidene Halides: These are typically compounds where a carbon-carbon double bond is adjacent to a halogenated carbon.

Gem-dibromide: This term specifically describes a molecule where two bromine atoms are attached to the same carbon atom, resulting in a geminal dihalide structure. In a typical reaction, an alkylidene halide (such as an alkene with a halogen attached to one of the carbons) can react with another halogenating agent to form a gem-dibromide.

For example, if you start with an alkylidene halide like \(R-C(Br)=CH_2\) and react it with another bromine source, you can form a gem-dibromide:

In this case, R-C(Br)_2-CH_2 is a gem-dibromide where two bromine atoms are attached to the same carbon atom.

So, alkylidene halides can lead to the formation of gem-dibromides through specific reactions where both halogen atoms end up on the same carbon.

(D) Dehydrohalogenation of haloalkanes (IV) Saytzeff rule:

The dehydrohalogenation of haloalkanes is a reaction where a haloalkane (an alkyl halide) is treated with a strong base to eliminate a hydrogen halide (HX) and form an alkene. The reaction typically follows the Saytzeff rule (also known as the Zaitsev rule), which predicts the major product of the elimination reaction based on the structure of the starting material. Here is a detailed explanation:

Dehydrohalogenation of Haloalkanes:

A haloalkane  reacts with a strong base (like KOH or NaOH) to undergo elimination of HX (where X is a halogen such as Cl, Br, or I). The result is the formation of an alkene. For example:

Mechanism: The base abstracts a proton from a carbon adjacent to the carbon bearing the halogen, leading to the formation of a double bond and the elimination of HX.

Saytzeff Rule: The Saytzeff rule states that in an elimination reaction, the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) is typically the major product. This is because more substituted alkenes are generally more stable due to hyperconjugation and inductive effects from the alkyl groups. Therefore, in the dehydrohalogenation of haloalkanes, the major product will often be the more substituted alkene.

For example, if you have 2-bromo-2-methylbutane and treat it with a strong base, the elimination will yield 2-methylbut-2-ene (a more substituted alkene) as the major product, following the Saytzeff rule. So, in the context of dehydrohalogenation of haloalkanes, the Saytzeff rule is used to predict the major product of the reaction, which is typically the more substituted alkene.