OZONOLYSIS OF ALKENE

 

The reaction of (cyclohexylidenemethyl)benzene with ozone followed by zinc metal is known as the Ozonolysis reaction. This reaction involves the oxidative cleavage of the carbon-carbon double bond in the alkene, resulting in the formation of two carbonyl compounds: cyclohexanone and benzaldehyde.

In the basic condition the cross-aldol condensation reaction between cyclohexanone and benzaldehyde involves the condensation of the carbonyl group of one compound with the α-carbon of the other compound, resulting in the formation of a β-hydroxyketone. 3-benzylidene-cyclohexanone

ANSWER B

TAUTOMERISM

 

Tautomerism is a chemical phenomenon where a molecule can exist in two different forms that can rapidly switch back and forth. These different forms are called tautomers. The switch between tautomers happens because a hydrogen atom moves around within the molecule.

The most common type of tautomerism involves a molecule changing between a form with a certain arrangement of atoms (called the keto form) and a form with a slightly different arrangement of atoms (called the enol form). This change usually involves the movement of a hydrogen atom.

These different forms can have different chemical properties and behaviors. For example, they may react differently with other substances or have different levels of acidity. The balance between the different tautomeric forms depends on factors like temperature, the type of solvent, and the pH of the solution.

Tautomerism can occur under specific conditions, such as:

1. Presence of functional groups: Tautomerism is commonly observed in compounds that contain certain functional groups, such as carbonyl (C=O) and hydroxyl (OH) groups.

2. Proton transfer: Tautomerism involves the migration of a hydrogen atom or proton within the molecule. This transfer can occur when there are appropriate acidic or basic sites within the molecule.

3.  Favorable thermodynamics: Tautomerism is influenced by factors such as temperature and energy differences between the tautomeric forms. The conversion between tautomers typically occurs when it is thermodynamically favorable.

4.  Solvent effects: The choice of solvent can impact tautomerism. Different solvents can stabilize or destabilize specific tautomeric forms, leading to a shift in the equilibrium between them.

5.  pH dependence: Tautomeric equilibria can be pH-dependent. For example, in the case of keto-enol tautomerism, the enol form is typically favored under acidic conditions, while the keto form is more stable under basic conditions.

One specific example of tautomerism is the interconversion between the keto form and the enol form of a compound called tautomeric aldehydes or ketones.

Aldehydes and ketones are organic compounds that contain a carbonyl group (C=O). Tautomeric aldehydes or ketones exhibit tautomeric behavior due to the presence of certain functional groups and the ability to undergo proton transfer.

In the keto form, tautomerone has a carbonyl group (C=O) where the carbon atom is bonded to an oxygen atom. In the enol form, the carbonyl group is converted to a hydroxyl group (-OH) adjacent to a double bond.

The interconversion between the keto and enol forms occurs through the migration of a hydrogen atom. The process involves the transfer of a proton from the carbon atom adjacent to the carbonyl group (the α-carbon) to the oxygen atom of the carbonyl group, resulting in the formation of a double bond and the hydroxyl group.

Tautomeric aldehydes or ketones exist as a dynamic equilibrium mixture of the keto and enol tautomers. The ratio between the two forms is influenced by various factors, including temperature, solvent, and pH. These factors affect the stability and energy difference between the tautomeric forms.

ANSWER  (d)

REACTION OF KETONE WITH GRIGNARD REAGENT

 

When a ketone reacts with a Grignard reagent, a reaction known as a Grignard reaction occurs. The reaction proceeds through a nucleophilic addition mechanism, resulting in the formation of an alcohol.

The Grignard reagent, which is an organomagnesium compound, acts as a strong nucleophile and attacks the carbonyl carbon of the ketone. The resulting intermediate is an alkoxide ion, which then protonates to form the corresponding alcohol.

ANSWER (a)

FREE RADICAL REACTION OF ALKENE POSSESING TERTIARY HYDROGEN

 A free radical reaction involving an alkene typically refers to a reaction where a radical species (a molecule or atom with an unpaired electron) reacts with an alkene to form a new product. One common example of such a reaction is the addition of a halogen to an alkene, known as halogenation.

In halogenation, a halogen molecule (such as chlorine or bromine) breaks its covalent bond to form two halogen radicals. One of these halogen radicals then reacts with the alkene, forming a new carbon-halogen bond and generating an alkyl radical. The alkyl radical can then react with another halogen molecule to produce the final product.

When an alkene containing a tertiary hydrogen (a hydrogen atom bonded to a tertiary carbon atom) undergoes a free radical reaction, it can result in the formation of alkyl radicals and subsequent radical rearrangements. One common example is the reaction known as tertiary hydrogen abstraction.

In this reaction, a radical species abstracts a hydrogen atom from the tertiary carbon of the alkene, creating an alkyl radical and leaving behind a radical species on the alkene. The alkyl radical can then participate in various reactions depending on the reaction conditions and the nature of the radical species present.

ANSWER (d)

E2 REACTION (Part 1)

The E2 reaction involves the simultaneous removal of a leaving group and a hydrogen atom from adjacent carbon atoms to form a double bond. It occurs through a concerted mechanism, and its rate depends on the concentrations of the substrate and the base/nucleophile. Factors such as the strength of the base, stability of the leaving group, steric hindrance, and solvent effects influence the reaction.

When a bulky base reacts with a secondary alkyl halide, the reaction can proceed through E2 elimination. The bulky base abstracts a proton from the beta carbon adjacent to the halide, causing the leaving group to depart simultaneously. This results in the formation of a double bond. However, steric hindrance around the reacting carbon atoms may affect the rate and selectivity of the E2 reaction.

When a secondary alkyl halide reacts with a bulky base, the product outcome can vary in terms of regioselectivity, resulting in either the Zaitsev or Hofmann product. Zaitsev product is typically favored in most E2 reactions, there are cases where the Hofmann product can be the major or exclusive product, especially when steric hindrance is significant. The actual product distribution will depend on the specific reaction conditions and the substrate used.

When 2-bromobutane reacts with tert-butoxide (t-BuO-), a strong and bulky base, an E2 elimination reaction can occur.

In this reaction, the tert-butoxide acts as a base and abstracts a proton from the beta carbon (adjacent to the bromine) in 2-bromobutane. At the same time, the bromine atom serves as the leaving group. This leads to the formation of a double bond, resulting in the production of 2-butene. Additionally, tert-butyl bromide is generated as a byproduct.

The use of tert-butoxide as a bulky base can influence the regioselectivity of the reaction. The steric hindrance provided by the bulky tert-butoxide can hinder the approach to the more substituted beta carbon. As a result, the reaction may exhibit some degree of Hofmann product selectivity, leading to the formation of the less substituted alkene in some cases.

ANSWER (b)

E1 REACTION (Part 1)

 E1 reaction, also known as Elimination-Unimolecular reaction, is a type of chemical reaction that involves the removal of a leaving group and a proton from adjacent carbon atoms in a molecule, resulting in the formation of a double bond.

The reaction mechanism involves a two-step process: in the first step, a leaving group departs from the molecule, generating a carbocation intermediate. In the second step, a proton is removed from an adjacent carbon atom, resulting in the formation of a double bond and the regeneration of a protonated leaving group.

E1 reactions typically occur in the presence of a strong base or heat, and they are most commonly observed in reactions involving secondary or tertiary alkyl halides. The rate of an E1 reaction depends only on the concentration of the substrate, as the reaction involves the formation of a carbocation intermediate that can be stabilized by neighboring groups.

Zaitsev and Hoffman products refer to two possible products that can be formed during an elimination reaction, particularly when a base is used to remove a proton from a beta-carbon atom in a molecule.

The Zaitsev product is the more stable and predominant product, which is formed when the elimination reaction occurs through the transition state that leads to the most substituted alkene. This product is also known as the "Saytzeff" product.

On the other hand, the Hoffman product is the less stable and less substituted product, which is formed when the elimination reaction occurs through the transition state that leads to the least substituted alkene. This product is also known as the "anti-Zaitsev" product.

The preference for the formation of Zaitsev or Hoffman products depends on the reaction conditions, the nature of the substrate, and the strength of the base used. Generally, Zaitsev products are favored in reactions involving strong bases and substrates that can stabilize the negative charge of the alkene intermediate through resonance or inductive effects. Meanwhile, Hoffman products are favored in reactions involving weaker bases or substrates that cannot stabilize the negative charge of the alkene intermediate.

It is important to note that while Zaitsev products are generally more stable and predominant, there are some instances where Hoffman products may be preferred, such as when steric hindrance around the beta-carbon atom makes it difficult for the base to approach and remove the proton from that position.

ANSWER (a)

Dehydration of Tertiary Alcohol

The dehydration of tertiary alcohols follows the E1 mechanism (elimination unimolecular), which involves the formation of a carbocation intermediate.

In the first step of the reaction, a proton from the beta-carbon adjacent to the hydroxyl group is removed by a strong acid, such as sulfuric acid (H₂SO₄) or phosphoric acid (H₃PO₄), resulting in the formation of a carbocation intermediate. Tertiary carbocations are relatively stable due to the presence of three alkyl groups that can stabilize the positive charge of the carbocation.

In the second step, a water molecule acts as a base and removes a proton from the beta-carbon, leading to the formation of a double bond and the release of a protonated water molecule (H3O+). The resulting product is an alkene, which is typically the major product of the reaction.

The (c) is the major product, (b) and (d) are the minor product. The (a) is not the product because the proton that has been remove is not in beta position. Furthermore, the hydride shift do not occur because its already tertiary carbocation.

JABIR IBN HAYYAN: Bapa Kimia Dunia Islam

 

Jabir ibn Hayyan, juga dikenali sebagai Geber di dunia Barat, adalah seorang ahli alkimia, ahli kimia, dan ahli falsafah terkemuka yang hidup pada abad ke-8 di Iran kini. Beliau dianggap sebagai salah satu tokoh paling penting dalam sejarah kimia dan alkimia, dan karyanya mempunyai impak yang signifikan terhadap perkembangan kedua-dua bidang tersebut.

Jabir ibn Hayyan dilahirkan di bandar Tus di Iran, sekitar tahun 721 Masihi. Beliau menerima pendidikan di Kufa, yang merupakan pusat pembelajaran pada masa itu, dan kemudian berpindah ke Baghdad, di mana beliau meneruskan kajian dan eksperimennya. Jabir ibn Hayyan dikenal pasti dengan pendekatan eksperimentalnya dalam alkimia, yang melibatkan pengujian teorinya melalui eksperimen praktikal dan pemerhatian.

Jabir ibn Hayyan menulis lebih daripada 3,000 karya dalam pelbagai subjek, termasuk kimia, alkimia, falsafah, perubatan, dan astrologi. Beliau dikenal pasti dengan sumbangan terhadap perkembangan pelbagai proses kimia, seperti penyulingan alkohol dan penyediaan asid. Beliau juga berminat dalam transmutasi logam, dan beliau percaya bahawa ia mungkin untuk menukar logam asas menjadi emas.

Karya Jabir ibn Hayyan diterjemahkan ke bahasa Latin semasa zaman Pertengahan, dan idea-ideanya mempunyai impak yang signifikan terhadap perkembangan alkimia dan kimia Barat. Pendekatan beliau terhadap eksperimen dan penekanan pada pemerhatian dan pengalaman praktikal adalah instrumental dalam membentuk kaedah sains, yang masih digunakan hari ini.

Secara keseluruhannya, sumbangan Jabir ibn Hayyan terhadap bidang kimia dan alkimia, serta pengaruhnya terhadap perkembangan sains dan falsafah, menjadikannya salah satu tokoh yang paling penting dalam sejarah sains.