Test: Alkenes - NEET MCQ

Mercuration is the reaction which is not undergone by the alkenes

• Mercuration is not a reaction that alkenes undergo. While it is a type of reaction that involves the use of mercury (Hg), it's not typically associated with alkenes. Mercuration is more often used in the context of organic synthesis where mercury is used as a catalyst to facilitate reactions or as a reagent to introduce mercury into an organic molecule.
• Oxymercuration, on the other hand, is a reaction that alkenes can undergo. It is a process where the alkene reacts with mercuric acetate in an aqueous solution to form an alcohol. This reaction is regiospecific and follows Markovnikov's rule, which means that the hydrogen (from water) attaches to the carbon with the most hydrogen atoms, and the OH group attaches to the other carbon.
• Hydroboration is another reaction that alkenes can undergo. In this reaction, borane (BH₃) or a borane derivative is added to an alkene to produce an organoborane compound, which can then be oxidized to form an alcohol. This reaction is also regiospecific, but it follows anti-Markovnikov's rule, which means that the hydrogen (from borane) attaches to the carbon with the least hydrogen atoms, and the boron attaches to the other carbon.
• Halogenation is a reaction where a halogen (like bromine or chlorine) is added to an alkene to form a dihalide. This reaction can occur in the presence of a solvent like carbon tetrachloride (CCl₄) or in the absence of a solvent. The halogenation reaction is stereospecific, meaning that the addition of the halogen atoms occurs with a specific orientation in space.

In conclusion, of the options provided, Mercuration is not a reaction typically undergone by alkenes.

The correct answer is option B

Option a) Diazene(N₂H₂) is a hydrogenating agent. So, there will be no reaction.
Option b) Al₂O₃+CrO₃ acts as a dehydrogenation catalyst and so an alkene is formed. (Here 1-propene is formed)
Option c) This reaction is Wolff Kishner reaction. Here, acetone would be converted to alkane.
Option d) Zn/CH₃COOH substitutes Cl with H and an alkane is formed.

Alkenes do not undergo mercuration, indeed they undergo oxymercuration , a process in which an alkene is converted into an alcohol.

A Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate or barium sulphate which is then poisoned with various forms of lead or sulphur. It is used for the hydrogenation of alkynes to cis alkenes (i.e. without further reduction into alkanes) When the triple bond is (2-butyne) hydrogenated over the Lindlar’s catalyst i.e.Pd/CaCO₃
 it gives predominantly cis alkene(2-butene). 

Zn reacts with 2,3 - dibromobutane to give alkene. As trans 2-butene is more stable than cis 2-butene(owing tosterric hinderancce), former is major product.

In ‘z’ mechanism, the compounds with higher priority will be located opposite to each other of the double bond, in ‘E’ mechanism the compounds with high priority will be located in z corners and hence 3-methylpent-2-ene is the one which shows E-Z mechanism in which the priority group is CH₃ and CH₂CH₃.

The above reaction between Ethene and bromine is known as electrophilic halogenation reaction and the products usually formed are ethylene dihalides.

The correct option is B trans-1,2-dichloroethylene

• Trans has no (or zero) dipole moment than cis isomer.
• Cis isomer has more dipole moment than trans isomer because it has two similar groups on same side of double bond. So the dipole gets added, thus cis isomer is more polar than trans.
• In the trans isomer, the terminal groups are on the opposite sides of the double bond. So, here the dipole moments of the trans isomer is zero. Hence, trans-1,2-dichloroethylene has zero dipole moment.
• 1,1-dichloroethylene also has some dipole moment because it has two similar groups on same side of double bond. So the dipole gets added.

Methene compound does not exist according to the formula C_nH_2n and also due to the lack of C=C.

To determine which compound reacts most readily with gaseous bromine, we need to analyze the given options based on their structure and the type of bonds they contain: 

Step 1: Identify the Types of Compounds
- C₃H₆ is propene (an alkene).
- C₂H₂ is acetylene (an alkyne).
- C₄H₁₀ is butane (an alkane).
- C₂H₄ is ethene (an alkene).

Step 2: Understand Electrophilic Addition Reaction
Electrophilic addition reactions occur when compounds with high electron density react with electrophiles. Alkenes (compounds with double bonds) are more reactive in these reactions compared to alkanes (single bonds) and alkynes (triple bonds).

Step 3: Analyze Reactivity of Each Compound
- C₃H₆ (Propene): This is an unsymmetrical alkene. It has a double bond, which provides a site for electrophilic attack.
- C₂H₂ (Acetylene): This is an alkyne with a triple bond. Although it can react with bromine, it is less reactive than alkenes in electrophilic addition.
- C₄H₁₀ (Butane): This is an alkane with only single bonds, making it the least reactive towards bromine.
- C₂H₄ (Ethene): This is a symmetrical alkene and also has a double bond, allowing for electrophilic addition.

Step 4: Compare the Alkenes
Among the alkenes (C₃H₆ and C₂H₄), propene (C₃H₆) is unsymmetrical, which generally makes it more reactive in electrophilic addition reactions compared to symmetrical alkenes like ethene (C₂H₄).

Conclusion:
The compound that will react most readily with gaseous bromine is C₃H₆ (Propene).

Explanation of the Reaction of Propene with HBr in the Presence of Peroxide

The reaction of propene with hydrogen bromide in the presence of peroxide follows the rule of anti-Markovnikov addition. This rule states that the hydrogen (H) from HBr will add to the carbon with the most hydrogen atoms already attached, and the bromine (Br) will add to the other carbon. Peroxide promotes this anti-Markovnikov addition.

In the case of propene (CH₃-CH=CH₂), the hydrogen from HBr will add to the terminal carbon, which already has two hydrogens. The bromine will add to the middle carbon.

Here are the steps of the reaction:

The stability of an alkene is determined by the steric hindrance and the extent of hyperconjugation.

• Steric Hindrance: Steric hindrance is the interference between two bulky groups that are so close together that their electron clouds experience a repulsion. This increases the energy levels making the molecule less stable.
• Hyperconjugation: Hyperconjugation is a stabilizing interaction that involves the delocalization of σ electrons. It is an intramolecular process, which involves the interaction of the electrons in a σ-bond with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system.

Comparison of the given alkenes:

• Cis-2-Butene: In Cis-2-Butene, the two methyl groups are on the same side of the double bond, causing a higher steric hindrance making the molecule less stable.
• Trans-2-Butene: In Trans-2-Butene, the two methyl groups are on opposite sides of the double bond, resulting in a lower steric hindrance than Cis-2-Butene. Also, it has more hyperconjugation structures, making it more stable.
• 1-Butene: 1-Butene has less hyperconjugation than Trans-2-Butene. Therefore, it is less stable than Trans-2-Butene.

Conclusion: From the above explanation, we can conclude that Trans-2-Butene is the most stable alkene among the given options due to its lower steric hindrance and higher extent of hyperconjugation.

Reaction of Ethylene Bromide with Zinc
Ethylene bromide, also known as 1,2-dibromoethane, is a halogenated hydrocarbon. When it is treated with zinc, an alkene is formed as a result. This reaction can be detailed as follows:

Why not Alkyne or Alkane?
An alkyne would require the removal of two pairs of hydrogen and bromine atoms, which does not occur in this reaction. An alkane would not have any double bonds, and the reaction with zinc specifically creates a double bond.

In conclusion, the correct answer is B: Alkene, because the reaction of ethylene bromide with zinc results in the formation of an alkene, specifically ethene. You can learn more about organic chemistry reactions on the EduRev platform.

Explanation of Alkene Reactions

Alkenes are a type of hydrocarbon that contains a carbon-carbon double bond. This double bond makes alkenes particularly reactive. The most common type of reaction that alkenes undergo is known as the addition reaction.

Addition Reactions

• An addition reaction in alkenes is where the carbon-carbon double bond is broken and atoms or groups of atoms are added to each carbon.
• These reactions are characterized by the breaking of the pi bond (double bond) and the formation of two new single bonds (sigma bonds).
• Common examples of addition reactions are hydrogenation (addition of hydrogen), halogenation (addition of halogens), and hydration (addition of water).

Elimination and Substitution Reactions

• Elimination and substitution reactions are less common in alkenes. In fact, these reactions are more often observed in alkanes or alkyl halides.
• In elimination reactions, atoms or groups of atoms are removed from the molecule, leading to the formation of a double or triple bond.
• In substitution reactions, one atom or group of atoms is replaced by another atom or group of atoms.

• Superposition is not a type of chemical reaction. Instead, it is a concept from physics that refers to the ability of waves to add up to form a new wave when they overlap.

Therefore, the most common reaction in alkenes is the addition reaction.