Test: Alkenes - NEET MCQ

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.

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 reaction between ethylene and bromine is called an electrophilic halogenation reaction. It typically produces a compound known as an ethylene dihalide.
C₂​H₄​+Br₂​→C_2​H₄​Br_2​

• Ethylene reacts with bromine to form a dihalide.
• In this process, bromine atoms add across the double bond of ethylene.
• The product is usually a dibromoethane, where each carbon atom of the original double bond gets a bromine atom.

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.

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:

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.

Key Points

•  C₂H₅Br + KOH → C₂H₄ + KBr +H₂O
• The above reaction is the reaction between alkyl halide with alcoholic alkali.
• When ethyl bromide reacts with alcoholic potassium hydroxide (KOH), ethene (C₂H₄) is formed.
• This reaction is called an Elimination or dehydrohalogenation reaction.

Additional Information C₂H₂

• C₂H₂ is the chemical formula of acetylene.
• It contains a triple bond between two carbon atoms.

C₂H₆

• C₂H₆ is the chemical formula of ethane.
• It is the second member of the alkane group.
• It is a straight-chain alkane containing two carbon atoms having single covalent bonds.

CH₄

• CH₄ is the chemical formula of methane.
• It is the first member of the alkane group.
• It is a gaseous compound.

C₃H₆

• C₃H₆ is the chemical formula of propene and cyclopropane.
• Propene is an alkene having double bonds.
• Cyclopropane is a cyclic ring alkane containing 3 carbon atoms.

Ozonides are formed by dissolving alkenes in an inert solvent such as CHCl3/CCl4 and passing ozone through it, on reduction of which carbonyl compound are obtained this ozonide is a reactive compound and with H2/Pd. On reduction, they react with explosion, they are also reduced with H2O/Zn.

The given alkenes for their products upon ozonolysis:

• CH₂=CH₂ (Ethene): Ozonolysis of ethene would yield formaldehyde (HCHO) as the product:
  CH₂=CH₂ + O₃ → 2HCHO
• CH₃-CH=CH-CH₃ (But-2-ene): Ozonolysis of but-2-ene would yield two molecules of acetaldehyde (CH₃CHO) as the product:
  CH₃-CH=CH-CH₃ + O₃ → 2CH₃CHO
  
• (CH₃)₂C=C(CH₃)₂ (2,3-Dimethylbut-2-ene): Ozonolysis of 2,3-dimethylbut-2-ene would yield two molecules of acetone (CH₃COCH₃) as the product:
  (CH₃)₂C=C(CH₃)₂ + O₃ → 2CH₃COCH₃
• CH₃-CH=CH₂ (Propene): Ozonolysis of propene would yield acetaldehyde (CH₃CHO) and formaldehyde (HCHO) as the products:
  CH₃-CH=CH₂ + O₃ → CH₃CHO + HCHO

Among the given alkenes, only 2,3-dimethylbut-2-ene ((CH₃)₂C=C(CH₃)₂) yields acetone (CH₃COCH₃) upon ozonolysis.

Key Points

• Chloroethene is commonly known as vinyl chloride.
• Vinyl refers to a specific functional group (CH₂=CH^-) attached to a carbon atom.
• It is a colourless gas with the chemical formula C₂H₃Cl.
• It is a key industrial chemical used to produce polyvinyl chloride (PVC), which is used in a wide range of plastic products.

Additional Information

• Methylene Chloride
  • Also known as dichloromethane (DCM), it is a volatile, colorless liquid with a sweet odor.
  • Used as a solvent in paint strippers, degreasers, and in the production of pharmaceuticals.
  • Exposure can affect the central nervous system, causing symptoms such as dizziness and headaches.
• Chloroform
  • Also known as trichloromethane, it is a colorless, sweet-smelling organic compound.
  • Used historically as an anesthetic, though it is no longer commonly used due to its potential toxicity.
  • Currently used in laboratories and as a solvent in the pharmaceutical industry.
• Carbon Tetrachloride
  • A colorless, volatile liquid with a characteristic sweet odor.
  • Used primarily in the past as a cleaning agent and in fire extinguishers.
  • Exposure can cause liver and kidney damage, and it is considered a probable human carcinogen.

ANTI MARKOVNIKOV RULE: 

Addition of halo acid or protic acid to an alkene in presence of peroxide in such a way that proton makes a bond with most substituted carbon atom of alkene or in simple words proton will form a bond with carbon which has less no. of hydrogens and nucleophile will form a bond with carbon which has more no. of hydrogens.

Mechanism: Peroxide effect proceeds via the free radical chain mechanism

Key Points

• Unsaturated hydrocarbons such as alkenes add Hydrogen in the presence of catalysts such as Ni to form saturated hydrocarbons. This is called an addition reaction.
• The reaction is commonly used in the hydrogenation of vegetable oils using a Nickel catalyst.
• The reaction is:

Key Points

• Hydrocarbon fuel is the fuel that is derived from a hydrocarbon, which includes gasoline and jet fuel.
• Unsaturated hydrocarbons give a yellow flame with lots of black smoke while saturated hydrocarbons give a clean flame.
• Unsaturated hydrocarbons are hydrocarbons that have double or triple covalent bonds between adjacent carbon atoms.  For instance- Alkenes (C_nH_2n) and Alkynes (C_nH_2n−2).
• The color of the flame is yellow because of the presence of unburnt carbon particles.
• Saturated hydrocarbon is a hydrocarbon in which all the carbon-carbon bonds are single bonds.

The compound but-3-ene have a structural formula different from the above three and hence it is not a structural isomer of C₄H₈.

Alkenes are not less reactive than alkanes, indeed they are very reactive compared with them due to the presence of C=C.