Problem 10
Question
The addition of \(\mathrm{HCl}\) to \(3,3.3\)-trichloropropene gives (a) \(\mathrm{Cl}_{3} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{Cl}_{3} \mathrm{CCH}(\mathrm{Cl}) \mathrm{CH}_{3}\) (c) \(\mathrm{Cl}_{2} \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CL}_{2} \mathrm{CHCH}_{2} \mathrm{CHCl}_{2}\)
Step-by-Step Solution
Verified Answer
(b) \(\mathrm{Cl}_3\mathrm{CCH}(\mathrm{Cl})\mathrm{CH}_3\) is the product formed.
1Step 1: Identify Reactant and Reagent
The reactant given is 3,3,3-trichloropropene, which is an alkene with the structure \ \(\mathrm{Cl}_3\mathrm{CCH} =\mathrm{CH}_2\). The reagent added is hydrochloric acid (HCl). We aim to predict the product formed when HCl adds to this alkene.
2Step 2: Understand Markovnikov's Rule
The addition of hydrogen halides (2HCl2 in this case) to alkenes generally follows Markovnikov's rule. According to this rule, the hydrogen atom from HCl will add to the less substituted carbon atom in the double-bonded system, while the chlorine atom will add to the more substituted carbon.
3Step 3: Apply Markovnikov's Rule to the Alkene
In the case of \ \(\mathrm{Cl}_3\mathrm{CCH} =\mathrm{CH}_2\), the carbon atom of \ \(\mathrm{CH}_2\) is less substituted compared to the carbon in \ \(\mathrm{CCH}\) that bears more chlorine atoms. Therefore, the hydrogen from HCl will add to the \ \(\mathrm{CH}_2\) and the chlorine will attach to the more substituted \ \(\mathrm{CCH}\).
4Step 4: Draw the Expected Product Structure
Based on Markovnikov's addition, the product will be \ \(\mathrm{Cl}_3\mathrm{CCH}(\mathrm{Cl})-\mathrm{CH}_3\). This corresponds to chlorine adding to the carbon that already has chlorine atoms and hydrogen adding to the other carbon.
5Step 5: Identify the Correct Option
Upon comparing the anticipated product structure \ \(\mathrm{Cl}_3\mathrm{CCH}(\mathrm{Cl})\mathrm{CH}_3\) with the provided options, it matches exactly with option (b). Option (b) is \ \(\mathrm{Cl}_3\mathrm{CCH}(\mathrm{Cl})\mathrm{CH}_3\).
Key Concepts
Hydrogen halide additionAlkene reactionsOrganic chemistry mechanisms
Hydrogen halide addition
The term "hydrogen halide addition" refers to a specific reaction where hydrogen halides such as HCl, HBr, or HI are added to alkenes — hydrocarbons characterized by at least one carbon-carbon double bond. This process is fundamentally important in organic chemistry as it transforms alkenes into more reactive and functionalized molecules. When a hydrogen halide reacts with an alkene, the addition occurs across the double bond to form a haloalkane.
This reaction is typically electrophilic, meaning the halide portion of the molecule (like Cl from HCl) attacks the electron-rich region of the alkene double bond, leading to the formation of a carbocation intermediate. This new structure temporarily holds the positive charge on a carbon atom.
What's crucial is the direction in which the halide and hydrogen atoms add across the double bond. Following chemist Markovnikov's observations, the hydrogen will preferably attach itself to the less substituted carbon atom in the double bond, while the halide will go to the more substituted, resulting in the most stable product.
This reaction is typically electrophilic, meaning the halide portion of the molecule (like Cl from HCl) attacks the electron-rich region of the alkene double bond, leading to the formation of a carbocation intermediate. This new structure temporarily holds the positive charge on a carbon atom.
What's crucial is the direction in which the halide and hydrogen atoms add across the double bond. Following chemist Markovnikov's observations, the hydrogen will preferably attach itself to the less substituted carbon atom in the double bond, while the halide will go to the more substituted, resulting in the most stable product.
Alkene reactions
Alkene reactions are central to transforming simple hydrocarbons into more complex compounds in organic chemistry.
Some typical alkene reactions include:
Some typical alkene reactions include:
- Hydrogenation: Addition of hydrogen (H₂) across the double bond to turn the alkene into an alkane, typically requiring a metal catalyst such as Pd or Pt.
- Halogenation: Addition of halogens, such as Br₂ or Cl₂, transforming alkenes into dihaloalkanes through a clear mechanism involving the formation of a cyclic intermediate.
- Hydration: Introduction of water across the double bond, often in the presence of an acid catalyst, leading to alcohol formation.
- Polymerization: Linking of alkene monomers into long-chain polymers, fundamental in producing plastics like polyethylene.
Organic chemistry mechanisms
Understanding organic chemistry mechanisms is essential for grasping how reactions occur at the molecular level. The statement "mechanism" refers to the step-by-step sequence of elementary reactions by which a chemical change occurs.
There are several fundamental concepts to grasp different mechanisms:
There are several fundamental concepts to grasp different mechanisms:
- Electron Movement: Electrons often move from areas of high electron density (nucleophiles) towards regions of low electron density (electrophiles).
- Intermediates: Many reactions proceed through fleeting intermediates, such as carbocations, which can dictate the pathway and stability of the reaction.
- Transition States: Represent the highest energy point that must be overcome for a reaction to proceed, often providing insights into reaction kinetics and energy barriers.
- Regioselectivity and Stereoselectivity: These dictate the preferred orientation and arrangement of products, with regioselectivity guiding where a reaction occurs and stereoselectivity affecting the spatial arrangement of atoms.
Other exercises in this chapter
Problem 8
Which among the following alkenes will be most stable? (a) Ithene (b) 2-Methylpropene (c) 2. 3. Dimethyl-but-2-ene (d) 2.Butene
View solution Problem 9
Ethylene reacts with \(\mathrm{O}_{2}\) at \(200^{\circ} \mathrm{C}\) in the presence of silver powder to produce (a) ethanol (b) cthanal (c) cthanoic acid (d)
View solution Problem 12
2-Butyne is treated with sodium in liquid \(\mathrm{NH}_{3}\). The major product formed is (a) cis-2-butene (b) trans-2-butene (c) butane (d) \(1.3\)-butadiene
View solution Problem 14
An alkene on reductive ozonolysis gives 2 molecules of \(\mathrm{CH}_{2}(\mathrm{CHO})_{2}\). The compound is (a) 2, 4 -hexadiene (b) 1, 3-cyclohexadiene (c) \(
View solution