Haloalkanes And Haloarenes MCQ Quiz in मल्याळम - Objective Question with Answer for Haloalkanes And Haloarenes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

By direct halogenation of benzene ring:

if X = Cl then the reaction is as follows:

CONCEPT:

Dipole Moment of Haloalkanes

• The dipole moment (μ) is a measure of the polarity of a bond, calculated as the product of the charge difference (δ) between two atoms and the bond distance (d).
• In haloalkanes, the dipole moment depends on two factors:
  • Electronegativity: The difference in electronegativity between carbon and the halogen atom. Higher electronegativity means a stronger dipole moment.
  • Bond length: As the bond length increases, the dipole moment increases if other factors remain constant.
• The trend in dipole moment for haloalkanes is influenced by the balance between electronegativity (which decreases as you go down Group 17 in the periodic table) and bond length (which increases down the group).

EXPLANATION:​

• Although fluorine has the highest electronegativity, the short C–F bond length reduces the overall dipole moment.
• Chlorine, having a longer bond length and moderate electronegativity, results in a higher dipole moment compared to CH₃F.
• Bromine's lower electronegativity and longer bond length reduce its dipole moment below that of CH₃Cl.
• Iodine, with the lowest electronegativity and longest bond length, results in the smallest dipole moment.

SUMMARY:

CONCLUSION:

The correct increasing order of dipole moments is: CH₃I < CH₃Br < CH₃F < CH₃Cl

Concept:

Due to partial double bond character of C-halogen bond, halogen leaves with great difficulty, if at all it does, hence vinyl halides do not undergo nucleophilic substitution easily. So, assertion is correct.

Explanation:

The given reaction is - \(C_7H_8\xrightarrow{3CI_2/\Delta}A\xrightarrow{Br_2Fe}B\xrightarrow{Zn/HCI}C\)

Reactant is toluene in the above reaction.

The reaction contains three steps -

1. Free radical substitution
2. Electrophillic substitution
3. Simple substitution Reaction

⇒Free radical substitution - 

• It occurs in absence of any hydrogen carrier.
• In this step free radical substitution reaction of chlorine takes place and all three H of -CH₃ side groups in toluene are replaced by the chlorine one by one.
• Benzochloride is obtained as the product of this step.

So, A in the given reaction is benzochloride.

⇒ Electrophillic substitution - 

• Benzochloride in presence of Br₂/Fe give an electrophilic substitution reaction.
• Br2/Fe generates a free radical Br^- which attacks on the meta position.
• m-bromo benzochloride is obtained as a product of the second step.

So, B will be m-bromo benzochloride.

⇒ Simple substitution reaction - 

• As HCl is a polar protic solvent, it replaces the Cl of side chain -CCl₃ and changes it into -CH₃.
• m-bromo toluene is obtained as the product of this step.

So, C will be m-bromotoluene.

The overall reaction is written as -

∴ final product or C in the given reaction is m-bromotoluene.

Hence, the correct option is (1).

Concept:

• During organic chemical reactions, covalent bonds are broken and formed.
• This leads to the formation of various reaction intermediates like carboniums, carbanions, free radicals, carbenes etc.
• Carbanions: In carbanions, a carbon center is generally bonded to three groups and bears a negative charge.
• Carbonium ions: In carbanions, a carbon center is generally bonded to five groups and bears a positive charge.
• Free Radicals: These are neutral species that have single lone electrons on their heads and are highly reactive.
• The covalent bonds can be broken by two processes of fission : 
  • Homolysis
  • Heterolysis

​Explanation:

​​Homolysis and Heterolysis:

• So when the Carbon - Chlorine bond is broken heterolytically, a cation and an anion are formed.
• An example is shown below:

Additional Information

• An example of homolysis is shown below:

Explanation:-

Step 1: From \(CH _3 -CH _2 -CH _2 -Br \)to the intermediate product
We start with a primary alkyl halide, \(CH _3 -CH _2 -CH _2 -Br \)(1-bromopropane). 

Possible reaction with X (reagent):
- Using concentrated alcoholic NaOH at 80°C will result in an elimination reaction, where the bromine atom and a hydrogen atom are removed, forming an alkene (propene).

Reaction:
\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{Br} \xrightarrow{\text{conc. alc. NaOH, 80°C}} \text{CH}_2=\text{CH}-\text{CH}_3 + \text{HBr} \)

So, after reaction with X (conc. alc. NaOH, 80°C), we obtain propene.

Step 2: From the intermediate product (propene) to \(CH _3 -CH(Br)-CH _3\)
Possible reaction with Y (reagent):
- Addition of HBr to propene will follow Markovnikov's rule, where the bromine atom attaches to the carbon with the most hydrogen atoms (the more substituted carbon), leading to 2-bromopropane.

Reaction:
\(\text{CH}_2=\text{CH}-\text{CH}_3 + \text{HBr} \rightarrow \text{CH}_3-\text{CH(Br)}-\text{CH}_3 \)

So, the correct transformation involves the addition of HBr to propene, resulting in 2-bromopropane.

Conclusion
Based on the steps and the required reagents:
- X should be concentrated alcoholic NaOH at 80°C to perform the elimination reaction.
- Y should be HBr to perform the addition reaction.

Thus, the correct set of reagents or reaction conditions is \( \text{ X = conc. alc. NaOH, 80°C; Y = HBr/acetic acid} \)

CONCEPT:

Elimination Reaction (E2 Mechanism)

• In elimination reactions, a β-hydrogen is removed along with a leaving group (such as Br) to form a double bond, resulting in the formation of an alkene.
• The reaction proceeds via an E2 mechanism, which is a single-step process where the base abstracts a proton while the leaving group departs, leading to the formation of the alkene.
• According to Zaitsev's rule, the most substituted alkene (the one with the greatest number of alkyl groups attached to the double bond) is the major product.

EXPLANATION:

• In the given reaction, the elimination of HBr occurs in the presence of a strong base (ethoxide ion) in ethanol, leading to the formation of the most stable alkene.
• The base abstracts a proton from the β-carbon, resulting in the formation of a double bond between the α and β carbons, following Zaitsev's rule.
• The major product is the more substituted alkene, which is 1-methylcyclopentene, corresponding to option (3).

The correct answer is option (3).

CONCEPT:

Electrolysis of Aqueous Sodium Chloride Solution Containing Ethyl Alcohol

• Electrolysis involves passing an electric current through a solution to cause a chemical reaction that would not normally occur.
• In the case of an aqueous solution of sodium chloride containing ethyl alcohol, the primary reaction involves the formation of chlorine gas at the anode and hydrogen gas at the cathode.
• Chlorine gas, when produced in the presence of ethyl alcohol, can react to form different chlorinated organic compounds depending on the conditions.

EXPLANATION:

During the electrolysis of the given solution, the following primary reactions occur:

• At the anode:

  2Cl⁻ → Cl2 + 2e⁻

• At the cathode:

  2H2O + 2e⁻ → H2 + 2OH⁻

• Chlorine gas (Cl₂) produced at the anode can react with ethyl alcohol (C₂H₅OH):

  C2H₅OH + Cl₂ → CH₃CHO + 2HCl

• This acetaldehyde can then react further:

  CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl

• Finally, chloral (CCl₃CHO) can react with water:

  CCl₃CHO + H₂O → CHCl₃ + HCOOH

• This leads to the formation of chloroform (CHCl₃) and formic acid (HCOOH).

CONCLUSION:

• The primary organic product of this series of reactions is chloroform (CHCl₃).

CONCEPT:

Elimination vs Substitution Reactions in t-butyl Chloride

• The reaction of t-butyl chloride (a tertiary alkyl halide) with sodium ethoxide (a strong base) primarily leads to elimination (E₂) rather than substitution due to steric hindrance.
• In the presence of a strong base, tertiary alkyl halides favor elimination to form alkenes, as substitution reactions are hindered by the bulky structure.
• In this case, the elimination product formed is 2-methylprop-1-ene (an alkene) via the E₂ mechanism.

EXPLANATION:

• Sodium ethoxide acts as a strong base, abstracting a proton from the β-carbon of t-butyl chloride.
• Due to the steric hindrance around the tertiary carbon, the reaction proceeds via an E₂ mechanism, leading to the formation of an alkene rather than an ether.
• As a result, 2-methylprop-1-ene (isobutene) is the major product formed, while the possibility of substitution or ether formation is minimal.

REACTION:

• The reaction of t-butyl chloride with sodium ethoxide proceeds via the E₂ elimination mechanism:
  
  Here, sodium ethoxide abstracts a proton from the β-carbon, leading to the formation of 2-methylprop-1-ene (isobutene), along with sodium chloride and ethanol as by-products.

CONCLUSION:

• The main product formed is 2-methylprop-1-ene, as elimination is favored in the reaction of t-butyl chloride with sodium ethoxide.

Explanation:-

sp³ hybridized carbon atom show fast nucleophilic substitution reaction than sp² hybridized carbon atom.