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

CONCEPT:

Saytzeff's Rule in Elimination Reactions

• Saytzeff's (Zaitsev's) rule states that in elimination reactions, the more substituted alkene (one with more alkyl groups attached to the double bond) is generally the major product.
• This rule is applicable to reactions like dehydrohalogenation and dehydration where the base or reagent promotes the formation of an alkene.
• However, some bulky bases or specific reaction conditions may lead to exceptions, producing the less substituted alkene as the major product (anti-Saytzeff or Hoffman product).

EXPLANATION:

• Reaction a: This involves dehydration using concentrated sulfuric acid, which follows Saytzeff's rule, leading to the formation of the more substituted alkene.
• Reaction b: Alcoholic KOH, in a dehydrohalogenation reaction, also typically follows Saytzeff's rule, forming the more substituted alkene.
• Reaction c: In the presence of a bulky base like tert-butoxide (CH3)3CO-, the less substituted alkene (anti-Saytzeff or Hoffman product) is formed.
• Reaction d: This reaction involves heat, leading to elimination that produces an unsaturated aldehyde. This does not follow the standard Saytzeff mechanism.

Conclusion:-

The reactions that do not produce a Saytzeff product are reaction c (due to bulky base) and reaction d (due to special conditions). Therefore, the correct answer is only 1.

CONCEPT:

Reactions with 2-Phenyl-2-bromopropane

• 2-Phenyl-2-bromopropane is a tertiary alkyl halide.
• In the presence of alcoholic KOH, elimination (E2) reaction occurs, leading to the formation of an alkene.
• The product [A] will be the corresponding alkene formed via the E2 mechanism.
• When [A] reacts with HBr in the presence of benzoyl peroxide ((C₆H₅CO)₂O), an anti-Markovnikov addition of HBr occurs, forming an alkyl bromide [B].

EXPLANATION:

• Starting compound: 2-Phenyl-2-bromopropane (C₆H₅-C(CH₃)₂-Br)
• Reaction with alcoholic KOH:

  The E2 elimination reaction removes a hydrogen and the bromide to form the most stable (more substituted) alkene:

  C₆H₅-C(CH₃)₂-Br + KOH (alcoholic) → C₆H₅-C(CH₃)=CH₂ ([A]) + KBr + H₂O

  Therefore, [A] is 2-Phenylpropene

• Reaction with HBr in the presence of benzoyl peroxide:

  Anti-Markovnikov addition of HBr to [A] forms:

  C₆H₅-CH=CH₂ + HBr → C₆H₅-CH₂-CH₂-Br ([B])

  Therefore, [B] is 2-Phenyl-1-bromopropane

Therefore, the correct identification of [A] and [B] is [A] = 2-Phenylpropene, [B] = 2-Phenyl-1-bromopropane

CONCEPT:

Grignard Reagents (RMgX)

• Grignard reagents are organomagnesium compounds with the formula RMgX, where R is an organic group and X is a halogen.
• They are commonly used in organic synthesis to form carbon-carbon bonds.
• Grignard reagents are highly reactive and must be handled with care.

EXPLANATION:

• Preparation of Grignard Reagents:
  • Grignard reagents are prepared by the addition of an alkyl or aryl halide (RX) to an excess of suspended magnesium (Mg) in a dry, nonprotic solvent such as diethyl ether.
  • Equation: RX + Mg → RMgX
• Solvent:
  • The solvent used must be nonprotic (e.g., diethyl ether or THF) to prevent the protonation and degradation of the Grignard reagent.
• Storage:
  • Grignard reagents cannot be stored like any other chemical because they are highly reactive and sensitive to moisture and air.
  • They must be stored under an inert atmosphere (e.g., nitrogen or argon) to prevent decomposition.
• Importance:
  • Grignard reagents are among the most important organometallic compounds of magnesium used in organic synthesis.

Therefore, the untrue statement about RMgX is It can be stored like any other chemical.

CONCEPT:

Wurtz Reaction and Kolbe's Electrolytic Method

• Wurtz Reaction: This reaction involves the coupling of alkyl halides in the presence of sodium metal in dry ether. It is used to prepare alkanes with an even number of carbon atoms.

• Kolbe's Electrolytic Method: This method involves the electrolysis of salts of carboxylic acids (like sodium acetate) to produce alkanes. It is also limited to the formation of alkanes with an even number of carbon atoms.
• Both methods are unsuitable for the synthesis of alkanes with an odd number of carbon atoms or methane.
• 

EXPLANATION:

• In the case of ethane and butane:
  • Both Wurtz Reaction and Kolbe's Electrolytic Method can be used as these alkanes have an even number of carbon atoms.
• In the case of propane and methane:
  • These alkanes cannot be synthesized using Wurtz Reaction because the reaction only works for alkanes with even numbers of carbon atoms.
  • Kolbe's Electrolytic Method also fails for methane, as it requires two carboxylate ions to couple, which results in an even number of carbon atoms.

Therefore, Wurtz Reaction and Kolbe's Electrolytic Method cannot be used for the preparation of methane and propane. The correct answer is Option 3: Methane.

So, the reactivity of \(CH_3Cl\) will be the most.

And as the steric hinderance increases rate of \( {S_N}^2 \) will decrease. And hinderance increases with crowding on the central atom.

Hence the correct order is option D.

Diazotization of p-toluidine with a cold solution of sodium nitrite and \(HCl\) gives 4-methyl benzene diazonium chloride (compound \(D\)).

On heating with \(CuCN\) and \(KCN\), the diazonium group is replaced with cyano group to obtain 4-methyl benzonitrile (compound \(E\)).

Option C is correct.

Alkyl bromides or alkyl chlorides are heated in presence of metallifluorides such as \(AgF, CoF_2, SbF_5 \text{ or } Hg_2F_2\) to obtain alkyl fluorides.

\(CH_3-Cl + AgF \rightarrow CH_3-F + AgCl\)

Allylic bromination of 4-tert-butylcyclohex-1-ene with NBS/ hv gives 3-bromo-5-tert-butylcyclohex-1-ene. Allylic H atom is replaced with Br. Next step is hydrolysis of \(C-Br\) bond to form \(C-OH\) bond.

Hence, the option (A) is the correct answer.

\(3-\text{Methyl-pent}-2-\text{ene}\) on reaction with \(HBr\) in presence of \(\text{peroxide}\) forms an addition product.

A molecule of \(HBr\) is added to \(C=C\) double bond. The addition follows anti-Markowikoff's rule. \(Br\) is added to less substituted C atom.

The number of possible stereoisomers for the products is four. The product has 2 chiral C atoms and is unsymmetrical.

Hence, 4 stereoisomers are possible.