Haloalkanes And Haloarenes MCQ Quiz - Objective Question with Answer for Haloalkanes And Haloarenes - Download Free PDF

CONCEPT:

Conversion of Alkene to Benzene (Aromatic Compound Formation)

• Alkenes like ethene undergo halogenation with Br₂ to form vicinal dihalides.
• These vicinal dihalides, when heated with alcoholic KOH, undergo dehydrohalogenation (elimination) to yield alkynes.
• Ethene (H₂C=CH₂) forms ethyne (HC≡CH) after two dehydrohalogenations.
• Three moles of ethyne, when passed through a red-hot copper tube, undergo trimerization (polymerisation) to form benzene (C₆H₆).

EXPLANATION:

• Given: Alkene [A] undergoes bromination → vicinal dibromide [B].
• [B] with hot alcoholic KOH → elimination → forms gas [C] (ethyne).
• 3 moles of ethyne → polymerise in presence of Cu-tube → form [D] (benzene).

Therefore, the IUPAC name of compound [D] is Benzene.

CONCEPT:

Substituents and Acidity in Benzene

• Substituents on a benzene ring can be classified as activating or deactivating based on their electronic effects.
• Activating substituents donate electron density to the benzene ring, stabilizing the negative charge on conjugate bases, which decreases acidity.
• Deactivating substituents withdraw electron density from the benzene ring, destabilizing the negative charge on conjugate bases, which increases acidity.

EXPLANATION:

• Acidity in benzene derivatives is influenced by the substituents attached to the ring.
• Deactivating substituents, such as nitro (-NO₂) or halogens, pull electron density away from the ring, making the molecule more acidic.
• Activating substituents, such as alkyl groups or amino (-NH₂), donate electron density to the ring, making the molecule less acidic.
• Thus:
  • A more deactivating substituent increases acidity.
  • A more activating substituent decreases acidity.

Correct Answer: Option 2 - The more activating substituent decreases the acidity.

CONCEPT:

Reaction of Alkenes with Hydrogen Halides (HBr):

• When an alkene reacts with hydrogen halides (e.g., HBr), the addition occurs via electrophilic addition mechanism.
• The Markovnikov's rule governs the addition of HBr to asymmetrical alkenes. According to the rule:
  • The hydrogen atom (H) adds to the carbon of the double bond that already has more hydrogen atoms.
  • The bromine atom (Br) adds to the carbon of the double bond with fewer hydrogen atoms.
• If the alkene is symmetrical, only one product is formed. If the alkene is asymmetrical, the addition follows Markovnikov's rule and only the major product is considered.

EXPLANATION:

• The given alkene is (CH₃)₂C=CH₂, which is asymmetrical.
• When HBr is added, the hydrogen atom (H) attaches to the carbon in the double bond with more hydrogen atoms (CH₂ group).
• The bromine atom (Br) attaches to the carbon in the double bond with fewer hydrogen atoms (C(CH₃)₂ group).
• The product formed is CH3-CH(CH3)(Br)-CH3

Therefore, the correct answer is CH3-CH(CH3)(Br)-CH3

CONCEPT:

Selectivity in Halogenation Reactions

• Halogenation of hydrocarbons involves the substitution of hydrogen atoms with halogen atoms (chlorine or bromine).
• Chlorination: Chlorine is less selective but reacts faster due to its higher reactivity.
• Bromination: Bromine is more selective but reacts slower because it is less reactive compared to chlorine.
• The selectivity of a halogenation process refers to its ability to preferentially substitute hydrogens at specific positions (e.g., tertiary, secondary, or primary hydrogens).

EXPLANATION:

• In chlorination:
  • The reaction is less selective, and chlorine can substitute hydrogens at primary, secondary, or tertiary positions without much preference.
  • This is because the transition state in chlorination has lower energy differences for different types of hydrogens.
    
• In bromination:
  • The reaction is more selective because bromine reacts preferentially with tertiary hydrogens over secondary and primary hydrogens.
  • This is due to the higher activation energy required for bromination, which makes the transition state more sensitive to the stability of the intermediate radical.
    
• Radical Stability
  • Stability of Radicals: The stability of the radicals formed during the reaction influences selectivity. Bromine radicals are less reactive and therefore favor the formation of more stable radicals, whereas chlorine radicals can react with hydrogen atoms from less stable positions (like primary hydrogens), resulting in a mix of products.
• Selectivity Factor
  • The selectivity of bromination is often quantified by the selectivity factor (S), which indicates the preference for forming products from the most stable radicals. The selectivity for bromination is higher because the energy barrier for bromination at less stable positions is significantly higher than that for chlorination

Therefore, bromination is more selective than chlorination.

Correct Answer is Bromination is more selective than Chlorination.

CONCEPT:

Conversion of Alcohols to Alkyl Halides using Thionyl Chloride (SOCl₂)

• Thionyl chloride (SOCl₂) is a commonly used reagent in organic chemistry to convert alcohols into alkyl halides, specifically alkyl chlorides.
• The reaction proceeds via a substitution mechanism, where the hydroxyl group (-OH) of the alcohol is replaced by a chlorine atom (-Cl).
• The reaction is typically carried out in the presence of a base like pyridine to neutralize the HCl produced as a byproduct.

REACTION:

• The general reaction is as follows:

  R-OH + SOCl₂ → R-Cl + SO₂ + HCl

• Here:
  • R-OH represents the alcohol.
  • R-Cl represents the alkyl halide (alkyl chloride).
  • SO₂ (sulfur dioxide) and HCl (hydrogen chloride) are byproducts.

EXPLANATION:

• Alcohols have a poor leaving group (the -OH group), making them less reactive in substitution reactions.
• SOCl₂ helps to convert the hydroxyl group (-OH) into a better leaving group by forming an intermediate, which then facilitates the substitution with a chlorine atom.
• This reaction is highly efficient and often used in laboratory synthesis to prepare alkyl chlorides from alcohols.

Therefore, by using SOCl₂, alcohols are converted into alkyl halides (Option 1).

The correct answer is 10. 

Key Points

Chloropropane:

• A chemical structure of a molecule includes the arrangement of atoms and the chemical bonds that hold the atoms together.
• The  1-Chloropropane or 2- Chloropropane molecule contains a total of 10 bonds.

• There are 3 non-H bonds.

Important Points

• Chloropropane appears as a colorless liquid with a chloroform-like odor.
• Vapors are heavier than air and less dense than water.
• May irritate the skin and eyes and be narcotic in high concentrations.
• A fire and explosion risk.
• Used to make other chemicals.

Explanation:

• The IUPAC name of Gammexene is benzene hexachloride or, 1,2,3,4,5,6- hexachlorocyclohexane.
• The trade names are Gammaxene, Lindane, and 666.
• It is prepared by the addition of chlorine to benzene in the presence of U.V sunlight.

• It is used as a pesticide in agriculture.
• Nine stereomers are possible for this compound. Among these seven meso compounds and two are enantiomers which exist as a racemic mixture.

Hence, Gammaxane is benzene hexachloride.

Important Points

•  Chlorobenzene is a monohalide of benzene that can be prepared by chlorination of benzene in presence of fecl3 as a catalyst. Its chemical formula is C6H5Cl.

C6H6  + Cl2  → C6H5Cl + HCl

• Bromobenzene, C6H5Br, can also be prepared in a similar way.
• Chloro benzene when treated with oxidizing agents, yields chlorobenzene acid, as the side chain goes oxidation.

Concept:

• Electrophilic aromatic substitution (EAS) is where benzene acts as a nucleophile to replace a substituent with a new electrophile. That is, benzene needs to donate electrons from inside the ring. An electrophile attacks the region of high electron density.
• Activating group is that which increases the rate of an electrophilic aromatic substitution reaction, relative to hydrogen.CH₃ is a perfect example of an activating group; when we substitute a hydrogen on benzene for CH₃, the rate of nitration is increased.
• A deactivating group, on the other hand, decreases the rate of an electrophilic aromatic substitution reaction, relative to hydrogen. The trifluoromethyl group, F₃, drastically decreases the rate of nitration when substituted for a hydrogen on benzene.
• This definition is ultimately based on experimental reaction rate data.  It doesn’t tell us why each group accelerates or decreases the rate. “Activating” and “deactivating” just refers to the effect of each substituent on the rate, relative to H.
• CH₃ is an activating group because of its +I effect, −Cl and −NO₂ are deactivating due to their −I and −M effect. −CH₃ group, having +I effect, increases the electron density in the benzene ring whereas −Cl group having −I effect decreases the electron density the benzene ring.

Calculation:

→ Order of strength for electrophilic substitution

+ M > + I > - I > - M

→ SE ∝ EDG

Where, EDG is Electron Drawing Group,

EWG is Electron Withdrawing Group

→ Thus, the increasing order of reactivity of the given compounds towards aromatic electrophilic substitution reaction is

D (-M)

Explanation:

• On heating with an ethanolic solution of silver nitrite, alkyl halides yield nitroalkanes.
• Some of the alkyl nitrites are also formed because nitrite is an ambidentate ligand.
• It has a lone pair of electrons on nitrogen as well as oxygen.
• Thus it can be bonded to an alkyl group of alkyl halide through oxygen to form alkyl nitrite and via Nitrogen to form nitroalkane.
• The reactions take place as follows:

• Hence, the treatment of ethyl bromide with alcoholic silver nitrite gives Nitroethane.

​ 

• Alkyl halides when treated with sodium or potassium salt of nitrite give alkyl nitrite as the major product.

​

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:

The neopentyl group has four carbons, and all the carbons are bonded to a hydrogen atom except for one, which is bonded to a chlorine atom. The neopentyl group is attached to a propane chain, which has three carbons.

To name this compound using IUPAC nomenclature, we follow the following steps:

Identify the longest carbon chain: In this case, it is a propane chain with three carbons.

Number the carbon chain: We start numbering from the end nearest to the substituent, which is the neopentyl group. The carbon atom in the neopentyl group that is attached to the chlorine atom is assigned the lowest possible number. Therefore, it is numbered as carbon 1.

Identify and name the substituents: The neopentyl group is a substituent in this compound. The substituent name is derived from the name of the parent hydrocarbon, neopentane. The neopentyl group has four carbon atoms, so its prefix is "neo-".

Assign locants to the substituents: The substituent is attached to carbon 2 of the propane chain. Therefore, it is named as 2-neopentyl.

Name the functional group: The functional group in this compound is a halogen (chlorine) and is named as chloro.

Complete the name: Putting it all together, we get the IUPAC name of neopentyl chloride as 1-chloro-2,2-dimethylpropane.

1-chloro-2, 2-dimethylpropane

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.

An intermediate carbocation is not stabilized by loosely held π electron overlap with p-orbitals of π bond. So, the reason is wrong.

Explanation:-

Step 1: From to the intermediate product
We start with a primary alkyl halide, (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:

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

Step 2: From the intermediate product (propene) to 
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:

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 

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).

Explanation:-

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