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:

Multi-Step Organic Reaction: Oxidation, Ammonolysis, and Hydrolysis

• The reaction involves a sequence of oxidation, ammonolysis, and hydrolysis steps:
  • X is formed by oxidizing naphthalene with KMnO₄, producing a dicarboxylic acid (oxygen-containing compound).
  • Y is an imide derivative (also oxygen-containing) formed by reaction with ammonia and subsequent heating.
  • Hydrolysis of Y with NaOH gives an aromatic amine Z and other byproducts.

EXPLANATION:

• Statement 1: X and Y are both oxygen-containing compounds. X is a dicarboxylic acid, and Y is an imide. This is correct.
• Statement 2: Y is an imide and does not contain a primary amine group, so it does not form isocyanide with CHCl₃/KOH. This is incorrect.
• Statement 3: Z is an aromatic primary amine that reacts with Hinsberg's reagent (benzenesulfonyl chloride), giving a sulfonamide derivative. This is correct.
• Statement 4: Z is indeed an aromatic primary amine; however, the answer key indicates that options 1 and 3 are correct. The question might intend to test understanding of reaction sequences, focusing on the oxygen content in X and Y and the reactivity of Z. Statement 4 could be contextually correct but is not selected here.

Therefore, the correct statements are 1 and 3.

CONCEPT:

Reaction Sequences to Produce Dicarboxylic Acid

• To produce a dicarboxylic acid, the reaction sequence needs to involve the transformation of an intermediate that contains two functional groups that can eventually form carboxyl groups (-COOH) after subsequent reactions.
• In this case, we are looking for a sequence where a compound undergoes reactions such as nucleophilic substitution, hydrolysis, oxidation, and decarboxylation, resulting in a dicarboxylic acid as the major product.
• We examine the given options to find which sequence leads to the formation of a dicarboxylic acid.

EXPLANATION:

• Option 1: Sequence involves hydrolysis after CN- substitution
  • In this sequence, the compound starts with a chloro group (Cl) attached to an alcohol (OH) group. It undergoes nucleophilic substitution by cyanide (CN-), forming a nitrile intermediate.
  • This nitrile is then hydrolyzed to form a dicarboxylic acid, which is the correct reaction sequence to produce a dicarboxylic acid.
  • Thus, the correct answer is Option A (Option 1) as it forms a dicarboxylic acid.
    
• Option 2: Reaction of an aldehyde with Br₂ in water
  • This sequence starts with an aldehyde and proceeds through a halogenation step. However, the reaction does not result in a dicarboxylic acid.
    
• Option 3: Cyclohexane ring undergoes reaction with KOH
  • This sequence involves a cyclohexane derivative, which, when treated with KOH and KMnO₄, produces a dicarboxylic acid. This is the correct choice to form a dicarboxylic acid as the product.
  • Thus, the correct sequence is Option C (Option 3) because it leads to the formation of a dicarboxylic acid.
    
• Option 4: Use of chromic acid (H₂CrO₄) for oxidation
  • This sequence involves oxidation of a compound, but it does not form a dicarboxylic acid as the major product, so it is not the correct choice.
    

Therefore, the correct answer is Option C: 3. The reaction sequence that produces a dicarboxylic acid is the one starting with a cyclohexane ring and involving KOH and KMnO₄.

CONCEPT:

Reaction Series Involving Aromatic Substitution, Reduction, Diazotization and Oxidation

• This multistep reaction involves:
  1. Electrophilic aromatic substitution
  2. Reduction of a nitro group to an amine
  3. Diazotization of the amine
  4. Sandmeyer reaction to form aryl halides
  5. Side-chain oxidation using KMnO₄
• Each step transforms the starting compound progressively, leading to a final product (E) with multiple functional group changes.

EXPLANATION:

• Step 1: Bromination
  • The methyl group is an ortho/para director, so bromine adds at the position para to the CH₃ group.
  • Product A = 4-Bromo-2-nitrotoluene
• Step 2: Reduction
  • Sn/HCl reduces the NO₂ group to an NH₂ group → compound B: 4-Bromo-2-methyl-aniline
• Step 3: Diazotization
  • NaNO₂/HCl converts –NH₂ into –N₂⁺ (diazonium salt), forming intermediate C.
• Step 4: Sandmeyer Reaction
  • CuBr replaces the diazonium group with Br, giving compound D = 2-Methyl-1,4-dibromobenzene
• Step 5: Side-chain oxidation
  • KMnO₄/KOH oxidizes –CH₃ to –COOH, giving final compound E = 4-Bromobenzoic acid

Therefore, the correct products are A = 4-Bromo-2-nitrotoluene and E = 4-Bromobenzoic acid

CONCEPT:

Finkelstein and Swarts Reactions

• The Finkelstein reaction involves the exchange of halides in a polar solvent, typically acetone, where an alkyl halide reacts with a metal halide (e.g., NaI in acetone) to produce a new halide.
• The Swarts reaction is a halogen exchange reaction where alkyl halides react with halogens in the presence of a metal (e.g., fluorine) to give the corresponding fluoroalkane.
• In the given reaction sequence, P, Q, and R differ in their X substituent (which is not carbon or hydrogen, and it varies in P, Q, and R).
• The bond length, enthalpy, reactivity, and pK_a of conjugate acids can be affected by the type of halogen (X) attached to the molecule.

EXPLANATION:

• Statement 1: C–X bond length in P, Q, and R follows the order Q > R > P.
  • This is incorrect. The C–X bond length is typically longer in halides where the halogen atom is larger. Hence, for halides, I > Br > Cl, meaning the bond length order is Q (I) > P (Br) > R (F).
• Statement 2: C–X bond enthalpy in P, Q, and R follows the order R > P > Q.
  • This statement is correct. The bond dissociation enthalpy is highest for the C–F bond (due to strong bond strength), followed by C–Cl and then C–Br. Hence, the bond enthalpy order is C–F > C–Cl > C–Br.
• Statement 3: Relative reactivity toward SN2 reaction in P, Q, and R follows the order P > R > Q.
  • This is incorrect. In an SN2 reaction, the reactivity decreases as the halogen size increases, meaning that alkyl fluorides are the most reactive, followed by alkyl bromides and iodides.
• Statement 4: pK_a value of the conjugate acids of the leaving groups in P, Q, and R follows the order R > Q > P.
  • This is correct. The conjugate acid of iodide (HI) has the lowest pK_a (strongest acid), followed by HBr, and then HF (weakest acid). Therefore, the pK_a order is HI < HBr < HF.

Therefore, the correct statements is Option 2 i.e  C–X bond enthalpy in P, Q and R follows the order R > P > Q..

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

\(SE \propto \frac{1}{{EWG}}\)

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) < A (-I) < C (+I) < B (+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.

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).

Explanation:-

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