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

CONCEPT:

SN2 Reaction and Leaving Group

• SN2 (Substitution Nucleophilic Bimolecular) reaction is a single-step reaction where the nucleophile attacks the substrate and the leaving group departs simultaneously.
• The rate of SN2 reaction depends on factors such as steric hindrance, the strength of the nucleophile, and the nature of the leaving group.
• A good leaving group facilitates the reaction. A good leaving group is one that can stabilize the negative charge after leaving the substrate.
• In halides, the leaving group ability increases with size due to better stabilization of the negative charge. Thus, Iodine (I^-) is a better leaving group than Bromine (Br^-).

EXPLANATION:

• Assertion (A): "undergoes SN₂ reaction faster than ." This statement is true because iodine is a better leaving group compared to bromine due to its larger size and ability to stabilize the negative charge more effectively.
• Reason (R): "Iodine is a better leaving group because of its large size." This statement is true and correctly explains the assertion.

In SN₂ reactions, the leaving group plays a crucial role in determining the reaction rate. Since iodine is a better leaving group than bromine, the substrate with iodine will undergo the SN₂ reaction faster.

Therefore, the correct answer is Both A and R are true and R is the correct explanation of A.

CONCEPT:

Chirality and Optical Activity

• A molecule is optically active if it:
  • Lacks an internal plane of symmetry
  • Contains one or more chiral centers
  • Exists in non-superimposable mirror images (enantiomers)
• Enantiomers are a pair of optically active isomers that are mirror images of each other.
• If a molecule has two chiral centers, the total number of possible stereoisomers is 2ⁿ (n = number of chiral centers), but the number of optically active ones excludes meso forms (which are optically inactive due to internal symmetry).

EXPLANATION:

• The compound shown is 1,2-dichlorocyclopentane.
• There are two chiral centers (at carbon-1 and carbon-2).
• Total stereoisomers = 2² = 4
• Among them:
  • One pair are enantiomers with both Cl atoms in trans configuration (mirror images, optically active)
  • Another pair are also enantiomers with cis configuration (optically active if no symmetry)
  • But for cyclopentane, the cis isomer is not a meso compound (unlike cyclohexane), so it is also chiral
• Thus, all 4 isomers are optically active (2 pairs of enantiomers).

Therefore, the number of optically active isomers of dichlorocyclopentane is: 4

Concept:

• A vinylic halide is an organic compound in which a halogen atom (X) is directly attached to a carbon atom that is part of a double-bonded (vinylic) carbon-carbon (C=C) bond.

Explanation:-
\(\text{CH}_2 = \text{CH} - \text{CH}_2 - \text{X}\)

• The halogen is attached to a carbon that is not part of the double bond.
• \(\text{CH}_3 - \text{CH} = \text{CH} - \text{X}\)
  • The halogen is attached directly to a carbon that is part of the C=C double bond, making it a vinylic halide.
• \( \text{CH}_3 - \text{C} \equiv \text{C} - \text{X}\)
  • The halogen is attached to a carbon that is part of a triple bond (C≡C), not a double bond.
• 
  \( \text{C}_6\text{H}_5 - \text{CH}_2 - \text{X}\)
  • The halogen is attached to a carbon that is part of a benzene ring and a single bond, not a C=C double bond.

Conclusion:

Therefore, the correct answer is: \( \text{CH}_3 - \text{CH} = \text{CH} - \text{X}\)

CONCEPT:

Markovnikov's Rule

• Markovnikov's rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom (H) will attach to the carbon with the greater number of hydrogen atoms already attached, and the halogen (Br) will attach to the carbon with fewer hydrogen atoms.
• This is because the formation of the more stable carbocation intermediate is favored during the reaction mechanism.

EXPLANATION:

• In the given reaction:

  F₃C - CH = CH₂ + HBr →

  • The double bond (C=C) between the two carbon atoms is broken, and new bonds are formed with the hydrogen (H) and bromine (Br) from HBr.
  • According to Markovnikov's rule, the hydrogen atom will attach to the carbon with more hydrogen atoms (the CH₂ group), and the bromine atom will attach to the carbon with fewer hydrogen atoms (the CH group).
  • 
  • Therefore, the product will be F₃C - CH(Br) - CH₃.

Therefore, the major product of the reaction is F₃C - CH(Br) - CH₃.

CONCEPT:

Nucleophilic Substitution Reaction with Aqueous KOH

• When an alkyl halide reacts with aqueous KOH, the reaction typically proceeds through a nucleophilic substitution mechanism (SN2 reaction).
• The hydroxide ion (OH⁻) from the aqueous KOH acts as a nucleophile, attacking the carbon atom bonded to the halogen, replacing the halogen with an -OH group.
• This results in the formation of an alcohol.

EXPLANATION:

CH₃CH₂CH₂Br + KOH (aq) → CH₃CH₂CH₂OH + KBr

When 1-bromopropane (CH₃CH₂CH₂Br) is treated with aqueous KOH, it undergoes a nucleophilic substitution reaction to form 1-propanol (CH₃CH₂CH₂OH).

CONCLUSION:

The correct option is: 1-Propanol

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.