Halkoarenes & Haloarenes MCQs for NEET — Chemistry Questions with Answers

The NCERT states, 'However, haloalkanes tend to dissolve in organic solvents because the new intermolecular attractions between haloalkanes and solvent molecules have much the same strength as the ones being broken in the separate haloalkane and solvent molecules.'

According to the NCERT text, 'When a haloalkane with $\beta$-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom.' Thus, the presence of a $\beta$-hydrogen atom is essential for elimination.

The NCERT text states, 'When a haloalkane with $\beta$-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom.' Alcoholic KOH functions as a strong base, promoting the elimination of a $\beta$-hydrogen and a halogen, leading to an alkene.

As per the NCERT definition, 'Carbon on which halogen atom is directly attached is called $\alpha$-carbon and the carbon atom adjacent to this carbon is called $\beta$-carbon.' In $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$, the bromine is attached to C1, making it the $\alpha$-carbon. The carbon adjacent to C1 (C2) is the $\beta$-carbon.

For an elimination reaction to occur, there must be at least one $\beta$-hydrogen atom. Bromomethane ($\text{CH}_3\text{Br}$) has only an $\alpha$-carbon (the carbon bonded to bromine) but no $\beta$-carbon, and therefore no $\beta$-hydrogen atoms. All other options have $\beta$-hydrogens.

The NCERT text lists 'Nucleophilic substitution' and 'Elimination reactions' as two main categories of reactions for haloalkanes, often occurring simultaneously or in competition depending on reaction conditions (e.g., strength of base/nucleophile, temperature, solvent).

The NCERT states, 'When a haloalkane with $\beta$-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom.' This process leads to the formation of a carbon-carbon double bond, thus producing an alkene.

2-chloropropane has $\beta$-hydrogens on both adjacent methyl groups. Heating with alcoholic KOH will cause an elimination reaction (dehydrohalogenation), removing a chlorine atom and a hydrogen atom from an adjacent carbon, resulting in the formation of propene.

While both aqueous and alcoholic KOH can act as a base/nucleophile, alcoholic KOH is a stronger base and a weaker nucleophile due to the steric hindrance of the alkoxide ion formed in alcohol, favoring elimination (E2) over substitution (SN2). Aqueous KOH, being a strong nucleophile, would primarily lead to nucleophilic substitution (formation of alcohol).

In an elimination reaction, the $\alpha$-carbon (bearing the halogen) and the $\beta$-carbon (losing a hydrogen) both change their hybridization state. The $\alpha$-carbon, initially sp3 hybridized, becomes part of a carbon-carbon double bond, thus transitioning to sp2 hybridization.