Chemistry MCQs for NEET — Practice Questions with Answers

Statement I is true because 4-Methylphenol (an aromatic alcohol) and phenylmethanol (benzyl alcohol) are functional isomers, which means they have the same molecular formula but different functional groups. Statement II is false because isomers with different functional groups are called functional isomers, but this difference in chemical properties is not the reason they are functional isomers. They are functional isomers simply because they contain different functional groups.

Carbon free radicals contain an unpaired electron, which makes them paramagnetic. Paramagnetic substances are attracted to magnetic fields due to the presence of unpaired electrons.

The nucleophilicity order $CH_3– > NH_2– > OH– > F–$ is determined by the basicity and the polarizability of the nucleophiles. Methyl anion ($CH_3–$) is the strongest nucleophile due to its high electron density and low electronegativity of carbon. Amide ion ($NH_2–$) comes next, followed by hydroxide ion ($OH–$), and finally fluoride ion ($F–$) is the weakest nucleophile because of its high electronegativity and low polarizability.

Heterolysis refers to the cleavage of a bond in a molecule where both electrons of the bond are taken by one of the atoms, resulting in the formation of a cation (positively charged ion) and an anion (negatively charged ion). In the case of the carbon-chlorine bond, heterolysis produces a carbon cation (carbocation) and a chlorine anion (Cl-).

The reaction $(CH_3)_3C—Br -> (CH_3)_3C^+ + Br^-$ is an example of heterolytic fission. In heterolytic fission, the bond breaks unevenly, with one atom retaining both of the bonding electrons, resulting in the formation of a cation (positively charged ion) and an anion (negatively charged ion). Here, the bond between carbon and bromine breaks heterolytically to produce a carbocation $(CH_3)_3C^+$ and a bromide anion $Br^-$.

In aromatic nitration, the electrophile is $NO_2^+$. Electrophiles are species that accept an electron pair during a reaction. The nitration of benzene involves the generation of the nitronium ion ($NO_2^+$) as the electrophile, which then attacks the electron-rich aromatic ring to form the nitrobenzene compound.

The least stable carbanion among the given options is $ (CH_3)_3 C^- $. This is because the carbanion is highly destabilized by the inductive effect of the three methyl groups which donate electron density to the negatively charged carbon, increasing electron-electron repulsion and thus making it less stable.

The most stable free radical among the given options is $C_6 H_5 CHCH_3 $. This is an allylic radical stabilized by resonance with the benzene ring. The unpaired electron can be delocalized over the aromatic ring, providing extra stability compared to the other given free radicals.