NEET Practice Questions (MCQs) with Answers & Solutions

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.

The stability of carbocations depends on the number of alkyl groups attached to the positively charged carbon. More alkyl groups increase the stability due to the electron-donating inductive effect and hyperconjugation. In this case, $(CH_3)_3C^+$ (tertiary carbocation) is the most stable because it has three alkyl groups stabilizing the positive charge.

The –I (inductive) effect is the electron-withdrawing effect of a substituent. The order of decreasing –I power for the given groups is $NO_2 > CN > F > –NH_2$. The $NO_2$ group has the strongest electron-withdrawing effect, followed by $CN$, then $F$, and lastly $–NH_2$.

The reaction $C_2H_5Br + KOH \rightarrow C_2H_5OH + KBr$ involves the substitution of the bromine atom in ethyl bromide ($C_2H_5Br$) with a hydroxyl group ($OH^-$) from potassium hydroxide (KOH). This is a classic example of a nucleophilic substitution reaction where the nucleophile $OH^-$ attacks the electrophilic carbon atom bonded to the bromine.

The dehydration of ethanol ($C_2H_5OH \rightarrow C_2H_4 + H_2O$) is an example of an elimination reaction. In this reaction, a molecule of water is eliminated from ethanol to form ethene ($C_2H_4$). This type of reaction is characterized by the removal of a small molecule (in this case, water) from a larger molecule.

Propyne ($C_3H_4$) and propene ($C_3H_6$) can be distinguished using $AgNO_3$ in ammonia. Propyne, being a terminal alkyne, reacts with $AgNO_3$ in ammonia to form a precipitate of silver acetylide, whereas propene, being an alkene, does not react with $AgNO_3$ in ammonia. This difference in reactivity is used to distinguish between the two compounds.

Trans-2-butene has the smallest heat of hydrogenation per mole among the given options. This is because trans isomers are generally more stable than their cis counterparts due to less steric hindrance between the bulky groups on the double bond. The stability of a compound is inversely related to its heat of hydrogenation. Therefore, trans-2-butene, being the most stable, has the smallest heat of hydrogenation.

The inductive effect is the electron-withdrawing or electron-donating effect transmitted through sigma bonds. The order of inductive effect is based on the alkyl groups' ability to donate electrons through sigma bonds. The correct order of increasing inductive effect for the given groups is $CH_3–CH_2 – < (CH_3)_2CH – < (CH_3)_3C –$ because the inductive effect increases with the number of alkyl groups attached to the carbon atom.

The basic nature of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation. In ammonia ($NH_3$), there is no alkyl group attached to the nitrogen, so its basicity is the lowest. In $CH_3NH_2$ (methylamine), the electron-donating effect of the methyl group increases the electron density on the nitrogen, making it more basic than ammonia. In $ (CH_3)_2NH$ (dimethylamine), two electron-donating methyl groups further increase the electron density on nitrogen, making it the most basic among the three. Therefore, the correct order of increasing basic nature is $ NH_3 < CH_3NH_2 < (CH_3)_2NH$.