NEET Practice Questions (MCQs) with Answers & Solutions

The NCERT text explains: 'Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond.' For geometrical isomerism, each carbon of the double bond must be attached to two different groups. In but-2-ene ($CH_3-CH=CH-CH_3$), both carbons of the double bond are attached to a hydrogen and a methyl group, allowing for cis and trans isomers. Pent-1-ene ($CH_2=CH-CH_2-CH_2-CH_3$) has one carbon of the double bond attached to two hydrogen atoms, preventing geometrical isomerism.

The NCERT text states: 'Functional group isomerism: Two or more compounds having the same molecular formula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism. For example, the molecular formula C3H6O represents an aldehyde and a ketone.' Propanal is an aldehyde and propanone is a ketone, both having the formula $C_3H_6O$ but different functional groups.

The NCERT text defines isomerism: 'The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Such compounds are called as isomers.' This difference in properties arises due to differences in their structures.

The NCERT defines carbon types: 'Tertiary carbon is attached to three carbon atoms and neo or quaternary carbon is attached to four carbon atoms.' In 2,2-dimethylpropane ($C(CH_3)_4$), the central carbon atom is bonded to four other carbon atoms (making it a quaternary or 4° carbon), and each of the four methyl groups has a carbon atom bonded to only one other carbon atom (making them primary or 1° carbons).

The problem in the NCERT text (Problem 9.1) asks to 'Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14. Also write their IUPAC names.' The solution lists five isomers for $C_6H_{14}$: n-Hexane, 2-Methylpentane, 3-Methylpentane, 2,3-Dimethylbutane, and 2,2-Dimethylbutane. These are all chain isomers.

The NCERT text explains: 'Rotation around C=C bond is not free. It is restricted... This illustrates that the restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries of such compounds. The stereoisomers of this type are called geometrical isomers.' This restricted rotation is a characteristic feature of the π-bond.

The context says, 'Structures I and II possess same molecular formula but differ in their boiling points and other properties... Structures I and II are isomers of butane... Since difference in properties is due to difference in their structures, they are known as structural isomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers.' Structure I ($CH_3-CH_2-CH_2-CH_3$) is n-butane (continuous chain) and Structure II ($CH_3-CH(CH_3)-CH_3$) is 2-methylpropane (branched chain), both are chain isomers of $C_4H_{10}$.

The NCERT text under 9.3 Alkenes states: 'If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is $C_nH_{2n}$'.

The NCERT text states: 'The double bond is shorter in bond length (134 pm) than the C–C single bond (154 pm).' Therefore, a C-C single bond is longer than a C=C double bond.

The NCERT problem 9.9 provides examples of isomers for $C_5H_{10}$ including Pent-1-ene ($CH_2=CH-CH_2-CH_2-CH_3$) and Pent-2-ene ($CH_3-CH=CH-CH_2-CH_3$). These compounds have the same carbon chain but differ in the position of the double bond, which is a characteristic of position isomerism. The section on structural isomerism also defines: 'Position isomerism: When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers'.