Hydrocarbons MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: A saturated hydrocarbon contains only carbon–carbon single bonds and carbon–hydrogen single bonds. The described structure contains five carbon atoms joined into a closed ring without any carbon–carbon multiple bond. Each carbon atom forms two single bonds with neighbouring carbon atoms and therefore requires two hydrogen atoms to complete its tetravalency. The resulting molecular formula is \(\mathrm{C_5H_{10}}\), and the compound is cyclopentane. Its cyclic structure makes it an alicyclic hydrocarbon rather than an acyclic hydrocarbon. The absence of carbon–carbon multiple bonds establishes its saturated character. Although cyclopentane has fewer hydrogen atoms than the corresponding open-chain alkane, pentane, this difference arises from ring formation rather than unsaturation. Alicyclic compounds are not automatically aromatic merely because their carbon atoms form rings. The important distinction is that saturation depends on the types of carbon–carbon bonds, whereas acyclic and cyclic classifications depend on carbon-skeleton connectivity. Therefore, cyclopentane is a saturated alicyclic hydrocarbon.

Explanation: Propene has the structure \(\mathrm{CH_3-CH=CH_2}\), containing one carbon–carbon double bond. Consequently, propene is an unsaturated open-chain hydrocarbon belonging to the alkene family. Cyclopropane contains three carbon atoms connected in a three-membered ring exclusively through carbon–carbon single bonds. It is therefore a saturated alicyclic hydrocarbon. Both compounds have the molecular formula \(\mathrm{C_3H_6}\), but their carbon connectivity and bonding patterns are different. The general formula of an acyclic monoalkene is \(\mathrm{C}_{n}\mathrm{H}_{2n}\). A saturated monocyclic hydrocarbon containing only carbon–carbon single bonds also follows \(\mathrm{C}_{n}\mathrm{H}_{2n}\). The reduction of two hydrogen atoms relative to \(\mathrm{C}_{n}\mathrm{H}_{2n+2}\) may therefore arise from either one double bond or one ring. Molecular formula alone cannot distinguish these two structural possibilities. This also demonstrates why a hydrocarbon should not be classified as unsaturated merely because its formula differs from that of the corresponding acyclic alkane.

Explanation: \(\textbf{Homologous relationship:}\) Successive members of the same homologous series differ in molecular composition by one \(\mathrm{-CH_2-}\) unit. An acyclic hydrocarbon containing exactly one carbon–carbon triple bond belongs to the monoalkyne series. Its general molecular formula is: \[ \mathrm{C}_{n}\mathrm{H}_{2n-2} \] For X, \(n=4\), giving \(\mathrm{C_4H_6}\). The next higher homologue therefore has five carbon atoms and the formula: \[ \mathrm{C_5H_8} \] \(\textbf{Carbon-skeleton bonding:}\) An open-chain structure containing five carbon atoms has four carbon–carbon connections, provided the carbon skeleton is connected and contains no ring. Each connection contributes one carbon–carbon \(\sigma\) bond, irrespective of whether it represents a single, double or triple bond. Thus, the total number of carbon–carbon \(\sigma\) bonds is four. A carbon–carbon triple bond contains one \(\sigma\) bond and two \(\pi\) bonds. Because Y retains exactly one triple bond and has no additional multiple bonds, it contains two carbon–carbon \(\pi\) bonds. The homologous-series relationship changes the carbon and hydrogen counts but preserves the defining triple-bond functionality. Hence, Y contains four carbon–carbon \(\sigma\) bonds and two \(\pi\) bonds.

Explanation: The general formula of a saturated open-chain hydrocarbon containing seven carbon atoms is \(\mathrm{C_7H_{16}}\). The given hydrocarbon contains four fewer hydrogen atoms than this saturated acyclic reference. \(\textbf{Hydrogen-deficiency relation:}\) For a hydrocarbon containing only carbon and hydrogen: \[ \mathrm{DBE}=\frac{2C+2-H}{2} \] Substituting the given molecular formula: \[ \mathrm{DBE}=\frac{2(7)+2-12}{2} \] \[ \mathrm{DBE}=\frac{16-12}{2}=2 \] A carbon–carbon triple bond contributes two degrees of hydrogen deficiency. Therefore, an acyclic hydrocarbon containing one triple bond can have the formula \(\mathrm{C_7H_{12}}\). A ring contributes one degree of hydrogen deficiency, and a carbon–carbon double bond contributes another. Consequently, a monocyclic monoalkene also has two degrees of hydrogen deficiency. Both families are therefore compatible with the given molecular formula. By contrast, an acyclic monoalkene or a saturated monocyclic hydrocarbon has only one degree of hydrogen deficiency and follows \(\mathrm{C_7H_{14}}\). The decisive principle is that molecular hydrogen deficiency reflects the combined effects of rings and multiple bonds rather than uniquely identifying a particular hydrocarbon family.

Explanation: \(\textbf{Saturated reference formula:}\) An acyclic saturated hydrocarbon containing six carbon atoms follows: \[ \mathrm{C}_{n}\mathrm{H}_{2n+2} \] Therefore: \[ \mathrm{C_6H_{2(6)+2}=C_6H_{14}} \] \(\textbf{Hydrogen deficiency:}\) One carbon–carbon double bond reduces the hydrogen count by two relative to a saturated structure. One carbon–carbon triple bond reduces the hydrogen count by four. The combined hydrogen deficiency is: \[ \Delta H=2+4=6 \] Thus, the original molecular formula is: \[ \mathrm{C_6H_{14-6}=C_6H_8} \] \(\textbf{Hydrogenation requirement:}\) Complete hydrogenation of one carbon–carbon double bond consumes one mole of \(\mathrm{H_2}\) per mole of that bond. Complete hydrogenation of one carbon–carbon triple bond consumes two moles of \(\mathrm{H_2}\). The total hydrogen requirement is: \[ n(\mathrm{H_2})=1+2=3\ \mathrm{mol} \] The overall reaction is: \[ \mathrm{C_6H_8+3H_2\xrightarrow{Ni}C_6H_{14}} \] \(\textbf{Consistency check:}\) The product contains six more hydrogen atoms than the original hydrocarbon, exactly matching the addition of three \(\mathrm{H_2}\) molecules. Both the original molecular formula and the hydrogen requirement are therefore consistent with the stated bonding pattern. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_6H_8};\ 3\ \mathrm{mol\ H_2\ mol^{-1}\ hydrocarbon}} \]

Explanation: The molecular formula of the corresponding saturated open-chain six-carbon hydrocarbon is \(\mathrm{C_6H_{14}}\). The starting formula \(\mathrm{C_6H_{10}}\) therefore represents two degrees of hydrogen deficiency. These may arise from one triple bond, two double bonds, one ring together with one double bond, or two rings. The consumption of exactly one mole of hydrogen indicates that the starting hydrocarbon contains one hydrogenatable carbon–carbon double bond under the stated complete-hydrogenation conditions. The resulting formula is: \[ \mathrm{C_6H_{10}+H_2\longrightarrow C_6H_{12}} \] Because the product contains no carbon–carbon multiple bonds, its remaining hydrogen deficiency cannot be attributed to a double or triple bond. The remaining deficiency must therefore arise from a ring. Cyclohexene contains one six-membered ring and one carbon–carbon double bond, satisfying both requirements. Hexa-1,3-diene and hex-1-yne each contain two hydrogenatable \(\pi\)-bond equivalents and require two moles of hydrogen for complete saturation. Cyclohexane already has the formula \(\mathrm{C_6H_{12}}\) and contains no multiple bond requiring hydrogenation. The important deduction combines the original formula, measured hydrogen consumption and preservation of the cyclic carbon skeleton.

Explanation: Statement I is correct because benzene and cyclohexene both possess closed carbon frameworks but differ in their fundamental bonding characteristics and classification. Benzene is an aromatic hydrocarbon with six carbon atoms and a delocalised \(\pi\)-electron system. Its molecular formula is \(\mathrm{C_6H_6}\), and its bonding cannot be represented adequately by treating three localised double bonds as independent ordinary alkene bonds. Cyclohexene has the formula \(\mathrm{C_6H_{10}}\) and contains one ordinary carbon–carbon double bond within a six-membered ring. It belongs to the unsaturated alicyclic hydrocarbon category. Statement II is incorrect because the presence of a cyclic structure and a carbon–carbon double bond does not by itself establish aromaticity. Aromaticity requires a suitable cyclic, continuously conjugated system with the appropriate delocalisation of \(\pi\) electrons. Cyclohexene does not possess a continuous cyclic conjugated system because four of its ring carbons are \(sp^3\)-hybridised. The presence of one carbon–carbon double bond therefore cannot make cyclohexene aromatic. The correct distinction is between ordinary cyclic unsaturation and the special electronic structure associated with aromatic hydrocarbons. Hence, Statement I is correct whereas Statement II is incorrect.

Explanation: \(\textbf{Carbon atoms from combustion:}\) Complete combustion converts every carbon atom of the hydrocarbon into carbon dioxide. For \(0.10\) mol of X: \[ n(\mathrm{CO_2})=0.50\ \mathrm{mol} \] Hence, the number of carbon atoms per molecule is: \[ C=\frac{0.50}{0.10}=5 \] \(\textbf{Hydrogen atoms from combustion:}\) Every water molecule contains two hydrogen atoms. The number of moles of hydrogen atoms produced is: \[ n(\mathrm{H\ atoms})=2(0.40)=0.80\ \mathrm{mol} \] Therefore, the number of hydrogen atoms per molecule of X is: \[ H=\frac{0.80}{0.10}=8 \] The molecular formula of X is thus: \[ \mathrm{X=C_5H_8} \] \(\textbf{Hydrogenation evidence:}\) The hydrogen consumed per mole of X is: \[ \frac{0.20}{0.10}=2\ \mathrm{mol\ H_2\ mol^{-1}\ X} \] Complete hydrogenation therefore gives: \[ \mathrm{C_5H_8+2H_2\longrightarrow C_5H_{12}} \] The product has the saturated open-chain alkane formula \(\mathrm{C_5H_{12}}\). Because carbon-skeleton cleavage is excluded, the starting compound must possess two hydrogenatable \(\pi\)-bond equivalents in an acyclic structure. Pent-1-yne contains one triple bond and has the molecular formula \(\mathrm{C_5H_8}\). Its triple bond consumes two moles of hydrogen per mole of alkyne during complete catalytic hydrogenation. Cyclopentene has the correct starting formula but consumes only one mole of hydrogen without breaking its ring. Pent-1-ene and cyclopentane both have the formula \(\mathrm{C_5H_{10}}\), so they fail the combustion-based formula check. Thus, pent-1-yne satisfies both the combustion evidence and the hydrogenation requirement.

Explanation: Constitutional isomers have the same molecular formula but different atom connectivity. The formula \(\mathrm{C_5H_{12}}\) corresponds to a saturated acyclic hydrocarbon containing five carbon atoms. A continuous arrangement of five carbon atoms gives pentane. Placing one methyl branch on a four-carbon parent chain gives 2-methylbutane. Placing two methyl branches on the central carbon of a three-carbon parent chain gives 2,2-dimethylpropane. These three structures have different carbon skeletons while preserving the same total number of carbon and hydrogen atoms. A structure described as 3-methylbutane is not a fourth isomer. Numbering its four-carbon parent chain from the opposite end places the methyl group at carbon 2, so it is identical to 2-methylbutane. Similarly, changing the orientation of a structural formula on paper does not create a new molecule. The correct enumeration requires checking both connectivity and symmetry-related duplicate descriptions. Pentane, 2-methylbutane and 2,2-dimethylpropane therefore exhaust the three constitutional isomers of \(\mathrm{C_5H_{12}}\).

Explanation: A tertiary carbon atom is directly bonded to three other carbon atoms. A quaternary carbon atom is directly bonded to four other carbon atoms. These classifications depend on carbon–carbon connectivity, not on the total number of hydrogen atoms in the molecule. \(\textbf{Structure of P:}\) The condensed structure of 2-methylpentane is: \[ \mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH_3} \] The second carbon atom is directly attached to three carbon atoms, making it tertiary. No carbon atom in this structure is attached to four other carbon atoms. Thus, 2-methylpentane satisfies the stated condition for P. \(\textbf{Structure of Q:}\) The condensed structure of 2,3-dimethylbutane is: \[ \mathrm{CH_3-CH(CH_3)-CH(CH_3)-CH_3} \] Both central carbon atoms are independently attached to three other carbon atoms. Therefore, the molecule contains exactly two tertiary carbon atoms. Neither central carbon is quaternary because each still carries one hydrogen atom. Hexane has no tertiary carbon, whereas 2,2-dimethylbutane contains a quaternary carbon and no tertiary carbon. The pair in option A alone satisfies the specified carbon-connectivity conditions for both P and Q. Both assigned structures also contain six carbon atoms and fourteen hydrogen atoms, confirming their constitutional-isomer relationship.

Explanation: Each displayed structural formula contains six carbon atoms connected exclusively through carbon–carbon single bonds. Their common molecular formula is therefore \(\mathrm{C_6H_{14}}\). To determine the number of distinct constitutional isomers, carbon connectivity must be compared rather than the left-to-right appearance of the condensed formulas. \(\textbf{Identify the skeletons:}\) Structure I has a five-carbon parent chain with one methyl substituent at carbon 2. Its IUPAC name is 2-methylpentane. Structure II represents the same connectivity written from the opposite end of the parent chain. Numbering Structure II from the end nearer its methyl substituent again produces 2-methylpentane. Thus, Structures I and II are identical molecules. Structure III is 3-methylpentane, with its methyl branch at the central carbon of the five-carbon chain. Structure IV is 2,2-dimethylbutane, containing two methyl branches on the same carbon of a four-carbon parent chain. Structure V is 2,3-dimethylbutane, containing methyl branches on two adjacent carbons of a four-carbon parent chain. These last three structures are constitutionally different from one another and from 2-methylpentane. The five formulas therefore represent four distinct carbon skeletons. This isomer-counting method prevents a reversed drawing or alternative numbering direction from being mistaken for a new chemical substance.

Explanation: The molecular formulas of P, Q and R are all \(\mathrm{C_4H_8}\). They are therefore candidates for isomerism, but the type of isomerism depends on their structural differences. P is but-1-ene, containing a straight four-carbon chain with the double bond between carbons 1 and 2. Q is but-2-ene, containing the same straight four-carbon skeleton with the double bond between carbons 2 and 3. Because the carbon skeleton remains unchanged while the position of the multiple bond changes, P and Q are position isomers. R is 2-methylpropene, which possesses a branched carbon skeleton. Q and R have the same molecular formula but differ in the arrangement of their carbon skeletons. They are therefore chain isomers. Geometrical isomerism would require the same atomic connectivity with a different spatial arrangement across a restricted carbon–carbon double bond. The connectivity changes between P and Q and between Q and R, so neither comparison represents geometrical isomerism. A further distinction is that but-2-ene can have cis and trans forms, whereas 2-methylpropene cannot because one doubly bonded carbon carries two identical methyl groups. The correct relationship is therefore position isomerism for P–Q and chain isomerism for Q–R.

Explanation: Geometrical isomerism in an ordinary alkene requires restricted rotation about the carbon–carbon double bond. It also requires each carbon atom participating in that double bond to be attached to two different substituents. But-1-ene has the structure \(\mathrm{CH_2=CH-CH_2-CH_3}\). Its terminal doubly bonded carbon carries two hydrogen atoms, so it cannot exhibit geometrical isomerism. Pent-2-ene has the structure \(\mathrm{CH_3-CH=CH-CH_2-CH_3}\). Both doubly bonded carbon atoms carry two different substituents, allowing distinct geometrical arrangements. The structure of 2-methylbut-2-ene is \(\mathrm{CH_3-C(CH_3)=CH-CH_3}\). One doubly bonded carbon is attached to two identical methyl groups, preventing geometrical isomerism. The structure of 3-methylpent-2-ene is \(\mathrm{CH_3-CH=C(CH_3)-CH_2-CH_3}\). One double-bonded carbon carries hydrogen and methyl, whereas the other carries methyl and ethyl. Both carbons therefore satisfy the different-substituent condition. Consequently, pent-2-ene and 3-methylpent-2-ene can each exist as geometrical isomers. The presence of a double bond alone is insufficient; the substituent identities on both double-bonded carbons must also be checked.

Explanation: Arrangements A and B correspond to cis-but-2-ene and trans-but-2-ene, respectively. They have identical molecular formulas and identical atom connectivity. Their difference lies in the spatial arrangement of substituents across the carbon–carbon double bond. Rotation about a carbon–carbon double bond is restricted because the \(\pi\) bond would have to be disrupted for free interconversion. Consequently, A and B represent distinct geometrical isomers rather than ordinary rotational arrangements. Arrangements P and Q correspond to two conformations of ethane. In P, the front and rear carbon–hydrogen bonds overlap when viewed along the carbon–carbon bond, giving the eclipsed conformation. In Q, the rear bonds occupy positions between the front bonds, giving the staggered conformation. Ethane contains a carbon–carbon single bond, about which rotation is possible without breaking a \(\pi\) bond. Eclipsed and staggered ethane therefore represent conformational forms of the same compound. The difference between geometrical isomerism and conformational variation is the restriction on interconversion and the nature of the bond involved. Changing the spatial arrangement about a double bond is fundamentally different from rotation about a single bond. Hence, A and B are geometrical isomers, while P and Q are conformational forms.

Explanation: Each pair must first be checked for molecular-formula equality, followed by an examination of carbon connectivity and spatial arrangement. Row I is correct because but-1-ene and cyclobutane both have the molecular formula \(\mathrm{C_4H_8}\). But-1-ene is an unsaturated open-chain hydrocarbon, whereas cyclobutane is a saturated cyclic hydrocarbon. They differ in atomic connectivity and are therefore constitutional isomers, specifically exhibiting ring–chain isomerism. Row II is correct because but-1-ene and but-2-ene have the same unbranched four-carbon skeleton. Their difference is the position of the carbon–carbon double bond, making them position isomers. Row III is incorrect because cis-but-2-ene and trans-but-2-ene possess identical carbon skeletons and identical bonding connectivity. Their difference is the relative spatial arrangement of substituents across the restricted double bond. They are geometrical isomers, not chain isomers. Row IV is correct because but-1-ene has a straight four-carbon skeleton, whereas 2-methylpropene has a branched carbon skeleton. Both possess the molecular formula \(\mathrm{C_4H_8}\), but their carbon connectivity differs. They are therefore chain isomers. The correct classification requires separating changes in carbon skeleton, changes in multiple-bond position, and changes in spatial arrangement. Consequently, Rows I, II and IV are correctly matched.

Explanation: \(\textbf{Parent-chain selection:}\) The first step in naming the compound is to identify the longest continuous carbon chain. The displayed structure contains a continuous chain of five carbon atoms. The remaining two carbon atoms belong to methyl groups attached as substituents. Therefore, the parent hydrocarbon is pentane rather than hexane or heptane. \(\textbf{Substituent positions:}\) Numbering the five-carbon chain from either end gives methyl substituents at positions 2 and 4. The appropriate substituent prefix is therefore 2,4-dimethyl. The complete IUPAC name is 2,4-dimethylpentane. A name using locants 3 and 5 would not follow the lowest-locant requirement and would not correctly describe the displayed connectivity. \(\textbf{Molecular-formula verification:}\) The total carbon count is: \[ C=5+2=7 \] The compound is an acyclic saturated hydrocarbon, so it follows: \[ \mathrm{C}_{n}\mathrm{H}_{2n+2} \] Substituting \(n=7\): \[ H=2(7)+2=16 \] Thus, its molecular formula is \(\mathrm{C_7H_{16}}\). This formula can also be verified directly by counting the hydrogen atoms shown in the condensed structure. The compound contains no ring or carbon–carbon multiple bond, so there is no hydrogen deficiency relative to the alkane formula. Option D correctly satisfies the longest-chain, substituent-numbering and molecular-formula requirements.

Explanation: The first requirement is to identify the longest continuous chain of carbon atoms. The given structure contains nine carbon atoms in total, but its longest continuous carbon chain contains six carbon atoms. Therefore, hexane is the parent hydrocarbon. The remaining carbon atoms form one ethyl substituent and one methyl substituent. Numbering from the left gives methyl at carbon 3 and ethyl at carbon 4. Numbering from the right gives ethyl at carbon 3 and methyl at carbon 4. Both numbering directions produce the same set of locants, namely 3 and 4. Consequently, the ordinary lowest-set-of-locants comparison does not distinguish the two directions. In such a tie, the substituent cited first alphabetically receives the lower locant. Ethyl precedes methyl alphabetically, so ethyl receives locant 3 and methyl receives locant 4. The prefixes are also written alphabetically in the completed name, not in an arbitrary order. The nine-carbon total is consistent with a six-carbon parent chain containing ethyl and methyl branches. Therefore, the preferred IUPAC name is 3-ethyl-4-methylhexane.

Explanation: \(\textbf{Reconstruct the named carbon skeleton:}\) The parent name heptane represents a continuous chain of seven carbon atoms. The two methyl groups at carbon 2 contribute two additional carbon atoms. The ethyl group at carbon 4 contributes another two carbon atoms. Hence, the total number of carbon atoms is: \[ C=7+2(1)+2=11 \] The corresponding condensed structure is: \[ \mathrm{CH_3-C(CH_3)_2-CH_2-CH(CH_2CH_3)-CH_2-CH_2-CH_3} \] \(\textbf{Determine the molecular formula:}\) The structure is saturated and acyclic, so its molecular formula follows: \[ \mathrm{C}_{n}\mathrm{H}_{2n+2} \] For \(n=11\): \[ H=2(11)+2=24 \] Therefore: \[ \mathrm{Molecular\ formula=C_{11}H_{24}} \] \(\textbf{Classify the substituted carbon atoms:}\) Carbon 2 is bonded directly to carbon 1, carbon 3 and two methyl substituents. It therefore has four carbon neighbours and is quaternary. Carbon 4 is bonded directly to carbon 3, carbon 5 and the ethyl substituent. It has three carbon neighbours and is therefore tertiary. No other carbon atom in the molecule is attached to three or four carbon atoms. Thus, the molecule contains exactly one tertiary carbon and one quaternary carbon. The independent formula check and carbon-connectivity check both support option C.

Explanation: The molecule contains two carbon–carbon double bonds separated by a saturated carbon atom. In naming an unsaturated hydrocarbon, selection of the parent chain must account for the inclusion of the maximum number of multiple bonds. The continuous chain joining the two terminal \(\mathrm{CH_2}\) groups contains five carbon atoms and includes both double bonds. Its structural sequence is: \[ \mathrm{CH_2=CH-CH-CH=CH_2} \] The central carbon of this chain carries a three-carbon propyl substituent. Although a six-carbon continuous chain can be selected by entering the propyl branch, that alternative includes only one of the two double bonds. It is therefore not the preferred parent chain for this molecule. The five-carbon chain containing both double bonds is selected as the parent. The double bonds begin at positions 1 and 4, producing the parent name penta-1,4-diene. The propyl substituent is attached to carbon 3 irrespective of the numbering direction. The full name is consequently 3-propylpenta-1,4-diene. As an independent structural check, the molecule contains eight carbon atoms and two carbon–carbon double bonds. Its molecular formula is \(\mathrm{C_8H_{14}}\), which agrees with an acyclic diene. The crucial naming distinction is that a longer chain containing fewer multiple bonds is not automatically preferable to a chain containing all the relevant unsaturation.

Explanation: Both structures contain five carbon atoms arranged in unbranched open chains. Each possesses one carbon–carbon double bond and one carbon–carbon triple bond. Their common molecular formula is therefore \(\mathrm{C_5H_6}\). \(\textbf{Numbering hydrocarbon X:}\) Numbering X from the double-bond end gives the double bond locant 1 and triple bond locant 4. Numbering from the triple-bond end reverses these positions. The combined multiple-bond locant set is 1,4 in either direction. When the two numbering directions give equivalent multiple-bond locant sets, the double bond receives the lower locant. Hence, X is named pent-1-en-4-yne. \(\textbf{Numbering hydrocarbon Y:}\) Numbering Y from the triple-bond end gives locants 1 and 3 for the triple and double bonds, respectively. Numbering from the opposite end gives locants 2 and 4. The set 1,3 is lower than 2,4, so numbering must begin at the triple-bond end. The double-bond tie-breaking preference does not override this earlier lowest-locant-set decision. Thus, Y is named pent-3-en-1-yne. \(\textbf{Identify their relationship:}\) X and Y have identical molecular formulas and the same unbranched five-carbon skeleton. However, their carbon–carbon double and triple bonds occupy different positions. Their connectivity therefore differs, making them constitutional isomers of the position-isomerism type. They cannot be geometrical isomers because geometrical isomerism requires identical connectivity.