201. Assertion: Gauche butane is less stable than anti butane.
Reason: In gauche butane, the two \( \mathrm{CH_3} \) groups are closer than in the anti conformation.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Gauche and anti butane are both staggered conformations, so neither has the strong torsional strain of an eclipsed form. Their difference lies mainly in the relative positions of the two methyl groups. In gauche butane, the \( \mathrm{CH_3} \) groups are \(60^\circ\) apart and experience more steric repulsion. In anti butane, they are \(180^\circ\) apart and are farther from each other. The Reason directly explains why anti butane is more stable than gauche butane.
202. A potential-energy graph for rotation about the central \( \mathrm{C-C} \) bond of butane has the highest peak when
ⓐ. the \( \mathrm{CH_3} \) groups are fully eclipsed
ⓑ. the \( \mathrm{CH_3} \) groups are anti
ⓒ. the \( \mathrm{CH_3} \) groups are gauche
ⓓ. all bonds are staggered and well separated
Correct Answer: the \( \mathrm{CH_3} \) groups are fully eclipsed
Explanation: The highest point on a conformational-energy graph corresponds to the least stable conformation. In butane, this occurs when the two methyl groups directly eclipse each other. That arrangement has both torsional strain and strong methyl-methyl steric repulsion. Ordinary eclipsed forms with \( \mathrm{CH_3-H} \) overlap are also high in energy, but not as high as methyl-methyl eclipsing. The graph peak is therefore linked to the fully eclipsed conformation.
203. A butane conformation record gives these relative energies:
| Conformation | Relative energy |
|---|
| anti | \(0.0\,\text{kJ mol}^{-1}\) |
| gauche | \(3.8\,\text{kJ mol}^{-1}\) |
| fully eclipsed | \(19.0\,\text{kJ mol}^{-1}\) |
The energy difference between fully eclipsed and anti butane is
ⓐ. \(19.0\,\text{kJ mol}^{-1}\)
ⓑ. \(3.8\,\text{kJ mol}^{-1}\)
ⓒ. \(15.2\,\text{kJ mol}^{-1}\)
ⓓ. \(22.8\,\text{kJ mol}^{-1}\)
Correct Answer: \(19.0\,\text{kJ mol}^{-1}\)
Explanation: \( \textbf{Reference conformation:} \) Anti butane is given as \(0.0\,\text{kJ mol}^{-1}\).
\( \textbf{Higher-energy conformation:} \) Fully eclipsed butane is given as \(19.0\,\text{kJ mol}^{-1}\).
\( \textbf{Energy difference needed:} \) Fully eclipsed minus anti.
\[
19.0-0.0=19.0\,\text{kJ mol}^{-1}
\]
\( \textbf{Meaning of the value:} \) This is the energy rise from the most stable listed conformation to the least stable listed conformation.
\( \textbf{Final answer:} \) The difference is \(19.0\,\text{kJ mol}^{-1}\).
The \(3.8\,\text{kJ mol}^{-1}\) value compares gauche with anti, not fully eclipsed with anti.
204. In comparing ethane and butane conformations, butane has an extra stability factor because
ⓐ. butane has no \( \mathrm{C-C} \) sigma bond
ⓑ. methyl positions add steric effects
ⓒ. ethane contains an aromatic benzene ring
ⓓ. butane cannot rotate around any single bond
Correct Answer: methyl positions add steric effects
Explanation: Ethane conformational analysis mainly compares torsional strain between \( \mathrm{C-H} \) bonds. Butane also has rotation around a central \( \mathrm{C-C} \) bond, but the terminal \( \mathrm{CH_3} \) groups add an important steric factor. When these methyl groups are close or eclipsed, the energy increases. When they are anti, the steric repulsion is lowest. This makes butane a richer conformational example than ethane.
205. A learner says, “All staggered conformations of butane must have identical stability because staggered ethane does.” The best correction is that
ⓐ. staggered butane always contains a \( \mathrm{C=C} \) bond
ⓑ. ethane has no staggered conformations
ⓒ. butane conformations require breaking the central bond
ⓓ. anti and gauche forms differ in methyl separation
Correct Answer: anti and gauche forms differ in methyl separation
Explanation: In ethane, all staggered conformations are equivalent because all substituents are hydrogen atoms. Butane has two larger \( \mathrm{CH_3} \) groups when viewed along its central \( \mathrm{C-C} \) bond. In the anti form, these methyl groups are \(180^\circ\) apart, while in the gauche form they are \(60^\circ\) apart. Both are staggered, but they are not equal in energy. The difference comes from steric repulsion between methyl groups.
206. In a Newman projection of butane, an eclipsed form with \( \mathrm{CH_3-H} \) eclipsing is generally
ⓐ. lower in energy than anti butane
ⓑ. identical in energy to anti butane
ⓒ. between gauche and fully eclipsed butane in energy
ⓓ. impossible because butane has no hydrogen atoms
Correct Answer: between gauche and fully eclipsed butane in energy
Explanation: Gauche butane is staggered, so it avoids the direct bond alignment that causes torsional strain in eclipsed forms. An eclipsed form with \( \mathrm{CH_3-H} \) overlap has torsional strain and some steric interaction, making it less stable than gauche. The fully eclipsed form with \( \mathrm{CH_3-CH_3} \) overlap is still less stable because the bulky methyl groups directly eclipse. Anti butane remains the lowest-energy arrangement. The type of eclipsing interaction matters, not merely the word “eclipsed.”
207. Alkenes are hydrocarbons that contain at least one
ⓐ. \( \mathrm{C-C} \) single bond only
ⓑ. \( \mathrm{C\equiv C} \) triple bond only
ⓒ. \( \mathrm{C=C} \) double bond
ⓓ. \( \mathrm{C-O} \) bond
Correct Answer: \( \mathrm{C=C} \) double bond
Explanation: Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond. They still contain only carbon and hydrogen atoms, so they remain hydrocarbons. The double bond gives alkenes a different reaction pattern from alkanes, especially addition reactions. A triple bond belongs to alkynes, while a \( \mathrm{C-O} \) bond would introduce oxygen and make the compound no longer a hydrocarbon. The \( \mathrm{C=C} \) group is the defining structural feature.
208. The general formula of an open-chain monoalkene is
ⓐ. \( \mathrm{C_nH_{2n+2}} \)
ⓑ. \( \mathrm{C_nH_{2n}} \)
ⓒ. \( \mathrm{C_nH_{2n-2}} \)
ⓓ. \( \mathrm{C_nH_{n+2}} \)
Correct Answer: \( \mathrm{C_nH_{2n}} \)
Explanation: An open-chain monoalkene has one carbon-carbon double bond and no ring. Compared with the corresponding open-chain alkane \( \mathrm{C_nH_{2n+2}} \), it has two fewer hydrogen atoms. Therefore, its formula becomes \( \mathrm{C_nH_{2n}} \). For example, ethene is \( \mathrm{C_2H_4} \), and propene is \( \mathrm{C_3H_6} \). This formula must be used with the condition “open-chain monoalkene” because cycloalkanes can also follow \( \mathrm{C_nH_{2n}} \).
209. In a carbon-carbon double bond of an alkene, the bonding consists of
ⓐ. one \( \sigma \)-bond and one \( \pi \)-bond
ⓑ. two \( \sigma \)-bonds
ⓒ. two \( \pi \)-bonds only
ⓓ. one ionic bond and one metallic bond
Correct Answer: one \( \sigma \)-bond and one \( \pi \)-bond
Explanation: A carbon-carbon double bond is made of one \( \sigma \)-bond and one \( \pi \)-bond. The \( \sigma \)-bond is formed by head-on overlap along the internuclear axis. The \( \pi \)-bond is formed by sideways overlap of unhybridised \(p\)-orbitals. The \( \pi \)-bond is more exposed and is responsible for many addition reactions of alkenes. A double bond should not be thought of as two identical single bonds.
210. The carbon atoms directly involved in the \( \mathrm{C=C} \) bond of an alkene are generally
ⓐ. \(sp^3\)-hybridised and tetrahedral
ⓑ. \(sp\)-hybridised and linear
ⓒ. unhybridised and ionic
ⓓ. \(sp^2\)-hybridised and trigonal planar
Correct Answer: \(sp^2\)-hybridised and trigonal planar
Explanation: Each double-bonded carbon in an alkene is generally \(sp^2\)-hybridised. Three \(sp^2\) orbitals form sigma bonds in a trigonal planar arrangement. The remaining unhybridised \(p\)-orbital on each carbon overlaps sideways to form the \( \pi \)-bond. This gives an approximate bond angle of \(120^\circ\) around the double-bonded carbon. The planar arrangement is an important structural feature of alkenes.
211. A hydrocarbon has formula \( \mathrm{C_4H_8} \) and is known to be open-chain with one double bond. Its family is
ⓐ. alkene
ⓑ. alkane
ⓒ. alkyne
ⓓ. arene
Correct Answer: alkene
Explanation: The formula \( \mathrm{C_4H_8} \) fits the pattern \( \mathrm{C_nH_{2n}} \) for \(n=4\). The question also states that the compound is open-chain and has one double bond. That structural condition removes the cycloalkane possibility, which can share the same formula. An open-chain hydrocarbon with one \( \mathrm{C=C} \) bond is an alkene. Formula and structure together give a safer classification than formula alone.
212. A table compares local bonding in hydrocarbons.
| Hydrocarbon site | Hybridisation at the carbon site | Geometry |
|---|
| P. Alkane carbon with four single bonds | \(sp^3\) | tetrahedral |
| Q. Alkene carbon of \( \mathrm{C=C} \) | \(sp^2\) | trigonal planar |
| R. Alkyne carbon of \( \mathrm{C\equiv C} \) | \(sp\) | linear |
| S. Alkene carbon of \( \mathrm{C=C} \) | \(sp^3\) | tetrahedral |
The row that needs correction is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: An alkane carbon with four single bonds is \(sp^3\)-hybridised and tetrahedral. An alkene carbon involved in a double bond is \(sp^2\)-hybridised and trigonal planar. An alkyne carbon in a triple bond is \(sp\)-hybridised and linear. Row S repeats the alkene double-bond site but assigns \(sp^3\) hybridisation and tetrahedral geometry, which belong to saturated alkane carbon. The geometry follows the bonding type at the carbon atom.
213. The suitable name for \( \mathrm{CH_3CH=CHCH_3} \) is
ⓐ. but-\(1\)-ene
ⓑ. butane
ⓒ. but-\(2\)-ene
ⓓ. but-\(1\)-yne
Correct Answer: but-\(2\)-ene
Explanation: The structure \( \mathrm{CH_3CH=CHCH_3} \) has a continuous chain of \(4\) carbon atoms, so the root is \( \mathrm{but} \). The presence of a carbon-carbon double bond requires the suffix \( \mathrm{-ene} \). The double bond lies between carbon \(2\) and carbon \(3\), so the locant is \(2\). The name is therefore but-\(2\)-ene. But-\(1\)-ene would have the double bond at the end of the chain, while butane would contain only single bonds.
214. In naming an alkene, the parent chain must be chosen so that it
ⓐ. contains the carbon-carbon double bond
ⓑ. contains the maximum number of hydrogen atoms
ⓒ. avoids all substituents even if the double bond is excluded
ⓓ. is always the shortest visible chain
Correct Answer: contains the carbon-carbon double bond
Explanation: For alkenes, the parent chain should include the \( \mathrm{C=C} \) bond. The double bond is the defining feature of the alkene, so it must be part of the named parent structure. After selecting the suitable chain, numbering is done to give the double bond the lowest possible locant. A chain that is long but excludes the double bond is not the correct parent for alkene naming. The name must describe the unsaturation clearly, not just the longest carbon path.
215. A five-carbon alkene can be numbered from two ends, giving possible double-bond locants \(1\) and \(4\). The preferred locant is
ⓐ. \(4\), because higher numbers are preferred for double bonds
ⓑ. either \(1\) or \(4\), because alkene locants are unused
ⓒ. \(1\), because the double bond gets the lowest locant
ⓓ. \(5\), because the chain contains five carbon atoms
Correct Answer: \(1\), because the double bond gets the lowest locant
Explanation: In alkene nomenclature, the chain is numbered to give the carbon-carbon double bond the lowest possible number. If the same chain gives locants \(1\) and \(4\) from opposite directions, \(1\) is selected. The locant indicates where the double bond begins in the parent chain. It is not chosen to match the total number of carbon atoms. This rule prevents the same alkene from receiving two different names.
216. A structure is written as \( \mathrm{CH_2=CHCH_2CH_2CH_3} \). Its name is
ⓐ. pent-\(2\)-ene
ⓑ. pentane
ⓒ. pent-\(1\)-ene
ⓓ. but-\(1\)-ene
Correct Answer: pent-\(1\)-ene
Explanation: The structure has a continuous chain of \(5\) carbon atoms, so the root is \( \mathrm{pent} \). The \( \mathrm{C=C} \) bond makes the compound an alkene, so the suffix is \( \mathrm{-ene} \). Numbering from the double-bond end gives the double bond at carbon \(1\). Numbering from the other end would give a higher locant, which is not preferred. The correct name is pent-\(1\)-ene because the carbon chain and double-bond position are both shown.
217. The pair \( \mathrm{CH_2=CHCH_2CH_3} \) and \( \mathrm{CH_3CH=CHCH_3} \) is best described as
ⓐ. chain isomers
ⓑ. position isomers
ⓒ. homologues differing by \( \mathrm{-CH_2-} \)
ⓓ. different molecular formulas
Correct Answer: position isomers
Explanation: Both compounds contain \(4\) carbon atoms and one carbon-carbon double bond, so they have the same molecular formula \( \mathrm{C_4H_8} \). In the first structure, the double bond begins at carbon \(1\). In the second structure, the double bond begins at carbon \(2\). The carbon skeleton remains a four-carbon chain in both cases. The difference is the position of the double bond, so the pair shows position isomerism.
218. A table of alkene names and structures is shown below.
| Row | Name | Structure |
|---|
| P | ethene | \( \mathrm{CH_2=CH_2} \) |
| Q | propene | \( \mathrm{CH_3CH=CH_2} \) |
| R | but-\(2\)-ene | \( \mathrm{CH_3CH=CHCH_3} \) |
| S | but-\(1\)-ene | \( \mathrm{CH_3CH=CHCH_3} \) |
The row that needs correction is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Ethene is correctly written as \( \mathrm{CH_2=CH_2} \). Propene is correctly written as \( \mathrm{CH_3CH=CH_2} \), where numbering gives the double bond the lowest locant. But-\(2\)-ene is correctly represented by \( \mathrm{CH_3CH=CHCH_3} \). Row S is wrong because the structure shown has the double bond between carbon \(2\) and carbon \(3\), not at carbon \(1\). But-\(1\)-ene should be written as \( \mathrm{CH_2=CHCH_2CH_3} \) or an equivalent condensed form.
219. Consider these statements about alkene nomenclature.
I. The suffix \( \mathrm{-ene} \) indicates a carbon-carbon double bond.
II. The parent chain should include the \( \mathrm{C=C} \) bond.
III. The double bond is numbered to receive the highest possible locant.
ⓐ. I and III only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is correct because \( \mathrm{-ene} \) is the suffix used for alkenes. Statement II is also correct because the double bond must be included in the parent chain selected for naming. Statement III is false because the double bond should receive the lowest possible locant, not the highest. This rule is used before deciding many substituent positions in simple alkene names. The name must make the position of the unsaturation as clear and low-numbered as possible.
220. A hydrocarbon has molecular formula \( \mathrm{C_5H_{10}} \) and is known to be an open-chain monoalkene. Two possible structures are \( \mathrm{CH_2=CHCH_2CH_2CH_3} \) and \( \mathrm{CH_3CH=CHCH_2CH_3} \). Their relation is
ⓐ. chain isomerism due to different carbon skeletons
ⓑ. position isomerism due to different double-bond positions
ⓒ. geometrical isomerism only, with no position change
ⓓ. homology because one has one extra \( \mathrm{CH_2} \) unit
Correct Answer: position isomerism due to different double-bond positions
Explanation: Both structures have the same five-carbon chain and the same molecular formula \( \mathrm{C_5H_{10}} \). In the first structure, the double bond is at carbon \(1\). In the second structure, the double bond is at carbon \(2\). Since the carbon skeleton remains the same but the multiple-bond position changes, the isomerism is position isomerism. They are not homologues because they do not differ by \( \mathrm{-CH_2-} \).
