201. Tautomerism, at the introductory level, refers to a dynamic equilibrium between structural forms that mainly differ in:
ⓐ. only the isotope of carbon used
ⓑ. total number of atoms in the molecular formula
ⓒ. colour of the structural formula
ⓓ. position of a hydrogen atom and a double bond
Correct Answer: position of a hydrogen atom and a double bond
Explanation: Tautomerism is a special type of structural relationship in which two forms are in dynamic equilibrium. The common introductory example is keto-enol tautomerism. In such forms, a hydrogen atom and a double bond shift position in a related way. The molecular formula remains the same, but the bonding arrangement changes. It should not be treated as a random drawing difference or a change in elemental composition.
202. In keto-enol tautomerism, the keto form contains \( \mathrm{C=O} \), while the enol form contains:
ⓐ. \( \mathrm{C=C} \) and \( \mathrm{-OH} \)
ⓑ. \( \mathrm{C\equiv C} \) and \( \mathrm{-CN} \)
ⓒ. only \( \mathrm{C-C} \) single bonds and no oxygen
ⓓ. \( \mathrm{-COOH} \) and \( \mathrm{-NH_2} \)
Correct Answer: \( \mathrm{C=C} \) and \( \mathrm{-OH} \)
Explanation: A keto form contains a carbonyl group, \( \mathrm{C=O} \). In the corresponding enol form, a hydrogen shifts and a carbon-carbon double bond appears along with an \( \mathrm{-OH} \) group. The word enol itself suggests an alkene-like \( \mathrm{C=C} \) part and an alcohol-like \( \mathrm{-OH} \) part. The molecular formula is preserved during the tautomeric relationship. The change is not the formation of a nitrile, amide, or carboxylic acid group.
203. A learner says, “Tautomers are just resonance forms.” The best correction is:
ⓐ. tautomerism changes atoms; resonance changes electrons
ⓑ. tautomers always have different molecular formulas
ⓒ. resonance forms are formed by moving atoms and sigma bonds freely
ⓓ. tautomers cannot contain oxygen
Correct Answer: tautomerism changes atoms; resonance changes electrons
Explanation: Tautomeric forms are structural forms that differ in the position of a hydrogen atom and a double bond. This means atom connectivity changes in a limited but real way. Resonance forms, on the other hand, keep the same atomic positions and differ only in electron distribution. The two ideas are therefore not identical. Treating tautomers as resonance forms hides the important role of hydrogen transfer in tautomerism.
204. Assertion: Keto and enol forms can be related by tautomerism.
Reason: In keto-enol tautomerism, only the total molecular formula changes while all bonding positions remain fixed.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Assertion is true, but Reason is false
Explanation: The Assertion is true because keto and enol forms are the common introductory pair used to explain tautomerism. The Reason is false because the molecular formula does not change in tautomerism. Instead, the position of a hydrogen atom and a double bond changes. The bonding arrangement changes, but the number of each type of atom remains the same. This is why tautomerism is a structural relationship rather than a change in composition.
205. The pair below is described without drawing full structures:
Case 1: a carbonyl-containing form with \( \mathrm{C=O} \)
Case 2: a related form with \( \mathrm{C=C} \) and \( \mathrm{-OH} \)
What relationship is most likely being described?
ⓐ. chain isomerism of alkanes
ⓑ. metamerism of ethers
ⓒ. geometrical isomerism of alkenes
ⓓ. keto-enol tautomerism
Correct Answer: keto-enol tautomerism
Explanation: A form containing \( \mathrm{C=O} \) is a keto form when discussed in this context. A related form containing \( \mathrm{C=C} \) and \( \mathrm{-OH} \) is an enol form. The relationship between these two forms is keto-enol tautomerism. The change involves a shift in hydrogen position and double-bond position while preserving the molecular formula. It is not chain isomerism or metamerism because the comparison is not about carbon skeleton branching or alkyl distribution around a polyvalent group.
206. Stereoisomerism is best described as a relationship in which compounds have:
ⓐ. different molecular formulas and identical spatial arrangement
ⓑ. the same molecular formula but no covalent bonds
ⓒ. same connectivity, different spatial arrangement
ⓓ. different functional groups with no relation to atom arrangement
Correct Answer: same connectivity, different spatial arrangement
Explanation: Stereoisomers have the same molecular formula and the same order of atom connectivity. Their difference lies in the three-dimensional arrangement of atoms or groups in space. This makes stereoisomerism different from structural isomerism, where connectivity changes. Geometrical and optical isomerism are introductory examples of stereoisomerism. A flat formula may sometimes hide spatial differences, so the same connectivity does not always mean the same three-dimensional form.
207. Geometrical isomerism in simple alkenes is mainly possible because a carbon-carbon double bond:
ⓐ. allows completely free rotation like a \( \mathrm{C-C} \) single bond
ⓑ. contains no \( \sigma \)-bond
ⓒ. restricts rotation due to the \( \pi \)-bond
ⓓ. changes the molecular formula during rotation
Correct Answer: restricts rotation due to the \( \pi \)-bond
Explanation: A \( \mathrm{C=C} \) double bond contains one \( \sigma \)-bond and one \( \pi \)-bond. The \( \pi \)-bond is formed by sideways overlap of orbitals. Rotation around the double bond would disturb this sideways overlap, so free rotation is restricted. Because groups cannot freely interchange positions, different spatial arrangements can persist. This restricted rotation is the basic reason why geometrical isomerism is possible in suitable alkenes.
208. Study the conditions for geometrical isomerism around a \( \mathrm{C=C} \) bond.
| Case | Groups on left double-bond carbon | Groups on right double-bond carbon |
|---|
| P | \( \mathrm{H} \), \( \mathrm{CH_3} \) | \( \mathrm{H} \), \( \mathrm{CH_3} \) |
| Q | \( \mathrm{H} \), \( \mathrm{H} \) | \( \mathrm{CH_3} \), \( \mathrm{Cl} \) |
| R | \( \mathrm{Cl} \), \( \mathrm{CH_3} \) | \( \mathrm{Br} \), \( \mathrm{H} \) |
Which cases can show geometrical isomerism?
ⓐ. P and Q only
ⓑ. P and R only
ⓒ. Q and R only
ⓓ. P, Q, and R
Correct Answer: P and R only
Explanation: For geometrical isomerism around a \( \mathrm{C=C} \) bond, each double-bonded carbon must have two different groups attached. In Case P, each double-bonded carbon has \( \mathrm{H} \) and \( \mathrm{CH_3} \), so geometrical isomerism is possible. In Case Q, the left carbon has two identical \( \mathrm{H} \) atoms, so the condition fails. In Case R, each double-bonded carbon has two different groups, so geometrical isomerism is possible. The condition must be checked separately on both carbons of the double bond.
209. The pair \( \mathrm{cis\text{-}but\text{-}2\text{-}ene} \) and \( \mathrm{trans\text{-}but\text{-}2\text{-}ene} \) differs mainly in:
ⓐ. spatial arrangement around the \( \mathrm{C=C} \)
ⓑ. position of the double bond along the chain
ⓒ. molecular formula
ⓓ. functional group family
Correct Answer: spatial arrangement around the \( \mathrm{C=C} \)
Explanation: \( \mathrm{cis\text{-}but\text{-}2\text{-}ene} \) and \( \mathrm{trans\text{-}but\text{-}2\text{-}ene} \) have the same molecular formula and the same connectivity. The double bond is between the same pair of carbon atoms in both forms. The difference is whether similar groups lie on the same side or opposite sides of the \( \mathrm{C=C} \) bond. This makes the pair geometrical isomers. The locant \(2\) is unchanged, so the pair is not position isomerism.
210. Ethene, \( \mathrm{CH_2=CH_2} \), does not show geometrical isomerism because:
ⓐ. it has a carbon-carbon double bond
ⓑ. it contains a \( \pi \)-bond
ⓒ. it is an organic compound
ⓓ. identical H atoms on each alkene carbon
Correct Answer: identical H atoms on each alkene carbon
Explanation: A carbon-carbon double bond alone is not enough for geometrical isomerism. Each carbon of the \( \mathrm{C=C} \) bond must have two different groups attached. In ethene, each double-bonded carbon is attached to two identical hydrogen atoms. Interchanging identical hydrogens does not create a distinct spatial form. This is why restricted rotation is necessary but not sufficient by itself.
211. Assertion: \( \mathrm{but\text{-}2\text{-}ene} \) can show geometrical isomerism.
Reason: In \( \mathrm{but\text{-}2\text{-}ene} \), each double-bonded carbon is attached to one \( \mathrm{H} \) and one \( \mathrm{CH_3} \) group.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. 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
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: \( \mathrm{but\text{-}2\text{-}ene} \) has the structure \( \mathrm{CH_3CH=CHCH_3} \). Each double-bonded carbon is attached to two different groups, \( \mathrm{H} \) and \( \mathrm{CH_3} \). The double bond restricts rotation, so the same-side and opposite-side arrangements of the \( \mathrm{CH_3} \) groups can be distinct. The Reason gives the exact structural condition needed for geometrical isomerism. If either double-bonded carbon had two identical groups, the conclusion would not follow.
212. Read the case below and answer the question.
Two molecules have formula \( \mathrm{C_4H_8} \). Molecule P is \( \mathrm{CH_2=CHCH_2CH_3} \). Molecule Q is \( \mathrm{CH_3CH=CHCH_3} \). A note says that Q may have two spatial forms, while P does not.
What is the best reason for the note?
ⓐ. P has no carbon-carbon double bond
ⓑ. P and Q have different molecular formulas
ⓒ. Q contains oxygen, while P does not
ⓓ. Q only; P has identical alkene H atoms
Correct Answer: Q only; P has identical alkene H atoms
Explanation: Molecule P is \( \mathrm{but\text{-}1\text{-}ene} \), where one double-bonded carbon is \( \mathrm{CH_2} \). That carbon has two identical hydrogen atoms, so geometrical isomerism is not possible. Molecule Q is \( \mathrm{but\text{-}2\text{-}ene} \), where each double-bonded carbon is attached to \( \mathrm{H} \) and \( \mathrm{CH_3} \). Since each double-bonded carbon has two different groups, Q can exist in cis and trans forms. The difference is not the molecular formula, but the group arrangement around the double bond.
213. A pair of compounds has the same formula \( \mathrm{C_4H_8} \). Pair I is \( \mathrm{but\text{-}1\text{-}ene} \) and \( \mathrm{but\text{-}2\text{-}ene} \). Pair II is \( \mathrm{cis\text{-}but\text{-}2\text{-}ene} \) and \( \mathrm{trans\text{-}but\text{-}2\text{-}ene} \). The correct comparison is:
ⓐ. Pair I shows geometrical isomerism, while Pair II shows position isomerism
ⓑ. both pairs show only chain isomerism
ⓒ. Pair I shows position isomerism, while Pair II shows geometrical isomerism
ⓓ. both pairs are identical compounds written twice
Correct Answer: Pair I shows position isomerism, while Pair II shows geometrical isomerism
Explanation: \( \mathrm{but\text{-}1\text{-}ene} \) and \( \mathrm{but\text{-}2\text{-}ene} \) differ in the position of the \( \mathrm{C=C} \) bond along the same four-carbon chain. That is position isomerism. \( \mathrm{cis\text{-}but\text{-}2\text{-}ene} \) and \( \mathrm{trans\text{-}but\text{-}2\text{-}ene} \) have the double bond at the same position and the same connectivity. Their difference is the spatial arrangement around the restricted \( \mathrm{C=C} \) bond. Locants describe position, while cis/trans describes spatial arrangement.
214. A graph is described with rotation about a carbon-carbon bond on the horizontal axis and preservation of effective sideways overlap on the vertical axis. For a \( \mathrm{C=C} \) bond, the curve would drop sharply on rotation because:
ⓐ. the \( \pi \)-bond requires parallel sideways overlap
ⓑ. the \( \sigma \)-bond disappears at all angles
ⓒ. hydrogen atoms change into carbon atoms during rotation
ⓓ. the molecular formula changes continuously
Correct Answer: the \( \pi \)-bond requires parallel sideways overlap
Explanation: The \( \pi \)-bond in a carbon-carbon double bond depends on sideways overlap of parallel orbitals. Rotation around the double bond would make these orbitals less parallel. As the sideways overlap is disturbed, the \( \pi \)-bond is weakened or broken. This is why free rotation is not normally allowed around a \( \mathrm{C=C} \) bond. A \( \sigma \)-bond alone would not create the same restriction because its overlap is along the internuclear axis.
215. Use the arrangement described below: each carbon of a \( \mathrm{C=C} \) bond has one \( \mathrm{H} \) and one \( \mathrm{Cl} \) attached. In one form, the two \( \mathrm{Cl} \) atoms are on the same side of the double bond; in another, they are on opposite sides. What relationship is described?
ⓐ. chain isomerism
ⓑ. functional group isomerism
ⓒ. geometrical isomerism
ⓓ. tautomerism
Correct Answer: geometrical isomerism
Explanation: The two described forms have the same atom connectivity around the \( \mathrm{C=C} \) bond. The difference is the spatial arrangement of the \( \mathrm{Cl} \) atoms relative to the double bond. Same-side and opposite-side arrangements correspond to cis/trans-type geometrical isomerism at this level. This is possible because each double-bonded carbon has two different groups, \( \mathrm{H} \) and \( \mathrm{Cl} \). No functional group is changed, and no hydrogen-double-bond shift like tautomerism is involved.
216. Optical isomerism is introduced through molecules that contain a chiral carbon. A chiral carbon is usually a carbon atom attached to:
ⓐ. four identical groups
ⓑ. two double bonds
ⓒ. four different groups
ⓓ. only hydrogen atoms
Correct Answer: four different groups
Explanation: A chiral carbon is commonly identified as a tetrahedral carbon attached to four different groups. Such an arrangement can give non-superimposable mirror-image forms. These mirror-image forms are related to optical isomerism. If two or more groups attached to the carbon are identical, that carbon is not chiral by this simple test. The idea depends on three-dimensional arrangement, not merely on the presence of carbon.
217. The carbon marked by \( * \) in \( \mathrm{CH_3CH^*(OH)COOH} \) is attached to \( \mathrm{H} \), \( \mathrm{OH} \), \( \mathrm{CH_3} \), and \( \mathrm{COOH} \). What is the best conclusion?
ⓐ. it is not chiral because it contains oxygen
ⓑ. four different groups make it chiral
ⓒ. it must be \(sp\)-hybridised because it is marked
ⓓ. it cannot show any spatial arrangement
Correct Answer: four different groups make it chiral
Explanation: The marked carbon is bonded to four different groups: \( \mathrm{H} \), \( \mathrm{OH} \), \( \mathrm{CH_3} \), and \( \mathrm{COOH} \). A tetrahedral carbon with four different groups is chiral at the introductory level. Such a carbon can give rise to non-superimposable mirror-image forms. The presence of oxygen in two of the groups does not make the groups identical, because \( \mathrm{OH} \) and \( \mathrm{COOH} \) are different structural groups. Chirality is decided by the complete groups attached to carbon.
218. A tetrahedral carbon is attached to \( \mathrm{H} \), \( \mathrm{H} \), \( \mathrm{CH_3} \), and \( \mathrm{Cl} \). The molecule does not have a chiral carbon at that atom because:
ⓐ. chlorine is too heavy to be attached to carbon
ⓑ. tetrahedral carbon can never be chiral
ⓒ. all compounds with hydrogen are achiral
ⓓ. two attached groups are identical
Correct Answer: two attached groups are identical
Explanation: The simple test for a chiral carbon requires four different groups attached to a tetrahedral carbon. In the given case, two attached groups are both \( \mathrm{H} \). Because two groups are identical, the carbon does not satisfy the four-different-groups condition. The presence of \( \mathrm{Cl} \) does not automatically create chirality. A tetrahedral carbon may be chiral only when the four attached groups are distinguishable.
219. Consider the following statements about optical isomerism.
Statement I: Optical isomerism is related to non-superimposable mirror-image forms.
Statement II: A carbon attached to four different groups is a common source of chirality.
Statement III: Optical isomerism is the same as position isomerism of a double bond.
Which statements are valid?
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is valid because optical isomerism is associated with non-superimposable mirror-image forms. Statement II is also valid because a tetrahedral carbon attached to four different groups is a common chiral centre. Statement III is not valid because position isomerism is a structural isomerism involving different positions of a group or multiple bond. Optical isomerism is a stereoisomeric relationship and depends on spatial arrangement. Moving a double bond along a chain is not the same as forming mirror-image stereoisomers.
220. Two mirror-image models of a molecule cannot be placed exactly on top of each other in the same orientation. This description is most closely connected with:
ⓐ. chain isomerism
ⓑ. optical isomerism
ⓒ. metamerism
ⓓ. acid-ester functional group isomerism
Correct Answer: optical isomerism
Explanation: Non-superimposable mirror images are the central introductory idea behind optical isomerism. Such forms often arise when a molecule contains a chiral carbon attached to four different groups. Chain isomerism involves different carbon skeletons, not mirror-image forms. Metamerism involves different alkyl-group distribution around a polyvalent atom or group. The mirror-image language points to stereochemistry rather than ordinary structural isomerism.
