101. The molecule \(\mathrm{CH_3CH_2CH(Cl)CH_2CH_3}\) is achiral because the chlorine-bearing carbon:
ⓐ. is \(\mathrm{sp^2}\)-hybridised
ⓑ. is bonded to two chlorine atoms
ⓒ. has two identical ethyl groups
ⓓ. has no hydrogen atom
Correct Answer: has two identical ethyl groups
Explanation: The chlorine-bearing carbon in \(3\)-chloropentane is attached to chlorine, hydrogen, and two ethyl groups. The two \(\mathrm{CH_2CH_3}\) groups are identical. Therefore the carbon does not have four different substituents and is not stereogenic. A tetrahedral carbon may still be achiral when two of its attached groups are the same. The position of chlorine at the centre of a symmetrical chain creates this duplication.
102. Appraise the following statements.
Statement I: A chiral haloalkane may exist as two non-superimposable mirror images.
Statement II: Every secondary haloalkane is chiral.
Statement III: A tetrahedral carbon with two identical substituents is not a stereogenic centre.
ⓐ. Statements I and II only
ⓑ. Statements I and III only
ⓒ. Statements II and III only
ⓓ. Statements I, II and III
Correct Answer: Statements I and III only
Explanation: Chiral molecules may occur as enantiomers, which are mirror images that cannot be superimposed. Statement I is therefore true. A secondary haloalkane is not automatically chiral because the halogen-bearing carbon may carry two identical carbon groups, as in \(3\)-chloropentane. Statement II is consequently false. Statement III is true because four different groups are required at a simple tetrahedral stereogenic carbon.
103. The relationship between the two stereoisomers of \(2\)-bromobutane is:
ⓐ. enantiomeric
ⓑ. constitutional
ⓒ. geometrical
ⓓ. conformational only
Correct Answer: enantiomeric
Explanation: \(2\)-Bromobutane contains one stereogenic carbon bonded to bromine, hydrogen, methyl, and ethyl groups. Two different spatial arrangements are possible at this centre. These arrangements are mirror images of each other and cannot be superimposed. They have the same atom connectivity, so they are not constitutional isomers. Their relationship is enantiomeric rather than geometrical because no carbon–carbon double-bond configuration is involved.
104. Enantiomers of a chiral haloalkane differ most directly in:
ⓐ. molecular formula
ⓑ. carbon–halogen bond count
ⓒ. connectivity of the carbon skeleton
ⓓ. optical rotation direction
Correct Answer: optical rotation direction
Explanation: Enantiomers possess the same molecular formula and the same bonding connectivity. In an achiral environment, many of their ordinary physical properties are identical. Their characteristic distinction is that they rotate plane-polarised light through equal magnitudes in opposite directions. One form is dextrorotatory and the other is levorotatory under the same measurement conditions. The sign of optical rotation must not be inferred directly from an \(\mathrm{R}\) or \(\mathrm{S}\) configuration label.
105. A sample contains \(50\%\) of one enantiomer of \(2\)-chlorobutane and \(50\%\) of its mirror-image form. Its net optical rotation is expected to be:
ⓐ. zero
ⓑ. positive and doubled
ⓒ. negative and doubled
ⓓ. dependent only on the chlorine isotope
Correct Answer: zero
Explanation: \( \textbf{Composition of the sample:} \) The two enantiomers are present in equal proportions.
\( \textbf{Rotation by the first enantiomer:} \) Let its optical rotation be \(+\alpha\).
\( \textbf{Rotation by the second enantiomer:} \) Its rotation under the same conditions is \(-\alpha\).
\( \textbf{Net rotation:} \)
\[
\alpha_{\text{net}}=(+\alpha)+(-\alpha)
\]
\( \textbf{Cancellation:} \)
\[
\alpha_{\text{net}}=0
\]
\( \textbf{Nature of the mixture:} \) Equal amounts of the two enantiomers form a racemic mixture.
\( \textbf{Final answer:} \) The net optical rotation is zero because the opposite rotations cancel exactly.
106. Match each stereochemical term in Column I with its meaning in Column II.
| Column I | Column II |
|---|
| P. Inversion | 1. Equal mixture of two enantiomers |
| Q. Retention | 2. Configuration at the reacting centre is preserved |
| R. Racemisation | 3. Opposite spatial arrangement forms at the reacting centre |
| S. Enantiomers | 4. Non-superimposable mirror-image molecules |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-3, Q-1, R-4, S-2
ⓓ. P-4, Q-2, R-3, S-1
Correct Answer: P-3, Q-2, R-1, S-4
Explanation: Inversion means that the spatial arrangement at the reacting stereogenic centre is reversed, so P matches 3. Retention means that the arrangement is preserved, giving Q-2. Racemisation produces equal amounts of two enantiomeric forms, so R matches 1. Enantiomers are non-superimposable mirror images and therefore S matches 4. These terms describe stereochemical relationships or outcomes rather than changes in molecular formula.
107. Use the following arrangement around a tetrahedral carbon:
P. \(\mathrm{Cl}\)
Q. \(\mathrm{H}\)
R. \(\mathrm{CH_3}\)
S. \(\mathrm{CH_2CH_3}\)
Interchanging any two groups once produces:
ⓐ. the same configuration without any spatial change
ⓑ. a constitutional isomer
ⓒ. the opposite configuration
ⓓ. a compound with a different molecular formula
Correct Answer: the opposite configuration
Explanation: The central carbon is stereogenic because P, Q, R, and S are four different groups. A single interchange of any two groups reverses the handedness of the tetrahedral arrangement. The product has the same molecular formula and the same bonding connectivity. It is therefore not a constitutional isomer. The new arrangement corresponds to the opposite configuration at that centre and is the mirror-image form when no other stereogenic element is present.
108. The statement that best separates molecular chirality from reaction stereochemistry is:
ⓐ. chirality is molecular; retention, inversion, and racemisation are reaction outcomes
ⓑ. a chiral substrate must undergo racemisation regardless of the reaction mechanism
ⓒ. inversion means that an achiral molecule is converted into an ionic intermediate
ⓓ. retention means that only the molecular formula, not spatial arrangement, is preserved
Correct Answer: chirality is molecular; retention, inversion, and racemisation are reaction outcomes
Explanation: Chirality is a structural property of a molecule before or after reaction. It depends on whether the molecular arrangement is superimposable on its mirror image. Retention, inversion, and racemisation describe stereochemical outcomes at a reacting centre. A chiral substrate may undergo different outcomes under different mechanisms and conditions. Preserving the molecular formula alone says nothing about whether the three-dimensional configuration has been retained.
109. In a carbon–halogen bond of a haloalkane, the usual partial-charge distribution is:
ⓐ. \(\mathrm{C^{\delta-}-X^{\delta+}}\)
ⓑ. \(\mathrm{C^{-}-X^{+}}\)
ⓒ. \(\mathrm{C^{\delta+}-X^{\delta-}}\)
ⓓ. \(\mathrm{C^+-X^-}\) with complete ionic separation
Correct Answer: \(\mathrm{C^{\delta+}-X^{\delta-}}\)
Explanation: Halogens are more electronegative than carbon and attract the shared electron pair toward themselves. The halogen therefore develops a partial negative charge, represented by \(\delta-\). Carbon becomes electron-deficient and carries a partial positive charge, represented by \(\delta+\). The bond remains polar covalent rather than fully ionic. These partial charges help explain why nucleophiles are attracted toward the carbon atom.
110. A nucleophile normally attacks the carbon atom of a haloalkane because that carbon:
ⓐ. contains a complete negative charge
ⓑ. carries partial positive character
ⓒ. is always tertiary
ⓓ. is more electronegative than the halogen
Correct Answer: carries partial positive character
Explanation: Polarisation of the carbon–halogen bond shifts electron density toward the halogen atom. The bonded carbon consequently becomes electron-deficient and carries \(\delta+\) character. A nucleophile donates an electron pair and is attracted to this electrophilic centre. The halogen-bearing carbon need not be tertiary; methyl, primary, secondary, and tertiary centres can all show bond polarisation. The direction of nucleophilic attack follows electron deficiency rather than carbon-chain length.
111. An arrow showing the carbon–halogen bond dipole in \(\mathrm{R-Cl}\) should point:
ⓐ. from chlorine toward carbon because carbon is more electronegative
ⓑ. from carbon to chlorine, the more electronegative atom
ⓒ. equally in both directions because the bond is non-polar
ⓓ. away from both atoms because no charge separation exists
Correct Answer: from carbon to chlorine, the more electronegative atom
Explanation: The dipole arrow points toward the more electronegative end of a polar bond. Chlorine attracts the shared electrons more strongly than carbon, so the negative end lies at chlorine. Carbon carries the positive end of the bond dipole. The arrow is therefore directed from carbon toward chlorine, with the positive marker at the carbon end. Reversing the arrow would incorrectly place greater electron density on carbon.
112. The increase in carbon–halogen bond length from \(\mathrm{C-F}\) to \(\mathrm{C-I}\) is mainly caused by:
ⓐ. decreasing carbon atomic size
ⓑ. increasing carbon electronegativity
ⓒ. increasing halogen atomic size
ⓓ. decreasing number of valence electrons
Correct Answer: increasing halogen atomic size
Explanation: Fluorine, chlorine, bromine, and iodine belong to the same group but have successively more occupied electron shells. Their atomic radii therefore increase down the group. As the halogen atom becomes larger, the distance between the carbon nucleus and the halogen nucleus in the bonded pair increases. This produces the order \(\mathrm{C-F\lt C-Cl\lt C-Br\lt C-I}\). The carbon atom is being held comparable in the series, so the changing halogen radius controls the trend.
113. A graph plots carbon–halogen bond length on the vertical axis and the halogen sequence \(\mathrm{F}\), \(\mathrm{Cl}\), \(\mathrm{Br}\), \(\mathrm{I}\) on the horizontal axis for similar alkyl halides. The expected trend is:
ⓐ. a horizontal line because all carbon–halogen bonds have the same length
ⓑ. a decreasing trend from fluorine to iodine
ⓒ. an irregular trend with chlorine giving the longest bond
ⓓ. an increasing trend from fluorine to iodine
Correct Answer: an increasing trend from fluorine to iodine
Explanation: \( \textbf{Horizontal-axis sequence:} \) The halogens progress from \(\mathrm{F}\) to \(\mathrm{I}\).
\( \textbf{Underlying periodic trend:} \) Halogen atomic size increases down Group \(17\).
\( \textbf{Bond-length consequence:} \) A larger bonded atom produces a greater internuclear distance.
\( \textbf{Expected order:} \)
\[
\mathrm{C-F\lt C-Cl\lt C-Br\lt C-I}
\]
\( \textbf{Graph interpretation:} \) The plotted points should rise as the sequence moves from fluorine to iodine.
\( \textbf{Final answer:} \) The graph shows an increasing trend because carbon–halogen bond length follows the increasing size of the halogen atom.
114. Check the following statements.
Statement I: The carbon–iodine bond is generally longer than the carbon–chlorine bond in comparable haloalkanes.
Statement II: A longer carbon–halogen bond is necessarily stronger than a shorter one.
Statement III: Carbon–halogen bond-length comparisons should preferably use similar carbon skeletons.
ⓐ. Statements I and II only
ⓑ. Statements I and III only
ⓒ. Statements II and III only
ⓓ. Statements I, II and III
Correct Answer: Statements I and III only
Explanation: Iodine is larger than chlorine, so the carbon–iodine bond is generally longer in comparable structures. Statement I is therefore true. Bond length and bond strength commonly vary in opposite directions in this series, making Statement II false. Structural differences such as carbon hybridisation and substitution can also influence bond length. Statement III correctly limits the comparison to related compounds so that the halogen effect is not confused with a change in carbon environment.
115. A reliable comparison of the \(\mathrm{C-Cl}\) and \(\mathrm{C-Br}\) bond lengths should preferably be made between:
ⓐ. chloroethane and bromoethane
ⓑ. chlorobenzene and methyl bromide
ⓒ. vinyl chloride and tert-butyl bromide
ⓓ. benzyl chloride and bromobenzene
Correct Answer: chloroethane and bromoethane
Explanation: Chloroethane and bromoethane have the same ethyl skeleton and the same type of halogen-bearing carbon. Their main structural difference is the identity of the halogen. This makes the comparison suitable for observing the effect of halogen size on bond length. Bromine is larger than chlorine, so the \(\mathrm{C-Br}\) bond is expected to be longer. Comparisons involving different hybridisation or different carbon frameworks introduce additional variables that can obscure the periodic trend.
116. Approximate bond lengths for a series of similar alkyl halides are shown below.
| Bond | Approximate length |
|---|
| P | \(139\,pm\) |
| Q | \(178\,pm\) |
| R | \(194\,pm\) |
| S | \(214\,pm\) |
The bond represented by S is most reasonably:
ⓐ. \(\mathrm{C-F}\)
ⓑ. \(\mathrm{C-Cl}\)
ⓒ. \(\mathrm{C-Br}\)
ⓓ. \(\mathrm{C-I}\)
Correct Answer: \(\mathrm{C-I}\)
Explanation: The longest value in the table is \(214\,pm\). Carbon–halogen bond length increases as the halogen changes from fluorine to chlorine, bromine, and iodine. Iodine is the largest atom in this set and therefore forms the longest bond with carbon. The values can consequently be assigned in the order P as \(\mathrm{C-F}\), Q as \(\mathrm{C-Cl}\), R as \(\mathrm{C-Br}\), and S as \(\mathrm{C-I}\). The numerical order follows atomic size rather than electronegativity.
117. In comparable alkyl halides, the \(\mathrm{C-Cl}\) and \(\mathrm{C-I}\) bond lengths are \(178\,pm\) and \(214\,pm\), respectively. The percentage by which the \(\mathrm{C-I}\) bond is longer than the \(\mathrm{C-Cl}\) bond is closest to:
ⓐ. \(16.8\%\)
ⓑ. \(18.0\%\)
ⓒ. \(20.2\%\)
ⓓ. \(36.0\%\)
Correct Answer: \(20.2\%\)
Explanation: \( \textbf{Given bond lengths:} \) \(\mathrm{C-Cl}=178\,pm\) and \(\mathrm{C-I}=214\,pm\).
\( \textbf{Required comparison:} \) Percentage increase relative to the \(\mathrm{C-Cl}\) bond.
\( \textbf{Increase in bond length:} \)
\[
\Delta l=214-178=36\,pm
\]
\( \textbf{Percentage-increase relation:} \)
\[
\%\text{ increase}=\frac{\text{increase}}{\text{original value}}\times100
\]
\( \textbf{Substitution:} \)
\[
\%\text{ increase}=\frac{36}{178}\times100
\]
\( \textbf{Calculation:} \)
\[
\%\text{ increase}=20.22\%\approx20.2\%
\]
\( \textbf{Final answer:} \) The \(\mathrm{C-I}\) bond is approximately \(20.2\%\) longer than the \(\mathrm{C-Cl}\) bond; using \(214\,pm\) as the denominator would reverse the reference basis.
118. Assertion: In comparable haloalkanes, carbon–halogen bond length increases from \(\mathrm{C-F}\) to \(\mathrm{C-I}\).
Reason: Halogen atomic size increases from fluorine to iodine.
ⓐ. 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: The covalent radius of the halogen increases down Group \(17\). As the bonded halogen becomes larger, the internuclear distance between carbon and halogen also increases. Thus the bond-length order is broadly \(\mathrm{C-F\lt C-Cl\lt C-Br\lt C-I}\) for comparable carbon skeletons. The Assertion and Reason are both true. The increase in halogen size directly explains the stated increase in carbon–halogen bond length.
119. The broad order of carbon–halogen bond dissociation enthalpy in similar alkyl halides is:
ⓐ. \(\mathrm{C-I\gt C-Br\gt C-Cl\gt C-F}\)
ⓑ. \(\mathrm{C-Cl\gt C-F\gt C-I\gt C-Br}\)
ⓒ. \(\mathrm{C-Br\gt C-I\gt C-F\gt C-Cl}\)
ⓓ. \(\mathrm{C-F\gt C-Cl\gt C-Br\gt C-I}\)
Correct Answer: \(\mathrm{C-F\gt C-Cl\gt C-Br\gt C-I}\)
Explanation: Bond dissociation enthalpy measures the energy needed to break a particular bond homolytically in the gaseous state. The short carbon–fluorine bond has the greatest strength and therefore the highest bond dissociation enthalpy. As halogen size increases, the carbon–halogen bond becomes longer and orbital overlap becomes less effective. The energy required for cleavage consequently decreases toward iodine. The bond-strength order is therefore the reverse of the bond-length order.
120. Iodide is generally a better leaving group than fluoride because:
ⓐ. iodide is smaller and forms the strongest carbon–halogen bond
ⓑ. iodide is polarizable and the \(\mathrm{C-I}\) bond is weak
ⓒ. fluoride has no lone pairs and cannot exist as an anion
ⓓ. iodide is more electronegative than fluoride
Correct Answer: iodide is polarizable and the \(\mathrm{C-I}\) bond is weak
Explanation: Iodide has a large electron cloud that can spread out and stabilise negative charge. It is therefore more polarizable and less strongly solvated in some nucleophilic environments than fluoride. The carbon–iodine bond is also longer and weaker than the carbon–fluorine bond. These factors allow iodide to depart more readily during substitution. Fluoride is more electronegative, but the very strong \(\mathrm{C-F}\) bond makes its release difficult.
