201. The following table records proposed products of alkene halogenation.
| Row | Reaction | Proposed product |
|---|
| P | Ethene with \(\mathrm{Br_2}\) | \(1,2\)-dibromoethane |
| Q | Propene with \(\mathrm{Cl_2}\) | \(1,1\)-dichloropropane |
| R | Cyclohexene with \(\mathrm{Br_2}\) | \(1,2\)-dibromocyclohexane |
| S | Ethene with \(\mathrm{Cl_2}\) | \(1,2\)-dichloroethane |
The row requiring correction is:
ⓐ. P only
ⓑ. Q only
ⓒ. R only
ⓓ. S only
Correct Answer: Q only
Explanation: Ethene gives \(1,2\)-dibromoethane with bromine and \(1,2\)-dichloroethane with chlorine, so P and S are suitable. Cyclohexene gives a vicinal dibromide with bromine, making R suitable. Propene also undergoes addition across its double bond, placing one chlorine on each of the two alkene carbons. Its product is \(1,2\)-dichloropropane, not \(1,1\)-dichloropropane. Row Q incorrectly converts a vicinal-addition product into a geminal dihalide.
202. A \(5.60\,g\) sample of ethene reacts completely with excess bromine to form \(1,2\)-dibromoethane.
\[
\mathrm{CH_2{=}CH_2+Br_2\rightarrow BrCH_2CH_2Br}
\]
Taking the molar masses of ethene and \(1,2\)-dibromoethane as \(28.0\,g\,mol^{-1}\) and \(188\,g\,mol^{-1}\), respectively, the theoretical product mass is:
ⓐ. \(18.8\,g\)
ⓑ. \(28.0\,g\)
ⓒ. \(37.6\,g\)
ⓓ. \(75.2\,g\)
Correct Answer: \(37.6\,g\)
Explanation: \( \textbf{Ethene mass:} \)
\[
m(\mathrm{C_2H_4})=5.60\,g
\]
\( \textbf{Ethene molar mass:} \)
\[
M(\mathrm{C_2H_4})=28.0\,g\,mol^{-1}
\]
\( \textbf{Moles of ethene:} \)
\[
n=\frac{5.60}{28.0}=0.200\,mol
\]
\( \textbf{Reaction ratio:} \)
\[
\mathrm{C_2H_4:C_2H_4Br_2}=1:1
\]
\( \textbf{Moles of product:} \)
\[
n(\mathrm{C_2H_4Br_2})=0.200\,mol
\]
\( \textbf{Product molar mass:} \)
\[
M(\mathrm{C_2H_4Br_2})=188\,g\,mol^{-1}
\]
\( \textbf{Theoretical mass:} \)
\[
m=nM=(0.200)(188)
\]
\( \textbf{Calculation:} \)
\[
m=37.6\,g
\]
\( \textbf{Final answer:} \) Complete addition forms \(37.6\,g\) of \(1,2\)-dibromoethane; the product mass exceeds the alkene mass because an entire \(\mathrm{Br_2}\) molecule is incorporated.
203. In dry acetone, the Finkelstein reaction is correctly represented by:
ⓐ. \(\mathrm{R-I+NaCl\xrightarrow{water}R-Cl+NaI}\)
ⓑ. \(\mathrm{R-Cl+AgF\rightarrow R-F+AgCl}\)
ⓒ. \(\mathrm{R-Br+KOH\xrightarrow{alcohol}alkene+KBr+H_2O}\)
ⓓ. \(\mathrm{R-Cl+NaI\rightarrow R-I+NaCl\downarrow}\)
Correct Answer: \(\mathrm{R-Cl+NaI\rightarrow R-I+NaCl\downarrow}\)
Explanation: The Finkelstein reaction is a halogen-exchange reaction used to prepare alkyl iodides. An alkyl chloride or alkyl bromide is treated with sodium iodide. Dry acetone is used as the solvent. Sodium iodide is soluble in acetone, whereas sodium chloride and sodium bromide are poorly soluble and precipitate. Removal of the sodium halide from solution shifts the reaction toward the alkyl iodide. The transformation is therefore driven by both substitution and precipitation.
204. The principal organic product formed when \(1\)-bromobutane is heated with \(\mathrm{NaI}\) in dry acetone is:
ⓐ. butan-\(1\)-ol
ⓑ. \(1\)-iodobutane
ⓒ. but-\(1\)-ene
ⓓ. \(2\)-iodobutane
Correct Answer: \(1\)-iodobutane
Explanation: Iodide ion replaces bromide at the carbon bearing the halogen. The four-carbon skeleton and the position of the functional group remain unchanged. The reaction is:
\[
\mathrm{CH_3CH_2CH_2CH_2Br+NaI\rightarrow CH_3CH_2CH_2CH_2I+NaBr\downarrow}
\]
The product is therefore \(1\)-iodobutane. Formation of \(2\)-iodobutane would require a change in the position of the halogen, which is not part of the ordinary exchange reaction. The precipitated \(\mathrm{NaBr}\) helps drive the conversion forward.
205. Assertion: Precipitation of \(\mathrm{NaCl}\) or \(\mathrm{NaBr}\) favours alkyl iodide formation in the Finkelstein reaction.
Reason: Removing one product from the reaction mixture shifts the equilibrium toward further product formation.
ⓐ. 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 halogen-exchange reaction can be treated as a reversible ionic substitution. Sodium chloride or sodium bromide formed during the process is poorly soluble in dry acetone. Its precipitation lowers the concentration of dissolved products. According to the equilibrium principle, the system responds by forming more alkyl iodide and sodium halide. The Reason therefore states the precise equilibrium effect responsible for the observation in the Assertion.
206. A Finkelstein-reaction record contains three proposed outcomes.
P. \(\mathrm{CH_3CH_2Cl+NaI}\) in dry acetone gives \(\mathrm{CH_3CH_2I+NaCl\downarrow}\).
Q. \(\mathrm{CH_3CH_2CH_2Br+NaI}\) in dry acetone gives \(\mathrm{CH_3CH_2CH_2I+NaBr\downarrow}\).
R. \(\mathrm{C_6H_5Cl+NaI}\) in dry acetone rapidly gives \(\mathrm{C_6H_5I}\).
Select the appropriate evaluation.
ⓐ. Only R is correctly represented
ⓑ. P and R are correctly represented
ⓒ. P and Q are correctly represented
ⓓ. P, Q, and R are all correctly represented
Correct Answer: P and Q are correctly represented
Explanation: Ethyl chloride and \(1\)-bromopropane are suitable alkyl halides for iodide substitution under Finkelstein conditions. Their reactions form alkyl iodides and precipitated sodium chloride or sodium bromide. Chlorobenzene is an aryl halide with chlorine bonded directly to an \(\mathrm{sp^2}\)-hybridised ring carbon. Its carbon–chlorine bond has partial double-bond character and does not undergo the ordinary substitution readily. Row R therefore applies an alkyl-halide reaction incorrectly to an aryl halide.
207. The substrate expected to undergo the Finkelstein reaction most readily is:
ⓐ. \(1\)-bromopropane
ⓑ. bromobenzene
ⓒ. vinyl chloride
ⓓ. tert-butyl chloride
Correct Answer: \(1\)-bromopropane
Explanation: The Finkelstein reaction commonly proceeds through a nucleophilic substitution pathway involving iodide ion. A primary alkyl bromide offers relatively little steric hindrance and has bromide as a suitable leaving group. Bromobenzene and vinyl chloride contain halogen bonded to \(\mathrm{sp^2}\)-hybridised carbon and resist ordinary substitution. A tertiary halide is strongly hindered for a direct backside-attack pathway and may favour competing processes. \(1\)-Bromopropane therefore provides the most suitable structure among the choices.
208. A learner predicts that chlorobenzene will rapidly give iodobenzene with \(\mathrm{NaI}\) in dry acetone. The prediction is unsuitable mainly because:
ⓐ. iodide ion is unable to form any carbon–iodine bond
ⓑ. the aryl \(\mathrm{C-Cl}\) bond resists Finkelstein substitution
ⓒ. chlorobenzene contains no chlorine atom
ⓓ. acetone converts chlorobenzene into benzene before iodide attack
Correct Answer: the aryl \(\mathrm{C-Cl}\) bond resists Finkelstein substitution
Explanation: In chlorobenzene, chlorine is attached directly to an aromatic \(\mathrm{sp^2}\)-hybridised carbon. Resonance gives the carbon–chlorine bond partial double-bond character. The bond is therefore shorter and stronger than a typical alkyl carbon–chlorine bond. The geometry also prevents the ordinary backside substitution pathway from occurring readily at the aromatic carbon. Finkelstein conditions are consequently most useful for suitable alkyl chlorides and bromides rather than chlorobenzene.
209. Match each item in Column I with the appropriate description in Column II.
| Column I | Column II |
|---|
| P. \(\mathrm{NaI}\) | 1. Common Finkelstein solvent |
| Q. Dry acetone | 2. Desired organic product |
| R. \(\mathrm{NaCl}\) or \(\mathrm{NaBr}\) | 3. Iodide source |
| S. \(\mathrm{R-I}\) | 4. Precipitated inorganic product |
ⓐ. P-3, Q-1, R-4, S-2
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-1, S-3
ⓓ. P-2, Q-4, R-3, S-1
Correct Answer: P-3, Q-1, R-4, S-2
Explanation: Sodium iodide supplies the iodide ion, so P matches 3. Dry acetone is the characteristic solvent, giving Q-1. Sodium chloride or sodium bromide separates as the precipitated inorganic product, so R matches 4. The exchanged organic compound is the alkyl iodide, making S-2. These four components together describe the reagent, condition, driving force, and target product of the Finkelstein reaction.
210. A graph shows the amount of \(\mathrm{NaBr}\) precipitate on the vertical axis and reaction time on the horizontal axis during a Finkelstein reaction with a fixed amount of \(1\)-bromopropane. The most reasonable curve:
ⓐ. begins at a maximum and decreases steadily below zero
ⓑ. remains horizontal at zero throughout the reaction
ⓒ. oscillates repeatedly as \(\mathrm{NaBr}\) converts into bromine gas
ⓓ. rises to a plateau when the limiting reactant is consumed
Correct Answer: rises to a plateau when the limiting reactant is consumed
Explanation: Sodium bromide forms as the substitution reaction proceeds. Its amount therefore begins near zero and increases as more alkyl iodide is produced. The precipitate does not normally disappear because it is poorly soluble in dry acetone. Once the limiting reactant has been consumed, no additional sodium bromide can form. The curve consequently becomes horizontal at a maximum value. The plateau corresponds to completion under the stated reagent amounts.
211. Which set contains reagents associated with the Swarts reaction?
ⓐ. \(\mathrm{NaI}\), \(\mathrm{KI}\), and dry acetone
ⓑ. \(\mathrm{PCl_3}\), \(\mathrm{PCl_5}\), and \(\mathrm{SOCl_2}\)
ⓒ. \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\), and ultraviolet light
ⓓ. \(\mathrm{AgF}\), \(\mathrm{SbF_3}\), and \(\mathrm{CoF_2}\)
Correct Answer: \(\mathrm{AgF}\), \(\mathrm{SbF_3}\), and \(\mathrm{CoF_2}\)
Explanation: Metallic fluorides are used to introduce fluorine during the Swarts reaction. Common textbook examples include silver fluoride, mercurous fluoride, cobalt fluoride, and antimony trifluoride. These reagents replace chlorine or bromine in suitable alkyl halides. Sodium iodide in acetone belongs to the Finkelstein reaction. Chlorine or bromine under light supports radical halogenation, while phosphorus chlorides and thionyl chloride prepare alkyl chlorides from alcohols.
212. Direct fluorination of an alkane is generally avoided as a controlled laboratory method for preparing a specific alkyl fluoride because:
ⓐ. fluorination is too vigorous for controlled preparation
ⓑ. fluorine cannot form a covalent bond with carbon
ⓒ. molecular fluorine is completely inert toward hydrocarbons
ⓓ. every carbon–fluorine bond decomposes immediately
Correct Answer: fluorination is too vigorous for controlled preparation
Explanation: Molecular fluorine is the most reactive halogen. Its direct reaction with hydrocarbons can be extremely vigorous and difficult to control. Multiple substitution, fragmentation, or combustion-like behaviour may occur rather than selective formation of one desired fluoroalkane. The Swarts reaction offers a safer and more selective exchange route. Carbon–fluorine bonds are not unstable; they are actually very strong once formed. The difficulty lies in controlling direct fluorination, not in maintaining the product bond.
213. Which equation best represents a Swarts reaction using silver fluoride?
ⓐ. \(\mathrm{R-F+AgCl\rightarrow R-Cl+AgF}\)
ⓑ. \(\mathrm{R-Cl+NaI\rightarrow R-I+NaCl}\)
ⓒ. \(\mathrm{R-Cl+AgF\rightarrow R-F+AgCl}\)
ⓓ. \(\mathrm{R-OH+AgF\rightarrow R-H+AgOH}\)
Correct Answer: \(\mathrm{R-Cl+AgF\rightarrow R-F+AgCl}\)
Explanation: Silver fluoride supplies fluoride for replacement of chlorine in the alkyl halide. The organic product is the corresponding alkyl fluoride. Silver chloride is formed as the inorganic product. The carbon skeleton remains unchanged during the exchange. The equation using sodium iodide represents the Finkelstein reaction rather than the Swarts reaction. The defining feature here is replacement of chlorine or bromine by fluorine.
214. A laboratory must prepare compound P, an alkyl iodide, and compound Q, an alkyl fluoride, from the corresponding alkyl bromides. The appropriate methods are:
ⓐ. Swarts reaction for P and Finkelstein reaction for Q
ⓑ. radical chlorination for P and bromine addition for Q
ⓒ. thionyl chloride treatment for P and Lucas reagent for Q
ⓓ. Finkelstein reaction for P and Swarts reaction for Q
Correct Answer: Finkelstein reaction for P and Swarts reaction for Q
Explanation: The Finkelstein reaction introduces iodine into a suitable alkyl chloride or bromide using \(\mathrm{NaI}\) in dry acetone. It is therefore appropriate for preparing compound P. The Swarts reaction uses a metallic fluoride to replace chlorine or bromine by fluorine. It is the appropriate route for compound Q. Both reactions are halogen-exchange methods, but they differ in the halogen introduced and in their reagents. Selecting the named reaction requires identifying the target halogen rather than only the starting haloalkane.
215. In the presence of anhydrous \(\mathrm{FeCl_3}\), benzene chlorination is represented by:
ⓐ. \(\mathrm{C_6H_6+Cl_2\rightarrow C_6H_6Cl_2}\)
ⓑ. \(\mathrm{C_6H_6+Cl_2\rightarrow C_6H_5Cl+HCl}\)
ⓒ. \(\mathrm{C_6H_6+HCl\rightarrow C_6H_5Cl+H_2}\)
ⓓ. \(\mathrm{C_6H_6+Cl^-\rightarrow C_6H_5Cl+H^-}\)
Correct Answer: \(\mathrm{C_6H_6+Cl_2\rightarrow C_6H_5Cl+HCl}\)
Explanation: Benzene reacts with chlorine by electrophilic aromatic substitution. One hydrogen atom on the aromatic ring is replaced by chlorine. The organic product is chlorobenzene, \(\mathrm{C_6H_5Cl}\). The removed hydrogen combines with the other chlorine atom to form \(\mathrm{HCl}\). Anhydrous \(\mathrm{FeCl_3}\) activates the chlorine molecule and helps generate a sufficiently strong electrophilic chlorinating species. The aromatic ring is restored at the end of the substitution.
216. Direct chlorination of benzene gives substitution rather than stable addition mainly because:
ⓐ. deprotonation restores the aromatic pi system
ⓑ. benzene contains no pi electrons available for reaction
ⓒ. chlorine is unable to form two carbon–chlorine bonds
ⓓ. addition would increase the number of delocalised pi electrons
Correct Answer: deprotonation restores the aromatic pi system
Explanation: Electrophilic attack initially produces a non-aromatic sigma complex. This intermediate has temporarily lost the complete cyclic delocalisation of benzene. Removal of the proton from the carbon bearing the incoming chlorine restores the continuous aromatic pi system. The final product therefore contains one chlorine atom in place of one ring hydrogen. A stable addition product would retain the loss of aromatic stabilisation. Aromaticity restoration strongly favours substitution over addition.
217. Arrange the following events in the appropriate order for electrophilic chlorination of benzene.
P. The sigma complex loses a proton.
Q. Benzene pi electrons attack the electrophilic chlorinating species.
R. Anhydrous \(\mathrm{FeCl_3}\) activates and polarises \(\mathrm{Cl_2}\).
S. Aromaticity is restored and chlorobenzene is obtained.
ⓐ. Q, R, P, S
ⓑ. R, P, Q, S
ⓒ. R, Q, P, S
ⓓ. P, R, S, Q
Correct Answer: R, Q, P, S
Explanation: The halogen molecule must first be activated by the Lewis acid catalyst. Benzene then uses its pi electrons to attack the electrophilic chlorinating species. This step forms a sigma complex in which one ring carbon is temporarily \(\mathrm{sp^3}\)-hybridised. The sigma complex subsequently loses a proton. Proton loss re-establishes the aromatic pi system. Chlorobenzene is obtained after the sequence R, Q, P, S.
218. Assertion: Chlorination of benzene generally requires a Lewis acid such as anhydrous \(\mathrm{FeCl_3}\).
Reason: The Lewis acid helps generate a chlorinating species strong enough to attack the aromatic pi system.
ⓐ. 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: Benzene is unusually stable because of aromatic delocalisation. Molecular chlorine alone is not generally electrophilic enough for rapid ring substitution under mild conditions. Anhydrous \(\mathrm{FeCl_3}\) interacts with \(\mathrm{Cl_2}\) and increases its electrophilic character. The activated species can then be attacked by the benzene ring. The catalyst is regenerated after loss of the ring proton. The Reason directly accounts for the catalyst requirement stated in the Assertion.
219. A reaction-coordinate graph for benzene chlorination shows two pathways with the same reactant and product energy levels. The curve for the \(\mathrm{FeCl_3}\)-catalysed pathway has a lower maximum than the uncatalysed curve. The graph indicates that \(\mathrm{FeCl_3}\):
ⓐ. increases the enthalpy change of the reaction
ⓑ. lowers the activation energy without altering \(\Delta H\)
ⓒ. changes chlorobenzene into the thermodynamic reactant
ⓓ. raises the energy of the products above that of the uncatalysed products
Correct Answer: lowers the activation energy without altering \(\Delta H\)
Explanation: The maximum point on a reaction-coordinate curve represents the energy barrier associated with the activated state. A lower maximum therefore indicates a lower activation energy. The catalysed and uncatalysed curves begin and end at the same energy levels because the reactants and products are unchanged. Thus, the catalyst provides an alternative pathway but does not change the overall enthalpy difference. A lower activation barrier increases the reaction rate. The graph does not imply that the catalyst changes the equilibrium identities of benzene and chlorobenzene.
220. The reagent system suitable for converting benzene into bromobenzene is:
ⓐ. \(\mathrm{HBr}\) with an organic peroxide
ⓑ. \(\mathrm{Br_2/FeBr_3}\)
ⓒ. \(\mathrm{NaBr}\) in dry acetone
ⓓ. \(\mathrm{PBr_3}\) in water
Correct Answer: \(\mathrm{Br_2/FeBr_3}\)
Explanation: Bromination of benzene is an electrophilic aromatic substitution reaction. Molecular bromine supplies the bromine atom introduced into the ring. Anhydrous \(\mathrm{FeBr_3}\) acts as a Lewis acid and activates \(\mathrm{Br_2}\). Benzene attacks the electrophilic brominating species, and loss of a proton restores aromaticity. \(\mathrm{HBr}\) with peroxide is associated with radical addition to suitable alkenes. Sodium bromide and phosphorus tribromide do not provide the required direct aromatic bromination conditions.
