301. A student records the following observations after treating suitable haloalkanes with ethanolic \(\mathrm{AgNO_3}\).
| Sample | Precipitate colour | Inferred halogen |
|---|
| P | White | Chlorine |
| Q | Cream | Bromine |
| R | Yellow | Iodine |
The appropriate evaluation is:
ⓐ. only P and Q are correct
ⓑ. only Q and R are correct
ⓒ. all three are correct
ⓓ. only P and R are correct
Correct Answer: all three are correct
Explanation: Chloride ions form white \(\mathrm{AgCl}\). Bromide ions form cream-coloured \(\mathrm{AgBr}\). Iodide ions form yellow \(\mathrm{AgI}\). The recorded colours therefore match the inferred halogens in all three cases. The test detects the halide released during reaction rather than directly displaying the colour of the original haloalkane.
302. Three colourless liquids are separately treated with ethanolic \(\mathrm{AgNO_3}\). Sample P gives an immediate cream precipitate, sample Q gives the same cream precipitate after several minutes, and sample R shows little change at room temperature. If all three are bromoalkanes, the most reasonable assignment is:
ⓐ. P is primary, Q is tertiary, and R is secondary
ⓑ. P is secondary, Q is primary, and R is tertiary
ⓒ. P, Q, and R must all have identical structures
ⓓ. P is tertiary, Q is secondary, and R is primary
Correct Answer: P is tertiary, Q is secondary, and R is primary
Explanation: All three samples give or are expected to give cream \(\mathrm{AgBr}\), so their halogen identity is the same. The difference lies in the time required for halide release. A tertiary bromoalkane ionises most readily and gives the earliest precipitate. A secondary bromoalkane responds more slowly. A primary bromoalkane generally shows the least rapid response under these conditions, supporting the assignment P tertiary, Q secondary, and R primary.
303. Chlorobenzene and vinyl chloride usually fail to give a rapid silver chloride precipitate with ethanolic \(\mathrm{AgNO_3}\) because:
ⓐ. their \(\mathrm{C_{sp^2}-Cl}\) bonds resist ionisation
ⓑ. both substances contain no chlorine atoms
ⓒ. chloride ion forms a highly soluble silver chloride complex immediately
ⓓ. their carbon–chlorine bonds are weaker than those in tertiary alkyl chlorides
Correct Answer: their \(\mathrm{C_{sp^2}-Cl}\) bonds resist ionisation
Explanation: In chlorobenzene and vinyl chloride, chlorine is bonded directly to an \(\mathrm{sp^2}\)-hybridised carbon. The carbon–chlorine bond is shorter and stronger than a typical alkyl carbon–chlorine bond. In chlorobenzene, resonance also gives the bond partial double-bond character. Formation of phenyl or vinyl carbocations is highly unfavourable. These compounds therefore do not release chloride readily under ordinary ethanolic silver nitrate conditions.
304. Two equimolar bromoalkanes eventually form the same amount of \(\mathrm{AgBr}\), but sample P reaches its final precipitate amount much sooner than sample Q. On a graph of precipitate amount against time, this means that:
ⓐ. P must have a lower final plateau than Q
ⓑ. P rises faster initially; both curves reach the same plateau
ⓒ. Q forms a different-coloured precipitate because it reacts more slowly
ⓓ. both curves must be horizontal from the start
Correct Answer: P rises faster initially; both curves reach the same plateau
Explanation: Equal initial moles of compounds containing one bromine atom can ultimately produce equal moles of \(\mathrm{AgBr}\). Their final plateau amounts may therefore be the same. A faster-reacting haloalkane releases bromide more rapidly. Its graph rises more steeply and reaches the plateau earlier. Reaction rate changes the time profile but not the maximum stoichiometric amount when both reactions go to completion.
305. Use the graph description below. A reaction-coordinate diagram has reactants and products separated by a single energy maximum, with no minimum between them. This profile is most consistent with:
ⓐ. a two-step reaction containing a carbocation intermediate
ⓑ. a radical chain containing initiation and propagation stages
ⓒ. an elimination reaction that passes through two isolated intermediates
ⓓ. a concerted \(\mathrm{S_N2}\) reaction with one transition state
Correct Answer: a concerted \(\mathrm{S_N2}\) reaction with one transition state
Explanation: A local maximum on an energy profile represents a transition state. A local minimum between two maxima would represent an intermediate. The described graph has only one maximum and no intermediate minimum. This matches the one-step nature of an ideal \(\mathrm{S_N2}\) reaction. The single barrier includes both nucleophile–carbon bond formation and carbon–leaving-group bond breaking.
306. In the transition state of an \(\mathrm{S_N2}\) reaction involving \(\mathrm{Nu^-}\) and \(\mathrm{R-X}\):
ⓐ. the carbon is bonded fully to both \(\mathrm{Nu}\) and \(\mathrm{X}\) as a stable compound
ⓑ. the carbon–halogen bond has broken completely before the carbon–nucleophile bond begins forming
ⓒ. the nucleophile remains unbonded until a carbocation has formed
ⓓ. the carbon–nucleophile and carbon–halogen bonds are both partial
Correct Answer: the carbon–nucleophile and carbon–halogen bonds are both partial
Explanation: A transition state represents the highest-energy arrangement along the reaction pathway. In an \(\mathrm{S_N2}\) transition state, the incoming nucleophile has begun to bond to carbon. At the same time, the bond between carbon and the leaving group has not yet broken completely. Both bonds are therefore partial. This fleeting arrangement cannot be isolated as a stable molecule.
307. A secondary haloalkane gives an unrearranged substitution product under conditions known to favour direct backside attack. The absence of a rearranged carbon skeleton supports:
ⓐ. a concerted \(\mathrm{S_N2}\) pathway
ⓑ. a free-carbocation \(\mathrm{S_N1}\) pathway
ⓒ. a radical-chain substitution
ⓓ. an electrophilic aromatic substitution
Correct Answer: a concerted \(\mathrm{S_N2}\) pathway
Explanation: Direct backside attack does not require prior formation of a carbocation. Without a carbocation, hydride and alkyl shifts cannot occur through the usual rearrangement pathway. The nucleophile replaces the leaving group at the same carbon in one elementary event. An unrearranged product is therefore consistent with \(\mathrm{S_N2}\) conditions. Product structure alone may not prove a mechanism in every case, but it agrees with the stated direct-attack evidence.
308. The following table compares proposed features of an \(\mathrm{S_N2}\) reaction.
| Row | Feature | Proposed description |
|---|
| P | Number of elementary steps | One |
| Q | Intermediate | No free carbocation |
| R | Direction of attack | Opposite the leaving group |
| S | Bond changes | Formation and cleavage occur together |
The appropriate evaluation is:
ⓐ. only P and Q are correct
ⓑ. only Q, R, and S are correct
ⓒ. all four are correct
ⓓ. only P, R, and S are correct
Correct Answer: all four are correct
Explanation: Row P correctly identifies the mechanism as a single elementary step. Row Q correctly excludes a free carbocation intermediate. Row R describes backside attack, which occurs from the side opposite the leaving group. Row S reflects the concerted nature of bond formation and bond cleavage. Together, the four rows provide a consistent description of an ideal \(\mathrm{S_N2}\) pathway.
309. The ideal rate law for an \(\mathrm{S_N2}\) reaction between a haloalkane \(\mathrm{R-X}\) and a nucleophile \(\mathrm{Nu^-}\) is:
ⓐ. \(\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]\)
ⓑ. \(\mathrm{rate}=k[\mathrm{R-X}]\)
ⓒ. \(\mathrm{rate}=k[\mathrm{Nu^-}]^2\)
ⓓ. \(\mathrm{rate}=k[\mathrm{R-X}]^2[\mathrm{Nu^-}]\)
Correct Answer: \(\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]\)
Explanation: The rate-determining event is the single elementary collision involving both the haloalkane and the nucleophile. The rate is therefore first order in \(\mathrm{R-X}\) and first order in \(\mathrm{Nu^-}\). Adding the concentration exponents gives an overall order of two. Changing either reactant concentration changes the reaction rate. The bimolecular label refers to the two reacting species involved in the elementary substitution event.
310. For the rate law \(\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]\), the standard unit of \(k\) when concentration is measured in \(\mathrm{mol\,L^{-1}}\) and time in seconds is ______.
ⓐ. \(\mathrm{s^{-1}}\)
ⓑ. \(\mathrm{mol\,L^{-1}\,s^{-1}}\)
ⓒ. \(\mathrm{L\,mol^{-1}\,s^{-1}}\)
ⓓ. \(\mathrm{L^2\,mol^{-2}\,s^{-1}}\)
Correct Answer: \(\mathrm{L\,mol^{-1}\,s^{-1}}\)
Explanation: The rate has the unit \(\mathrm{mol\,L^{-1}\,s^{-1}}\). The concentration product has the unit:
\[
(\mathrm{mol\,L^{-1}})(\mathrm{mol\,L^{-1}})=\mathrm{mol^2\,L^{-2}}
\]
Therefore:
\[
k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol^2\,L^{-2}}}
\]
Cancelling powers gives:
\[
k=\mathrm{L\,mol^{-1}\,s^{-1}}
\]
This is the standard unit for an overall second-order rate constant. The unit \(\mathrm{s^{-1}}\) belongs to a first-order rate law such as the ideal \(\mathrm{S_N1}\) expression.
311. Use the graph description below. For an ideal \(\mathrm{S_N2}\) reaction, rate is plotted on the vertical axis against \([\mathrm{Nu^-}]\) on the horizontal axis while \([\mathrm{R-X}]\) and temperature are fixed. The graph is a straight line through the origin. Its slope is:
ⓐ. \(k[\mathrm{R-X}]\)
ⓑ. \(k[\mathrm{Nu^-}]\)
ⓒ. \(\frac{k}{[\mathrm{R-X}]}\)
ⓓ. \([\mathrm{R-X}][\mathrm{Nu^-}]\)
Correct Answer: \(k[\mathrm{R-X}]\)
Explanation: The rate law is:
\[
r=k[\mathrm{R-X}][\mathrm{Nu^-}]
\]
With \([\mathrm{R-X}]\) fixed, the quantities \(k\) and \([\mathrm{R-X}]\) are constants. The equation can be written as:
\[
r=\left(k[\mathrm{R-X}]\right)[\mathrm{Nu^-}]
\]
This has the linear form \(y=mx\). The vertical variable is rate, and the horizontal variable is nucleophile concentration. The slope is therefore \(k[\mathrm{R-X}]\), while the zero intercept reflects zero ideal rate when no nucleophile is present.
312. An ideal \(\mathrm{S_N2}\) reaction has an initial rate of \(2.4\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\) when \([\mathrm{R-X}]=0.20\,\mathrm{mol\,L^{-1}}\) and \([\mathrm{Nu^-}]=0.30\,\mathrm{mol\,L^{-1}}\). The rate constant is:
ⓐ. \(1.44\times10^{-5}\,\mathrm{L\,mol^{-1}\,s^{-1}}\)
ⓑ. \(8.0\times10^{-4}\,\mathrm{L\,mol^{-1}\,s^{-1}}\)
ⓒ. \(1.2\times10^{-3}\,\mathrm{L\,mol^{-1}\,s^{-1}}\)
ⓓ. \(4.0\times10^{-3}\,\mathrm{L\,mol^{-1}\,s^{-1}}\)
Correct Answer: \(4.0\times10^{-3}\,\mathrm{L\,mol^{-1}\,s^{-1}}\)
Explanation: \( \textbf{Rate law:} \)
\[
r=k[\mathrm{R-X}][\mathrm{Nu^-}]
\]
\( \textbf{Solve for the rate constant:} \)
\[
k=\frac{r}{[\mathrm{R-X}][\mathrm{Nu^-}]}
\]
\( \textbf{Insert the data:} \)
\[
k=\frac{2.4\times10^{-4}}{(0.20)(0.30)}
\]
\( \textbf{Concentration product:} \)
\[
(0.20)(0.30)=0.060
\]
\( \textbf{Calculation:} \)
\[
k=\frac{2.4\times10^{-4}}{0.060}=4.0\times10^{-3}
\]
\( \textbf{Unit:} \)
\[
k=4.0\times10^{-3}\,\mathrm{L\,mol^{-1}\,s^{-1}}
\]
The concentration product belongs in the denominator because both reactants appear in the rate law.
313. Initial-rate data for a substitution reaction are shown below.
| Experiment | \([\mathrm{R-X}]\) in \(\mathrm{mol\,L^{-1}}\) | \([\mathrm{Nu^-}]\) in \(\mathrm{mol\,L^{-1}}\) | Initial rate in \(\mathrm{mol\,L^{-1}\,s^{-1}}\) |
|---|
| P | \(0.10\) | \(0.10\) | \(2.0\times10^{-3}\) |
| Q | \(0.20\) | \(0.10\) | \(4.0\times10^{-3}\) |
| R | \(0.20\) | \(0.20\) | \(8.0\times10^{-3}\) |
The rate law supported by the data is:
ⓐ. \(\mathrm{rate}=k[\mathrm{R-X}]^2\)
ⓑ. \(\mathrm{rate}=k[\mathrm{Nu^-}]\)
ⓒ. \(\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]\)
ⓓ. \(\mathrm{rate}=k[\mathrm{R-X}]^2[\mathrm{Nu^-}]\)
Correct Answer: \(\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]\)
Explanation: \( \textbf{Compare P and Q:} \) The nucleophile concentration stays at \(0.10\,\mathrm{mol\,L^{-1}}\).
\[
[\mathrm{R-X}]\text{ doubles from }0.10\text{ to }0.20
\]
The rate also doubles from \(2.0\times10^{-3}\) to \(4.0\times10^{-3}\).
\( \textbf{Conclusion for haloalkane:} \) The reaction is first order in \(\mathrm{R-X}\).
\( \textbf{Compare Q and R:} \) The haloalkane concentration stays at \(0.20\,\mathrm{mol\,L^{-1}}\).
\[
[\mathrm{Nu^-}]\text{ doubles from }0.10\text{ to }0.20
\]
The rate doubles from \(4.0\times10^{-3}\) to \(8.0\times10^{-3}\).
\( \textbf{Conclusion for nucleophile:} \) The reaction is first order in \(\mathrm{Nu^-}\).
\( \textbf{Combined rate law:} \)
\[
\mathrm{rate}=k[\mathrm{R-X}][\mathrm{Nu^-}]
\]
The data therefore describe second-order kinetics consistent with an ideal \(\mathrm{S_N2}\) pathway.
314. A claim states: “Changing nucleophile concentration cannot alter the rate of an \(\mathrm{S_N2}\) reaction because the nucleophile attacks only after the slow step.” The claim is:
ⓐ. correct, because only the haloalkane appears in the rate law
ⓑ. incorrect; the nucleophile enters the rate-determining step
ⓒ. correct, because the nucleophile is merely a catalyst
ⓓ. incorrect, because the rate is independent of both reactants
Correct Answer: incorrect; the nucleophile enters the rate-determining step
Explanation: An ideal \(\mathrm{S_N2}\) mechanism has only one elementary step. The nucleophile is directly involved in that step together with the haloalkane. Its concentration therefore appears in the rate law. Increasing nucleophile concentration increases the frequency of productive nucleophile–substrate encounters. The claim incorrectly describes the stepwise timing associated with a mechanism in which nucleophile capture follows prior ionisation.
315. The term “bimolecular” in \(\mathrm{S_N2}\) refers to:
ⓐ. formation of two products from every substrate molecule
ⓑ. presence of two transition states in the energy profile
ⓒ. use of two different solvents in the reaction mixture
ⓓ. two reactants participate in the rate-determining event
Correct Answer: two reactants participate in the rate-determining event
Explanation: Molecularity counts the reacting species involved in an elementary step. The \(\mathrm{S_N2}\) substitution event involves one haloalkane molecule or ion and one nucleophile. Two reacting species are therefore present in the concerted elementary event. This gives the designation bimolecular nucleophilic substitution. It does not refer to the number of products, solvents, or energy maxima.
316. In an ideal \(\mathrm{S_N2}\) reaction, the haloalkane concentration is decreased by \(40\%\), while the nucleophile concentration is increased by \(50\%\). Temperature remains constant. The ratio of the new rate to the original rate is:
ⓐ. \(0.90\)
ⓑ. \(0.60\)
ⓒ. \(1.10\)
ⓓ. \(1.50\)
Correct Answer: \(0.90\)
Explanation: \( \textbf{Rate relation:} \)
\[
r=k[\mathrm{R-X}][\mathrm{Nu^-}]
\]
\( \textbf{Haloalkane change:} \) A decrease of \(40\%\) leaves \(60\%\) of the original concentration.
\[
[\mathrm{R-X}]_2=0.60[\mathrm{R-X}]_1
\]
\( \textbf{Nucleophile change:} \) An increase of \(50\%\) gives \(150\%\) of the original concentration.
\[
[\mathrm{Nu^-}]_2=1.50[\mathrm{Nu^-}]_1
\]
\( \textbf{Rate ratio:} \)
\[
\frac{r_2}{r_1}
=
\frac{k(0.60[\mathrm{R-X}]_1)(1.50[\mathrm{Nu^-}]_1)}
{k[\mathrm{R-X}]_1[\mathrm{Nu^-}]_1}
\]
\( \textbf{Cancellation:} \)
\[
\frac{r_2}{r_1}=(0.60)(1.50)
\]
\( \textbf{Calculation:} \)
\[
\frac{r_2}{r_1}=0.90
\]
The opposing concentration changes nearly compensate, leaving the new rate at \(90\%\) of the original value.
317. A tertiary haloalkane is generally a very poor substrate for an \(\mathrm{S_N2}\) reaction mainly because:
ⓐ. its carbon–halogen bond has no polarity
ⓑ. it cannot contain a leaving group
ⓒ. tertiary carbon atoms cannot form four covalent bonds
ⓓ. steric crowding blocks backside attack
Correct Answer: steric crowding blocks backside attack
Explanation: The nucleophile in an \(\mathrm{S_N2}\) reaction must approach along the direction opposite the carbon–halogen bond. In a tertiary haloalkane, three bulky alkyl groups surround the electrophilic carbon. These groups block close approach to the required backside region. The resulting steric repulsion greatly raises the activation barrier for substitution. The carbon–halogen bond is still polar, but polarity alone cannot overcome the severe geometric obstruction. Tertiary substrates therefore tend to follow other pathways under suitable conditions.
318. Three primary bromoalkanes are treated separately with the same concentration of a strong nucleophile in the same polar aprotic solvent.
Sample P: \(\mathrm{CH_3CH_2CH_2Br}\)
Sample Q: \(\mathrm{(CH_3)_2CHCH_2Br}\)
Sample R: \(\mathrm{(CH_3)_3CCH_2Br}\)
The most reasonable \(\mathrm{S_N2}\) rate order is:
ⓐ. \(\mathrm{R\gt Q\gt P}\)
ⓑ. \(\mathrm{P\gt Q\gt R}\)
ⓒ. \(\mathrm{Q\gt R\gt P}\)
ⓓ. \(\mathrm{P\gt R\gt Q}\)
Correct Answer: \(\mathrm{P\gt Q\gt R}\)
Explanation: All three compounds are primary because the carbon bearing bromine is attached to only one other carbon. Primary classification alone, however, does not make their steric environments identical. Sample P has the least branching near the reacting carbon and permits relatively easy backside attack. Sample Q has additional branching at the neighbouring carbon and is more hindered. Sample R is neopentyl bromide, whose bulky tert-butyl group strongly blocks access to the \(\mathrm{CH_2Br}\) carbon. The rate order therefore decreases as nearby branching increases.
319. Steric hindrance at four alkyl bromides is assessed for an ideal \(\mathrm{S_N2}\) reaction.
P. \(\mathrm{CH_3Br}\) — minimal hindrance
Q. \(\mathrm{CH_3CH_2Br}\) — low hindrance
R. \(\mathrm{(CH_3)_2CHBr}\) — moderate hindrance
S. \(\mathrm{(CH_3)_3CBr}\) — severe hindrance
Select the appropriate evaluation.
ⓐ. all four are reasonable
ⓑ. only P and Q are reasonable
ⓒ. only Q, R, and S are reasonable
ⓓ. only P, R, and S are reasonable
Correct Answer: all four are reasonable
Explanation: Methyl bromide offers essentially unrestricted access to the reacting carbon. Bromoethane is primary and has relatively low steric demand. \(2\)-Bromopropane is secondary, so two carbon groups partially obstruct the backside approach. Tert-butyl bromide has three alkyl groups surrounding the reaction centre and is extremely hindered. The entries therefore show the progressive decrease in \(\mathrm{S_N2}\) accessibility from methyl to tertiary substrate.
320. Under otherwise identical \(\mathrm{S_N2}\) conditions, increasing steric crowding near the carbon bearing the leaving group generally:
ⓐ. increases the rate because bulky groups attract the nucleophile
ⓑ. decreases the rate by obstructing backside attack
ⓒ. first decreases and then increases the rate without limit
ⓓ. has no effect because substrate structure is absent from the rate law
Correct Answer: decreases the rate by obstructing backside attack
Explanation: An \(\mathrm{S_N2}\) reaction requires the nucleophile to approach the electrophilic carbon from the side opposite the leaving group. Increasing steric crowding makes this approach more difficult. Greater repulsion must be overcome to reach the transition-state geometry, so the activation energy rises and the reaction rate falls. The effect is especially pronounced on moving from primary to secondary and tertiary substrates. Substrate structure therefore influences the rate constant even though the concentration term in the rate law is written simply as \([\mathrm{R-X}]\).
