401. A reaction record describes the behaviour of \(2\)-bromopropane under two conditions.
Case P: dilute aqueous hydroxide at lower temperature gives mainly propan-\(2\)-ol.
Case Q: concentrated alcoholic hydroxide with heating gives mainly propene.
The best interpretation is:
ⓐ. the substrate structure changes from secondary to primary in Case Q
ⓑ. bromide acts as a nucleophile in P and as an acid in Q
ⓒ. both cases follow exactly the same mechanism and differ only in product name
ⓓ. aqueous medium favours substitution; hot alcoholic base favours elimination
Correct Answer: aqueous medium favours substitution; hot alcoholic base favours elimination
Explanation: The same secondary haloalkane can follow competing reaction pathways. In aqueous medium, hydroxide acts effectively as a nucleophile and replaces bromide to form the alcohol. In concentrated alcoholic medium with heat, its basic character becomes more important and beta-hydrogen removal gives propene. The carbon skeleton remains secondary in both vessels; only the reaction environment changes. This comparison demonstrates why a reagent formula such as \(\mathrm{KOH}\) is incomplete without its solvent and temperature.
402. A \(0.40\,mol\) sample of \(2\)-bromobutane is heated with alcoholic alkali. Suppose \(75\%\) of the sample undergoes elimination, and \(80\%\) of the resulting alkene mixture is but-\(2\)-ene. If the molar mass of but-\(2\)-ene is \(56.0\,g\,mol^{-1}\), the mass of but-\(2\)-ene formed is:
ⓐ. \(13.44\,g\)
ⓑ. \(16.80\,g\)
ⓒ. \(17.92\,g\)
ⓓ. \(22.40\,g\)
Correct Answer: \(13.44\,g\)
Explanation: \( \textbf{Initial haloalkane amount:} \)
\[
n_0=0.40\,mol
\]
\( \textbf{Fraction undergoing elimination:} \)
\[
75\%=0.75
\]
\( \textbf{Total alkene amount:} \)
\[
n_{\text{alkenes}}=(0.40)(0.75)=0.30\,mol
\]
\( \textbf{Fraction that is but-}2\textbf{-ene:} \)
\[
80\%=0.80
\]
\( \textbf{But-}2\textbf{-ene amount:} \)
\[
n_{\text{but-2-ene}}=(0.30)(0.80)=0.24\,mol
\]
\( \textbf{Mass relation:} \)
\[
m=nM
\]
\( \textbf{Substitution:} \)
\[
m=(0.24)(56.0)
\]
\( \textbf{Calculation:} \)
\[
m=13.44\,g
\]
The theoretical mass from complete conversion to but-\(2\)-ene would be larger because both the incomplete elimination and product distribution must be included.
403. \(2\)-Bromobutane is subjected to the two reaction conditions below.
Case P: \(\mathrm{NaCN}\) in a polar aprotic solvent at moderate temperature
Case Q: concentrated sodium ethoxide in ethanol with heating
The most reasonable dominant pathways are:
ⓐ. P mainly \(\mathrm{E1}\); Q mainly \(\mathrm{S_N1}\)
ⓑ. P mainly \(\mathrm{S_N2}\); Q mainly \(\mathrm{E2}\)
ⓒ. P mainly radical substitution; Q mainly electrophilic addition
ⓓ. P mainly \(\mathrm{E2}\); Q mainly \(\mathrm{S_N2}\)
Correct Answer: P mainly \(\mathrm{S_N2}\); Q mainly \(\mathrm{E2}\)
Explanation: Cyanide is a strong, relatively compact nucleophile, and a polar aprotic solvent supports its direct attack on the secondary carbon. Although secondary substrates can show competition, the moderate temperature and nucleophilic reagent make \(\mathrm{S_N2}\) a reasonable dominant assignment for Case P. Sodium ethoxide is a strong base, and heating in alcoholic medium strongly promotes beta-hydrogen removal in Case Q. The elimination produces mainly the more substituted alkene under ordinary ethoxide conditions. The comparison shows how changing reagent role and temperature can redirect the same substrate from substitution toward elimination.
404. The general equation for the Wurtz reaction is:
ⓐ. \(\mathrm{R-X+Mg\xrightarrow{dry\ ether}R-MgX}\)
ⓑ. \(\mathrm{2R-X+2Na\xrightarrow{dry\ ether}R-R+2NaX}\)
ⓒ. \(\mathrm{R-X+KOH_{(alc)}\xrightarrow{\Delta}alkene+KX+H_2O}\)
ⓓ. \(\mathrm{R-X+2[H]\rightarrow R-H+HX}\)
Correct Answer: \(\mathrm{2R-X+2Na\xrightarrow{dry\ ether}R-R+2NaX}\)
Explanation: The Wurtz reaction couples two alkyl groups derived from alkyl halide molecules. Sodium metal is used in dry ether, and sodium halide is formed as the inorganic product. The new bond formed in the organic product is a carbon–carbon bond. The product \(\mathrm{R-R}\) is an alkane containing the combined carbon skeletons of the two alkyl groups. Magnesium would form a Grignard reagent, while alcoholic alkali would favour elimination.
405. An alkyl bromide containing four carbon atoms undergoes Wurtz coupling with itself. The coupled alkane contains:
ⓐ. eight carbons
ⓑ. six carbon atoms
ⓒ. four carbon atoms
ⓓ. ten carbon atoms
Correct Answer: eight carbons
Explanation: Wurtz coupling combines two identical alkyl groups without removing carbon atoms from either group. Each starting alkyl bromide contributes a four-carbon skeleton. The product carbon count is therefore:
\[
4+4=8
\]
The reaction doubles the carbon number when a single alkyl halide is coupled with itself. The exact structure may be straight or branched depending on the starting alkyl group.
406. Assertion: The Wurtz reaction is most suitable for preparing symmetrical alkanes.
Reason: Using two different alkyl halides can produce two symmetrical coupling products and one cross-coupled product.
ⓐ. 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: A single alkyl halide gives only one kind of alkyl group for coupling. The resulting alkane therefore has identical groups on the two sides of the newly formed carbon–carbon bond. With two different alkyl halides, each group can couple with itself or with the other group. Several alkanes can consequently be produced. The Reason directly explains why the preparation of a pure unsymmetrical alkane is a major limitation of the Wurtz method.
407. Review the following Wurtz-reaction product assignments.
| Alkyl halide | Coupled product |
|---|
| P. \(\mathrm{CH_3Br}\) | Ethane |
| Q. \(\mathrm{C_2H_5Br}\) | Butane |
| R. \(\mathrm{CH_3CH_2CH_2Br}\) | Hexane |
The appropriate evaluation is:
ⓐ. only Q and R are correct
ⓑ. only P and Q are correct
ⓒ. all three are correct
ⓓ. only P and R are correct
Correct Answer: all three are correct
Explanation: Two methyl groups couple to give the two-carbon alkane ethane. Two ethyl groups produce the four-carbon alkane butane. Two normal propyl groups produce the six-carbon straight-chain alkane hexane. In each row, the carbon number of the product is twice that of the alkyl group in the starting halide. The table applies the same carbon-skeleton rule to three different homologues.
408. A laboratory preparation produces \(0.25\,mol\) of butane by Wurtz coupling of bromoethane. If sodium is used in the exact stoichiometric amount and its molar mass is \(23.0\,g\,mol^{-1}\), the required mass of sodium is:
ⓐ. \(5.75\,g\)
ⓑ. \(23.0\,g\)
ⓒ. \(2.88\,g\)
ⓓ. \(11.5\,g\)
Correct Answer: \(11.5\,g\)
Explanation: \( \textbf{Balanced Wurtz equation:} \)
\[
\mathrm{2C_2H_5Br+2Na\rightarrow C_4H_{10}+2NaBr}
\]
\( \textbf{Mole relation:} \)
\[
1\,mol\ \mathrm{C_4H_{10}}\text{ requires }2\,mol\ \mathrm{Na}
\]
\( \textbf{Butane amount:} \)
\[
n(\mathrm{C_4H_{10}})=0.25\,mol
\]
\( \textbf{Sodium amount required:} \)
\[
n(\mathrm{Na})=(0.25)(2)=0.50\,mol
\]
\( \textbf{Mass relation:} \)
\[
m=nM
\]
\( \textbf{Substitution:} \)
\[
m=(0.50)(23.0)
\]
\( \textbf{Calculation:} \)
\[
m=11.5\,g
\]
The factor of two must be retained because two moles of sodium are consumed for every mole of coupled alkane.
409. Which alkyl halide can give normal hexane as the single symmetrical hydrocarbon product in a Wurtz reaction?
ⓐ. \(2\)-bromopropane
ⓑ. \(1\)-bromopropane
ⓒ. \(1\)-bromobutane
ⓓ. bromoethane
Correct Answer: \(1\)-bromopropane
Explanation: Normal hexane can be divided symmetrically at its central carbon–carbon bond into two normal propyl groups. The required starting material is therefore a normal propyl halide such as \(1\)-bromopropane. Coupling two isopropyl groups from \(2\)-bromopropane would form \(2,3\)-dimethylbutane. Bromoethane gives butane, while \(1\)-bromobutane gives octane. Carbon count alone is insufficient unless the branching pattern is also preserved.
410. An aryl halide and an alkyl halide are heated with sodium in dry ether to form an alkyl-substituted aromatic hydrocarbon. This coupling is called:
ⓐ. the Wurtz reaction
ⓑ. the Fittig reaction
ⓒ. the Wurtz–Fittig reaction
ⓓ. Grignard-reagent formation
Correct Answer: the Wurtz–Fittig reaction
Explanation: The Wurtz–Fittig reaction couples one aryl group with one alkyl group using sodium in dry ether. The aryl halide supplies the aromatic fragment, while the alkyl halide supplies the side chain. Wurtz coupling uses two alkyl halides and usually gives an alkane. Fittig coupling uses two aryl halides and gives a biaryl compound. Grignard-reagent formation instead uses magnesium and produces an organomagnesium halide rather than the coupled hydrocarbon.
411. A learner claims that the Fittig reaction is the standard sodium-coupling method for converting chlorobenzene and chloromethane into toluene. The claim is:
ⓐ. correct, because toluene is a biaryl compound
ⓑ. correct, because every sodium coupling involving an aryl halide is called Fittig
ⓒ. incorrect, because chlorobenzene cannot participate in any sodium coupling
ⓓ. incorrect; this is Wurtz–Fittig coupling
Correct Answer: incorrect; this is Wurtz–Fittig coupling
Explanation: The Fittig reaction specifically involves coupling of two aryl halide molecules. Chlorobenzene and chloromethane provide one aryl group and one alkyl group. Their cross-coupling product is toluene, an alkylbenzene rather than a biaryl compound. This combination is classified as the Wurtz-Fittig reaction. The presence of an aryl halide alone is not sufficient to assign the name Fittig.
412. Two dry-ether reaction vessels contain sodium metal.
Vessel P contains bromobenzene only.
Vessel Q contains bromobenzene and bromoethane.
The expected principal coupling products are:
ⓐ. P gives biphenyl, while Q gives ethylbenzene
ⓑ. P gives ethylbenzene, while Q gives biphenyl
ⓒ. P gives benzene, while Q gives butane
ⓓ. P gives toluene, while Q gives benzene
Correct Answer: P gives biphenyl, while Q gives ethylbenzene
Explanation: Vessel P contains only an aryl halide, so two phenyl groups couple by the Fittig reaction. Its product is biphenyl. Vessel Q contains an aryl halide and an ethyl halide, permitting Wurtz-Fittig formation of a phenyl–ethyl bond. The intended product is ethylbenzene. The product identity follows the carbon groups supplied by the reacting halides rather than the identity of sodium or ether.
413. An aryl halide containing six carbon atoms is coupled by the Wurtz-Fittig reaction with a three-carbon alkyl halide. The intended alkylbenzene product contains:
ⓐ. six carbon atoms
ⓑ. twelve carbon atoms
ⓒ. nine carbon atoms
ⓓ. three carbon atoms
Correct Answer: nine carbon atoms
Explanation: The aryl halide contributes a six-carbon aromatic ring. The alkyl halide contributes its complete three-carbon alkyl group. Wurtz-Fittig coupling forms a new bond between these two groups without removing carbon atoms. The total number of carbon atoms is:
\[
6+3=9
\]
The product is a propyl-substituted benzene skeleton, although its precise branching depends on the structure of the three-carbon alkyl halide.
414. Which equation correctly represents the formation of a Grignard reagent from an alkyl halide?
ⓐ. \(\mathrm{R-X+2Na\xrightarrow{dry\ ether}R-R+2NaX}\)
ⓑ. \(\mathrm{R-X+Mg\xrightarrow{dry\ ether}R-MgX}\)
ⓒ. \(\mathrm{R-X+KOH_{(aq)}\rightarrow R-OH+KX}\)
ⓓ. \(\mathrm{R-X+H_2\rightarrow R-H+HX}\)
Correct Answer: \(\mathrm{R-X+Mg\xrightarrow{dry\ ether}R-MgX}\)
Explanation: A Grignard reagent is an organomagnesium halide with the general formula \(\mathrm{R-MgX}\). It is prepared by allowing an alkyl halide to react with magnesium metal in dry ether. Magnesium becomes inserted between the carbon group and the halogen. The reaction must be protected from moisture because the product reacts rapidly with compounds containing acidic hydrogen. Sodium in dry ether would instead be associated with Wurtz-type coupling.
415. Assertion: All apparatus used for preparing a Grignard reagent should be thoroughly dry.
Reason: Water protonates the carbon bonded to magnesium and converts the Grignard reagent into a hydrocarbon.
ⓐ. 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 carbon–magnesium bond gives carbon strong carbanion-like character. This carbon readily removes a proton from water. As a result, \(\mathrm{R-MgX}\) is converted into the hydrocarbon \(\mathrm{R-H}\), and the useful organomagnesium reagent is consumed. Moist glassware or wet solvent can therefore reduce or completely prevent successful preparation. The Reason directly explains why strictly dry conditions are required.
416. Assess the following features of Grignard-reagent formation.
| Row | Feature | Description |
|---|
| P | Organic halide | Provides the carbon group and halogen |
| Q | Magnesium | Becomes inserted into the carbon–halogen linkage |
| R | Dry ether | Provides a non-protic coordinating medium |
| S | Moisture | Destroys the organomagnesium reagent |
The appropriate evaluation is:
ⓐ. only P and Q are correct
ⓑ. only Q, R, and S are correct
ⓒ. P, Q, R, and S are correct
ⓓ. only P, R, and S are correct
Correct Answer: P, Q, R, and S are correct
Explanation: The organic halide supplies both the carbon skeleton and the halogen represented by \(\mathrm{X}\). Magnesium becomes bonded to carbon and halogen in the product \(\mathrm{R-MgX}\). Dry ether stabilises the reagent without donating a proton. Moisture reacts with the strongly basic carbon centre and forms a hydrocarbon. Every row therefore describes a necessary component or consequence of the preparation.
417. A student states, “The carbon atom of \(\mathrm{R-MgX}\) is electrophilic because it is bonded to a metal.” The best correction is:
ⓐ. the carbon is neutral and never participates in bond formation
ⓑ. the carbon behaves as a radical because magnesium removes one electron
ⓒ. the carbon is positively charged and accepts electron pairs
ⓓ. the \(\mathrm{C-Mg}\) bond polarises carbon nucleophilically
Correct Answer: the \(\mathrm{C-Mg}\) bond polarises carbon nucleophilically
Explanation: Magnesium is much more electropositive than carbon. The shared electrons of the carbon–magnesium bond lie closer to carbon. This gives carbon partial negative charge and carbanion-like reactivity. It can donate electron density to an electrophilic centre and form a new carbon–carbon bond. The presence of a metal does not automatically make the attached carbon electron deficient.
418. A student attempts to prepare ethylmagnesium bromide using freshly distilled bromoethane but a flask that was rinsed with water and not dried. The most likely outcome is:
ⓐ. much of the Grignard reagent is converted into ethane
ⓑ. ethylmagnesium bromide is converted selectively into butane
ⓒ. bromoethane forms ethanol without consuming magnesium
ⓓ. water acts only as an inert solvent
Correct Answer: much of the Grignard reagent is converted into ethane
Explanation: Any ethylmagnesium bromide formed encounters water remaining on the flask surface. Its carbanion-like ethyl carbon removes a proton from water. The organic product of this protonation is ethane, \(\mathrm{C_2H_6}\). The useful Grignard reagent is therefore consumed as quickly as it forms. A reaction vessel can appear clean yet still be unsuitable if it contains traces of moisture.
419. Methylmagnesium bromide reacts with water to form:
ⓐ. methanol
ⓑ. methane
ⓒ. ethane
ⓓ. methanoic acid
Correct Answer: methane
Explanation: The methyl carbon of \(\mathrm{CH_3MgBr}\) has strong carbanion-like character. It accepts a proton from water. Protonation converts the methyl group into \(\mathrm{CH_4}\). The organic product is therefore methane rather than methanol. The reaction demonstrates why Grignard reagents cannot be handled in aqueous media.
420. A reaction mixture intended to contain propylmagnesium bromide is accidentally exposed to moist air. Analysis later detects propane and very little organomagnesium compound. The observation is best explained by:
ⓐ. moisture protonated the propyl group
ⓑ. Wurtz coupling of two propyl groups by oxygen
ⓒ. oxidation of propane into the Grignard reagent
ⓓ. substitution of bromide by atmospheric nitrogen
Correct Answer: moisture protonated the propyl group
Explanation: Moist air supplies small amounts of water vapour. Propylmagnesium bromide reacts rapidly with this moisture because its propyl carbon is strongly basic. Proton transfer forms propane and destroys the organomagnesium reagent. Oxygen or nitrogen is not required to explain the main product. The observation is therefore evidence of inadequate exclusion of moisture during handling.
