401. Aniline, \(18.60\ \mathrm{g}\), is diazotised with \(90.0\%\) conversion. Sandmeyer cyanation converts \(80.0\%\) of the diazonium salt formed into benzonitrile. Hydrolysis converts \(75.0\%\) of that benzonitrile into benzoic acid. Calculate the benzoic acid mass and the volume of nitrogen evolved during the cyanation step at STP. Use \(M(\mathrm{aniline})=93.0\ \mathrm{g\,mol^{-1}}\), \(M(\mathrm{benzoic\ acid})=122\ \mathrm{g\,mol^{-1}}\), and \(22.4\ \mathrm{L\,mol^{-1}}\) at STP.
ⓐ. \(17.57\ \mathrm{g}\) and \(4.03\ \mathrm{L}\)
ⓑ. \(13.18\ \mathrm{g}\) and \(2.42\ \mathrm{L}\)
ⓒ. \(9.88\ \mathrm{g}\) and \(3.23\ \mathrm{L}\)
ⓓ. \(13.18\ \mathrm{g}\) and \(3.23\ \mathrm{L}\)
Correct Answer: \(13.18\ \mathrm{g}\) and \(3.23\ \mathrm{L}\)
Explanation: \( \textbf{Initial aniline amount:} \)
\[
n_0=\frac{18.60}{93.0}=0.200\ \mathrm{mol}
\]
Diazotisation conversion is \(90.0\%\):
\[
n(\mathrm{diazonium\ salt})=0.200\times0.900
\]
\[
n(\mathrm{diazonium\ salt})=0.180\ \mathrm{mol}
\]
Cyanation conversion is \(80.0\%\):
\[
n(\mathrm{benzonitrile})=0.180\times0.800
\]
\[
n(\mathrm{benzonitrile})=0.144\ \mathrm{mol}
\]
One mole of cyanated diazonium salt releases one mole of nitrogen.
\[
n(\mathrm{N_2})=0.144\ \mathrm{mol}
\]
\[
V(\mathrm{N_2})=0.144\times22.4
\]
\[
V(\mathrm{N_2})=3.23\ \mathrm{L}
\]
Hydrolysis conversion is \(75.0\%\):
\[
n(\mathrm{benzoic\ acid})=0.144\times0.750
\]
\[
n(\mathrm{benzoic\ acid})=0.108\ \mathrm{mol}
\]
\[
m=0.108\times122
\]
\[
m=13.176\ \mathrm{g}\approx13.18\ \mathrm{g}
\]
402. Gattermann bromination of benzenediazonium chloride is represented by:
ⓐ. copper powder with \(\mathrm{HBr}\), giving bromobenzene and \(\mathrm{N_2}\)
ⓑ. \(\mathrm{CuBr/HBr}\), classified specifically as the Gattermann reaction
ⓒ. \(\mathrm{KI}\), giving bromobenzene without nitrogen evolution
ⓓ. warm water, giving bromobenzene through hydrolysis
Correct Answer: copper powder with \(\mathrm{HBr}\), giving bromobenzene and \(\mathrm{N_2}\)
Explanation: In the Gattermann reaction, copper powder assists replacement of the arenediazonium group. Hydrobromic acid supplies the bromine introduced into the aromatic ring. The diazonium group departs as stable molecular nitrogen. Benzenediazonium chloride is therefore converted into bromobenzene with evolution of \(\mathrm{N_2}\). Cuprous bromide with hydrobromic acid instead identifies the Sandmeyer bromination conditions.
403. The accurately named reagent–reaction pair is:
ⓐ. Sandmeyer chlorination: copper powder with \(\mathrm{HCl}\); Gattermann chlorination: \(\mathrm{CuCl/HCl}\)
ⓑ. Sandmeyer bromination: copper powder with \(\mathrm{HBr}\); Gattermann bromination: \(\mathrm{CuBr/HBr}\)
ⓒ. Sandmeyer chlorination: \(\mathrm{CuCl/HCl}\); Gattermann chlorination: copper powder with \(\mathrm{HCl}\)
ⓓ. Sandmeyer chlorination: \(\mathrm{KI}\); Gattermann chlorination: \(\mathrm{H_3PO_2/H_2O}\)
Correct Answer: Sandmeyer chlorination: \(\mathrm{CuCl/HCl}\); Gattermann chlorination: copper powder with \(\mathrm{HCl}\)
Explanation: Sandmeyer chlorination uses cuprous chloride in hydrochloric acid. Gattermann chlorination uses copper powder with hydrochloric acid. The analogous bromination distinction is \(\mathrm{CuBr/HBr}\) for Sandmeyer and copper powder with \(\mathrm{HBr}\) for Gattermann. Potassium iodide is used for iodide replacement, while hypophosphorous acid replaces the diazonium group by hydrogen. The reagent identity, not merely the final halogen, determines the named reaction.
404. Why is a copper(I) catalyst generally unnecessary when an arenediazonium salt is converted into an iodoarene?
ⓐ. Iodide ion first converts the aromatic ring into an alkane
ⓑ. Iodide replaces the diazonium group directly under usual conditions
ⓒ. Copper(I) compounds react only with aliphatic diazonium ions
ⓓ. Molecular iodine is formed before any aromatic substitution can occur
Correct Answer: Iodide replaces the diazonium group directly under usual conditions
Explanation: Chlorine, bromine, and cyanide replacements are commonly carried out using copper(I) salts in Sandmeyer reactions. Iodide ion, however, can react directly with the arenediazonium ion. Loss of stable \(\mathrm{N_2}\) provides a strong driving force for the substitution. The aromatic carbon then becomes bonded to iodine. The absence of a copper catalyst is a useful reagent-level distinction between iodination and the usual Sandmeyer halogenations.
405. Assertion: Potassium iodide converts benzenediazonium chloride into iodobenzene with evolution of nitrogen gas.
Reason: The diazonium group is replaced by iodide ion, while its two nitrogen atoms leave as stable \(\mathrm{N_2}\).
ⓐ. 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 reaction replaces the complete \(\mathrm{-N_2^+}\) group by iodine. Iodide ion is supplied by potassium iodide. The nitrogen atoms do not remain in the aromatic product but depart together as molecular nitrogen. Formation and escape of stable \(\mathrm{N_2}\) strongly favour the reaction. The Reason therefore accounts for both iodobenzene formation and gas evolution.
406. The reagents that give iodobenzene, chlorobenzene, and bromobenzene from benzenediazonium salt, respectively, are:
ⓐ. \(\mathrm{KI}\); \(\mathrm{CuCl/HCl}\); \(\mathrm{CuBr/HBr}\)
ⓑ. \(\mathrm{CuCN}\); \(\mathrm{KI}\); \(\mathrm{H_3PO_2/H_2O}\)
ⓒ. heated \(\mathrm{BF_4^-}\) salt; \(\mathrm{CuBr/HBr}\); \(\mathrm{CuCl/HCl}\)
ⓓ. warm water; \(\mathrm{CuCl/HCl}\); \(\mathrm{KI}\)
Correct Answer: \(\mathrm{KI}\); \(\mathrm{CuCl/HCl}\); \(\mathrm{CuBr/HBr}\)
Explanation: Potassium iodide replaces the diazonium group by iodine and gives iodobenzene. Cuprous chloride with hydrochloric acid gives chlorobenzene by the Sandmeyer reaction. Cuprous bromide with hydrobromic acid similarly gives bromobenzene. Heating an arenediazonium tetrafluoroborate gives fluorobenzene, while warm water gives phenol. The three reagents in option A therefore match the requested products in order.
407. The Balz–Schiemann reaction is used to prepare:
ⓐ. phenol from an arenediazonium salt and warm water
ⓑ. iodobenzene from an arenediazonium chloride and potassium iodide
ⓒ. benzonitrile from an arenediazonium salt and cuprous cyanide
ⓓ. fluorobenzene from an arenediazonium tetrafluoroborate
Correct Answer: fluorobenzene from an arenediazonium tetrafluoroborate
Explanation: The Balz–Schiemann reaction is a characteristic method for preparing aryl fluorides. An arenediazonium salt is first converted into its tetrafluoroborate salt. On heating, the diazonium tetrafluoroborate decomposes and forms the aryl fluoride. Molecular nitrogen and boron trifluoride are released. The reaction is especially useful because direct introduction of fluorine into an aromatic ring is difficult to control.
408. Match each stage in Column I with its corresponding role or product in Column II.
| Column I | Column II |
|---|
| P. Diazotisation of aniline | 1. Formation of benzenediazonium tetrafluoroborate |
| Q. Addition of \(\mathrm{HBF_4}\) | 2. Formation of benzenediazonium chloride |
| R. Heating the tetrafluoroborate salt | 3. Formation of fluorobenzene |
| S. Gaseous by-products | 4. \(\mathrm{N_2}\) and \(\mathrm{BF_3}\) |
ⓐ. P-4, Q-2, R-1, S-3
ⓑ. P-1, Q-3, R-4, S-2
ⓒ. P-3, Q-4, R-2, S-1
ⓓ. P-2, Q-1, R-3, S-4
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: Diazotisation converts aniline into benzenediazonium chloride, so P matches \(2\). Fluoroboric acid supplies \(\mathrm{BF_4^-}\) and forms the tetrafluoroborate salt, matching Q with \(1\). Heating this salt produces fluorobenzene, so R matches \(3\). Molecular nitrogen and boron trifluoride are released during decomposition, matching S with \(4\). The matches reproduce the complete Balz–Schiemann sequence.
409. Aniline, \(18.60\ \mathrm{g}\), is diazotised with \(90.0\%\) conversion. Of the diazonium salt formed, \(80.0\%\) is converted into the tetrafluoroborate salt. On heating, \(75.0\%\) of that salt produces fluorobenzene, which is isolated in \(80.0\%\) yield. Calculate the isolated fluorobenzene mass and the volume of nitrogen formed during thermal decomposition at STP. Use \(M(\mathrm{aniline})=93.0\ \mathrm{g\,mol^{-1}}\), \(M(\mathrm{fluorobenzene})=96.0\ \mathrm{g\,mol^{-1}}\), and \(22.4\ \mathrm{L\,mol^{-1}}\) at STP.
ⓐ. \(8.29\ \mathrm{g}\) and \(1.94\ \mathrm{L}\)
ⓑ. \(10.37\ \mathrm{g}\) and \(1.94\ \mathrm{L}\)
ⓒ. \(8.29\ \mathrm{g}\) and \(2.42\ \mathrm{L}\)
ⓓ. \(12.96\ \mathrm{g}\) and \(2.42\ \mathrm{L}\)
Correct Answer: \(8.29\ \mathrm{g}\) and \(2.42\ \mathrm{L}\)
Explanation: The initial amount of aniline is
\[
n_0=\frac{18.60}{93.0}=0.200\ \mathrm{mol}
\]
Diazotisation gives
\[
n(\mathrm{diazonium\ salt})=0.200\times0.900
\]
\[
n(\mathrm{diazonium\ salt})=0.180\ \mathrm{mol}
\]
Tetrafluoroborate formation gives
\[
n(\mathrm{ArN_2^+BF_4^-})=0.180\times0.800
\]
\[
n(\mathrm{ArN_2^+BF_4^-})=0.144\ \mathrm{mol}
\]
Only \(75.0\%\) undergoes the desired thermal decomposition:
\[
n(\mathrm{fluorobenzene\ formed})=0.144\times0.750
\]
\[
n(\mathrm{fluorobenzene\ formed})=0.108\ \mathrm{mol}
\]
The isolated amount is
\[
n_{\mathrm{isolated}}=0.108\times0.800
\]
\[
n_{\mathrm{isolated}}=0.0864\ \mathrm{mol}
\]
\[
m(\mathrm{fluorobenzene})=0.0864\times96.0
\]
\[
m(\mathrm{fluorobenzene})=8.29\ \mathrm{g}
\]
Nitrogen is formed during thermal decomposition before isolation:
\[
n(\mathrm{N_2})=0.108\ \mathrm{mol}
\]
\[
V(\mathrm{N_2})=0.108\times22.4
\]
\[
V(\mathrm{N_2})=2.42\ \mathrm{L}
\]
410. Use the following passage.
A student prepares benzenediazonium chloride at \(273\text{–}278\ \mathrm{K}\). One portion is kept in an ice bath, while a second portion is gradually warmed in water. The cold portion shows little immediate change, but the warm portion releases gas and gives an aromatic compound that reacts with aqueous sodium hydroxide.
The aromatic product in the warm portion is:
ⓐ. benzene
ⓑ. chlorobenzene
ⓒ. aniline
ⓓ. phenol
Correct Answer: phenol
Explanation: The cold portion retains the diazonium salt because low temperature slows its decomposition. Warming the second portion causes the diazonium group to leave as molecular nitrogen. Water replaces it by hydroxyl. The product is phenol, which reacts with aqueous sodium hydroxide because it is weakly acidic. Chlorobenzene and benzene would not show the same acid–base behaviour with sodium hydroxide under ordinary conditions.
411. A graph plots moles of phenol formed against moles of benzenediazonium chloride warmed in excess water. The hydrolysis is quantitative. The graph shows:
ⓐ. The curve rises with slope \(2\) because two nitrogen atoms leave
ⓑ. The curve rises with slope \(1\) from the origin
ⓒ. The curve remains at zero because water cannot react with a diazonium salt
ⓓ. The curve falls as more diazonium salt is warmed
Correct Answer: The curve rises with slope \(1\) from the origin
Explanation: The hydrolysis equation shows a \(1:1\) relationship:
\[
\mathrm{ArN_2^++H_2O\rightarrow ArOH+N_2+H^+}
\]
One mole of diazonium salt gives one mole of phenol.
Therefore,
\[
n(\mathrm{phenol})=n(\mathrm{diazonium\ salt\ hydrolysed})
\]
The graph passes through the origin.
Its slope is
\[
\frac{\Delta n(\mathrm{phenol})}{\Delta n(\mathrm{diazonium\ salt})}=1
\]
The presence of two nitrogen atoms affects the identity of the gaseous product, not the phenol mole ratio.
412. A cold aqueous solution contains (14.05\ \mathrm{g}) of benzenediazonium chloride. On warming with excess water, (80.0%) of the diazonium salt is converted into phenol, and the phenol is isolated in (75.0%) yield. Calculate the isolated phenol mass and the volume of nitrogen formed at STP. Use (M(\mathrm{C_6H_5N_2Cl})=140.5\ \mathrm{g,mol^{-1}}), (M(\mathrm{phenol})=94.0\ \mathrm{g,mol^{-1}}), and (22.4\ \mathrm{L,mol^{-1}}) at STP.
ⓐ. (7.52\ \mathrm{g}) and (1.34\ \mathrm{L})
ⓑ. (5.64\ \mathrm{g}) and (1.34\ \mathrm{L})
ⓒ. (5.64\ \mathrm{g}) and (1.79\ \mathrm{L})
ⓓ. (7.52\ \mathrm{g}) and (1.79\ \mathrm{L})
Correct Answer: (5.64\ \mathrm{g}) and (1.79\ \mathrm{L})
Explanation: The initial amount of benzenediazonium chloride present in the cold solution is
[
n_0=\frac{14.05}{140.5}
]
[
n_0=0.100\ \mathrm{mol}
]
Only (80.0%) undergoes hydrolysis:
[
n_{\mathrm{reacted}}=0.100\times0.800
]
[
n_{\mathrm{reacted}}=0.0800\ \mathrm{mol}
]
The hydrolysis relation is
[
\mathrm{C_6H_5N_2^+ + H_2O\longrightarrow C_6H_5OH+N_2+H^+}
]
One mole of reacted diazonium salt therefore produces one mole of phenol and one mole of nitrogen:
[
n(\mathrm{phenol\ formed})=n(\mathrm{N_2})=0.0800\ \mathrm{mol}
]
Only (75.0%) of the phenol formed is isolated:
[
n(\mathrm{phenol\ isolated})=0.0800\times0.750
]
[
n(\mathrm{phenol\ isolated})=0.0600\ \mathrm{mol}
]
Its mass is
[
m(\mathrm{phenol})=0.0600\times94.0
]
[
m(\mathrm{phenol})=5.64\ \mathrm{g}
]
Nitrogen is formed during the chemical conversion, so its amount is based on the (0.0800\ \mathrm{mol}) that reacted:
[
V(\mathrm{N_2})=0.0800\times22.4
]
[
V(\mathrm{N_2})=1.79\ \mathrm{L}
]
Loss of phenol during isolation does not alter the amount of nitrogen already released.
413. Four stages are listed below.
P: Dissolve aniline in hydrochloric acid and cool it.
Q: Add sodium nitrite solution at \(273\text{–}278\ \mathrm{K}\).
R: Warm the resulting aqueous diazonium solution.
S: Isolate phenol after nitrogen evolution.
The conversion order from aniline to phenol is:
ⓐ. R \(\rightarrow\) P \(\rightarrow\) S \(\rightarrow\) Q
ⓑ. P \(\rightarrow\) Q \(\rightarrow\) R \(\rightarrow\) S
ⓒ. Q \(\rightarrow\) S \(\rightarrow\) P \(\rightarrow\) R
ⓓ. S \(\rightarrow\) R \(\rightarrow\) Q \(\rightarrow\) P
Correct Answer: P \(\rightarrow\) Q \(\rightarrow\) R \(\rightarrow\) S
Explanation: Aniline is first converted into its soluble acidic form and cooled. Sodium nitrite then generates nitrous acid in the reaction mixture and produces the diazonium salt. The cold condition must be maintained during diazotisation. Warming is introduced only after diazonium formation is complete. Hydrolysis then gives phenol with evolution of nitrogen gas.
414. Why is diazonium-group replacement by hydrogen synthetically useful?
ⓐ. It reduces the complete benzene ring to cyclohexane
ⓑ. It permits temporary use and removal of an amino director
ⓒ. It removes every substituent from the aromatic ring at the same time
ⓓ. It adds an extra carbon atom to the ring
Correct Answer: It permits temporary use and removal of an amino director
Explanation: The amino group strongly activates an aromatic ring and directs incoming electrophiles mainly to ortho and para positions. This directing power can be used to install other substituents at selected positions. The amino group can then be converted into a diazonium group. Reduction with hypophosphorous acid replaces that diazonium group by hydrogen. The amino group therefore serves as a temporary synthetic directing group rather than remaining in the final product.
415. Assertion: Treatment of benzenediazonium chloride with hypophosphorous acid gives benzene.
Reason: Hypophosphorous acid reduces the diazonium group and replaces it by hydrogen without reducing the aromatic ring.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The reduction affects the diazonium substituent rather than the benzene \(\pi\)-system. The \(\mathrm{-N_2^+}\) group departs as stable molecular nitrogen. Hydrogen occupies the aromatic carbon that previously carried the diazonium group. The product from benzenediazonium chloride is therefore benzene. The Reason correctly distinguishes deamination from hydrogenation of the aromatic ring.
416. The table compares several diazonium transformations.
| Row | Condition applied to benzenediazonium salt | Principal organic product |
|---|
| P | \(\mathrm{H_3PO_2/H_2O}\) | Benzene |
| Q | Ethanol | Benzene with oxidation of ethanol |
| R | Warm water | Phenol |
| S | \(\mathrm{H_3PO_2/H_2O}\) | Cyclohexane |
The valid rows are:
ⓐ. P and S only
ⓑ. Q and S only
ⓒ. P, Q and R only
ⓓ. P, R and S only
Correct Answer: P, Q and R only
Explanation: Hypophosphorous acid replaces the diazonium group by hydrogen and gives benzene. Ethanol can produce the same parent aromatic hydrocarbon while being oxidised, commonly to ethanal. Warm water replaces the diazonium group by hydroxyl and gives phenol. Hypophosphorous acid does not hydrogenate the benzene ring to cyclohexane. Row S is therefore incorrect.
417. Use the following passage.
Aniline is treated with excess bromine water to form \(2,4,6\)-tribromoaniline. The product is diazotised at low temperature, and the resulting diazonium salt is treated with hypophosphorous acid.
The final aromatic product is:
ⓐ. bromobenzene
ⓑ. \(2,4,6\)-tribromophenol
ⓒ. \(1,2,3\)-tribromobenzene
ⓓ. \(1,3,5\)-tribromobenzene
Correct Answer: \(1,3,5\)-tribromobenzene
Explanation: The amino group of aniline directs bromination to both ortho positions and the para position. This gives \(2,4,6\)-tribromoaniline. Diazotisation changes only the amino group into a diazonium group and leaves the bromine positions unchanged. Hypophosphorous acid then replaces the diazonium group by hydrogen. After removal of the group at carbon \(1\), the three bromine substituents occupy alternating ring positions, giving \(1,3,5\)-tribromobenzene.
418. Replacement of the diazonium group by a nitro group can be achieved at recognition level by treating a suitable arenediazonium species with:
ⓐ. potassium iodide under the usual aqueous substitution conditions
ⓑ. copper-assisted substitution by nitrite
ⓒ. hypophosphorous acid
ⓓ. warm dilute alkali only
Correct Answer: copper-assisted substitution by nitrite
Explanation: A diazonium group can be replaced by several different substituents. Under suitable copper-assisted conditions, nitrite introduces the \(\mathrm{-NO_2}\) group and forms a nitroarene. Molecular nitrogen is released as the diazonium group leaves. Potassium iodide gives an iodoarene, while hypophosphorous acid replaces the group by hydrogen. Warm water gives phenol rather than a nitro compound.
419. Assertion: Conversion of an aromatic amino group into a diazonium group can help prepare a nitroarene whose substitution pattern is difficult to obtain by direct nitration.
Reason: The diazonium group can be replaced by \(\mathrm{-NO_2}\) without changing the positions of other ring substituents.
ⓐ. 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: Diazotisation changes the amino substituent into a diazonium group at the same aromatic carbon. Other substituents already present on the ring normally retain their positions. A suitable nitro-replacement reaction then exchanges \(\mathrm{-N_2^+}\) for \(\mathrm{-NO_2}\). This permits the amino group to act as a temporary synthetic handle. The Reason directly explains the positional advantage stated in the Assertion.
420. Compare the diazonium conversions below.
| Row | Diazonium replacement | Carbon-count consequence |
|---|
| P | \(\mathrm{-N_2^+}\) replaced by \(\mathrm{-CN}\) | One carbon added |
| Q | \(\mathrm{-N_2^+}\) replaced by \(\mathrm{-NO_2}\) | Carbon count unchanged |
| R | \(\mathrm{-N_2^+}\) replaced by \(\mathrm{-H}\) | One carbon removed |
| S | \(\mathrm{-N_2^+}\) replaced by \(\mathrm{-OH}\) | Aromatic ring destroyed |
The valid rows are:
ⓐ. P and Q only
ⓑ. P and R only
ⓒ. Q and S only
ⓓ. P, Q and R only
Correct Answer: P and Q only
Explanation: Cyanide contributes its own carbon atom, so aryl nitrile formation increases the organic carbon count by one. A nitro group contains no carbon, so replacement by \(\mathrm{-NO_2}\) leaves the carbon count unchanged. Replacement by hydrogen does not remove a ring carbon; it removes only the diazonium substituent. Hydroxyl replacement also preserves the aromatic six-carbon framework. Only P and Q correctly describe the carbon-count consequences.
