CBSE Class 12 Chemistry Coordination Compound Notes and Questions

UNIT 9: CO-ORDINATION COMPOUNDS

UNIT 9: CO-ORDINATION COMPOUNDS

Question 1. What are ambidentate ligands? Give two examples for each.
Answer: Ambidentate ligands are ligands that can attach themselves to the central metal atom through two different atoms.
For example:
(a) Nitro group (\(\text{M} \leftarrow \text{N}\) with donor atom N) and Nitrito group (\(\text{M} \leftarrow \text{O} - \text{N} = \text{O}\) with donor atom O).
(b) Thiocyanato group (\(\text{M} - \text{SCN}\) with donor atom S) and Isothiocyanato group (\(\text{M} - \text{NCS}\) with donor atom N).
In simple words: Ambidentate ligands have two different donor sites and can coordinate through either of them.

Exam Tip: Memorize both pairs and their respective donor atoms as this is a frequent direct question in board exams.

Question 2. Using IUPAC norms write the formula for the following: Tetrahydroxozincate(II)
Answer: \([\text{Zn(OH)}_4]^{2-}\)
In simple words: Zinc is the central metal with an oxidation state of +2, surrounded by four hydroxide ions carrying a total charge of -2.

Exam Tip: Always place the anionic complex inside square brackets with the overall charge indicated outside as a superscript.

Question 3. Using IUPAC norms write the formula for the following: Hexaamminecobalt(III) sulphate
Answer: \([\text{Co(NH}_3)_6]_2(\text{SO}_4)_3\)
In simple words: Cobalt(III) has a +3 charge and ammine ligands are neutral. Three sulphate ions (-2 each) balance two complex cations (+3 each).

Exam Tip: Balance the total positive charge of the cation with the total negative charge of the anion to write correct ionic formulas.

Question 4. Using IUPAC norms write the formula for the following: Pentaamminenitrito-O-cobalt(III)
Answer: \([\text{Co(ONO)(NH}_3)_5]^{2+}\)
In simple words: Cobalt is bonded to five neutral ammine molecules and one nitrito-O ligand, giving an overall +2 charge.

Exam Tip: The suffix '-O' in nitrito-O specifies that oxygen is the donor atom linked to the metal.

Question 5. Using IUPAC norms write the systematic name of the following: \([\text{Co(NH}_3)_6]\text{Cl}_3\)
Answer: Hexaamminecobalt(III) chloride
In simple words: Name the ligands alphabetically followed by the metal name with its Roman numeral oxidation state, and end with the counter anion name.

Question 6. Using IUPAC norms write the systematic name of the following: \([\text{Pt(NH}_3)_2\text{Cl(NH}_2\text{CH}_3)]\text{Cl}\)
Answer: Diamminechlorido(methylamine)platinum(II) chloride
In simple words: List ammine, chlorido, and methylamine ligands in alphabetical order, followed by platinum(II) and the chloride counter ion.

Exam Tip: Multi-word ligand names like methylamine are enclosed in parentheses when preceded by numerical prefixes or listed in complex names.

Question 7. Using IUPAC norms write the systematic name of the following: \([\text{Co(en)}_3]^{3+}\)
Answer: Tris(ethane-1,2-diamine)cobalt(III) ion
In simple words: Since ethane-1,2-diamine already contains a numerical prefix in its name, the prefix 'tris' is used for three such ligands.

Exam Tip: Use bis, tris, tetrakis for chelating or complex ligands containing numbers in their names.

Question 8. Draw the structures of optical isomers of: \([\text{Cr(C}_2\text{O}_4)_3]^{3-}\)
Answer: The complex exhibits two non-superimposable mirror images corresponding to the dextro (d) and laevo (l) optical isomers.
In simple words: Octahedral complexes with three symmetrical bidentate ligands exist as chiral enantiomers that rotate plane-polarized light.

Exam Tip: Clearly draw the octahedral metal center connected to three bidentate oxalate (ox) rings and show the dashed mirror plane.

Question 9. What is meant by the chelate effect? Give an example.
Answer: When a di- or polydentate ligand attaches to the metal ion in a manner that forms a ring, the metal-ligand association is found to be more stable. This is known as the chelate effect. For example, \([\text{Co(en)}_3]^{3+}\) is exceptionally stable due to ring formation by ethylenediamine.
In simple words: Cyclic complexes formed by multidentate ligands are more stable than those formed by monodentate ligands.

Exam Tip: Mention entropy changes; chelation increases entropy, making the formation thermodynamically favorable.

2 / 3 MARK QUESTIONS

Question 1. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (\(\Delta_0\)) values. Weak field ligands have relatively low crystal-field splitting energy and generally form high-spin complexes (e.g., \(\text{I}^-, \text{Br}^-, \text{Cl}^-\)). Strong field ligands have high crystal-field splitting energy and cause pairing of electrons, forming low-spin complexes (e.g., \(\text{CN}^-, \text{CO}\)).
In simple words: Spectrochemical series orders ligands by their ability to split d-orbitals. Weak ligands don't force electron pairing, while strong ligands do.

Exam Tip: Memorize key reference points in the series: halides are weak fields, whereas \(\text{CN}^-\) and \(\text{CO}\) are very strong fields.

Question 2. \([\text{Co(NH}_3)_6]^{3+}\) is paramagnetic while \([\text{Ni(CN)}_4]^{2-}\) is diamagnetic. Explain why?
Answer: In \([\text{Co(NH}_3)_6]^{3+}\), Co is in the +3 oxidation state with a \(d^6\) configuration. However, due to crystal field considerations or specific complex properties, coordination number and ligand interactions dictate its magnetic behavior. In \([\text{Ni(CN)}_4]^{2-}\), Ni is in the +2 oxidation state with a \(d^8\) configuration. \(\text{CN}^-\) is a strong field ligand that forces complete pairing of electrons in the 3d orbitals, resulting in \(dsp^2\) hybridization with zero unpaired electrons, making it diamagnetic.
In simple words: Strong field ligands like cyanide pair up all d-electrons leading to diamagnetism, whereas weaker setups retain unpaired electrons.

Exam Tip: Always determine the metal oxidation state and electron configuration first before applying ligand field strength rules.

Question 3. A solution of \([\text{Ni(H}_2\text{O})_6]^{2+}\) is green but a solution of \([\text{Ni(CN)}_4]^{2-}\) is colourless. Explain.
Answer: In \([\text{Ni(H}_2\text{O})_6]^{2+}\), water is a weak field ligand, leaving unpaired electrons in the nickel d-orbitals, allowing \(d-d\) electronic transitions upon absorbing visible light, which imparts a green color. In \([\text{Ni(CN)}_4]^{2-}\), \(\text{CN}^-\) is a strong field ligand that pairs all electrons, preventing any \(d-d\) transitions, rendering the solution colorless.
In simple words: Unpaired electrons allow light absorption and color via d-d transitions; paired electrons make the complex colorless.

Exam Tip: Mention \(d-d\) transition explicitly as the core reason for color in transition metal complexes.

UNIT 13: AMINES

Question 1. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \((\text{CH}_3)_2\text{CHNH}_2\)
Answer: 1-Methylethanamine (Primary / \(1^\circ\) amine)
In simple words: The amino group is attached to a carbon bonded to two other carbons, and has two hydrogens on nitrogen.

Exam Tip: Classify amines based on how many alkyl or aryl groups are directly attached to the nitrogen atom.

Question 2. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \(\text{CH}_3(\text{CH}_2)_2\text{NH}_2\)
Answer: Propan-1-amine (Primary / \(1^\circ\) amine)
In simple words: A straight-chain three-carbon primary amine.

Exam Tip: Number the carbon chain starting from the end closest to the amino functional group.

Question 3. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \(\text{CH}_3\text{NHCH}(\text{CH}_3)_2\)
Answer: \(N\)-Methyl-2-methylethanamine (Secondary / \(2^\circ\) amine)
In simple words: Nitrogen is connected to a methyl group and an isopropyl group.

Exam Tip: Use the prefix \(N-\) to designate substituents attached directly to the nitrogen atom.

Question 4. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \((\text{CH}_3)_3\text{CNH}_2\)
Answer: 2-Methylpropan-2-amine (Primary / \(1^\circ\) amine)
In simple words: A tertiary-butyl group attached to an amino group.

Exam Tip: The classification depends on nitrogen substitution, not the carbon skeleton attached to it.

Question 5. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \(\text{C}_6\text{H}_5\text{NHCH}_3\)
Answer: \(N\)-Methylbenzenamine or \(N\)-methylaniline (Secondary / \(2^\circ\) amine)
In simple words: Aniline where one hydrogen on nitrogen is replaced by a methyl group.

Exam Tip: Benzenamine is the IUPAC preferred name for aniline.

REASONING QUESTIONS

Question 1. Account for the following: \(pK_b\) of aniline is more than that of methylamine.
Answer: In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring due to resonance, making it less available for protonation. In contrast, methylamine experiences an electron-releasing \(+I\) effect from the methyl group, which increases electron density on nitrogen and enhances basicity. Lower \(pK_b\) means higher basicity, hence aniline has a higher \(pK_b\) value than methylamine.
In simple words: Resonance in aniline locks up the lone pair, making it less basic than methylamine.

Exam Tip: Remember the inverse relationship: Stronger base = Lower \(pK_b\) value.

Question 2. Account for the following: Ethylamine is soluble in water whereas aniline is not.
Answer: Ethylamine molecules form intermolecular hydrogen bonds with water molecules easily due to the small size of the ethyl group. Aniline contains a large hydrophobic phenyl group (\(-\text{C}_6\text{H}_5\)) that hinders effective hydrogen bonding with water, making it insoluble.
In simple words: Small alkyl amines dissolve via hydrogen bonding, while the large benzene ring in aniline blocks water interaction.

Exam Tip: Hydrophobic alkyl/aryl size determines aqueous solubility for organic derivatives.

Question 3. Account for the following: Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Answer: Methylamine is more basic than water and accepts a proton from water molecules, releasing hydroxide ions (\(\text{OH}^-\)). These hydroxide ions react with \(\text{Fe}^{3+\) ions from ferric chloride to precipitate hydrated ferric oxide (\(\text{Fe}_2\text{O}_3 \cdot 3\text{H}_2\text{O}\)).
In simple words: Methylamine generates basic hydroxide ions in water, which precipitate iron as hydrated oxide.

Question 4. Account for the following: Although amino group is \(o-\) and \(p-\) directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of \(m\)-nitroaniline.
Answer: Nitration is carried out in a strongly acidic medium where aniline gets protonated to form the anilinium ion (\(-\text{NH}_3^+\)). The anilinium ion exerts a strong electron-withdrawing \(-I\) effect, which is meta-directing, leading to the formation of a significant amount of \(m\)-nitroaniline.
In simple words: Acidic nitration media protonates aniline into an anilinium ion, shifting substitution to the meta position.

Exam Tip: To get pure ortho/para isomers in nitration, the amino group must first be protected via acetylation.

Question 5. Account for the following: Aniline does not undergo Friedel-Crafts reaction.
Answer: Aniline is a Lewis base, while anhydrous aluminum chloride (\(\text{AlCl}_3\)), used as a catalyst in Friedel-Crafts reactions, is a Lewis acid. They react to form a stable salt, deactivating the benzene ring by placing a positive charge on the nitrogen atom, which stops further electrophilic substitution.
In simple words: Aniline reacts with the Lewis acid catalyst and blocks the reaction.

Exam Tip: Mention salt formation between the amine base and Lewis acid catalyst.

Question 6. Account for the following: Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Answer: Aromatic diazonium ions are stabilized by resonance involving the benzene ring, which delocalizes the positive charge. Aliphatic diazonium ions lack this resonance stabilization and readily lose nitrogen gas to form carbocations.
In simple words: Resonance stabilization keeps aromatic diazonium salts stable at low temperatures.

Exam Tip: Draw resonance structures showing charge delocalization onto the aromatic ring to score full marks.

Question 7. Account for the following: Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer: Gabriel phthalimide synthesis yields pure primary aliphatic amines exclusively, without forming secondary or tertiary amine impurities, making it the preferred method of preparation.
In simple words: It produces clean, pure primary amines without unwanted side-products.

Exam Tip: Note that this method cannot be used to prepare aromatic primary amines because aryl halides do not undergo nucleophilic substitution.

The compounds which contain dative bonds between metal atom and surrounding species is called co-ordination compounds.

The branch of inorganic chemistry which deals with the study of preparation properties of coordination compound is called co-ordination chemistry.

\( \text{K}_4[\text{Fe}(\text{CN})_6] \), \( [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \)

POINTS TO REMEMBER:

1. Coordination compounds:
Coordination compounds are compounds in which a central metal atom or ion is linked to a number of ions or neutral molecules by coordinate bonds or which contain complex ions.
Examples: \( \text{K}_4[\text{Fe}(\text{CN})_6] \); \( [\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \); \( \text{Ni}(\text{CO})_4 \)

2. The main postulates of Werner’s theory of coordination compounds:

• i) In coordination compounds metals show two types of linkages or valencies—Primary and Secondary.
• ii) The primary valencies are ionisable and are satisfied by negative ions.
• iii) The secondary valencies are non-ionisable and are satisfied by neutral molecules or negative ions. The secondary valence is equal to the C.N and is fixed for a metal.
• iv) The ions or groups bound by secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination nos.

3. Difference between a double salt and a complex:
Both double salts as well as complexes are formed by the combination of two or more stable compounds in stoichiometric ratio. However, double salts such as carnallite, \( \text{KCl}\cdot\text{MgCl}_2\cdot 6\text{H}_2\text{O} \), Mohr‘s salt, \( \text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot 6\text{H}_2\text{O} \), potash alum, \( \text{KAl}(\text{SO}_4)_2\cdot 12\text{H}_2\text{O} \), etc. dissociate into simple ions completely when dissolved in water. However, complex ions such as \( [\text{Fe}(\text{CN})_6]^{4-} \) of \( \text{K}_4[\text{Fe}(\text{CN})_6] \), do not dissociate into \( \text{Fe}^{2+} \) and \( \text{CN}^- \) ions.

IMPORTANT TERMINOLOGY

• (i) Coordination entity: It constitutes the central metal ion or atom bonded to a fixed number of ions or molecules represented within a square bracket.
• (ii) Central atom/ ion: In a coordination entity, the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it, is called the central atom or ion.
• (iii) Ligands: The neutral or negative ions bound to the central metal or ion in the coordination entity. These donate a pair/s of electrons to the central metal atom /ion. Ligands may be classified as:
  • a) Monodentate/Unidentate: Ligands bound to the central metal atom/ion through a single donor atom. Ex: \( \text{Cl}^- \); \( \text{H}_2\text{O} \); \( \text{NH}_3 \); \( \text{NO}_2^- \).
  • b) Didentate: Ligates through two donor atoms. Ex: \( \text{C}_2\text{O}_4^{2-} \) (ox); \( \text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 \) (en).
  • c) Polydentate: which ligates through two or more donor atoms present in a single ligand. Ex: \( (\text{EDTA})^{4-} \).
  • d) Chelating ligands: Di- or polydentate ligands that uses two or more donor atoms to bind to a single metal ion to form ring-like complexes. (ox); (edta).
  • e) Ambidentate ligand: A ligand that can ligate through two different atoms, one at a time. Ex: \( \text{NO}_2^- \); \( \text{SCN}^- \).

  5. NAMING OF MONONUCLEAR COORDINATION COMPOUNDS

  The principle of additive nomenclature is followed while naming the coordination compounds. The following rules are used:

  • i. The cation is named first in both positively and negatively charged coordination entities.
  • ii. The ligands are named in an alphabetical order before the name of the central atom/ion.
  • iii. The name of the anionic ligands end in –o, those of neutral and cationic ligands are the same except aqua for \( \text{H}_2\text{O} \), ammine for \( \text{NH}_3 \), carbonyl for \( \text{CO} \) and nitrosyl for \( \text{NO} \). These are placed within enclosing marks.
  • iv. When the prefixes mono, di, tri, etc., are used to indicate the number of the individual ligands in the coordination entity. When the names of the ligands include a numerical prefix, then the terms bis, tris, tetrakis are used, the ligand to which they refer being placed in parenthesis.
  • v. Oxidation state of the metal in cation, anion, or neutral coordination entity is indicated by roman numeral in parenthesis.
  • vi. If the complex ion is a cation, the metal is same as the element.
  • vii. The neutral complex molecule is named similar to that of the complex cation.

  6. NAMES OF SOME COMMON LIGANDS

  NEGATIVE LIGANDS	NAME	CHARGE	NEUTRAL LIGANDS	NAME	CHARGE
  \( \text{CN}^- \)	Cyano	-1	\( \text{NH}_3 \)	Ammine	0
  \( \text{Cl}^- \)	Chlorido	-1	\( \text{H}_2\text{O} \)	Aqua/aquo	0
  \( \text{Br}^- \)	Bromido	-1	\( \text{NO} \)	Nitrosyl	0
  \( \text{F}^- \)	Fluoride	-1	\( \text{CO} \)	Carbonyl	0
  \( \text{SO}_4^{2-} \)	Sulphato	-2	\( \text{PH}_3 \)	Phosphine	0
  \( \text{C}_2\text{O}_4^{2-} \)	Oxalato	-2	\( \text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 \)	(1,2-Ethanediamine)	0
  \( \text{NH}_2^- \)	Amido	-1	POSITIVE LIGANDS
  \( \text{NH}^{2-} \)	Imido	-2	\( \text{NH}_2\text{-NH}_3^+ \)	Hydrazinium	+1
  \( \text{ONO}^- \)	Nitrito-O	-1	\( \text{NO}^+ \)	Nitrosonium	+1
  \( \text{NO}_2^- \)	Nitro	-1	\( \text{NO}_2^+ \)	Nitronium	+1
  \( \text{NO}_3^- \)	Nitrato	-1
  \( \text{SCN}^- \)	Thiocyanato	-1
  \( \text{NCS}^- \)	Isothiocyanato	-1
  \( \text{CH}_2(\text{NH}_2)\text{COO}^- \)	Glycinato	-1
  \( \text{OH}^- \)	Hydroxo	-1

  7. ISOMERISM IN COORDINATION COMPOUNDS

  Two or more substances having the same molecular formula but different spatial arrangements are called isomers and the phenomenon is called isomerism. Coordination compounds show two main types of isomerism:

  A) Structural Isomerism
  B) Stereoisomerism

  STRUCTURAL ISOMERISM: It arises due to the difference in structures of coordination compounds. It is further subdivided into the following types:

  • 1) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionization isomers \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br} \) and \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \).
  • 2) Hydrate or solvate isomerism: This form of isomerism is known as ‘hydrate isomerism’ in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex \( [\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3 \) (violet) and its solvate isomer \( [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O} \) (grey-green).
  • 3) Linkage Isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, \( \text{NCS}^- \), which may bind through the nitrogen to give \( \text{M–NCS} \) or through sulphur to give \( \text{M–SCN} \).
  • 4) Coordination isomerism: It arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. Example: \( [\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6] \) & \( [\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6] \).

  STEREOISOMERISM: Stereo isomers have the same chemical formula and chemical bonds but they have different spatial arrangement. They are of two kinds:

  A. Geometrical isomerism
  B. Optical isomerism

  GEOMETRICAL ISOMERISM: This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula \( [\text{MX}_2\text{L}_2] \) (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer. In \( [\text{MABXL}] \)—where A, B, X, L are unidentates—two cis- and one trans- isomers are possible.

  Another type of geometrical isomerism occurs in octahedral coordination entities of the type \( [\text{Ma}_3\text{b}_3] \) like \( [\text{Co}(\text{NH}_3)_3(\text{NO}_2)_3] \). If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.

  OPTICAL ISOMERISM: Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. The molecules or ions that cannot be superimposed are called chiral. The two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involving didentate ligands. In a coordination entity of the type \( [\text{CoCl}_2(\text{en})_2]^+ \), only the cis-isomer shows optical activity.

  TYPES OF HYBRIDISATION

  Coordination number	Type of hybridisation	Acquired geometry
  4	\( \text{sp}^3 \)	Tetrahedral
  4	\( \text{dsp}^2 \)	Square planar
  5	\( \text{sp}^3\text{d} \)	Trigonal bipyramidal
  6	\( \text{sp}^3\text{d}^2 \)	Octahedral
  6	\( \text{d}^2\text{sp}^3 \)	Octahedral

  8. CRYSTAL FIELD THEORY:

  • 1. The metal-ligand bond is ionic arising purely from electrostatic interactions between the metal ion and the ligand.
  • 2. Ligands are treated as point charges or dipoles in case of anions and neutral molecules.
  • 3. In an isolated gaseous metal atom or ion the five d-orbitals are degenerate.
  • 4. Degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal /ion.
  • 5. In a complex the negative field becomes asymmetrical and results in splitting of the d-orbitals.

  A) CRYSTAL FIELD SPLITTING IN OCTAHEDRAL COORDINATION ENTITIES

  For \( \text{d}^4 \) ions, two possible patterns of electron distribution arise:

  • (i) If \( \Delta_o < P \), the fourth electron enters one of the \( e_g \) orbitals giving the configuration \( t_{2g}^3 e_g^1 \). Ligands for which \( \Delta_o < P \) are known as weak field ligands and form high spin complexes.
  • (ii) If \( \Delta_o > P \), it becomes more energetically favourable for the fourth electron to occupy a \( t_{2g} \) orbital with configuration \( t_{2g}^4 e_g^0 \). Ligands which produce this effect are known as strong field ligands and form low spin complexes.

  B) CRYSTAL FIELD SPLITTING IN TETRAHEDRAL COORDINATION ENTITIES

  • 1. The four surrounding ligands approach the central metal atom/ion along the planes between the axes.
  • 2. The \( t_{2g} \) orbitals are raised in energy \( (2/5)\Delta_t \).
  • 3. The two \( e_g \) orbitals are lowered in energy \( (3/5)\Delta_t \).
  • 4. The splitting is smaller as compared to octahedral field splitting, \( \Delta_t = (4/9)\Delta_o \).
  • 5. Pairing of electrons is rare and thus complexes have generally high spin configurations.

  BONDING IN METAL CARBONYLS

  The metal-carbon bond in metal carbonyls possess both \( \sigma \) and \( \pi \) character. The \( \text{M–C} \) \( \sigma \) bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The \( \text{M–C} \) \( \pi \) bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding \( \pi^* \) orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between \( \text{CO} \) and the metal.

  SOLVED QUESTIONS

  1 MARK QUESTIONS

  Question. What are ambidentate ligands? Give two examples for each.
  Answer: Ambidentate ligands are ligands that can attach themselves to the central metal atom through two different atoms. For example:
  (a) Nitro group: \( \text{M}-\text{NO}_2 \) (The donor atom is N) and Nitrito group: \( \text{M}-\text{ONO} \) (The donor atom is oxygen).
  (b) Thiocyanate: \( \text{M}-\text{SCN} \) (The donor atom is S) and Isothiocyanate: \( \text{M}-\text{NCS} \) (The donor atom is N).

  Question. Using IUPAC norms write the formula for the following: Tetrahydroxozincate(II)
  Answer: \( [\text{Zn}(\text{OH})_4]^{2-} \)

  Question. Using IUPAC norms write the formula for the following: Hexaamminecobalt(III) sulphate
  Answer: \( [\text{Co}(\text{NH}_3)_6]_2(\text{SO}_4)_3 \)

  Question. Using IUPAC norms write the formula for the following: Pentaamminenitrito-O-cobalt(III)
  Answer: \( [\text{Co}(\text{ONO})(\text{NH}_3)_5]^{2+} \)

  Question. Using IUPAC norms write the systematic name of the following: \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \)
  Answer: Hexaamminecobalt(III) chloride

  Question. Using IUPAC norms write the systematic name of the following: \( [\text{Pt}(\text{NH}_3)_2\text{Cl}(\text{NH}_2\text{CH}_3)]\text{Cl} \)
  Answer: Diamminechlorido(methylamine)platinum(II) chloride

  Question. Using IUPAC norms write the systematic name of the following: \( [\text{Co}(\text{en})_3]^{3+} \)
  Answer: Tris(ethane-1,2-diamine)cobalt(III) ion

  Question. Draw the structures of optical isomers of: \( [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-} \)
  Answer: The optical isomers are dextro and laevo forms of \( [\text{Cr}(\text{ox})_3]^{3-} \), which are non-superimposable mirror images of each other.

  Question. What is meant by the chelate effect? Give an example.
  Answer: When a ligand attaches to the metal ion in a manner that forms a ring, then the metal-ligand association is found to be more stable. For example, complexes formed by \( \text{en} \) (ethane-1,2-diamine) such as \( [\text{Ni}(\text{en})_3]^{2+} \) are more stable than those formed by monodentate ligands like \( [\text{Ni}(\text{NH}_3)_6]^{2+} \).

  2 / 3 MARK QUESTIONS

  Question. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
  Answer: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values:
  \( \text{I}^- < \text{Br}^- < \text{S}^{2-} < \text{SCN}^- < \text{Cl}^- < \text{N}_3^- < \text{F}^- < \text{OH}^- < \text{C}_2\text{O}_4^{2-} \sim \text{H}_2\text{O} < \text{NCS}^- \sim \text{H}^- < \text{CN}^- < \text{NH}_3 < \text{en} \sim \text{SO}_3^{2-} < \text{NO}_2^- < \text{phen} < \text{CO} \)
  Ligands with small \( \Delta_o \) values are weak field ligands and form high spin complexes (as \( \Delta_o < P \)). Ligands with large \( \Delta_o \) values are strong field ligands and cause pairing of electrons to form low spin complexes (as \( \Delta_o > P \)).

  Question. \( [\text{Cr}(\text{NH}_3)_6]^{3+} \) is paramagnetic while \( [\text{Ni}(\text{CN})_4]^{2-} \) is diamagnetic. Explain why?
  Answer: Cr is in the +3 oxidation state i.e., \( \text{d}^3 \) configuration. Also, \( \text{NH}_3 \) is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital. Therefore, it undergoes \( \text{d}^2\text{sp}^3 \) hybridization and the 3 electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
  In \( [\text{Ni}(\text{CN})_4]^{2-} \), Ni exists in the +2 oxidation state i.e., \( \text{d}^8 \) configuration. \( \text{CN}^- \) is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, \( \text{Ni}^{2+} \) undergoes \( \text{dsp}^2 \) hybridization with no unpaired electrons. Hence, it is diamagnetic.

  Question. A solution of \( [\text{Ni}(\text{H}_2\text{O})_6]^{2+} \) is green but a solution of \( [\text{Ni}(\text{CN})_4]^{2-} \) is colourless. Explain.
  Answer: In \( [\text{Ni}(\text{H}_2\text{O})_6]^{2+} \), \( \text{H}_2\text{O} \) is a weak field ligand. Therefore, there are unpaired electrons in \( \text{Ni}^{2+} \). In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, \( [\text{Ni}(\text{H}_2\text{O})_6]^{2+} \) is coloured.
  In \( [\text{Ni}(\text{CN})_4]^{2-} \), the electrons are all paired as \( \text{CN}^- \) is a strong field ligand. Therefore, d-d transition is not possible in \( [\text{Ni}(\text{CN})_4]^{2-} \). Hence, it is colourless. As there are no unpaired electrons, it is diamagnetic.

  Question. Draw all the isomers (geometrical and optical) of:
  (i) \( [\text{CoCl}_2(\text{en})_2]^+ \)
  (ii) \( [\text{Co}(\text{NH}_3)\text{Cl}(\text{en})_2]^{2+} \)
  (iii) \( [\text{Co}(\text{NH}_3)_2\text{Cl}_2(\text{en})]^+ \)
  Answer:
  (i) For \( [\text{CoCl}_2(\text{en})_2]^+ \): Trans-isomer is symmetrical and optically inactive (superimposable mirror images). Cis-isomer is unsymmetrical and resolves into dextro and laevo optical isomers (non-superimposable mirror images). In total, three isomers are possible.
  (ii) For \( [\text{Co}(\text{NH}_3)\text{Cl}(\text{en})_2]^{2+} \): Cis and trans geometrical isomers are possible; cis-isomer is optically active.
  (iii) For \( [\text{Co}(\text{NH}_3)_2\text{Cl}_2(\text{en})]^+ \): Several geometrical isomers are possible, and those lacking a plane of symmetry are optically active.

  Question. Write all the geometrical isomers of \( [\text{Pt}(\text{NH}_3)(\text{Br})(\text{Cl})(\text{py})] \) and how many of these will exhibit optical isomers?
  Answer: Three geometrical isomers are possible for square planar complex \( [\text{Pt}(\text{NH}_3)(\text{Br})(\text{Cl})(\text{py})] \). None of these will exhibit optical isomers because square planar complexes possess a plane of symmetry and are achiral.

  Question. What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.
  Answer: The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.
  \( \text{M} + 3\text{L} \longleftrightarrow \text{ML}_3 \)
  Stability constant, \( \beta = \frac{[\text{ML}_3]}{[\text{M}][\text{L}]^3} \)
  For this reaction, the greater the value of the stability constant, the greater is the proportion of \( \text{ML}_3 \) in the solution.
  Factors governing stability include:
  1. Charge on the central metal ion (higher charge gives greater stability).
  2. Size of the central metal ion (smaller size gives greater stability).
  3. Basic strength and nature of the ligand.
  4. Chelate effect (chelating ligands form more stable complexes).

  5 MARKS QUESTIONS

  Question. (a) Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
  (i) \( [\text{Fe}(\text{CN})_6]^{4-} \)
  (ii) \( [\text{FeF}_6]^{3-} \)
  (iii) \( [\text{Co}(\text{C}_2\text{O}_4)_3]^{3-} \)
  (iv) \( [\text{CoF}_6]^{3-} \)
  Answer:
  (i) \( [\text{Fe}(\text{CN})_6]^{4-} \): Iron is in +2 oxidation state (\( 3\text{d}^6 \)). \( \text{CN}^- \) is a strong field ligand and causes pairing of electrons in 3d orbitals, leaving two vacant 3d orbitals. It undergoes \( \text{d}^2\text{sp}^3 \) hybridization. Geometry is octahedral and it is diamagnetic.
  (ii) \( [\text{FeF}_6]^{3-} \): Iron is in +3 oxidation state (\( 3\text{d}^5 \)). \( \text{F}^- \) is a weak field ligand and does not cause pairing of electrons. Outer 4d orbitals are used, undergoing \( \text{sp}^3\text{d}^2 \) hybridization. Geometry is octahedral and it is strongly paramagnetic (5 unpaired electrons).
  (iii) \( [\text{Co}(\text{C}_2\text{O}_4)_3]^{3-} \): Cobalt is in +3 oxidation state (\( 3\text{d}^6 \)). In presence of oxalate ligand, pairing occurs to give \( \text{d}^2\text{sp}^3 \) hybridization. Geometry is octahedral and diamagnetic.
  (iv) \( [\text{CoF}_6]^{3-} \): Cobalt is in +3 oxidation state (\( 3\text{d}^6 \)). Fluoride is a weak field ligand, electrons remain unpaired, undergoing \( \text{sp}^3\text{d}^2 \) hybridization. Geometry is octahedral and paramagnetic (4 unpaired electrons).

  Question. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
  (i) \( \text{K}[\text{Cr}(\text{H}_2\text{O})_2(\text{C}_2\text{O}_4)_2]\cdot 3\text{H}_2\text{O} \)
  (ii) \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \)
  Answer:
  (i) IUPAC Name: Potassium diaquadioxalatochromate(III) trihydrate.
  Oxidation state of chromium = +3
  Electronic configuration: \( 3\text{d}^3: t_{2g}^3 \)
  Coordination number = 6
  Shape: octahedral
  Stereochemistry: Cis and trans isomers. Cis form is optically active.
  Magnetic moment, \( \mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ BM} \sim 4\text{ BM} \)
  (ii) IUPAC Name: Pentaamminechloridocobalt(III) chloride
  Oxidation state of Co = +3
  Electronic configuration: \( \text{d}^6: t_{2g}^6 \)
  Coordination number = 6
  Shape: octahedral
  Stereochemistry: Does not show geometrical or optical isomerism.
  Magnetic Moment = 0 (diamagnetic).

  LEVEL 1

  Question. Why do tetrahedral complex not show geometrical isomerism?

  Question. Why does the colour changes on heating \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \).

  Question. \( [\text{Fe}(\text{H}_2\text{O})_6]^{3+} \) is strongly paramagnetic whereas \( [\text{Fe}(\text{CN})_6]^{3-} \) is weakly paramagnetic. Explain.

  Question. What happens when potassium ferrocyanide solution is added to a ferric salt solution?

  LEVEL 2

  Question. A coordination compound has a formula \( (\text{CoCl}_3\cdot 4\text{NH}_3) \). It does not liberate \( \text{NH}_3 \) but precipitates chloride ion as \( \text{AgCl} \). Give the IUPAC name of the complex and write its structural formula.

  Question. Write the correct formula for the following co-ordination compounds: \( \text{CrCl}_3\cdot 6\text{H}_2\text{O} \) (Violet, with 3 Chloride ions/ Unit formula) and \( \text{CrCl}_3\cdot 6\text{H}_2\text{O} \) (Light green colour with 2 Chloride ions/ unit formula).

  Question. Give the electronic configuration of the d-orbitals of Ti in \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \) ion in an octahedral crystal field.

  Question. Co(II) is stable in aqueous solution but in the presence of strong ligands and air, it can get oxidized to Co(III). (Atomic Number of cobalt is 27). Explain.

  Question. Give a chemical test to distinguish between \( [\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 \) and \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br} \). Name the type of isomerism exhibited by these compounds.

  Question. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is that no precipitate of copper sulphide is obtained when \( \text{H}_2\text{S} \) (g) is passed through this solution?

  LEVEL 3

  Question. Aqueous copper sulphate solution (blue in colour) gives a green precipitate with aqueous potassium fluoride, a bright green solution with aqueous potassium chloride. Explain these experimental results.

  Question. A metal complex having the composition \( \text{Cr}(\text{NH}_3)_4\text{Cl}_2\text{Br} \) has been isolated in two forms, A and B. The form A reacts with \( \text{AgNO}_3 \) solution to give a white precipitate readily soluble in dilute aqueous ammonia whereas B give a pale yellow precipitate soluble in concentrated ammonia solution. Write the formulae of A and B and write their IUPAC names.

  Question. Explain the following:
  i. All octahedral complexes of \( \text{Ni}^{2+} \) must be outer orbital complexes.
  ii. \( \text{NH}_4^+ \) ion does not form any complex.
  iii. \( (\text{SCN})^- \) ion is involved in linkage isomerism in co-ordination compounds.

  Question. A metal ion \( \text{M}^{n+} \) having \( \text{d}^4 \) valence electronic configuration combines with three didentate ligands to form complexes. Assuming \( \Delta_o > P \), Draw the diagram showing d orbital splitting during this complex formation. Write the electronic configuration of the valence electrons of the metal \( \text{M}^{n+} \) ion in terms of \( t_{2g} \) and \( e_g \). What type of the hybridization will \( \text{M}^{n+} \) ion have? Name the type of isomerism exhibited by this complex.

  Question. The coordination no. of \( \text{Ni}^{2+} \) is 4.
  \( \text{NiCl}_2 + \text{KCN(excess)} \rightarrow \text{A (a cyano complex)} \)
  \( \text{A} + \text{Conc HCl(excess)} \rightarrow \text{B (a chloro complex)} \)
  i) Write IUPAC name of A and B
  ii) Predict the magnetic nature of A and B
  iii) Write hybridization of Ni in A and B

  Question. Explain the following:
  i. \( \text{Cu}(\text{OH})_2 \) is soluble in ammonium hydroxide but not in sodium hydroxide solution.
  ii. EDTA is used to cure lead poisoning.
  iii. Blue coloured solution of \( [\text{CoCl}_4]^{2-} \) changes to pink on reaction with \( \text{HgCl}_2 \).

  PRACTICE QUESTIONS (1 MARK)

  Question. Write the formula for the following coordination compound: Tetraamineaquachloridocobalt(III) chloride

  Question. Write the IUPAC name of the following coordination compound: \( [\text{CoCl}_2(\text{en})_2]\text{Cl} \)

  Question. Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion ?

  Question. Out of the following two coordination entities which is chiral (optically active)?
  (a) cis-\([\text{CrCl}_2(\text{ox})_2]^{3-}\)
  (b) trans-\([\text{CrCl}_2(\text{ox})_2]^{3-}\)

  Question. The spin only magnetic moment of \( [\text{MnBr}_4]^{2-} \) is 5.9 BM. Predict the geometry of the complex ion?

  Question. \( [\text{NiCl}_4]^{2-} \) is paramagnetic while \( [\text{Ni}(\text{CO})_4] \) is diamagnetic though both are tetrahedral. Why?

  PRACTICE QUESTIONS (2 MARKS)

  Question. Draw structures of geometrical isomers of \( [\text{Fe}(\text{NH}_3)_2(\text{CN})_4]^- \)

  Question. Indicate the type of isomerism exhibited by the following complex and draw the structures for these isomers: \( [\text{Co}(\text{en})_3]\text{Cl}_3 \)

  Question. Give evidence that \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \) and \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) are ionization isomers.

  Question. Calculate the overall complex dissociation equilibrium constant for the \( [\text{Cu}(\text{NH}_3)_4]^{2+} \) ion, given that \( \beta_4 \) for this complex is \( 2.1 \times 10^{13} \).

  Question. What is meant by unidentate ligand? Give two examples.

  Question. What is meant by didentate ligand? Give two examples.

  Question. What is meant by ambidentate ligands? Give two examples.

  Question. Draw the structures of optical isomers of: \( [\text{Cr}(\text{C}_2\text{O}_4)_3]^{3-} \)

  Question. Discuss the nature of bonding in metal carbonyls.

  Question. What is meant by the chelate effect? Give an example.

  Question. Draw the structures of: (i) \( \text{Ni}(\text{CO})_4 \) (ii) \( \text{Fe}(\text{CO})_5 \)

  PRACTICE QUESTIONS (3 MARKS)

  Question. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
  (i) \( [\text{Fe}(\text{CN})_6]^{4-} \)
  (ii) \( [\text{FeF}_6]^{3-} \)
  (iii) \( [\text{Co}(\text{C}_2\text{O}_4)_3]^{3-} \)
  Also predict their magnetic behaviour.

  Question. What is crystal field splitting energy? Draw figure to show the splitting of d orbitals in an octahedral crystal field. How does the magnitude of \( \Delta_o \) decide the actual configuration of d orbitals in a coordination entity?

  Question. Discuss briefly giving an example in each case the role of coordination compounds in:
  (i) biological systems
  (ii) medicinal chemistry
  (iii) analytical chemistry

  UNIT 13: AMINES

  2.

  Amines

  1. Ammonolysis of alkyl halides, Gabriel Phthalimide synthesis, Hoffmann Bromamide Degradation. 2. Basic character of Amines (\(\text{p}K_b\)) and comparisons in gaseous and aqueous phase. 3. Carbylamine Reaction, Hinsberg’s Test. 4. Electrophilic substitution. 5. Diazonium salts—reactions.

  IUPAC NOMENCLATURE OF AMINES

  Amine	IUPAC name
  \( \text{CH}_3\text{-CH}_2\text{-NH}_2 \)	Ethanamine
  \( \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-NH}_2 \)	Propan-1-amine
  \( \text{CH}_3\text{-CH}(\text{NH}_2)\text{-CH}_3 \)	Propan-2-amine
  \( \text{CH}_3\text{-NH-CH}_2\text{-CH}_3 \)	N-Methylethanamine
  \( \text{CH}_3\text{-N}(\text{CH}_3)_2 \)	N,N-Dimethylmethanamine
  \( \text{C}_2\text{H}_5\text{-N}(\text{C}_2\text{H}_5)\text{-CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \)	N,N-Diethylbutan-1-amine
  \( \text{NH}_2\text{-CH}_2\text{-CH=CH}_2 \)	Prop-2-en-1-amine
  \( \text{NH}_2\text{-(CH}_2)_6\text{-NH}_2 \)	Hexane-1,6-diamine
  \( \text{C}_6\text{H}_5\text{NH}_2 \)	Aniline or Benzenamine
  \( \text{C}_6\text{H}_4(\text{CH}_3)(\text{NH}_2) \) (2-methylaniline)	2-Aminotoluene
  \( \text{C}_6\text{H}_4(\text{Br})(\text{NH}_2) \) (4-bromoaniline)	4-Bromobenzenamine or 4-Bromoaniline
  \( \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2 \)	N,N-Dimethylbenzenamine

  NAME REACTIONS

  1. Gabriel phthalimide synthesis:
  Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
  \( \text{Phthalimide} \xrightarrow{\text{KOH}} \text{Potassium phthalimide} \xrightarrow{\text{R-X}} \text{N-Alkylphthalimide} \xrightarrow{\text{NaOH(aq)}} \text{Sodium phthalate} + \text{R-NH}_2 \text{ (1}^\circ\text{ amine)} \)

  2. Hoffmann bromamide degradation reaction:
  Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. The amine so formed contains one carbon less than that present in the amide.
  \( \text{R-CO-NH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O} \)

  3. Carbylamine reaction:
  Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.
  \( \text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{Heat}} \text{R-NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)

  4. Hinsberg Test:
  Benzenesulphonyl chloride (\( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \)), which is also known as Hinsberg‘s reagent, reacts with primary and secondary amines to form sulphonamides.
  (a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.
  (b) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. Since N,N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.
  (c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines.

  5. Sandmeyer Reaction:
  The \( \text{Cl}^- \), \( \text{Br}^- \) and \( \text{CN}^- \) nucleophiles can easily be introduced in the benzene ring of diazonium salts in the presence of Cu(I) ion.
  \( \text{ArN}_2^+\text{X}^- \xrightarrow{\text{CuCl/HCl}} \text{ArCl} + \text{N}_2 \)
  \( \text{ArN}_2^+\text{X}^- \xrightarrow{\text{CuBr/HBr}} \text{ArBr} + \text{N}_2 \)
  \( \text{ArN}_2^+\text{X}^- \xrightarrow{\text{CuCN/KCN}} \text{ArCN} + \text{N}_2 \)

  6. Gatterman Reaction:
  Chlorine or bromine can be introduced in the benzene ring by treating the diazonium salt solution with corresponding halogen acid in the presence of copper powder.
  \( \text{ArN}_2^+\text{X}^- \xrightarrow{\text{Cu/HCl}} \text{ArCl} + \text{N}_2 + \text{CuX} \)
  \( \text{ArN}_2^+\text{X}^- \xrightarrow{\text{Cu/HBr}} \text{ArBr} + \text{N}_2 + \text{CuX} \)

  7. Coupling reactions:
  The azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N=N– bond. These compounds are often coloured and are used as dyes. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction.
  \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH} + \text{Cl}^- + \text{H}_2\text{O} \) (orange dye)
  Similarly the reaction of diazonium salt with aniline yields p-aminoazobenzene.
  \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-NH}_2 + \text{Cl}^- + \text{H}_2\text{O} \) (yellow dye)

  DISTINCTION BETWEEN PAIRS OF COMPOUNDS

  Question. Give one chemical test to distinguish between the following pairs of compounds:
  (i) Methylamine and dimethylamine
  (ii) Secondary and tertiary amines
  (iii) Ethylamine and aniline
  (iv) Aniline and benzylamine
  (v) Aniline and N-methylaniline.
  Answer:
  (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Methylamine (being an aliphatic primary amine) on heating with \( \text{CHCl}_3 \) and ethanolic \( \text{KOH} \) gives foul-smelling methyl isocyanide, but dimethylamine does not.
  \( \text{CH}_3\text{-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{CH}_3\text{-NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
  (ii) Secondary and tertiary amines can be distinguished using Hinsberg‘s reagent (\( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \)). Secondary amines react to form N,N-dialkylbenzenesulphonamide which is insoluble in alkali. Tertiary amines do not react with Hinsberg‘s reagent.
  (iii) Ethylamine and aniline can be distinguished using the azo-dye test. Aniline forms a stable diazonium salt at 273-278 K which on coupling with alkaline 2-naphthol forms a bright orange-red dye. Ethylamine reacts with nitrous acid with brisk effervescence of \( \text{N}_2 \) gas and forms ethanol.
  (iv) Aniline and benzylamine can be distinguished by reaction with nitrous acid (\( \text{HNO}_2 \)). Aniline forms stable diazonium salt at 0-5°C which gives azo dye with 2-naphthol. Benzylamine forms unstable diazonium salt which decomposes with evolution of \( \text{N}_2 \) gas to yield benzyl alcohol.
  (v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Aniline (primary amine) gives a positive carbylamine test with a foul smell, whereas N-methylaniline (secondary amine) does not react.

  REASONING QUESTIONS

  Question. Account for the following:
  (i) \( \text{p}K_b \) of aniline is more than that of methylamine.
  (ii) Ethylamine is soluble in water whereas aniline is not.
  (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
  (iv) Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
  (v) Aniline does not undergo Friedel-Crafts reaction.
  (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
  (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
  Answer:
  (i) In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring through resonance, making it less available for protonation. In methylamine, the +I effect of the methyl group increases electron density on nitrogen. Hence, aniline is less basic and its \( \text{p}K_b \) is higher.
  (ii) Ethylamine forms strong intermolecular hydrogen bonds with water due to smaller size of ethyl group. In aniline, the large non-polar hydrophobic phenyl group (\( -\text{C}_6\text{H}_5 \)) opposes hydrogen bonding, making it insoluble in water.
  (iii) Due to +I effect of methyl group, methylamine is more basic than water and accepts a proton from water to produce \( \text{OH}^- \) ions:
  \( \text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \longrightarrow \text{CH}_3\text{NH}_3^+ + \text{OH}^- \)
  These \( \text{OH}^- \) ions react with \( \text{Fe}^{3+} \) ions of \( \text{FeCl}_3 \) to precipitate brown hydrated ferric oxide:
  \( 2\text{Fe}^{3+} + 6\text{OH}^- \longrightarrow \text{Fe}_2\text{O}_3\cdot 3\text{H}_2\text{O} \)
  (iv) Nitration is carried out in strongly acidic medium (\( \text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4 \)). In acidic medium, aniline gets protonated to form anilinium ion (\( -\text{NH}_3^+ \)), which is strongly deactivating and meta-directing. Hence, 47% m-nitroaniline is formed.
  (v) Aniline forms a Lewis acid-base complex/salt with the catalyst \( \text{AlCl}_3 \) used in Friedel-Crafts reaction (\( \text{C}_6\text{H}_5\text{NH}_2\cdot\text{AlCl}_3 \)). Due to this, the nitrogen acquires a positive charge and strongly deactivates the benzene ring towards electrophilic attack.
  (vi) Arenediazonium ions are stabilized by resonance dispersal of the positive charge over the benzene ring. In aliphatic diazonium salts, no such resonance stabilization is possible, and they rapidly lose \( \text{N}_2 \) to form carbocations.
  (vii) Gabriel phthalimide synthesis gives only pure \( 1^\circ \) amines because the monoalkylated product cannot undergo further alkylation. Secondary and tertiary amines are not formed as by-products.

  Question. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
  Answer: Gabriel phthalimide synthesis involves an \( \text{S}_\text{N}2 \) nucleophilic attack by phthalimide anion on an alkyl halide. Aryl halides do not undergo nucleophilic substitution under normal conditions due to partial double bond character of the C–X bond and steric/electronic repulsion by the benzene ring.

  Question. Give plausible explanation for each of the following:
  (i) Why are amines less acidic than alcohols of comparable molecular masses?
  (ii) Why do primary amines have higher boiling point than tertiary amines?
  (iii) Why are aliphatic amines stronger bases than aromatic amines?
  Answer:
  (i) Loss of a proton from an amine gives amide ion (\( \text{R-NH}^- \)), while from alcohol gives alkoxide ion (\( \text{R-O}^- \)). Since oxygen is more electronegative than nitrogen, it accommodates the negative charge more effectively. Hence, alkoxide ion is more stable than amide ion, making alcohols more acidic than amines.
  (ii) Primary amines have two hydrogen atoms bonded to nitrogen, allowing extensive intermolecular hydrogen bonding. Tertiary amines have no hydrogen on nitrogen and cannot form intermolecular hydrogen bonds. More energy is required to boil primary amines.
  (iii) In aliphatic amines, alkyl groups show electron-donating +I effect which increases electron density on nitrogen. In aromatic amines, the lone pair on nitrogen is delocalized into the benzene ring by resonance (-R effect), making it less available for donation.

  SOLVED 1 MARK QUESTIONS (AMINES)

  Question. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \( (\text{CH}_3)_2\text{CHNH}_2 \)
  Answer: 1-Methylethanamine or Propan-2-amine (\( 1^\circ \) amine)

  Question. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \( \text{CH}_3(\text{CH}_2)_2\text{NH}_2 \)
  Answer: Propan-1-amine (\( 1^\circ \) amine)

  Question. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \( \text{CH}_3\text{NHCH}(\text{CH}_3)_2 \)
  Answer: N-Methyl-2-methylethanamine or N-Methylpropan-2-amine (\( 2^\circ \) amine)

  Question. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \( (\text{CH}_3)_3\text{CNH}_2 \)
  Answer: 2-Methylpropan-2-amine (\( 1^\circ \) amine)

  Question. Give the IUPAC name of the compound and classify into primary, secondary or tertiary amines: \( \text{C}_6\text{H}_5\text{NHCH}_3 \)
  Answer: N-Methylbenzamine or N-methylaniline (\( 2^\circ \) amine)

  Question. Write short notes on diazotization
  Answer: Aromatic primary amines react with nitrous acid (prepared in situ from \( \text{NaNO}_2 \) and a mineral acid such as \( \text{HCl} \)) at low temperatures (273-278 K) to form diazonium salts. This conversion is known as diazotization.
  \( \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \)

  Question. Write short notes on ammonolysis
  Answer: Cleavage of the carbon-halogen bond in alkyl or benzyl halides by reaction with an ethanolic solution of ammonia at 373 K is known as ammonolysis. The primary amine formed reacts further with alkyl halides to yield secondary, tertiary amines and finally quaternary ammonium salts.
  \( \text{NH}_3 + \text{R-X} \rightarrow \text{R-NH}_3^+\text{X}^- \xrightarrow{\text{NaOH}} \text{R-NH}_2 + \text{H}_2\text{O} + \text{NaX} \)

  Question. Write short notes on acetylation.
  Answer: The process of introducing an acetyl group (\( \text{CH}_3\text{CO-} \)) into primary and secondary amines by nucleophilic substitution with acid chlorides or anhydrides in the presence of a stronger base like pyridine is called acetylation.
  \( \text{C}_2\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \xrightarrow{\text{Pyridine}} \text{C}_2\text{H}_5\text{NHCOCH}_3 + \text{HCl} \)

  Question. Why are amines basic in character?
  Answer: The nitrogen atom in amines has an unshared lone pair of electrons which it can donate to electron-deficient species (Lewis base character) or accept a proton from an acid (Bronsted-Lowry base).

  Question. Arrange the following in decreasing order of their basic strength: \( \text{C}_6\text{H}_5\text{NH}_2 \), \( \text{C}_2\text{H}_5\text{NH}_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \), \( \text{NH}_3 \)
  Answer: \( (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3 > \text{C}_6\text{H}_5\text{NH}_2 \)

  SOLVED EXAMPLES (2 MARKS)

  Question. Write chemical equations for the following reactions:
  (i) Reaction of ethanolic \( \text{NH}_3 \) with \( \text{C}_2\text{H}_5\text{Cl} \).
  (ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of \( \text{CH}_3\text{Cl} \).
  Answer:
  (i) \( \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{NH}_3} \text{C}_2\text{H}_5\text{NH}_2 \xrightarrow{\text{C}_2\text{H}_5\text{Cl}} (\text{C}_2\text{H}_5)_2\text{NH} \xrightarrow{\text{C}_2\text{H}_5\text{Cl}} (\text{C}_2\text{H}_5)_3\text{N} \)
  (ii) \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{NH}_3} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \xrightarrow{2\text{CH}_3\text{Cl}} \text{C}_6\text{H}_5\text{CH}_2\text{N}(\text{CH}_3)_2 \)

  Question. Write chemical equations for the following conversions:
  (i) \( \text{CH}_3\text{–CH}_2\text{–Cl} \) into \( \text{CH}_3\text{–CH}_2\text{–CH}_2\text{–NH}_2 \)
  (ii) \( \text{C}_6\text{H}_5\text{–CH}_2\text{–Cl} \) into \( \text{C}_6\text{H}_5\text{–CH}_2\text{–CH}_2\text{–NH}_2 \)
  Answer:
  (i) \( \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{ethanolic NaCN}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{H}_2/\text{Ni or LiAlH}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \)
  (ii) \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{ethanolic NaCN}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2 \)

  Question. Write structures and IUPAC names of
  (i) the amide which gives propanamine by Hoffmann bromamide reaction.
  (ii) the amine produced by the Hoffmann degradation of benzamide.
  Answer:
  (i) Starting amide has four carbons: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2 \) (Butanamide).
  (ii) Amine produced from benzamide (\( \text{C}_6\text{H}_5\text{CONH}_2 \)) is Aniline or Benzenamine (\( \text{C}_6\text{H}_5\text{NH}_2 \)).

  Question. How will you convert 4-nitrotoluene to 2-bromobenzoic acid?
  Answer:
  1. 4-Nitrotoluene \( \xrightarrow{\text{Br}_2/\text{Fe}} \) 2-bromo-4-nitrotoluene.
  2. \( \xrightarrow{\text{Sn/HCl}} \) 2-bromo-4-aminotoluene.
  3. \( \xrightarrow{\text{NaNO}_2/\text{HCl, 273-278 K}} \) diazonium salt.
  4. \( \xrightarrow{\text{H}_3\text{PO}_2/\text{H}_2\text{O}} \) 2-bromotoluene (deamination).
  5. \( \xrightarrow{\text{KMnO}_4/\text{OH}^-, \Delta} \) 2-bromobenzoic acid.

  Question. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
  Answer:
  (i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 + \text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O} \) (forms stable benzenediazonium chloride).
  (ii) \( \text{R-NH}_2 + \text{HNO}_2 \xrightarrow{\text{NaNO}_2+\text{HCl}} [\text{R-N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \text{ROH} + \text{N}_2 \uparrow + \text{HCl} \) (forms unstable diazonium salt which gives alcohol and nitrogen gas).

  Question. How will you convert:
  (i) Ethanoic acid into methanamine
  (ii) Hexanenitrile into 1-aminopentane
  Answer:
  (i) \( \text{CH}_3\text{COOH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{COCl} \xrightarrow{\text{NH}_3(\text{excess})} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2 \)
  (ii) \( \text{C}_5\text{H}_{11}\text{CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{C}_5\text{H}_{11}\text{COOH} \xrightarrow{\text{SOCl}_2} \text{C}_5\text{H}_{11}\text{COCl} \xrightarrow{\text{NH}_3} \text{C}_5\text{H}_{11}\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_5\text{H}_{11}\text{NH}_2 \)

  Question. How will you convert:
  (i) Methanol to ethanoic acid
  (ii) Ethanamine into methanamine
  Answer:
  (i) \( \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COOH} \)
  (ii) \( \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{CH}_3\text{NH}_2 \)

  Question. How will you convert
  (i) Ethanoic acid into propanoic acid
  (ii) Methanamine into ethanamine
  Answer:
  (i) \( \text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{NaCN}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{COOH} \)
  (ii) \( \text{CH}_3\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{NaCN}} \text{CH}_3\text{CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{NH}_2 \)

  Question. How will you convert
  (i) Nitromethane into dimethylamine
  (ii) Propanoic acid into ethanoic acid?
  Answer:
  (i) \( \text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{CH}_3\text{NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH, }\Delta} \text{CH}_3\text{NC} \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{NHCH}_3 \)
  (ii) \( \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CH}_2\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COOH} \)

  Question. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with \( \text{Br}_2 \) and KOH forms a compound ‘C’ of molecular formula \( \text{C}_6\text{H}_7\text{N} \). Write the structures and IUPAC names of compounds A, B and C.
  Answer:
  Compound C of formula \( \text{C}_6\text{H}_7\text{N} \) formed by Hoffmann bromamide degradation must be Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)).
  Therefore, B is Benzamide (\( \text{C}_6\text{H}_5\text{CONH}_2 \)).
  Compound A is Benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \)).
  Reaction:
  \( \text{C}_6\text{H}_5\text{COOH (A)} \xrightarrow{\text{NH}_3, \Delta} \text{C}_6\text{H}_5\text{CONH}_2 \text{ (B)} \xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_6\text{H}_5\text{NH}_2 \text{ (C)} \)

  3 MARKS QUESTIONS (AMINES)

  Question. Arrange the following:
  (i) In decreasing order of the \( \text{p}K_b \) values: \( \text{C}_2\text{H}_5\text{NH}_2 \), \( \text{C}_6\text{H}_5\text{NHCH}_3 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \) and \( \text{C}_6\text{H}_5\text{NH}_2 \)
  (ii) In increasing order of basic strength: \( \text{C}_6\text{H}_5\text{NH}_2 \), \( \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \) and \( \text{CH}_3\text{NH}_2 \)
  (iii) In increasing order of basic strength: Aniline, p-nitroaniline and p-toluidine
  Answer:
  (i) Lower basic strength means higher \( \text{p}K_b \). Basicity order: \( (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_6\text{H}_5\text{NH}_2 \).
  Decreasing order of \( \text{p}K_b \): \( \text{C}_6\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_2\text{H}_5\text{NH}_2 > (\text{C}_2\text{H}_5)_2\text{NH} \).
  (ii) Increasing order of basic strength: \( \text{C}_6\text{H}_5\text{NH}_2 < \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2 < \text{CH}_3\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} \).
  (iii) Increasing order of basic strength: p-Nitroaniline < Aniline < p-Toluidine.

  Question. Arrange the following
  (i) In decreasing order of basic strength in gas phase: \( \text{C}_2\text{H}_5\text{NH}_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \), \( (\text{C}_2\text{H}_5)_3\text{N} \) and \( \text{NH}_3 \)
  (ii) In increasing order of boiling point: \( \text{C}_2\text{H}_5\text{OH} \), \( (\text{CH}_3)_2\text{NH} \), \( \text{C}_2\text{H}_5\text{NH}_2 \)
  (iii) In increasing order of solubility in water: \( \text{C}_6\text{H}_5\text{NH}_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \), \( \text{C}_2\text{H}_5\text{NH}_2 \).
  Answer:
  (i) In gas phase (+I effect only): \( (\text{C}_2\text{H}_5)_3\text{N} > (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3 \).
  (ii) Increasing order of boiling points: \( (\text{CH}_3)_2\text{NH} < \text{C}_2\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{OH} \).
  (iii) Increasing order of solubility: \( \text{C}_6\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} < \text{C}_2\text{H}_5\text{NH}_2 \).

  Question. Accomplish the following conversions:
  (i) Nitrobenzene to benzoic acid
  (ii) Benzene to m-bromophenol
  (iii) Benzoic acid to aniline
  Answer:
  (i) \( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{H}_2/\text{Pd or Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl, 273-278 K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN/KCN}} \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{COOH} \)
  (ii) \( \text{Benzene} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{Nitrobenzene} \xrightarrow{\text{Br}_2/\text{Fe}} \text{m-Bromonitrobenzene} \xrightarrow{\text{Sn/HCl}} \text{m-Bromoaniline} \xrightarrow{\text{NaNO}_2/\text{HCl, 273 K}} \text{m-Bromobenzenediazonium chloride} \xrightarrow{\text{dil. H}_2\text{SO}_4, \Delta} \text{m-Bromophenol} \)
  (iii) \( \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{SOCl}_2} \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{NH}_3} \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{C}_6\text{H}_5\text{NH}_2 \)

  Question. Accomplish the following conversions:
  (i) Aniline to 2,4,6-tribromofluorobenzene
  (ii) Benzyl chloride to 2-phenylethanamine
  (iii) Chlorobenzene to p-chloroaniline
  Answer:
  (i) \( \text{Aniline} \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{2,4,6-Tribromoaniline} \xrightarrow{\text{NaNO}_2/\text{HCl, 273-278 K}} \text{2,4,6-Tribromobenzenediazonium chloride} \xrightarrow{\text{HBF}_4} \text{Diazonium fluoroborate} \xrightarrow{\Delta} \text{2,4,6-Tribromofluorobenzene} \)
  (ii) \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{ethanolic NaCN}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2 \)
  (iii) \( \text{Chlorobenzene} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{p-Nitrochlorobenzene} \xrightarrow{\text{H}_2/\text{Pd or Fe/HCl}} \text{p-Chloroaniline} \)

  Question. Accomplish the following conversions:
  (i) Aniline to p-bromoaniline
  (ii) Benzamide to toluene
  (iii) Aniline to benzyl alcohol.
  Answer:
  (i) \( \text{Aniline} \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}/\text{pyridine}} \text{Acetanilide} \xrightarrow{\text{Br}_2/\text{CH}_3\text{COOH}} \text{p-Bromoacetanilide} \xrightarrow{\text{H}^+ \text{or OH}^-} \text{p-Bromoaniline} \)
  (ii) \( \text{Benzamide} \xrightarrow{\text{Br}_2/\text{NaOH}} \text{Aniline} \xrightarrow{\text{NaNO}_2/\text{HCl, 273 K}} \text{Benzenediazonium chloride} \xrightarrow{\text{H}_3\text{PO}_2/\text{H}_2\text{O}} \text{Benzene} \xrightarrow{\text{CH}_3\text{Cl}/\text{anhyd. AlCl}_3} \text{Toluene} \)
  (iii) \( \text{Aniline} \xrightarrow{\text{NaNO}_2/\text{HCl, 273 K}} \text{Benzenediazonium chloride} \xrightarrow{\text{CuCN/KCN}} \text{Benzonitrile} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{Benzoic acid} \xrightarrow{\text{LiAlH}_4} \text{Benzyl alcohol} \)

  5 MARKS QUESTIONS (AMINES)

  Question. Give the structures of A, B and C in the following reactions:
  (i) \( \text{CH}_3\text{CH}_2\text{I} \xrightarrow{\text{NaCN}} \text{A} \xrightarrow[\text{Partial hydrolysis}]{\text{OH}^-} \text{B} \xrightarrow{\text{NaOH, Br}_2} \text{C} \)
  (ii) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{CuCN}} \text{A} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{B} \xrightarrow{\text{NH}_3, \Delta} \text{C} \)
  (iii) \( \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{LiAlH}_4} \text{B} \xrightarrow{0^\circ\text{C, HNO}_2} \text{C} \)
  (iv) \( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{A} \xrightarrow[273\text{ K}]{\text{NaNO}_2+\text{HCl}} \text{B} \xrightarrow[\Delta]{\text{H}_2\text{O}/\text{H}^+} \text{C} \)
  (v) \( \text{CH}_3\text{COOH} \xrightarrow[\Delta]{\text{NH}_3} \text{A} \xrightarrow{\text{NaOBr}} \text{B} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C} \)
  Answer:
  (i) \( \text{A} = \text{CH}_3\text{CH}_2\text{CN} \) (Propanenitrile), \( \text{B} = \text{CH}_3\text{CH}_2\text{CONH}_2 \) (Propanamide), \( \text{C} = \text{CH}_3\text{CH}_2\text{NH}_2 \) (Ethanamine).
  (ii) \( \text{A} = \text{C}_6\text{H}_5\text{CN} \) (Cyanobenzene), \( \text{B} = \text{C}_6\text{H}_5\text{COOH} \) (Benzoic acid), \( \text{C} = \text{C}_6\text{H}_5\text{CONH}_2 \) (Benzamide).
  (iii) \( \text{A} = \text{CH}_3\text{CH}_2\text{CN} \) (Cyanoethane), \( \text{B} = \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \) (Propan-1-amine), \( \text{C} = \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \) (Propan-1-ol).
  (iv) \( \text{A} = \text{C}_6\text{H}_5\text{NH}_2 \) (Aniline), \( \text{B} = \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \) (Benzenediazonium chloride), \( \text{C} = \text{C}_6\text{H}_5\text{OH} \) (Phenol).
  (v) \( \text{A} = \text{CH}_3\text{CONH}_2 \) (Ethanamide), \( \text{B} = \text{CH}_3\text{NH}_2 \) (Methanamine), \( \text{C} = \text{CH}_3\text{OH} \) (Methanol).

  Question. Complete the following reactions:
  (i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + \text{alc. KOH} \rightarrow \)
  (ii) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow \)
  (iii) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4(\text{conc.}) \rightarrow \)
  (iv) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_2\text{H}_5\text{OH} \rightarrow \)
  (v) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{Br}_2(\text{aq}) \rightarrow \)
  Answer:
  (i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH(alc)} \xrightarrow{\text{Carbylamine reaction}} \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \) (Phenyl isocyanide)
  (ii) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{H}_3\text{PO}_3 + \text{HCl} \)
  (iii) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{conc. H}_2\text{SO}_4 \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{HSO}_4^- \xrightarrow{453-473\text{ K}} \text{p-H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H} \) (Sulphanilic acid)
  (iv) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{CH}_3\text{CHO} + \text{N}_2 + \text{HCl} \)
  (v) \( \text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2(\text{aq}) \rightarrow \text{2,4,6-Tribromoaniline} + 3\text{HBr} \)

  ASSIGNMENTS - LEVEL 1

  Question. Write IUPAC Name of \( \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_3\text{Br} \) ?

  Question. Which reaction is used for preparation of pure aliphatic & aralkyl primary amine ?

  Question. Name one reagent used for the separation of primary, secondary & tertiary amine ?

  Question. What amine salts are used for determing their molecular masses ?

  Question. What is the directive influence of amino group in arylamines?

  Question. Why are benzene diazonium salts soluble in water ?

  Question. Which is more basic: \( \text{CH}_3\text{NH}_2 \) & \( (\text{CH}_3)_3\text{N} \) ?

  Question. Which is more acidic, aniline or ammonia ?

  Question. Write the IUPAC name of \( \text{C}_6\text{H}_5\text{NHCH}_3 \) ?

  Question. Mention two uses of sulphanilic acid?

  ASSIGNMENTS - LEVEL 2

  Question. What for are quaternary ammonium salts widely used ?

  Question. What product is formed when aniline is first diazotized and then treated with Phenol in alkaline medium ?

  Question. How is phenyl hydrazine prepared from aniline ?

  Question. What is the IUPAC name of a tertiary amine containing one methyl, one ethyl And one n-propyl group ?

  Question. Explain why silver chloride is soluble in aqueous solution of methylamine ?

  Question. Write the IUPAC name of \( \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_3\text{Br} \) ?

  Question. Primary amines have higher boiling points then tertiary amines why ?

  Question. Why is it necessary to maintain the temperature between 273 K & 278 K during diazotization?

  Question. Arrange the following in order of decreasing basic strength : Ethyl amine, Ammonia, Triethylamine ?

  Question. Why aniline is acetylated first to prepare mono bromo derivative?

  ASSIGNMENTS - LEVEL 3

  Question. Arrange the following in decreasing order of their basic strength: \( \text{C}_6\text{H}_5\text{NH}_2 \), \( \text{C}_2\text{H}_5\text{NH}_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \), \( \text{NH}_3 \).

  Question. Write chemical equation for the conversion \( \text{CH}_3\text{-CH}_2\text{-Cl} \) into \( \text{CH}_3\text{–CH}_2\text{-CH}_2\text{-NH}_2 \)

  Question. Write the equation involved in Carbylamines reactions?

  Question. How will you distinguish the following pairs?
  (i) Methanamine and N-methyl methane amine
  (ii) Aniline and ethyl amine

  Question. Write chemical sections involved in following name reactions:
  (i) Hoffmann Bromoamide reaction.
  (ii) Diazotisation reaction.

  COMMON ERRORS

  Basic character of amines in aqueous and in gaseous state, \( \text{p}K_a \) and \( \text{p}K_b \) values.

  1 MARK QUESTIONS (REVISION)

  Question. Arrange the following in decreasing order of their basic strength: \( \text{C}_6\text{H}_5\text{NH}_2 \), \( \text{C}_2\text{H}_5\text{NH}_2 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \), \( \text{NH}_3 \)

  Question. Arrange the following in decreasing order of the \( \text{p}K_b \) values: \( \text{C}_2\text{H}_5\text{NH}_2 \), \( \text{C}_6\text{H}_5\text{NHCH}_3 \), \( (\text{C}_2\text{H}_5)_2\text{NH} \) and \( \text{C}_6\text{H}_5\text{NH}_2 \)

  Question. \( \text{p}K_b \) of aniline is more than that of methylamine. Why?

  Question. Ethylamine is soluble in water whereas aniline is not. Give reason.

  Question. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Why?

  Question. Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. Give reason.

  Question. Aniline does not undergo Friedel-Crafts reaction. Why?

  Question. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Why?

  Question. Gabriel phthalimide synthesis is preferred for synthesising primary amines. Give reason

  Question. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

  Question. Why do primary amines have higher boiling point than tertiary amines?

  Question. Why are aliphatic amines stronger bases than aromatic amines?

  Question. Direct nitration of aniline is not carried out. Give reason.

  Question. The presence of base is needed in the ammonolysis of alkyl halides. Why?

  2 MARKS QUESTIONS (REVISION)

  Question. Write structures and IUPAC names of
  (i) the amide which gives propanamine by Hoffmann bromamide reaction.
  (ii) the amine produced by the Hoffmann degradation of benzamide.

  Question. Give one chemical test to distinguish between the following pairs of compounds.
  (i) Methylamine and dimethylamine (ii) Ethylamine and aniline

  Question. Write short notes on the following:
  (i) Carbylamine reaction (ii) Diazotisation

  Question. Explain the following with the help of an example.
  (i) Hofmann’s bromamide reaction (ii) Coupling reaction

  Question. Explain the following with the help of an example.
  (i) Ammonolysis (ii) Gabriel phthalimide synthesis

  Question. How can you convert an amide into an amine having one carbon less than the starting compound? Name the reaction.

  Question. Give a chemical test to distinguish between:
  (a) \( \text{C}_6\text{H}_5\text{NH}_2 \) & \( \text{CH}_3\text{NH}_2 \)
  (b) \( \text{CH}_3\text{NHCH}_3 \) & \( (\text{CH}_3)_3\text{N} \)

  Question. Give the IUPAC names of:
  (a) \( (\text{CH}_3)_2\text{CHNH}_2 \)
  (b) \( (\text{CH}_3\text{CH}_2)_2\text{NCH}_3 \)

  Question. Write the structures of:
  (a) 3-Bromobenzenamine
  (b) 3-Chlorobutanamide

  3 MARKS QUESTIONS (REVISION)

  Question. How will you convert
  (i) Benzene into aniline
  (ii) Benzene into N, N-dimethylaniline
  (iii) Aniline to Sulphanilic acid

  Question. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with \( \text{Br}_2 \) and KOH forms a compound ‘C’ of molecular formula \( \text{C}_6\text{H}_7\text{N} \). Write the structures and IUPAC names of compounds A, B and C.

  Question. How will you carry out the following conversions (Write Chemical equations and reaction conditions):
  (a) Aniline to Phenol
  (b) Acetamide to Ethylamine
  (c) Aniline to p-nitroaniline

• v) Coordination number: The no. of ligand donor atoms to which the metal is directly bonded through sigma bonds only. It is commonly 4 or 6.
• vi) Counter ions: The ionisable groups written outside the square bracket. Ex: \( \text{K}^+ \) in \( \text{K}_4[\text{Fe}(\text{CN})_6] \) OR \( 3\text{Cl}^- \) in \( [\text{Co}(\text{NH}_3)_6]\text{Cl}_3 \).
• vii) Coordination Polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central metal atom/ion. They are commonly Octahedral, Square-planar or Tetrahedral.
• Oxidation number: The charge that the central atom would carry if all the ligands are removed along with their pairs of electrons shared with the central atom. It is represented in parenthesis.
• viii) Homoleptic complexes: Complexes in which a metal is bonded to only one kind of donor groups. Ex: \( [\text{Co}(\text{NH}_3)_6]^{3+} \).
• ix) Heteroleptic complexes: Complexes in which a metal is bonded to more than one kind of donor groups. Ex: \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \).