Question Set A

CBSE 12th Standard Chemistry Coordination Compounds HOT Questions Study Material - QB365

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Questions:

Generic 2

6 x 2 = 12

Give reasons:
(i) Electrophilic substitution in aromatic amines takes place more readily than benzene.
(ii) CH₃CONH₂ is a weaker base than CH₃CH₂.NH₂.

How will you bring about the following conversions?
(i) Nitrobenzene to Phenol.
(ii) Aniline to Chlorobenzene.

How will you distinguish 1^oand 2^o hydroxyl groups present in glucose? Explain with reactions.

Why does compound(A) given below not form an oxime?

CoSO₄Cl. 5NH₃ exists in ywo isomeric forms 'A' and 'B' . Isomer 'A' reacts with AgNO₃ to give white precipitate, but does not react with BaCl₂. Isomer 'B' gives white precipitate with BaCl₂ but does not react with AgNO₃. Answer the following questions.
(i) Identify 'A' and 'B' and write their structural formulae.
(ii) Name the type of isomerism involved.
(iii) Give the IUPAC name of 'A' and 'B'.

What is the relationship between observaed colour of the complex and the wavelength of light absorbed by the complex?

Generic 3

8 x 3 = 24

Label the glucose and fructose units in the following disaccharide and identify anomeric carbons atoms in these units. Is the sugar reducing in nature? Explain.

Write chemical equations involved when aniline is treated with the following reagents:
(i) Br₂(aq)
(ii) CHCI3 + KOH
(iii) HCI

(i) Draw the geometrical isomers of complex [Pt(en)₂Cl₂]²⁺. Which of them is optically inactive?
(ii) On the basis of crystal field theory write the electronic configuration for d⁴ ion, if \(\triangle\)₀ > p.
(iii) Write the hybridization type and magnetic behaviour of the complex [Fe(H₂O)₆]²⁺.

10.

Explain the shape and magnetic behaviour of [Fe(CN)₆]^4-.

11.

Write the structures and names of all the stereo isomers of the following compounds
(i) [Co(en)₂] Cl₂]⁺
(ii) [Pt (NH₃)₂CI₂]
(iii) [Fe(NH₃)4CI₂]CI

12.

Give the structures of A, B and C in the following reaction:
(a) \({ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }\overset { Fe/HCI }{ \longrightarrow } A\overset { { HNO }_{ 2 };273K }{ \longrightarrow } B\overset { { C }_{ 6 }{ H }_{ 5 }{ OH } }{ \longrightarrow } C\)
(b) \({ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }CI\overset { CuCN }{ \longrightarrow } A\overset { { H }_{ 2 }O/{ H }^{ + } }{ \longrightarrow } B\overset { { NH }_{ 3 };\triangle }{ \longrightarrow } C\)

13.

(i) Write the structural difference between starch and cellulose.
(ii) What type of linkage is present in Nucleic acids?
(iii) Give one example each for fibrous protein and globular protein.

14.

Answer the following queries about proteins:
(i) How are proteins related to amino acids?
(ii) How are oligopeptides different from polypeptides?
(iii) When is a protein said to be denaturated?

Multiple Choice Question

12 x 1 = 12

15.

The correct name of the given reaction is
\(\mathrm{Ar}-\mathrm{N}_2^{+} \mathrm{X}^{-} \xrightarrow[\text { Cu powder }]{\mathrm{HBr}} \mathrm{Ar}-\mathrm{Br}+\mathrm{N}_2\)

(a)

Hofmann bromamide degradation reaction

(b)

Gabriel phthalimide synthesis

(c)

Carbylamine reaction

(d)

Gattermann reaction

16.

\(^{\prime} \mathrm{A}^{\prime}\stackrel{\text { Reduction }}{\longrightarrow}{ }^{\prime} \mathrm{B}^{\prime} \stackrel{\mathrm{HNO}_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
The compound' A' is

(a)

propane nitrile

(b)

ethane nitrile

(c)

nitro methane

(d)

methyl isocyanate

17.

Correct increasing order of wavelength of absorption in visible region for cornplex of Co³⁺ is

(a)

\( {\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}} \)

(b)

\({\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}} \)

(c)

\({\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+},\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}} \)

(d)

\({\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_{6}\right)_{6}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}} \)

18.

Which of the following vitamin is responsible for beri-beri disease?

(a)

(b)

B₁

(c)

(d)

19.

Which of the following amine will form stable diazonium salt at 273-278K?

(a)

C₂H₅NH₂

(b)

C₆H₅NH₂

(c)

C₆H₅CH₂NH₂

(d)

C₆H₅N(CH₃)₂

20.

Which of the following is not a fat soluble vitamin?

(a)

Vitamin B complex

(b)

Vitamin D

(c)

Vitamin E

(d)

Vitamin A

21.

Method by which aniline cannot be prepared is

(a)

degradation of benzamide with bromine in alkaline solution

(b)

reduction of nitrobenzene with H₂/Pd in ethanol

(c)

potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

(d)

hydrolysis of phenylisocyanide with acidic solution

22.

Hydrolysis of sucrose is called

(a)

inversion

(b)

esterification

(c)

hydration

(d)

saponificatio

23.

Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present ?

(a)

5' and 3'

(b)

1' and 5'

(c)

5' and 5'

(d)

3' and 3'

24.

Optical isomerism is exhibited by (ox = oxalateanion, en = ethylene diamine)

(a)

cis-[CrCl₂(ox)₂]^3-

(b)

[Co(en)₃]³⁺

(c)

trans-[CrCl₂(ox)₂]^3-

(d)

[Co(ox)(en)₂]⁺

25.

The non-existent metal carbonyl among the following is

(a)

Cr(CO)₆

(b)

Mn(CO)₅

(c)

Ni(CO)₄

(d)

Fe(CO)₅

26.

Which of the following is an outer orbital complex?

(a)

[Fe(CN)₆]⁴^-

(b)

[FeF₆]^3-

(c)

[Co(NH₃)₆]³⁺

(d)

[Co(CN)₆]²⁺

Generic 5

6 x 5 = 30

27.

(a) Write the structures of the main products of the following reactions:

(b) Give a simple chemical test to distinguish between Aniline and N,N-dimethylaniline.
(c) Arrange the following in the increasing order of their pK_b values:
C₆H₅NH₂ , C₂H₅NH₂ , C₆H₅NHCH₃

28.

(a) Write the structures of main products when benzene diazonium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}\right)\)reacts with the following reagents:
(i) HBF₄/\(\triangle\)
(ii) Cu/HBr
(b) Write the structures of A, Band C in the following reactions:
\(\text {(i) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2} \stackrel{\mathrm{Sn} / \mathrm{HCl}}{\longrightarrow} \mathrm{A} \frac{\mathrm{NaNO}_{2}+\mathrm{HCl}}{273 \mathrm{~K}} \longrightarrow \mathrm{B} \frac{\mathrm{H}_{2} \mathrm{O}}{\Delta} \mathrm{C}\)
\(\text {(ii) } \mathrm{CH}_{3} \mathrm{Cl} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{B} \frac{\mathrm{HNO}_{2}}{273 \mathrm{~K}} \mathrm{C}\)

29.

(i) Define:
(a) Metal carbonyls
(b) Chelate
(ii) (a) Give one chemical test as an evidence to show that [Co(NH₃)₅CI]SO₄ and [Co(NH₃)₅(SO₄)] Cl are ionisation isomers.
(b) [NiCl₄]^2- is paramagnetic while [Ni(CO₄] is diamagnetic though both are tetrahedral. Why? (Atomic no. of Ni = 28)
(c) Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field ligand, and (ii) weak field ligand. (Atomic no. of Fe = 26)

30.

Account for the following:
(i) pK_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.

31.

An aromatic compound 'A' of molecular formula C₇H₆O₂ undergoes a series of reactions as shown below. Write the structures of A, B, C,D and E in the following reactions.

32.

Co-ordination compounds have an important role in the field of medicines. Out of these, the compound cis-platin is quite effective against cancer. It inhibits the growth of tumors leading to cancer.
(i) What is the chemical formula and name of the complex?
(ii) How does it behave as an anti cancer agent?
(iii) Why is trans-isomer not effective?
(iv) What is the value associated with the use of cis platin?

Assertion and reason

4 x 1 = 4

33.

Assertion : [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless.
Reason : d-d transition is not possible in [Sc(H₂O)₆]³⁺.
Codes:
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

34.

In the following questions. an Assertion (A) is followed by a corresponding Reason (R) Use the following keys to choose the appropriate answer.
Assertion (A) Fresh tomatoes are better source of vitamin C than those which have been stored for sometime.
Reason (R) On prolonged exposure to air, vitamin C is destroyed due to aerial oxidation.
Codes:
(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A).
(c) (A) is correct; (R) is incorrect.
(d) (A) is incorrect; (R) is correct.

35.

In the following questions. an Assertion (A) is followed by a corresponding Reason (R) Use the following keys to choose the appropriate answer.
Assertion (A) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.
Reason (R) Aryl halides do notundergo electrophilic substitution with anion formed by phthalimide.
Codes:
(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A).
(c) (A) is correct; (R) is incorrect.
(d) (A) is incorrect; (R) is correct.

36.

In the following questions. an Assertion (A) is followed by a corresponding Reason (R) Use the following keys to choose the appropriate answer.
Assertion (A) Oxidation number of Cr in \(\begin{equation} \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right] \mathrm{Cl}_{3} \end{equation}\) is same as the charge of the complex ion, +3.
Reason (R) All the ligands are neutral molecules in this compound.
Codes:
(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A).
(c) (A) is correct; (R) is incorrect.
(d) (A) is incorrect; (R) is correct.

Case Study Questions

2 x 4 = 8

37.

Amines are usually formed from nitro compounds, halides, amides, imides, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing,steric and hydrogen bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines.
In aromatic amines, electron releasing and withdrawing groups, respectively increase and decrease their basic character. Influence of the number of hydrogen atoms at nitrogen atom on the type of reactions and nature of products is responsible foridentification and distinction between primary, secondary and tertiary amines.
Presence of amino group in aromatic ring enhances reactivity of the aromatic amines. Aryl diazonium salts provide advantageous methods for producing aryl halides, cyanides, phenols and arencs by reductive removal of the diazo group.
Answer the following questions
(i) Arrange the following in the increasing order of their pK_b values in aqueous solutions:
\(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}\)
(ii) Aniline on nitration gives a substantial amount of m-nitroaniline though amino group is o/p directing. Why?
(iii) An aromatic compound 'A' of molecular formula C₇H₆O₂ on treatment with aqueous ammonia and heating forms compound 'B'. Compound 'B' on heating with Br₂ and aqueous KOH gives a compound 'C' of molecular formula C₆H₇N. Write the structures of A, B and C.
Or
(iii) Complete the following reactions giving main products:

38.

Carbohydrates are polyhydroxy aldehydes or ketones and are also called saccharides, Glucose is an example of monosaccharides. GIucose (C₆H₁₂O₆) is an aldohexose and its open chain structure was assigned on the basis of many reactions as evidences like presence of carbonyl group, presence of straight chain, presence of five -OH groups etc.
Glucose is correctly named as D-(+)-glucose. Glucose is found to exist in two different crystalline forms which are named as \(\alpha\) and \(\beta\). Despite having the aldehyde group, glucose does not give 2, 4-DNP test.
(i) What is the correct structure of D-(+)-glucose?
(ii) Glucose on oxidation with HNO₃ gives a dicarboxylic acid called saccharic acid. What does this result indicate?
(iii) The pentaacetate of glucose does not react with H₂N-OH What does this result indicate?
(iv) What is the difference between \(\alpha\)-D-glucose and \(\beta\)-D-glucose?
Or
Give the possible explanation for the following:
Glucose doesn't give 2,4-DNP test.

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Answers Key:

Generic 2

6 x 2 = 12

(i) 一NH₂ group of aromatic amines strongly activates the aromatic ring through delocalization of the lone pair of electrons of the N-atom over the aromatic ring. Due to the strong activating effect of the 一NH₂ group, aromatic amines undergo electrophilic substitution reactions readily than. benzene.
(ii) Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is delocalized over the keto group.
As a result, electron density on the N-atom in CH₃CONH₂ decreases. On the other hand, in C₂H₅NH₂ , due to + I effect of the ethyl group, the electron density on the N-atom increases consequently, CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.

Glucose consists of 5 - OH groups. Among these, the - OH group present on the terminal carbon atom (i.e. C-6 atom) is called 1° hydroxyl group while all the four remaining -OH groups present on C-2, C-3, C-4 and C-5 atoms are called 2° hydroxyl groups. 1° hydroxyl groups are easily oxidised to carboxylic acid group, while 2° hydroxyl groups undergo oxidation only under drastic conditions.
e.g. On oxidation with nitric acid, glucose gives a dicarboxylic acid, saccharic acid (also called glycaric acid) having the same number of carbon atoms as glucose. This indicates that glucose contains one primary (1°) alcoholic or hydroxyl group.

It is because it does not have free aldehyde group, therefore it does not orm oxime.

\((i)\ \left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }{ SO }_{ 4 } \right] Cl+{ AgNO }_{ 3 }(aq)\longrightarrow AgCl(s)+\left[ Co{ \left( { NH }_{ 3 } \right) }{ SO }_{ 4 } \right] { NO }_{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 'A'\quad \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad white\ ppt\)
\( \left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { SO }_{ 4 }+{ BaCl }_{ 2 }\longrightarrow { BaSO }_{ 4 }(s)+\left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 'B'\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad white\ ppt\)
(ii) Ionisation isomerism
(iii) 'N' is pentaamrnine chloridocobalt(III) sulphate.
'B' is pentaammine sulphatocobalt(III) chloride.

When white light falls on the complex, some part of it is absorbed, Higher \({ \triangle }_{ o }\) (crystal field splitting energy) lower will be the wavelength absorbed by the complex. The colour of the complex is the colour from the wavelength left over (complementry colour).

Generic 3

8 x 3 = 24

C-1 of glucose unit and C-2 of fructose unit are anomeric carbon atoms in the given disaccharide. The disaccharide is non-reducing sugar because -OH groups attached to anomeric carbon atoms are involved in the formation of glycosidic bond.

(i)

(ii) \(\triangle\)_o > P
\(\therefore \quad \mathrm{t}_{2 g}^{4} \mathrm{e}_{g}^{0}\)
(iii) Ni(28): [Ar]4s²3d⁸
Ni²⁺: [Ar] 4s^o3d⁸
CN^- is a strong field ligand, will cause pairing of electrons in [Ni(CN)₄]^2-.

It has dsp² hybridization and diamagnetic in nature.

10.

Hybridization : d² sp³
Shape : Octahedral Magnetic
character : Diamagnetic.

11.

(i) [Co(en)₃] Cl₃
IUPAC name:
Tris - (ethane - l,2- diamine) cobalt (III) chloride.

(ii) Pt(NH₃)₂CI₂
(ill) [Fe(NH₃)₄CI₂]Cl
IUPACname
Tetraamminedichloridoiron (II) chloride
Isomers Geometrical isomers (cis and trans)

12.

(i) \(A-{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 },\quad B-{ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }^{ + }C{ I }^{ - },C-{ C }_{ 6 }{ H }_{ 5 }{ -N }_{ 2 }-{ C }_{ 6 }{ H }_{ 4 }OH\)
(ii) \(A-{ C }_{ 6 }{ H }_{ 5 }CN,\quad B-{ C }_{ 6 }{ H }_{ 5 }NCOOH,\quad C-{ C }_{ 6 }{ H }_{ 5 }{ CONH }_{ 2 }\)

13.

(i) In starch, the glucose monomers are in alpha configuration while in cellulose the glucose monomers are in beta configuration. Starch is a polymer consisting of amylose and amylopectin while cellulose is a long chain composed only of
β-D-glucose units.
(ii) Phosphodiester linkage between the 5' and 3'atoms is present in nucleic acids.
(iii) Example of fibrous protein-Collagen, keratin, myosin.
Example of globular protein-Insulin, haemoglobin, egg albumin.

14.

(i) Proteins are biopolymers of amino acids which condense together forming peptide bonds.
(ii) Oligopeptides on hydrolysis give 3 to 12 \(\alpha \)-amino acids, where as polypeptides give large number of \(\alpha \)-amino acids on hydrolysis.
(iiI) When tertiary structure of protein is ruptured by,protein p^H or on heating, protein is said to be denatured.It lose its biological activity.

Multiple Choice Question

12 x 1 = 12

15.

(d)
Gattermann reaction

16.

(b)
ethane nitrile

17.

(d)
\({\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{NH}_{6}\right)_{6}\right]^{3+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}} \)

18.

(b)
B₁

19.

(b)
C₆H₅NH₂

20.

(a)
Vitamin B complex

21.

Aniline cannot be prepared by Gabriel phthalimide method because chlorobenzene (i.e.,aryl halide) does not undergo nucleophilic substitution reaction with potassium phthalimide in aqueous NaOH solution.

22.

(a)
inversion

23.

(a)
5' and 3'

24.

(a)
cis-[CrCl₂(ox)₂]^3-

25.

(b)
Mn(CO)₅

26.

(b)
[FeF₆]^3-

Generic 5

6 x 5 = 30

27.

(b) Carbylamine test: Add CHCl₃and KOH to each of them, aniline will give offensive smelling compound whereas N, N-Dimethyl aniline will not react
(c) C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂ is increasing order of their pKb due to decreasing order of basic character.

28.

29.

(i) (a) Metal carbonyls: Metal carbonyls are the compounds in which carbon monoxide (CO) acts as the ligand. The metal-carbon bond in metal carbonyls possesses both \(\sigma\) and \(\pi\) characters, e.g. N(CO)₄[nickeltetracarbonyl]. Its IUPAC name is tetra carbonyl nickel(0).
(b) Chelate: An inorganic metal complex in which there is a close ring of atoms caused by attachment of a ligand to a metal atom at two points. An example is the complex ion formed between ethylene diamine and cupric ion,[Cu(NH₂CH₂CH₂NH₂)₂]²⁺.

(ii) (a) Add BaCl₂ solution. [Co(NH₃)]₅Cl]SO₄ will give white ppt. of BaSO₄.
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}+\mathrm{BaCl}_{2}(a q) \)\( \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}+\mathrm{BaSO}_{4} \downarrow \)
  \(\text { (White ppt.) }\)
Add AgNO₃(aq) solution [Co(NH₃)₅SO₄] Cl will give white ppt. of AgCl.
\( {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Cl}+\mathrm{AgNO}_{3}(a q)} \) \(\longrightarrow \mathrm{AgCl}+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{NO}_{3} \)
  \(\text { (White ppt) }\)
(c) Fe³⁺ (4s⁰3d⁵) In strong field \( \Delta_{0}>P t_{2 g}^{5} e g^{0} \)
In weak field \(\Delta_{0}

30.

(i) pKb of aniline is more than that of methylamine.
In aniline, the lone pair of electrons on N atom is in resonance with benzene ring. Hence, it cannot be easily donated to an acid. This decreases its basicity. In methyl amine, the +I effect of methyl group increases the electron density on N atom so that the lone pair of electrons on N atom can be easily donated to an acid. Hence, methylamine is more basic than aniline. Higher is the basicity, lower is the pKb and vice versa.
(ii) Ethylamine is soluble in water whereas aniline is not. With increase in the molecular weight, the solubility decreases. Aniline has higher molecular weight than ethylamine.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Due to the +I effect of −CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH− ions by accepting H+ ions from water. OH− ions react with ferric chloride to precipitate hydrated ferric oxide.
(iv) The nitration of aniline is carried out using conc. HNO₃ and H₂SO₄. However, in the presence of conc. H₂SO₄, aniline forms aniline hydrogen sulphate in which the anilinium ion, C₆H₅NH₃⁺ is meta directing because the positive charge on the nitrogen attracts electrons from the benzene ring.
(v) Aniline does not undergo Friedel craft's reactions because the reagent AlCl₃(the Lewis acid which is used as a catalyst in friedel crafts reaction), being electron deficient acts as a Lewis base. and attacks on the lone pair of nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceed.
(vi) In aromatic diazonium salts, due to resonance, there is dispersal of positive charge on benzene ring.
But in aliphatic diazonium salts, resonance is not possible, so aliphatic diazonium salts are less stable than aromatic diazonium salt.
(vii) The reaction between alkyl halide (R- X) and ammonia gives mixture, of 1, 2, and 3 amines with tetra alkyl ammonium halide. The mixture is not separable. To get pure 1 amine, Gabriel phthalimide synthesis is preferred.

31.

32.

(i)The chemical formula of the compound is shown. Its name is cis-diammine dichlorido platinum (II).
(ii) It behaves as an anti cancer agent by taking part in the chelate formation or chelation.
(iii) In the trans isomer the similar groups are on opposite sides and chelate formation does not take place.
(iv) Cis-platin saves many valueable lives by forming co-ordinate complex. So it is life saving.

Assertion and reason

4 x 1 = 4

33.

(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

34.

(a) Both (A) and (R) are correct and (R) is correct explanation of (A).

35.

(c) Aromatic primary amines cannot be prepared by the Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide. Thus, (A) is correct but (R) is incorrect

36.

(a) Oxidation number of Cr in \(\begin{equation} \left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\right] \mathrm{Cl}_{3} \end{equation}\) is same as the charge ofthe complex ion, i.e. +3 because all the ligands are neutral molecules in this compound. Both (A) and (R) are correct and (R) is correct explanation of (A).

Case Study Questions

2 x 4 = 8

37.

(i) \(\begin{aligned} \left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \end{aligned}\)

38.

(ii) Glucose on oxidation with HNO₃ gives saccharic acid. This indicates the presence of a one primary alcoholic (-OH) group in glucose.
(iii) The pentaacetate of glucose does not react with hydroxylamine which shows the absence of free aldehydic (-CHO) group.
(iv) In \(\alpha\)-D-glucose, the -OH group at C₁ is towards right, whereas in \(\beta\)-D-glucose, the -OH group at C₁ is towards left. Such a pair of stereoisomers which differ in the configuration only at C₁ are called anomers.

Or
Actually, glucose exists in the cyclic hemiacetal form (hence the aldehyde group is not free) with only a small amount (<0.05%) of the open chain form.
Since, the concentration of the open chain form is low and its reaction with 2,4-DNP is reversible, therefore, formation of 2,4-DNP derivative cannot disturb the equilibrium to regenerate more open chain form from the cyclic hemiacetal form and hence, does not give this test.