Question Set C

cbse 12th Standard Chemistry Biomolecules Model Questions Papers Study Material - QB365

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Questions:

2 Marks

How will you convert 4-nitrotoluene to 2-bromobenzoic acid?

Give the IUPAC names of the following compounds:
(i) Ph CH₂CH₂COOH
(ii) (CH₃)₂C=CHCOOH
(iii)

(iv)

When liquid 'A' is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline the solid on treatment with sodium hydrogensulphite. Liquid 'B' also forms a white crystalline solid with sodium hydrogensulphite but it does not give test with ammoniacal silver nature. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.

What is glycogen ? How is it different from starch?

What is the rol of HNO₃ in the nitrating mixture used for nitration of benzene?

3 Marks

Give one chemical test each to distinguish between the compounds in the following pairs:
(i) Methyleamine and dimethylamine
(ii) Aniline and benzylamine
(iii) Ethylamine and aniline

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Define the following as related to proteins.
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give their common names.
(i) \(\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}\)
(ii) \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\)
(iii) \( \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}\)
(iv) \( \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\)

(vi) PhCOPh

10.

Account for the following.
(i) Cl-CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.

11.

Write the major products:

12.

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.

Multiple Choice Question

13.

Which is most acidic?

(a)

CCl₃COOH

(b)

CHCl₂COOH

(c)

CH₂ClCOOH

(d)

CH₃COOH

14.

In the following reaction, X is \(X\overset { Bromination }{ \longrightarrow } Y\overset { NaNO_{ 2 }\quad /HCI }{ \longrightarrow } Z\overset { Boiling }{ \underset { { C }_{ 2 }{ H }_{ 5 }OH }{ \longrightarrow } } Tribromobenzene\)

(a)

Benzoic acid

(b)

Salicylic acid

(c)

Phenol

(d)

Aniline

15.

Butan-2-one can be converted to propionic acid by which of the following :

(a)

NaOH. NaIH⁺

(b)

Fehling solution

(c)

NaOH,I₂/H⁺

(d)

Tollens'reagent

16.

DNA and RNA contain four bases each. which of the following bases is not present in RNA ?

(a)

Adenine

(b)

Uracil

(c)

Thymine

(d)

Cytosine

17.

The number of tripeptides formed by three different amino acids are

(a)

Three

(b)

Four

(c)

Five

(d)

Six

18.

Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known as

(a)

an enamine

(b)

a Schiff's base

(c)

an amine

(d)

an imine

19.

Which of the following statements about primary amines is 'False' ?

(a)

Alkylamines are stronger base than ammonia.

(b)

Alkylamines are stronger bases than arylamines

(c)

Alkylamines react with nitrous acid to produce alcohols

(d)

Arylamines react with nitrous acid to produce pheols

20.

Which of the following carboxylic acid can undergo decarboxylation easily?

(a)

(b)

(c)

(d)

C₆H₅COCOOH

21.

Which of the following is a polysaccharide?

(a)

Glucose

(b)

Maltose

(c)

Glycogen

(d)

Lactose

Fill up / 1 Marks

22.

Hell-Volhard Zelinsky reaction involves the replacement of an -------- atom from the alkyl group of a monocarboxylic acid by a ------ atom

23.

Vinegar is a dilute solution of --------

24.

Aniline on treatment with bromine water gives................................ .

25.

Ethanamine reacts with benzenesulphonyl chloride to form.........................which deissolves in.................. .

26.

Denaturation involves conversion of ........... proteins to ....... proteins

27.

Three types of RNA are _____________

5 Marks

28.

The K_a and K_b values of \(\alpha\) - amino acids are very low. Explain.

29.

(a) Write the products of the following reactions:

(b) Which acid of each pair shown here would you expect to be stronger?

30.

(a) Write the structure of the main products when benzene diazonium chloride reacts with the following reagents.
(i) Kl
(ii) CH₃CH₂OH
(iii) Cu/HCl
(b) Arrange the following in the increasing order of their basic character in aqueous solution: CH₃NH₂ , (CH₃ )₂NH, (CH₃ )₃N
(c) Give a simple test to distinguish between the following pair of compounds:
C₆H₅NH₂ and CH₃NH₂.

Passage Based Questions

31.

Aldehydes and ketones containing atleast one \(\alpha\)- H atom undergo a reaction in the presence of dilute alkali as catalyst to form \(\beta\)-hydroxy ketones. \(\beta\)- hydroxy aldehydes are called aldols while \(\beta\)-hydroxy ketones are collectively called ketols. When two different aldehydes or ketones combine then mixture of four products are formed. This reaction is called cross-aldol condensation.
Write the products formed in the following reaction.

32.

Carboxylic acids evolve hydrogen with metals and form salts with alkalies similar to phenols. However, unlike phenols, they react with weaker bases like sodium carbonate and hydrogen carbonate to evolve carbon dioxide. In aqueous solution, carboxylic acids ionise and exist in dynamic equilibrium between the resonance stabilised carboxylate ions and the hydronium ions.

Resonance stabilisation of carboxylate anion is more than that of undissociated carboxylic acid. Therefore, greater stability of carboxylate ion is responsible for the acidic character of carboxylic acids. Carboxylic acids are more acidic than alcohols because carboxylate anions are more stable than alkoxide ions, so carboxylic acids have strong tendency to release a proton.
Give reason, why monochloro ethanoic acid has a higher pK_a than dichloroethanoic acid?

33.

Amines are very reactive due to the difference in electro negativity between nitrogen and hydrogen atoms and due to the presence of unshared pair of electrons over N-atom. The number of hydrogen atoms attached to the N -atorn decides the course of reactions of amine, that is why amines differ in many reactions. In aromatic amines like aniline, electron density at ortho and para-positions with respect to -NH₂ group is high. Therefore, this group is ortho or para directing and a powerful activating group.
Aromatic amines are weaker bases than ammonia. Give reasons?

34.

Glucose is an aldohexose. It can occur freely as well as in combined form in the nature. It is present in sweet fruits and honey. It is also present in quantities in ripe grapes. As glucose is an aldohexose, it consists of six C-atoms and an aldehyde group. It is the most abundant organic compound on the earth and used as an immediate source of energy for all metabolic reactions in the animals.
It was found that glucose forms a six membered ring in which -OH at C-5 is involved in the ring formation. The cyclic six membered structure of glucose is known as pyranose structure. It is analogous to pyran which is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring.
What happens when glucose is treated with HI?

35.

Glucose is an aldohexose. It can occur freely as well as in combined form in the nature. It is present in sweet fruits and honey. It is also present in quantities in ripe grapes. As glucose is an aldohexose, it consists of six C-atoms and an aldehyde group. It is the most abundant organic compound on the earth and used as an immediate source of energy for all metabolic reactions in the animals.
It was found that glucose forms a six membered ring in which -OH at C-5 is involved in the ring formation.The cyclic six membered structure of glucose is known as pyranose structure. It is analogous to pyran which is a cyclic organic compound with one oxygen atom and five carbon atoms in the ring.
Write the structure of open-chain glucose.

Case Study Questions

36.

Read the passage given below and answer the following questions:
Carboxylic acids having an a-hydrogen atom when treated with chlorine or bromine in the presence of small amount of red phosphorus gives a-halo carboxylic acids. The reaction is known as Hell- Volhard-Zelinsky reaction.

When sodium salt of carboxylic acid is heated with soda lime it loses carbon dioxide and gives hydrocarbon with less number of C-atoms.

In these questions (i - iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
(i) Assertion: (CH₃)₃CCOOH does not give H.V.Z reaction.
Reason: (CH₃)₃CCOOH does not have \(\alpha\)-hydrogen atom.
(ii) Assertion: H.V.Z. reaction involves the treatment of carboxylic acids having \(\alpha\)-hydrogens with Cl₂ or Br₂in presence of small amount of redphosphorus.
Reason : Phosphorus reacts with halogens to form phosphorus trihalides.
(iii) Assertion: C₆H₅COCH₂COOH undergoes decarboxylation easily than C₆H₅COCOOH.
Reason : C₆H₅COCH₂COOH is a 13-ketoacid.
(iv) Assertion: On heating 3-methylbutanoic acid with soda lime, isobutane is obtained.
Reason: Soda lime is a mixture of NaOH + CaO in the ratio 3 : 1.

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Answers Key:

2 Marks

i) PhCH₂CH₂COOH
3-Phenylpropanoic acid
ii) (CH₃)₂C = CHCOOH
3-Methylbut-2-enoic acid
iii) 2-Methylcyclopentane carboxylic acid
iv) 2,4,6-Trinitrobenzoic acid

Since, the liquid A reduces ammoniacal silver nitrate (Tollen's reagent), hence A is aldehyde and B does not give test with ammnoniacal silver nitrate, thus, B is a ketone.
Further, B forms a white crystalline solid on treatment with sodium hydrogen sulphite. This suggests that B is a methyl ketone.

Starch is not a single compound but is a mixture of two components-a water soluble component called amylose (15-20%) and water insoluble component called amylopectin (80-85%). Amylose is a linear polymer of \(\alpha \)-D-glucose.But both glycogen and amylopectin are branched polymers of \(\alpha \)-D-glucose; rather glycogen is more highly branched than amylopectin. Whereas amylopectin chains consist of 20-25 glucose units, glycogen chains consist of 10-14 glucose units.

HNO acts as a base in the nitrating mixture (conc.HNO₃+conc.H₂SO₄). H₂SO₄ acts on HNO₃ to generate the electrophile, \({ NO }_{ 2 }^{ + }\) (nitronium ion)

3 Marks

(i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.
(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsbergs reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl). Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent amines, however, do not react with Hinsberg’s reagent.
(iii) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

Hinsberg's test is used for the identification of primary, secondary, and tertiary amines.
Hinsberg's reagent is benzenesulphonyl chloride (C₆H₅SO₂Cl).
It reacts differently with primary, secondary, and tertiary amines.
(i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.
Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.
(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.
Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.
(iii) Hinsberg's reagent does not react with tertiary amines.

(i) Peptide linkage. Peptide bond is formed by the condensation of two or more same or different n-amino acids. The condensation occurs between amino acids with the elimination of water. In this case, the carboxyl group of one amino acid and amino group of another amino acid get condensed with the elimination of water molecule.
The resulting \(\quad O\\ \quad \parallel \\ -C-NH-\) linkage is called peptide linkage. The formation of a dipeptide and the peptide

(ii) Primary structure. The primary structure of proteins gives the sequence in which the amino acids are linked in one or more polypeptide chains of proteins. This is shown below:

(iii) Denaturation. A process that changes the physical and biological properties of proteins without affecting the chemical composition of a protein is called denaturation. The denaturation is caused by certain physical or chemical treatments such as changes in pH, temperature, presence of some salts or certain chemical agents.

S.No.	Structure	IUPAC name	Common name
(i)	\(\mathrm{CH}_{3} \mathrm{CO}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}\)	Heptane- 2-one	Methyl n-pentyl ketone
(ii)	\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\)	4-Bromo-2-methylhexanal	⋎-bromo- ∝.-methyl caproaldehyde
(iii)	\( \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{CHO}\)	Heptanal	Heptanaldehyde
(iv)	\( \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\)	3-phenylprop -2-enal	β-phenylacrolein
(v)		Cyclopentane carbaldehyde	Cyclopentane carbaldehyde
(vi)	PhCOPh	Diphenylmethanone	Benzophenone

10.

(i) CI-CH₂COOH is a stronger acid than CH₃COOH. It is because --CI group exhibits - I-effect which makes the carboxylate ion more stable. Higher the stability of carboxylate ion, easier is the removal of proton from the carboxylic acid and stronger is the acid. In CH₃COOH, --CH₃ group has +I- effect which destabilised it. Hence, CH₃COOH is a weaker acid.
(ii) Carboxylic acids contain carbonyl group but do not show the reactions of carbonyl group such as nucleophilic addition reaction like aldehydes and ketones. Because of the presence of lone pair of electrons on the O-atom of OH group, the electrophilic character of carbonyl carbon by resonance decreases, hence the partial positive charge on carbonyl carbon atom is reduced and therefore, they do not show nucleophilic addition reactions.

11.

12.

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.

Multiple Choice Question

13.

(a)
CCl₃COOH

14.

(d)
Aniline

15.

16.

17.

(d)
Six

18.

An enamine

19.

Arylamines react with NHO₂ to form diazonium salts.

20.

21.

Fill up / 1 Marks

22.

()
alpha, chlorine or bromine

23.

()
acetic or ethanoic acid

24.

()
2, 4, 6-tribromoaniline

25.

()
N-ethylbenzenesulphonamide, aqueous NaOH solurion

26.

()
globular, fibrous

27.

()
m-RNA (Messenger RNA), t-RNA (Transfer RNA), r-RNA (Ribosomal RNA).

5 Marks

28.

The Ka and Kb values of a-amino acids are very low because, in a-amino acids, the acidic group is \({ NH }_{ 3 }\) instead of -COOH in carboxylic acids and basic group is -\({ COO }^{ - }\) instead of -\({ NH }_{ 2 }\) group in aliphatic amines. For example, the Ka and Kb values of glycine are \(1.6\times { 10 }^{ -10 }\) and \(2.5\times { 10 }^{ -12 }\) respectively.

29.

b) (i) F-CH₂COOH is stronger acid.
(ii) CH₃COOH is stronger acid.

30.

(b) (CH₃)₂NH > CH₃NH2 > (CH₃)₂N
(c) Add NaNO₂ and conc. HCI. Cool it to 0-5^o C. Add alkaline solution of phenol. C₆H₅NH₂ gives orange azo dye, whereas CH₃NH₂ does not.

Passage Based Questions

31.

32.

This is because dichloroethanoic acid is a stronger acid than mono chloroethanoic acid.

33.

Aromatic amines are weaker bases than ammonia due to electron withdrawing group nature of aryl group.

34.

n-hexane is formed.

35.

Open chain strcuture of glucose is

Case Study Questions

36.

(i) (a)
(ii) (c): Phosphorus converts a little .of the acid into acid chloride which is more reactive than the parent carboxylic acid. Thus, it is the acid chloride, not the acid itself, that undergoes chlorination at the \(\alpha\)-carbon.
(iii) (a): \(\beta\)-ketoacids are unstable acids. They readily undergo decarboxylation through a cyclic transition state.

(i) (b): All aliphatic aldehydes give red ppt. with
Fehling's solution, but ketones do not reduce Fehling's
solution.
(ii) (e): Aliphatic aldehydes reduce Fehling's solution,
but aromatic aldehydes do not.