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Biomolecules

Question 1. Glucose or succrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:

• A glucose molecule contains five -OH groups while a sucrose molecule contains eight -OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water.
• Hence, they are soluble in water.
• But, cyclohexane and benzene do not contain -OH groups. Hence, they cannot undergo II- bonding with water and as a result, they are insoluble in water.

Question 2. What are the expected products of the hydrolysis of lactose?
Answer:

Lactose is composed of β-D galactose and β-D glucose. Tims, on hydrolysis, gives β-D galactose and β-D glucose.

Question 3. How do you explain the absence of an aldehyde group in the pentaacctate of D-glucose?
Answer:

D-glucose reacts with hydroxylamine (NH₂, OH) to form an oxime because of the presence of an aldehydic (-CHO) group of carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH₂, OH to give an oxime.

But pentaacctate of D-glucose does not react with NH₂, OH. This is because pentaacctate does not form an open chain structure.

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Question 4. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer:

Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept, thus giving rise to a dipolar ion known as a zwitter ion.

• Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. However, halo-acids do not exhibit such dipolar behaviour.
• For this reason, the melting points and the solubility of amino acids in water are higher than those of the corresponding halo acids.

Question 5. Where does the water present in the egg go after boiling the egg?
Answer:

When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg. the water present in it is absorbed by the coagulated protein through H-bonding.

Question 6. Why cannot vitamin C be stored in our bodies?
Answer: Vitamin C cannot be stored in our body because it is water-soluble. As a result, it is readily excreted in the urine.

Question 7. What products would be formed when a nucleotide from DMA containing thymine is hydrolysed?
Answer: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine. β-D-2- deoxyribose and phosphoric acid are obtained as products.

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:

• A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine.
• Therefore, on hydrolysis of DNA. the quantity of adenine produced is equal of that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
• But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.

Question 9. What are monosaccharides?
Answer:

• Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified based on a number of carbon atoms and the functional group present in them.
• Monosaccharides containing an aldehyde group are known as aldose and those containing a keto group are known as ketose.
• Monosaccharides are further classified as triose, tetrose, pentose, Itexose, and heptose according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3  carbon atoms is called aldotriose.

Question 10. What are reducing sugars?
Answer: Reducing sugars are carbohydrates that reduce l’ehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.

Question 11. Write two main functions of carbohydrates in plants.
Answer:

Two main functions of carbohydrates in plants are:

1. Polysaccharides such as starch serve as storage molecules.
2. Cellulose, a polysaccharide, is used to build the cell wall.

Question 12. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose. maltose, galactose, fructose and lactose
Answer:

1. Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
2. Disaccharides: Maltose, lactose

Question 13. What do you understand by the term glycosidic linkage?
Answer:

• Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
• For example, in a sucrose molecule, two monosaccharide units, α- glucose and β- fructose, are joined together by a glycosidic linkage.

Question 14. What is glycogen? How is it different from starch?
Answer:

Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components – amylose (15-20%) and amylopectin (80-85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.

Question 15. What are the hydrolysis products of (1) sucrose and (2) lactose?
Answer: On hydrolysis, sucrose gives one molecule of α-D glucose and one molecule of β-D-fructose.

The hydrolysis of lactose gives β-D-galactose and β-D-glucose.

Question 16. What is the basic structural difference between starch and cellulose?
Answer:

Starch consists of two components – amylose and atm lopectin. Amylose is a long linear chain of α-D-(+)-glucose units joined by C₁– C₄ glycosidic linkage (α-link).

Amylopectin is a branched-chain polymer of α—D—glucose units, in which the chain is formed by C₁ – C₄ glycosidic linkage and the branching occurs by C₁ – C₆, glycosidic linkage.

On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁ – C₄ glycosidic linkage (β-link).

Question 17. What happens when D-glucose is treated with the following reagents?

1. HI
2. Bromine water
3. HNO₃

Answer:

When D-glucose is heated with III for a long time, n-hexane is formed.

When D-glucose is treated with Br₂ water, D-gluonic acid is produced.

On being treated with HNO₃, D-glucose gets oxidised to give saeeliaric acid.

Question 18. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

1. Aldehydes give 2, 4-DNP test. Schiff’s test, and react with NaHSO₄ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free-CHO group is absent from glucose.
3. Glucose exists in two crystalline forms -α and β. The α-form (m.p. 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p.=423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Question 19. What are essential and non-essential amino acids? Give two examples of each type.
Answer:

Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food. For example: valine and leucine Non-essential amino acids are also required by the human body, but they can be synthesised in the body.

For example: Glycine, and Alanine

1. Peptide linkage
2. Primary structure
3. Denaturation

Question 20. Define the following as related to proteins

1. Peptide linkage
2. Primary structure
3. Denaturation

Answer:

Peptide linkage: The amide for pied between the -COOH group of one molecule of an amino acid and the -NH₂, group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.

Primary structure: The primary structure of a protein refers to the specific sequence in which various amino acids are present in it. i.e. the sequence of linkage between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.

Denaturation: In a biological system, a protein is found to have a unique 3- dimensional structure and a unique biological activity. In such a situation, the protein is called native protein.

• However, when the native protein is subjected to physical changes such as changes in temperature or chemical changes such as changes in pH. its H-bonds are disturbed.
• This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation.
• During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.  One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.

Question 21. What are the common types of secondary structures of proteins?
Answer:

There are two common types of secondary .structure of proteins:

1. α- helix structure
2. β-pleated sheet structure

α- Helix Structure: In this structure, the Nil group of an amino acid residue forms H-bonds with the group of the adjacent turn of the right-handed screw (α-helix).

β-pleated Sheet Structure: This structure is called so because it looks like the pleated folds of draper). In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.

Question 22. What type of bonding helps in stabilising the a -helix structure of proteins?
Answer: The H-bonds formed between the -Ml group of each amino acid residue and the group of
the adjacent turns of the α-helix help in stabilising the helix.

Question 23. Differentiate between globular and fibrous proteins.
Answer:

Question 24. How do you explain the amphoteric behaviour of amino acids?
Answer: In an aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as a zwitter ion.

Therefore, in Zwitter ionic form, the amino acid can act both as an acid and as a base.

Thus, amino acids show amphoteric behaviour.

Question 25. What are enzymes?
Answer:

• Enzymes are proteins that catalyse biological reactions. They are very specific and catalyse only a particular reaction for a particular substrate. Enzymes are usually named after a particular substrate or class of substrate and sometimes after a particular reaction.
• For example, enzymes used to catalyse the hydrolysis of maltose into glucose are named as maltase.

Again, the enzymes used to catalyse the oxidation of one substrate with the simultaneous reduction of another substrate are named oxidoreductase enzymes, the name of an enzyme ends with ’-ase’.

Question 26. What is the effect of denaturation on the structure of proteins?
Answer:

• As a result of denaturation. globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation.
• secondary and tertiary-structured proteins get converted into primary-structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 27. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer:

Based on their solubility in water or fat. vitamins are classified into two groups.

1. Fat-soluble Vitamins: Vitamins that are soluble in fat and oils, but not in ‘water, belong to this group.
   • For example: Vitamins A, D, E, and K
2. Water-soluble Vitamins: Vitamins that are soluble in water belong to this group,
   • For example: B group vitamins (B₁, B₂, B₆, B₁₂, etc.) and vitamin C

However, biotin or vitamin H is neither soluble in water nor fat. Vitamin K is responsible for the coagulation of blood.

Question 28. Why are vitamin A and vitamin C essential to us? Give their important sources,
Answer:

• The deficiency of vitamin A leads to xerophthalmia (hardening of the cornea of the eye) and night blindness. The deficiency of vitamin C leads to scurvy (bleeding gums).
• The source of vitamin A is fish liver oil, carrots, butter, and milk. The sources of vitamin C are citrus fruits. amla, and green leafy vegetables.

Question 29. What are nucleic acids? Mention their two important functions.
Answer:

Nucleic acids are biomolecules found in the nuclei of all living cells, as one of the constituents of chromosomes. There are mainly two types of nucleic acids – deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Nucleic acids are also known as polynucleotides as they are long-chain polymers of nucleotides.

Two main functions of nucleic acids are:

1. DNA is responsible for the transmission of inherent characteristics from one generation to the next. This process of transmission is called heredity.
2. Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell, even though the proteins are synthesised by the various RNA molecules in a cell, the message for the synthesis of a particular protein is present in DNA.

Question 30. What is the difference between a nucleoside and a nucleotide?
Answer:

A nucleoside is formed by the attachment of a base to the 1′ position of the sugar.

Nucleoside = Sugar + Base

On the other hand, all three basic components of nucleic acids (i.e. pentose sugar, phosphoric acid, and base) are present in a nucleotide.

Question 31. The two strands in DNA are not identical bill are complementary. Explain.
Answer:

In the helical structure of DNA. the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosin forms hydrogen bonds with guanine, while adenine forms hydrogen bonds with thymine. As a result, the two strands are complementary to each other.

Question 32. Write the important structure and functional difference between DNA and RNA.
Answer: The structural differences between DNA and RNA are as follows:

The functional differences between DNA and RNA are as follows:

Question 33. What are the different types of RNA found in the cell?
Answer:

1. Messenger RNA (m-RNA)
2. Ribosomal RNA (rRNA)
3. Transfer RNA (t-RNA)

Question 34. What are hormones and write to its functions?
Answer:

Hormones are molecules that act as intercellular messengers. These are produced by endocrine glands in the body and are poured directly into the bloodstream which transports them to the site of action.

In terms of chemical nature, some of these are steroids, for example., estrogens and androgens: some are poly peptides for example insulin and endorphins and some others are amino acid derivatives such as epinephrine and norepinephrine.

1. Insulin keeps the blood glucose level within a narrow limit. Hormone glucagon tends to increase the glucose level in the blood. The two hormones together regulate the glucose level in the blood.
2. Epinephrine and norepinephrine mediate responses to external stimuli. Growth hormones and sex hormones play a role in growth and development. Thyroxine produced in the thyroid gland is an iodinated derivative of the amino acid tyrosine.
3. Steroid hormones are produced by the adrenal cortex and gonads (testes in males and ovaries in females).
4. Glucocorticoids control carbohydrate metabolism, modulate inflammatory reactions and are involved in reactions to stress.
5. Testosterone is the major sex hormone produced in males. It is responsible for the development of secondary male characteristics (deep voice, facial hair, general physical constitution) and estradiol is the main female sex hormone. It’s responsible for the development of secondary female characteristics and participates in the control of the menstrual cycle.
6. Progesterone is responsible for preparing the uterus for the implantation of a fertilised egg.

Amine

Question 1. Classify the following amines as primary, secondary, or tertiary:

Answer:

1. Primary: (1) and (3)
2. Secondary: (4)
3. Tertiary: (2)

Question 2. Write structures of different isomeric amines corresponding to the molecular formula. C₄H₁₁N

1. Write the IUPAC names of all the isomers.
2. What type of isomerism is exhibited by different pairs of amines?

Answer: (1),(2) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula. C₄H₁₁N are given below:

(3) The pairs (1) and (2). (5) and (7) exhibit position isomerism. The pairs (1) and (3); (2) and (d) exhibit chain isomerism. The pairs (5) (6) and (7) exhibit metamerism.

All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa.

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Question 3. How will you convert

1. Benzene into aniline
2. Benzene into N, N-dimethylaniline
3. Cl—(CH₂)₄-Cl into hexan-1, 6-diamine?

Answer:

Question 4. Arrange the following in increasing order of their basic strength:

1. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 \text { and }\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)
3. \(\mathrm{CH}_3 \mathrm{NH}_2,\left(\mathrm{CH}_3\right)_2 \mathrm{NH},\left(\mathrm{CH}_3\right)_3 \mathrm{~N}, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

Answer:

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
3. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_3 \mathrm{~N}<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{CH}_3\right)_2 \mathrm{NH}\)

Question 5. Complete the following acid-base reactions and name the products:

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2+\mathrm{HCl} \longrightarrow\)
2. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}+\mathrm{HCl} \longrightarrow\)

Answer:

Question 6. Write reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.

Answer: Aniline reacts with methyl iodide to produce N. N-dimethylaniline.

With excess methyl iodide, in the presence of Na₂CO₃ solution. N, N-dimethylaniline produces N, N, N-trimethylanilinum carbonate.

Question 7. Write the chemical reaction of aniline with benzyl chloride and write the name of the product obtained.
Answer:

Question 8. Write the structure of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers that will liberate nitrogen gas on treatment with nitrous acid.
Answer:

The structure of different isomers corresponding to the molecular formula, C₃H₉N are given below:

1 amine i.e., (1) propan-1-amine, (2) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid.

Question 9. Convert

1. 3-Methylaniline into 3-nitrotoluene.
2. Aniline into 1,3,5-tribromobenzene

Answer:

Question 10. Write chemical equations for the following reactions:

1. Reaction ofelhanolie NH₃ with C₂H₅CI.
2. Ammonolysis of benzyl chloride and the reaction of amine so formed with two moles of CH₃CI.

Answer:

Question 11. Write chemical equations for the following conversions:

1. \(\mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{CH}_3 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{Cl} \text { into } \mathrm{C}_6 \mathrm{H}_5 \cdot \mathrm{CH}_2 \cdot \mathrm{CH}_2 \cdot \mathrm{NH}_2\)

Answer:

Question 12. Write structures and IUPAC names of

1. The amide gives propanamide by Hoffmann bromamide reaction.
2. The amine produced by the Hoffmann degradation of ben/amide.

Answer:

Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. The structure and IUPAC name of the starling amide with four carbon atoms are given below:

Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is an aromatic primary amine-containing six carbon atoms.

Question 13. Arrange the following mg in decreasing order of their basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{NH}_3\)

Answer: The decreasing order of basic strength of the above amines and ammonia follows the following order:

⇒ \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

Question 14. How will you convert 4-nitrotoluene to 2-bromobenzoic acid?

Question 15. Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH} \mathrm{N} \mathrm{H}_2\)
2. \(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_2 \mathrm{NH}_2\)
3. \(\mathrm{CH}_3 \mathrm{NHCH}\left(\mathrm{CH}_3\right)_2\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{C} \mathrm{NH}_2\)
5. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3\)
6. \(\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NCH}_3\)
7. \(\mathrm{m}-\mathrm{BrCC}_6 \mathrm{H}_4 \mathrm{NH}_2\)

Answer:

1. Propan – 2 – amine (p)
2. Propan – I – amine(p)
3. N – methyl – 2 – propanamide (s)
4. 2 – methyl propane – 2 – amine (p)
5. N – methyl aniline (s)
6. N – ethyl – N – methyl ethanolamine (t)
7. 3 – Bromo benzydamine or 3 – Bromoaniline (p)

Question 16. Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amines
3. Ethylamine and aniline
4. Aniline and benzylamine
5. Aniline and N – methylaniline

Answer: Methylamine gives an arylamine reaction on heating with CHCl₃ and alcoholic KOH. Gives foul smelling (Carbyl amine)

whereas Dimethylamine does not give this reaction.

Secondary amines were given Kinsberg’s test. It gives insoluble substance with Ginsberg’s reagent (C₆H₅SO₂Cl) which is not affected by base.

• Whereas tertiary amine does not react with Hinsberg’s reagent.
• Aniline gives azo dye test : Dissolve Aniline in cone. HCl and add ice – a cold solution of HNO₂ (NaNO₂ + dil HCl) at 273 K and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aromatic amine (ie Aniline).
• Whereas aliphatic amine (ie ethylamine) does not form a dye. It will give brisk effervescence due to the evolution of N₂ but the solution remains clear.

Nitrous acid test: Benzvlainine reacts with nitrous acid (HNO₂) to form a diazonium salt which being unstable even at low temperatures, decomposes with the evolution of N₂ gas.

Aniline reacts with HNO₂ to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N₂ gas.

Carbylamine test: Aniline being a primary amine gives a carbylamine test. i.e. when heated with an alcoholic solution of KOH and CHCl₃. it gives an offensive smell of phenyl isocyanide. In contrast, N-methyl aniline. being secondary amine does not give this test.

Question 17. Account for the following:

1. pK_b of aniline is more than that of methylamine.
2. Ethylamine is soluble in water whereas aniline is not.
3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
4. Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
5. Aniline does not undergo Friedel-Crafts reaction.
6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
7. Gabrial phthalimide synthesis is preferred for synthesizing primary amines,

Answer:

1. In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene ring. Resulting, the electron density of the nitrogen decreases. On the other hand, in CH₃NH₂ +1 effect of CH₃ increases the electron density of the N-atom. Thus, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.

2. Ethylamine dissolves in water because it forms hydrogen bonds with water molecules. In aniline due to the large, hydrocarbon part, the extent of H-bonding decreases considerably, and hence aniline is insoluble in water.

3. Methylaminc being more basic than water, accepts a proton from water liberating OH^– ions.

These OHQ ions combine with Fe³⁺ ions present in H₂O to form a brown precipitate of hydrated ferric oxide.

⇒ \(\mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}^{+3}+3 \mathrm{Cl}^{-}\)

4. Nitration is usually carried out with a mixture of cones. HNO₃, and cone. H₂SO₄. In the presence of these acids, most of the aniline gets protonated to form anilingus ion. Thus in the presence of acids, the reaction mixture consists of aniline and anilinium ion. Now -NH₂ groups in aniline are o, p-directing, and activating while the NH⁺₃ group in anilinium ion is m-directing.

5. Aniline being a Lewis base, reacts with Lewis acid AlCl₃ to form a salt.

As a result, N of aniline acquires a positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction, consequently, aniline does not undergo Friedel – Crafts reaction.

6. The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to the dispersal of the positive charge on the benzene ring as shown below.

7. Gabriel’s phthalimide reaction gives pure primary amines without any contamination of secondary and tertiary amines. Therefore It is preferred for synthesising (aliphatic) primary amines.

Question 18. Arrange the following:

1. In decreasing order of the pKi, values :

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\)

2. In increasing order of basic strength:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} \text { and } \mathrm{CH}_3 \mathrm{NH}_2\)

3. In increasing order of basic strength:

1. Aniline, p-nitroaniline and p-toluidinc
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3, \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)

4. In decreasing order of basic strength in the gas phase:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 .\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} .\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N} \text { and } \mathrm{NH}_3\)

5. In increasing order of boiling point:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} .\left(\mathrm{CH}_3\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

6. In increasing order of solubilit> in water:

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \cdot\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH} . \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Answer:

1. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}-\mathrm{CH}_3>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
2. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2<\mathrm{CH}_3 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}\)
3. \(p \text {-nitroaniline }<\text { aniline }<p \text { – toluidine }\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3<\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2\)
5. \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3\)
6. \(\left(\mathrm{CH}_3\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
7. \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2\)

Question 19. How will you convert:

1. Ethanoic acid into methenamine
2. Hexamentirile into 1-aminopentane
3. Methanol to ethanoic acid
4. Ethanamine into methamine
5. Ethanoic acid into propionic acid
6. Methenamine into ethanolamine
7. Nitromethane into dimethylamine
8. Propanic acid into ethanoic acid

Answer:

Question 20. Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:

Benzene sulphonyl chloride (C₆H₅SO₂Cl), Which is also known as llinsberg’s reagent, reacts with primary and secondary amine to form sulphonamide.

The hydrogen attached to nitrogen in sulphone amide is strongly acidic due to the presence of a strong electron-withdrawing sulphonyl group. Hence it is soluble in alkali.

Tertiary amines do not react with benzene sulphonyl chloride.

This property of amines reacting with benzene sulphonyl chloride differently is used for the distinction of primary, secondary, and tertiary amines.

Question 21. Write short notes on the following:

1. Carbylamine reaction
2. Diazotisation
3. Hafmann’s bromamide reaction
4. Coupling reaction
5. Ammonolysis
6. Acetylation
7. Gabriel phthalimide synthesis

Answer:

Carbylamine Reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanide or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction, this reaction is known as the carby lamine reaction or idiocy and test and is used as a least for primary amine.

Diazotization reaction: When a cold solution of a primary aromatic amine in a dilute mineral acid (MCI or IfS()4) is treated with a cold solution of nitrous acid at 273 – 278 K. arene diazonium salt is formed. This reaction is called the diazotization reaction.

For example: \(\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow \mathrm{HNO}_2+\mathrm{NaCl}\)

Hoffmann’s Bromamide Reactions: The conversions of a primary amide to a primary amine-containing one carbon atom less than the original amide on heating with a mixture of Br₂ in the presence of NaOH or KOH is called Haffmann’s bromamide reaction, for example.

This reaction is extremely useful for converting a higher homolog to the next lower ho mo log lie.

Coupling Reaction: The reaction of diazonium salts with phenols and aromatic amines to form a/o compounds of the general formula Ar — N = N — Ar is called a coupling reaction. In this reaction, the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while with amines it occurs under faintly acidic conditions.

Animonolysis: The process of cleavage of the C – X bond by ammonia molecule is known as ammonolysis.

Acetylation: The process of introducing an acetyl group into a molecule is called acetylation. Common acetylating agents used arc acetyl chloride and acetic anhydride.

Gabriel Phthalimide Synthesis: In this reaction, phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N alkylphthalimide which is hydrolyzed to phthalic acid and primary amine by heating with HCl or KOH solution.

aromatic primary amines cannot be prepared by this method.

Question 22. Accomplish the following conversions:

1. Nirobenzene to benzoic acid
2. Benzene to m-bromophenol
3. Benzoic acid to aniline
4. Aniline to 2.4.6-tribromolluorobenxcne
5. Benzyl chloride to 2-phenylethanolamine
6. Chlorobenzene to p-chloroaniJine
7. Aniline to p-bromoaniline
8. Benzamide to toluene
9. Aniline to benzyl alcohol

Answer:

Question 23. Give the structure of A, B, and C in the following reaction:

Answer:

Question 24. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound C of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:

Question 25. Complete the following reactions:

Answer:

Question 26. Win aromatic primary amines can not be prepared by Gabriel phthalimide synthesis.
Answer:

The success of Gabriel’s phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.

As aryl halides do not undergo nucleophilic substitution reaction easily, aromatic primary amines cannot be prepared by Gabriel phthalimide reaction.

Question 27. Write the reactions of (1) aromatic and (2) aliphatic primary amines with nitrous acid.
Answer:

Question 28. Give a plausible explanation for each of the following:

1. Why are amines less acidic than alcohols of comparable molecular masses?
2. Why do primary amines have a higher boiling point than tertiary amines?
3. Why do aliphatic amines arc stronger bases than aromatic amines?

Answer:

Loss of a proton from an amine gives an amide ion (N^–H) while loss of a proton from alcohol gives an alkoxide ion (O^–R) as shown below.

As O is more electronegative than N, So negative charge on a more electronegative atom is more stable. As R—O is more stable than RNH^–. Therefore amines are less acidic than alcohols.

Due to the presence of two H-atoms on N-atoms of 1° amines, they undergo extensive intermolecular H-bonding while at 3° amines due to the absence of H-atoms on N-atom, there is no hydrogen bonding takes place. So primary amines have higher b.p. than tertiary amines of comparable molecular mass.

Inter Molecular Hydrogen Bonding In 1° Amines.

Aliphatic amines are stronger bases than aromatic amines because:

Due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less available for protonation.

The aryl amine ions have lower stability than the corresponding aliphatic amines i.e. protonation of aromatic amines is not favored.

Aldehyde Ketones And Carboxylic Acid

Question 1. Write the structure of the following reactions:

Answer:

Question 2. Arrange the following compounds in the increasing order of their boiling points. CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃

Answer:

Question 3. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Enthanal, Propanal, Propanone,, Butanone,
2. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Hint: Consider the steric effect and electronic effect.

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Answer: Reactivity of carbonyl compound α (+)ve change

The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

If the +I effect increases then (+)ve change on sp² decreases.

So the reactivity order is: Butanone < Propanone < Propanal < Ethanal

-I group on >C = O in reuses (+)ve charge so reactivity order is Acetophenone < p-tolualdchyde < Benzaldehvde < p-Nitrohenzaldehyde

Question 4. Predict the product of the following reactions:

Answer:

Question 5. Show how each of the following compounds can be converted to benzoic acid.

1. Ethylbenzene
2. Acetophenone
3. Bromobenzene
4. Phenylethane (Styrene)

Answer:

Question 6. Which of each pair shown here would you expect to be stronger?

1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { or } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { or } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { or } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
4. 

Answer: Acidic Strength

1. \(\mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H}_2 \text { < } \mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H}\)
2. \(\mathrm{CH}_2 \mathrm{FCO}_2 \mathrm{H} \text { > } \mathrm{CH}_2 \mathrm{ClCO}_2 \mathrm{H}\)
3. \(\mathrm{CH}_2 \mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{CO}_2 \mathrm{H} \text { < } \mathrm{CH}_3 \mathrm{CHFCH}_2 \mathrm{CO}_2 \mathrm{H}\)
4. 

Question 7. Give names of the reagents to bring about the following transformations:

1. Hexan-1-ol to hexanal
2. Cyclohexanol to cyclohexanone
3. p-FluorotoIuene to p-fluorobenzaldchydc
4. Ethanenitrile to ethanal
5. Allyl alcohol to propenal
6. But-2-ene to ethanal

Answer:

1. C₅H₅NH+CrO₃Cl(PCC)
2. K₂Cr₂O₇ in an acidic medium
3. CrO in the presence of acetic anhydride 1. CrO₂Cl₂ 2. HOH
4. (Diisobutyl) aluminium hydride (DIBAL-H)
5. PCC
6. O₃/H₂O-Zn dust

Question 8. Arrange the following compounds in the increasing order of their boiling points:

⇒\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}, \mathrm{H}_5 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_3, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

• The molecular masses of these compounds are in the range of 72 to 74. Since only butane-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest.
• Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former, n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows :

⇒\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3<\mathrm{H}_3 \mathrm{C}_2-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5\)<\(\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\)

Question 9. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Answer:

The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Question 10. An organic compound (A) with molecular formula CsHsO forms an orange-red precipitate with 2,4-DNP reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens’ or Foldings’ reagent nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.
Answer:

(A)forms 2,4-DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ or Fehling reagent, (A) must be a ketone. (A) responds to the iodoform test. Therefore, it should be a methyl ketone.

• The molecular formula of (A) indicates a high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent.
• This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid.
• The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Question 11. Write chemical reactions to affect the following transformations:

1. Butan-1-ol to butanoic acid
2. Benzyl alcohol to phenylethanoid acid
3. 3-Nilrobromobenzene to 3-nitrobenzoic acid
4. 4-Methylacetophenone to benzene-1,4-dicarboxylic acid
5. Cyelohexene to hexane-1,6-dioic acid
6. Butanal to butanoic acid.

Answer:

Question 12. What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Inline
9. 2,4- DNP-derivative
10. Schiffs base

Answer: Cyanohydrin: The organic compounds have -CN and -OH groups in the structure.

Acetal: Acetals are gem-dial koxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

Semicarbazone: Semicarbazone,s are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones with aH in the presence of a base.

Hemiacetal: Hemiacetal are a-alkoxyalcohols Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas.

Oxime: Oximes are a class of organic compounds having a general formula.

On treatment with hydroxyl amine in a weakly acidic medium, aldehydes or ketones form oximes.

Ketal: Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

Imine:

Iminces Imines are produced when aldehydes and ketones react with R- NH₂.

2, 4-DNP-derivative: 2, 4-dinitrophenyIhydrazones are 2, 4-DNP-derivatives, which are produced when aldehydes or ketones react with 2. 4-dinitrophenylhydrazine in a weakly acidic medium.

To identify and characterize aldehydes and ketones, 2, 4-DNP derivatives are used.

Sehiffs Base: Is >c = NR (R is not Hydrogen) (R may be -Ph)

Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of a trace of an acid yield a Schiffs base.

Question 13. Draw structures of the following derivatives.

1. The 2, 4-dinitrophenylhydrazone of benzaldehyde
2. Cyclopropane oxime
3. Acctaldehydedimcthylacctal
4. The semicarbazone of cyclobutanone
5. The ethylene ketal of hexane-3-one
6. The methyl hemiacetal of formaldehyde

Answer:

Question 14. Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents.

1. PhMgBr and then H₃O⁺
2. Tollen’s reagent
3. Semicarbazide and weak acid
4. Excess ethanol and acid
5. Zinc amalgam and dilute hydrochloric acid

Answer:

Question 15. Which of the following compounds would undergo aldol condensation, which is the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

1. Methanal
2. 2-Methylpentanal
3. Benzaldehyde
4. Benzophenone
5. Cyclohexanone
6. 1-Phenylpropanone
7. Phenylacetaldehyde
8. Bulan-1-ol
9. 2,2-Dimethylbutanal

Answer: Aldehydes and ketones having at least one a-hydrogen undergo aldol condensation (2), (5), (6), and (7) give aldol condensation.

• Aldehydes (only) having no a-hydrogen undergo Cannizzaro reactions, (1), (3), (9) give Cannizzaro reaction.
• Compound (4) is a ketone having no a-hydrogen atom and compound (8) Bulan-l-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or Cannizzaro reactions.

Aldol condensation:

Cannizaro Reaction:

Question 16. How will you convert cthanal into the following compounds?

1. Butane-1,3-diol
2. But-2-enal
3. But-2-enoic acid

Answer:

When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.

Question 17. Write structural formulas and names of four possible aids from propanal and butanal. In each case, indicate which aldehyde acts as a nucleophile and which is an electrophile.
Answer:

Take two molecules of propanal. one which acts as a nucleophile and the other as an electrophile.

Taking two molecules of butanal. one which acts as a nucleophile and the other as an electrophile.

Take one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal act as an electrophile.

Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Question 18. An organic compound (A) (molecular formula (C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-cue. Write equations for the reactions involved.
Answer: Hydrolysis of ester gives acid and alcohol

Question 19. Arrange the following compounds in increasing order of their property as indicated:

Acetaldehyde, Acetone, Di-tert-butyl ketone. Methyl tert-butyl ketone

⇒ \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{Br})\mathrm{COOH}\), \(\mathrm{CH}_3\mathrm{CH}(\mathrm{Br})\mathrm{CH}_2\mathrm{COOH},\left(\mathrm{CH}_3\right)_2\mathrm{CHCOOH},\mathrm{CH}_3 \mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}\)

Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid

⇒ \(\mathrm{CH}_3 \mathrm{COCl}, \mathrm{CH}_3 \mathrm{CONH}_2, \mathrm{CH}_3 \mathrm{COOCH}_3,\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}\)

Answer: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

4-Methoxybezo

Question 20. Give simple chemical tests to distinguish between the following pairs of compounds.

1. Propanal and Propanonc
2. Acetophenone and Bcnzophenonc
3. Phenol and Benzoic acid
4. Benzoic acid and Ethyl benzoate
5. Pentan-2-one and Pentan-3-one
6. Benzaldehyde and Acetophenone
7. Ethanal And Propanal

Answer: 1. By Tollen’s test

Propanal reduces Tollen’s reagent, Fehling solution, and Benedict solution but propanonc (ketone) does not. reduces. Iodoform test: Propanone gives an iodoform test but propanal (CH₃-CH₂-CHO) does not.

2. Acetophenone and Benzophenone can be distinguished using the iodoform test. Acetophenone gives an iodoform test but benzophenone does not.

3. Phenol and benzoic acid can be distinguished by the ferric chloride test and NaHCO₃, test.

1. Phenol gives a violet colour with FeCl₃,
2. Benzoic acid gives CO₂ with NaHCO₃ while phenol does not.

4. Benzoic acid and Ethyl benzoate can be distinguished by the sodium bicarbonate test. Sodium bicarbonate test: Benzoic acid reacts with NaHCO₃ to produce brisk effervescence due to the evolution of CO₂ gas but ethylbenzoate does not.

5. Pentan-2-one and pentan-3-one can be distinguished by iodoform lest. Pentan-2-one gives an iodoform test but pentan-3-one does not.

6. Benzaldehyde and acetophenone can be distinguished by (1) Tollen’s test and (2) the Iodoform test. Benzaldehyde gives Tollen’s test whereas acetophenone gives an iodoform test.

7. Ethanal and propanal can be distinguished by the iodoform test. Ethanal gives an iodoform test but propanal does not.

Question 21. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic regent having not more than one carbon atom

1. Methoxybenzoic
2. M-Nitrobenzoic acid
3. P-Nitrobenzoic acid
4. Phenylacetic acid
5. P-Nitrobenzaldehyde

Answer:

Question 22. How will you about the following conversions in more than two steps?

1. Propanone to Propene
2. Benzoic acid to Benzaldehyde
3. Ethanol to 3-Hydroxybutyanal
4. Benzene to m- Nitroacetophenone
5. Benzaldehyde to Benzophenone
6. Bromobenze to 1-Phenylethanol
7. Benzaldehyde to 3-Phenylpropan-1-ol
8. Benzaldehyde to α-Hydroxyphenylacetic acid

Answer:

Question 23. Describe the following:

1. Acetylation
2. Carnnizzaro reaction
3. Cross aldol condensation
4. Decarboxylation

Answer: Acetylation: To introduce acetyl groupin an organic compound.

Carnnizzaro reaction: The self-oxidation-reduction (disproportionation) reaction of aldehydes having no ex-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

Cross-aldol condensation: The reaction between two different aldehydes, or two different ketones, or an aldehyde and a ketone (having α-H) in the presence of a base is called a cross-aldol condensation.

Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Question 24. Complete each synthesis by giving missing starting material, reagent or products.

Answer:

Question 25. Give a plausible explanation for each of the following:

1. Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6-trimethyl cyclohexanone does not.
2. There are two, groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
3. During the preparation of esters, a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Answer: Cyclohexanones form cyanohydrins according to the following equation.

In 2, 2. 6 – trimethyl cyclohexanone, methyl groups at α-positions offer steric hindrance and as a result.

CN^– cannot attack effectively.

2, 2, 6 – trimethyl cyclohexanone,

For this reason, it does not form a cyanohydrin.

The electron density on the -NH₂, group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other -NH₂, the group is not involved in resonance: it can act as a nucleophile and can attack carbonyl – carbon atoms of aldehydes and ketones to produce semicarbazone.

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

Question 26. An organic compound contains 69.77% carbon. 11.63% hydrogen and the rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an additional compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:

% of carbon = 69.77%

% of hydrogen = 11.63%

% of oxygen = [100-(69.77 + 11.63)]% = 18.6%

Thus, the ratio of the number of carbon, hydrogen and oxygen atoms in the organic compound can be given as:

⇒ \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{69.77}{12}: \frac{11.63}{1}: \frac{18.6}{16}=5.81: 11.63: 1.16=5: 10: 1\)

Therefore, the empirical formula of the compound is C₅H₁₀O. Now, the empirical formula mass of the compound can be given as:

5 × 12 +10 × 1 + 1 × 16 = 86

Molecular mass of the compound = 86 (given)

Therefore, the molecular formula of the compound is given by C₅H₁₀O.

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.

Hence, the given compound is plan-2-one.

The given reactions can be explained by the following equations:

Question 27. Although phenoxide ion has more number of resonating structures than carboxylate ion. carboxylic acid is a stronger acid than phenol. Why?
Answer: Resonance structures of phenoxide ion are:

Unequivalent resonating structures

Equivalent resonating structures are more stable.

Alcohol Phenol And Ether

Question 1. Classify the following as primary, secondary, and tertiary alcohols:

Answer:

1. Primary alcohol > (1), (2), (3)
2. Secondary alcohol > (4), (5)
3. Tertiary alcohol > (6)

Question 2. Identify allylic alcohols in the above examples.
Answer:

• Allylic that means C = C – C – OH
• The alcohols given in (2) and (6) are allylic.

Question 3. Name the following compounds according to the IUPAC system.

Answer:

1. 3-Chloromethyl-2-isopropylpentan-1-ol
2. 2, 5 – Dimelhylhcxane – 1,3- cliol
3. 3-Bromocyclohexanol
4. Hex-1 -en-3-ol
5. 2-Bromo-3-methylbut-2-en-1 -ol

Question 4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal.

Answer:

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 5. Write structures of the products of the following reactions:

Answer:

Question 6. Give structures of the products you would expect when each of the following alcohols reacts with

1. HCl-ZnCl₂
2. HBr and
3. SOCl₂
4. Bulan-1-ol
5. 2-Methylbutan-2-ol

Answer:

Tertiary alcohols react immediately with Lucas’ reagent.

Question 7. Predict the major product of acid-catalyzed dehydration of

1. 1-methylcyclohexane and
2. Butan-1-ol

Answer:

Question 8. Ortho para nitrophenols are more than phenol. DFrawn the resonance structure of the corresponding phenoxide ions.
Answer:

(Resonance structure of the phenoxide ion).

(Resonance structure of p-nitro phenoxide ion)

(Resonance structure of o-nitro phenoxide ion)

It can be observed that the presence of the nitro group increases the stability of phenoxide ions.

Question 9. Write the equations involved in the following reactions:

1. Reimer – Tiemann reaction
2. Kolbe – Schmidt reaction

Answer:

Question 10. Write the reactions of Williamson synthesis of 2-ethoxy-3-inethylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4- nitrobenzene and why?

Answer: Set (2) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzcne.

In set (1), the Bromine atom is stable by resonance.

Question 12. Predict the products of the following reactions:

Answer:

Question 13. Give IUPAC names of the following compounds:

Answer:

1. 4-ChIoro-2,3-dimethylpentan-1-oI
2. 2-Ethoxy propane
3. 2,6-Diinethylphenol
4. 1-Elhoxy-2-nilrocyclohexane

Question 14. You will notice that the reaction produces a primary alcohol with methanol, a secondary alcohol with other aldehydes, and a tertiary alcohol with ketones. Give the structures and IUPAC names of the products expected from the following reactions:

1. Catalytic reduction of butane.
2. Hydration of propene in the presence of dilute sulphuric acid.
3. Reaction of propanone with methyl magnesium bromide followed by hydrolysis.

Answer:

Question 15. Arrange the following sets of compounds in order of their increasing boiling points:

1. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
2. Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Answer:

1. Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
2. n-Butane, ethoxyethane, pentanal, and pentan-1-ol.

Question 16. Arrange the following compounds in increasing order of their acid strength: Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.
Answer: Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 17. Write the structures of the major products expected from the following reactions:

1. Mononitration of 3-methyl phenol
2. Denitration of 3-methyl phenol
3. Mononitration of phenyl methanoate.

Answer: The combined influence of -OH and -CH₃ groups determines the position of the incoming group.

Question 18. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

1. What would be the major product of this reaction?
2. Write a suitable reaction for the preparation of t-butyl ethyl ether.

Answer:

1. The major product of the given reaction is 2-methyl prop-1-one.

This is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

2. Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.

Question 19.

Answer:

Question 20. Write IUPAC names of the following compounds:

Answer:

1. 2, 2, 4-Trimethylpcntan-3-ol
2. 5-Ethylheplane-2, 4-diol
3. Butane-2, 3-diol
4. Propane-1, 2, 3-triol
5. 2-Methylphenol
6. 4-Methylphenol
7. 2. 5-Dimcthylphenol
8. 2, 6-Dimethylphenol
9. 1-Methoxy-2-methylpropane
10. Ethoxyben/ene
11. 1-Phenoxyheptane
12. 2-Elhoxvbutane

Question 21. Write structures of the compounds whose IUPAC names are as follows:
Answer:

1. 2-Methylbutan-2-ol
2. 1-Phenyl-2-ol
3. 3,5- Dimethylhexene-1,3,5-triol
4. 2,3-Diethylphenol
5. 1-Ethoxypropane
6. 2-Ethoxy-3-methyl pentane
7. Cyclohexylmethanol
8. 3-Cyclohexylpentan-3-ol
9. Cyclopent-3-en-1-ol
10. 3-Chloromethylpentan-1-ol

Answer:

Question 22. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O and give their IUPAC names. Classify them as primary, secondary, and tertiary alcohols.
Answer:

1. Primary alcohols – (1), (2)(3), and (4)
2. Secondary alcohols – (5),(6) and (7)
3. Tertiary alcohol – (8)

Question 23. Explain why propanol has a higher boiling point than that of the hydrocarbon, butane.
Answer:

Propanol undergoes intermolecular H-bonding because of the presence of the -OH group. On the other hand, butane does not

Question 24. Explain why are alcohols comparatively more soluble in water than the corresponding hydrocarbons.
Answer:

Hydrogen bonding exists between alcohol and water molecules while hydrocarbons are nonpolar, they can not form an H-bond with water molecules.

Question 25. What is meant by hydroboration oxidation reaction? Illustrate it with an example.
Answer:

Question 26. Give the structures and IUPAC names of monohydric phenols of molecular formula C₇H₈O.
Answer:

Question 27. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile. Give reason.
Answer: O-nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Question 28. Give the equation of reaction of preparation of phenol from cumene.
Answer:

Question 29. Write a chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification.

Question 30. You are given a benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Question 31. Show how will you synthesize:

1. 1-phenylethanol from a suitable alkane.
2. CyelohexyI methanol using an alkyl halide by an S_N2 reaction
3. Pentan-1-ol using a suitable alkyl halide?

Answer:

Question 32. Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:

The acidic nature of phenol can be represented by the following two reactions.

1. Phenol reacts with sodium to give sodium phenoxide, liberating H₂

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-products.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas the ethoxide ion does not.

Question 33. Explain why is ortho-nitro phenol more acidic than ortho-methoxy phenol.
Answer: Electron withdrawing effect of nitro group and electron releasing effect of methoxy group.

Question 34. Explain how the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution.
Answer: Electron density at the o- and p-position increases and thus the electrophile attacks easily.

Question 35. Give equations of the following reactions:

1. Oxidation of propane-1 -ol with KMnO₄ solution.
2. Bromine is CS₂ with phenol.
3. Dilute HNO₃ with phenol.
4. Treating phenol with chloroform in the presence of aqueous NaOH.

Answer:

Question 36. Explain the following with an example.

1. Williamson ether synthesis.
2. Unsymmetrical ether.

Answer:

1. Williamson’s ether synthesis:

Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxide.

This reaction involves an S_N2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in the case of primary alkyl halides.

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

2. Unsymmetrieal ether:

An unsymmetrical ether is an ether where two alkyl groups on the two sides of an oxygen atom differ. For example: ethyl methyl ether (CH₃ -O-CH₂CH₃).

Question 37. How are the following conversions carried out?

1. Propene → Propan-2-ol
2. Benzyl chloride → Benzyl alcohol
3. Ethyl magnesium chloride → Propan-1-ol.
4. Methyl magnesium bromide → 2-Methylpropan-2-ol,

Answer:

Question 38. Give a reason for the higher boiling point of ethanol in comparison to methoxyrnethane.
Answer:

• Ethanol undergoes intermolecular H-bonding due to the presence of the -OH group, resulting in the association of molecules.
• On the other hand, methoxyrnethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxyrnethane.

Question 39. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer: The reaction of Williamson synthesis involves an S_N2 attack of an alkoxide ion on a primary alkyl
halide.

• But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced.
• This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.

Question 40. How is 1-propoxypropane synthesized from propan- 1-ol? Write the mechanism of this reaction.

Answer:

The mechanism of this reaction involves the following three steps:

Step 1: Protonation

Step 2: Nucleophilic attack

Step 3: Deprotonation

Question 41. The preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

• The formation of ethers by dehydration of alcohol is a bimolecular reaction (S_N2) involving the attack of an alcohol molecule on a protonated alcohol molecule.
• In this method, the alkyl group should be unhindered. In the case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of others, alkenes are formed.

Question 42. Write the equation of the reaction of hydrogen iodide with:

1. 1-propoxypropane
2. Methoxybcnezcnc and
3. Benzyl ethyl ether

Answer:

Question 43. Explain the fact that in aryl alkyl ethers

1. The alkoxy group activates the benzene ring towards electrophilic substitution and
2. It directs the incoming substituents to ortho and para positions in the benzene ring.

Answer:

In arly alkyl ethers, due to the + R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the alkoxy group.

2. It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 44. Write the mechanism of the reaction of III with methoxymethyl.
Answer:

The mechanism of the reaction of III with methoxymethane involves the following steps:

Step 1: Protonation of methoxymelthane:

Step 2: Nucleophilic attack of I^–

Question 45. Write equations of the following reactions:

1. Friedel-Crafts reaction-alkylation of anisole.
2. Nitration of anisole.
3. Bromination of anisole in ethanoic acid medium.
4. Friedcl-Craft’s acetylation of anisole.

Answer:

Question 46. When 3-methyl butane-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.

Answer: The mechanism of the given reaction involves the following steps:

Step 1: Protonation

Step 2: Formation of 2° carbonation by the elimination of a water molecule

Step 3: Re-arrangement by the hydride-ion shift

Step 4: Nucleophilic attack

Haloalkanes And Haloarenes

Question 1. Write The structure of the following compounds:

1. 2-Cliloro-3-methylpentane
2. 1-Chloro-4-methylcyclohexane
3. 4-terl. ButyI-3-iodohcplane
4. 1, 4-DibmmobiU-2- one
5. 1-Bromo-4-sec, butyl-2-melhylbenzene

Answer:

Question 2. Draw the structure of major monohalo products in each of the following reactions:

Answer:

Question 3. Arrange each set of increasing boiling points.

1. Bromomethane, Bromoform, Chloromethane, Dibromethane.
2. 1-chloride, Isopropyl chloride, 1-chlorobutane.

Answer:

Read and Learn More Class 12 Chemistry with Answers Chapter Wise

Question 4. Which alkyl halide from the following pairs would you accept to react more rapidly by an S_N2 mechanism? Explain your answer.

Answer:

Question 5. In the following pairs of halogen compounds, which compound undergoes a faster S_N1 reaction?

Answer:

Question 6. Identify A, B, C, D, E, R, and R¹ in the following:

Answer:

Question 7. Write the IUPAC name of the following:

Answer:

1. 4-Bromopent-2-ene
2. 3-Bromo-2-methylbut-1-ene
3. 4-Bromo-3-methyl pent-2-ene
4. 1-Bromo-2-methylbut-2-ene
5. 1-Bromobut-2-ene
6. 3-Bromo-2-methylpropene

Question 8. Write the product of the following reactions:

Answer:

Question 9. Haloalkanes react with KCN to form alkyl cyanides as the main product while AgCN forms isocyanides as the chief product. Explain.
Answer:

• KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond.
• However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

Question 10. In the following pairs of hydrogen compounds, which would undergo SN2 reaction faster?

Answer:

• It is a primary halide and therefore undergoes S_N2 reaction faster.
• As iodine is a better-leaving group because of its large size, it will be released at a faster rate in the presence of income nucleophiles.

Question 11. Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions:

1. The four isomeric bromobutan.es
2. C₆H5CH₂Br, C₆H₅CH(C₆H₅)BR, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}<\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3<\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{SN}_{\mathrm{N}} 1\right)\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Br}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{Br}_2 \mathrm{CH}_3>\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}\left(\mathrm{S}_{\mathrm{N}} 2\right)\right.\)

Question 12. Although chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why?
Answer: Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Question 13. Name the following halides according to the IUPAC system and classify them as alkyl, alkyl, benzene(primary, secondary, tertiary), vinyl, or aryl halide:

1. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_3\)
2. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{Cl}\)
3. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{I}\)
4. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{C}_6 \mathrm{H}_5\)
5. \(\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CH}_2 \mathrm{Br}\)
7. \(\mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{CH}_2 \mathrm{CH}_3\)
8. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}(\mathrm{Cl}) \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
9. \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2\)
10. \(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
11. \(\mathrm{m}-\mathrm{ClCH}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_3\)
12. \(\mathrm{O}-\mathrm{Br}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:

Question 14. Give the IUPAC names of the following compounds

1. \(\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
2. \(\mathrm{CHF}_2 \mathrm{CBrClF}\)
3. \(\mathrm{ClCH}_2 \mathrm{C} \equiv \mathrm{CCH}_2 \mathrm{Br}\)
4. \(\left(\mathrm{CCl}_3\right)_3 \mathrm{CCl}\)
5. \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4\right)_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
6. \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}=\mathrm{CClC}_6 \mathrm{H}_4 \mathrm{I}-\mathrm{p}\)

Answer:

Question 15. Write the structures of the following organic halogen compounds.

1. 2-Chloro-3-methyl pentane
2. p-Bromochlorobenzene
3. l-Chloro-4-ethylcyelohexane
4. 2-(2-Chlorophenyl)-1-iodooctane
5. 2-Bromobutane
6. 4-tert-butyl-3-iodoheptane
7. 1-Bromo-4-sec-butyl-2-methyl benzene
8. 1,4-Dibromobut-2-ene

Answer:

2-Chloro-3-methyl pentane

p-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane

2-(2-ChlorophenyI)-1-idodooctane

2-Bromobutane CH

4-Tert-butyl-3-iodoheptane

1-Bromo-4-sec-butyl-2-methyl benzene

1,4-Dibromobut-2-ene

Question 16. Which of the following has the highest dipole moment?

1. CH₂Cl₂
2. CHCl₃
3. CCl₄

Answer:

CCl₄ < CHCl₃ < CH₂Cl₂

Question 17. A hydrocarbon C₅H₁₀ does not react with chlorine in the dark but gives a single monochrome compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:

The reactions involved in the question are:

Question 18. Write the isomers of the compound having the formula C₄H₉Br.
Answer: There are four isomers of the compound having the formula Chi I- Hr. These isomers are given below.

7. Write the equations for the preparation of 1-iodobutane from

1. 1-butanol
2. 1-chlorobutane
3. but-1-ene

Answer:

Question 19. What are ambident nucleophiles? Explain with an example.
Answer: Ambident nucleophiles are nucleophiles having two nucleophilic sites. For example, nitrite ion is an ambident nucleophile.

Question 20. Which compound in each of the following pairs will react faster in S_N2 reaction with Oil?

1. CH₃Br or CH₃I
2. (CH₃)₃CCl or CH₃Cl

Answer: CH₃-I is more reactive because I⁺ is a better-leaving group than Br⁺

In the case of (CH₃)₃CCl. the attack of the nucleophile at the carbon atom is hindered by the presence of bulk)- substituents on that carbon atom bearing the leaving group.

Question 21. Predict all the alkanes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.

1. 1-Brorao-1-methylcyclohexane
2. 2-Chloro-2-methyl butane
3. 2,2, 3-Trirpethyl-3-bromopentane

Answer:

Question 22. How will you bring about the following conversion?

1. Ethanol to but-1-yne
2. Ethane to broiuoethene
3. Propene to I-nitropropane
4. Toluene to benzyl alcohol
5. Propene to propyne
6. Ethanol to ethyl fluoride
7. Bromomethane to propanone
8. But-1-cue to but-2-ene
9. 1-Chlorobutane to n-octane
10. Benzene to biphenyl

Answer:

Question 23. Explain why

1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
2. Alkyl halides, though polar, are immiscible with water?
3. Grignard reagents should be prepared under anhydrous conditions.

Answer:

+M of -Cl and I of -Cl are in opposite directions so the dipole moment is decreased in chlorobenzene.

1. No H-bonding of halide with water.
2. Grignard reagents in the presence of moisture, react with H₂O to give alkanes.

Question 24. Give the uses of freon 12. DDT, carbon tetrachloride, and iodoform.
Answer:

• Uses of Freon-12 Freon-12 (dichlorodifluoromethane. CF₂Cl₂) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.
• The use of DDT (p-p-dichlorodiphenyltrichloroethane) is one of the best-known insecticides.

Uses of carbonletrachloride (CCI₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cans.
2. It is used as a solvent in the manufacture of pharmaceutical products.
3. It is used as a fire extinguisher.

Uses of iodoform (CHI₃) Iodoform was used earlier as an antiseptic. The antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 25. Write the structure of the major organic product in each of the following reactions.

Answer:

Question 26. Write the mechanism of the following reaction:

Answer: The given reaction is:

The given reaction is an S_N2 reaction. In this reaction. CM acts as the nucleophile and attacks the carbon atom to which Br is attached. CN ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 27. Arrange the compounds of each set in order of reactivity towards S_N2 displacement:

1. 2-Bromo-2-metln Ihutane. 1-Bromopentane, 2-Bromopentane
2. 1-Bromo-3-methyl butane.2-Bromo-2-methyl butane.3-Bromo-2-inethyIhutane
3. 1-Bromobutane. l-Bromo-2. 2-dimethyIpropane. 1-Bromo-2-methylbutane. 1-Bromo 3-methyIhutane

Answer:

(more branching at a nearer distance)

Hence, the increasing order of reactivity of the given compounds towards S_N2 is:

1-Bromo-2,2-dimethy Ipropane < 1-Bromo-2-methy Ihutane < 1-Bromo-3-methylbutane < 1-Bromobutane.

Question 28. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolyzed by aqueous KOH?
Answer: C₆H₅CHClC₆H₅ (more reactive)

(more stable intermediate)

Question 29. p-Dichlorobenzene has higher m.p. and lower solubility than those of o-and m-isomers. Discuss.
Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, molecules of p-dichlorobenzene are more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.

Question 30. How the following conversions can be carried out?

1. Propene to propane-1-ol
2. Ethanol to but-1-yne
3. 1-Bromopropane to 2-bromopropane
4. Toluene to benzyl alcohol
5. Benzene to 4-bronitobenzene
6. Benzyl alcohol to 2-phenylethanoid acid
7. Ethanol to propane nitrile
8. Aniline to chloro benzene
9. 2-Chlorobutane to 3, 4-dimethyl hexane
10. 2-Methyl-1-propane to 2-chloro-2-methyl propane
11. Ethyl chloride to propanoic acid
12. But-1-ene to n-butyliodide
13. 2-chloropropane to 1-propanol
14. Isopropyl alcohol to iodoform
15. Chlorobenzene to p-nitrophenol
16. 2-Bromopropane to 1-bromopropane
17. Chloroethane to butane
18. Benzene to diphene
19. tert-Butyl bromide to isobutyl bromide
20. Aniline to phenyl isocyanide

Answer:

Question 31. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products, Explain.
Answer:

OH^– ion is highly solvated in an aqueous solution and as a result, the basic character of OH^– ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon. So

⇒ \(\mathrm{R}-\mathrm{Cl}+\mathrm{Aq} \cdot \mathrm{KOH} \longrightarrow \mathrm{R}-\mathrm{OH}+\mathrm{KCl}\)

On the other hand, an alcoholic solution of KOFI contains an alkoxide (RO^–) ion. which is a strong base. Thus, it can abstract hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

Question 32. Primary alkyl halide C₄H₉Br(a) reacted with alcoholic KOH to given compound (b). Compound (b) is reacted with lIBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:

Two primary alkyl halides have the formula. C₄H₉Br. They are n-butyl bromide and isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide.

Now, compound (a) reacts with Na metal to give compound (h) of molecular formula. C₈H₁₈. Which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence. compound (a) must be isobutyl bromide.

Thus, compound (d) is 2. 5-Dimethylhexane.

It is given that compound (a) reacts with alcoholic KOH to give compound (b). compound (b) is 2-methyl propene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence compound (e) is -2-Bromo-2-methy propane.

Question 33. What happens when

1. n-butyl chloride is treated with alcoholic KOH?
2. Broinohen/ene is treated with Mg in the presence of dry ether.
3. Chlorobenzene is subjected to hydrolysis.
4. Ethyl chloride is treated with aqueous KOH.
5. Methyl bromide is treated with sodium in the presence of dry ether.
6. Methyl chloride is treated with KCN.

Answer:

Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.