CBSE Class 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 14 Biomolecules. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 14 Important Extra Questions Biomolecules

Biomolecules Important Extra Questions Very Short Answer Type

Question 1.
Write the structure of the product obtained when glucose is oxidised with nitric acid. (CBSE 2012)
Answer:
Saccharic acid (or glucaric acid) is formed.

Question 2.
Write a reaction that shows that all the carbon atoms in glucose are linked in a straight chain. (CBSE 2012)
Answer:
When glucose is heated with HI and red P at 100°C for a long period, it gives n-hexane.

The formation of n-hexane suggests that all six carbon atoms in glucose are arranged in a straight chain.

Question 3.
What are the hydrolysis products of sucrose? (CBSE 2019C)
Answer:
Glucose and fructose

Question 4.
Write the name of the linkage joining two amino acids. (CBSE 2013)
Answer:
Peptide linkage

Question 5.
What are the hydrolysis products of lactose? (CBSE2019C)
Answer:
Lactose on hydrolysis gives β – D – galactose and β – D – glucose.

Question 6.
What are three types of RNA molecules which perform different functions? (CBSE Delhi 2013)
Answer:
(i) messenger RNA: m – RNA
(ii) ribosomal RNA: r – RNA
(iii) transfer RNA: t – RNA

Question 7.
What type of bonding helps in stabilising the a-helix structure of proteins? (CBSE Delhi 2013)
Answer:
Hydrogen bonding between -NH and -C = 0 groups of peptide bonds.

Question 8.
What is a glycosidic linkage? (CBSE Delhi 2013)
Answer:
Two monosaccharide units are linked to each other by a bond called glycosidic linkage.

Question 9.
Which of the two components of starch is water-soluble? (CBSE Delhi 2014)
Answer:
Amylose

Question 10.
Which component of starch is a branched polymer of a-glucose and insoluble in water? (CBSE Delhi 2014)
Answer:
Amylopectin

Question 11.
What are the products of hydrolysis of maltose? (CBSE 2014)
Answer:
Glucose

Question 12.
What is the basic structural difference between glucose and fructose?
OR
Write the products obtained after hydrolysis of lactose. (CBSE Delhi 2019)
Answer:
Glucose contains an aldehyde group (-CHO) and is aldose, which is present at the end of the carbon chain, i.e. C₁.

Fructose contains a keto group (>C = 0) and is called ketose, which is present at a carbon atom next to the terminal, i.e. C₂.
OR
Hydrolysis of lactose gives β – D – glucose and β – D – galactose.

Question 13.
What is the difference between a glycosidic linkage and a peptide linkage?
OR
What is the difference between Nucleotide and Nucleoside? (CBSE AI 2019)
Answer:
Glycosidic linkage is the linkage that joins two monosaccharides through an oxygen atom (-0-). Peptide linkage is the linkage that joins two amino acids through -CONH- bond.
OR
The molecules in which one of the nitrogen bases (purines or pyrimidine) is bonded with a sugar molecule (ribose or deoxyribose) is a nucleoside. The nucleoside linked with the phosphate group is called a nucleotide.

Question 14.
What is meant by the inversion of sugar? (CBSE Sample Paper 2011)
Answer:
The change of specific rotation of sugar from dextrorotatory to laevorotatory is called inversion of sugar.

Question 15.
Glucose does not give 2, 4-DNP test and Schiff’s test. Why? (CBSE Sample Paper 2011)
Answer:
Glucose has a cyclic structure in which the -CHO group is not free because it forms a hemiacetal linkage with the -OH group at C-5. Therefore, it does not give 2, 4-DNP test although it has -CHO group.

Question 16.
Write any two reactions of glucose that could not be explained by an open-chain structure of the glucose molecule.
Answer:
The open-chain structure of a glucose molecule cannot explain the following:

1. Glucose does not react with sodium bisulphite (NaHS0₃) to form an addition product though it has an aldehyde group,
2. Glucose does not give the Schiffs test and 2,4-DNP test like other aldehydes.

Question 17.
Write the product when D-glucose reacts with the cone. HNO₃.
Answer:

Question 18.
(i) Which vitamin deficiency causes rickets?
Answer:
Vitamin D

(ii) Name the base that is found in the nucleotide of RNA only. (CBSE Sample Paper 2017-18)
Answer:
Uracil

Question 19.
Write the Zwitter ion structure of glycine. (CBSE Sample Paper 2011)
Answer:

Question 20.
Name a carbohydrate present in the liver, muscles and brain. (CBSE 2019C)
Answer:
Glycogen

Question 21.
Name the species formed when an aqueous solution of amino acid is dissolved in water. (CBSE Sample Paper 2019)
Answer:
Zwitter ion/dipolar ion

Question 22.
Name the unit formed by the attachment of a base to the 1′ position of sugar in a nucleoside. (CBSE Sample Paper 2019)
Answer:
Nucleotide

Biomolecules Important Extra Questions Short Answer Type

Question 1.
Name the bases present in RNA. Which one of these is not present in DNA? (CBSE Delhi 2011)
Answer:

• The bases present in RNA are guanine, adenine, cytosine and uracil.
• Uracil is not present in DNA. Instead, it contains thymine.

Question 2.
What is meant by biocatalysts? (CBSE Delhi 2012)
Answer:
Biocatalysts. Biocatalysts or enzymes are produced by living cells which catalyse the biochemical reactions occurring in living cells.

Question 3.
Name one fibrous protein and one globular protein.
Answer:
Fibrous: Keratin Globular: Albumin

Question 4.
What is the chemical name of vitamin A and which disease is caused by its deficiency?
Answer:
Retinol, deficiency disease: Xerophthalmia.

Question 5.
What is the chemical name of vitamin C and which disease is caused by its deficiency?
Answer:
Ascorbic acid, deficiency disease: Scurvy.

Question 6.
Which enzyme is present in saliva? What is its function?
Answer:
The enzyme present in the saliva is amylase. It hydrolyses starch into maltose.

Question 7.
What is the difference between DNA and RNA on the basis of the bases they contain?
Answer:
Both DNA and RNA contain two bases derived from purine: guanine and adenine and one base derived from pyrimidine: cytosine. However, they have a fourth different base; DNA contains thymine whereas RNA contains uracil.

Question 8.
Shanti, a domestic helper of Mrs Anuradha, fainted while mopping the floor. Mrs Anuradha immediately took her to the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron-rich diet and multivitamins supplement to her. Mrs Anuradha supported her financially to get the medicines. After a month, Shanti was diagnosed to be normal.

After reading the above passage, answer the following questions:
(i) What values are displayed by Mrs Anuradha?
Answer:
Mrs Anuradha has shown generosity and a caring attitude for the health and life of her domestic helper. Timely help, kindness and financial support saved her domestic helper from a serious health problem.

(ii) Name the vitamin whose deficiency causes ‘pernicious anaemia’.
Answer:
B₁₂

(iii) Give an example of a water-soluble vitamin. (CBSE 2013)
Answer:
Vitamin C

Biomolecules Important Extra Questions Long Answer Type

Question 1.
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and Glycosidic linkage
(iii) Fibrous proteins and Globular proteins
OR
Write chemical reactions to show that the open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as carbonyl group (CBSE Delhi 2019)
Answer:
(i) Amylose and Amylopectin:

(ii) Peptide linkage and Glycosidic linkage:

(iii) Fibrous and Globular proteins:

OR
(i) Glucose on prolonged heating with HI and red P at 100 °C gives n-hexane, which shows the straight-chain structure of glucose.

(ii) Glucose reacts with acetic anhydride in the presence of anhydrous zinc chloride to form glucose pentaacetate. The formation of pentaacetate confirms the presence of five -OH groups in the glucose molecule.

(iii) Glucose reacts with hydrogen cyanide forming glucose cyanohydrin. This shows that glucose contains a carbonyl group (>C=0).

Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on treatment with a mild oxidising agent like Br2 water. This indicates that the carbonyl group present in glucose is an aldehyde (-CHO) group.

Question 2.
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins
OR
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br₂ water
(ii) H₂N-OH
(iii) (CH₃CO)₂0 (CBSE Delhi 2019)
Answer:
(i) Oligosaccharides. These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis, for example, disaccharides such as sucrose, lactose, maltose (C₁₂H₂₂O₁₆), trisaccharides such as raffinose (C₁₈H₃₂O₁₆).

(ii) Deriaturatlon of proteins: A process that changes the physical and biological properties of proteins without affecting the chemical composition of the protein is called denaturation. It is caused by certain physical change like change In temperature or chemical change like change In pH presence of electrolytes, etc., for example, coagulation of albumin present in the white of an egg.

(iii) Vitamins: The organic compounds which cannot be produced by the body and must be supplied in small amounts in the diet to perform specific biological functions for the normal health, growth and maintenance of the body are called vitamins, for example, vitamin A, B, C, D, etc.
OR

Question 3.
Define the following terms with a suitable example of each:
(a) Anomers
Answer:
Anomers: Carbohydrates that differ in configuration at the glycosidic carbon (i.e. C₁ in aldoses and C₂ in ketoses) are called anomers. For example, α-D-glucose and β-D- glucose.

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet are called essential amino acids. For example, valine, leucine, etc.

(c) Denaturation of protein (CBSE AI 2019)
Answer:
A process that changes the native conformation of a protein is called denaturation. The denaturation can be caused by changes in pH, temperature, presence of salts of certain chemical agents. The denatured protein will lose its biological activity. During denaturation, the protein molecule uncoils from an ordered and specific conformation into a more random conformation and protein precipitates from the solution. For example, when an egg is boiled in water, the globular proteins present in it change to a rubber-like insoluble mass.

Question 4.
Define the following terms with a suitable example of each:
(a) Tertiary structure of the protein
Answer:
The tertiary structure of proteins: Tertiary structure of proteins arises because of the folding, coiling or bending of polypeptide chains producing a three-dimensional structure. This structure gives the overall shape of proteins. For example, fibrous proteins such as silk, collagen and α-keratin have large helical structures

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be synthesised in our body and must be supplied through our diet are called essential amino acids, for example, valine, leucine, etc.

(c) Disaccharides (CBSE AI 2019)
Answer:
Disaccharides: The carbohydrates which on hydrolysis give two same or different monosaccharides are called disaccharides. Sucrose is an example of a disaccharide, which on hydrolysis gives glucose and fructose.

Question 5.
Describe what do you understand by primary structure and secondary structure of proteins. (CBSE Delhi 2011)
Answer:
Primary structure:
The sequence in which the amino acids are linked in one or more polypeptide chains of a protein is called the primary structure.

The primary structure is usually determined by its successive hydrolysis with enzymes or mineral acids.

Secondary structure: The secondary structure gives the manner in which the polypeptide chains are folded or arranged. Therefore, it gives the shape or conformation of the protein molecule. This arises from the plane geometry of the peptide bond and hydrogen bond between the —C=O and N—H groups of different peptide bonds. Pauling and Corey studied that there are two common types of structures:

(i) α – Helix structure: In this structure, the formation of hydrogen bonding between amide groups within the same chain causes the peptide chains to coil up into a spiral structure (Fig. 1). This is called the a-helix.

(ii) β-pleated sheet structure: In this structure, all polypeptide chains are stretched out to nearly maximum extension and then laid side by side in a zigzag manner to form a flat sheet. Each chain is held to the two neighbouring chains by a hydrogen bond. These sheets are stacked one upon another to form a three-dimensional structure called
β-pleated sheet structure (Fig. 2).

Question 6.
ExplaIn what is meant by the pyranose structure of glucose. (CBSE 2011)
Answer:
The six-membered cyclic structure of glucose is catted pyranose structure similar to the pyran heterocyclic compound. For structure,

Question 7.
Write the main structural difference between RNA and DNA. Of the four bases, name those which are common to both DNA and RNA. (CBSE 2011)
Answer:
Structural differences:

Common bases present in both: adenine, cytosine and guanine.

Question 8.
(i) Deficiency of which vitamin causes night blindness? (CBSE DelhI 2014)
(ii) Name the base that is found in the nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does It suggest about the structure of glucose?
Answer:
(i) Deficiency of Vitamin A causes night blindness.
(ii) Uracil
(iii) Glucose reacts with HI to give n-hexane.

This suggests that alt the six carbon atoms in glucose are arranged in a straight chain structure of glucose.

Question 9.
(i) DefIciency of which vitamin causes rickets? (CBSE Delhi 2014)
Answer:
Vitamin D.

(ii) Give an example for each of fibrous protein and globular protein.
Answer:
Fibrous protein: Keratin Globular protein: Albumin

(iii) Write the product formed on reaction of D-glucose with Br2 water.
Answer:

Question 10.
(i) Deficiency of which vitamin causes scurvy? (CBSE DelhI 2014)
Answer:
Vitamin C.

(ii) What type of linkage is responsible for the formation of proteins?
Answer:
Peptide Linkage.

(iii) Write the product formed when glucose is treated with HI.
Answer:

Question 11.
Define the following terms: (CBSE 2014)
(i) Gtycosidic linkage
Answer:
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Answer:
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Answer:
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C₁₂H₂₂O₁₁.
Trisaccharide: Raffinose (C₁₈H₃₂O₁₆).
Tetrasaccharides: Stachyose (C₂₄H₄₂O₂₁ ).

Question 12.
Define the following terms: (CBSE 2014)
(i) Nucleotide
Answer:
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Answer:
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 13.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Answer:
Difference between fibrous protein and globútar protein:

(b) Essential amino acids and Non-essential amino acids
Answer:
Difference between essential amino acids and non-essential amino acids:

(C) Amylose and Amylopectin (CBSE AI 2019)
Answer:
Difference between amylose and amylopectin:

Question 14.
(i) Which one of the following is a polysaccharide: (CBSE 2015, 2014)
Starch, Maltose, Fructose, Glucose
Answer:
Starch

(ii) What is the difference between native protein and denatured protein?
Answer:
Native protein is the protein found in a biological system with a unique three-dimensional structure and biological activity.
Denatured protein is the protein which loses its biological activity by certain physical or chemical treatment such as changes in pH, temperature, presence of certain salts or chemical agent, known as denaturation.

(iii) Write the name of the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K.

Question 15.
(i) Write one reaction of D-glucose which cannot be explained by its open-chain structure. (CBSE 2016)
Answer:
Despite having an aldehydic (-CHO) group, glucose does not react with NaHS0₃ to form an addition product.

(ii) What type of linkage is present in nucleic acids?
Answer:
Phosphodiester linkage

(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins.
Answer:
Water-soluble vitamin: vitamin B Fat-soluble vitamin: vitamin A

Question 16.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
Answer:
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C₁ atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = +111° and the β-form of glucose has m.p. 423 K and |α|_D = +19.2°.

Question 17.
(a) What happens when D-glucose is treated with the following reagents:
(i) HI
Answer:
When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight line.

(ii) cone. HN03
Answer:
Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.

(b) What is the basic structural difference between starch and cellulose? (CBSE 2019C)
Answer:
Starch consists of two components: amylose and amylopectin. Amylose is a long linear polymer of 200-1000 a-D-(+)-glucose units held by C₁ – C₄ glycosidic linkages. It is soluble in water. Amylopectin is a branched-chain polymer of a-D-(+)-glucose linkages whereas branching occurs by C₁ – C₆ glycosidic linkage. It is insoluble in water.

On the other hand, cellulose is a straight-chain polysaccharide composed only of (β – D – (+)-glucose units which are formed by the glycosidic linkage between C₁ of one glucose unit and C₄ of the next glucose unit.

Question 18.
(1) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (CBSE Delhi 2016)
Answer:
Glucose, Galactose

(ii) Why Vitamin C cannot be stored in our body?
Answer:
Vitamin C is water-soluble and hence cannot be stored in our body.

(iii) What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains only two basic components of nucleic acids, namely a pentose sugar and a nitrogenous base.

A nucleotide contains all the three basic components of nucleic acids, namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

Question 19.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO₃.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
Answer:
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine
OR
(i) Saccharic acid / COOH-(CHOH)₄-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 20.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Answer:
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
Answer:
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.
Answer:

Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 21.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
Answer:
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = + 111° and the β-form of glucose has m.p. 423 K and |α|_D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:

Question 22.
Which sugar is called invert sugar? Why is it called so?
Answer:
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]_D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.

So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 23.
How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.

This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 24.
Describe the term D- and L-configuration used for sugars with examples.
Answer:
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:

The D-configuration has —OH attached to the carbon adjacent to —CH₂OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH₂OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 13 Amines. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 13 Important Extra Questions Amines

Amines Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in decreasing order of solubility in water:
(CH₃)₃N, (CH₃)₂H, CH₃NH₂ (CBSE Delhi 2019)
Answer:
CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N

Question 2.
Arrange the following compounds in increasing order of their solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂ (CBSE Delhi 2011, CBSE 2013)
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Question 3.
Write the structure of n-methylhexanamine. (CBSE 2013)
Answer:
CH₃ – NH – CH₂CH₃

Question 4.
Write the structure of 2-amino toluene. (CBSE 2013)
Answer:

Question 5.
Write the IUPAC name of the given compound: (CBSE Delhi 2016)

Answer:
2, 4, 6-Tribromoaniline

Question 6.
Arrange the following in increasing order of their basic strength in an aqueous solution:
CH₃NH₂, (CH₃)₃N, (CH₃)₂NH (CBSE Delhi 2013)
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Question 7.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:

Question 8.
Arrange the following in increasing order of base strength in the gas phase:
(C₂H₅)₃N, C₂H₅NH₂, (C₂H₅)₂NH (CBSE Delhi 2019)
Answer:
(C₂H₅)₃N > (C₂H5)₂NH > C₂H₅NH₂

Question 9.
Arrange the following in increasing order of boiling points:
(CH₃)₃N, C₂H₅OH, C₂H₅NH₂ (CBSE Delhi 2019)
Answer:
(CH₃)₃N < C₂H₅NH₂ < C₂H₅OH

Question 10.
Arrange the following in the increasing order of their pKb values.
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃ (CBSE AI 2018)
Answer:
C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Question 11.
Write IUPAC name of the following compound:
CH₃NHCH(CH₃)₂ (CBSE Delhi 2017)
Answer:
N-Methylpropan-2-amine

Question 12.
Write IUPAC name of the following compound: (CH₃CH₂)₂NCH₃ (CBSE Delhi 2017)
Answer:
N-Ethyl-N-methylhexanamine

Question 13.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
Aniline gives azo dye test while ethylamine does not give azo dye test.

Question 14.
Arrange the following in increasing order of their basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₆H₅)₂NH, and CH₃NH₂ (CBSE AI 2011)
Answer:
(C₆H₅)₂ NH < C₆H₅NH₂ < C₆H₅N (CH3)₂ < CH₃NH₂

Question 15.
Write the IUPAC name of the compound: (CBSE AI 2016)

Answer:
N-Methyl-2-methyl propanamine

Amines Important Extra Questions Short Answer Type

Question 1.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 2.
Complete the following reaction equations:
(i) C₆H₅N₂CI + H₃PO₂ + H₂0 →
Answer:

(ii) C₆H₅NH₂ + Br₂(oq) → (CBSE 2012)
Answer:

Question 3.
Give the structures of products A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 4.
Write the structures of A, B, and C in the following reactions: (CBSE 2016)

Answer:

Answer:

Question 5.
Write the structures of A, B, and C In the following: (CBSE Delhi 2016)

Answer:

Answer:

Question 6.
Give a chemical test to distinguish between ethylamine and aniline. (CBSE AI 2011)
Answer:
These can be distinguished by the azo dye test. Dissolve the compound in a cone. HCl and add an ice-cold solution of HNO₂ (NaNO₂ + dil. HCl) and then treat it with an alkaline solution of 2-naphthol. The appearance of brilliant orange or red dye indicates aniline.


Ethylamine does not form a dye. It will give brisk effervescence due to the evolution of N₂ but the solution remains clear.

Question 7.
How will you distinguish between the following? Give one chemical test:
(i) Aniline and benzylamine
Answer:
These can be distinguished by the azo dye test. Aniline reacts with HNO₂ (NaNO₂ + dil. HCl) at 273-278K to form stable benzene diazonium chloride which on treatment with an alkaline solution of 2-naphthol gives an orange dye (as given above).
Benzylamine does not give azo dye tests.

(ii) Aniline and N-methylaniline. (CBSE AI 2010)
Answer:
These can be distinguished by carbylamine test. Aniline being primary amines gives carbylamine test, i.e. when heated with an alcoholic solution of KOH and CHCl₃, it gives the foul smell of phenyl isocyanide.

Question 8.
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. (CBSE Sample Paper 2011)
Answer:
(i) Since the compound C of molecular formula C₆H₇N is formed from B on treatment with Br₂ and KOH (Hoffmann bromamide reaction), therefore, the compound ‘B’ must be an amide and ‘C’ must be an amine. The only aromatic amine having molecular formula C₆H₇N is C₆H₅NH₂ (aniline).

(ii) Since ‘C’ is aniline, the amide from which it is formed must be benzamide (C₆H₅CONH₂).

Thus, B is benzamide.

(iii) Since B is formed from A with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

Thus, A = C₆H₅COOH, B = C₆H₅CONH₂, C = C₆H₅NH₂.

Question 9.
Account for the following:
(a) Gabriel phthalimide synthesis is not preferred for preparing aromatic primary amines.
Answer:
Gabriel phthalimide synthesis is not preferred for preparing aryl amines because aryl halides do not undergo nucleophilic substitution reaction with phthalimide.

(b) On reaction with benzene sulphonyl chloride, primary amine yields product soluble in alkali whereas secondary amine yields product insoluble in alkali. (CBSE Al 2019)
Answer:
Answer:
The sulphonamide formed by the reaction of secondary amine and benzene sulphonyl chloride does not contain any hydrogen atom attached to the N atom, so it is not acidic.
Therefore, it is insoluble in alkali.

Due to the presence of ‘H’ on nitrogen, it is soluble in alkali.

Since there is no ‘H’ on nitrogen, it is insoluble in alkali.

Amines Important Extra Questions Long Answer Type

Question 1.
Write equations of the following reactions:
(i) Acetylation of Aniline
Answer:
Acetylation of Aniline:

(ii) Coupling reaction
Answer:
Coupling reaction:

(iii) Carbylamine reaction (CBSE Delhi 2019)
Answer:
Carbylamine reaction:

Question 2.
An aromatic compound ‘A’ on heating with Br₂ and KOH forms a compound ‘B’ of molecular formula C₆H₇N which on reacting with CHCl₃ and alcoholic KOH produces a foul-smelling compound ‘C’.
Write the structures and IUPAC names of compounds A, B, and C. ((SSE Delhi 2019)
Answer:

A: C₆ H₅CONH₂: Benzamide B: C₆H₅NH₂: Benzenamine C: C₆H₅NC: Phenyt isocyanide

Question 3.
Write the structures of main products when benzene diazonium chloride reacts with the following reagents: (CBSE Delhi 2019)
(i) CuCN
Answer:

(ii) CH₃CH₂OH
Answer:

(iii) Kl
Answer:

Question 4.
(a) Write the product formed when
(i) 2-chioropropane is treated with aic. KOH.
Answer:

(ii) Aniline reacts with conc. H₂SO₄ at 453 – 473 K.
Answer:
Sulphonation: Sulphonation of aniline is carried out by heating aniline with sulphuric acid. The product formed is anilinium hydrogen sulfate which on heating gives sulphanilic acid.

The sulphanilic acid exists as a dipolar ion (structure II) which has acidic and basic groups in the same molecule. Such ions are called Zwitter ions or inner salts.

(b) When aniline is heated with CHCl₃ and aic. KOH, a foul-smelling compound is formed. What is this compound? (CBSE 2019C)
Answer:
Pheriyt isocyanide is formed.

Question 5.
(a) Identify ‘A’ and ‘B’ In the following reaction:

Answer:

(b) Why does aniline not undergo Friedel-Crafts reaction? (CBSE 2019C)
Answer:
Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Hence aniline does not undergo Friedel-Crafts reaction.

Question 6.
Complete the following reactions: (CBSE 2013)
(i) CH₃CH₂NH₂ + CHCl₃ + ale. KOH →
Answer:

(ii) C₆H₅N₂+Cl^– →
Answer:

Answer:

Question 7.
Write the main products of the following reactions: (CBSE 2013, CBSE AI 2013)

Answer:

Answer:

Answer:

Question 8.
Write the main products of the following reactions: (CBSE 2013)

Answer:

Answer:

Answer:

Question 9.
Account for the following:
(i) Primary amines (R-NH₂) have a higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel — Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
OR
Give the structures of A, B, and C in the following reactions: (CBSE 2014)

Answer:
(i) Primary amines (RNH₂) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.

Tertiary amines (R₃N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.

(ii) Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Hence aniline does not undergo Friedel Crafts reaction.

(iii) Due to the presence of lone pair of electrons on the N atom, amines are basic in nature. The methyl group is the electron releasing group (+I inductive effect) and therefore, it increases the electron density on the N atom, and therefore, basic character increases, so that (CH₃)₃N should be more basic than (CH₃)₂NH. But tertiary ammonium ion formed from tertiary amines is less hydrated than secondary ammonium ion formed from secondary amine. Therefore, (CH₃)₃N has less tendency to form ammonium ion, and consequently, it is less basic than (CH₃)₂NH. Thus, (CH₃)₂NH is more basic than (CH₃)₃N due to the combined effect of inductive effect and hydration effect.
OR

Question 10.
Give the structures of A, B, and C in the following reactions:

OR
How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethylamine
(Write the chemical equations involved.) (CBSE Delhi 2014)
Answer:

OR

Question 11.
Write chemical equations for the following conversions:
(i) Nitrobenzene to benzoic acid.
Answer:

(ii) Benzyl chloride to 2-phenytethanamine.
Answer:

(iii) Aniline to benzyl alcohol. (CBSE Delhi 2012)
Answer:

Question 12.
(a) Identify ‘A’ and ‘B’ in the following reaction:

Answer:

(b) Why is ethylamine soluble in water whereas aniline is not?
Answer:
Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:

However, because of the large hydrophobic part (i.e. hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

Question 13.
(a) Identify X and Y in the following:

Answer:
X =
benzene diazonium ch(orlde,
Y =
Cyanobenzene

(b) Amino group is o, p-directing for aromatic electrophilic substitution reactions. Why does aniline on nitration give m-nitroaniline? (CBSE 2019C)
Answer:
Under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion having an NH₃⁺ group. This group is an m-directing group, therefore, m-nitro aniline is also obtained along with o- and p-products.

Question 14.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 15.
Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in an aqueous solution:
CH₃NH₃, (CH₃)₃N, (CH₃)₂NH.
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂ NH

(ii) Identify ‘A’ and ‘B’:

Answer:
A: C6H5N2+ Cl- B: C6H5OH

(iii) Write the equation of carbylamine reaction. (CBSE 2018C)
Answer:

Question 16.
(i) Illustrate the following reactions giving a suitable example in each case:
(a) Hoffmann bromamide degradation reaction
(b) Diazotisation
(c) Gabriel phthalimide synthesis
(ii) Distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) (CH₃)₂NH and (CH₃)₃N
OR
(i) Write the structures of main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following reagents:
(a) CuCN/KCN
(b) H₂0
(c) CH₃CH₂OH
(ii) Arrange the following:
(a) C₂H₅NH₂, C₂H₅OH, (CH₃)₃N – in the increasing order of their boiling point.
(b) Aniline, p-nitroaniline, p-methyl aniline – in the increasing order of their basic strength. (CBSE Delhi 2015)
Answer:
(i) (a) Hoffmann bromamide degradation reaction: Primary amines can be prepared from amides by treatment with Br₂ and KOH solution. The amine formed contains one carbon atom less than the parent amide.

(b) Diazotisation: The reaction of aniline or other aromatic amines, with nitrous acid at 0-5 °C to form diazonium salts is called diazotization. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO₂.

(c) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with an alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to arylamines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

(ii) (a) Add an alcoholic solution of KOH and CHCl3 to the compounds. Aniline gives the foul smell of isocyanide whereas N-methyl aniline does not give a foul smell.

(b) When treated with Hinsberg’s reagent (benzene sulphonyl chloride, C₆H₅SO₂CI), dimethylamine, (CH₃)₂NH gives precipitate which is insoluble in aqueous KOH.

(CH₃)₃N does not react with Hinsberg’s reagent.
Or

(ii) (a) (CH₃)₂ N < C₂H₅NH₂ < C₂H₅OH
(b) p-nitroaniline < aniline < p-methylaniline

Question 17.
Give the IUPAC names of the following compounds:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Delhi 2017)
Answer:

(CBSE Delhi 2017)
Answer:

Answer:

Question 18.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

Question 19.
Write the main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following:
(i) CuCN/KCN
Answer:

(ii) H₂0
Answer:

(iii) CH₃CH₂OH (CBSE AI 2015, 2018)
Answer:

(iv) Copper powder/HCI
Answer:

Question 20.
Complete the following chemical equations:

(CBSE Delhi 2011)
Answer:

(CBSE Delhi 2011)
Answer:

(CBSE Delhi 2011)
Answer:

(CBSE AI 2013)
Answer:

(CBSE AI 2013)
Answer:

(CBSE AI 2013)
Answer:

Answer:

Question 21.
Give the structures of A, B, and C In the following reactions:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

(CBSE Delhi 2016)
Answer:

(CBSE Delhi 2016)
Answer:

(CBSE AI 2016)
Answer:

(CBSE AI 2016)
Answer:

Answer:

Question 22.
An organic compound A’ with molecular formula C₇H₇NO reacts with Br₂/aq KOH to give compound B’, which upon reaction with NaNO₂ and HCI at OC gives C’. Compound C’ on heating with CH₃CH₂OH gives a hydrocarbon D’. Compound B’ on further reaction with Br₂ water gives a white precipitate of compound E’. Identify the compounds A, B, C, D, and E; also justify your answer by giving relevant chemical equations. (CBSE Sample Paper 2019)
OR
(a) How will you convert:
(i) Aniline into Fluorobenzene?
(ii) Benzamide into Benzylamine?
(iii) Ethanamine into N, N-Diethylethanamine?
(b) Write the structures of A and B in the following:

Answer:

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 12 Important Extra Questions Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. (CBSE Delhi 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 2.
Which of the following compounds would undergo the Cannizzaro reaction? Benzaldehyde, Cyclohexanone, 2- Methylpentanal. (CBSE Sample Paper 2019)
Answer:
Benzaldehyde

Question 3.
Draw the structure of semicarbazone of cyclopentanone. (CBSE 2019C)
OR
Draw the structure of the product formed when propanal is treated with zinc amalgam and concentrated hydrochloric acid.
Answer:

OR

Question 4.
Write the IUPAC name of the following: (CBSE AI 2012)

Answer:
Pent-2-enal.

Question 5.
Write the structure of 3-methylbutanal. (CBSE Delhi 2013)
Answer:

Question 6.
Write the structure of the p-Methylbenzal- dehyde molecule. (CBSE Delhi 2013, 2014)
Answer:

Question 7.
Draw the structure of the compound whose IUPAC name is 4-chloropentan 2-one. (CBSE Delhi 2008, CBSE AI 2011, 2013, 2014)
Answer:

Question 8.
Draw the structural formula of a 1-phenyl propane-2-one molecule. (CBSE 2010)
Answer:

Question 9.
Draw the structure of 3-methylbutanal. (CBSE 2011, CBSE Delhi 2013)
Answer:

Question 10.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
ethanal, propanal, propanone, butanone. (CBSE 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
Answer:
3-Phenyl prop-2-en-al.

Question 12.
Rearrange the following compounds in the increasing order of their boiling points:
CH₃ – CHO, CH₃ – CH₂ – OH, CH₃ – CH₂ – CH₃ (CBSE 2013)
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Answer:
Ethanol is soluble in water because of hydrogen bonding between the polar carbonyl group of ethanal and water molecules:

Question 14.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Hydroxybutanoic acid

Question 15.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:

Question 16.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Answer:
3-Aminobutanal

Question 17.
Write the IUPAC name of the following compound: (CBSE AI2019)

Answer:
But-3-en-2-one

Question 18.
Write the IUPAC name of the following compound: (CBSE AI 2019)

Answer:
1-Phenylbutan-2-one.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Short Answer Type

Question 1.
Write structures of compounds A and B in each of the following reactions: (CBSE Delhi 2019)

Answer:

Answer:

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
Answer:
The ammoniacal solution of silver nitrate is called Tollen’s reagent. It is used as an oxidizing reagent and helps to distinguish between aldehydes and ketones. Aldehydes are oxidized by Tollen’s reagent whereas ketones are not.

Question 3.
Name the reagents used in the following reactions: (CBSE Delhi 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer
Alkaline potassium permanganate (KMnO₄), KOH

Question 4.
Distinguish between
C₆H₅CH = CH – COCH₃ and C₆H₅CH = CH CO CH₂ CH₃ (CBSE AI 2016)
Answer:
Heat both the compounds with NaOH and l₂. C₆H₅CH = CHCOCH₃ gives yellow ppt of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.

Question 5.
Arrange the following in the increasing order of their reactivity in nucleophilic addition reactions.
CH₃CHO, C₆H₅CHO, HCHO (CBSE AI 2016)
Answer:
C₆H₅CHO < CH₃CHO < HCHO

Question 6.
Write the product in the following reaction: (CBSE AI 2017)

Answer:

Question 7.
Write the structure of the products formed: (CBSE Sample Paper 2019)
(a) CH₃CH₂COOH
Answer:
CH₃ – CH(Cl) – C00H

(b) C₆H₅COCl
Answer:
C₆H₅CHO

Question 8.
Name the reagents in the following reactions: (CBSE AI 2015)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer:
Alkaline potassium permanganate (KMnO₄), KOH

Answer:
CH₃ Mg Br, H₃O⁺

Answer:
Cl₂, P (Hell-Volhard-Zelinsky reaction)

Question 9.
Predict the organic products of the following reactions:

Answer:

Answer:

Answer:

Answer:

Question 10.
Describe how the following conversions are carried out:
(i) Toluene to benzoic acid
Answer:

(ii) Bromobenzene to benzoic acid (CBSE AI 2013)
Answer:

(iii) Ethylcyanide to ethanoic acid
Answer:

(iv) Butan-1-ol to butanoic acid (CBSE Delhi 2010)
Answer:

Question 11.
Write the structures of A, B, C, and D in the following reaction: (CBSE AI 2016)

Answer:

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Answer:
Aromatic carboxylic acids do not undergo Friedel Crafts reaction because -COOH group is deactivating and the catalyst aluminum chloride (Lewis acid) gets bonded to the carboxyl group.

Question 13.
pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Answer:
Due to the presence of a strong electron-withdrawing (-NO₂) group in 4-nitrobenzoic acid, it stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more acidic than benzoic acid and its pK_a value is lower.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Long Answer Type

Question 1.
Complete the following reactions:

Or
Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)₂.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCI
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄. (CBSE Delhi 2019)
Answer:



Or

Question 2.
Predict the products of the following B reactions: (CBSE Delhi 2015)

Answer:

Answer:

Answer:

Question 3.
Predict the products of the following reactions: (CBSE 2015)

Answer:

Answer:

Answer:

Question 4.
What happens when
(a) Salicylic acid Is treated with (CH₃CO)₂O/H⁺?
Write a chemical equation in support of your answer. (CBSE AI 2019)
Answer:
Salicylic acid is treated with acetic anhydride In the presence of H⁺ to give aspirin.

(b) Phenol is oxidized with Na₂Cr₂O₇/H⁺?
Answer:
Phenol in the presence of acidified sodium dichromate gives benzoquinone.

(C) Anisole Is treated with CH₃Cl/anhydrous AlCl₃?
Answer:

Question 5.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
Answer:

(ii) 4-Chlocopentan-2-one
Answer:

(iii) 4-Methylpent-3-en-2-one
Answer:

(iv) p-Methyl benzaldehyde (CBSE AI 2014)
Answer:

Question 6.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene (CBSE Delhi 2017)
Answer:
Propanone to propene

(ii) Propanal to butanone
Answer:
Propanal to butanone

(iii) Benzaldehyde to benzophenone
Answer:
Benzaldehyde to benzophenone

(iv) Benzaldehyde to 3-phenyl propan – 1-ol
Answer:
Benzaldehyde to 3- phenyl propan -1 -ol

(v) Benzaldehyde to α-hydroxyphenyl acetic acid
Answer:
Benzaldehyde to α-hydroxyphenyl acetic acid

(vi) Ethanol to 3-hydroxybutanal
Answer:
Ethanol to 3-hydroxybutanal

Question 7.
Convert the following:
(i) Ethanal to propanone. (CBSE AI 2018)
Answer:
Ethanal to propanone

(ii) Ethanal to lactic acid.
Answer:
Ethanal to lactic acid

(iii) Ethanal to 2-hydroxy-3-butenoic acid.
Answer:
Ethanal to 2-hydroxy-3-butenoic acid

(iv) Acetaldehyde to formaldehyde.
Answer:
Acetaldehyde to formaldehyde

(v) Formaldehyde to acetaldehyde.
Answer:
Formaldehyde to acetaldehyde

(vi) Acetaldehyde to crotonic acid.
Answer:
Acetaldehyde to crotonic acid

Question 8.
A compound ‘X’ (C₂H₄O) on oxidation gives ‘Y’ (C₂H₄O₂). ‘X’ undergoes haloform reaction. On treatment with HCN ‘X’ forms a product ‘V which on hydrolysis gives 2-hydroxy propanoic acid.
(i) Write down structures of ‘X’ and ‘Y’.
(ii) Name the product when ‘X’ reacts with dil NaOH.
(iii) Write down the equations for the reactions involved. (CBSE Sample Paper 2007)
Answer:
Compound ‘X’ (C₂H₄O) is oxidized to ‘Y’ (C₂H₄O₂). Since it undergoes a haloform reaction, it must be acetaldehyde.
(i) X = CHCHO Y = CH₃COOH
On treatment with HCN, X gives cyanohydrin which on hydrolysis gives 2-hydroxypropanoic acid.

(ii) When ‘X’ reacts with dil. NaOH, undergoes an aldol condensation reaction forming aldol which on heating gives but-2-enal.

(iii) Equations for reactions

Other equations are given above.

Question 9.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Sample Paper 2017 – 2018)
Answer:

(CBSE Delhi 2013)
Answer:

(CBSE AI 2013)
Answer:

Question 10.
(i) An organic compound (A) has a characteristic odor. On treatment with NaOH, it forms two compounds (B) and (C). Compound (B) has molecular formula C₇H₈0 which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda lime it yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved.
Answer:
The compound A is C₆H₅CHO, benzaldehyde having a characteristic odor. The reactions are:

(ii) Complete each synthesis by filling the missing starting materials, reagents, or products. (X and Z).
(a) C₆H₅CHO + CH₃CH₂CHO NaOH X
Answer:

(b) CH₃(CH₂)₉ COOC₂H₅ CH₃(CH₂),CHO
Answer:

(iii) How will you bring about the following conversions in not more than two steps?
(a) Toluene to benzaldehyde
Answer:

(b) Ethylcyanide to 1-phenylpropanoid. (CBSE Sample Paper 2011)
Answer:

Question 11.
(i) How do you convert the following?
(a) Ethanal to propanone
(b) Toluene to benzoic acid
OR
(ii) (A), (B), and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCI gives the same product (D).
(a) Write the structures of (A), (B), (C), and (D).
(b) Out of (A), (B), and (C) isomers, which one is least reactive towards the addition of HCN? (CBSEAI 2018)
Answer:

OR
(ii) (a) Out of A, B, and C, (A) and (C) give positive Tollens’ test, and therefore, these are aldehydes. (B) does not give Tollens’ test and therefore, it is a ketone, with a group because it gives positive iodoform test. Thus, the three isomers are:

Question 12.
Predict the products of the following reactions:

(CBSE Delhi 2015)
Answer:

(CBSE Delhi 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

Question 13.
Write the structures of A, B, C, D, and E in the following reactions: (C8SE Delhi 2016)

Answer:

Question 14.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid.
OR
Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal. (CBSE Sample Paper 2019)
Answer:
(a) (i)

Acetaldehyde (Ethanal) and Acetone (Propanone)
(i) Acetaldehyde gives silver mirror with Tollen’s reagent.

Acetone does not give this test.

(ii) Acetaldehyde gives red ppt with Fehling solution.

(ii) Pentan-2-one and Pentan-3-one

Pentan-3-one and Pentan-2-one
(i) Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH₃COCH₂CH₂CH₃ + 3I₂+ 4NaOH → CH₃CH₂CH₂COONa + CHI₃ + 3H₂0 + 3NaI

(ii) Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.

Or any other suitable test.

(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Effect of substituents on acidic strength of acids. The substituents have a marked effect on the acidic strength of carboxylic acids. The nature of substituents affects the stability of the conjugate base (carboxylate ion) and hence affects the acidity of the carboxylic acids.

In general, electron-withdrawing groups (EWG) increase the stability of the carboxylate ion by delocalizing the negative charge and hence increase the acidity of the carboxylic acid. Conversely, electron-donating groups (EDG) decrease the stability of the carboxylate ion by intensifying the negative charge and hence decrease the acidity of the carboxylic acid.
OR
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in ethanol. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence less reactive than ethanal.

Aromatic Aldehydes and Ketones
In general, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogs. For example, benzaldehyde is less reactive than aliphatic aldehydes. This can be easily understood from the resonating structures of benzaldehyde as shown below:

It is clear from the resonating structures that due to the electron releasing (+I effect) of the benzene ring, the magnitude of the positive charge on the carbonyl group decreases, and consequently it becomes less susceptible to the nucleophilic attack. Thus, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes and ketones.

However, amongst aromatic aldehydes and ketones, aromatic aldehydes are more reactive than alkyl aryl ketones which in turn are more reactive than diaryl ketones. Thus, the order of reactivity of aromatic aldehydes and ketones is:

Question 15.
(i) Illustrate the following name reaction giving a suitable example:
(a) Clemmensen reduction
(b) Hell-Volhard-Zelinsky reaction

(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid
(b) Butan-1-ol to butanoic acid
(c) Benzoic acid to m-bromobenzoic acid .
OR
(i) Illustrate the following reactions giving a suitable example for each.
(a) Cross aldol condensation
(b) Decarboxylation

(ii) Give simple tests to distinguish between the following pairs of compounds:
(a) Pentan-2-one and Pentan-3-one
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid (CBSE Delhi 2012)
Answer:
(i) (a) Clemmensen reduction

(b) Hell-Volhard-Zelinsky reaction

(ii)

OR
(i) (a) Cross Aldol condensation

(b) Decarboxylation

(ii) (a) Pentan-2-one and pentan-3-one can be distinguished by iodoform test

Pentan-3-one does not give this test.

(b) Benzaldehyde and acetophenone: Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent).

Acetophenone does not react with Tollen’s reagent.

(c) Phenol and Benzoic acid: Benzoic acid reacts with NaHC03 to give effervescence due to the evolution of CO₂.

Phenol does not give effervescence.

Question 16.
(i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid.

(ii) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Toilen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
OR
(i) Give chemical tests to distinguish between (a) Propanol and propanone
(b) Benzaldehyde and acetophenone

(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(b) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:

(ii) The given compound forms a 2,4-DNP derivative. Therefore, it is an aldehyde or ketone. Since it reduces Tollen’s reagent, it must be an aldehyde. The compound undergoes Cannizzaro’s reaction, so it does not contain hydrogen. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, which means that it must be containing an alkyl group at 2-position with respect to the CHO group on the benzene ring.
The molecular formula suggests it should be
(2- Ethytbenza(dehyde)

Answer:
(i) (a) Propanone gives iodoform test but propanol does not. When propanone is heated with aqueous sodium carbonate and iodine solution, a yellow precipitate of iodoform is obtained.

Propanol does not give this test.

(b) Acetophenone forms yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test). Benzaldehyde does not react.

(ii) (a) Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(b) 4-Methoxybenzoic acid < Benzoic add < 3, 4-Dinitrobenzoic acid.
(c) (CH₃)₂CHCOOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question 17.
(i) Although phenoxide ion has more resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons.
(ii) How will you bring about the following conversions?
(a) Propanone to propane
(b) Benzoyl chloride to benzaldehyde
(c) Ethanal to but-2-enal
OR
(i) Complete the following reactions:

(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol (CBSE AI 2013)
Answer:
(i) The carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrid structures of carboxylate ion and phenoxide ion.

The resonance hybrids may be represented as:

(a) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in the carboxylate ion as compared to only one oxygen atom in the phenoxide ion.

(b) Carboxylate ion is stabilized by two equivalent resonance structures in which negative charge is on more electronegative 0 atoms. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on a less electronegative carbon atom.

Therefore, the carboxylate ion is relatively more stable as compared to the phenoxide ion. Thus, the release of H⁺ ions from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

OR

(ii) (a) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

(b) Benzoic acid reacts with NaHC03 to give effervescence due to the liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂0 + CO₂
Phenol does not give effervescence.
Phenol gives violet color with FeCl₃ solution but benzoic acid does not give such color.

Question 18.
(i) How will you convert the following:
(a) Propanone to Propan-2-ol
(b) Ethanal to 2-hydroxypropanoic acid
(c) Toluene to benzoic acid

(ii) Give a simple chemical test to distinguish between:
(a) Pentan-2-one and Pentan-3-one
(b) Ethanal and Propanal
OR
(i) Write the products of the following reactions:

(ii) Which acid of each pair shown here would you expect to be stronger? (CBSE AI 2013)
(a) F – CH₂ – COOH or Cl – CH₂ – COOH OH

Answer:

(ii) (a) Pentan-2-one forms yellow ppt. with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.

(b) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

OR

(ii) (a) FCH₂COOH
(b) CH₃COOH

Question 19.
(a) Predict the main product of the following reactions:

(b) Give a simple chemical test to distinguish between

(c) Why is alpha (a) hydrogen of carbonyl compounds acidic in nature?
OR
(a) Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH₃OH in presence of dry HCI
(ii) Dilute NaOH
(iii) H₂N – NH₂ followed by heating with KOH in ethylene glycol

(b) Arrange the following compounds in increasing order of their property as indicated:
(i) F – CH₂COOH, O₂N – CH₂COOH, CH₃COOH, HCOOH – acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards the addition of HCN (CBSE AI 2019)
Answer:

(b) On adding NaOH/l2 and heat, acetophenone gives yellow ppt of iodoform but benzophenone does not.

(c) α – hydrogen of carbonyl compounds is acidic because of the electron-withdrawing nature of the carbonyl group.
OR

(b) (i) CH₃COOH < HCOOH < FCH₂COOH < NO₂-CH₃ COOH
(ii) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde

Question 20.
(a) An organic compound with the molecular formula C₇H₆0 forms a 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes the Cannizzaro reaction. On oxidation, it gives benzoic acid. Identify the compound and state the reactions involved.
(b) Give chemical tests to distinguish between the following pair of compounds:
(i) Phenol and propanol
(ii) Benzoic acid and benzene
OR
(a) Predict the products of the following:

(b) Arrange the following in increasing order of acidic character: (CBSE 2019C)
HCOOH, CF₃COOH, ClCH₂COOH, CCl₃COOH
Answer:
(a) Compound = Benzaldehyde or C₆H₅CHO
Reaction
Reaction with 2,4-DNP

With Tollens reagent
RCHO + 2[Ag(NH₃)₂ ]⁺ + 30H^– → RCOO^– + 2 Ag + 2H₂O + 4NH₃ (where R = – C₆H₅)
Cannizzaro

(b) (i) FeCl₃ test: Add 2 drops of neutral FeCl₃ to both the compounds separately. Phenol gives violet color.

(ii) Sodium bicarbonate test: Add NaHCO₃ to both the compounds: Benzoic acid reacts with NaHCO₃ to give effervescence due to liberation of CO₂.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂
OR
(a) A = CH₃COOH
B = CH₃COCl
C = CH₃CONH₂
D = CH₃NH₂

(b) HCOOH < ClCH₂COOH < CCl₃COOH < CF₃COOH

• the electron-withdrawing substituents disperse the negative charge on carboxylate ion and stabilize it and thus, increase acidity.
• the electron-releasing substituents intensify the negative charge of the carboxylate ion, destabilize it and thus, decrease the acidity.

Question 21.
(i) Give chemical tests to distinguish between:
(a) Propanal and propanone
(b) Benzaldehyde and acetophenone

(ii) How would you obtain:
(a) But-2-enal from the ethanal
(b) Butanoic acid from butanol
(c) Benzoic acid from ethylbenzene? (CBSE 2011)
Answer:
(i) (a) Propanal gives a silver mirror with Tollen’s reagent.

Propanone does not give this test.

(b) Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent). Acetophenone does not react.

Question 22.
(i) Describe the following giving linked chemical equations:
(a) Cannizzaro reaction
Answer:
Aldehydes that do not contain any a-hydrogen atom (e.g. benzaldehyde, formaldehyde) undergo self- oxidation and reduction reaction on treatment with cone, solution of caustic alkali. In this reaction, one molecule is oxidized to acid while another molecule is reduced to an alcohol.

(b) Decarboxylation
Answer:
The process of removal of a molecule of C02 from a carboxylic acid is called decarboxylation. It is usually carried out by heating a mixture of carboxylic acid or its sodium salt with soda lime (NaOH + CaO).

(ii) Complete the following chemical equations: (CBSE Delhi 2011)

Answer:

Answer:

Answer:

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
Answer:
(a) When treated with NaHCO₃ solution, benzoic acid gives brisk effervescence while ethyl benzoate does not

(b) Ethyl benzoate on boiling with an excess of NaOH gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.

(ii) Complete each synthesis by giving missing reagents or products in the following:

Answer:

Answer:

(CBSE 2011)
Answer:

Question 24.
(i) Write the products formed when CH₃CHO reacts with the following reagents:
(a) HCN
(b) H2N – OH
(C) CH₃CHO in the presence of dilute NaOH
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Phenol
(b) Propanal and Propanone
OR
(i) Account for the following:
(a) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(b) Carboxylic acids do not give reactions of the carbonyl group.
(ii) Write the chemical equations to illustrate the following name reactions:
Rosenmund reduction
(iii) Out of CH₃CH₂—CO—CH₂ – CH₃ and CH₃CH₂ — CH₂ — CO — CH₃, which gives iodoform test? (CBSE 2014)
Answer:

(ii) (a) Add neutral FeCl₃ in both the solutions, phenol gives violet color with FeCl₃ solution, but benzoic acid does not give such color.
(b) Add an ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions, propanal gives silver mirror whereas propanone does not.

OR
(0) Chlorine is an electron-withdrawing atom (-I effect). It withdraws the electrons from carbon to which it is attached and therefore, this effect is transmitted throughout the chain. As a result, electrons are withdrawn more strongly towards oxygen of 0—H bond and promotes the release of the proton.

Consequently, ClCH₂COOH is a stronger acid than CH₃COOH.

(b) Carboxylic acids do not give the characteristic reactions of the carbonyl group (>C = 0) as given by aldehydes and ketones. In carboxylic acids, the carbonyl group is involved in resonance, as follows:

Therefore, it is not a free group. But no resonance is possible in aldehydes and ketones. They give the characteristic reactions of the group.

(ii) Rosenmund reduction. Acid chlorides are converted to corresponding aldehydes by catalytic reduction. The reaction is carried out by passing through a hot solution of the acid chloride in the presence of palladium deposited over barium sulfate (partially poisoned with sulfur or quinoline).

The poisoning of palladium catalyst decreases its activity and it does not allow the further reduction of an aldehyde into alcohol.

(iii) CH₃CH₂CH₂COCH₃ gives iodoform test

Question 25.
(i) Write the structures of A, B, C, and D in the following reactions:

(ii) Distinguish between:
(a) C₆H₅ – CH = CH – COCH₃ and C₆H₅ – CH = CH – CO CH₂CH₃
(b) CH₃CH₂COOH and HCOOH

(iii) Arrange the following in the increasing order of their boiling points:
CH₃CH₂OH, CH₃COCH₃, CH₃COOH
OR
(i) Write the chemical reaction involved in the Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH₃ – CHO, C₆H₅COCH₃, HCHO
(iii) Why pKa of Cl-CH₂-COOH is lower than the pK_a of CH₃COOH?
(iv) Write the product in the following reaction.

(v) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and l₂, isomer A forms a yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. (CBSE 2016)
Answer:

(ii) (a) Heat both the compounds with NaOH and l₂, C₆H₅ CH = CHCOCH₃ gives yellow ppt. of iodoform. C₆H₅CH = CHCOCH₂CH₃ does not give yellow ppt. of iodoform.
(b) Add ammoniacal AgNO₃ solution (Tollen’s reagent), HCOOH gives silver mirror while CH₃CH₂COOH does not.

(iii) CH₃COCH₃ < CH₃CH₂OH < CH₃COOH
OR
(i) Etard’s reaction

(ii) C₆H₅COCH₃ < CH₃CHO < HCHO

(iii) Chlorine is an electron-withdrawing group (-I inductive effect).
It withdraws the electrons from the C to which it is attached and this effect is transmitted throughout the chain. As a result, the electrons are withdrawn more strongly towards 0 of the O-H bond and promote the release of the proton. Consequently, Cl — CH₂COOH is more acidic than acetic acid.

Therefore, pKa of ClCH₂COOH is less than that of CH₃COOH.

(iv) CH₃ CH₂ CH = CH CH₂ CN CH₃ CH₂ CH = CH CH₂ CHO

(v) Two isomers are CH₃COCH₃ and CH₃CH₂CHO
Since A forms yellow ppt. of iodoform on heating with NaOH and l₂, it is CH₃COCH₃.
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3Nal + 3H₂O
B does not give iodoform, it is CH₃CH₂CHO.
A: CH₃COCH₃ B: CH₃CH₂CHO

Question 26.
Write the structures of A, B, C, D, and E in the following reactions:

OR
(i) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(ii) Draw the structure of the semicarbazone of ethanal.
(iii) Why pKa of F-CH₂-COOH is lower than that of Cl – CH₂ – COOH?
(iv) Write the product in the following reaction:
CH₃ – CH = CH – CH₂CN
(v) How can you distinguish between propanal and propanone? (CBSE Delhi 2016)
Answer:

OR
(i) Cannizzaro’s reaction: Aldehydes that do not contain a-hydrogen undergo self-oxidation and reduction on treatment with cone. KOH.


(iii) Fluorine is more electronegative than Cl and therefore, will have a greater electron-withdrawing effect (-I effect). As a result, the carboxylate ion will be stabilized to a larger extent and therefore fluoro acetic acid will be a stronger acid than chloroacetic acid. Hence, the pK_a of FCH₂COOH will be lower than that of ClCH₂COOH.

(iv) CH₃ — CH = CH — CH₂ CN CH₃ CH = CH CH₂ CHO

(v) Propanal gives red ppt with Fehling solution but propanone does not.

Question 27.
(i) Give reasons :
(a) HCHO is more reactive than CH₃-CHO towards the addition of HCN. (CBSE 2018C)
(b) pK_a of O₂N—CH₂—COOH is lower than that of CH₃—COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
OR
(i) Write the structure of the product(s) formed :
(a) CH₃ — CH₂ — COOH
(b) C₆H₅COCl
(c) 2HCHO

(ii) How will you bring the following conversions in not more than two steps :
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
Answer:
(i) (a) Due to + I effect of the methyl group is CH₃CHO.
(b) Due to – I effect of the nitro group in nitroacetic acid.
(c) Due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) (a) Add NaOH and l₂ to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and l₂ to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.
OR

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 11 Alcohols, Phenols and Ethers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 11 Important Extra Questions Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following in increasing order of their boiling point:
CH₃CH₂OH, CH₃CHO, CH₃—O—CH₃ (CBSE AI 2019)
Answer:
CH₃—O—CH₃ < CH₃CHO < CH₃CH₂OH

Question 2.
Arrange the following in increasing order of their acidic character:
Ethanol, Phenol, Water (CBSE AI 2019)
Answer:
Ethanol <Water < Phenol

Question 3.
How will you convert ethanol to ethene? (CBSE 2011)
Answer:

Question 4.
Draw the structural formula of the 2-methylpropan-2-ol molecule. (CBSE Delhi 2012)
Answer:

Question 5.
Which of the following isomers is more volatile: (CBSE Delhi 2014)
o-nitrophenol or p-nitrophenol?
Answer:
o-nitrophenol

Question 6.
Write the IUPAC name of the given compound: (CBSE 2015)

Answer:
2-Methylpropane-1, 3-diol

Question 7.
Write the IUPAC name of the given compound: (CBSE Delhi 2015)

Answer:
2, 5- Dinitrophenol

Question 8.
Arrange the following in increasing order of their acidic character: Benzoic acid, Phenol, Cresol (CBSE Al 2019)
Answer:
Cresol < Phenol < Benzoic acid

Question 9.
Rearrange the following compounds in the increasing order of their boiling points: (CBSE Al 2013)
CH₃-CHO, CH₃-CH₂-OH, CH₃-CH₂-CH₃
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 10.
An alkoxide is a stronger base than a hydroxide ion. Justify. (CBSE Sample Paper 2012)
Answer:
Due to the presence of electron-donating alkyl group, there is high electron density in alkoxide ion as compared to hydroxide ion. Therefore, the alkoxide ion is more basic than the hydroxide ion.

Question 11.
Why is (±) butan-2-ol optically inactive? (CBSE Delhi 2013, CBSE AI 2013)
Answer:
(±)-Butan-2-ol represents a racemic mixture of (+)-butan-2-ol and (-)-2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, (±) compound is optically inactive.

Question 12.
Write the IUPAC name of the following compound: (CBSE Al 2017)

Answer:
2-Methoxy-2-methylpropane

Question 13.
Write the IUPAC name of the following compound: (CBSE AI 2017)

Answer:
3-Phenylprop-2-en-1-ol

Question 14.
Write the IUPAC name of the following: (CBSEAI 2018)

Answer:
3, 3-Dimethylpentan-2-ol

Alcohols, Phenols and Ethers Important Extra Questions Short Answer Type

Question 1.
Predict the major product obtained when t-butyl bromide reacts with sodium methoxide. Also, give its IUPAC name.
Or
(a) Show the chemical reaction with bond movements and arrows for the nucleophilic attack of water on carbocation in acid catalysed hydration of alkenes.
(b) Give IUPAC name for the following: (CBSE 2019C)

Answer:

OR

Mechanism: hydration of alkenes
The mechanism of the reaction involves the following steps:
Step 1. Alkene gets protonated to form a carbocation by the electrophilic attack of H₃O⁺:

Step 2. Nucleophile (H₂0) attacks the carbocation forming protonated alcohol.

Step 3. Loss of H+ from oxygen (deprotonation) to form alcohol

(b) 2,6-dimethylphenol

Question 2.
How are the following conversions carried out?
(i) Propene to propan-2-ol
Answer:

(ii) Ethyl magnesium chloride to propan-1-ol (CBSE Delhi 2010)
Answer:

Question 3.
How would you obtain:
(i) Picric add (2, 4, 6-trlnitrophenol) from phenol.
Answer:

(ii) 2-Methyl propene from 2-methyl propanol? (CBSE Delhi 2011)
Answer:

Question 4.
Write the chemical equations involved in the following reactions:
(i) Kolbe’s reaction
(ii) Friedel-Crafts acetylation of anisole (CBSE Delhi 2016)
OR
How do you convert:
(i) Phenol to toluene
(ii) Formaldehyde to ethanol?
Answer:
(i) Kolbe’s reaction. Sodium phenoxide reacts with CO₂ under pressure (6-7 atm) at 400 K to form sodium salicylate which upon acidification with HCL gives salicylic acid.

(ii) Friedel-Crafts acetylation of anisole. On reacting anisole with acetyl chloride in the presence of anhyd. AlCl₃, 2-methoxy acetophenone and 4-methoxy acetophenone are formed.

(i) Phenol to toluene

(ii) Formaldehyde to ethanol

Question 5.
Explain the mechanism of the following reaction: (CBSE Delhi 2013, 2016)

Answer:
The mechanism for the formation of diethyl ether from ethanol at 413 K is given below:
(i) Ethyl alcohol gets protonated in the presence of H⁺.

(ii) Nucleophilic attack by another alcohol (unprotonated) molecule occurs on the protonated alcohol with the elimination of a molecule of water.

(iii) Oxonium ion [oses a proton to form an ether.

Question 6.
How will you convert:
(i) Propene to Propan-2-ol?
Answer:

(ii) Phenol to 2, 4, 6 – trinitrophenol? (CBSE Delhi 2013)
Answer:

Question 7.
How will you convert:
(i) Propene to propan-1 -ol?
Answer:

(ii)Ethanal to propan-2-ol?
Answer:

Question 8.
Give chemical tests to distinguish between (CBSE Sample Paper 2011)
(i) Isopropyl alcohol and n-propyl alcohol
Answer:
Isopropyl alcohol gives the iodoform test. On heating with NaOH/l₂ or NaOl, isopropyl alcohol forms a yellow precipitate of iodoform (CHI₃).

(ii) Phenol and alcohol.
Answer:
Phenol reacts with neutral FeCl₃ solution to give red-violet complex, whereas alcohol does not give this test.

Question 9.
(i) Arrange the following compounds in the increasing order of their acid strength:
p-cresol, p-nitrophenol, phenol
Answer:
p-cresol < phenol < p-nitrophenol

(ii) Write the mechanism (using curved arrow notation) of the following reaction:
(a) CH₂ = CH₂ CH₃-CH₂ + H₂O
OR
Write the structures of the products when butan-2-ol reacts with the following: (CBSE AI 2017)
(i) CrO₃
(ii) SOCl₂
Answer:

Or

Question 10.
Alcohols are easily protonated in comparison to phenols. Explain. (CBSE AI 2016)
Answer:
In alcohols, the electron-releasing inductive effect (+ I effect) of the alkyl group attached to the carbon having the -OH group increases the electron density on the oxygen atom. Therefore, alcohols are easily protonated.

On the other hand, in the case of phenol, the oxygen atom acquires a partial positive charge due to resonance.
Thus, it is not protonated.

Question 11.
Explain why ortho nitrophenol is more acidic than ortho methoxy phenol. (CBSE AI 2015)
Answer:
This is because -NO₂ (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH₃ group is an electron-releasing group and will decrease +ve charge on oxygen making it less acidic as the O-H bond will not break easily.

Question 12.
Anisole on reaction with HI gives phenol and CH₃I as main products and not iodobenzene and CH₃OH. Why? (CBSE AI 2016)
Answer:
It this reaction protonated anisole, i.e. methyl phenyl oxonium ion, is first formed and then the halide ion attacks this protonated anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.

Question 13.
The boiling points of ethers are lower than their corresponding isomeric alcohols. Explain. (CBSE AI 2012)
Answer:
Ethers have low polarity and therefore, do not show any association by intermolecular hydrogen bonding. On the other hand, their isomeric alcohols have strong intermolecular hydrogen bonding and therefore, their boiling points are high.

Question 14.
Butan-1-ol has a higher boiling point than diethyl ether. Give reason.
Answer:
Butan-1-ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and a large amount of energy is required to break these bonds and hence its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

Question 15.
Write the mechanism of acid dehydration of ethanol to yield ethene. (CBSE Sample Paper 2018)
Answer:
The reaction is believed to occur as follows:
(i) Formation of protonated alcohol: Alcohol combines with a proton to form a protonated alcohol.

(ii) Formation of the carbocation: The protonated alcohol loses a water molecule to form a carbocation.

(iii) Elimination of a proton to form alkene: The carbocation then eliminates a proton and undergoes rearrangement of electrons to form the alkene.

The acid used in step (i) is released in step (iii). To drive the equilibrium reaction in the forward direction, ethene is removed as soon as it is formed.

Alcohols, Phenols and Ethers Important Extra Questions Long Answer Type

Question 1.
What happens when
(a) Sodium phenoxide is treated with CH₃Cl?
Answer:

(b) CH₂ = CH – CH₂ – OH is oxidised by PCC?
Answer:

(c) Phenol is treated with CH₃COCI/anhydrous AlCl₃?
Write chemical equations in support of your answer.
Answer:

Question 2.
(a) How will you convert the following:
(i) Phenol to benzoquinone
Answer:

(ii) Propanone to 2-methyl propane-2-ol
Answer:

(b) Why does propanol have a higher boiling point than that butane? (CBSE 2019C)
Answer:
The molecules of propanol are held together by intermolecular hydrogen bonding while butane molecules have only weak van der Waals forces of attraction. Since hydrogen bonds are stronger than van der Waals forces, therefore, propanol has a higher boiling point than butane.

Question 3.
Identify the product formed when propan-1 -ol is treated with a cone. H₂S0₄ at 413 K. Write the mechanism involved for the above reaction. (CBSE Sample Paper 2019)
Answer:
(a) 1-Propoxypropane is formed.
Mechanism involved:
Step 1: Formation of protonated alcohol. Propanol gets protonated in the presence of H⁺.

Step 2: Nucleophilic attack. Due to the presence of a +ve charge on the oxygen atom, the carbon of the CH₂ part becomes electron deficient. As a result, a nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a molecule of water.

Step 3: Deprotonation. Oxonium ion loses a proton to form an ether.

Question 4.
Write the equations involved in the following reactions: (CBSE 2013, 2014)
(i) Reimer-Tiemann reaction
Answer:
Reimer-Tiemann reaction: When phenol is refluxed with chloroform in the presence of aqueous caustic alkali at 340 K, an aldehydic group (CHO) gets introduced in the ring at a position ortho to the phenolic group. Ortho hydroxy benzaldehyde or salicylaldehyde is formed as the product of the reaction.

(ii) Williamson’s ether synthesis
Answer:
Williamson’s ether synthesis. This is used to prepare symmetrical and unsymmetrical ethers by treating alkyl halide with either sodium alkoxide or sodium phenoxide.

Aryl halides cannot be used for the preparation of alkyl-aryl ethers because of their low reactivity.

Question 5.
Draw the structure and name the product formed if the following alcohols are oxidised. Assume that an excess of the oxidising agent is used. (CBSE Delhi 2012)
(i) CH3CH2CH2CH2OH
Answer:

(ii) 2-butenol
Answer:

(iii) 2-methyl-1-propanol
Answer:

Question 6.
Write the main product(s) in each of the following reactions: (CBSE Delhi 2016)

Answer:

Answer:

Answer:

Question 7.
Give reasons for the following: (CBSE Delhi 2016, Outside Delhi)
(i) Protonation of phenols Is difficult whereas ethanol easily undergoes protonation.
Answer:
In phenols, the oxygen atom acquires a partial positive charge due to resonance and therefore, it is not easily protonated. On the other hand, in ethanol, the alkyl group is an electron-releasing group and increases the electron density on 0 atoms. Therefore, ethanol is easily protonated.

(ii) Bolting point of ethanol is higher than that of dimethyl ether.
Answer:
The molecules of ethanol are held together by intermolecular hydrogen bonding while dimethyl molecules have only weak van der Waals forces of attractions. Since hydrogen bonds are stronger than van der Waals forces, ethanol has a higher boiling point than dimethyl ether.

(iii) Anisole on reaction with Hl gives phenol and CH₃-I as main products and not iodobenzene and CH₃OH.
Answer:
This is because, during the reaction, the attack of halide ion occurs to the protonated anisole, i.e. methyl phenyl oxonium ion, which is formed during the protonation of anisole. Due to steric hindrance of the bulky phenyl group, the attack preferably occurs to the alkyl group forming methyl iodide and phenol.

Question 8.
How do you convert the following: (CBSE Delhi 2015, Delhi)
(i) Phenol to anisole
(ii) Propan-2-ol to 2-methylpropan-2-ol
(iii) Aniline to phenol?
OR
(i) Write the mechanism of the following reaction:
2CH₃CH₂OH CH₃CH₂—O—CH₂CH₃
(ii) Write the equation involved In the acetylation of salicylic acid.
Answer:
(i) Phenol to anisole

(iii) Aniline to phenol

Or
(i) The mechanism for the formation of ether from ethanol at 413 K is a nucleophilic bimolecular reaction as given below:
(a) Ethyl alcohol gets protonated in the presence of H⁺

(b) Due to the presence of a +ve charge on the oxygen atom, the carbon of CH₂ part of CH₃CH₂ becomes electron deficient. As a result, nucleophilic attack by another alcohol molecule (unprotonated) occurs on the protonated alcohol with the elimination of a

(c) Oxonium ion loses a proton to form an ether.

Question 9.
Give reasons for the following: (CBSE 2015, Outside Delhi)
(i) o-nitrophenol is more acidic than o-methoxyphenyl.
Answer:
This is because —NO₂ (nitro group) is an electron-withdrawing group and will increase the +ve charge on oxygen to make it more acidic. On the other hand, the -OCH₃ group is an electron¬releasing group and will decrease +ve charge on oxygen making it less acidic as O-H bond will not break easily.

(ii) Butan-1-oi has a higher boiling point than diethyl ether.
Answer:
Butan-1 -ol has intermolecular hydrogen bonding between their molecules. Therefore, it exists as associated molecules and large amount of energy is required to break these bonds and hence, its boiling point is high. But diethyl ether does not show any association by intermolecular hydrogen bonding. Hence, its boiling point is low.

(iii) (CH₃)₃C – O – CH₃ on reaction with HI gives (CH₃)₃C – I and CH₃ – OH as the main products and not (CH₃)₃C – OH and CH₃ – I.
Answer:
The reaction:

gives (CH₃)₃C – I and CH₃OH as the main products and not (CH₃)₃COH and CH₃I. This is because the reaction occurs by the S_N1 mechanism and the formation of products is governed by the stability of carbocation formed from the cleavage of the C-0 bond in the protonated ether. Since tert. butyl carbocation, (CH₃)₃C⁺ is more stable than methyl carbocation, CH₃, the cleavage of C-0 gives a more stable carbocation, [(CH₃)₃C]⁺ and methanol. Then, iodide ion, I^– attacks this tert. butyl carbocation to form tert. butyl iodide.

Question 10.
(i) Write the mechanism of the following reaction: (CBSE 2015, Outside Delhi)

Answer:
HBr → H⁺ + Br^–
(a) H+ attacks oxygen of O-H to form protonated alcohol

(b) Protonated alcohol Loses a molecule of water to form a carbocation.

(C) Br attacks the carbocation to form bromoatkane
CH₃CH₂+ + Br^– → CH₃CH₂Br (Bromoethane)

(ii) Write the equation involved in the Reimer-Tiemann reaction.

Question 11.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈0. Isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl, give the same product (D).
(i) Write the structures of (A), (B), (C) and (D).
Answer:
A = CH₃CH₂CH₂CHO
B = CH₃COCH₂CH₃
C = (CH₃)₂CHCHO
D = CH₃CH₂CH₂CH₃

(ii) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? (CBSE 2018 C)
Answer:
B

Question 12.
(a) How will you convert the following:
(i) Phenol to benzene
Answer:

(ii) Propene to propanol
Answer:

(b) Why is ortho-nitrophenol more acidic than ortho-methoxy phenol? (CBSE 2018C) OH
Answer:
Due to the strong -R and -I effect of the -NO₂ group, the electron density in the O-H bond decreases and hence the loss of a proton becomes easy.

Moreover, the o-nitrophenoxide formed after the Loss of a proton is stabilised by resonance.

o-nitrophenoxide ion is stabilised by resonance and hence o-nitrophenol is a stronger acid. On the other hand, due to the +R effect of the -OCH3 group, the electron density in the O-H bond increases and this makes the loss of proton difficult.

Furthermore, after the Loss of proton o-methoxyphenoxide ion left is destabilised by resonance.

The two negative charges repel each other and therefore, destabilise the o-methoxyphenoxide ion. Thus, o-nitrophenol is more acidic than o-methoxyphenyl.

Question 13.
(a) How will you convert the following:
(i) Ethanal to propan-2-ol
Answer:

(ii) Phenol to 2-hydroxy acetophenone
Answer:

(b) Why is phenol more acidic than ethanol? (CBSE 2019C)
Answer:
It is because phenoxide ion is more stable than ethoxide ion. Phenols are more acidic than alcohols. The greater acidic character of phenoLs as compared to alcohols can be explained on the basis of resonance. Phenol is a resonance hybrid of the following structures:

It is clear that three structures of phenol (III, IV and V) have a +ve charge on the oxygen of the —OH group. This oxygen attracts the electron pair of O—H bond strongly towards itself, weakens the O—H bond and, therefore, facilitates the release of W. Similarly, the phenoxide ion is resonance stabilised as follows:

Thus, we observe that both phenol and phenoxide ion are stabilised by resonance. Now if we carefully observe the resonance structures, we observe that phenoxide ion is more resonance stabilised than phenol. In phenol three contributing structures (III, IV and V) have both positive and negative charges and therefore, these will be unstable. These structures will require energy to separate the charge and therefore, will be unstable.

On the other hand, there is no structure in phenoxide ion which requires charge separation. Thus, the resonance hybrid of phenol is less stable than phenoxide ion and the reaction is very much in favour of the phenoxide ion. Therefore, the phenol is more acidic.

On the other hand, in the case of alcohols, neither the alcohol nor the alkoxide ion is stabilised by resonance.

Thus, phenols are more acidic than alcohols.

Question 13.
(a) How do you convert the following?
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write the mechanism of the following reaction:

(c) Why phenol undergoes electrophilic substitution more easily than benzene?
OR
(a) Account for the following:
(i) o-Nitrophenol is more steam volatile than p-nitrophenol.
(ii) Tert-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of tert-butyl methyl ether.

(b) Write the reaction involved in the following:
(i) Reimer-Tiemann reaction
(ii) Friedel-Crafts Alkylation of Phenol

(c) Give a simple chemical test to distinguish between Ethanol and Phenol. (CBSE Delhi 2019)
Answer:
(i)
(ii) Ethanol to Propan-2-ol

(b) Mechanism CH₃
(i) Ethanol combines with a proton to form protonated alcohol.

(ii) Protonated alcohol loses a water molecule to form a carbocation.

(iii) Carbocation eliminates a proton and undergoes rearrangement of electrons to form an alkene.

(c) In phenol, the -OH group is an activating group and therefore, electrophilic substitution reaction undergoes more easily in phenol than benzene.
OR
(a) (i) o-Nitrophenol is more steam volatile than p-nitrophenol because of chelation due to intramolecular hydrogen bonding.

(ii) Tert-Butyl chloride reacts with sodium methoxide to give 2-methylpropene.

This is because alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to an elimination reaction.

(b) Reimer-liemann reaction

(C) Phenols react with neutral Fecl₃ to give violet colour, while ethanol does not give any colour.

Question 14.
(i) Write steps to carry out the conversion of phenol to aspirin.
Answer:

(ii) What happens when benzene diazonium chloride Is heated with water?
Answer:

(iii) Write the structures of the Isomers of alcohols with molecular formula C₄H₁₀O. Which of these exhibits optical activity?
Answer:

(ii) is optically active.

Question 15.
How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol
Answer:

(ii) Methyl magnesium bromide to 2-methyl propane-2-ol
Answer:

(iii) Propene to propan-2-ol
Answer:

(iv) Ethylmagnesium chloride to propan-1-ol. (CBSE Delhi 2010, 2013)
Answer:

Question 16.
How will you convert:
(i) Phenol to 2, 4, 6-trlnftrophenot
Answer:

(ii) Propan-2-ol to propanone
Answer:

(iii) Phenol to 2, 4, 6-trlbromophenol
Answer:

(iv) Propene to propan-1-ol
Answer:

(v) Ethanal to propan-2-ol? (CBSE DelhI 2013)
Answer:

Question 17.
How will you convert
(i) Propan-2-ol to 2-methylpropan-2-ol (CBSE Delhi 2015)
Answer:

(ii) Aniline to phenol (CBSE Delhi 2015)
Answer:

(iii) Ethanol to propane nitrile (CBSE AI 2015)
Answer:

(iv) Phenol to toluene (CBSE AI 2016)
Answer:

(v) Formaldehyde to ethanol? (CBSE AI 2016)
Answer:

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 10 Haloalkanes and Haloarenes.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 10 Important Extra Questions Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Important Extra Questions Very Short Answer Type

Question 1.
Define ambident nucleophile with an example. (CBSE Delhi 2019)
Answer:
The nucleophile which has two sites through which it can attack is called ambident nucleophile, for example, nitrite ion (NO₂^–). It can attack either through 0 (-O^– -N = O) or through N

Question 2.
What happens when bromine attacks CH₂ = CH – CH₂ -C ≡ CH? (CBSE AI2012)
Answer:
The Colour of bromine gets discharged.

Question 3.
What happens when CH₃ – Br is treated with KCN? (CBSE Delhi 2013)
Answer:
Ethane nitrile is formed.
CH₃Br + KCN → CH₃C ≡ N + KBr (Ethane nitrile)

Question 4.
Write the IUPAC name of

Answer:
4-Bromo-4-methylpent-2-ene

Question 5.
What happens when ethyl chloride is treated with aqueous KOH? (CBSE Delhi 2013)
Answer:
Ethyl alcohol is formed.
CH₃CH₂Cl + KOH(aq) → CH₃CH₂OH + KCl Ethyl alcohol

Question 6.
Write the IUPAC name of (CH₃)₂CH.CH(Cl)CH₃. (CBSE Delhi 2013)
Answer:
2-Chloro-3-methyl butane

Question 7.
Which compound in the following pair undergoes a faster S_N1 reaction? (CBSE Delhi 2013)

Answer:

Question 8.
Write the IUPAC name of the following compound: (CBSE AI 2013)

Answer:
3-Chloro-2, 2-dimethylbutane

Question 9.
Which aerosol depletes the ozone layer? (CBSE AI 2013)
Answer:
Chlorofluoro compounds of methane (CCl₂F₂) called freons.

Question 10.
Write the IUPAC name of the following compound: (CBSE AI 2013)

Answer:
2-Bromo-4-chloropentane

Question 11.
Why is CH₂ = CH – CH₂ – Cl more easily hydrolysed than CH₃ – CH₂ – CH₂ – Cl? (CBSE AI 2019)
Answer:
CH₂ = CH – CH₂ – Cl gives allylic carbocation as intermediate, which is resonance stabilised.

Therefore, CH₂ = CH – CH₂ – Cl is more readily hydrolysed.

Question 12.
Write the IUPAC name of the following compound:
CH₂ = CHCH₂Br (CBSE 2011)
Answer:
3-Bromoprop-1-ene

Question 13.
Write the IUPAC name of the following compound: (CH₂ )₃CCH₂Br (CBSE 2011)
Answer:
1 -Bromo 2,2-dimethylpropane

Question 14.
Write the IUPAC name of the following: (CBSE 2012)

Answer:

3-Bromo- 2-methylpropene

Question 15.
Identify the chiral molecule in the following pair: (CBSE 2014)

Answer:

2-Chlorobutane is a chiral molecule.

Question 16.
Which would undergo S_N2 reaction faster in the following pair and why? (CBSE Delhi 2015)

Answer:
CH₃CH₂Br would undergo S_N2 reaction faster because of less steric hindrance.

Question 17.
Out and which is more reactive towards S_N1 reaction and why? (CBSE Delhi 2016)
Answer:
is more reactive.
The S_N1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.


Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the S_N1 reaction.

Question 18.
Write the structure of the product formed when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether. (CBSE Al 2018)
Answer:

Question 19.
Which of the following two reactions is S_N2 and why? (CBSE AI2016)

Answer:
Reaction (i) is S_N2 because it proceeds by inversion of configuration.

Question 20.
Why is chloroform kept in dark coloured bottles? (CBSE AI 2019)
Answer:
Chloroform reacts with atmospheric oxygen in presence of light to produce an extremely poisonous gas called phosgene.

Chloroform is stored in dark coloured bottles so that it is not exposed to light and the formation of phosgene is prevented.

Question 21.
Write one stereochemical difference between S_N1 and S_N2 reactions. (CBSE Delhi 2019)
Answer:
S_N1 reaction forms a racemic mixture while S_N2 reaction results in inversion of configuration.

Question 22.
A solution of KOH hydrolyses CH₃CHClCH₂CH₃ and CH₃CH₂CH₂CH₂Cl. Which one of these is more easily hydrolysed? (CBSE Delhi 2010)
Answer:
CH₃CHClCH₂CH₃

Question 23.
Write the structure of 2, 4-dinitrochlorobenzene. (CBSE Delhi 2017)
Answer:

Question 24.
Write the structure of 1-Bromo-4- chlorobut-2-ene. (CBSE Delhi 2017)
Answer:

Question 25.
Write the structure of 3-Bromo-2- methylprop-1 -ene. (CBSE Delhi 2017)
Answer:

Question 26.
Write IUPAC name of the given compound: (CBSE Delhi 2019)

Answer:
4-Chlorobenzene sulphonic acid

Question 27.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? (CBSE Sample paper 2018)
Answer:
Benzyl chloride;
Due to resonance, stable benzyl carbocation is formed.

Haloalkanes and Haloarenes Important Extra Questions Short Answer Type

Question 1.
Which one of the following compounds will undergo hydrolysis at a faster rate by S_N1 mechanism? Justify. (CBSE Sample Paper 2019)

Answer:
The following compound will undergo S_N1 faster:

Greater the stability of the carbocation, greater will be its ease of formation from the corresponding halide and faster will be the rate of reaction. The benzylic carbocation formed gets stabilised through resonance.

CH₃CH₂CH₂Cl forms 1° carbocation, which is less stable than benzylic carbocation.

Question 2.
How would you differentiate between S_N1 and S_N2 mechanisms of substitution reactions? Give one example of each. (CBSE 2010)
Answer:
The S_N1 reaction occurs in two steps and the reaction is of the first order.
The S_N2 reaction occurs in one step and the reaction is of second order. These can also be distinguished as:

In S_N1 reaction retention of the configuration takes place while in S_N2 reaction inversion of the configuration takes place.

S_N2 reaction:

S_N1 reaction:

Question 3.
(i) State one use each of DDT and iodoform.
Answer:
DDT is used as an insecticide for killing mosquitoes and other insects. chloroform is used as an antiseptic.

(ii) Which compound in the following couples will react faster in S_N2 displacement and why?
(a) I -Bromopentane or 2-bromopentane
Answer:
1-Bromopentane (1°) will react faster than 2-bromopentane (2°) by S_N2 displacement because It is Less stericaLly hindered in the transition state.

(b) I-Bromo-2-methyl butane or 2-Bromo-2-methyl butane (CBSE Delhi 2010)
Answer:
1 -Bromo-2-methyl butane (1°) reacts faster than 2-bron,o-2-methyl butane (2°) because being primary, it will have Less steric hindrance.

Question 4.
Although chlorine Is an electron-withdrawing group, yet It Is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why Is It so? (CBSE 2012)
Answer:
Chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in eLectrophiLic aromatic substitution reactions due to eLectron reLeasing resonance effect (+ R-effect).

As can be seen from the above resonating structures that -Ct increases electron density at ortho- and para-positions due to the + R-effect. Therefore, the attack of electrophile takes pLace at ortho- and paro- positions and not at meta- positions.

Question 5.
(i) Which alkyl halide from the following pair Is chiral and undergoes a faster S_N2 reaction? (CBSE Delhi 2014)

Answer:
is chiral

(ii) Out of S_N1 and S_N2 which reaction occurs with
(a) Inversion of configuration
Answer:
S_N2

(b) Racemisation
Answer:
S_N1

Question 6.
Draw the structure of major moon halo product In each of the following reactions: (CBSE Delhi 2014)

Answer:

Answer:

Question 7.
Propose the mechanism of the reaction taking place when
(i) (-) – 2 – Bromooctane reacts with sodium hydroxide to form (+) – octane-2-ol.
Answer:

(ii) 2-Bromopentane is heated with (ale.) KOH to form alkenes. (CBSE Sample Paper 2011)
Answer:

Question 8.
Out of and which is more reactive towards S_N1 reaction
Answer:
is more reactive.
The S_N1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.

Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the S_N1 reaction.

Haloalkanes and Haloarenes Important Extra Questions Long Answer Type

Question 1.
(i) Out of (CH₃)₃C-Br, and (CH₃)₃C-I, which one is more reactive towards S_N1 and why?
Answer:
(CH₃)₂ C-I, because it can readily form stable carbocation because the C – l bond is weaker than the C – Br bond.

(ii) Write the product formed when p-nitro chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
Answer:

(iii) Why are dextro- and laevorotatory isomers of Butan-2-ol difficult to separate by fractional distillation? (CBSE Delhi 2019)
Answer:
Dextro- and laevorotatory isomers of butan-2-ol are enantiomers of each other and both have the same boiling point. Hence, these cannot be separated by fractional distillation.

Question 2.
Give reasons for the following:
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
Answer:
Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide because I- ion is a better leaving group than Br- ion.

(ii) (±) 2 – Butanol is optically inactive.
Answer:
(±) 2 – Butanol represents a racemic mixture of (+) 2-butanol and (-) 2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, the (±) compound is optically inactive.

(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃ — X. (CBSE 2013)
Answer:
In halobenzene, there is delocalisation of electrons due to resonance. For example, chlorobenzene is considered to be a resonance hybrid of the following structures:

It is evident that the contribution of structures III, IV and V imparts a partial double bond character to the carbon-chlorine bond. Therefore, the C-X bond length in halobenzene is less than the C-X bond length in CH₃-X, which has only a single C-X bond.

Question 3.
(i) Draw the structures of major moon halo products in each of the following reactions:

Answer:

Answer:

(ii) Which halogen compound in each of the following pairs will react faster in SN2 reaction:
(a) CH₃Br or CH₃I
Answer:
CH₃I

(b) (CH₃)₃C – Cl or CH₃ – Cl (CBSE 2014)
Answer:
CH₃Cl

Question 4.
How would you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile
OR
Write the main products when
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
(iii) methyl chloride is treated with AgCN.
Answer:

OR

Question 5.
Give reasons:
(i) C – Cl bond length in chlorobenzene is shorter than C — Cl bond length in CH₃ — Cl.
Answer:
In chlorobenzene, there is delocalisation of electrons due to resonance.
It is considered to be a resonance hybrid of the following structures.

The contribution of structure III, IV and V imparts a partial double bond character to the C – Cl bond.

This is confirmed by X-ray analysis which shows that the C – Cl bond length in chlorobenzene is 1.69 whereas the C – Cl bond length in ethyl chloride is 1.82 A. There is no resonance in CH₃Cl and hence no partial double bond character.

(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Answer:
In chlorobenzene, the C of C-Cl bond is sp²-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp³-hybridised.

Therefore, the sp²-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp³-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp³ hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride. Thus, chlorobenzene is less polar than cyclohexyl chloride.

(iii) S_N1 reactions are accompanied by racemisation in optically active alkyl halides. (CBSE 2016)
Answer:
The S_N1 reactions proceed through the formation of carbocations. In the case of optically active alkyl halides, the product formed is a racemic mixture. This is because the intermediate carbocation is planar species, therefore, the attack of the nucleophile, OFT ion can take place from both the faces (front and rear) with equal ease. As a result, a 50: 50 mixture of the two enantiomers (laevo and dextro) is formed. Thus, the product formed is a racemic mixture (±) which is optically inactive.

For example, hydrolysis of optically active 2-bromobutane results in the formation of a racemic mixture (±)-butan-2-ol.

Question 6.
The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards the S_N2 reaction.
Answer:
1-Bromopentane

(ii) Write the compound which is optically active.
Answer:
2-Bromopentane

(iii) Write the compound which is most reactive towards the S_N2 elimination reaction.
Answer:
2-Bromo-2-methyl butane

Question 7.
Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE Delhi 2016)
Answer:
In chlorobenzene, the C of C-Cl bond is sp²-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.

Therefore, the sp²-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp³-hybrid C of cyclohexyl chloride. As a result, the sp² hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp³ hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride.

Thus, chlorobenzene is less polar than cyclohexyl chloride. In other words, the magnitude of negative charge (δ-) is less on the Cl atom of chlorobenzene than in cyclohexyl chloride. Further, due to the delocalisation of the lone pair of electrons of the Cl atom over the benzene ring due to resonance, the C-Cl bond in chlorobenzene acquires some double-bond character. On the other hand, the C-Cl bond in cyclohexyl chloride is a pure single bond.

Since dipole moment is a product of charge and distance, therefore, chlorobenzene has a lower dipole moment than cyclohexyl chloride due to the lower magnitude of charge (δ-) on Cl atom and small C-Cl distance.

(ii) alkyl halides, though polar, are immiscible with water,
Answer:
Alkyl halides are polar molecules and therefore, their molecules are held together by dipole-dipole forces. On the other hand, the molecules of H₂0 are held together by hydrogen bonds. When alkyl halides are added to water, the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides are immiscible (not soluble) in water.

(iii) Grignard reagents should be prepared under anhydrous conditions? (CBSE Sample Paper 2011)
Answer:
Grignard reagents are very reactive. They react with the moisture present in the apparatus or the starting materials (RX or Mg).
RMgX + HOH → R-H + Mg(OH)X
Therefore, Grignard reagents must be prepared in anhydrous conditions.

Question 8.
Which one of the following compounds will undergo a faster hydrolysis reaction by the S_N1 mechanism? Justify your answer.

OR
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism. (CBSE Sample paper 2018-19)
Answer:
C₆ H₅CH₂Cl will undergo S_N1 reaction faster.

The carbocation formed by C₆ H₅CH₂Cl gets stabilised through resonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.

OR

The following isomer will exhibit enantiomerism:

IUPAC name: 1,2-Dichloropropane.

Question 9.
(i) Identify the chiral molecule in the following pair:
and
Answer:

(ii) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
Answer:

(iii) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1- methylcyclohexane with alcoholic KOH. (CBSE 2018)
Answer:

Question 10.
(a) Out of and which one is more reactive towards S_N2 reaction and why?
Answer:
because it is a primary halide and has less steric hindrance and therefore, undergoes S_N2 action faster.

(b) Out of and which one is more reactive towards nucleophilic substitution reaction and why?
Answer:
is more reactive because of the presence of electron-withdrawing – NO₂ group.

(c) Out of and which one is optically active and why? (CBSE AI 2019)
Answer:
is optically active because of the presence of chiral carbon. OH

Question 11.
Write the products of the following reactions:
(i) CH₃ CH₂ -CH = CH₂ + HCl →
Answer:

(ii) Me₂CHCH₂OH
Answer:
Me₂CHCH₂OH MeCHCH₂Cl

Answer:

(iv) Me₂C = CMe₂
Answer:

(v) (CBSE AI 2016)
Answer:

Question 12.
(i) In the following pairs of the halogen compounds, which would undergo S_N2 faster?
Answer:
S_N2 reaction proceeds through the formation of a transition state involving the bonding of carbon to five atoms or groups. The reactivity is decided by the stability of the transition state on the basis of steric hindrance. The reactivity follows the order: CH₃ > 1° > 2° > 3° halide.
Therefore,

Answer:
is primary alkyl halide and hence undergoes S_N2 reaction faster.

Answer:
I will undergo S_N2 reaction faster because iodine is a better leaving group due to its large size and hence it will be released at a faster rate in the presence of incoming nucleophile.

Answer:
(1 -Bromo-2, 2-dimethylpentane) reacts faster because it is 1 ° alkyl halide.

(CBSE AI 2009)
Answer:
(2-Methyl-1-bromopropane) reacts faster because it has lesser steric hindrance in the transition state.

Answer:
CH₃ CH₂ Br would undergo S_N2 reaction faster because of less steric hindrance.

(ii) Which one of the following pairs undergoes S_N1 substitution reaction faster and why?
Answer:
S_N1 reaction occurs through the formation of a carbocation. Therefore, the greater the stability of the carbocation, the faster is the rate of an S_N1 reaction. Since the stability of carbocation follows the order 3° > 2° > 1° > CH₃, therefore,

Answer:
Out of or reacts faster because it invoLves the formation of 3° -carbocation while involves the formation of V -carbocation

(CBSE AI 2009)
Answer:
Out of or reacts faster because it forms 2° carbocation while involves 1° carbocation.

(CBSE AI 2015)
Answer:
undergoes SN1 reaction faster. This is because the carbocation derived from CH₃(CH₃)₃ C Br is 3° carbocation and is more stable than carbocation (1°) obtained from CH₃ CH₂ Br.

Question 13.
(i) How would you convert the following:
(a) Prop-1-ene to 1-fluoropropane
Answer:

(b) Chlorobenzene to 2-chlorotoluene
Answer:

(ii) Write the main products when
(a) n-butyl chloride is treated with alcoholic KOH.
Answer:

(b) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
Answer:

(c) methyl chloride is treated with AgCN. (CBSE AI2015)
Answer:

Question 14.
Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
Answer:
The molecular formulae of isomers of C4H9Br are CH3

Since the rate of reaction of the compound ‘A’ (C4H9Br) with aqueous KOH depends upon the concentration of compound ‘A’ only, therefore, the reaction occurs by SN1 mechanism and compound ‘A’ is tertiary bromide, i.e. 2-Bromo-2-methylpropane.
(CH₃)₃CBr + KOH(aq) → (CH₃ )₃COH + KBr rate = k[(CH₃)₃CBr]

(ii) Out of these two compounds, which one will be converted to the product with an inverted configuration. (NCERT Exemplar)
Answer:
Since compound ‘B’ is optically active and is an isomer of compound ‘A’ (C₄H₉Br), therefore, compound ‘B’ must be 2-bromobutane. Since the rate of reaction of compound ‘B’ with aqueous KOH depends upon the concentration of compound ‘B’ and KOH, therefore, the reaction occurs by S_N2 mechanism and the product of hydrolysis will have inverted configuration.

Compound ‘B’ will be converted with an inverted configuration.

Question 15.
(i) Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.

Answer:
It is clear from resonating structures that there is an increase in electron density at ortho and para positions. Therefore, the functional group ‘X’ is ortho and para directing.

(ii) Classify the following compounds as primary, secondary and tertiary halides.
(a) 1-Bromobut-2-ene
Answer:
Primary

(b) 4-Bromopent-2-ene
Answer:
secondary

(c) 2-Bromo-2-methylpropane (NCERT Exemplar)
Answer:
tertiary

Question 16.
tert-Butylbromide reacts with aq. NaOH by S_N1 mechanism while n-butyl bromide reacts by an S_N2 mechanism. Why?
Answer:
In general, the SN1 reaction proceeds through the formation of a carbocation. The tert-butyl bromide readily loses Br^– ion to form stable 3° carbocation.
Therefore, it reacts with aqueous KOH by S_N1 mechanism as:

On the other hand, n-butyl bromide does not undergo ionisation to form n-butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an S_N2 mechanism, which occurs in one step through a transition state involving the nucleophilic attack of OH^– ion from the backside with simultaneous expulsion of Br^– ion from the front side.

S_N1 mechanism follows the reactivity order as 3° > 2° > 1° while the S_N2 mechanism follows the reactivity order as 1° >2° >3°
Therefore, tert-butyl bromide (3°) reacts by SN1 mechanism while n-butyl bromide (1°) reacts by the S_N2 mechanism.