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Explore the comprehensive NCERT Solutions for Class 12 Science Biology Chapter 1: “Reproduction In Organisms.” These solutions, accompanied by clear step-by-step explanations, are highly favored by Biology students at the 12th-grade level. Whether you need assistance with homework or effective exam preparation, these solutions for Reproduction In Organisms prove to be a valuable resource. All questions and answers from Chapter 1 of the NCERT Book for class 12 Science Biology are conveniently provided here for free, ensuring a convenient and accessible study aid for students. Page No 17: Question 1: Why is reproduction essential for organisms? ANSWER: Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offspring’s similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct. Page No 17: Question 2: Which is a better mode of reproduction sexual or asexual? Why? ANSWER: Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt to constantly changing and challenging environments. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves. Page No 17: Question 3: Why is the offspring formed by asexual reproduction referred to as clone? ANSWER: A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones. Page No 17: Question 4: Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true? ANSWER: Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival. However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction. Page No 17: Question 5: How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction? ANSWER: Progeny formed from asexual reproduction Progeny formed from sexual reproduction 1. Asexual reproduction does not involve the fusion of the male and the female gamete. Organisms undergoing this kind of reproduction produce offspring’s that are morphologically and genetically identical to them. Sexual reproduction involves the fusion of the male and the female gamete of two individuals, typically one of each sex. Organisms undergoing this kind of reproduction produce offspring’s that are not identical to them. 2. Offsprings thus produced do not show variations and are called clones. Offspring’s thus produced show variations from each other and their parents. Page No 17: Question 6: Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction? ANSWER: Sexual reproduction Asexual reproduction 1 It involves the fusion of the male and female gamete. It does not involves the fusion of the male and the female gamete 2. It requires two (usually) different individuals. It requires only one individual. 3. The individuals produced are not identical to their parents and show variations from each other and also, from their parents. The individuals produced are identical to the parent and are hence, called clones. 4. Most animals reproduce sexually. Both sexual and asexual modes of reproduction are found in plants. Asexual modes of reproduction are common in organisms having simple organizations such as algae and fungi. 5. It is a slow process. It is a fast process. Vegetative propagation is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction. Page No 18: Question 7: What is vegetative propagation? Give two suitable examples. ANSWER: Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants. Examples of vegetative reproduction are: 1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant. 2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil. Page No 18: Question 8: Define (a) Juvenile phase, (b) Reproductive phase, (c) Senescent phase. ANSWER: (a) Juvenile phase: It is the period of growth in …

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Here are the NCERT Solutions for Class 12 Science Chemistry Chapter 7, titled “Chemistry In Everyday Life.” These solutions offer clear and straightforward explanations, making them highly favored by class 12 Science students. They prove to be invaluable for efficiently completing homework assignments and preparing for exams. Free access to all questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 7 is provided here, ensuring a helpful resource for students. Page No 448: Question 16.1: Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor, Why? ANSWER: Most drugs when taken in doses higher than recommended may cause harmful effects and sometimes, may even lead to death. Hence, a doctor should always be consulted before taking any medicine. Page No 448: Question 16.2: With reference to which classification has the statement, ‘ranitidine is an antacid” been given? ANSWER: The given statement refers to the classification of pharmacological effects of the drug. This is because any drug that is used to counteract the effects of excess acid in the stomach is called an antacid. Page No 450: Question 16.3: Why do we require artificial sweetening agents? ANSWER: A large number of people are suffering from diseases such as diabetes and obesity. These people cannot take normal sugar i.e., sucrose as it is harmful for them. Therefore, artificial sweetening agents that do not add to the calorie intake of a person are required. Saccharin, aspartame, and alitame are a few examples of artificial sweeteners. Page No 453: Question 16.4: Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below. (i)  (ii)  ANSWER: (i) (ii) Page No 453: Question 16.5: Following type of nom-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group (s) present in the molecule. ANSWER: Functional groups present in the molecule are: (i) Ether, and (ii) primary alcoholic group Page No 454: Question 16.1: Why do we need to classify drugs in different ways? ANSWER: The classification of drugs and the reasons for classification are as follows: (i) On the basis of pharmacological effect: This classification provides doctors the whole range of drugs available for the treatment of a particular type of problem. Hence, such a classification is very useful to doctors. (ii) On the basis of drug action: This classification is based on the action of a drug on a particular biochemical process. Thus, this classification is important. (iii) On the basis of chemical structure: This classification provides the range of drugs sharing common structural features and often having similar pharmacological activity. (iv) On the basis of molecular targets: This classification provides medicinal chemists the drugs having the same mechanism of action on targets. Hence, it is the most useful to medicinal chemists. Page No 454: Question 16.2: Explain the term target molecules or drug targets as used in medicinal chemistry. ANSWER: In medicinal chemistry, drug targets refer to the key molecules involved in certain metabolic pathways that result in specific diseases. Carbohydrates, proteins, lipids, and nucleic acids are examples of drug targets. Drugs are chemical agents designed to inhibit these target molecules by binding with the active sites of the key molecules. Page No 454: Question 16.3: Name the macromolecules that are chosen as drug targets. ANSWER: The macromolecules chosen as drug targets are carbohydrates, lipids, proteins, and nucleic acids. Page No 454: Question 16.4: Why should not medicines be taken without consulting doctors? ANSWER: A medicine can bind to more than one receptor site. Thus, a medicine may be toxic for some receptor sites. Further, in most cases, medicines cause harmful effects when taken in higher doses than recommended. As a result, medicines may be poisonous in such cases. Hence, medicines should not be taken without consulting doctors. Page No 454: Question 16.5: Define the term chemotherapy. ANSWER: The use of chemicals for therapeutic effect is called chemotherapy. For example: the use of chemicals in the diagnosis, prevention, and treatment of diseases. Page No 454: Question 16.6: Which forces are involved in holding the drugs to the active site of enzymes? ANSWER: Either of the following forces can be involved in holding drugs to the active sites of enzymes. (i) Ionic bonding (ii) Hydrogen bonding (iii) Dipole − dipole interaction (iv) van der Waals force Page No 454: Question 16.7: While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other? ANSWER: Specific drugs affect particular receptors. Antacids and anti-allergic drugs work on different receptors. This is the reason why antacids and anti-allergic drugs do not interfere with each other’s functions, but interfere with the functions of histamines. Page No 454: Question 16.8: Low level of noradrenaline is the cause of depression. What types of drugs are needed to cure this problem? Name two drugs. ANSWER: Anti-depressant drugs are needed to counteract the effect of depression. These drugs inhibit enzymes catalysing the degradation of the neurotransmitter, noradrenaline. As a result, the important neurotransmitter is slowly metabolised and then it can activate its receptor for longer periods of time. Two anti-depressant drugs are: (i) Iproniazid (ii) Phenelzine Page No 454: Question 16.9: What is meant by the term ‘broad spectrum antibiotics’? Explain. ANSWER: Antibiotics that are effective against a wide range of gram-positive and gram-negative bacteria are known as broad spectrum antibiotics. Chloramphenicol is a broad spectrum antibiotic. It can be used for the treatment of typhoid, dysentery, acute fever, pneumonia, meningitis, and certain forms of urinary infections. Two other broad spectrum antibiotics are vancomycin and ofloxacin. Ampicillin and amoxicillin −synthetically modified from penicillin − are also broad spectrum antibiotics. Page No 454: Question 16.10: How do antiseptics differ from disinfectants? Give one example of each. ANSWER: Antiseptics and disinfectants are effective against micro-organisms. However, antiseptics are applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, …

NCERT Solutions for Class 12 Science Chemistry Chapter 7 – Chemistry In Everyday Life Read More »

Here are step-by-step explanations for NCERT Solutions for Class 12 Science Chemistry Chapter 6 on Polymers. These solutions are highly favored by Chemistry students for completing homework and exam preparation. All questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 6, Polymers, are available here for free use. Page No 428: Question 15.1: What are polymers? ANSWER: Polymers are high molecular mass macromolecules, which consist of repeating structural units derived from monomers. Polymers have a high molecular mass (103 − 107u). In a polymer, various monomer units are joined by strong covalent bonds. These polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers. Page No 428: Question 15.2: How are polymers classified on the basis of structure? ANSWER: Polymers are classified on the basis of structure as follows: 1. Linear polymers: These polymers are formed of long straight chains. They can be depicted as: For e.g., high density polythene (HDP), polyvinyl chloride, etc. 2. Branched chain polymers: These polymers are basically linear chain polymers with some branches. These polymers are represented as: For e.g., low density polythene (LDP), amylopectin, etc. 3. Cross-linked or Network polymers: These polymers have many cross-linking bonds that give rise to a network-like structure. These polymers contain bi-functional and tri-functional monomers and strong covalent bonds between various linear polymer chains. Examples of such polymers include bakelite and melmac. Page No 428: Question 15.3: Write the names of monomers of the following polymers: ANSWER: (i) Hexamethylenediamine and adipic acid  (ii) (iii) Tetrafluoroethene  Page No 428: Question 15.4: Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride, Polythene. ANSWER: Addition polymers: Polyvinyl chloride, polythene Condensation polymers: Terylene, bakelite Page No 435: Question 15.5: Explain the difference between Buna-N and Buna-S. ANSWER: Buna − N is a copolymer of 1, 3−butadiene and acrylonitrile. Buna − S is a copolymer of 1, 3−butadiene and styrene. Page No 435: Question 15.6: Arrange the following polymers in increasing order of their intermolecular forces. (i) Nylon 6, 6, Buna-S, Polythene. (ii) Nylon 6, Neoprene, Polyvinyl chloride. ANSWER: Different types of polymers have different intermolecular forces of attraction. Elastomers or rubbers have the weakest while fibres have the strongest intermolecular forces of attraction. Plastics have intermediate intermolecular forces of attraction. Hence, the increasing order of the intermolecular forces of the given polymers is as follows: (i) Buna − S < polythene < Nylon 6, 6 (ii) Neoprene < polyvinyl chloride < Nylon 6 Page No 437: Question 15.1: Explain the terms polymer and monomer. ANSWER: Polymers are high molecular mass macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (103 − 107u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6, 6 are examples of polymers. Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example, ethene, propene, styrene, vinyl chloride. Page No 437: Question 15.2: What are natural and synthetic polymers? Give two examples of each type. ANSWER: Natural polymers are polymers that are found in nature. They are formed by plants and animals. Examples include protein, cellulose, starch, etc. Synthetic polymers are polymers made by human beings. Examples include plastic (polythene), synthetic fibres (nylon 6, 6), synthetic rubbers (Buna − S). Page No 437: Question 15.3: Distinguish between the terms homopolymer and copolymer and give an example of each. ANSWER: Homopolymer  Copolymer  The polymers that are formed by the polymerization of a single monomer are known as homopolymers. In other words, the repeating units of homopolymers are derived only from one monomer. For example, polythene is a homopolymer of ethene. The polymers whose repeating units are derived from two types of monomers are known as copolymers. For example, Buna−S is a copolymer of 1, 3-butadiene and styrene. Page No 437: Question 15.4: How do you explain the functionality of a monomer? ANSWER: The functionality of a monomer is the number of binding sites that is/are present in that monomer. For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene and adipic acid is two. Page No 437: Question 15.5: Define the term polymerisation. ANSWER: Polymerization is the process of forming high molecular mass (103 − 107u) macromolecules, which consist of repeating structural units derived from monomers. In a polymer, various monomer units are joined by strong covalent bonds. Page No 437: Question 15.6: Is , a homopolymer or copolymer? ANSWER:  is a homopolymer because it is obtained from a single monomer unit, NH2−CHR−COOH. Page No 437: Question 15.7: In which classes, the polymers are classified on the basis of molecular forces? ANSWER: On the basis of magnitude of intermolecular forces present in polymers, they are classified into the following groups: (i) Elastomers (ii) Fibres (iii) Thermoplastic polymers (iv) Thermosetting polymers Page No 437: Question 15.8: How can you differentiate between addition and condensation polymerisation? ANSWER: Addition polymerization is the process of repeated addition of monomers, possessing double or triple bonds to form polymers. For example, polythene is formed by addition polymerization of ethene. Condensation polymerization is the process of formation of polymers by repeated condensation reactions between two different bi-functional or tri-functional monomers. A small molecule such as water or hydrochloric acid is eliminated in each condensation. For example, nylon 6, 6 is formed by condensation polymerization of hexamethylenediamine and adipic acid. Page No 437: Question 15.9: Explain the term copolymerisation and give two examples. ANSWER: The process of forming polymers from two or more different monomeric units is called copolymerization. Multiple units of each monomer are present in a copolymer. The process of forming polymer Buna−S from 1, 3-butadiene and styrene is an example of copolymerization Nylon 6, 6 is also a copolymer formed by hexamethylenediamine and adipic acid. Page No 437: Question 15.10: Write the free radical mechanism for the polymerisation of ethene. ANSWER: Polymerization of ethene to polythene consists of heating or exposing to …

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Explore the highly sought-after NCERT Solutions for Class 12 Science Chemistry Chapter 5: Biomolecules. These step-by-step solutions offer clear explanations, making them a favorite among Chemistry students. Whether you’re completing homework assignments or gearing up for exams, the Biomolecules Solutions for Class 12 Science provide a valuable resource. All questions and answers from Chapter 5 of the NCERT Book for Class 12 Science Chemistry are readily available here for free, ensuring a convenient and effective study experience. Page No 412: Question 14.1: Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain. ANSWER: A glucose molecule contains five −OH groups while a sucrose molecule contains eight −OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water. But cyclohexane and benzene do not contain −OH groups. Hence, they cannot undergo H-bonding with water and as a result, are insoluble in water. Page No 412: Question 14.2: What are the expected products of hydrolysis of lactose? ANSWER: Lactose is composed of β-D galactose and β-D glucose. Thus, on hydrolysis, it gives β-D galactose and β-D glucose. Page No 412: Question 14.3: How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? ANSWER: D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure. Page No 417: Question 14.4: The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain. ANSWER: Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion. Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour. For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids. Page No 417: Question 14.5: Where does the water present in the egg go after boiling the egg? ANSWER: When an egg is boiled, the proteins present inside the egg get denatured and coagulate. After boiling the egg, the water present in it is absorbed by the coagulated protein through H-bonding. Page No 422: Question 14.6: Why cannot vitamin C be stored in our body? ANSWER: Vitamin C cannot be stored in our body because it is water soluble. As a result, it is readily excreted in the urine. Page No 422: Question 14.7: What products would be formed when a nucleotide from DNA containing thymine is hydrolysed? ANSWER: When a nucleotide from the DNA containing thymine is hydrolyzed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products. Page No 422: Question 14.8: When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA? ANSWER: A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine. But when RNA is hydrolyzed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded. Page No 423: Question 14.1: What are monosaccharides? ANSWER: Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone. Monosaccharides are classified on the bases of number of carbon atoms and the functional group present in them. Monosaccharides containing an aldehyde group are known as aldoses and those containing a keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose. Page No 423: Question 14.2: What are reducing sugars? ANSWER: Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars. Page No 423: Question 14.3: Write two main functions of carbohydrates in plants. ANSWER: Two main functions of carbohydrates in plants are: (i) Polysaccharides such as starch serve as storage molecules. (ii) Cellulose, a polysaccharide, is used to build the cell wall. Page No 423: Question 14.4: Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose ANSWER: Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose Disaccharides: Maltose, lactose Page No 423: Question 14.5: What do you understand by the term glycosidic linkage? ANSWER: Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule. For example, in a sucrose molecule, two monosaccharide units, ∝-glucose and β-fructose, are joined together by a glycosidic linkage. Page No 423: Question 14.6: What is glycogen? How is it different from starch? ANSWER: Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen. Starch is a carbohydrate consisting of two components − amylose (15 − 20%) and amylopectin (80 − 85%). However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin. Page No 423: Question 14.7: What are the hydrolysis products of (i) sucrose and (ii) lactose? ANSWER: (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β- D-fructose. (ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose. Page No 423: Question 14.8: What is the basic structural difference between starch and cellulose? ANSWER: Starch consists of two components …

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Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 4 on Amines, complete with easy-to-follow step-by-step explanations. Widely favored among Class 12 Science students, these Chemistry Amines Solutions are invaluable for efficiently tackling homework assignments and gearing up for exams. Free of charge, you can access all questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Chemistry, ensuring a convenient resource for your academic needs. Page No 384: Question 13.1: Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH ANSWER: Primary: (i) and (iii) Secondary: (iv) Tertiary: (ii) Page No 384: Question 13.2: (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? ANSWER: (i), (ii) The structures and their IUPAC names of different isomeric amines corresponding to the molecular formula, C4H11N are given below: (a) CH3-CH2-CH2-CH2-NH2 Butanamine (10) (b)  Butan-2-amine (10) (c)  2-Methylpropanamine (10) (d) 2-Methylpropan-2-amine (10) (e) CH3-CH2-CH2-NH-CH3 N-Methylpropanamine (20) (f) CH3-CH2-NH-CH2-CH3 N-Ethylethanamine (20) (g) N-Methylpropan-2-amine (20) (h) N,N-Dimethylethanamine (3°) (iii) The pairs (a) and (b) and (e) and (g) exhibit position isomerism. The pairs (a) and (c); (a) and (d); (b) and (c); (b) and (d) exhibit chain isomerism. The pairs (e) and (f) and (f) and (g) exhibit metamerism. All primary amines exhibit functional isomerism with secondary and tertiary amines and vice-versa. Page No 387: Question 13.3: How will you convert? (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl−(CH2)4−Cl into hexan-1, 6-diamine? ANSWER: (i) (ii) (iii) Page No 396: Question 13.4: Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. ANSWER: (i) Considering the inductive effect of alkyl groups, NH3, C2H5NH2, and (C2H5)2NH can be arranged in the increasing order of their basic strengths as: Again, C6H5NH2 has proton acceptability less than NH3. Thus, we have: Due to the −I effect of C6H5 group, the electron density on the N-atom in C6H5CH2NH2 is lower than that on the N-atom in C2H5NH2, but more than that in NH3. Therefore, the given compounds can be arranged in the order of their basic strengths as: (ii) Considering the inductive effect and the steric hindrance of the alkyl groups, C2H5NH2, (C2 H5)2NH2, and their basic strengths as follows: Again, due to the −R effect of C6H5 group, the electron density on the N atom in C6H5 NH2 is lower than that on the N atom in C2H5NH2. Therefore, the basicity of C6H5NH2 is lower than that of C2H5NH2. Hence, the given compounds can be arranged in the increasing order of their basic strengths as follows: (iii) Considering the inductive effect and the steric hindrance of alkyl groups, CH3NH2, (CH3)2NH, and (CH3)3N can be arranged in the increasing order of their basic strengths as: In C6H5NH2, N is directly attached to the benzene ring. Thus, the lone pair of electrons on the N−atom is delocalized over the benzene ring. In C6H5CH2NH2, N is not directly attached to the benzene ring. Thus, its lone pair is not delocalized over the benzene ring. Therefore, the electrons on the N atom are more easily available for protonation in C6H5CH2NH2 than in C6H5NH2 i.e., C6H5CH2­NH2 is more basic than C6H5NH2. Again, due to the −I effect of C6H5 group, the electron density on the N−atom in C6H5CH2NH2 is lower than that on the N−atom in (CH3)3N. Therefore, (CH3)3N is more basic than C6H5CH2NH2. Thus, the given compounds can be arranged in the increasing order of their basic strengths as follows. Page No 396: Question 13.5: Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl  (ii) (C2H5)3N + HCl  ANSWER: (i)  (ii) (C2H5)3N Triethylamine+ HCl → (C2H5)3N+HCl−TriethylammoniumchlorideC2H53N Triethylamine+ HCl → C2H53N+HCl-Triethylammoniumchloride Page No 396: Question 13.6: Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. ANSWER: Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, N−trimethylanilinium carbonate. Page No 396: Question 13.7: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. ANSWER: Page No 396: Question 13.8: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. ANSWER: The structures of different isomers corresponding to the molecular formula, C3H9N are given below: (a)  Propan-1-amine (10) (b) Propan-2-amine (10) (c) (d) N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-amine will liberate nitrogen gas on treatment with nitrous acid. Page No 399: Question 13.9: Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene. ANSWER: (i) (ii) Page No 400: Question 13.1: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2 CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii) m−BrC6H4NH2 ANSWER: (i) 1-Methylethanamine (10 amine) (ii) Propan-1-amine (10 amine) (iii) N−Methyl-2-methylethanamine (20 amine) (iv) 2-Methylpropan-2-amine (10 amine) (v) N−Methylbenzamine or N-methylaniline (20 amine) (vi) N-Ethyl-N-methylethanamine (30 amine) (vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine) Page No 400: Question 13.2: Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. ANSWER: (i) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not. (ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl). Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent. (iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions. ​ (iv) Aniline …

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Find comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 3 on Aldehydes, Ketones, and Carboxylic Acids. These step-by-step explanations are highly favored by Chemistry students for quick completion of homework and effective exam preparation. All answers to the questions in NCERT Book of Class 12 Science Chemistry Chapter 3 are available here at no cost, making them a valuable resource for students. Page No 353: Question 12.1: Write the structures of the following compounds. (i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec-butyl ketone (vi) 4-Fluoroacetophenone ANSWER: (i) (ii) (iii) (iv) (v) (vi) Page No 356: Question 12.2: Write the structures of products of the following reactions; (i) (ii) (iii) (iv) ANSWER: (iv) Page No 358: Question 12.3: Arrange the following compounds in increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3 ANSWER: The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3 and so CH3CHO has stronger intermolecular dipole − dipole attraction than CH3OCH3⋅ CH3CH2CH3 has only weak van der Waals force. Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by: CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH Page No 365: Question 12.4: Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i)Ethanal, Propanal, Propanone, Butanone. (ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. Hint:Consider steric effect and electronic effect. ANSWER: (i) The +I effect of the alkyl group increases in the order: Ethanal < Propanal < Propanone < Butanone The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is: Butanone < Propanone < Propanal < Ethanal (ii) The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating −CH3 group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO2 group. Hence, the increasing order of the reactivities of the given compounds is: Acetophenone < p-tolualdehyde < Benzaldehyde< p-Nitrobenzaldehyde Page No 365: Question 12.5: Predict the products of the following reactions: (i) (ii) (iii) (iv) ANSWER: (i) (ii) (iii) (iv) Page No 367: Question 12.6: Give the IUPAC names of the following compounds: (i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH (iii)  (iv)  ANSWER: (i) 3-Phenylpropanoic acid (ii) 3-Methylbut-2-enoic acid (iii) 2-Methylcyclopentanecarboxylic acid (iv)2,4,6-Trinitrobenzoic acid Page No 370: Question 12.7: Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene) ANSWER: (i) (ii) (iii) (iv) Page No 376: Question 12.8: Which acid of each pair shown here would you expect to be stronger? (i) CH3CO2H or CH2FCO2H (ii)CH2FCO2H or CH2ClCO2H (iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv) ANSWER: (i) The +I effect of −CH3 group increases the electron density on the O−H bond. Therefore, release of proton becomes difficult. On the other hand, the −I effect of F decreases the electron density on the O−H bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H. (ii) F has stronger −I effect than Cl. Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is stronger acid than CH2ClCO2H. (iii) Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H is stronger acid than CH2FCH2CH2CO2H. (iv) Due to the −I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +I effect of −CH3 group. Hence, (A) is a stronger acid than (B). Page No 377: Question 12.1: What is meant by the following terms? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base ANSWER: (i) Cyanohydrin: Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′ can be alkyl or aryl groups. Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydrin. These reactions are known as cyanohydrin reactions. Cyanohydrins are useful synthetic intermediates. (ii) Acetal: Acetals are gem−dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom. One bond is connected to an alkyl group while the other is connected to a hydrogen atom. When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal. (iii) Semicarbarbazone: Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide. Semicarbazones are useful for identification and characterization of aldehydes and ketones. (iv) Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base. (v) Hemiacetal: Hemiacetals are α−alkoxyalcohols General structure of a hemiacetal Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas. (vi) Oxime: Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is either hydrogen or an organic side chain. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime. On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes. (vii) Ketal: Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups. Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals. (viii) Imine: Imines are chemical compounds containing a carbon …

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Explore the comprehensive NCERT solutions for Class 12 Science Chemistry Chapter 2 – “Alcohols, Phenols, and Ethers” with straightforward, step-by-step explanations. These solutions have gained immense popularity among Chemistry students in class 12 Science, serving as valuable resources for completing homework efficiently and preparing for exams. All the questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 2 are available here at no cost, providing a convenient and accessible aid for students. Page No 317: Question 11.1: Classify the following as primary, secondary and tertiary alcohols: (i) (ii)  (iii)  (iv) (v) (vi) ANSWER: Primary alcohol → (i), (ii), (iii) Secondary alcohol → (iv), (v) Tertiary alcohol → (vi) Page No 317: Question 11.2: Identify allylic alcohols in the above examples. ANSWER: The alcohols given in (ii) and (vi) are allylic alcohols. Page No 320: Question 11.3: Name the following compounds according to IUPAC system. (i) (ii) (iii) (iv) (v) ANSWER: (i) 3-Chloromethyl-2-isopropylpentan-1-ol (ii) 2, 5-Dimethylhexane-1, 3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol Page No 325: Question 11.4: Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal? (i) (ii) ANSWER: (i) (ii) Page No 325: Question 11.5: Write structures of the products of the following reactions: (i) (ii) (iii) ANSWER: (i)   (ii) (iii) Page No 335: Question 11.6: Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl2 (b) HBr and (c) SOCl2. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol ANSWER: (a) (i) Primary alcohols do not react appreciably with Lucas’ reagent (HCl-ZnCl2) at room temperature. (ii) Tertiary alcohols react immediately with Lucas’ reagent. (b) (i) (ii) (c) (i) (ii) Page No 335: Question 11.7: Predict the major product of acid catalysed dehydration of (i) 1-Methylcyclohexanol and (ii) Butan-1-ol ANSWER: (ii) Page No 335: Question 11.8: Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions. ANSWER: Resonance structure of the phenoxide ion Resonance structures of p-nitrophenoxide ion Resonance structures of o-nitrophenoxide ion It can be observed that the presence of nitro groups increases the stability of phenoxide ion. Page No 335: Question 11.9: Write the equations involved in the following reactions: (i) Reimer-Tiemann reaction (ii) Kolbe’s reaction ANSWER: Page No 342: Question 11.10: Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol. ANSWER: In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an SN2 reaction. In the reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol. Page No 342: Question 11.11: Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why? (i) (ii) ANSWER: Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene. In set (i), sodium methoxide (CH3ONa) is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction. Page No 343: Question 11.12: Predict the products of the following reactions: (i)  (ii) (iii) (iv)  ANSWER: (i) (ii) (iii) (iv) Page No 344: Question 11.1: Write IUPAC names of the following compounds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x)  (xi)  (xii) ANSWER: (i) 2, 2, 4-Trimethylpentan-3-ol (ii) 5-Ethylheptane-2, 4-diol (iii) Butane-2, 3-diol (iv) Propane-1, 2, 3-triol (v) 2-Methylphenol (vi) 4-Methylphenol (vii) 2, 5-Dimethylphenol (viii) 2, 6-Dimethylphenol (ix) 1-Methoxy-2-methylpropane (x) Ethoxybenzene (xi) 1-Phenoxyheptane (xii) 2-Ethoxybutane Page No 344: Question 11.2: Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane −1, 3, 5-triol (iv) 2,3 − Diethylphenol (v) 1 − Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 3-Chloromethylpentan-1-ol. ANSWER: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) Page No 344: Question 11.3: (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names. (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols. ANSWER: (i) The structures of all isomeric alcohols of molecular formula, C5H12O are shown below: (a)  Pentan-1-ol (1°) (b) 2-Methylbutan-1-ol (1°) (c) 3-Methylbutan-1-ol (1°) (d) 2, 2-Dimethylpropan-1-ol (1°) (e) Pentan-2-ol (2°) (f) 3-Methylbutan-2-ol (2°) (g) Pentan-3-ol (2°) (h) 2-Methylbutan-2-ol (3°) (ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol; 3-Methylbutan-1-ol; 2, 2−Dimethylpropan-1-ol Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol; Pentan-3-ol Tertiary alcohol: 2-methylbutan-2-ol Page No 344: Question 11.4: Explain why propanol has higher boiling point than that of the hydrocarbon, butane? ANSWER: Propanol undergoes intermolecular H-bonding because of the presence of −OH group. On the other hand, butane does not Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane. Page No 344: Question 11.5: Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact. ANSWER: Alcohols form H-bonds with water due to the presence of −OH group. However, hydrocarbons cannot form H-bonds with water. As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Page No 344: Question 11.6: What is meant by hydroboration-oxidation reaction? Illustrate it with an example. ANSWER: The addition of borane followed by oxidation is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide. Page No 344: Question 11.7: Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O. ANSWER: Page No 344: Question 11.8: While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason. ANSWER: Intramolecular H-bonding is present in o-nitrophenol. In p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, o-nitrophenol is steam volatile. Page No 344: Question 11.9: Give the equations of reactions for the preparation of phenol from cumene. ANSWER: To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide. Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products. Page No 344: Question 11.10: Write chemical reaction for the preparation of phenol from chlorobenzene. ANSWER: …

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Here, you can find NCERT solutions for Class 12 Science Chemistry Chapter 1 – Haloalkanes and Haloarenes. These solutions offer clear and straightforward explanations, making them highly favored among Chemistry students. Whether you need assistance with homework or preparing for exams, these solutions for Haloalkanes and Haloarenes are a valuable resource. All questions and answers from the NCERT Book of Class 12 Science Chemistry Chapter 1 are available here for free, providing a convenient and effective way to complete assignments and study for examinations. Page No 285: Question 10.1: Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec. butyl-2-methylbenzene ANSWER: (i) 2-Chloro-3-methyl pentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4- tert-Butyl-3-iodoheptane (iv) 1, 4-Dibromobut-2-ene (v) 1-Bromo-4-sec-butyl-2-methylbenzene Page No 289: Question 10.2: Why is sulphuric acid not used during the reaction of alcohols with KI? ANSWER: In the presence of sulphuric acid (H2SO4), KI produces HI Since  is an oxidizing agent, it oxidizes HI (produced in the reaction to I2). As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4 is used. Page No 289: Question 10.3: Write structures of different dihalogen derivatives of propane. ANSWER: There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below. (i) 1, 1-Dibromopropane (ii) 2, 2-Dibromopropane (iii) 1, 2-Dibromopropane (iv) 1, 3-Dibromopropane Page No 289: Question 10.4: Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides. ANSWER: (i) To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane. Neopentane (ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane. (iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-methylbutane. Page No 289: Question 10.5: Draw the structures of major monohalo products in each of the following reactions: (i) (ii) (iii) (iv) (v) (vi) ANSWER: (i) (ii) (iii) (iv) (v) (vi) Page No 291: Question 10.6: Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. ANSWER: (i) For alkyl halides containing the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom. Since the atomic mass of Br is greater than that of Cl, the boiling point of bromomethane is higher than that of chloromethane. Further, for alkyl halides containing the same alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of bromoform. Hence, the given set of compounds can be arranged in the order of their increasing boiling points as: Chloromethane < Bromomethane < Dibromomethane < Bromoform. (ii) For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane. Further, the boiling point decreases with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than that of 1-chloropropane. Hence, the given set of compounds can be arranged in the increasing order of their boiling points as: Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane Page No 307: Question 10.7: Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i) (ii) (iii) ANSWER: (i) 2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism. (ii) 2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism. (iii) Both the alkyl halides are primary. However, the substituent −CH3 is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism. Page No 307: Question 10.8: In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction? (i) (ii) ANSWER: (i) SN1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e. 2−chloro-2-methylpropane, undergoes faster SN1 reaction than (II) i.e., 3-chloropentane. (ii) The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane. Page No 307: Question 10.9: Identify A, B, C, D, E, R and R1 in the following: ANSWER: Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is Therefore, the compound R − Br is When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz …

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