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Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance:
Molecular Basis of Inheritance
The Search for Genetic Material
Regulation of Gene Expression
Human Genome Project
QUESTIONS FROM TEXTBOOK SOLVED
1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine. Ans: Nitrogenous Bases – Adenine, Uracil and Cytosine, Thymine; Nucleosides – Cytidine, guanosine.
2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA. Ans: In a DNA molecule, the number of cytosine molecule is equal to guanine molecules & the number of adenine molecules are equal to thymine molecules. As a result, if a double stranded DNA has 20% of cytosine, it has 20% of guanine. The remaining 60% includes both adenine & thymine which are in equal amounts. So, the percentage of adenine is 30%.
3. If the sequence of one strand of DNA is written as follows: 5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of complementary strand in 5′ —> 3′ direction. Ans: If the sequence of one strand of DNA is written as follows: 5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′ The sequence of the complementary strand in 5′ —> 3′ direction will be: 5′ – GCATGCATGCATGCATGCATGCATGCAT – 3′
4. If the sequence of the coding strand in a transcription unit is written as follows: 5-ATGCATGCATGCATGCATGCA TGCATGC-3′ Write down the sequence of mRNA. Ans: mRNA: 5′ -A U G CAU G CAU G C AU G CA UGCAUGCAUGC-3′.
5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain Ans: The antiparallel, double-stranded nature of the DNA molecule led Watson and Crick to hypothesise semi-conservative mode of DNA replication. They suggested that the two strands of DNA molecule uncoil and separate, and each strand serves as a template for the synthesis of a new (complementary) strand alongside it. The template and its complement, then form a new DNA double strand, identical to the original DNA molecule. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G, and G with C. Thus, two daughter DNA molecules identical to the parent molecule are formed and each daughter DNA molecule consists of one old (parent) strand and one new strand. Since only one parent strand is conserved in each daughter molecule, this mode of replication is said to be semiconservative. Meselson and Stahl and Joseph Taylor, later proved it by experiments.
6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases. Ans: (i) DNA dependent DNA polymerase – synthesis. (ii) DNA dependent RNA polymerase – synthesis. (iii) RNA dependent DNA polymerase – Retroviral nucleic acid. (iv) RNA dependent RNA polymerase – cDNA synthesis.
7. How did Hershey and Chase differentiate between DNA and protein in their experiment white proving that DNA is the genetic material? Ans: Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 for their experimental material. They grew some viruses on a medium that contained radioactive phosphorus (p32) and some others on medium that contained radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protem does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radio’active DNA because DNA does not contain sulphur. Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. , Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
8. Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNAand tRNA (c) Template strand and Coding strand Ans: (a) The main differences between repetitive DNA and satellite DNA are as following: (b) The main difference between mRNA and tRNA are as following: (c) The main difference between template strand and coding strand are as follows :
9. List two essential roles of ribosome during translation. Ans: Two essential roles of ribiosomes during translation are ;o (i) they provide surface for binding of mRNA in the groove of smaller sub unit of ribosome. (ii) As larger sub unit of ribosome has peptidy transferase on its ‘P’ site, therefore, it helps in joining amino acids by forming peptide bonds. .
10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then why does lac operon shut down some time after addition of lactose in the medium? Ans: Lac operon is switched on, on adding lactose in medium, as lactose acts as inducer and makes repressor inactive by binding with it. When the lac operon system is switched on, β-galactosidase is formed, which converts lactose into glucose and galactose. As soon as all the lactose is consumed, repressor again becomes active and causes the system to switch off (shut down).
11. Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons Ans: Promoter: It is one of the three components of a transcription unit that takes part in transcription. It is located at the start 5′ end and provides site for attachment of transcription factors (TATA Box) and RNA polymerase. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome for their taking part in protein formation. Exons: In eukarytoes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are transcribed and translated both.
12. Why is the Human genome project called a mega project? Ans: Human genome project is called a mega project because (i) it required bioinformatics data basing and other high speed computational devices for analysis, storage and retrieval of information. (ii) it generated lot of information in the form of sequence annotation. (iii) it was carried out in number of labs and coordinated on extensive scale.
13. What is DNA fingerprinting? Mention its application. Ans: DNA fingerprinting or DNA typing is a technique of determining nucleotide sequences of certain areas (VNTRs) of DNA which are unique to each individual. Each person has a unique DNA fingerprint. Unlike a conventional fingerprint that occurs only on the fingertips and can be altered by surgery, a DNA fingerprint is the same for every cell, tissue and organ of a person. It cannot be changed by any known treatment. Applications of DNA fingerprinting are as follows:
Paternity disputes can be solved by DNA fingerprinting.
DNA fingerprinting technique is being used to identify genes connected with hereditary diseases.
It is useful in detection of crime and legal pursuits.
It can identify racial groups, their origin, historical migrations and invasions.
14. Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics Ans: Transcription : It is DNA directed synthesis of RNA in which the RNA is transcribed on 3*—>5’ template strand of DNA in 5’—>3’ direction. Polymorphism: Variation at genetic level arisen due,to mutation, is called polymorphism. Such variations are unique at particular site of DNA, forming satellite DNA. The polymorphism in DNA sequences is the basis of genetic mapping and DNA finger printing. Translation : Protein synthesis from mRNA, tRNA, rRNA. Bioinformatics : Computational method of handling and analyzing biological databases.
Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation:
PRINCIPLES OF INHERITANCE AND VARIATION
Mendel’s Laws of Inheritance
Inheritance of One Gene
Inheritance of Two Genes
QUESTIONS FROM TEXTBOOK SOLVED
1.Mention the advantages of selecting pea plant for experiment by Mendel Ans: Mendel select garden pea (Pisum Sativum) for the following reasons. (i) It is an annual plant with short life span and gives results within 3 months. (ii) The plant is grown easily and does not require after care except at the time of pollination. (iii) F1 hybrids are fertile. (iv) Seven pairs of contrasting characters easily detectable. (v) True breeding self pollination.
2. Differentiate between the following – (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid. Ans: (a)The difference between dominance and recessive me:
(b) Differences between homozygous and heterozygous individuals :
(c) In breeding experiments when a cross is made between the individuals considering their same single character, it is called mono¬hybrid cross, while a cross is done consid¬ering two characters at fee same time is called dihybrid cross (Yellow Round * Green Wrinkled).
3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced? Ans: For a diploid organism, which is heterozygous for 4 loci, then 24 i.e. 2 x 2 x 2 x 2 = 16 types of gametes can be produced if the genes are not linked because for each heterozygous pair of genes there are two possibilities. So, for 4 pair the number of combination will be 16 gametes.
4. Explain the Law of Dominance using a monohybrid cross. Ans: When two different factors (genes) or a pair of contrasting forms of a character are present in an organism, only one expresses itself in the F, generation and is termed as dominant while the other remains unexpressed and called recessive factors (gene). A tall (TT) true breeding plant is crossed with a dwarf (tt) plant. The character of height is represented by‘T’ for tall‘t’ for dwarf are the alternate form as character of height. The Fj hybrid ‘Tt’ is Tall, showing that tall is dominant over dwarf while dwarf remains unexpressed in F, offspring due to phenomenon of dominance by tall factor or gene.In this Tt heterozygous has tall phenotype showing T is dominant over t allele.
5. Define and design a test-cross. Ans: When an individual is crossed with the homozygous recessive parent. It is called test cross.
Test cross helps in establishing hetero/ homozygosity of dominant trait.
6. Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for single locus. Ans:
Phenotype: All tall Genotype ratio : TT: Tt 2:2 or 1:1
7. When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be (a)tali and green. (b)dwarf and green. Ans: A cross between tall plant with yellow seeds (TtYy) & tall plant with green seed (Ttyy) is given below.
8. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross? Ans: Consider 2 characters Blue (B), long (L) seeds of a plant – both characters linked.
9. Briefly mention the contribution of T.H. Morgan in genetics. Ans: Thomas Hunt Morgan (1866-1945), an American geneticist and Nobel Prize winner of 1933, is considered as “Father of experimental genetics” for his work on and discovery of linkage, crossing over, sex linkage, criss cross inheritance, linkage maps, mutability of genes, etc. He is called fly man of genetics because of selecting fruit fly (Drosophila melanogaster) as research, material in experimental genetics. It was largely due to his book, “The Theory of Gene”, that genetics was accepted as a distinct branch of biology. In 1910, he discovered linkage and distinguished linked and unlinked genes. Morgan and Castle (1911) proposed “Chromosome Theory of Linkage” showing that genes are located on the chromosomes and arranged in linear order. Morgan and Sturtevant (1911) found that frequency of crossing over (recombination) between two linked genes is directly proportional to the distance between the two. 1% recombination is considered to be equal to 1 centi Morgan (cM) or 1 map unit. He worked on sex linked inheritance and reported a white eyed male Drosophila in a population of red eyed and proved that gene of eye colour is located on X-chromosome. The male passed its genes on X-chromosomes to the daughter while the son gets genes on X-chromosome from the female (mother): It is called criss-cross inheritance.
10. What is pedigree analysis? Suggest how such an analysis, can be useful. Ans: Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual. Pedigree analysis helps- (i) in analysis of transmission of character in family over generation. (ii) in genetic counselling of disease like haemophilia. (iii) to identify whether a particular genetic disease is due to recessive gene or a dominant gene. (iv) to identify the possible origin of the defective gene in the family or in a population.
11. How is sex determined in human beings? Ans: Sex determination refers to the mechanisms employed by organisms to produce offsprings that are of two different sexes. The sex of an individual is determined by the genetic information present in the individual’s sex chromosomes. Sex determination in human is done by XY type chromosome. In humans, females have two XX chromosomes and males have two different chromosomes (XY).
12. A child has blood group O. If the father has blood group A and mother of blood group B, work out the genotypes of the parents and the possible genotypes of the other off springs. Ans:
13. Explain the following terms with example (a) Co-dominance (b) Incomplete dominance Ans: (a) Codominance : Codominance is the phenomenon of two contrasting alleles of the same gene lacking dominant recessive ratio and expressing themselves simultaneously when present together. E.g. ABO blood group system – Human blood group AB is formed when alleles of blood groups A and B are present together (IAIB). Such RBCs carry both antigen A & B showing that both die alleles are expressing their effect phenotypically & codominant. (b) Incomplete dominance : Incomplete dominance may be defined as the partial /expression of both alleles in a heterozygote so that the phenotype is intermediate between those of two homozygotes. In this none of the allele is completely dominant, e.g., Pink colour flower in dog flower. In Mirabilis jalapa & Snapdragon or dog flowers, there are two types of flower colour in pure state-red & white. When the two types of plant are crossed, the hybrid or plants of Fj generation have pink flowers. The pink colour apparently appears either due to mixing of red & white colours (incomplete dominance) or expression of a single gene for pigmented flower which produces only pink colour.
14. What is point mutation? Give one example. Ans: Mutations arising due to change in single base pair of DNA is called point mutation. Eg., sickle cell anaemia, haemophilia.
15. Who had proposed the chromosomal theory of the inheritance? Ans: Chromosomal theory of inheritance was proposed by Sutton and Boveri independently in 1902. The two workers found a close similarity between the transmission of Mendelian hereditary factors (genes) and behaviour of chromosomes during gamete formation and fertilisation. They proposed that chromosomes were the carriers of the Mendelian factors. It is the chromosome and not genes which segregate and assort independently during meiosis and recombine at the time of fertilisation in the zygote. Chromosomal theory of inheritance was expanded by Morgan, Sturtevant and Bridges.
16. Mention any two autosomal genetic disorders with their symptoms. Ans: Sickle cell anaemia : Haemoglobin has less 02 transport, sickle shaped RBCs etc. Phenylketonuria : Mental retardation (due to accumulation of phenylalanine in brain), hypopigmentation of skin & hair, eczema etc.
Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health:
Reproductive Health – Problems and Strategies
Population Explosion and Birth Control
Medical Termination of Pregnancy
Sexually Transmitted Diseases
QUESTIONS FROM TEXTBOOK SOLVED
1. What do you think is the significance of reproductive health in a society? Ans: Reproductive health in a society is significant because the people are aware of (i) birth control methods and advantages of small family, (ii) sexually transmitted diseases and methods to avoid them, (iii) importance breast feeding and post natal care of the mother and baby and (iv) equal opportunities for the male and female children.
2. Suggest the aspects of reproductive health which need to be given special attention in the present scenario. Ans: The aspects of reproductive health which need to be given special attention are :
Introduction of sex education in schools to give right information to the young minds about reproductive organs, accessory organs of reproduction, secondary sexual characters, adolescence and related changes, safe and hygienic sexual practices, STDs etc.
Providing knowledge about available birth control methods, care of pregnant mothers, post-natal care of the mother and child, importance of breast feeding etc.
Creating awareness about consequences of uncontrolled population growth and social evils (sex abuses and sex-related crimes, use of drugs, tobacco and alcohol etc.) among young people.
3. Is sex education necessary in schools? Why? Ans: Yes, sex education is necessary in school.Because introduction of sex education in school encourage to provide the right or correct information to the young peoples so as to discourage children from believing in myths & having misconceptions about sex related aspects. Proper information about reproductive organs, safe & hygienic sexual practices, STD’s etc. would help people, those in the adolescent age group to lead a healthy reproductive life. In many countries, sexual education raises much contentious debate. Chief among the controversial points is whether covering child sexuality is valuable or detrimental; the use of birth control such as condoms and hormonal contraception, and the impact of such use on pregnancy, outside marriage, teenage pregnancy, and the transmission of STDs. Increasing support for abstinence – only sex education by conservative groups has been one of the primary causes of the controversies.
4. Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement Ans: Yes, in the last 50 years, reproductive health in our country has improved. Some such areas of improvement are (i) massive child immunization (ii) maternity and child health (iii) increasing use of contraceptives (iv) family planning. Bringing sexual and reproductive health services to the millions of people living in countries which still suffer from short life expectancies, high levels of child and maternal mortality, child lSS&ur and illiteracy and poor overall health remains a major challenge for governments and non government organizations.
5. What are the suggested reasons for population explosion? Ans: (i) Increased health facilities along with better living conditions had an explosive impact on the growth of population. (ii) Decline in Death rate (iii) Decline in Maternal Mortality Rate (iv) Decline in Infant Mortality Rate (v) Increase in the number of people in the reproductive age.
6. Is the use of contraceptives justified ? Give reasons. Ans: Population in India is increasing at a very fast rate and such a growth necessitated intense use of contraceptive methods to bring all the fertile couples under its cover. It will help in bringing the birth rate down and consequently check the population growth. Therefore, the use of contraceptives is justified. It is essential to mention here that the selection of a suitable contraceptive method should be practiced in consultation with qualified doctors. The government provides these facilities free at the family planning centres. Contraceptives are given free or at nominal prices at these centres to the couples of reproductive age desirous of preventing conception.
7. Removal of gonads cannot be considered as a contraceptive option. Why? Ans: Removal of gonads or its parts is a surgical method, of sterilization. Vasectomy and tubectomy are carried out in males and females respectively. It will lead to infertility & both male and female will be dependent on hormones in their remaining life to regulate functioning of reproductive organs. These are very effective but their irreversibility is very poor, so they are not good options for contraception.
8. Amniocentesis for sex determination is banned? in our country. Is this ban necessary? Comment Ans: Since Amniocentesis is misused to find out the sex of the foetus leading to female foeticides, it is necessary. It is the prenatal diagnostic technique, that helps to find out chromosomal (genetic) disorders, metabolic disorders of the foetus; in extreme cases where the foetus is found to be suffering from an incurable disorder Medical Termination of Pregnancy (MTP) is banned. The solution to the problem is not ‘ passing more laws, such as banning sex determination ultrasounds but raising the social & economic status of women,
9. Suggest some methods to assist infertile couples to have children. Ans: ART (Assisted Reproductive Technologies) is a term that describes several different methods used to help infertile couples. ART involves removing eggs from a woman’s body, mixing them with sperm in the laboratory and putting the embryos bath into a woman’s body. Success rates vary depending on many factors.Something that affects the success rate of ART includes age of the partners, reason for infertility,type of ART, if the egg is fresh or frozen. Various methods are now available to help such couples are — in vitro fertilization, gamete intra fallopian transfer, intracytoplasmic sperm injection & artificial insemination. (i) In vitro fertilization – Patient’s egg & her partner’s sperm are collected & mixed together in a laboratory to achieve fertilization outside the body. The embryo produced may then be transferred into the female patient. It is commonly knownas test tube baby programme. (ii) Gamete Intra Fallopian Transfer (GIFT) – A procedure in which eggs are retrieved from a woman, mixed with sperm & immediately replaced in one or other of the women’s fallopian tubes so that they fertilize inside the body (invivo). (iii) Intra Cytoplasmic Sperm Injection (ICSI)– In this method, sperm & eggs are retrieved from both the parents. A single sperm is injected directly into an egg, then the fertilized egg is implanted into the woman’s uterus. (iv)Artificial insemination – In this technique, the semen is collected either from husband or a healthy donor & is artificially introduced either into the vagina or into the uterus of the female (IUI-Intra uterine insemination).
10. What are the measures one has to take to prevent from contracting STDs? Ans: (i) Avoiding sex with unknown partner /multiple partners. (ii) Always use condoms during intercourse. (iii) In case of doubt, go for medical professional for early detection and get complete treatment if diagnosed with disease. (iv) Education & counselling of persons at risk on ways to adopt safer sexual behaviour.
11. State True/False with explanation (a) Abortions could happen spontaneously too. (True/False) (b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False) (c) Complete lactation could help as a natural method of contraception. (True/False) (d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False) Ans: (a) True. Due to internal factors like incompatibility, abortion could happen spontaneously. (b)False. It is due to abnormalities/defects in either male or female or both the partners. (c)True, but it is limited to period up to six months after parturition. (d)True. Creating awareness about sex-related aspects removes the myths and miconcep- tions about these problems.
12. Correct the following statements: (a) Surgical methods of contraception prevent gamete formation. (b )Ail sexually transmitted diseases are completely curable. (c) Oral pills are very popular contraceptives, among the rural women. (d) In E.T. techniques, embryos are always transferred into the uterus. Ans: (a) Surgical methods of contraception prevent gamete transport & thereby prevent conception. (b) Except for hepatitis-B, genital herpes, and HIV infections, other STD diseases are completely curable if detected early and treated properly. (c) Oral pill are very popular contraceptives among the educated urban women. (d) In E.T. techniques, embryos with 8 blastomeres are transferred into fallopian tube and more than 8 blastomeres are transferred into the uterus.
Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 3 Human Reproduction:
The Male Reproductive System
The Female Reproductive System
Fertilisation and Implantation
Pregnancy and Embryonic Development
Parturition and Lactation
QUESTIONS FROM TEXTBOOK SOLVED
1. Fill in the blanks: (a)Humans reproduce————— (asexually/sexually). (b)Humans are————— (oviparous, viviparous,ovoviviparous). (c)Fertilization is————— in humans (external/internal). (d)Male and female gametes are—————(diploid/haploid). (e)Zygote is————— (diploid/haploid). (f)The process of release of ovum from a mature follicle is called————— . (g)Ovulation is induced by a hormone called————— (h)The fusion of male and female gametes is called————— . (i)Fertilization takes place in————— (j)Zygote divides to form————— which is implanted in uterus. (k)The structure which provides vascular connection between foetus and uterus is called————— . Ans: (a) sexually (b) viviparous (c) internal (d)haploid (e)diploid (f)ovulation (g)LH (Luteinizing hormone) (h)fertilization (i)ampullary-isthmic junction (fallopian tube) (j)blastocyst (k)placenta (Umbilical cord)
2. Draw a labelled diagram of male reproductive system. Ans:
3. Draw a labelled diagram of female reproductive system. Ans:
4. Write two major functions each of testis and ovary. Ans: Testes are components of both the reproductive system (being gonads) and the endocrine system (being endocrine glands). The respective functions of the testes are – producing sperm (spermatozoa) by the process of spermatogenesis and producing male sex hormones, of which testosterone is the best-known. Testosterone stimulates development of testes and of male secondary sexual characteristics.
The ovaries have two major functions. One is the production of eggs or ova, and the second is the production of hormones or chemicals which regulate menstruation and other aspects of health and well-being, including sexual well-being. Estrogen and progesterone are the most important hormones which serve many functions like, they induce and maintain the physical changes during puberty and the secondary sex characteristics and they support maturation of the uterine endometrium in preparation for implantation for a fertilised egg, etc.
5. Describe the structure of a seminiferous tubule. Ans: The seminiferous tubule is a structural unit in the adult testis. The seminiferous tubules are situated in testicular lobules. Seminiferous tubule consists of two types of cells – Sertoli or supporting cells & spermatogenic cellsl Sertoli cells, are elongated and pyramidal & partially envelop the spermatogenic cells. The cells provide nourishment to the developing spermatogenic cells. Spermatogenic cells are stacked in 4-8 layers. These cells divide several times & differentiate to produce spermatozoa. Between seminiferous tubules lie the interstitial cells or leydig cells which produces testosterone hormone.
6. What is spermatogenesis? Briefly describe the process of spermatogenesis. Ans: Spermatogenesis is the process of producing sperms with half the number of chromosomes (haploid) as somatic cells. It occurs in seminiferous tubules. Sperm production begins at puberty continues throughout life with several hundred million sperms be ing produced each day. Once sperm are formed they move into the epididymis, where they mature and are stored. During spermatogenesis one spermatogonium produces 4 sperms. Spermatogenesis completes through the following phases – multiplicative phase, growth phase, maturation phase & spermiogenesis. In multiplicative phase the sperm mother cells divide by mitosis & produce spermatogonia. The spermatogonia grow in size to form large primary spermatocytes by getting nourishment from sertoli cells in growth phase. Maturation phase involves meiosis I in which primary spermatocytes divide to produce secondary spermatocyte and meiosis II which produces spermatids. Thus each primary spermatocyte gives rise to four haploid spermatids. Spermiogenesis or spermateliosis is process of formation of flagellated spermatozoa from spermatids. Spermiogenesis begins in the seminiferous tubules but usually completed in epididymis.
7. Name the hormones involved in regulation of spermatogenesis. Ans: The hormones involved in regulation of spermatogenesis are GnRH, LH, FSH and androgens.
Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin releasing hormone (GnRH). The increased levels of GnRH then acts at the anterior pituitary gland and stimulates secretion of two gonadotropins – luteinising hormone (LH) and follicle stimulating hormone (FSH). LH acts at the Leydig cells and stimulates synthesis and secretion of androgens. Androgens, in turn, stimulate the process of spermatogenesis. FSH acts on the Sertoli cells and stimulates secretion of some factors which help in the process of spermiogenesis.
8. Define spermiogenesis and spermiation. Ans: Spermiogenesis is the process of transformation of spermatids into mature flagellated spermatozoa (sperms).Spermiation is the process of release of mature spermatozoa. In this spermatozoa are shed into the lumen of seminiferous tubule for transport.
9. Draw a labelled diagram of sperm. Ans:
10.What are the major components of seminal plasma? Ans: Seminal plasma is the fluid in which sperm is ejaculated. Major components of seminal plasma are secretions from seminal vesicles, prostrate and bulbourethral gland and sperms from testis. It is rich in fructose and contains enzymes, citric acid, hormones like prostaglandins, calcium and clotting proteins.
11. What are the major functions of male accessory ducts and glands? Ans: Male accessory ducts include rete testis, vasa efferentia, epididymis and vas deferens. These ducts store and transport sperms from the testis to the outside through urethra. The male accessory glands include paired seminal vesicles, a prostate and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose, calcium and certain enzymes. The secretions of bulbourethral glands also helps in the lubrication of the penis.
12. What is oogenesis? Give a brief account of oogenesis. Ans: The process of formation of a mature female gamete (ovum) is called oogenesis. It occurs in the ovaries of female reproductive system. Oogenesis is a discontinuous process it begins before birth, stops in midprocess & only resumes after menarch. It occurs in three phases : Multiplicative phase (formation of oogonia mitotically from the primary germ cells), Growth phase (growth of oogonia into primary oocyte) & Maturation phase (formation of mature ova from primary oocyte through meiosis). Maturation phase produces two haploid cells – Larger one called secondary oocyte & the smaller one called polar bodies (1st polar body). Meiosis II of secondary oocyte results in the formation of functional egg or ovum and a second polar body: The first polar body may also divide to form two polar bodies of equal sizes which do not take part in reproduction & ultimately degenerates. First maturation division may be completed in the ovaries just prior to ovulation but second one (Final) is completed outside the ovary after fertilization. Secondary oocyte is female gamete in which the 1st meiotic division is completed & second meiotic division (Metaphase stage) has begin. The egg is released at secondary oocyte stage under the effect of LH.
13. Draw a labelled diagram of a section through ovary. Ans:
14. Draw a labelled diagram of a Graafian follicle. Ans:
15. Name the functions of the following: (a) Corpus luteum (b) Endometrium (c) Acrosome (d) Sperm tail (e) Fimbriae Ans: (a) Corpus luteum : The corpus luteum secretes large amounts of progesterone which is essential for maintenance of the endometriuip. (b) Endometrium is necessary for implantation of the fertilized ovum and other events of pregnancy. (c) The acrosome is filled with enzymes that help during fertilization of the ovum. (d) Sperm tail: Tail facilitates sperm motility which is essential for fertilization. (e) Fimbriae: Fimbriae help in collection of the ovum after ovulation.
16. Identify True/False statements. Correct each false statement to make it true. (a) Androgens are produced by Sertoli cells. (True/False) (b) Spermatozoa get nutrition from sertoli cells. (True/False) (c) Leydig cells are found in ovary. (True/ False) (d) Leydig cells synthesize androgens. (True/ False) (e) Oogenesis takes place in corpus luteum. (True/False) (i) Menstrual cycle ceases during pregnancy. (True/False) (g) Presence or absence of hymen is not a reliable indicator of virginity or sexual – experience. (True/False) Ans: (a) False, Androgens or male sex hormones (e.g, testosterone) are secreted by Leydig cells. (b) True. (c) False, Leydig cells are found in testis. (d) True. (e) False, Oogenesis takes place in ovary. (f) True. (g) True.
17. What is menstrual cycle? Which hormones regulate menstrual cycle? Ans: Menstrual cycle is the cyclic change( itf’the reproductive tract of primate female. This period is marked by a characteristic event repeated almost every month (28 days with minor variation) in the form of a menstrual flow (i.e. shedding of the endometrium of the uterus with bleeding. It may be temporarily stopped only in pregnancy. The hormones that regulates menstrual cycles are (i) FSH (Follicle stimulating hormone), (ii) LH (Luteinizing hormone), (iii) Oestrogens, (iv) Progesterone.
18. What is parturition ? Which hormones are involved in induction of parturition? Solution: Parturition (or labour) means child birth. Parturition is the sequence of actions by which a baby and the afterbirth (placenta) are expelled from the uterus at childbirth. The process usually starts spontaneously about 280 days after conception, but it may be started by artificial means.
The process of parturition is induced by a complex neuroendocrine mechanisms involving cortisol, estrogen and oxytocin.
19. In our society the women are often blamed for giving birth to daughters. Can you explain why this is not correct? Ans: The sex chromosome pattern in the human females is XX and that of male is XY. Therefore, all the haploid female gametes (ova) have the sex chromosome X, however, the haploid male gametes have either X or Y. Thus 50% of sperms carry the X-chromosome while the other 50% carry the Y-chromosome. After fusion of the male and female gametes, the zygote carries either XX or XY depending upon whether the sperm carrying X or Y fertilizes the ovum. The zygote carrying XX would be a female baby and XY would be a male baby. That is why it is correct to say that the sex of the baby is determined by the father.
20. How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal? Ans: One egg is released by human ovary in a month. Identical twins: Identical twins are formed when a single fertilized egg splits into two genetically identical parts. The twins share the same DNA set, thus they may share many similar attributes. However, since physical appearance is influenced by environmental factors and not just genetics, identical twins can actually look very different. Fraternal twins: These twins are formed when two fertilized eggs are formed. The twins share the different DNA set, thus they may share different attributes (dizygotic embryo).
21. How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies? Ans: Since dogs have multiple births, several eggs mature and are released at the same time. If fertilised, the egg will implant on the uterine wall. Dogs bear their litters roughly 9 weeks after fertilisation, although the length of gestation can vary from 56 to 72 days. An average litter consists of about six puppies, though this number may vary widely based on the breed of dog. On this basis 6 eggs were released by the ovary of a female dog which gave birth to 6 puppies.
Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants:
Sexual Reproduction in Flowering Plants
Flower – A Fascinating Organ of Angiosperms
Pre-fertilisation : Structures and Events
Post-fertilisation: Structures and Events
Apomixis and Polyembryony
QUESTIONS FROM TEXTBOOK SOLVED
1. Name the parts of anangiosperm flower in which development of male and female gametophytes take place. Ans: Development of male gametophyte (micro- gametogenesis) occurs in pollen sac of anther up to 2 – celled stage. The female gametophyte develops (megagametogenesis) in the nucellus of ovule.
2. Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events? Ans: Differences between microsporogenesis and megasporogenesis are as follows –
Each microspore mother cell and megaspore mother cell contain two sets of chromosomes and are therefore diploid. The diploid megaspore mother cell and microspore mother cell enlarges and undergo meiosis to produce, four haploid cells called megaspores and microspores respectively.The chromosome number is reduced by half and therefore megaspores and microspores are haploid. Microsporogenesis and megasporogenesis give rise to pollen grains and embryo sac respectively. Pollen grain is the male gametophyte and embryo sac represents the female gametophyte.
3. Arrange the following terms in the correct development sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes. Ans: The correct developmental sequence for the formation of male gametes is : Sporogenous tissue —» Pollen mother cell —» Microspore tetrad —» Pollen grain —» Male gametes.
4. With a neat, labelled diagram, describe the parts of a typical angiosperm ovule. Ans:
A typical angiospermic ovule is a small structure which is formed in the ovary. Ovule first develops as a projection on the placenta and composed of multilayered cellular tissue called the nucellus. The hypodermal cell of die nucellus enlarges and transformed into megaspore mother cell. This cell undergoes meiosis to produce four haploid cells only one of which develops & forms embryo sac (female gametophyte). An ovule may be surrounded by one or two protective layers called integuments, leaving a small opening at one end termed as micropyle which acts as passage for the entry of the pollen tube into the ovule. Thus, a typical ovule consists of a fully developed embryo sac with the nucellus and integuments.
5. What is meant by monosporic development of female gametophyte? Ans: In majority of flowering plants one of the megaspores is functional while the other three degenerate. Only the functional megaspore develops into the female gametophyte or embryo sac. This method of development of embryo sac from a single megaspore is called monosporic development.
6. With a neat diagram explain the 7-celled, 8- nucleate nature of the female gametophyte. Ans:
Embryo sac (or female gametophyte) is formed by three successive mitotic divisions that take place in the nucleus of megaspore. The nucleus of the functional megaspore divides meiotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac. Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac. After the 8-nucleate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo sac. Six of the eight nuclei are grouped together at micropylar and chalazal end and form the egg apparatus and antipodals respectively. The large central cell left over with two polar nuclei. Thus, a typical female gametophyte consists of 7 cells with 8 nucleus.
7. What are chasmogamous flowers? Can cross-pollination occur in deistogamous flowers? Give reasons for your answer Ans: Chasmogamous flowers are those flowers which are open with exposed anther and stigma.
Cleistogamous flowers are those flowers which do not open at all. In these flowers, the anthers and stigma lie close to each other, when anthers dehisce in the flower buds, the pollen grains come in the contact with stigma to effect pollination. So these flowers are invariably self-pollinated as the flowers remain closed and there is no chance of cross pollen landing on the stigma. Pollination and seed setting are assured even in the absence of pollinators
8. Mention two strategies evolved to prevent self-pollination in flowers. Ans: Continued self-pollination decreases the vigour and vitality of a particular race. Thus, flowering plants have developed many devices to discourage self-pollination and to encourage cross-pollination. Dichogamy and self-sterility are.two most common devices that ensure cross-pollination. Dichogamy – Maturation of anther and stigma at different times in a bisexual flower prevent self-pollination. Self-sterility (or self-incompatibility) – Due to the presence of self-sterile gene in some flowers, pollen grains do not germinate on the stigma of that flowers. e.g.,- tobacco, potato.
9. What is self-incompatiblility? Why does self-pollination not lead to seed formation in self-incompatible species? Ans: If a pistil carrying functional female gametes fails to set seeds following pollination with viable and fertile pollen, capable of bringing about fertilisation in another pistil, the two are said to be incompatible, and the phenomenon is known as sexual incompatibility. Sexual incompatibility may be interspecific (between individuals of different species) or intraspecific (between individuals of the same species). The latter is also called self-incompatibility. Self-incompatibility is a gene-physiological process. Incompatibility reactions are controlled by a single gene, called S-gene, which has several alleles. Pollen grains that possess the S-allele common to any one of the two alleles present in the cells of the pistil, will not be functional on that particular pistil. However, every pollen grain having no common S alleles with pistil would be functional on the pistil of a that plant. As self pollens are unable to fertilise the egg to form pmbryo, hence seeds are not formed in self-incompatible species.
10. What is bagging technique? How is it useful in a plant breeding programme? Ans: It is the covering of emasculated flowers (removal of anthers in bud condition from a bisexual flower by a bag of butter paper or polythene in their bud condition i.e., before anthesis) to prevent contamination of its stigmas with unwanted pollens. When the stigmas of emasculated flowers mature the bags are removed, stigmas are dusted with pollen grains of desired male . plants by means of a presterilized brush and flowers are rebagged till fruit develop. This technique is mainly used in artificial hybridization. Plant breeders often use this technique to prevent the contamination of stigma of the flowers from unwanted pollen grains.
11. What b triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion. Ans: Fusion of second male gamete with die two polar nuclei located in the central cell to form the triploid primary endosperm nucleus (PEN) is called triple fusion or vegetative fertilization. This process takes place in the embryo sac. After reaching the ovary, pollen tube enters into the embryo sac from the micropylar end. After penetration, the tip of the pollen tube ruptures releasing the two male gametes. The one male gamete fuses with the egg to form the diploid zygote. This process is called syngamy and the other male gamete fuses with the two polar nuclei to form the triploid primary endosperm & this process is known as triple fusion. These two events of fertilization constitute the process of double fertilization.
12.Why do you think the zygote is dormant for sometime in a fertiUsed ovule? Ans: The zygote after a period of rest develops into embryo. Most zygotes remain dormant till certain amount of endosperm forms. They do so, to provide assured nutrition to the developing embryo.
13. Differentiate between: (a)hy pocotyl and epicotyl; (b)coleoptile and coleorrhiza; (c)integument and testa; (d)perisperm and pericarp. Ans:
14. Why is apple called a false fruit? Which Part(s) of the flower forms the fruit? Ans: Botanically ripened ovary is called a true fruit. The fruits in which thalamus and other floral parts develop along with the ovary are called false fruits. For example – apple, strawberry, cashew etc. In apple the main edible portion of the fruit is the fleshy thalamus. Ovary forms the fruit after fertilization or without fertilization in parthenocarpic fruits.
15. What is meant by emasculation? When and why does a plant breeder employ this technique? Ans: Removal of stamens or anthers of a bisexual flower without affecting the female reproductive organs is called emasculation. This technique is used in artificial hybridisation. In such crossing experiments it is important to make sure that only the desired pollen grains are used for pollination and the stigma is protected from contamination from unwanted pollens. This is achieved by emasculation and bagging technique. This technique is used to obtain desired variety of seeds.
16. If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why? Ans: Parthenocarpic fruits are seedless. They develop from ovary without fertilization. Banana, grapes, oranges, Pineapple, Guava, Watermelon, lemon are selected because these seedless of units are of high economic importance. The fruits in which seeds or seed part form edible portion (e.g.,Pomegranate) are not selected to induce parthenocarpy.
17. Explain the role of tapetum in the formation of pollen grain wall. Ans: Tapetum is the innermost wall layer of anther. The cells of this layer have large nuclei and dense cytoplasm. This layer is of great physiological importance as most of the food material from outside passes through this layer. At maturity, these cells degenerate and provide nourishment to developing microspores or pollens inside. Tapetum is the layer, which secretes both enzymes and substances of hormonal nature. The main function of tapetum is to provide nutrition to pollens but it also secretes some substances of utmost importance like :
Qallase enzyme : Tapetum secretes callase enzyme which dissolves callose substances by which four pollens of a pollen tetrad are united, hence separating microspores or pollens of a tetrad.
Ubisch bodies : These bodies of lipid nature are also secreted by tapetum. Ubisch bodies get covered with sporopollenin and thus increase thickness of exine (i.e., outer layer of pollen wall). Ubisch bodies are spheroidal and have diameter of only few microns. These are produced only by glandular tapetum (not by amoeboid tapetum).
Pollen kit substances : Tapetum also secretes pollen kit, outer most oily, thick, viscous, sticky, electron dense homogeneous coating of pollen grains of many entomophilous plants.
18. What is apomixis and what is its importance? Ans: Apomixis is a mode of asexual reproduction that produces seeds without fertilization, e.g.- some species of Asteraceae and Grasses. This method is important in hybrid seed industry. Hybrids are extensively cultivated for increasing productivity. But the main drawback is that the hybrid seeds are to be produce every year because the seeds of the hybrid plants da not maintain hybrid characters for longer period due to segregation of characters. This can be avoided if apomixis can be introduced in hybrid seeds. For this reason scientists are trying hard to identify genes for apomixis.
Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms:
Reproduction in Organisms
QUESTIONS FROM TEXTBOOK SOLVED
1. Why is reproduction essential for organisms? Ans: Reproduction is the ability of living organisms to produce a young one similar to itself. It ensures continuity of a species generation after generation. Reproduction introduces variation in the organisms. Useful variations are essential for adaptation and evolution. Therefore, it is essential for organisms.
2. Which is a better mode of reproduction sexual or asexual? Why? Ans: Sexual reproduction is a better mode of reproduction because of the following reasons:
Variation : Since fusion of gametes from different parents occur during sexual reproduction, hence genetic recombination takes place causing variations.
Evolution : Variation being a major factor of natural selection, therefore, it plays an important role in evolution.
Adaptation : The offspring produced due to sexual reproduction adapt better to the changing environmental conditions.
Vigour and Vitality : Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.
3. Why is the offspring formed by asexual reproduction referred to as clone? Ans: Asexual reproduction is a type of reproduction in which a single individual is capable of producing offspring. These offspring are not only genetically and morphologically similar to one another but also similar to their parent. Clone is the term given to individuals that are genetically and morphologically similar. Thus the offspring produced by asexual reproduction are called clones.
4. Offsprings formed due to sexual reproduction have better chances of survival. Why? Is this statement always true? Ans: The offsprings obtained from sexual reproduction have better chances of survival because the genetic material of such organisms are formed from both the parents. Daughter organisms/offsprings show variation that leads to the evolution of species.
This statement is always true. The offspring produced due to sexual reproduction adapt better to the changing environmental conditions. Genetic recombination, interaction, etc. during sexual reproduction provide vigour and vitality to the offspring.
5. How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction? Ans: Production of offspring by a single parent without the formation and fiision of gametes is called asexual reproduction. It involves only mitotic cell division that gives rise the daughter cells which are genetically identical to the parent cell. Sexual reproduction is the production of offspring by two parents, male and female. It involves meiotic cell divisions producing haploid nuclei which on fusion produce offspring that are genetically different from their parents.
6. Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction? Ans: The difference between asexual and sexual reproduction are as follows :
In plants asexual reproduction is called vegetative reproduction because vegetative plant parts like rhizome, runner, sucker, tuber, bulb all are capable of producing off springs These parts give rise to daughter individuals without the involvement of two parents.
7. What is vegetative propagation? Give two suitable examples. Ans: Vegetative propagation is the formation of new plants from vegetative units. In plants, the units of vegetative propagation are runner, rhizome, sucker, tuber, offset, bulb, etc. These are capable of producing new offsprings. These structures are called v vegetative propagules.
Modified tuberous roots of sweet potato, tapioca, yam, Dahlia and Tinospora can be propagated vegetatively when planted in soil. Small plants emerging from the buds (called eyes) of the potato tuber, from the rhizomes of banana and ginger are other examples.
8. Define: (a)Juvenile phase (b)Reproductive phase (c)Senescent phase. Ans:(a)Juvenile phase : All organisms have to reach a certain stage of growth and maturity in their life before they can reproduce sexually. That period of growth is called juvenile phase. However, this phase is known as vegetative phase in plants. This phase is of different durations in different organisms. (b)Reproductive phase: The end of juvenile/ vegetative phase marks the beginning of reproductive phase. During this phase, the organisms produce offspring. In higher plants, this phase can be easily seen when they come to flower but in animals, the juvenile phase is followed by morphological and physiological changes prior to active reproductive behaviour. The reproductive phase is also of variable period in different organisms like some plants, flower throughout the year while others show seasonal flowering. In animals like birds lay eggs seasonally “but when in captivity (as in poultry farms) can be made to lay eggs throughout the year. Placental female mammals, undergo cyclical changes in reproductive organs during this phase. (c) Senescent phase: It begins from the end of the reproductive phase. During this phase of life span, there is progressive deterioration in the body (like slowing of metabolism, etc.). Old age ultimately leads to death.
9. Higher organisms have resorted to sexual reproduction in spite of its complexity. Why? Ans: Higher organisms have resorted to sexual reproduction in spite of its complexity because sexual reproduction results in multiplication and perpetuation of species and also contributes to evolution of species by introducing variation much more faster than asexual reproduction in a particular population. Sexual reproduction enables higher organisms to survive during unfavourable conditions.
10. Explain why meiosis and gametogenesis are always interlinked? Ans: Gametogenesis is the process of formation of two types of haploid gametes (male and female). In gametogenesis, gametes are haploid in number and formed by meiosis so the chromosome number is haploid. Thus gametogenesis is always linked with meiosis.
11. Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n). (a)Ovary ———————— (b)Anther ———————— (c)Egg ———————— (d)Pollen ———————— (e)Male gamete ———————— (f)Zygote ———————— Ans: (a)2n (b)2n (c)n (d)n (e)n (f)2n
12. Define external fertilization. Mention its disadvantages. Ans: When fusion of the gametes takes place outside the body of the organisms, it is called external fertilization or external syngamy. The external medium like water is required for this form of fertilization. This form, is found in many aquatic animals like fishes, amphibians, majority of algae. In this, parents release eggs and sperms in the surrounding water, then fertilization and development of offspring occur externally. Disadvantages of external fertilization: (i)if occurs only in aquatic medium. (ii)A chance factor is involved requiring synchronous release of gametes nearby and absence of turbulence of water. (iii)There is no protection to young ones. They are vulnerable to a number of predators.
13. Differentiate between a zoospore and a zygote. Ans: The zoospore is flagellated, motile, haploid or diploid spore formed inside a zoosporangium. It is the result of asexual reproduction.
The zygote is always diploid and formed by the fusion of gametes. It is usually non- flagellated and non-motile or motile. It is the net result of sexual reproduction.
14. Differentiate between gametogenesis from embryogenesis. Ans: Differences between gametogenesis and embryogenesis are as follows :
15. Describe the post-fertilization changes in a flower. Ans: In sexual reproduction, events that occur after the formation of zygote are called post-fertilization events. In flowering plants, the zygote is formed inside the ovule. After fertilization the sepals, petals and stamens of the flower wither and fall off. But the pistil remains attached to the plant. The zygote develops into the embryo and the ovules develop into the seed. The ovary develops into die fruit that develops.a thick wall called pericarp which is protective in – function. After dispersal, seeds germinate under favourable conditions to produce new plants.
16. What is % bisexual flower? Collect five bisexual flowers from your neighbourhood and with the help of your teacher find out their common and scientific names. Ans: Flowers in which male and female sex organs (stamens and carpels) are borne on the same flowers are called bisexual flowers. You can observe following bisexual flowers in your kitchen and colony gardens : (i)Brassica (sarson) – Brassica campestris (ii)Onion – Allium cepa (iii)Garden Pea (Edible pea) – Pisum sativum (iv)Petunia – Petunia hybrida (v)China rose (shoe flower) – Hibiscus rosa- sinensis.
17. Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that bears unisexual flowers? Ans: The male or staminate flowers of cucurbits bear bright coloured petals and a prominent group of stamens. Male plants or staminate flowers do not bear fruits. The female or pistillate flowers bear fruits. In a fertilised young pistillate flower very small fruit is visible below petals and sepals. Some unisexual plants are : Papaya, Mulberry and Date-palm.
18. Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals? Ans: On the basis of the development of the zygote, animals are grouped into oviparous and viviparous. The oviparous animals such as reptiles and birds lay eggs. Their fertilised eggs are covered by hard calcareous shell and are laid in a safe place in the environment. After incubation period, young ones hatch out. In viviparous animals such as majority of mammals including human beings, the zygote develops into a young one inside the body of the female individual. After a certain growth period, the young ones are delivered by the female individual. Due to proper care and protection, the chances of survival of young ones are more in viviparous individuals. Oviparous offsprings are at a greater risk than viviparous ones.
Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 16 Chemistry in Everyday Life:
Chemistry in Everyday Life
Drugs and their Classification
Therapeutic Action of Different Classes of Drugs
Chemicals in Food
NCERT INTEXT QUESTIONS
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? Ans: Most of drugs taken in doses higher than recommended may produce harmful effects and act as poison and cause even death. Therefore, a doctor must always be consulted before taking the drug.
16.2. “Ranitidine is an antacid” With reference to which classification, has this statement been given? Ans: Ranitidine is labelled as antacid since it is quite effective in neutralising the excess of acidity in the stomach. It is sold in the market under the trade name Zintac.
16.3. Why do we require artificial sweetening agents? Ans: To reduce calorie intake and to protect teeth from decaying, we need artificial sweeteners.
16.4. Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structures of these compounds are given below: (i) (C15H31COO)3C3H5-Glyceryl palmitate (ii) (C17H32COO)3C3H5-Glyceryl oleate Ans:
16.5. Following type of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic part in the molecule. Identify the functional group (s) present in the molecule.
Functional groups present in the detergent molecule are: (i)ether (ii)1°alcoholic group
16.1. Why do we need to classify drugs in different ways? Ans: Drugs are classified in following different ways: (a) Based on pharmacological effect. (b) Based on action on a particular biochemical process. (c) Based on chemical structure. (d) Based on molecular targets. Each classification has its own usefulness. (а) Classification based on pharmacological effect is useful for doctors because it provides them the whole range of drugs available for the treatment of a particular disease. (b) Classification based on action on a particular biochemical proc*ess is useful for choosing the correct compound for designing the synthesis of a desired drug. (c) Classification based on chemical structure helps us to design the synthesis of a number of structurally similar compounds having different substituents and then choosing the drug having least toxicity. (d) Classification on the basis of molecular targets is useful for medical chemists so that they can design a drug which is most effective for a particular receptor site.
16.2. Explain the following as used in medicinal chemistry (a) Lead compounds (b) Target molecules or drug targets. Ans: (a) Lead compounds are the compounds which are effective in different drugs. They have specific chemical formulas and may be extracted either from natural sources (plants and animals) or may be synthesised in the laboratory.
(b) Target molecules or drug targets. An enzyme (E) functions by combining with the reactant (called substrate) denoted as ‘S’ to form an activated complex known as enzyme-substrate complex (E-S). The complex dissociates to form product and releases the enzyme for carrying out further activity.
16.3. Name the macro molecules that are chosen as drug targets. Ans: Proteins, carbohydrates, lipids and nucleic acids are chosen as drug targets.
16.4. Why the medicines should not be taken without consulting doctors? Ans: No doubt medicines are panacea for most of the body ailments. But their wrong choice and overdose can cause havoc and may even prove to be fatal. Therefore, it is of utmost importance that the medicines should not be given without consulting doctors.
16.5. Define the term chemotherapy. Ans: It is the branch of chemistry that deals with the treatment of diseases by using chemicals as medicines.
16.6. Which forces are involved in holding the drugs to the active site of enzymes? Ans: The following forces are involved in holding the drugs to the active site of enzymes: (a) Hydrogen bonding (b) Ionic bonding (c) Dipole-dipole interactions (d) van der Waals interactions
16.7. Antacids and antiallergic drugs interfere with the function of histamines but do not interfere with the function of each other. Explain. Ans: They donot interfere with the functioning of each other because they work on different receptors in the body.Histamine stimulates the secretion of pepsin and hydrochloric acid in the stomach. The drug cimetidine (antacid) was designed to prevent the interaction of histamine with the receptors present in the stomach wall. This resulted in release of lesser amount of acid. Antacid and antiallergic drugs work on different receptors.
16.8. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs. Ans: In case of low level of neurotransmitter, . noradrenaline, tranquilizer (antidepressant) drugs are required because low levels of noradrenaline leads to depression. These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, noradrenaline is slowly metabolized and can activate its receptor for longer periods of time thereby reducing depression. Two important drugs are iproniazid and phenylzine.
16.9. What is meant by the term broad spectrum antibiotics? Explain. Ans: Broad spectrum antibiotics are effective against several different types or wide range of harmful bacteria. For example, tetracycline, chloramphenicol and of loxacin. Chloramphenicol can be used in case of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.
16.10. How do antiseptics differ from disinfectants ? Give one example of each. Ans: Many times, the same substance can act as an antiseptic as well as disinfectant by changing the concentration of the solution used. For example, a 0.2 per cent solution of phenol acts as an antiseptic and its 1 percent solution is a disinfectant. Chlorine is used in India for making water fit for drinking at a concentration of 0.2 to 0.4 ppm (parts per million). Low concentration of sulphur dioxide is used for sterilizing squashes for preservation. A few points of distinction between antiseptics and disinfectants are listed.
1. Can kill or prevent the growth of micro-organisms.
1.Can kill micro-organisms.
2. Do not harm the living tissues. Therefore, these can be applied to the skin.
2. Toxic to the living tissues. Therefore, these cannot be applied to the skin.
3. These are used for the dressing of wounds, ulcers and in the treatment of diseased skin.
3. These are used for disinfecting floors, toilets, drains, instruments etc.
16.11. Why are cimetidine and ranitidine better antacids than sodium hydrogencarbonate or magnesium or aluminium hydroxide? Ans: If excess of NaHCO3 or Mg(OH)2 or Al(OH)3 is used, it makes the stomach alkaline and thus triggers the release of even more HCl which may cause ulcer in the stomach. In contrast, cimetidine and ranitidine prevent the interaction of histamine with the receptor cells in the stomach wall and thus release of HCl will be less as histamine stimulates the secretion of acid.
16.12. Name a substance which can be used as an antiseptic as well as disinfectant. Ans: 0.2% solution of phenol acts as antiseptic while 1% solution acts as a disinfectant.
16.13. What are the main constituents of dettol? Ans: Chloroxylenol .and α-terpineol in a suitable solvent.
16.14. What is tincture of iodine? What is its use? Ans: 2-3% solution of iodine in alcohol and water is called tincture of iodine. It is a powerful antiseptic. It is applied on wounds.
16.15. What are food preservatives? Ans: Preservation has a major role in the food products. Chemically preserved squashes and crushes can be kept for a fairly long time even after opening the seal of bottle. A preservative may be defined as the substance which is capable of inhibiting or arresting the process of fermentation, acidification or any other decomposition of food. Salting i.e. addition of table salt is a well known method for food preservation and was applied in ancient times for preserving raw mangoes, tamarind, meat, fish etc. Sugar syrup can also act as a preservative. Vinegar is a useful preservative for pickles. Apart from these, sulphur dioxide and benzoic acid can be employed for the preservation of food. The major source of sulphur dioxide is potassium metabisulphite (K2S2O5). It is fairly stable in neutral and alkaline medium but gets decomposed by weak acids such as carbonic, citric, tartane and malic acids. Benzoic acid is used either as such or in the form of sodium benzoate. However, sulphur dioxide has a better preservative action than sodium benzoate against bacteria and moulds. It also retards the development of yeast in juice but fails to arrest their multiplication once the number has reached a high value. Sorne salts of sorbic acid and propionic acid are also being used these days for the preservation of the food. The use of preservatives must be properly controlled as their indiscriminate use is likely to be harmful. The preservative should not be injurious to health and should be also non-irritant.
16.16. Why is the use of aspartame limited to cold foods and drinks? Ans: This is because it decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks as an artificial sweetener
16.17. What are artificial sweetening agents? Give two examples. Ans: Artificial sweeteners are chemical substances which are sweet in taste but do not add any calories to our body. They are excreted as such through urine. For example, saccharin, aspartame, alitame etc.
16.18. Name the sweetening agent used in the preparation of sweets for a diabetic patient. Ans: Saccharine, aspartame or alitame may be used in the preparation of sweets for a diabetic patient.
16.19. What problem arises in using alitame as artificial sweetener? Ans: Alitame is a high potency artificial sweetener.Therefore, it is difficult to control the sweetness of the food to which it is added.
16.20. How are synthetic detergents better than soaps? Ans: They can be used in hard water as well as in acidic solution. The reason being that sulphonic acids and their calcium and magnesium salts are soluble in water thus they do not form curdy white precipitate with hard water but the fatty acids and their calcium and magnesium salts of soaps are insoluble. Detergents also works in slightly acidic solution due to formation of soluble alkyl hydrogen sulphates. Soaps react with acidic solution to form insoluble fatty acids.
16.21. Explain the following terms with suitable examples: (i) cationic detergents (ii) anionic detergents and (iii) non-ionic detergents Ans: (i) Cationic detergents: These are quaternary ammonium salts, chlorides, acetates, bromides etc containing one or more long chain alkyl groups. For example, cetyltrimethyl ammonium chloride. (ii) Anionic detergents are called so because a large part of their molecules are anions. ‘These are of two types: (a) Sodium alkyl sulphates: For example, sodium lauryl sulphate, C11H23CH2OSO3 Na+. (b) Sodium alkylbenzenesulphonates.Vor example, sodium 4-(l-dodecyl) benzenesu Iphphonate (SDS).
(iii) Neutral or non-ionic detergents: These are esters of high molecular mass alcohols with fatty acids. These can also be obtained by treatment of long chain alcohols by with excess of ethylene oxide in presence of a base. For example, polyethylene glycol stearate,CH3(CH2)16COO (CH2CH2O)11 CH2CH2OH Polyethylene glycol stearate.
16.22. What are biodegradable and non-biodegradable detergents? Give one example of each. Ans: Detergents having straight chain hydrocarbons are easily degraded (or decomposed) by microorganisms and hence are called biodegradable detergents while detergents containing branched hydrocarbon chains are not easily degraded by the microorganisms find hence are called non-biodegradable detergents. Consequently, non-biodegradable detergents accumulate in rivers and water ways thereby causing severe water pollution. Examples of biodegradable detergents are sodium lauryl sulphate, sodium 4-(-l-dodecyl) benzenesulphonate and sodium 4-(2-dodecyl) benzenesulphonate. Examples of non-biodegradable detergents is sodium 4-(1, 3,5,7 – tetramethyloctyl) benzene sulphonate.
16.23. Why do soaps not work in hard water? (C.B.S.E. Outside Delhi 2009, 2011) Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid (C15H31COOH), oleic acid (C17H33COOH) and stearic acid (C17H35COOH). Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds. These being insoluble, get separated as curdy white precipitates resulting in wastage of soap.
16.24. Can you use soaps and synthetic detergents to check the hardness of water? Ans: Soaps get precipitated as insoluble calcium and magnesium soaps in hard water but detergents do not. Therefore, soaps but not synthetic detergents can be used to check the hardness of water.
16.25. Explain the cleansing action of soaps. Ans: Cleansing action of soaps : Soaps contain two parts, a large hydrocarbon which is a hydrophobic (water repelling) and a negative charged head, which is hydrophillic (water attracting). In solution water molecules being polar in nature, surround the ions & not the organic part of the molecule. When a soap is dissolved in water the molecules gather together as clusters, called micelles. The tails stick inwards & the head outwards. The hydrocarbon tail attaches itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface & dissociates into fragments. The solution now contains small globules of oil surrounded detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus the oily dirt is removed from the object.
16.26. If water contains dissolved calcium hydrogencarbonate, out of soaps and synthetic detergents, which one will you use for cleaning clothes? Ans: Calcium hydrogencarbonate makes water hard. Therefore, soap cannot be used because it gets precipitated in hard water. On the other hand, a synthetic detergent does not precipitate in hard water because its calcium salt is also soluble in water. Therefore, synthetic detergents can be used for cleaning clothes in hard water.
16.27. Label the hydrophilic and hydrophobic parts in the following compounds. (i)cCH3(CH2)10CH2OSO3–Na+ (ii) CH3(CH2)15 -N+(CH3)3Br– (iii) CH3(CH2)16C00(CH2CH2O)11CH2CH2OH Ans:
Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers:
Classification of Polymers
Types of Polymerisation Reactions
Molecular Mass of Polymers
Polymers of Commercial Importance
NCERT INTEXT QUESTIONS
15.1. What are polymers? Ans: Polymers are high molecular mass substances (103 — 107u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.
15.2. How are polymers classified on the basis of structure? Ans: On the basis of structure, polymers are classified into three types. These are linear chain polymers, branched chain polymers and crossedlinkedpolymers.
1. Linear chain polymers: In this case, the monomer units are linked to one another to form long linear chains. These linear chains are placed one above the other and are closely packed in space. The close packing results in high densities, tensile strength and also high melting and boiling points. High density polyethene is a very common example of this type. Nylon, polyesters and PVC are also linear chain polymers.
2. Branched chain polymers: In this type of polymers, the monomer units are linked to form long chains which have also side chains or branched chains of different Lengths attached to them. As a result of branching, these polymers are not closely packed in space. They have low densities, low tensile strength as well as low melting and boiling points. Some common examples of such polymers are ; low density polyethene, amylopectin, starch, glycogen etc.
3. Cross: linked polymers. In these polymers, also called net—work polymers, the monomer units are linked together to form three dimensionaL net—work as shown in the figure. These are expected to be quite hard, rigid and brittle. Examples of cross linked polymers are bakelite, glyptal. melamine formaldehyde polymer etc.
15.3. Write the names of the monomers of the following polymers:
Ans: (i) Hexamethylene diamine NH2-(CH2)6NH2 and adipic acid HOOC – (CH2)4 – COOH (ii) Caprolactum (iii) Tetrafluoroethene F2C = CF2
15.4. Classify the following as addition and condensation polymers: Terylene, Bakelite, Polyvinyl chloride,Polythene Ans: Addition polymers: Polyvinyl chloride, Polythene Condensation polymers : Terylene, bakelite.
15.5. Explain the difference between Buna-N and Buna-S. Ans: Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents. Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile. Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.
Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the ratio of 3 : 1 in the presence of sodium.
Buna-N (Nitrile rubber): h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:
15.6. Arrange the following polymers in increasing order of their intermolecuiar forces. (i) Nylon 6,6, Buna-S, Polythene (ii) Nylon 6, Neoprene, Polyvinyl chloride Ans: On the basis of intermolecuiar forces, polymers are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre. Thus, we have (i)Buns-S < Polythene < Nylon 6,6 (ii)Neoprene < Polyvinyl chloride < Nylon 6.
15.1. Explain the terms polymer and monomer. Ans: Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.
15.2. What are natural and synthetic polymers ? Give two examples of each. Ans: 1. Natural polymers: The polymers which occur in nature mostly in plants and animals are called natural polymers. A few common examples are starch, cellulose, proteins, rubber nucleic acids, etc. Among them, starch and cellulose are the polymers of glucose molecules. Proteins are formed from amino acids which may be linked in different ways. These have been discussed in detail in unit 15 on biomolecules. Natural rubber is yet another useful polymer which is obtained from the latex of the rubber tree. The monomer units are of the unsaturated hydrocarbon 2-methyl-i, 3-butadiene, also called isoprene. Example of natural polymers: Natural rubber, cellulose, nucleic acids, proteins etc.
2. Synthetic polymers: The polymers which are prepared in the laboraroiy are called synthetic polymers. These are also called man made polymers and have been developed in the present century to meet the ever increasing demand of the modem civilisation. Example of synthetic polymers: Dacron (or terylene), Bakelite, PVC, Nylon-66, Nylon-6 etc.
15.3. Distinguish between the terms homopolymer and copolymer and give an example of each. Ans: Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc. Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.
15.4. How do you explain the functionality of a monomer? Ans: Functionality of a monomer implies the number of bonding sites present in it. For example, monomers like propene, styrene, acrylonitrile have functionality of one which means that have one bonding site. Monomers such as ethylene glycol, hexamethylenediamine, adipic acid have functionality of two which means that they have two bonding sites.
15.5. Define the term polymerisation? Ans: It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.
15.6. Is (-NH — CHR—CO-)n a homopolymer or copolymer? Ans: It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH2—CHR—COOH.
15.7. In which classes, are the polymers classified on the basis of molecular forces? Ans: Polymers are classified into four classes on the basis of molecular forces. These are: elastomers, fibres, thermoplastic polymers and thermosetting polymers.
1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross – links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.
2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have aJso sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.
3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.
4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.
15.8. How can you differentiate between addition and condensation polymerisatiop? Ans: In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H2O, NH3 etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.
15.9. Explain the term copolymerisation and give two examples. Ans: When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.
15.10. Write the free radical mechanism for the polymerisation of ethene. Ans:
15.11. Define thermoplastics and thermo setting polymers with two examples of each Ans: Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene. Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.
15.12. Write the monomers used for gettingThe following polymers: (i) Polyvinylchloride (ii) Teflon (iii) Bakelite Ans:
15.13. Write the name and structure of one of the common initiators used in free radical addition polymerisation. Ans:
15.14. How does the presence of double bonds in rubber molecules influence their structure and reactivity? Ans: Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C2 and C3 of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.
15.15. Discuss the main purpose of vulcanisation of rubber. Ans: Natural rubber has the following disadvantages: (a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained. (b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion. (c)It is not resistant to the action of organic solvents. (d)It is easily attacked by oxygen and other oxidising agents. . To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.
15.16. What are the monomeric repeating units of Nylon-6 and Nylon 6,6? Ans:
15.17. Write the names and structures of the monomers of the following polymers: (i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene Ans:
15.18. Identify the monomer in the following polymeric structures:
15.19. How is dacron obtained from ethylene glycol and terephthalic acid? Ans: Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.
15.20. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester. Ans: Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.
Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules:
NCERT INTEXT QUESTIONS
14.1. Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membred ring compounds) are insoluble in water Explain. Ans: The .solubility of a solute in a given solvent follows the rule ‘ Like dissolves like’.Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. As a result of this extensive intermoleeular H-bonding, glucose and sucrose are soluble in water.On the other hand, benzene and cyclohexane do not contain -OH bonds and hence do not form H-bonds with water. Moreover, they are non-polar molecules and hence do not dissolve in polar water molecules.
14.2. What are the expected products of hydrolysis of lactose? Ans: Lactose being a disaccharide gives two molecules of monosaccharides Le. one molecule each of D-(+) – glucose and D-(+)-galactbse.
14.3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose? Ans: The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH2OH -to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the -OH group at C-l along with the other -OH groups at C-2, C-3, C-4 and C-6 form a pentaacetate. Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH2OH to form glucose oxime. The reactions are shown as:
14.4. The melting points and solubility in water of a-amino acids are generally higher than those of corresponding haloacids. Explain. Ans: a-amino acids as we all know, are dipolar in nature (N+H3-CHR-COO– ) and have strong dipolar interactions. As a result, these are high melting solids. These are also involved in intermolecular hydrogen bonding with the molecules of water and are therefore, water soluble. On the contrary, the haloacids RCH(X)COOH are not dipolar like a-amino acids. Moreover, only the carboxyl group of haloacids are involved in hydrogen bonding with the molecules of water and not the halogen atoms. These have therefore, comparatively less melting points and are also soluble in water to smaller extent.
14.5. Where does the water present in the egg go after boiling the egg? Ans: When egg is boiled, proteins first undergo denaturation and then coagulation and the water present in the egg gets absorbed in coagulated protein, probably through H- bonding
14.6. Why cannot Vitamin C be stored in our body? Ans: Vitamin C cannot be stored in the body because it is water soluble and is, therefore, easily excreted in urine.
14.7. Which products would be formed when a nucleotide from DNA containing thymine is hydrolysed? Ans: Upon hydrolysis, nucleotide from DNA would form 2-deoxyribose and phosphoric acid along-with thymine.
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? Ans: A DNA molecule has two strands in which the four complementary bases pair each other, i.e., cytosine (C) always pair with guanine (G) while thymine (T) always pairs with adenine (A). Thus, when a DNA molecule is hydrolysed, the molar amounts of cytosine is always equal to that of guanine and that of adenine is always equal to thymine.In RNA, there is no relationship between the quantities of four bases (C, G, A and U) obtained, therefore, the base pairing principle, i.e. A pairs with U and C pairs with G is not followed. Therefore, unlike DNA, RNA has a single strand.
14.1. What are monosaccharides ? Ans: Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH2O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) group are called ketoses. They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses according as they contain 3,4,5,6, and 7 carbon atoms respectively.For example.
14.2. What are reducing sugars? Ans: Reducing sugars are those which can act as reducing agents. They contain in them a reducing group which may be aldehydic (-CHO) or ketonic (>C=0) group. The characteristic reactions of reducing sugars are with Tollen’s reagent and Fehling solution. Non-reducing sugars donot give these reactions. For example, glucose, fructose, lactose etc. are reducing sugars. Sucrose is regarded as a non-reducing sugar because both glucose and fructose are linked through their aldehydic and ketonic groups by glycosidic linkage. Since these groups are not free, sucrose is a non-reducing sugar.
14.3. Write two main functions of carbohydrates in plants. Ans: Two major functions of carbohydrates in plants are following (a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls. (b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and act as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.
14.4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose. Ans: Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose. Disaccharides: Maltose and lactose.
14.5. What do you understand by the term glycosidic linkage? Ans: The ethereal or oxide linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage in maltpse molecule is shown below:
14.6. What is glycogen? How is it different from starch? Ans: The carbohydrates are stored in animal body as glycogen. It is also called animal starch and its structure is similar toamylopectin which means that it is a branched chain polymer of α-D-glucose units in which the chain is formed by C1 – C4 glycosidic linkage whereas branching occurs by the formation of C1– C6 glycosidic linkage. One main difference between glycogen and amylopectin is the length of the chain. In amylopectin, the chain consists of 20 – 25 α – D – glucose molecules whereas in glycogen, there are 10 -14 molecules of α – D – glucose present. Glycogen is more branched than amylopectin. It is present mainly in liver, muscles and also in brain. Glycogen gets converted into glucose when the body needs it with the help of certain enzymes present in the body. Glycogen has also been found to be present in yeast and fungi.
Starch is a major source of carbohydrates which are very much essential to the human body since they supply energy to the body. It occurs as granules mainly in seeds, fruits, tubers and also in the roots of the plants. The chief commercial sources of starch are wheat, maize, rice, potatoes etc.
14.7. What are the hydrolysis products of (i) sucrose, and (ii) lactose? Ans: Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.
14.8. What is the basic structural difference between starch and cellulose? Ans: Starch consists of amylose and amylopectin. Amylose is a linear polymer of α-D-glucose while cellulose is a linear polymer of β -D- glucose. In amylose, C -1 of one glucose unit is connected to C – 4 of the other through α-glycosidic linkage. However in cellulose, C – 1 of one glucose unit is connected to C-4 of the other through β – glycosidic linkage. Amylopectin on the other hand has highly branched structure.
14.9. What happens when D-glucose is treated with . the following reagents. (i) HI (ii) Bromine water (iii) HNO3 Ans:
14.10. Enumerate the reactions of D-glucose which cannot be explained with open chain structure. (C.B.S.E. Delhi 2008, C.B.S.E. Sample Paper 2011) Ans: Open chain structure of D-glucose contains a free aldehydic group (- CHO). However, it does not give the following reactions:
D(+) glucose does not react with 2, 4 D.N.P.
D(+) glucose does not react with NaHSO3.
D(+) glucose does not restore the pink colour to Schiff’s reagent.
Penia acetyl glucose formed by carrying acetylation with acetic anhydride does not react with hydroxyl amine (NH2OH) which is the characteristic reaction of all aldehydes.
D( +) glucose is found to exist in two different crystalline forms which are named as α and β. Both these forms have actually been isolated. For example, α form with m.p. 419 K is obtained by the crystallisation of the saturated solution of glucose prepared at 303 K. Similarly, β-form with m.p. 423 K is isolated by carrying out the crystallisation of the saturated solution of glucose prepared at 371 K. Apart from that the a-form has a specific rotation (α) equal to + 112° while the β- form has specific rotation (α) equal to + 19°.
In the light of the limitations stated above, Tollen stated that an open chain structure for D(+) glucose is probably not practicable. He proposed a cyclic structure which is a hemiacetal structure. In this structure, the aldehydic (CHO) group is involved in the form of a ring with the -OH group attached to C5 carbon. It is a six membered ring, often called ô- oxide ring. The ring structure accounts for the two isomeric forms a and shown below.
14.11. What are essential and non-essential amino acids? Give two examples of each type. Ans: α-Amino acids which are needed for good health and proper growth of human beings but are not synthesized by the human body are called- essential amino acids. For example, valine, leucine, phenylalanine, etc. On the other hand, α-amino acids which are needed for health and growth of human beings and are synthesized by the human body are called non-essential amino acids. For example, glycine, alanine, aspartic acid etc.
14.12. Define the following as related to proteins: (i) Peptide linkage (ii) Primary structure (iii) Denaturation Ans: (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between – COOH group of one α-amino acid and -NH-, group of the other α-amino acid by lo;ss of a molecule of water. For example,
(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of α-amino acids creates a different protein.
(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biologicalactivity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity, is called denaturation. During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.
14.13. What are the common types of secondary structure of proteins? Ans: Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :
β-pleated sheet structure.
Secondary Structure of Proteins: The long, flexible peptide chains of proteins are folded into the relatively rigid regular conformations called the secondary structure. It refers to the conformation which the polvpeptide chains assume as a result of hydrogen bonding between the > C= O and > N-H groups of different peptide bonds. The type of secondary structure a protein will acquire, in general depends upon the size of the R-group. If the size of the R-groups are quite large, the protein will acquire ct-helix structure. If on the other hand, the size of the R-groups are relatively smaller, the protein will acquire a β – flat sheet structure.
(a) α-Helix structure: If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group of one amino acid unit and the > N-H group of the fourth amino acid unit within the same chain. As such the polypeptide chain coils up into a spiral structure called right handed ct- helix structure. This type of structure is adopted by most of the fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be stretched. During this process, the weak hydrogen bonds causing the a – helix are broken. This tends to increase the length of the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.
(b) β—Flat sheet or β—Pleated sheet structure: If R-groups are relatively small, the peptide chains lie side by side in a zig zag manner with alternate R-groups on the same side situated at fixed distances apart. The two such neighbouring chains are held together by intermolecular hydrogen bonds. A number of such chains can be inter-bonded and this results in the formation of a flat sheet structure These chains may contract or bend a little in order to accommodate moderate sized R-groups. As a result, the sheet bends into parallel folds to form pleated sheet structure known as β – pleated sheet structure. These sheets are then stacked one above the other like the pages of a book to form a three dimensional structure. The protein fibrion in silk fibre has a β – pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its β – sheet structure. For example, silk is non-elastic since stretching leads to pulling the peptide covalent bonds. On the other hand, it can be bent easily like a stack of pages because during this process, the sheets slide over each other.
14.14. What types of bonding helps in stabilising the α-helix structure of proteins? Ans: α-helix structure of proteins is stabilised through hydrogen bonding. (a) α -Helix structure. If the size of the R-groups are quite large, the hydrogen bonding occurs between > C = O group of one amino acid unit and the > N- H group of the fourth amino acid unit within the same chain. As such the polypeptide chain coils up into a spiral structure called right handed a—helix structure. This type of structure is adopted by most of the fibrous structural proteins such as those present in wool, hair and muscles. These proteins are elastic i.e., they can be stretched. During this process, the weak hydrogen bonds causing the α-helix are broken. This tends to increase the length of the helix like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.
14.15: Differentiate between globular and fibrous proteins. Ans.(i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibres. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very’ strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH. Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles. (ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular protein are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH. This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.
14.16. How do you explain the amphoteric behaviour of amino acids? Ans: Amino acids contain an acidic (carboxyl group) and basic (amino group) group in the same molecule. In aqueous solution, they neutralize each other. The carboxyl group loses a proton while the amino group accepts it. As a result, a dipolar or zwitter ion is formed.
In zwitter ionjc form, a-amino acid show amphoteric behaviour as they react with both acids and bases.
14.17. What are enzymes? Ans: Enzymes are biological catalyst. Each biological reaction requires a different enzyme. Thus, as compared to conventional catalyst enzymes are very specific and efficient in their action. Each type of enzyme has its own specific optimum conditions of concentration, pH and temperature at which it works best.
14.18. What is the effect of denaturation on the structure of proteins? Ans: Denaturation of proteins is done either by change in temperature (upon heating) or by bringing a change in the pH of the medium. As a result, the hydrogen bonding is disturbed and the proteins lose their biological activity i.e., their nature changes. During the denaturation, both the tertiary and secondary structures of proteins are destroyed while the primary structures remain intact.
14.19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood. Ans: Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Water soluble vitamins: These include vitamin B-complex (B1, B2, B5, i.e., nicotinic acid,B6, B12, pantothenic acid, biotin, i.e., vitamin H and folic acid) and vitamin C. (ii) Fat soluble vitamins: These include vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues. Vitamin K is responsible for coagulation of blood.
14.20. Why are vitamin A and vitamin C essential to us? Give their important sources. Ans: Vitamin A is essential for us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness. Sources: Fish liver oil, carrots, butter, milk, etc. Vitamin C is essential for us because its deficiency causes scurvy (bleeding of gums) and pyorrhea (loosening and bleeding of teeth). Sources: Citrous fruits, amla, green leafy vegetables etc.
14.21. What are nucleic acids ? Mention their two important functions. Ans: Nucleic acids are biomolecules which are found in the nuclei of all living cell in form of nucleoproteins or chromosomes (proteins contains nucleic acids as the prosthetic group).
Nucleic acids are of two types: deoxyribonucleic acid (DNA) and ribonucleic acid.(RNA). The two main functions of nucleic acids are: (a) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to its unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells. (b) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually the proteins are synthesized by various RNA molecules (r-RNA, m-RNA) and t-RNA) in the cell but the message for the synthesis of a particular protein is coded in DNA.
14.22. What is the difference between a nucleoside and a nucleotide? Ans: A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. It is formed when 1- position of pyrimidine (cytosine, thiamine or uracil) or 9-position of purine (guanine or adenine) base is attached to C -1 of sugar (ribose or deoxyribose) by a β-linkage. Nucleic acids are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide. A nucleotide contains all the three basic . components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and a nitrogenous base. These are obtained by esterification of C5, – OH group of the pentose sugar by phosphoric acid.
14.23. The two strands in DNA are not identical but are complementary. Explain. Ans: The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C = G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A = T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and not identical.
14.24. Write the important structural and functional differences between DNA and RNA. Ans:
14.25. What are the different types of RNA found in the cell? Ans: There are three types of RNA: (a) Ribosomal RNA (r RNA) (b) Messenger RNA (m RNA) (c) Transfer RNA (t RNA)
Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 13 Amines:
Structure of Amines
Preparation of Amines
Method of Preparation of Diazonium Salts
Importance of Diazonium Salts in Synthesis of Aromatic Compounds
NCERT INTEXT QUESTIONS
13.1. Classify the following amines as primary, secondary and tertiary:
Ans: (i) 1° (ii) -3° (iii) 1° (iv) 2°
13.2. Write the structures of different isomeric amines corresponding to the molecular formula, C4H11N. (i) Write the IUPAC names of all the isomers (ii) What type of isomerism is exhibited by different types of amines? Ans: Eight isomeric amines are possible
Isomerism exhibited by different amines
Chain isomers: (i) and (ii) ; (iii) and (iv) ; (i) and (iv)
Position isomers: (ii) and (iii) ; (ii) and (iv)
Metamers: (v) and (vi) ; (vii) and (viii)
Functional isomers: All the three types of amines are the functional isomers of each other.
13.3. How will you convert: (i) Benzene into aniline (ii) Benzene into N,N-dimethylaniline (iii) Cl-(CH2)4-Cl into Hexane -1,6- diamine Ans:
13.4. Arrange the following in increasing order of their basic strength : (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6HsNH2, C6H5CH2NH2 Ans: In general, the basic character of ammonia (NH3) and the amines is linked with the availability of the lone electrons pair on the nitrogen atom. In other words, these are all Lewis bases. Amines act as Lewis bases due to the presence of lone electron paîr on the nitrogen atom. Since the nitrogen atom is sp³ hybridised, its electron attracting tendency is considerably reduced. It can readily lose its electron pair and acts as a base. For example, amines form hydroxides with water.
Here Kt is called dissociation constant for the base. Greater the Kb value stronger will be the base. The basic strength of amines can also be expressed as pKb value which is related to Kb as :
The Kb values are :
13.5. Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2+HCl ——–> (ii) (C2H5)3 N+HCl ——–> Ans:
13.6. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. Ans:
13.7. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. Ans:
13.8. Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberated N2 gas on treatment with nitrons acid. Ans: In ‘all, four structural isomers are possible. These are:
13.9. Convert: (i) 3-Methylanilineinto3-nitrotoluene (ii) Aniline into 1,3,5- Tribromo benzene Ans:
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)3 CNH2 (v) C6H5NHCH3(vi) (CH3CH2)2NCH3 (vii)m-BrC6H4NH2 Ans: (i) Propan-2-amine(1°) (ii) Propan-1-amine (1°), (iii) N-Methylpropan-2-amine (2°). (iv) 2-Methylpropan-2-amine(l°) (v) N-MethylbenzenamineorN-methylaniline(2°) (vi) N-Ethyl-N-methylethanamine (3°) (vii) 3-Bromobenzenamine or 3-bromoaniline (1°)
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. Ans:
13.3. Account for the following (i) pKb of aniline is more than that of methylamine (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines. Ans: (i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, electron density on the nitrogen . atom decreases. Whereas in CH3NH2,+ I-effect of -CH3 group increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine. (ii) Ethylamine dissolves in water due to intermolecular H-bonding. However, in case of aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding is very less therefore aniline is insoluble in water.
(iii) Methylamine being more basic than water, accepts a proton from water liberating OH– ions,
(iv) Nitration is usually carried out with a mixture of cone HNO3 + cone H2SO4. In presence of these acids, most of aniline gets protonated to form ahilinium ion. Therefore, in presence of acids, the reaction mixture consist of aniline and anilinium ion. Now, -NH2 group in aniline is activating and o, p-directing while the -+NH3 group in anilinium ion is deactivating and rw-directing: Nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.
13.4. Arrange the following: (i) In decreasing order of pKb values: C2H5NH2,C6H5NHCH3,(C2H5)2NH and C6H5NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and CH3NH2. (iii) In increasing order of basic strength: (а)Aniline,p-nitroaniline andp-toluidine (b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2 (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi) In increasing order of solubility in water: C6H5NH2,(C2H5)2NH,C2H5NH2 Ans: (i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C6H5NH2 and C6H5NHCH3 are far less basic than C2H5NH2 and (C2H,)2NH. Due to +I-effect of the -CH3 group, C6H5NHCH3 is little more basic that C6H5NH2. Among C2H5NH2 and (C2H5)2NH, (C2H5)2NH is more basic than C2H5NH2 due to greater+I-effect of two -C2H5 groups. Therefore correct order of decreasing pKb values is:
(ii) Among CH3NH2 and (C2H5)2NH, primarily due to the greater +I-effect of the two -C2H5 groups over one -CH3 group, (C2H5)2NH is more basic than CH3NH2.In both C6H5NH2 and C6H5N(CH3)2 lone pair of electrons present on N-atom is delocalized over the benzene ring but C6H5N(CH3)2 is more basic due to +1 effect of two-CH3 groups.
(iii) (a) The presence of electron donating -CH3 group increases while the presence of electron withdrawing -NO2 group decreases the basic strength of amines.
(b) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C6H5NH2 and C6H5NHCH3 are weaker base in comparison to C6H5CH2NH2. Among C6H5NH2 and C6H5NHCH3, due to +1 effect of-CH3 group C6H5NHCH3 is more basic.
(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,
(v) Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.
(vi) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease is the number of H-atoms on the N-atom which undergo H-bonding.
13.5. How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid. (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid? Ans:
13.6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved. Answer: The distinction in the three types of amines can be done by the following methods : (i) Hinsberg’s Test: This is a very useful test for the distinction of primary, secondary and tertiary amines. An amine is shaken with Hinsberg’s reagent (benzene suiphonyl chloride) in the presence of excess of aqueous KOH solution. The reactions taking place are given on the next page.
A primary amine forms N – alkyl benzene suiphonamide which dissolves in aqueous KOH solution to form potassium salt and upon acidification with dilute HCI regenerates the insoluble suiphonamide.
A secondary amine forms N, N – dialkylbenzene suiphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl
A tertiary amine does not react with benzene suiphonyl chloride and remains insoluble in aqueous KOH. However, on acidification with dilute HCI it gives a clear solution due to the formation of the ammonium salt.
(ii) Reaction with nitrous acid: All the three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.
(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium nitrite with dilute hydrochloric acid. For example,
The reaction is used as a tesijôr primary aliphadc amines as no other amine evolves nitrogen with nurous acid. (b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273 – 278 K) to form benzen diazonium salt. The reaction is known as diazotisation reaction.
in case, the temperature is allowed to rise above 278 K, benzene diazortium chloride is decomposed by water to form phenol.
Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are quite unstable and decompose to form a mixture of alcohols, alkenes and alkyl halides along with the evolution of N2 gas.
(c) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosoamines which separate as Yellow oily liquids.
(d) Tertiary aliphatic amines dissolve in a cold solution of nitrous acid to form salts which decompose on warming to give nitrosoamine and alcohol. For example,
(e) Tertiary aromatic amines react with nitrous acid to give a coloured nitrosoderivative. This reaction is called nitrosation and as a result, a hydrogen atom in the para position gets replaced by a nitroso (-NO) group. For example,
13.7. Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) ‘Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis Ans: (i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.
(ii) Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.
(iii) Hoffmann’s bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann’s bromamide degradation reaction.
(iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.
(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.
(vi) Acetylation: The process of introducing an acetyl (CH3CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.
(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCI or with alkali give primary amines.
13.8. Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Benzoic acid to aniline (iv) Aniline to 2,4,6-tribromofluorobenzene (v) Benzyl chloride to 2-phenylethanamine (vi) Chlorobenzene to p-Chloroaniline (vii) Aniline to p-bromoaniIine (viii)Benzamide to toluene (ix) Aniline to benzyl alcohol. Ans:
13.9. Give the structures of A,B and C in the following reaction:
13.10. 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. Ans: From the available information, we find that ‘B’ upon heating with Br2 and KOH forms a compound ‘C’. The compound ‘B’ is expected to be an acid amide. Since ‘B’ has been formed upon heating compound ‘A’ with aqueous ammonia, the compound ‘A’ is an aromatic acid. It is benzoic acid. The reactions involved are given as follows:
13.11. Complete the following reactions:
13.12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis? Ans: The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound. Since aryl halides do not undergo nucleophilic substitution reactions easily, therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.
13.13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid. Ans: Both aromatic and aliphatic primary amines react with HNO2 at 273-278 K to form aromatic and aliphatic diazonium salts respectively. But aliphatic diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds. Aromatic and aliphatic primary amines react with
13.14. Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines? Ans: (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion. R—NH2—>R—NH– +H+ R—O —H—>R— O– +H+ . Since O is more electronegative than N, so it wijl attract positive species more strongly in comparison to N. Thus, RO~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols. (ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass. (iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons: (a) Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines. (b) Aromatic amines arS more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.