CBSE Class 12

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara.

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara

Question 1.
What have been the methods used to study the ruins of Hampi over the last two centuries ? In what way do you think they would have complemented the information provided by the priests of the Virupaksha temple ?
Solution :
(a) The methods used to study the ruins of Hampi over the last two centuries were as given below:

• Colonel Colin Mackenzie, an engineer, antiquarian and an employee of the English East India Company discovered the ruins at Hampi in 1800. He prepared the first survey map of site.
• From 1856, photographers began to record the monuments which enabled scholars to study them.
• Since 1836, epigraphists began collecting several dozen inscriptions found at the temples at Hampi.
• Thereafter the historians collated the information from above sources with accounts of foreign travellers and other literature written in Telugu, Kannada, Tamil and Sanskrit.

(b) The above methods would have complemented the information provided by the priests of the Virupaksha temple because that was based on the memories of the priests. The earlier information was corroborated by the inscriptions, photographs, maps and accounts of foreign travellers and other material.

Question 2.
How were the water requirements of Vijayanagara met ?
Solution :
The water requirements of Vijayanagara were met in the following ways :

1. Its location is the natural basin formed by the river Tungabhadra which flows in a north-easterly direction. The stunning granite hills form a girdle around the city. A number of streams flow down to the river from these rocky outcrops.
2. The embankments were built along these streams to create reservoirs of varying
   sizes.
3. Tanks were built to store rainwater and conduct it to the city. The most important tank built is now called Kamalapuram tank. Water from this tank irrigated fields nearby as well as was also conducted through a channel to the “royal centre”.
4. The Hiriya canal drew water from a dam across the Tungabhadra and irrigated the cultivated valley that separated the “sacred centre” from the “urban core”.

Question 3.
What do you think were the advantages and disadvantages of enclosing agricultural land within the fortified area of the city?
Solution :
Advantages of enclosing agriculture land within fortified area:
(i) It had an elaborate canal system which drew water from the Tungabhadra to provide irrigation facilities.
(ii) It enclosed agricultural tracts, cultivated fields, gardens, and forests.
(iii) This enclosure saved crops from being eaten by wild animals.
(iv) In the medieval period, sieges were laid to starve the defending armies into submission. These sieges lasted for many months or many years. So the rulers of Vijayanagara adopted and elaborated a strategy to protect the agricultural belt and built large granaries.

Disadvantages
(i) This system was very expensive.
(ii) During adverse, circumstances this system proved inconvenient to the farmers.
(iii) The farmers had to seek the permission of gate-keeper to reach their field.
(iv) If enemy encircled the field the farmer could not look after their field.

Question 4.
Figure given below is an illustration of another pillar from the Virupaksha temple. Do you notice any floral motifs? What are the animals shown? Why do you think they are depicted? Describe the human figures shown.

Solution :
(a) There are many floral motifs.
(b) Horses and elephants have been shown in the pillar.
(c) The Vijayanagara kings competed with contemporary rulers including the Sultans of the Deccan and the Gajapati rulers of Orissa – for control of the fertile river valleys and resources generated by lucrative overseas trade. The kingdom remained in constant state of military preparedness. So, the kings paid attention to improve harbours and encouraged its commerce so that horses, elephants, precious gems, etc. are freely imported. Thus, due to the importance of horses and elephants in the warfare, these animals had been depicted on the pillars.
(d) The images of gods have been shown in the pillar. One devotee has also been shown before a Shiva linga.

Question 5.
What do you think was the significance of the rituals associated with the mahanavami dibba?
Solution :
The mahanavami dibba was one of the most impressive platforms in the “king’s palace”. It was located on one of the highest points in the city. Rituals associated with the structure probably coincided with mahanavami (literally the great ninth day) of the ten day Hindu festival during the autumn months of September and October, known variously as Dusehra (northern India), Durga Puja (in Bengal) and Navaratri or Mahanavami (in peninsular India). The Vijayanagara kings displayed their prestige, power and suzerainty on this occasion.

The ceremonies such as worship of the image, worship of the state horse, the sacrifice of buffaloes and other animals were performed on this occasion. Dances, wrestling matches, processions of caparisoned horses, elephants, chariots, soldiers and ritual presentations were held before the kings, guests, the chief nayakas were held. On the last day, the king inspected his army and the armies of the nayakas who brought rich gifts for the king as well as the stipulated tribute. Thus, there was great significance of the rituals associated with the mahanavami dibba.

Question 6.
Discuss whether the term “royal centre” is an appropriate description for the part of the city for which it is used.
Solution :
The term “royal centre” is an appropriate description for the part of the city for which it is used because the Royal center had more than 60 temples. Most of these temples were constructed by the ruler of Vijayanagara Empire to express their supremacy. The royal centre had 30 palaces. These were made of perishable material. A brief description of the building of Royal centre are as given below:
(i) One of the most beautiful buildings in the royal centre is the Lotus Mahal. It was named by British travellers in the nineteenth century. While the name is certainly romantic, historians are not quite sure what the building was used for. One suggestion, found in a map drawn by Mackenzie, is that it may have been a council chamber, a place where the king met his advisers.
(ii) Most temples were located in the sacred centre. One of the most spectacular of these is the Hazara Rama Temple. This was probably meant to be used only by the king and his family.

Question 7.
What does the architecture of buildings like the Lotus Mahal and elephant stables tell us about the rulers who commissioned them ?
Solution :
The Lotus Mahal had nine towers – a high central one, and eight along the sides. Although it is not clear for what the building was used for but according to Mackenzie, it may have been a council chamber, place where the king met his advisers. Elephant stables were located close to the Lotus Mahal.

The architecture of Lotus Mahal tells us that the rulers used to consult their advisers on various issues and problems and meetings were held in the council chamber i.e., Lotus Mahal. The construction of “elephant stables” shows that the rulers took interest in the trade of elephants as well as in keeping them properly because elephants were very important factor in the warfare. It is perhaps one of reasons that elephants and horses have been depicted on the panels of the Hazara Rama temple.

Question 8.
What are the architectural traditions that inspired the architects of Vijayanagara ? How did they transform these traditions ?
Solution :
(a) The architectural traditions that inspired the architects of Vijayanagara were as given below :

1. Prior to Vijayanagara, Cholas in Tamil Nadu and the Hoysalas in Karnataka had extended patronage to elaborate temples such as the Brihadishvara temple, Thanjavur and the
   Chennakeshava temple at Belur. The rulers of Vijayanagara built on these traditions and carried them literally to new heights.
2. Like Indo-Islamic architecture, there was an arch on the gateway leading into fortified settlement as well as dome over the gate.
3. The architecture of tombs and mosques located in the urban core resembles that of the mandapas found in the temples of Hampi.
4. The Pallavas, Chalukyas, Hoysalas and Cholas encouraged temple building as a means of associating themselves with the divine – the deity was identified with the king. The choice of the site Vijayanagara was perhaps inspired by the existence of the shrines of Virupaksha and Pampadevi.
5. The arches in the Lotus Mahal were inspired by Indo-Islamic technique.

(b) They transformed these traditions in the followings ways :

1. In the fortifications, according to Abdur Razzaq, no mortar or cementing agent was employed. The stone blocks were wedge shaped, which held them in place, and the inner portion of the walls was of earth packed with rubble.
2. Royal portrait sculpture was innovated and developed. It was displayed in temples and the king’s visits to temples were treated as important state occasions.
3. In temple architecture, new features were structures of immense scale best exemplified by the Raya gopurams or royal gateways. They often dwarfed the towers on the central shrines and signalled the presence of the temple from a great distance. Other features include mandapas or pavilions and long, pillared corridors.

Question 9.
What impressions of the lives of the ordinary people of Vij ay an agar a can you cull from the various descriptions in the chapter ?
Solution :
The various descriptions in this chapter give the following impression of the lives of the ordinary people of Vijayanagara :

1. Horses were imported from Arabia and Central Asia. This trade was done by the traders and local communities of merchants i.e., kudirai chettis or horse merchants.
2. There were markets dealing in spices, textiles and precious stones.
3. Vijayanagara boasted of a wealthy population that demanded high-value exotic goods.
4. Portuguese traveller Barbosa described the houses of ordinary people, which have not survived as “the other houses of the people are thatched, but nonetheless well-built and arranged according to occupations, in long streets with many open places”. Besides this there is little archaeological evidence of the houses of ordinary people.
5. There were numerous shrines and small temples which implies that there were variety of cults, supported by different communities.
6. There were wells, rainwater tanks, temple tanks which may have served as sources of water to the ordinary town dwellers.
7. Paes gives us a vivid description of a bazaar. He states that the provisions, such as rice, wheat, barley, etc. were available cheaply and abundantly. This means that the life of the ordinary people was good and they did not suffer for want of essential things.

We hope the NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Vistas Chapter 3 Journey to the end of the Earth. Students can get Class 12 English Journey to the end of the Earth NCERT Solutions, Questions and Answers designed by subject expert teachers.

Journey to the end of the Earth NCERT Solutions for Class 12 English Vistas Chapter 3

Journey to the end of the Earth NCERT Text Book Questions and Answers

Journey to the end of the Earth Reading with insight

Question 1.
How do geological phenomena help us to know about the history of humankind?
Answer:
Geological phenomena such as the drifting of land masses and their separating into countries help us to know about the history of humankind. A visit to Antarctica around which Gondwana once existed, is like going back to past as it gives us an understanding of evolution and extinction, ozone and carbon, where humankind came from, and where it is headed.

Question 2.
What are the indications for the future of humankind?
Answer:
All thoughtless activities of humankind such as increasing cities and megacities, cutting forests and turning those to concrete jungles, careless burning of fossil fuel, depleting ozone and increasing carbon dioxide, and global warming, melting ice caps and shields, our battle with other species for limited resources and other similar reckless activities point to a grim future for humankind. If concrete steps are not taken immediately, these drastic changes may lead to the end of the world.

Journey To The End Of The Earth Reading with Insight

Question 1.
‘The world’s geological history is trapped in Antarctica’. How is the study of this region useful to us?
Answer:
Antarctica holds half a million-year-old carbon track records in its layers of ice. It gives us an understanding of evolution and extinction, ozone and carbon. A visit to Antarctica, around which Gondwana once existed, is like going back to the past. Witnessing the geological phenomena, such as the drifting of land masses and their spreading into countries, help us to know about the history of humankind. These are visible signs of where humankind came from and it gives us a clear understanding of where human life is headed if we do not take care of the environment. Actually seeing with our own eyes all these changes, make us understand that global warming is a real threat.

Question 2.
What are Geoff Green’s reasons for including high school students in the Students on Ice Expedition?
Answer:
Geoff Green feels that students are the future generation of policy-makers. They should be provided an opportunity to have this life-changing experience at a young age in order to foster a new understanding and respect for our planet. It would help them to absorb, learn and act for the benefit of the planet. The youngsters still have the idealism to save the world and they need to understand that it belongs to them. So, to sensitize them, it is important to provide them the visible life changing experience.

Question 3.
‘Take care of the small things and the big things will take care of themselves.’ What is the relevance of the statement in the context of the Antarctica environment?
Ans. This statement means that if small things are taken care of, big things will take their own care. There are tall grasses, called phytoplankton, in the southern oceans that use the sun’s energy to assimilate • carbon and synthesize organic compounds by photosynthesis. Marine life and birds in the region sustain themselves on these tall grasses. Any disturbance in the environment in Antarctica might affect the activities of the phytoplankton, which, in turn, might affect the existence of the other life forms that depend on them. Small things like the phytoplankton are important in the food chain.

Question 4.
Why is Antarctica the place to go to understand the Earth’s present, past and future?
Answer:
The author states that to understand the earth’s present, past and future, Antarctica is the right place to go. Antarctica is relatively untouched in this respect as it has never had human population. It is relatively pristine. It holds in its ice cores half a million-year-old carbon records, trapped in the layers of ice. It embodies all that is pre-historic: cordilleran folds, pre-Cambrian granite shields ozone and carbon: evolution and extinction. The simple eco system and lack of biodiversity indicate how little changes in the environment can have big repercussions.

A visit to Antarctica and witnessing the geological phenomena, such as the drifting of land masses, glaciers receding and ice shelves collapsing makes us understand that global warming is a real threat. Hence, to study the earth’s past, present and future, these factors make Antarctica the best place to go.

Journey To The End Of The Earth Extra Questions and Answers

Journey To The End Of The Earth Short Answer Questions

Question 1.
When did the author start her journey to Antarctica and what had she to pass through?
Answer:
The author started her journey 13.09 degrees north of the Equator in Madras—she was on board a Russian research vessel—the Akademik Shokalskiy. She had to pass through nine time zones, six checkpoints, three bodies of water and at least as many ecospheres. After travelling over hundred hours in combination of a car, an aeroplane and a ship, she reached Antarctica.

Question 2.
What emotions did the author experience when she reached Antarctica at last?
Answer:
The author finally set foot on the Antarctica continent after travelling over 100 hours in combination of car, aeroplane and ship. Her first emotion on seeing the vast expansive white landscape and the blue horizon was of relief. She experienced the emotion of wonder at its immensity and isolation and its strange relationship with India.

Question 3.
How would you describe Gondwana?
Answer:
Gondwana was a giant amalgamated southern supercontinent, centering around present-day Antarctica. Humans had not arrived on the global scene. The climate was much warmer. There was a huge variety of flora and fauna. Gondwana thrived for 500 million years. When the age of the mammals got underway, the landmass was forced to separate into countries. Antarctica separated from the whole landmass shaping the globe as we know it today.

Question 4.
What is that thing that can happen in a million years and would be mind-boggling?
Answer:
The author says that in a million years India may push northwards, jamming against Asia. It will buckle its crust and form the Himalayas – South America may drift off to join North America. The Drake Passage may open up to create a cold circumpolar current. Antarctica may remain frigid, desolate and at the bottom of the world.

Question 5.
In what respect, Tishani Doshni’s encounter with Antarctica is a chilling prospect?
Answer:
The author remained there for two weeks. For a sun worshipper South Indian, being face to face with ninety per cent of earth’s total ice volume was a mind-boggling and chilling prospect. It was also a chilling experience for circulatory and metabolic functions and for imagination. It is like walking into a giant ping-pong ball with no human markers such as trees, billboards, and buildings.

Question 6.
What is the visual experience in Antarctica?
Answer:
In Antarctica the visual scale ranges from the microscopic to the mighty midgets and mites to blue whales and icebergs as big as countries. The writer refers to it as walking into a giant ping-pong ball devoid of any human markers, without trees, billboards, buildings. Days go on in 24 hours austral summer light. A ubiquitous silence, interrupted only by an occasional avalanche or calving ice sheet consecrates the place.

Question 7.
How, according to the author, has mankind etched its dominance over nature?
Answer:
According to the author, though civilizations have been around for barely a few seconds on the geological clock, yet they have created a ruckus by their various activities like exploiting the limited resources and careless burning of fossil fuels. In the short span of existence on the earth, they have already created a blanket of carbon dioxide and increased the average global temperature.

Question 8.
How has Antarctica sustained itself and managed to remain pristine?
Answer:
Antarctica, on account of being the coldest, windiest and driest continent in the world, has never sustained a human population and has thus managed to remain pristine. This has prevented man from being able to create ruckus in this part of the world by his thoughtless exploitation of the natural resources.

Question 9.
How is global temperature increasing? What are the immediate fears due to it?
Answer:
Global temperature is increasing due to the increasing burning of fossil fuels. It has now created a blanket of carbon dioxide around the world. This has given birth to questions like: Will the West Antarctica ice sheet melt entirely? Will the Gulf Stream Ocean current be disrupted? Will it be the end of the world as we know of? It may be. It may not be.

Question 10.
How is Antarctica a crucial element in the debate of climate change?
Answer:
Antarctica is a crucial element not because it has no human population but because it holds in its ice cores half a million year old carbon records. They are trapped in its layers of. ice. It will open up areas of knowledge about the past, present and future of the earth.

Question 11.
What are the reasons for the success of the Students on Ice programme?
Answer:
Sitting distant in the comfort zone of our houses, any talk about global warming looks so unreal and one can be unconcerned. But the visible experience of seeing glaciers retreating, ice caps melting and ice shelves collapsing makes one understand and realize what global warming is all about. The indications for the future of humankind become clear when one actually witnesses the geological phenomena.

Question 12.
The author says that her Antarctica experience was full of such epiphanies. What was that best epiphany that occurred there?
Answer:
The Akademik Shokalskiy got wedged into a thick white sheet of ice. The captain decided to turn around and asked the passengers to walk on the ocean. Underneath their feet they saw 180 metres of living, breathing salt water. Crab eater seals were stretching and sunning themselves on ice floes much like stray dogs under a banyan tree. It was a great epiphany, a revelation.

Question 13.
What is that beauty of balance that a trip to Antarctica unfolded to the author?
Answer:
The author was wonderstruck by the beauty of balance in play on our planet. Travelling across nine time zones, three bodies of water and as many ecospheres was an experience that unfolded a wide range of climate, geographical features, and flora and fauna. It was also a visible experience of the varied geographical phenomena.

Question 14.
Why does the author conclude the chapter by saying that a lot can happen in a million years, but what a difference a day makes?
Answer:
The author concludes the chapter by saying that much more can really happen in a million years as it happened in the case of Antarctica. But in this long period, changes even in a day make a great difference because global climate is changing. It is posing a threat to the beauty of balance on the earth.

Question 15.
What are phytoplanktons? What is their importance?
Answer:
Phytoplanktons, the grasses of the sea, are single-celled organisms living in the southern ocean. They nourish and sustain the entire ocean’s food chin, being first link in the food chain of ocean. Using sun’s energy, they assimilate carbon and synthesize organic compounds.
The diminishing number of these organisms due to the depletion of ozone layers affects other organisms of the ocean, finally leading to the extinction of life on earth.

Question 16.
Why does the author feel that the prognosis for the human beings is not healthy?
Answer:
The world is battling an ever increasing population, leading to burning of fossil fuels. This has created a blanket of carbon dioxide around the world thereby increasing global temperatures. All this is hazardous and life threatening for all flora and fauna. Hence the future of mankind in fact, all life on earth, is bleak. So, the author is correct in saying that the prognosis for man is not encouraging and healthy. . , j

Question 17.
Why is it necessary to remain fully equipped while walking on ice?
Answer:
While walking on ice, the troupe was fully kitted out in Gore-Tex (type of spiked boots that help in walking on ice) and glares. The spiked boots protect them from falling down on ice which might result in injury and the glares protect the eyes because the sunglasses can injure their eyes, particularly the ratina.

Question 18.
Do you think that programmes like the Students on Ice do more harm than good? Support your answer.
Answer:
I personally feel that such trips do more harm than good. We have ruined the earth as much as we could and as wide as we could go, because Antarctica was far away and extremely cold. But now we have so many reasons to go to this pristine continent. Let’s not encourage such trips. After all, what else do we have to learn about the earth than the fact that we have been running a business, not a service. Please spare Antarctica.

Student on Ice is an educational journey to Antarctica. It took high school students to Antarctica where they understood the seriousness of the threat that the end of the earth is quite near. By visiting Antarctica they would act their bit to save the planet from further deterioration. The educational youth of today is the hope for the earth and if they are more informed and more aware of the weakening strength of the earth, they will be able to steer the government machinery of their countries as they grow up.

Question 19.
Does the study of the lesson give you a feeling that man is his own great enemy?
Answer:
In his 12000-year-long stint on the earth so far man has caused untold harm to the planet, its environment and biodiversity. His activities in the name of development have spelt doom for the flora and fauna and his own existence is in danger. Man is to blame for all the havoc and ruckus created on earth. Thus it is quite right that man is his own great enemy.

Journey To The End Of The Earth Long Questions and Answers

Question 1.
What is the significance of the title ‘Journey to the End of the Earth’?
Answer:
The title ‘Journey to the End of the Earth’, has more than one meaning. It describes an educational journey to Antarctica undertaken by a group of high school students. To learn more about the real impact of global warming and future of the earth 52 students went to the coldest, driest, windiest continent in the world called Antarctica in Russian research vessel, the Akademik Shokalskiy.

The author calls it a journey to the end of the earth because it began 13:09 degrees North of Equator in Madras, involved crossing nine time zones, six checkpoints, three oceans and as many ecospheres. She travelled over 100 hours in combination of a car, an aeroplane and a ship. The journey being to the extreme south of the the earth, was really towards the end of it. Another meaning of this title is more significant as the warnings that Antarctica gives are shocking and much concerning the humanity and the millions of other species on the earth. The changes taking place in Antarctica are pointing a warning finger at the existence of of the earth; the earth is journeying to its end.

Question 2.
The author says, ‘It was nothing short of a revelation: everything does connect.’What does it mean?
Answer:
Antarctica is a perfect place to study how little changes in the environment can have big repercussions as far as Antarctica is concerned. Various human activities like exploiting the limited resources and careless burning of fossil fuel have already created a blanket of carbon dioxide, increased the average global temperatures and caused the retreating of glaciers, melting of ice caps and collapse of ice shelves as far as Antarctica. Global warming does not only change the geographical features, but also cause depletion in the ozone layer which will affect the activities of the phytoplanktons, the tall grasses which support the lives of marine animals and birds of the region. Hence, the author says everything does connect and all human activities are interlinked with the geological phenomena, whatever be the geological distance.

Question 3.
By whom and with what objective was Students on Ice programme started? How far has it achieved its goals?
Answer:
The Students on Ice programme was started by Canadian Geoff Green. He felt students are the future generation of policy-makers. They should be provided an opportunity to have this life¬changing experience at a young age in order to foster a new understanding and respect for our planet. It would help them to absorb, learn and, more importantly, act for the benefit of the planet.

Geoff Green was tired of taking celebrities and retired rich curiosity seekers who could only give back in a limited way. It means Geoff wanted something in return from his passengers to solve the problems relating to climate changes due to environmental pollution. It is difficult to imagine or be affected by the polar ice caps melting while sitting in our living rooms and so this visible life changing expence is important. Hence, this programme made the children learn that to save big things, small . things must be cared for.

Question 4.
What makes Antarctica an ideal subject of study?
Answer:
Antarctica is the only place in the world which has never sustained a human population. It thus remains relatively pristine in this respect. But, more importantly, it holds in its ice core, half a million- year-old carbon records trapped in its layers of life. Antarctica has a simple ecosystem and lack of biodiversity. It is, therefore, a perfect place to study how little changes in the environment can have big repercussions. Visiting Antarctica means knowing where we have come from and where we could possibly be heading. This place holds the key to know the geological evolution and it shall reveal the earth’s past, present and future.

Question 5.
The author states that her Antarctic experience was full of epiphanies, but the best occurred just short of the Antarctic Circle of 65-55 degrees south? Explain.
Answer:
Epiphanies is a Christian festival that celebrates the revelation or enlightenment. Here epiphanies are used metaphorically to suggest moments when the author suddenly becomes conscious of something that is very important to her.

The author experienced the rare of the rarest experiences there in Antarctica both in relation to beauty, wonder, and geological phenomena. Such masterly geological epiphany was experienced by her when the Akademik Shokalskiy got wedged into a thick white stretch of ice between the peninsula and Tadpole Island. The captain decided to turn around and asked the passengers to walk on the ocean. They kitted out in Gore-Tex and glares, walking on a white sheet of ice. Underneath their feet was a metre-thick ice pack. And underneath that, 180 metres of living breathing, saltwater lay before them. In the periphery, crabeater seals were stretching and sunning themselves on ice floes. They were doing so like stray clogs will do under the shade of a banyan tree. It was nothing short of revelation. The author saw in it that everything does indeed connect. This really proved to be the most wonderful experience of all experiences of Antarctica.

NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases

Question 1.
What are the various public health measures, which you would suggest as a safeguard against infectious diseases?
Solution:
Prevention and control of infectious diseases

I. For water-borne diseases like typhoid, amoebiasis, etc.
Practice personal and public hygienic measures.

a. Personal hygienic measures

• Keeping the body clean
• Consumption of clean drinking water
• Eating fresh food

b. Public hygienic measures

• Proper disposal of waste and excreta
• Periodic cleaning and disinfection of water reservoirs, pool, tank etc.

II. For air-borne diseases like common cold, pneumonia

• Avoid close contact with infected persons.
• Avoid the use of belongings of the infected persons.

III. For vector-borne diseases like malaria

• Control and eliminate the vectors and their breeding places
• Introducing larvivorous fishes like Gambusia in ponds that feed on the larvae of the mosquito
• Avoid stagnation of water around the residential area.
• Spraying of insecticides in ditches, drainage areas, etc.
• Protection from a mosquito bite. Use mosquito nets in the doors and windows to prevent the entry of mosquitoes. It is very important in the light of recently widespread diseases like dengue fever, chikungunya etc.

The use of vaccines and immunization programmes has enabled us to eradicate smallpox. Diseases like polio, diphtheria, tetanus etc. have been controlled to an extent by the use of vaccines. Nowadays biotechnology is focussing on the preparation of newer and safer vaccines. A large number of antibiotics are available to treat many infectious diseases.

Question 2.
In which way has the study of biology helped us to control infectious diseases?
Solution:
Study of biology has helped us to know about causes of diseases, carriers of diseases (vectors), effects of diseases on different body functions and above all, means to control diseases. Our immune system plays a major role in preventing diseases.

Question 3.
How does the transmission of each of the following diseases take place ?

1. Amoebiasis
2. Malaria
3. Ascariasis
4. Pneumonia

Solution:

1. Through contaminated food and water.
2. Through Anopheles mosquito.
3. Through contaminated food and water.
4. By inhaling the droplets or aerosols released by infected persons.

Question 4.
What measure would you take to prevent water-borne diseases?
Solution:
Water-borne diseases can be prevented by drinking clean water. Water should be free from contamination, suspended and dissolved substances. If water is contaminated it should be boiled and filtered before drinking. Periodic cleaning and disinfection of water reservoirs, pools, and tanks should be done.

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Solution:
‘A suitable gene’ means the gene which is able to produce antigenic polypeptides of the pathogen in bacteria and yeast. Using recombinant DNA technology, it is possible to produce vaccines in large scale for immunisation. Hepatitis B vaccine is produced using this technology.

Question 6.
Name the primary and secondary lymphoid organs.
Solution:
Primary lymphoid organs are bone marrow and thymus. Secondary lymphoid organs are the spleen, lymph nodes, tonsils, Peyer’s patches of the small intestine, and mucosa-associated lymphoid tissues (MALT).

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form.

1. MALT
2. CMI
3. AIDS
4. NACO
5. HIV

Solution:

1. MALT – Mucosal-associated lymphoid tissue.
2. CMI – Cell-Mediated Immunity
3. AIDS – Acquired Immuno Deficiency Syndrome
4. NACO – National AIDS Control Organisation
5. HIV – Human immunodeficiency virus.

Question 8.
Differentiate the following and give examples of each.

1. Innate and acquired immunity,
2. Active and passive immunity

Solution:

1. : Differences between innate and acquired immunity are as follows:
   
2. Differences between active and passive immunity are as follows:
   

Question 9.
Draw a well-labeled diagram of an antibody molecule.
Solution:

Question 10.
What are the various routes by which transmission of the human immunodeficiency virus takes place?
Solution:
Various routes of entry of ADDS virus are:

• Sexual contact with the infected person.
• Through placenta (from infected mother to foetus).
• Transfusion of infected blood or blood products.
• Sharing infected needles by drug abusers.

Question 11.
What is the mechanism by which the AIDS virus causes a deficiency of the immune system of the infected person?
Solution:
After getting into the body, the virus enters into the macrophages and converts its RNA genome into DNA with the help of a reverse transcriptase enzyme. The viral DNA takes and directs the infected cells to produce more virus particles i.e., the infected macrophages act like an HIV factory. Simultaneously, the HIV attack the T- lymphocytes and replicate and produce more viruses. Then they are released into the blood and attack other T-lymphocytes.

This will lead to a decrease in the number of T-lymphocytes and the patient begins to show the symptoms such as fever, diarrhea, weight loss etc. Subsequently, his immune system weakens and becomes more prone to infections of bacteria (like Mycobacterium), viruses, fungi and even parasites like Toxoplasma. Finally, he is unable to protect himself.

Question 12.
How is a cancerous cell different from a normal cell?
Solution:
Cancerous cell and normal cell are different in the following aspects:

Question 13.
Explain what is meant by metastasis.
Solution:
The rapid growth of cancerous tumour causes overcrowding and disruption of normal cells. It extends to neighbouring tissues. In the last stage, bits of tumour tissue break off and are carried by the circulating blood or lymphs to other parts of the body, where they invade new tissues and start new tumors called secondary tumors. This property is called metastasis. It is fated due to increasing interference with the body’s life processes.

Question 14.
List the harmful effects caused by alcohol/drug abuse.
Solution:
Harmful effects caused by alcohol/drug abuse are as follows:

• Among youth there is drop in academic performance, lack of interest in personal hygiene, isolation, depression, fatigue, aggressive and rebellious behavior, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, etc.
• Excessive dose of drugs leads to coma and death due to respiratory failure, heart failure or cerebral haemorrhage.
• Abusers become mental and cause financial distress to their entire family and friends.
• They may acquire serious infections like AIDS and hepatitis by taking drugs intravenously.
• Intake of alcohol/drugs damages nervous system, liver (cirrhosis) and kidney.
• Drug abuse adversely affects foetus in case of pregnancy, leading to Foetal Alcohol Syndrome (FAS).
• Continuous use of narcotics and stimulants cause impotency and chromosomal aberrations.
• Heavy drinking can cause an acute alcoholic myopathy characterised by painful and swollen muscles and high levels of serum creatine phosphokinase (CK). Chronic alcoholic men may show testicular atrophy with shrinkage of the seminiferous tubules and loss of sperm cells.
• Heavy drinking causes acute and chronic pancreatitis.
• Alcohol increases RBC size causing a mild anemia.
• Legal problems occur, such as arrest by police for obtaining and keeping drugs unlawfully.

Question 15.
Do you think that friends can influence one to take alcohol/drugs? If yes, how can one protect himself/herself from such an influence?
Solution:
Yes. This can be avoided by

• Choosing a good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
• Telling the program of an outing to family.
• Keeping contact with family while outside the home.

Question 16.
Why is it that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.
Solution:
Once a person starts taking alcohol or drugs, it is difficult to get rid of this habit because he becomes addicted to it. Addiction is a psychological attachment to certain effects such as euphoria and a temporary feeling of well-being. These drive people to consume drugs/alcohol even when these are not needed, or even when their use becomes self-destructive. With repeated use, the tolerance level of the receptors present in the body increases, which consequently leads to a higher dose of drugs/alcohol and addiction.

Thus, the addictive potential of drugs and alcohol pull the user into a vicious circle leading to their regular use from which he/she may not be able to get out.

Question 17.
In your view what motivates youngsters to take alcohol or drugs and how can this be avoided?
Solution:
There are many factors that motivate youngsters to take alcohol or drug. These include:

• Choosing a good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counselors
• Telling the programme of an outing to family.
• Keeping contact with family while outside the home.

This can be avoided by the following measures:

1. Education and counseling: Educating and counseling people to face problems and stresses, and to accept disappointments and failures as a part of life.
2. Seeking help from parents and peers: Help from parents and peers should be sought immediately so that they can guide appropriately. Help may even be sought from close and trusted friends.
3. Looking for danger signs: Alert parents and teachers to look for and identify the danger signs. Even friends, if they find someone using drugs or alcohol, should not hesitate to bring this to the notice of parents or teachers in the best interests of the person concerned.
4. Seeking professional and medical help: Lots of help is available in the form of highly qualified psychologists, psychiatrists, and de-addiction and rehabilitation programmes to help individuals who have unfortunately got in the quagmire of drug/alcohol abuse.
5. Cross-checking before prescribing and selling drugs: The physicians should prescribe the habituating drugs only to genuine persons and only for the essential duration. Pharmacists should not sell these drugs without the physician’s prescription.
6. Discipline: Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.
7. Communication: The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.
8. Appreciation: For even the smallest achievement, good behavior and other activities, the child should be appreciated.
9. Independent working: Giving responsibility to the child for small tasks and allowing him/her to perform independently. However, guidance should be provided where required.
10. Avoid undue pressure: Every child has a specific personality with certain preferences and choices. They should be taken care of and respected. No child should be asked to perform beyond threshold limits whether in studies, sports or extracurricular activities.

We hope the NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 8 Human Health and Diseases, drop a comment below and we will get back to you at the earliest.

Class 12 Chemistry NCERT Solutions Chapter 14 provides a detailed insight into the concepts related to the chapter “Biomolecules”. Chemistry is an important subject and the students need to be thorough with its concepts if they are preparing for boards or NEET and JEE.

NCERT Solutions also contains solutions to the questions provided in the textbook. The students can use these solutions to answer in the exam and score well.

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules

Biomolecules is a very important chapter and requires detailed understanding of the concepts. This topic is often asked in the examination and the students need to be thorough with the concepts.

The students can learn the structure, properties and classification of various biomolecules such as carbohydrates, nucleic acids, etc. These concepts are also taught in further studies. Therefore, the students need to be thorough with the basics.

NCERT INTEXT QUESTIONS

Question 1.
Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer:
Both glucose (C₆H₁₂O₆) and sucrose (C₁₂H₂₂O₁₁) are organic compounds and are expected to be insoluble in water. But quite surprisingly, they readily dissolve in water. This is due to the presence of a number of OH groups (five in the case of glucose and eight in sucrose) which are of polar nature. These are involved in the intermolecular hydrogen bonding with the molecules of H₂O (water). As a result, both of them readily dissolve in water.
Benzene (C₆H₆) and cyclohexane (C₆H₁₂) are hydrocarbons which don’t have any polar group. They, therefore, don’t dissolve in water since there is hardly any scope of hydrogen bonding in their molecules with those of H₂O (water).

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
The hydrolysis of lactose (disaccharide) can be done either with dilute HC1 or with enzyme emulsin. D-glucose and D-galactose are the products of hydrolysis. Both of them are monosaccharides with the molecular formula C6Hi₂0₆.

Question 3.
How do you explain the absence of an aldehydic group in the pentaacetate of D-glucose?
Answer:
Glucose, as we know is an aldohexose and it is expected to give the characteristic reactions of the aldehydic group e.g., action with NH₂OH, HCN, Tollen’s reagent, Fehling reagent etc. However, the pentadactyl glucose formed by the acylation of glucose with acetic anhydride does not give these reactions.

This means that the aldehydic group is either absent or is not available in the penta acetyl glucose for chemical reactions. In fact, the aldehydic group is a part of the hemiacetal structure which the penta acetyl derivative has. It is, therefore, not free or available to take part in these reactions.

Question 4.
The melting points and solubility of amino acids in water are generally higher than those of corresponding haloacids. Explain.
Answer:
The amino acids exist as zwitter ions, H₃N⁺ — CHR-COO-. Due to this dipolar salt like character, they have strong dipole-dipole attractions. Therefore, their melting points are higher than corresponding haloacids which do not have salt-like character.
Due to their salt-like character, amino acids interact strongly with water. As a result, their solubility in water is higher than corresponding haloacids which do not have a salt-like character.

Question 5.
Where does the water in the egg go after boiling the egg?
Answer:
Upon boiling the egg, denaturation of globular protein present in it occurs. Water present probably gets either absorbed or adsorbed during denaturation and disappears.

Question 6.
Explain why vitamin C can not be stored in the body.
Answer:
Vitamin C is mainly ascorbic acid which is water-soluble. It is readily excreted through urine and cannot be stored in the body.

Question 7.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer:
The products obtained are 2-deoxy-D-ribose, phosphoric acid, and thymine.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained? What does this fact suggest about the structure of RNA?
Answer:
As we know, a molecule of DNA has a double-strand structure, and the four complementary bases pair each other. Cytosine (C) always pairs up with guanine (G) while thymine (T) is paired up with adenine (A). Because of the presence of the double-strand structure, when a molecule of DNA is hydrolysed, in each pair the molar ratio of the bases remains the same. However, this is not seen when RNA is subjected to hydrolysis. This suggests that RNA has not a double-strand structure like DNA. It exists as a single strand.

NCERT Exercises

Question 1.
What are monosaccharides?
Answer:
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 (-CHO) group are called aldose and those which contain a keto (C = 0) group are called ketose. They are further classified as triose, tetrose, pentose etc. according to the no. of carbon atoms present (3, 4, 5 respectively).

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red precipitate of Cu₂0 or Tollen’s reagent to metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars. Thus, D – (+) – glucose, D-(-)-fructose, D – (+) – maltose and D – (+) – lactose are reducing sugars.

Question 3.
Write two major functions of carbohydrates in plants. (C.B.S.E. Delhi 2008)
Answer:

1. Structural material for plant cell walls: The polysaccharides cellulose acts as the chief structural material of the plant’s cell walls.
2. Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and acts as the reserve food material for the tiny plant till its capable of making food on its own by photosynthesis.

Question 4.
Classify the following into monosaccharides and disaccharides:
Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose.
Disaccharides: Maltose, lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
Glycosidic linkage is used to link different monosaccharides in disaccharides and polysaccharides through an oxygen atoms. For example, the glycosidic linkage is present in sucrose, lactose, maltose, etc. These are all disaccharides.

Question 6.
What is glycogen? How is it different from starch?
Answer:

1. The carbohydrates are stored in the animal body as glycogen. It is present in the liver, muscles, and brain. Enzymes break the glycogen down to glucose when the body needs glucose.
2. Glycogen is more highly branched than amylopectin (starch) glycogen chain consists of 10-14 glucose units, whereas amylopectin (starch) glycogen chain consists of 20-25 glucose units.

Question 7.
What are the hydrolysis products of starch and lactose?
Answer:
Starch upon hydrolysis gives α – D(+) glucose which is the constituent of both amylose and amylopectin. Lactose upon hydrolysis gives galactose and glucose.
Upon hydrolysis, cellulose gives only D(+) glucose. This means that only D(+) glucose units are present in cellulose but unlike starch these are -D(+) glucose molecules and not a – D(+) glucose molecules. The X – ray analysis has shown that there are large linear chains of 3 – D( +) glucose molecules lying side by side in the form of bundles held together by hydrogen bonding in the neighbouring hydroxyl groups. 

Question 8.
What is the basic structural difference between starch and cellulose?
Answer:
Starch is not a single component. It consists of amylose and amylopectin. In contrast, cellulose is a single compound. Amylose is a linear polymer of α – D glucose while cellulose is a linear polymer of β -D glucose. In amylose, C1 – C4 α- glycosidic linkage is present, whereas in cellulose C1 – C4 β- glycosidic linkage is present. Amylopectin has a highly branched structure.

Question 9.
What happens when D-glucose is treated with
(i) HI
(ii) Bromine water
(iii) HNO₃?
Answer:
(i) Reaction with HI

(ii) Reaction with bromine water

(iii) Reaction with HNO₃

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open-chain structure.
Answer:
The following reactions of D- glucose cannot be explained by its open-chain structure.
1. D – the glucose does not undergo certain characteristic reactions of aldehydes. For example, glucose does not form NaHSO₃ addition product, aldehyde-ammonia adduct, 2, 4, DNP derivative and does not respond to Schiff’s reagent test.

2. Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehyde group is absent in glucose pentaacetate.

3. D (+) – Glucose exists in two stereoisomeric forms ie. α – glucose and β- glucose, α – D (+) – glucose is obtained when a concentrated aqueous or alcoholic solution is crystallised at 303K. It has a melting point of 419K and has a specific rotation of +111° in a freshly prepared aqueous solution. However when glucose is crystallised from the water above 371 K β – D (+) glucose is obtained.

4. Both α – D glucose and β – D glucose undergoes mutarotation in an aqueous solution.

Question 11.
What are essential and non-essential amino acids?
Answer:
α – Amino acids which are needed for the health and growth of human beings but are not synthesised 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 the health and growth of human beings and are synthesised by the human body are called non-essential amino acids. For example glycine, alanine, aspartic acid etc.

Question 12.
Define the following as related to proteins
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide Linkage: Proteins are the polymers of a-amino acids which are connected to each other by peptide bond or peptide linkage. Chemically, peptide linkage is an amide formed between -COOH group and -NH2 group. The reaction between two molecules of similar or different amino acids, proceeds through the combination of the amino group of one molecule with the carboxyl group of the other This results in the elimination of a water molecule and formation of a peptide bond -CO-NH-. The product of the reaction is called a dipeptide because it is made up of two amino acids. For example, when carboxyl group of glycine combines with the amino group of alanine we get a dipeptide, glycylalanine.

(ii) Primary Structure: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein. Any change in this primary structure i.e.; the sequence of amine acids creates a different protein.

(iii) Denaturation: Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to physical change like change in temperature or chemical change in pH, the hydrogen bonds are disturbed. Due to this; globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. During denaturation 2° and 3° structures are destroyed but 1° structure remains intact. The coagulation of egg white on boiling is a common example of denaturation. Another example is the curdling of milk which is caused due to the formation of lactic acid by the bacteria present in milk.

Question 13.
What are the common types of secondary structures of proteins?
Answer:
Secondary structure of a protein refers to the shape in which a long polypeptide chain can exist. These are found to exist in two types :

• α-helix structure
• 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 polypeptide 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 a 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 is 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 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.

(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 a 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.

Question 14.
What type of bonding helps in stabilising the α-helix structure of proteins?
Answer:
The α-helix structure of proteins is stabilized by intramolecular H-bonding between C = O of one amino acid residue and the N – H of the fourth amino acid residue in the chain. This causes the polypeptide chain to coil up into a spiral structure called the right-handed α- helix structure.

Question 15.
Differentiate between globular proteins and fibrous proteins. (Jharkhand Board 2014; C.B.S.E. Delhi 2015)
Answer:

Question 16.
How do you explain the amphoteric behaviour of amino acids?
Answer:
Amino acids have basic (NH₂) and acidic (COOH) groups. These are, therefore, amphoteric in nature. However, they exhibit these characters in the amino acid molecule itself i.e., NH₂ group (basic) accepts a proton from COOH group (acidic) in the same molecule. Therefore, a-amino acid exists as a dipolar ion.

Question 17.
What are enzymes?
Answer:
We have learned in the study of carbohydrates that these are body fuels i.e., they provide the necessary energy to the body and
keeps it working. Actually, the human body is just like a furnace in which chemical reactions take place and are responsible for the digestion of food, absorption of appropriate molecules, and production of energy. The entire process involves a series of reactions that are catalyzed by biocatalysts known as enzymes. Thus, enzymes may be defined as:
biological or biocatalysts which catalyse the reactions in living beings.
All enzymes are basically globular proteins. Enzymes are very specific for a particular reaction as well as for a particular
substrate. These are generally named after the compounds or the class of substances with which they are linked or work.

Question 18.
What is the effect of denaturation on the structure of proteins?
Answer:
As a result of denaturation, globules get unfolded and helixes get uncoiled. Secondary and tertiary structures of the protein are destroyed, but the primary structures remain unaltered. It can be said that during denaturation, secondary and tertiary – structured proteins get converted into primary – structured proteins. Also, as the secondary and tertiary structures of a protein are destroyed, the enzyme loses its activity.

Question 19.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood? (C.B.S.E. Outside Delhi 2015)
Answer:
On the basis of their solubility in water or fat, vitamins are classified into two groups:
1. Fat-soluble vitamins:
Vitamins that are soluble in fat and oils, but not in the water, belong to their group.
For example Vitamins A, D, E, and K.

2. Water-soluble vitamins:
Vitamins that are soluble in water belong to their group.
For example, B group vitamins (B₁, B₂, B₆, B₁₂, etc) and vitamin C.
However, biotin or vitamin H is neither soluble in water nor in fat.
Vitamin K is responsible for the coagulation of blood.

Question 20.
Why are vitamin A and vitamin C essential to us? Mention their sources.
Answer:
vitamin A: Soluble in water but insoluble in oils and fas. Destroyed by cooking or prolonged exposure to air. it increases the resistance of the body towards diseases. Maintains healthy skin and helps in the healing of cuts and abrasions. It is available in

vitamin C:  Soluble in oils and fats but insoluble in water, stable to heat. Promotes growth and improves vision. ¡t also increases resistance to disease. It is available in Citrus fruits (oranges, lemon, grapefruit, lime, etc.), amia, cabbage. guava etc.

Question 21.
What are nucleic acids? Mention their two important functions.
Answer:
Nucleic acids are biologically important polymers which are present in all living cells.

• Nucleic acids play a vital role in the transmission of heredity characteristics.
• Nucleic acids help in the biosynthesis of proteins.

Two important biological functions of nucleic acids are (i) Replication and (ii) protein synthesis.
These are briefly discussed.
Replication: Replication may be defined as the process by which a single DNA molecule produces two identical copies of itself.

Replication is an enzyme-catalyzed process.  The process of replication starts with the partial unwinding of the two strands of the DNA double helix through the breaking of the hydrogen bonds between pairs of bases. Each strand then acts as the template (or pattern) for the synthesis of two new strands of DNA in the celllix environment. The specificity of base-pairing ensures that each new strand is complementary to its old template strand. As a result, two identical copies of DNA from the original DNA are produced. Each of these two copies is then passed on to the two new cells resulting from cell division. In this way, hereditary characters are transmitted from one cell to another.

Question 22.
What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in the nucleotide, a phosphoric acid component is also present.
Arrangement of constituents in Nucleic Acids. These are, intact, three building blocks in nucleic acid. A combination ol
base and sugar are known as a nucleoside. Similarly, base, sugar, and phosphates from nucleotides while nucleic acids are
polynucleotides which means that these are the polymers of nucleotides.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms a hydrogen bond with guanine, while adenine forms a hydrogen bond with thymine. As a result, the two strands are complementary to each other.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:

Question 25.
What are the different types of RNA found in the cell?
Answer:

• Messenger RNA (m – RNA)
• Ribosomal RNA (r – RNA)
• Transfer RNA (t – RNA)

We hope the NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules, drop a comment below and we will get back to you at the earliest.

Class 12 NCERT Solutions for Chemistry Chapter 13 contains solved answers for the questions provided in the textbook. These answers are provided by the subject experts and are accurate to the best of our knowledge. The students can use these answers along with the diagrammatic representation in the examination and score well.

NCERT Solutions are beneficial for the students appearing for UP board, MP Board, CBSE, Gujarat board, etc., and also for competitive exams such as NEET and JEE.

NCERT Solutions for Class 12 Chemistry Chapter 13 Amines

This chapter explains the importance, structure, physical and chemical properties, and the methods of preparation of amines. This chapter also explains the diazonium salts and the methods of their preparation. This will help you to understand the importance of these salts in the synthesis of aromatic compounds.

The students can use NCERT Solutions for better preparations of the concepts mentioned in this chapter. Basic understanding is very important for advanced concepts.

NCERT INTEXT QUESTIONS

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

Answer:
(i) primary
(ii) tertiary
(iii) primary
(iv) secondary.

Question 3.
(i)Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(ii) Write 1UPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:

Chain isomers: (a) and (c); (a) and (d); (b) and (c); (b) and (d)
Metamers: (e) and (g); (f) and (g)
Functional isomers: All 10 amines are functional isomers of 2° and 3° amines and vice-versa.

Question 3.
How will you convert:
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline
(iii) C1(CH₂)₆Cl into hexane-1, 6-diamine
Answer:
(i) Benzene into aniline:

(ii) Benzene into N, N-dimethylaniline:

(iii) Cl(CH₂)₆Cl into hexane-1, 6-diamine:

Question 4.
Arrange the following in increasing order of their basic strength:
(i) C₂H₅NH₂,C₆H₅NH₂,NH₃,C₆H₅CH₂NH₂ and(C₂H₅)₂ NH
(ii) C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N,C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ <  C₂H₅NH₂<(C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N <(C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂<(CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products
(i) CH₃CH₂CH₂NH₂ + HC1 →
(ii) (C₂H₅)₃N + HC1  →
Answer:

Question 6.
Write the reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.
Answer:
Aniline is a primary amine. It will react with excess methyl iodide to form quaternary ammonium salt as the final product. The reaction is known as Hoffmann’s ammonolysis.

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:
Aniline will undergo benzoylation to form a benzoyl derivative. The reaction will take place in the presence of aqueous alkali.

Question 8.
Write the structures of the different isomers corresponding to the molecular formula C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:
Four isomeric aliphatic amines are represented by the molecular formula C₃H₉N. These are:

Only the primary amines will evolve N₂ gas on reacting with nitrous acid (HONO) and form corresponding primary alcohols.

Question 9.
How will you convert:
(i) 3-Methylaniline into 3-nitrotoluene
(ii) Aniline into 1,3,5-tribromobenzene ?
Answer:
(i) 3-Methylaniline into 3-nitrotoluene

(ii) Aniline into 1,3,5-tribromobenzene

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH2)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCN₃
(vii) m-BrC₆H₄NH₂
Answer:

Question 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. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Methylamine on reaction with nitrous acid evolves N₂ gas with brisk effervescence while dimethylamine does not. Methylamine also gives carbylamine reaction upon warming with chloroform and alcoholic KOH while dimethylamine does not.
(ii) Secondary amines, both aliphatic and aromatic respond to Libermann’s nitroso reaction while tertiary amines do not.
(iii) Aniline responds to diazotisation and coupling reactions to form a dye while ethylamine does not.
(iv) Aniline gives diazotisation coupling reaction while benzylamine does not.
(v)  Aniline gives carbyl amine test with an extremely unpleasant smell while N-Methyl aniline does not.

Question 3.
Account for the following :
(i) pK_b of aniline is more than that of methylamine. (C.B.S.E. Delhi 2008, 2011)
(ii) Ethylamine is soluble in water whereas aniline is not. (C.B.S.E. Delhi 2008, 2011)
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (C.B.S.E. Delhi 2008)
(iv) Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (C.B.S.E. Sample Paper 2010)
(v) Aniline does not undergo Friedel Crafts reaction. (C.B.S.E. Delhi 2008, Sample Paper 2010, C.B.S.E. Outside Delhi 2015)
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
(i) pK_b of aniline is more than that of methylamine because aniline is less basic. In aniline, the electron pair on the nitrogen atom is involved in conjugation with the ring and is less available for protonation than in methylamine. Therefore, aniline has more pK_b.

(ii) Ethylamine is water-soluble due to hydrogen bonding. However, in aniline, the phenyl (C₆H₅) group is bulky in size and has -I effect. As a result, its hydrogen bonding with water is negligible and is therefore not soluble or miscible with water.

(iii) Methylamine forms a soluble hydroxide on reacting with water. The OH^– ions released by the hydroxide combine with  Fe³⁺ ions of ferric chloride to give ferric hydroxide or hydrated ferric oxide which is brown in colour.

(iv) Amino group (-NH₂) is an electron releasing or activating group when present on the benzene ring. It activates the ortho and para positions in the ring towards electrophilic substitution due to its +M or +R effect. The nitration of aniline carried by nitrating mixture (cone. HN0₃ + cone. H₂SO₄) is electrophilic in nature. The expected product of nitration is a mixture of ortho and para nitroaniline. However, in this case, a substantial amount of metanitroaniline is also formed. In fact, aniline being a base gets protonated in the acidic medium to form anilinium cation which is no longer activating. Rather, it is deactivating in nature and deactivates the ring. The substitution takes place at the meta position.

Thus, the nitration of aniline as such gives a significant amount of m-nitroaniline (47%). In addition to this, p-nitroaniline
is the major constituent (51%) while ortho isomer is in negligible amount (2%) mainly due to the reason that the ortho position
is sterically hindered because of the —NH2 group.

In order to check the activation of the ring by an amino group, the nitration of aniline is carried out indirectly by first
acetylating with acetic anhydride (or acetyl chloride) to form acetanilide. The compound formed is nitrated by the nitrating
mixture and the isomeric nitro derivatives are then hydrolysed in the acidic medium as discussed under halogenation.

(v) Aniline does not undergo Friedel Crafts reaction. Actually, aniline being a Lewis base forms a complex with AICI3 which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution i.e., alkylation or acylation which leads to Friedel Crafts reaction. Therefore, the reaction is not possible. The same problem arises in phenols as well.

(vi) The diazonium salts of aromatic amines are more stable than those of aliphatic amines because these are resonance stabilised while no such resonance stabilisation is possible in the corresponding diazonium salts of aliphatic amines.

(vii) Gabriel phthalimide synthesis is generally preferred over other methods for the synthesis of primary aliphatic amines. Potassium phthalimide formed by reacting phthalimide with alcoholic KOH reacts with an alkyl halide such as C₂H₅-I to form N-alkyl derivative which undergoes hydrolysis to form the primary amine. However, no reaction is possible with aryl halide such as C₆H₅-I. Therefore, primary aromatic amines are not formed in the reaction.

Question 4.
Arrange the following:
(a) In decreasing order of the pK_b values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

(b) In decreasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂ (C.B.S.E. Delhi 2011, Haryana Board 2013, C.B.S.E. Outside Delhi 2015)

(c) Increasing order of basic strength
Aniline, p-nitroaniline and p-toluidine

(d) Decreasing order of basic strength in gas phase
C₂H₅NH₂, (C₂H_s)₂NH, (C₂H5)₃N and NH₃

(e) Increasing order of boiling point
C₂H₅OH, (CH₃)₂NH, C₂H_sNH₂

(f) Increasing order of solubility in water
C6H₅NH₂, (C₂H_s)₂NH, C₂H5NH₂.
Answer:
From K_b  and PK_b values of some Amines:

(a) The decreasing order of pK_b values or increasing order of basic strength is:
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

(b) The decreasing order of basic strength is:
(C₂H₅)₂NH > CH₃NH₂ > C₆H₅N(CH₃)₂ > QH5NH2

(c) The increasing order of basic strength is:
p-nitroaniline < Aniline < p-Toluidine

(d) The decreasing order of basic strength in gaseous phase.
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

(e) Increasing order of boiling point is:
(C₂H₃)₂NH < C₂H₅NH₂ < C₂H₅OH

(f) Increasing order of solubility in water is :
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 5.
How will you convert
(i) Ethanoic acid to methanamfrie,
(ii)Hexanenitrile to pentan-l-amine,
(iii)Methanol to ethanoic acid,
(iv) thatfaminft to methanamine,
(v)Ethanoic acid to propanoic acid,
(vi)Methanamine to ethanamine,
(vii)Nitromethane into dimethylamine,
(viii)Propanoic acid into ethanoic acid ?
Answer:
(i) Ethanoic acid to methanamfrie

(ii) Hexanenitrile to pentan-l-amine

(iii) Methanol to ethanoic acid

(iv) thatfaminft to methanamine

(v) Ethanoic acid to propanoic acid

(vi) Methanamine to ethanamine

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid

Question 6.
Describe a method for the identification of primary, secondary, and tertiary amines. Also, write chemical equations of the reactions involved.
Answer:
The three type of amines can be distinguished by Hinsberg test. In this test, the amine is shaken with benzene sulphonyl chloride (C₆H₅SO₂Cl) in the presence of excess of aqueous NaOH or KOH. A primary amine reacts to give a clear solution, which on acidification yields an insoluble compound.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hoffmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi)Acetylation
(vii)Gabriel phthalimide synthesis. (C.B.S.E. Delhi 2011, C.B.S.E. Outside Delhi 2015)
Answer:
(i) Carbyl amine reaction:
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul-smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

(ii) Diazotisation:
Primary aromatic amines such as aniline react with nitrous acid under ice-cold conditions (273 – 278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium 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.

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.

(iii) Hoffmann’s bromamide reaction:
By Hoffmann degradation of Acid Amides. (Hoffmann Bromamide Reaction). When a primary acid amide is heated with an aqueous or ethanolic solution of sodium hydroxide and bromine, it gives a primary amine with one carbon
atom less.

The reaction is, therefore, regarded as a degradation reaction. For example.

(iv) Coupling reaction:
The reaction of diazonium salts with phenols and aromatic amines to form azo compounds having an extended conjugated system with both aromatic rings joined through the — N = N — bond, is called coupling reaction. In this reaction; the nitrogen atoms of the diazo group are retained in the product. The coupling with phenols takes place in a mildly alkaline medium while that with amines occurs under faintly acidic conditions. For example;

Coupling generally occurs at the p-position with respect to the hydroxyl or the amino group, if free, otherwise it takes place at the o-position.

(v) Ammonolysis:
The mechanism involves the nucleophilic attack of NH₃ molecule (through lone pair) on alkyl halide by an SN₂ mechanism. Amine salt is formed which reacts with ammonia to give primary amine and ammonium halide as follows:

The primary amine formed now acts as the nucleophile and reacts with another molecule of the alkyl halide to form secondary amine.

The reaction is repeated to form tertiary amine and quaternary ammonium salt as follows :

(vi) Acetylation:
Acylation of Amines Both aliphatic and aromatic amines form acyl derivatives (substituted acid amides) with reagents such as acid chlorides, esters, or acid anhydrides. The acylation is carried out in the presence of a base stronger than pyridine (e.g., NaOH) which can remove the acid formed in the reaction by neutralising it.
(a) Acylation of Aliphatic Amines: Both primary and secondary aliphatic amines from acyl derivatives as follows:

(b) Acylation of Aromatic Amines: Aromatic amines such as aniline can be acylated in the same manner with both acid
chloride and acid anhydride.

(vii) Gabriel’s phthalimide synthesis:
In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an N-alkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis

This synthesis is very useful for the preparation of pure aralkyl and aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.

Question 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-bromoaniline
(viii)Benzamide to toluene
(ix)Aniline to benzyl alcohol.
Answer:
(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-bromoaniline: 

(viii)Benzamide to toluene :

(ix)Aniline to benzyl alcohol:

Question 9.
Give the structures of A, B, and C in the following reactions:

Answer:

Question 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.
Answer:
Since the compound ‘C’ with molecular formula C₆H₇N is formed from compound ‘B’ on treatment with Br₂ KOH, therefore, compound ‘B’ must be an amide and ‘C’ must be an amine.
The only amine having the molecular formula C₆H₇N, i. e., C₆H₅NH₂ is aniline.
Since ‘C’ is aniline, therefore, the die amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide. Since compound ‘B’ is formed from compound ‘A’ with aqueous ammonia and heating, therefore, compound ‘A’ must be benzoic acid.

Question 11.
Complete the following reactions:

Answer:

Question 12.
Why cannot aromatic amines be prepared by Gabriel’s phthalimide reaction? (C.B.S.E. Sample Question Paper 2012, H.P. Board2017)
Answer:
In Gabriel phthalimide reaction, the potassium salt of phthalimide is formed. It readily reacts with an alkyl halide to form the corresponding alkyl derivative.

But it is not in a position to react with the aryl halide in case primary aromatic amine is to be prepared. Actually, the cleavage of C – X bond in haloarene or aryl halide is quite difficult due to partial double bond character. Therefore, aromatic amines cannot be prepared by this method.

Question 13.
How do aromatic and aliphatic primary amines react with nitrous acid?
Answer:
Reaction with nitrous acid. All three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form a 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 ¡s used as a rest for primary aliphatic amines as no other amine evolves nitrogen with nitrous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273—278 K) to form benzene diazonium salt. The reaction is known as diazotisation reaction.

In case, the temperature is allowed to rise above 278 K, benzene diazonium 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.

Question 14.
Give plausible explanation for each of the following :
(i) Why are amines less acidic than alcohols of comparable molecular masses ?
(ii) Why are primary amines higher boiling than tertiary amines ?
(iii) Why are aliphatic amines stronger bases than aromatic amines ? (H.P. Board 2008)
Answer:
(i) The acidic character in the both in cases is due to the release of H⁺ ion. Now, the anion in case of amine

has a negative charge on the nitrogen atom while the anion formed in case of alcohol has negative charge on the oxygen atom. Since oxygen is more electronegative than nitrogen atom, the negative charge can be accomodated easily on oxygen than on nitrogen in these anions. In other words, RO” ion is more stable than RNH“ ion. Consequently, alcohol is a stronger acid than amine. Please remember that even alcohols are very weakly acidic so much so that they donot turn blue litmus red.

(ii) Primary amines are higher boiling than tertiary amines due to the presence of intermolecular hydrogen bonding in their molecules. Since tertiary amines (R₃N) have no hydrogen atom present, these are not involved in any such hydrogen bonding. For example, the boiling point of n-butylamine (CH₃CH₂CH₂CH₂NH₂) is 322 K while that of trimethylamine (CH₃)₃ N is 276 K.

(iii) In the aromatic amines, (lie secondary and tertiary amines are more basic than aniline.
Actually, the basic strength or the electron releasing tendency of an amine depends upon the following factors.

1. The ability of the nitrogen atom to donate a pair of electrons.
2.  The stability of cation by accepting the pair of electrons.

Any factor which tends lo increase the electron releasing tendency of amine or increase the stability of the cation, will tend to increase the basic strength of amine.

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