Prasanna

These MP Board Class 11th Biology Notes for Chapter 10 Cell Cycle and Cell Division help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 10 Cell Cycle and Cell Division

→ Multicellular organism grows and develop by mitosis cell division, whereas some unicellular organisms reproduce by this division.

→ Body of some lower category of organisms are made up of haploid cells. Zygote of these organ-isms divides by meiosis division to form haploid individual.

→ Meiosis cell division helps in the maintenance of chromosome number in the species.

→ Spindles regulate movement of chromosomes during cell division.

→ Mitosis help to maintain genetic stability.

→ Nerve cells never divide.

→ Generally in normal condition cells complete cell cycle in 20-25 hours.

→ Duration of cell cycle depends upon type of cell and some external factors such as temperature, nutrition, availability of oxygen, etc.

→ Actual cell division occurs in M-phase of the cell cycle.

→ Duration of cell cycle in bacteria is 20 minutes, in epithelial cell is 8-10 hours and in root cells of onion is 20 hours.

→ Cell division is the basis of continuity of life.

→ The main difference between the cell division of plant cell and animal cell are in centriole mechanism and cytokinesis process.

These MP Board Class 11th Biology Notes for Chapter 9 Biomolecules help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 9 Biomolecules

→ Body of the living organisms is made up of many mineral elements.

→ Elements are found either in free form or in the form of compound in the living body.

→ Collection of various types of molecules in a cell is called as cellular pool.

→ Smaller molecules found in the cells are called as micromolecules and larger molecules found in the cells are called as macromolecules.

→ Sugar, amino acids, nucleotides, minerals and water are the important micromolecules found in the cells.

→ Carbohydrates are formed by the union of carbon, hydrogen and oxygen in the ratio of 1: 2 :1.

→ Fat and fat like substances are called as lipids.

→ Carbohydrates and fats provide energy to the body.

→ Some nucleotides and their derivatives act as coenzymes.

→ General formula of carbohydrates is C_x(H₂O)_y.

→ C, H and O forms 93% part of the body.

→ Approximately 22 types of amino acids are found in the cells of living organisms.

→ Proteins are formed by the polymerization of amino acids.

→ Nucleotides are the basic units of nucleic acids.

→ Minerals regulate metabolic activities of the body.

→ Fructose is the sweetest natural sugar.

→ Mammary glands synthesize lactose (milk sugar) from glucose and galactose.

→ Water soluble vitamins are vitamin C and vitamin B-complex.

→ Fat soluble vitamins are A, D, E and K.

→ Deficiency of iodine causes the enlargement of thyroid gland known as goitre. It can be pre-vented by taking iodized salt.

→ Deficiency of iron causes anaemia.

→ Proteins are the polymers of amino acids.

→ Ecdysone is a steroid hormone which promotes the transformation of larva to the pupal stage during insect metamorphosis.

→ Glycosidic bond is between alcoholic group of one and aldose / ketose group of another sugar.

→ Extracellular fluid : Most abundant is Na, followed by Cl, carbonate, Ca, K, P04 and Mg.

→ Intracellular pool: Most abundant is K followed by P04, Mg, C03, Na, Cl and Ca.

→ One molecule of cellulose has about 6,000 glucose residues.

→ Cellulose can be hydrolyzed into soluble sugars.

→ Artificial fibre, rayon is manufactured by dissolving cellulosic materials in alkali.

→ The vitreous humour of the eye and the synovial fluid also contains mucopolysaccharides (mucilages).

→ Polysaccharides are the high molecular weight polymers of monosaccharides.

→ In diabetic persons, excess amount of sugar is not converted into glycogen.

→ Cellulose is not digested in our body. Ruminants like cow and buffalo can digest it.

→ Chitin forms the exoskeleton in arthropods its periodical pealing off is known as moulting or ecdysis.

→ Nucleic acids are composed of four types of nucleotides.

→ DNA is known as the ‘master molecule of the body’.

→ DNA is the hereditary material known as gene. Genes are found in the nucleus located on chromosomes.

→ Proteins are the ‘building blocks’ of the living organism.

→ Tertiary and quaternary proteins are always globular in nature.

→ Antibody is a protein produced in our body which neutralizes antigen and protects our body.

→ In meat, percentage of proteins is higher than carbohydrate but in rice it is reverse.

→ Two types of pentose sugar are found in nucleic acids. Ribose (in RNA) and deoxyribose (in DNA). Deoxyribose has one oxygen atom less than ribose sugar.

→ Keratin is a fibrous, insoluble protein which is called as scleroprotein and found in ectoder¬mal cells, hair, nails and horns of animals.

→ DNA produces RNA by a process called transcription (heterocatalytic function).

→ Monellin is a protein which is the sweetest chemical.

→ All enzymes are proteins, but all proteins are not enzymes.

→ Enzymes are the biological catalysts.

→ Enzymes are manufactured in the living cells but they are non-living.

→ DNA of all cells act as carrier of message for synthesis of enzyme.

→ Enzyme substrate complex is short lived.

→ Enzymes are very specific in their mode of action.

→ Pepsin is secreted as proenzyme, in an inactive form known as pepsinogen.

→ Inactive pepsinogen is hydrolysed to active pepsin in the stomach by gastric hydrochloric acid or by the pre-existing pepsin.

→ Enzymes have two sites : One is called as active site and another is called as allosteric site. By active site enzymes combine with the substrate, whereas by allosteric site it combines with other substances such as activator or inhibitor.

→ Due to presence of three dimensional structure, enzymes are more active than catalysts.

→ High temperature causes unfolding of protein structure which is also called as denaturation.

→ Many vitamins form prosthetic group of important enzymes. Some of the such vitamins are thymine, nicotinic acid, riboflavin and pyridoxin of vitamin B-complex.

→ Enzymology : Study of enzymes, their actions and functions.

→ Allozymes : Similar enzymes formed by different genes.

→ Allosteric enzyme : Structure was studied by Monod et al (1965). The enzymes do not obey Michaelis-Menten or Km constant.

→ AIDS is tested by ELISA (Enzyme Linked Immunesor bent Assay).

These MP Board Class 11th Biology Notes for Chapter 8 Cell: The Unit of Life help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 8 Cell: The Unit of Life

→ The blue-green algae resembles bacteria in many details but most of them form multicellular colonies.

→ The smallest cells observed so far are of Pleuropneumonia like organism (PPLO) namely Mycoplasma gallisepticum.

→ Mature nerve cells never divide but may reach length of 90 cm.

→ In our body many red blood cells and outer cells of the epidermis of the skin are destroyed daily.

→ The Ostrich egg cell is considered as the largest cell, its diameter is about 175 mm.

→ Liver cells and muscle cells retain mitotic ability, but rarely divide.

→ Multicellular organisms are adapted for longer life span.

→ Tissue culture is a technique by which cells are cultured on an artificial culture medium. The technique was developed by White (1932).

→ Embryo developed outside the cell other than zygote is called embryoids.

→ Term ‘cell’ was used for the first time by Robert Hooke.

→ Rudolf Virchow states that cells are originated from pre-existing ones.

→ Viruses do not have cellular structure hence they are acellular bodies.

→ Centriole and ribosomes are non-membranous organelles.

→ Plant cell contains cell wall of cellulose which is absent in animal cells.

→ Prokaryotic cell: A cell which has an incipient nucleus and lacks membrane-bound organelles, is called as prokaryotic cell.

→ Eukaryotic cell: A cell which has a well defined nucleus and membrane-bound organelles, is called as eukaryotic cell.

→ Mesosomes : Mesosomes are extensions of plasma membrane into the cell in the form of vesicles, tubules and lamellae, usually in prokaryotic cells.

→ Lysozyme : It is an enzyme present in body secretions like tear, saliva and sweat; it lyses the cell wall of bacteria.

→ Transmembrane proteins : The membrane proteins which extend through the phospholipid bimolecular layer as a single helix are called transmembrane proteins.

→ Plasmodesmata : The cytoplasmic bridges in the primary cell wall between adjacent cells are called plasmodesmata.

→ Symplasm : The cytoplasm (living matter) present in the plasmodesmata and cells is called symplasm.

→ Apoplasm : The non-living substances present in the intercellular spaces (outside the cell) constitute apoplasm.

→ Pits : The unthickened areas in the secondary cell wall are called pits.

→ Tonoplast: The single semipermeable membrane boundary of vacuoles in plant cells is called tonoplast.

→ Dictyosomes : In plant cells, the Golgi apparatus is present in the form of unconnected units, called dictyosomes.

→ GERL (Golgi-Endoplasmic Reticulum-Lysosome): It is a region of golgi bodies, which is thought to be involved in the formation of lysosomes and the secretory granules.

→ Diffusion : It is the phenomenon of movement of ions, atoms or molecules of any substance (gas, liquid or solid) from a region of higher concentration to a region of lower concentration, until equilibrium is reached.

→ Osmosis : It is the phenomenon of movement of solvent/water molecules from a region of higher concentration or weaker solution to a region of lower concentration or stronger solution through a semi-permeable membrane, until equilibrium is reached.

→ Endocytosis : It is a process of active cellular intake of such materials (in bulk) that cannot pass through the unbroken plasma membrane.

→ Exocytosis : It is a process of active cellular extrusion/expulsion of such materials (in bulk) which cannot pass through the unbroken plasma membrane.

→ Pinocytosis : It is the process of active cellular intake of droplets of fluid along with submicroscopic particles.

→ Phagocytosis : It is the process of active cellular intake (ingestion) of solid particles.

→ Osmotic pressure: The hydrostatic pressure which balances and prevents any osmotic entry of water into the concerned solution, is called osmotic pressure.

→ Electrical gradient: The difference of electrical charges between two sides of a membrane is called as electrical gradient.

These MP Board Class 11th Biology Notes for Chapter 7 Structural Organisation in Animals help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 7 Structural Organisation in Animals

→ The term morphology refers to the study of external structure. Earthworm is an annelid that has segmented body and looks like a snake. It has about 100 to 120 segments with metameric segmentation.

→ The mouth is situated in the first segment. A single median female genital pore opens on the ventral side of the 14th segment. A pair of male genital pore is present on the ventral side of the 18th segment. A prominent band encircles 14th, 15th and 16th segments. This is known as clitellum.

→ Cockroach has the characteristic jointed legs and is nocturnal in its habit. The body is segmented externally and divisible into a number of segments under head, thorax and abdomen.

→ The head is somewhat pear shaped which ariculated with the thorax by flexible neck. It bears compound eyes, antennae, mouth parts. Thorax is three segmented bearing jointed appendages. There are ten segments in abdomen.

→ Alimentary canal is well developed and is divisible into foregut, midgut and hind gut, Malpighian tubules are present at the junction of fore and mid gut and help in excretion. Respiration occurs by trachea. The blood vascular system is of open type. Fertilization is internal.

→ Frog is an amphibian that lives in water or on land near water. Its skin is soft, smooth and moist. The male frog bears copulatory pads and well developed vocal sac on the ventral surface of body cavity.

→ Body is divisible into head and trunk. It can respire through skin in water and through lungs on land. Circulatory system is closed with single circulation. The male reproductive organ is a pair of testes. The female reproductive organ is a pair of ovaries.

→ Rat is a mammal and its body is divided into head, neck, trunk and tail. Integument is made up of epidermis, dermis and their derivatives. Closed and double circulatory system is present.

→ Heart is four chambered. Six to eight young ones are produced in a litter.

→ Glucose is the main nutrient in the blood.

→ The total volume of E.C.F. (extracellular fluid) in adult human being is about 15 litres i.e., about 45% of the total body water.

→ Spongy bone occurs in the deeper central parts of bones. It has no concentric organization like haversian system. It consists of a network of many fine irregular bony plates called trabeculae.

→ In whale thick layer of adipose tissue is called blubber.

→ Ligament is a strong band of elastin connecting the two bones at a joint and holds them in position preventing dislocation.

→ Tendon is a collagenous connective tissue which connects a muscle to a bone or cartilage.

→ The average lifespan of R.B.Cs. in human beings is 120 days.

→ Increase in the number of R.B.Cs. in our body is called as polycythemia.

→ The membrane which covers the muscle fibres is called sarcolemma.

→ Largest muscle of human body is buttock muscles (Gluteus maximus).

→ Smallest muscle of human body is stapedius. .

→ Riger mortis : The rigidity and non-elasticity of muscles after death. It is first seen in jaw muscles.

→ Myology : Study of muscles.

→ The process of the formation of blood corpuscles is known as haemopoiesis.

→ Decomposition of blood cells is called as haemolysis.

→ Collagen and elastin is the fibrous protein found in connective tissue.

→ The amount of haemoglobin is 15 gm per 100 ml of blood.

→ Abnormal decrease in the number of R.B.Cs. in the blood is called as anaemia.

→ Excessive stretching of ligaments is called sprain.

→ Haversian systems are absent in spongy bones of mammals.

→ Myoglobin protein is called as muscle haemoglobin.

→ Platelets occur in mammals only.

→ Muscular tissues are formed from the mesoderm layer of embryo.

→ Striated muscles contain acetylcholine which stimulates their contraction.

These MP Board Class 11th Biology Notes for Chapter 6 Anatomy of Flowering Plants help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 6 Anatomy of Flowering Plants

(A) Plant Tissue And Anatomy

→ A group of cells having a similar structure and function is called a tissue.

→ In plants two types of tissues are present: (i) Meristematic tissue and (ii) Permanent tissue.

→ The tissues that are capable of cell division are called meristematic tissue.

→ There are three groups of meristematic tissues : (i) Apical meristem, (ii) Intercalary meristem, (iii) Lateral meristem.

→ Apical meristem is found in the tips of shoot and roots, intercalary meristem occurs below the nodes and lateral meristem occurs on the lateral portion of the plant body.

→ Tissues which have no capacity to divide are called permanent tissues.

→ Callose is a polysaccharide which lines the pores of sieve plate.

→ The thickening of lignin and suberin on the inner tangential and radial walls of endodermis is called casparian strip.

→ The xylem having protoxylem towards the centre of the axis is called endarch.

→ The xylem having protoxylem towards periphery and metaxylem towards the centre is called
exarch.

→ The endodermal cells opposite the protoxylem groups that lack the thickening of lignin and suberin are called passage cells.

→ Tracheids are the elongated cells with tapering ends found in xylem of gymnosperms and angiosperms.

→ Vessel is a pipe like structure representing rows of cells, placed one above the other, in the xylem tissue.

→ Parenchyma and collenchyma are the living tissues of plant body.

→ The xylem of stem is endarch whereas it is exarch in roots.

→ Xylem is made up of tracheids, vessels, xylem fibres, and xylem parenchyma.

→ Xylem is the conductive tissue which plays a vital role in the conduction of water and minerals.

→ Phloem is made up of sieve tube, companion cell, phloem fibre and phloem parenchyma.

→ Phloem plays an important role in the conduction of organic food manufactured by leaves.

→ Xylem and phloem are the conductive tissues of plant which are collectively called as vascular bundle.

→ Simple tissues consist of single type of cells which are similar in origin, morphology and functions. Simple tissue may be (i) Parenchyma, (ii) Collenchyma, (iii) Sclerenchyma.

→ Parenchymatous cells act as storage tissue for food starch, fats, oil and proteins.

→ Collenchyma provides strength to the organs.

→ Sclerenchyma has thickened, usually lignified cell walls, which acts as supporting tissue.

→ Calyptrogen is a constituent of root apical meristem which occurs in monocots and is meant for producing the root cap.

→ The vascular bundle in which xylem and phloem are found in separate bundles present on different radii is called radial V.B. It is the characteristic of roots.

(B) Secondary Growth

→ Secondary growth is the increase in the girth of stems and roots.

→ Tissues formed from primary meristem are called as primary tissues, whereas tissues formed from secondary meristem are called as secondary tissues.

→ Increase in the girth of stems and roots due to secondary tissues is called as secondary growth.

→ Secondary growth is the characteristic feature of dicotyledonous stems and roots.

→ Except few cases, secondary growth is absent in monocotyledonous stems and roots.

→ Secondary growth takes place due to the activity of cambium and cork cambium.

→ Cambium present between xylem and phloem is called fascicular cambium.

→ Cambium present between two vascular bundles in ground tissue is called interfascicular cambium.

→ Fascicular and interfascicular cambium together constitute cambium ring.

→ Cambium ring cuts off secondary phloem on outer surface and secondary xylem on its inner surface.

→ During secondary growth, cells of the outermost layer of cortex become meristematic to produce cork cambium.

→ During secondary growth due to increase in the girth of stems and roots, the epidermis ruptures and cells become dead to produce bark.

→ Cork cambium cuts off cork cells on outer surface and secondary’ cortex on inner surface.

→ Cork cells and secondary cortex together constitute cork which is used for making plug of bottles.

→ Secondary xylem forms the maximum part of wood.

→ Lenticels are the pores on the bark of stem, composed of loosely arranged parenchymatous cells having intercellular spaces for gaseous exchange.

→ Secondary wood formed during a single year or one growth period is called annual ring.

→ The age of the plant can be calculated by the counting of annual ring.

→ The bladder like out growths of parenchyma, through the pores in the lateral walls of xylem vessel cause plugging of them and are called as tyloses.

→ Heartwood is the dark coloured wood in the centre and is dead. It contains oils, gums, resins, tannins, etc.

→ Heartwood is more durable and resistant to micro-organisms and insects. It is good for furniture.

→ Sapwood is the peripheral light coloured living wood responsible for conduction of materials.

→ Sapwood is less durable and resistant to micro-organisms and insects. Good for fuel.

→ Heartwood is a porous wood as it has vessels.

→ Early wood is formed during favourable seasons of spring and summer when cambium is active and produce large vessels or tracheids.

→ Late wood is produced during adverse conditions of climate (autumn and winter) in which tracheids and vessels are smaller.

→ Wood is the name of secondary xylem of the stems and main roots but the term wood can also be used for all types of xylem.

→ During leaf fall, cork cambium is formed in the leafbase which results in the formation of an abscission layer.

→ The cells formed on inner surface of the phellogen are called phelloderm or secondary cortex.

→ Dracena is a monocot plant, whose stem exhibit abnormal secondary growth.

These MP Board Class 11th Biology Notes for Chapter 4 Animal Kingdom help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 4 Animal Kingdom

→ All eukaryotic, multicellular, chlorophyll less, heterotrophic, cell wall less and holozoic organ¬isms are placed under animal kingdom.

→ Animals possessing the capacity of sensitivity and irritability.

→ Animals are the most advanced organisms of the living kingdom.

→ Porifera is the simplest and chordata is the advanced phylum of the kingdom animalia.

→ Animal body construction is characterized by : (i) Body plan, (ii) Symmetry, (iii) Body cavity, (iv) Body segmentation and appendages, (v) Body support and protection.

→ Invertebrates and chordates are differentiated on the basis of absence and presence of noto¬chord, respectively.

→ Animals may be diploblastic or triploblastic.

→ Sponges are primitive sessile pore bearing, mostly marine animals with cell aggregate body plan.

→ Animals of phylum coelenterata are diploblastic, radially symmetrical, bearing stinging cells (nematocysts) exhibit blind sac body plan.

→ Herdmania (Urochordate) show retrogressive metamorphosis, i.e., its larva is more developed than adult.

→ Whale and dolphins are aquatic animals. Their forelimbs are modified as flippers. Hind limbs are absent. .

→ Blue whale is the largest animal of the world.

→ Tusks of elephants are modified incisors.

→ Bats are able to detect objects in the dark with the help of echo location.

→ Roundworm (phylum nemathelminthes) are parasite in animal and plant or live in soil. These are triploblastic, pseudocoelomate and with a tube within tube body plan.

→ The nervous system of annelids comprises preoral ganglia jointed by circumoesophageal commissures to a double ventral ganglionated nerve cord.

→ Phylum arthropoda is the largest phylum of the animal kingdom. Their body is metamerically segmented.

→ The body of arthropods is differentiated into head, thorax and abdomen.

→ Prototheria the egg laying mammals like duck-billed platypus and echidna.

→ Eutheria are the true placental mammals.

→ The pouched mammals (marsupials) give birth to weak young one. These are kept in a marsu- pium present on the abdomen of the female.

→ The primates includes the prosimians such as lemurs, tarsiers and lorises and simians such as old world and new world monkeys.

→ Amphiblastula is the flagellated larva found in the porifera. At metamorphosis the flagellated cells move to the interior and become choanocytes.

→ Bipinnaria larva is the larval form found in asteroid echinoderms.

→ Choanocyte is the flagellated collar cell found as a lining of the internal cavities of the porifera.

→ Diploblastic : Animals having two layers of cells in the body wall.

→ Haemocoel is a cavity filled with blood which replace perivisceral coelom, specially in arthro-pods and molluscs.

→ Pelagic animals living in the surface waters of the sea.

→ Pericardium is a cavity in which heart is located.

→ Triploblastic animals having three layers of cells in the body wall.

→ Tube-feet is a tentacle like outpushings from the water vascular system which may be used for locomotion in the echinodermata.

→ Viviparous animals giving birth to living young.

→ About 75% animals of the animal kingdom belonging to phylum arthropoda.

→ Man {Homo sapiens) is the highly advanced organism of the animal kingdom.

→ Platyhelminthes are bilaterally symmetrical, dorsiventrally flattened acoelomate worm like animals.

→ The Mollusca are soft bodied, non-metameric, triploblastic coelomate animals consisting of anterior head, a ventral muscular foot and dorsal mass surrounded by a thin fleshy envelope.
The mantle generally sheltered in an external calcareous shell of their own secretion.

These MP Board Class 11th Biology Notes for Chapter 3 Plant Kingdom help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 3 Plant Kingdom

→ Annuals : Plants which live for one season and complete their life cycle in a single favourable season.

→ Algae : They are large assemblage of autotrophic plants which are predominantly aquatic.

→ Land plants are divided into two divisions, i.e., Bryophytes and tracheophytes.

→ Bryophytes are the amphibians of plant kingdom. In them, the gametophyte is the dominant phase and sporophyte is short-lived.

→ Tracheophytes include pteridophytes, gymnosperms and angiosperms.

→ Tracheophytes are characterized by the presence of vascular tissue.

→ Gymnosperms are woody plants with naked seed but without flowers.

→ Angiosperms are the flowering plants and their seeds are covered by the ovary wall.

→ The flower is a special modified reproductive shoot and is composed of four whorls namely calyx, corolla, androecium and gynoecium.

→ Bryophytes serve as a connecting link of aquatic algae and terrestrial pteridophytes.

→ Thallophyta have thallus like body and constitute algae and fungi.

→ Bryophytes may be thalose or foliose and found in moist and shady places.

→ In bryophytes, zygote develops into sporogonium which is further distinguished into food, seta and capsule.

→ On the basis of number of cotyledons, angiosperms are divided into two following major groups : (i) Monocotyledons : Bear only one cotyledon and (ii) Dicotyledons : Bear two cotyledons.

→ Heterospory : The production of spores of two different sizes and two different developmental patterns is called as heterospory. e.g., Selaginella.

→ Formation of fruit without fertilization is called parthenocarpy and the fruits are called parthenocarpic fruits.

→ Bicollateral vascular bundles : When both phloem and cambium occur twice once on outer side of xylem and then again on its inner side. The sequence is outer phloem, outer cambium, xylem, inner cambium and inner phloem.

→ Concentric vascular bundles : Vascular bundles in which xylem is covered by phloem or vice versa are called as concentric vascular bundles.

→ Radial vascular bundle : Vascular bundles in which xylem and phloem occur in separate patches on alternate radii as in roots are called as radial vascular bundle.

→ Polyembryony : Presence of more than two embryo in a seed is called as a polyembryony.

→ All eukaryotic, photosynthetic (autotrophs) multicellular organisms are included in kingdom plantae.

→ Plant kingdom is classified into three groups : (i) Algae, (ii) Bryophyta and (iii) Tracheophyta. Algae is divided into three subdivisions : (i) Rhodophyta, (ii) Phaeophyta and (iii) Chlorophyta.

→ Bryophyta and tracheophyta together are called as embryophyta.

→ Reproductive organs of embryophyta are multicellular.

→ Bryophytes are amphibian.

→ Higher vascular plants are evolved from bryophytes.

→ Flowering plants are called as angiosperms.

→ Flower of rafflesia is the largest flower.

→ Bryophytes serves as link between aquatic algae and terrestrial pteridophytes.

→ Main plant of bryophytes are gametophyte.

→ Main plant of pteridophytes are sporophyte.

→ Gymnosperms are naked seeded plants.

→ The flower is a special modified reproductive shoot and is composed of four whorls namely, calyx, corolla, androecium and gynoecium.

→ The ovary develops into fruit and the ovule develops into seed.

These MP Board Class 11th Biology Notes for Chapter 2 Biological Classification help students to get a brief overview of all the concepts.

MP Board Class 11th Biology Notes Chapter 2 Biological Classification

→ The bacteria constitute the most ancient, smallest, simplest organisms which are classified on the basis of structure and activities.

→ Bacteria often reproduce by binary fission.

→ Many eubacteria are disease producers in man, animals and plants. Some bacteria produce antibiotics.

→ Cyanobacteria are ancient photosynthetic prokaryotes. They liberate oxygen during photosynthesis.

→ Bacteria are photoautotrophic, chemoautotrophic or heterotrophic.

→ Archaebacteria are living fossils.

→ Plasmids are the extranuclear circular DNA found in bacteria.

→ In Moneran, 70S type ribosomes are found.

→ The cell wall of bacteria contain polysaccharide, lipids, proteins and mucopeptides.

→ Photosynthetic bacteria does not ‘split water’ to obtain reducing power. Instead of water they obtain H2 from H2S or other sulphur containing compounds.

→ Use of more concentration of salt or sugar in food preservation kills the bacteria by plasmolysis thus prevent the food from decaying.

→ Joker of plant kingdom : Mycoplasma is also known as Joker of plant kingdom.

→ Trichome : A filament without mucilage sheath is called trichome.

→ ‘Germ theory of disease’ is proposed by Louis Pasteur.

→ Louis Pasteur (1822-1895) is the ‘founder of bacteriology’ and regarded as ‘father of microbiology.

→ Endospore : A resting spore produced by few bacteria under unfavourable conditions.

→ Antibiotics : A chemical substance excreted by micro-organisms which inhibit the development and growth of other organisms.

→ Bacterial cell wall: It is made up of peptidoglycan.

→ Ehrich (1854-1915) is regarded as ‘father of chemotherapy’.

→ Fungi are heterotrophic thallophytes because of the absence of chlorophyll.

→ The body of fungi is made up of thin filaments. A single filament of fungi is called as hyphae.

→ The group of filaments forming a network is called as mycelium.

→ The cell wall of fungi is made up of chitin.

→ Glycogen is the stored food material of fungi.

→ Kingdom fungi is classified into five classes :

1. Class 1. Mastigomycotina or Eumycotina.
2. Class 2. Zygomycotina.
3. Class 3. Ascomycotina.
4. Class 4. Basidiomycotina.
5. Class 5. Deuteromycotina.

→ Asexual reproduction in fungi mostly takes place by spores or conidia, whereas the sexual reproduction may be isogamous or heterogamous. In lower forms, it takes place by conjuga¬tion.

→ The higher fungi are included under ascomycetes and basidiomycetes.

→ The most distinctive feature of ascomycetes is the formation of sac like ascus in which sexually produced ascospores develop. In some members, asci are borne in a fruiting body called the ascocarp.

→ An association of algae and fungi is called as Lichen.

→ Mushrooms are the fungi used as food.

→ Mycology is the branch of biology deals with study of fungi.

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons which are most likely to be asked in the exam.

MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons

Hydrocarbons Class 11 Important Questions Very Short Answer Type

Question 1.
Frans-alkene is formed as a result of reduction of alkyne in liquid ammonia by sodium. Will butene formed by such type reduction of 2-butyne show geometrical isomerism?
Answer:

Trans-2-butene formed by the reduction of 2-butyne shows geometrical isomerism.

Question 2.
In spite of the -I effect, halogen in haloarene compounds is o-(ortho) and p-(para) directive. Give reason.
Answer:
Halogen is a highly reactive group. Due to strong -I effect electron density decreases on benzene ring. But due to resonance, compared to o- and p-place, electron density is more at m-place due to which halogens are o- and ρ -directive.

Question 3.
Alkenes are more reactive than alkanes. Why?
Answer:
In alkanes there is only a-bond between C – C but in alkenes there C = C has one σ and one π-bond. Due to lateral overlapping, the π-bond is weaker than the σ-bond. Hence, alkenes are more reactive than alkanes. The bond energy of π-bond (25 kJ mol^-1) is lower than the bond energy of a -bond (347 kJ mol^-1). Due to this difference of bond energy alkenes are more reactive as compared to alkanes.

Question 4.
What is meant by asymmetric carbon atom?
Answer:
Asymmetric carbon atom: If all the four valencies of carbon atom in a regular tetrahedron compound are linked to four different groups, then the carbon atom of tetrahedron is referred as asymmetric carbon atom or chiral carbon.

Question 5.
Draw Newman’s projection figures of Staggered and Eclipsed conformations of Butane.
Answer:
structural formula of Butane :

Question 6.
What is meant by Chirality?
Answer:
Molecules which are non-superimposable on their mirror images are called chiral molecules and this property is known as chirality. Chiral molecules are optically active. Chirality in the molecule is due to asymmetrical carbon atom. Chiral molecules are optically active, but molecules with asymmetric carbon may be optically active or may not.

Question 7.
What are Alkanes? What type of bond is present in them?
Answer:
Alkanes are saturated hydrocarbons. They are also known as paraffins. They are less reactive and more stable. Their general formula is C_nH2_n+2.
In alkanes, single σ-bond is present between carbon-carbon and carbon-hydrogen bond. Each carbon in alkanes is sp³ hybridized.
Example : Methane CH₄, Ethane C₂H₆.

Question 8.
What are Alkenes? In which hybrid state is carbon present in them?
Answer:
Alkenes are unsaturated hydrocarbons. They are also known as olefins. They are more reactive than alkanes and less stable than them. They easily show addition reactions. Their general formula is C_nH2_n. Carbon-carbon double bond is present in alkenes and carbon is in sp2 hybridized state.
Example: Ethylene H₂C =CH₂, Propylene CH₃ -CH =CH₂.

Question 9.
What are Alkynes? What type of bond is present in them?
Answer:
Alkynes are unsaturated hydrocarbons. They are highly reactive and are very unstable. Their general formula is C_nH2_n-2.
Triple bond is present between carbon-carbon, of which one is σ and two are π bonds and both carbons are in sp hybridised state.
Example: Acetylene HC ≡ CH, Propyne CH3 – C ≡ CH.

Question 10.
Write the isomers of pentane.
Answer:
Pentane (C₅H₁₂) forms three isomers.

Question 11.
What is cis-trans isomerism ? Explain with example.
Answer:
cis-trans isomerism is also known as geometrical isomerism which is shown by such double bond compounds in which the double-bonded carbon atoms are linked to two different atoms or groups. When similar groups or hydrogen atoms are on the same side of double bond, the compound is cis isomer and if situated on opposite sides, then the compound
is trans isomer.
Example: CH₃CH =CH – CH₃ :

Question 12.
Why is benzene extraordinarily stable though it contains three double bonds?
Answer:
Due to Resonance and displacement of electrons, benzene molecule is extraordinarily stable.

In hybrid state, the dotted ring represents the 6 electrons displaced between 6 carbon atoms of benzene ring. Thus, due to delocalized 6 electrons benzene is stable.

Question 13.
How will you obtain nitrobenzene from acetylene?
Answer:

Question 14.
Electrophilic and Nucleophilic reactions occur between electron-deficient and electron-rich species. Thus, their tendency is to attack electron rich and electron-deficient nucleus. Classify the following species into electrophiles and nucleophiles :
(i) H3CO^-,
(ii)
(iii) Cl,
(iv) Cl₂C:,
(v) (CH₃)3C⁺,
(vi) Br^–,
(vii) CH₃OH,
(viii) R-NH-R.
Answer:
Electrophiles are electron deficient species. These are neutral or positively charged.
(iii) Cl, (iv) Cl₂C: and
(v) (CH₃)3C⁺ are electrophiles.
Nucleophiles are electron rich species. These are neutral or negatively charged species.
(i) CH3O^–,
(ii)
(vi) Br^–,
(vii) img and
(viii) RNHR are nucleophiles.

Question 15.
Cyclopropane is more reactive as compared to cyclohexane.
Answer:
The bond angle in cyclopropane C-C-C is 60° due to which the molecule is in more strain (less stable) and hence more reactive, whereas in cyclohexane, the bond angle is very close to 109°28′ due to which the molecule is in less strain and hence its reactivity decreases.

Question 16.
What is meant by functional isomerism? Explain with example.
Answer:
Functional isomerism: When two compounds have same molecular formula but have different functional groups are called functional isomers. For example, dimethyl ether and ethyl alcohol. Ether contains -0- group while alcohol contains -OH group. Molecular formula of both is C₂H₆O.

Question 17.
What is Resolution?
Answer:
Process of separation of racemic mixture into d and / forms or optically active forms, is called Resolution. Resolution can be done by biochemical or chemical methods.

Question 18.
Which type of isomerism is found in arenas? Explain with example.
Answer:
In Arenes, position isomerism is found.
Example: Linking of one or more methyl group to arenes gives rise to position isomerism. Xylene (C₈H₁₀) is a dimethyl disubstituted product of benzene.
(i) When two substituent groups are present 1,2 i.e., adjacent position, then it is called ortho (or o) isomers.
(ii) When the position is on 1 and 3 (alternate) then it is called meta (m).
(iii) When the position is on 1 and 4 (diagonally opposite to one another) then it is called para (p).

Question 19.
Write a reaction to show acidic nature of alkynes.
Answer:
Alkynes whose molecules possess a triple bond at the terminal position act as weak acid. Such alkynes release hydrogen when treated with reactive metals like Na, Ca and their derivatives are obtained which are known as acetylides.

Question 20.
What is Baeyer’s reagent? How is it used to test unsaturation?
Answer:
Alkaline KMnO₄ is known as Baeyer’s reagent. On reacting unsaturated hydrocarbons with Baeyer’s reagent, alkaline KMnO₄ decolourises. Therefore, it is used to test unsaturation.

Question 21.
Write action of Baeyer’s reagent on aikene. Give equation also.
Answer:
Alkenes react with Baeyer’s reagent (1 % basic KMnO₄) and form ethylene glycol. The colour of KMnO₄ disappears. This is also called as Baeyer test for unsaturation.

Question 22.
What do you understand by Peroxide effect? Explain with example.
Answer:
Peroxide effect: Addition of HBr to propene in presence of organic peroxides takes place against the Markownikoff’s rule and this effect of peroxide which changes the direction of reaction and proceeds against Markownikoff’s rule is called peroxide effect.

Di-tertiary butyl peroxide (CH₃)₃CO – OC(CH₃)₃ and benzoyl peroxide (C₆H₅COO)₂ are organic peroxides. It should be remembered that peroxide effect on alkenes takes place only during addition of HBr. Addition of HCl, Hl, H₂SO₄ and water are not affected by presence or absence of peroxide.

Question 23.
How can you differentiate between Ethane, Ethylene and Acetylene?
Answer:
Differences between Ethane, Ethylene and Acetylene

Question 24.
What is Berthelot synthesis?
Answer:
On passing a stream of hydrogen gas through an electric arc between two carbon electrodes, Acetylene is obtained.

Question 25.
What is Westron or Westrosol ?
Answer:
1. Compound formed by the addition of halogen on acetylene in the presence of CCl₄ is known as Westron.

2. When Westron is heated with alcoholic KOH, then due to dehydrohalogenation Westrosol is obtained.

Question 26.
What is Prototrophy?
Answer:
By the hydrolysis of Propyne, enol and keto forms are obtained. It represents tautomerism. This type of isomerism is shown by such compounds which contain at least one hydrogen which can transfer from one position to another. This is known as prototrophy.

Question 27.
What is conformational isomerism?
Answer:
Various spatial arrangements produced due to rotation of carbon-carbon double bond are called conformations and molecular orientation of conformers is known as conformational isomerism.

Question 28.
Explain polymerisation reaction with example.
Answer:
Polymerisation reactions: In polymerisation, many simple molecules of a substance combine together to form a huge molecule. The simple molecule is called monomer and the huge molecule as polymer or macromolecule. Rubber, nylon, bakelite, terylene and P.V.C. are examples of high polymers.
Polymerisation of alkenes takes place in presence of Lewis acid BF₃ or AlCl₃ or organic and inorganic peroxides. Organic peroxides used are benzoyl and acetyl peroxide.

Question 29.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Answer:
Due to the presence of delocalized 6 π-electrons, benzene behaves like an electron-rich source. Thus, it attracts electrophilic reagents (electron-deficient species) but repel nucleophilic reagents (electron-rich species). This is the reason that benzene undergoes electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty.

Question 30.
Alkenes exhibit Electrophilic addition reactions whereas arenes exhibit Electrophilic substitution reactions. Give reason.
Answer:
Alkenes are rich source of loosely bonded π-electrons due to which they exhibit electrophilic addition reactions. Huge energy change takes place in electrophilic addition reactions due to which they are more effective in energetic form rather than electrophilic substitution reactions.

In Arenes, during electrophilic addition reactions aromatic nature of benzene ring is destroyed whereas in electrophilic substitution reactions it remains constant. Electrophilic substitution reactions of arenes are energetically more effective than electrophilic addition reactions.

Question 31.
Write the necessary conditions for geometrical isomerism.
Answer:
The following conditions are required for geometrical isomerism:
1. At least one carbon-carbon double bond should be present.
2. The double-bonded carbon atoms should contain different atoms or groups linked to them.

Question 32.
Write name and structural formula of cyclic compounds which exhibit geometrical isomerism.
Answer:
Some cyclic compounds, due to restricted rotation of carbon-carbon single bond exhibit geometrical isomerism.
Example:

Question 33.
What is Saytzeff’s rule? Explain with example.
Answer:
The rule states that “During dehydrohalogenation of an alkyl halide, that alkene is formed in which the double-bonded carbon atom is more alkylated i. e., hydrogen is eliminated from that carbon atom which has lesser number of hydrogen.”

Question 34.
Write a note on octane number.
Answer:
The resistance of a sample of petrol to check knocking is measured in terms of octane number. For this purpose two substances viz iso-octane and n-heptane have selected and arbitrarily rated, n-heptane has no resistance to check knocking and has been rated as octane No. zero. Iso-octane has 100% resistance to check knocking and is rated as 100.

Octane number of a petrol is defined as percentage of iso-octane by volume in a mixture of iso-octane and n-heptane which gives the same knocking as oil under investiga¬tion operated in a standard engine and under standard conditions.

Question 35.
Compare paraffins, olefins and acetylene on the basis of the following points :
1. IUPAC name, 2. General formula, 3. Reactivity, 4. Reaction with Br2 water.
Answer:

Hydrocarbons Class 11 Important Questions Short Answer Type 

Question 1.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also, give reason for this behaviour.
Answer:
Hybridisation of carbon in the given compound is as follows :

Acidic character increases with the increase in 5-character of orbital. Thus, decreasing order of acidic behaviour of benzene, n-hexane and ethyne is as follows :
Ethyne > Benzene > n-hexane.

Question 2.
What effect does branching of alkane chain has on its boiling point?
Answer:
As a result of increase in branching in the chain, surface area decreases. Thus, due to increase in branching in the chain van der Waals’ force of attraction decreases due to which boiling point also decreases.

Question 3.
What are the necessary conditions for a system to be aromatic?
Answer:
The necessary conditions for a system to be aromatic are as follows :

• Molecule should be planar.
• Molecule should be cyclic with alternate single and double bonds, there should be complete delocalisation of n electrons in the ring.
• There should be (4n + 2) n electrons in the ring of the molecule where n- 0,1,2,3,……………………….. (Huckel’s rule)
  If the molecule does not satisfy anyone or more conditions, then it is non-aromatic.

Question 4.
What is Metamerism? Explain with example.
Answer:
Metamerism: Isomerism which exists among same class of compounds is known as metamerism. Alkyl groups attached to functional group are different.
Example :
(a) C₄H₁₀O :

(b) C₅H₁₀O:

Question 5.
What is Tautomerism? Explain with example.
Answer:
Tautomerism: It is a special type of functional isomerism in which same compound represents both of isomers which remain in equilibrium. It is due to migration of an H atom from one polyvalent atom to other. It involves change of bonds also.

Dyad system: It involves oscillation of hydrogen atom between only two polyva¬lent atoms e.g., Hydrocyanic acid:

Triad system: It involves the oscillation of hydrogen atom between three polyva¬lent atoms:

This is also known as Keto-Enol Tautomerism.

Question 6.
What is Newman’s Projection formula?
Answer:
If a single bond is present between two carbon atoms, then both the carbon atoms are represented by a circle. Hydrogen atoms linked to them are represented from the centre. If both the carbon atoms are one behind the other, then only front carbon atom is seen and its three hydrogens are joined from the centre and the hydrogens of the carbon atom behind are linked to the circumference.

Question 7.
State Markownikoff’s rule. Explain with example.
Answer:
Markownikoff’s rule: Markownikoffs rule is for addition of polar molecules to unsymmetrical alkenes.
According to this rule, in case of unsymmetrical olefines, the negative part of adding molecule goes to that carbon atom to which less number of hydrogen atoms are attached.
These reactions take place in presence of ether but in absence of organic peroxides. For example, when HBr reacts with propene, the main product is 2-bromopropane (isopropyl bromide). Small amount of 1 -bromopropane i.e., n-propyl bromide is formed.

Question 8.
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why?
Answer:
Hex-2-ene is CH₃ — CH₂ — CH₂ — CH = CH — CH₃ . Its cis and transform are as follows:

cis isomer is more polar than transform, therefore dipole moment of cis form is more than that of trans form. Thus, boiling point of cis form is higher than transform because dipole-dipole interaction is more, transform of hex-2-ene is non-polar.

Question 9.
An alkene, ‘A’ on ozonolysis gives a mixture of ethanal and pentane-3-one. Write structure and IUPAC name of ‘A’.
Answer:
The structures of the products formed by ozonolysis are written in such a way that their oxygen atoms point towards each other. After removing both the oxygen, both ends are joined by double bond. When structure of Alkane ‘A’ is obtained.

Question 10.
Explain Racemic mixture with example.
Answer:
Racemic mixture: When equal quantity of dextrorotatory and laevorotatory isomers are present in a sample, it will not show optical activity because both enantiomers rotate the plane-polarized light to the same extent in opposite direction and hence net rotation is zero. Such an optically inactive mixture is represented by dl or ±.
Example: Racemic lactic acid: It is an equimolar mixture of d and l forms. It has 50% d form and 50% l form. This form is optically inactive and it can be separated into d and / forms. It is represented by dl or ±.

Question 11.
Give substitution reaction of alkanes with example.
Answer:
The reaction in which any atom present in the compound (e.g., H or other atoms) is replaced by other atom or radical in such a way that there is no change in external structure is called substitution reaction. Examples of substitution reaction are halogenation, nitration, sulphonation, etc. e.g.,
Halogenation reaction: Mixture of mono, di, tri, tetra haloalkane is formed.

Question 12.
An alkene ‘A’ contains three C – C, eight C-H σ bonds and one C-C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44u. Write IUPAC name of ‘A’.
Answer:
Aldehyde with molar mass 44u is ethanal, CH₃CHO. Thus, form its two moles

Thus, in But-2-ene, there are three C – C, eight C – H σ-bonds and one C – C π-bond are present.

Question 13.
Propanal and pentane-3-one are the ozonolysis products of an alkene. What is the structural formula of the alkene?
Answer:

Question 14.
Suggest the name of a Lewis acid other than anhydrous AICl₃ which can be used during ethylation of benzene.
Answer:

In the ethylation of benzene, ethyl group gets linked with benzene ring. This is done by Friedel-Crafts ethylation reaction of benzene. Instead of anhydrous AlCl₃, anhydrous FeCl₃ or SnCl₄ can be used as catalyst (Lewis acid).

Question 15.
What is B.H.C. ? State its uses.
Answer:
On treating benzene with Cl₂ in the presence of sunlight B.H.C. is formed.

It is available in eight three-dimensional forms but only four isomeric forms α,β,γ and δ can be obtained in pure state. The various isomeric forms are obtained due to different arrangements of hydrogen and chlorine. γ isomeric form is more stable and a powerful insecticide. It is also known by the name of Lindane or Gammaxene or 666. It is used as an insecticide.

Question 16.
How will you convert the following? Give only chemical equations.
(i) Methane to Ethane,
(ii) Ethane to Methane
(iii) Acetylene to Benzene.
Answer:
(i) Methane to Ethane :

(ii) Ethane to Methane


(iii) Acetylene to Benzene.

Question 17.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking an example.
Answer:
For the preparation of alkane with odd number of carbon atoms, two different alkyl halides should be taken and two different alkyl halides will react in three ways as a result of which mixture of alkanes are obtained.
Example:

This is the reason that Wurtz reaction is not preferred for the preparation of Alkanes containing odd number of carbon atoms.

Question 18.
Out of benzene, m-dinitrobenzene and toluene which will undergo nitra-tion most easily and why ?
Answer:
Nitration is Electrophilic substitution reaction. Reactivity of benzene nucleus towards E+ decreases in the presence of electron attracting group whereas it increases in the presence of electron releasing group.
Electron attracting groups (-I) like (-NO₂) deactivate electrophilic substitution reaction.
Electron releasing groups (+I) like (-CH₃) activate benzene nucleus towards electrophilic reactions.
Thus, rate of nitration is as follows:

Thus, Toluene will undergo nitration easily.

Question 19.
How will you obtain (Give only chemical equations):
(i) B.H.C. from benzene
(ii) Acetophenone from benzene
(iii) P.V.C. from chloroethene
(iv) Teflon from tetrafluoroethene.
Answer:
(i) B.H.C. from benzene :

(ii) Acetophenone from benzene :

(iii) P.V.C. from chloroethene

(iv) Teflon from tetrafluoroethene :

Question 20.
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile E⁺ :
(a) Chlorobenzene, 2,4-dinitrochiorobenzene,p-nitrochlorobenzene.
(b) Toluene, p – H₃C – C₆H₄ – NO₂, p – O₂N – C₆H – NO₂.
Answer:
In the presence of electron releasing groups (or active group) in benzene nucleus, electron density increases. Thus, electrophilic reagents can easily attack on benzene nucleus.
Example : (-CH₃)
But electron attracting groups (like : NO₂) decrease the electron density on benzene nucleus. Thus, electrophilic reagents attack on benzene nucleus with difficulty.
Thus, decreasing order of relative reactivity of the above compounds towards electrophiles (E+) is as follows :
(a) Toluene > p -nitrochlorobenzene > 2,4-dinitrochlorobenzene.
(b) Toluene > p – H₃C – C₆H₄ – NO₂ > p – O₂N – C₆H – NO₂.

Question 21.
Write structures of all the alkenes which on hydrogenation give 2-methyl- butane.
Answer:
Structure of 2-methyl butane is as follows :

In this structure, double bond is introduced in various places and tetravalency of each carbon is satisfied by hydrogen atoms.

Question 22.
Explain geometrical isomerism with an example. Or, Geometrical isomerism is found in which type of compounds? Explain with example.
Answer:
Geometrical isomerism: This type of isomerism is exhibited by those compounds in which two carbon atoms are joined by a double bond. Both carbon atoms are linked by two different groups or atoms and their general formulae (a)(b)C=C(b)(a) can be represented by two forms. When identical groups are present on the same side then it is called cis type and when identical groups are present on the opposite sides then it is referred as trans type of isomerism.

Question 23.
What is Huckel’s Law?
Answer:
Huckel stated that “All planar cyclic compounds possess aromatic character, whose cycle contain (4n + 2) n electrons.”
Here n is a whole number whose value is 0,1,2,………………….
Thus, cyclic compounds which contain 6(n = 1), 10 (n = 2), 14 (n = 3) electrons possess aromatic character.
Aromatic compounds are Of two types :
1. Benzenoids: Compounds which contain one or more benzene rings are called benzenoids.

2. Non-benzenoids: Compounds which do not contain benzene ring.

Question 24.
Express Friedel-Crafts reaction with equation.
Answer:
Friedel-Crafts reaction: When benzene is treated with alkyl halide in presence of anhydrous AlCl₃, then alkyl benzene is formed.
Alkylation :

Acetylation: When benzene is treated with acetyl chloride in presence of anhydrous AlCl₃, acetophenone is formed. In this, H-atom of benzene is displaced by acyl group.

Question 25.
For the nomenclature of geometrical isomerism what are E and Z symbol?
Answer:
When double-bonded carbon atom has all four groups different, then geometrical isomers are represented by E and Z symbols. It is based on the following rules:
Rule I. All atoms or groups attached to doublebonded carbon atom is the same, the atom with higher atomic number gets highest priority and atom with lower atomic number gets less priority. If the first atom is same then second atom or third atom is considered.
Rule II. If the two higher priority groups are on opposite side of the double bond, the isomer is E and if the two higher priority groups are on the same side of the double bond, the isomer is called Z.
Example :

Question 26.
What are Dienes? How many types of dienes are there? Explain with example.
Answer:
Dienes are unsaturated hydrocarbons which contain double bond betweenm carbon-carbon atoms and based on the position of double bond, they are of three types:
1. Isolated dienes: Dienes in which one or more single bonds are present between two double bonds are isolated dienes.
CH₂=CH – CH₂ – CH=CH₂

2. Conjugated dienes: Dienes in which two double bonds are conjugated or in alter nate position are conjugated dienes.

3. Cumulated dienes: Dienes in which two double bonds are in close position.
CH₂=C= CH-CH₃
CH₃-CH= C = CH₂.

Question 27.
Write Diels-Alder reaction with example.
Answer:
When conjugated dienes are heated with an alkene or substituted alkene, six membered cyclic compound is formed, therefore this reaction is known as cyclic reaction, in it 4n electrons join to the system. This type of reaction is known as Diels-Alder reaction.

Question 28.
What is Lindlar’s catalyst? Write its uses.
Answer:
Mixture of Palladium supported on Barium sulphate or Calcium sulphate is called Lindlar’s catalyst. Here sulphur or quinoline acts as catalytic poison and reduce alkyne only up to alkene state.
Uses: Reduction of alkyne by hydrogen to alkene is done in presence of Lindlar’s catalyst. This catalyst does not allow alkene further to reduce to alkane.

Question 29.
Write four facts in favour of Kekule’s structure for benzene.
Answer:
Kekule’s structure for benzene: Kekule proposed a six-membered ring structure for benzene. In this ring, all the six carbon atoms occupy comers of a hexagon and hydrogen is linked with each carbon atom.
In order to meet the fourth valency of carbon, Kekule suggested presence of alternate single and double bonds.

Facts in favour of Kekule’s structure :
(i) Benzene reacts with three molecules of hydrogen to form cyclohexane which proves the presence of three double bonds.

(ii) Benzene gives addition reaction with 3 molecules of chlorine to form benzene hexachloride, hence presence of double bonds is confirmed.

(iii) With ozone, it forms a triozonide which on hydrolysis gives three molecules of glyoxal.

(iv) Three molecules of acetylene polymerize in a red hot tube to form benzene.

Question 30.
Give reasons :
1. Boiling point of branched compounds is less than straight chain compounds. Why?
2. Melting point of compounds with odd number carbon atoms is less than the compounds with even number carbon atoms.
Answer:
1. Surface area of Linear chain organic compounds is more and intermolecular attractive force is high. Whereas in case of branched chain carbon atoms, the atoms come closer and intermolecular attractive force between them is less, therefore boiling point of branched compounds is less than straight chain compounds.

2. In alkanes with odd number of carbon atoms, the end carbon atoms lie on the same side whereas in alkanes with even number of carbon atoms, the end carbon atoms lie in opposite sides. Packing of alkanes with even number of carbon atoms is more dense than that of odd number carbon atoms.
Therefore, melting point of odd number alkanes is less than that of even number alkanes.

Question 31.
Write chemical equation for the combustion reaction of the following hydrocarbons :
(i) Butane,
(ii) Pentene,
(iii) Hexyne,
(iv) Toluene.
Answer:
All hydrocarbons on complete combustion give CO₂ and H₂O.

Question 32.
Write notes on :
(i) Sabatier and Senderens reaction,
(ii) Wurtz reaction,
(iii) Duma reaction.
Answer:
(i) Sabatier and Senderens reactions: In this method alkanes are formed by hydrogenation of unsaturated hydrocarbons in presence of Ni as catalyst.

(ii) Wurtz reaction: When an alkyl halide (bromide or iodide) is heated with sodium in presence of dry ether, an alkane is obtained. The number of carbon atoms in alkane is double than that of alkyl halide. This is called Wurtz reaction.

This method is suitable for preparing alkanes containing even number of carbon atoms like CH₃-CH₃, C₂H₅-C₂H₅,etc.

(iii) Duma reaction: Sodium salt of carboxylic acid reacts with soda lime when due to decarboxylation, alkanes are obtained.

Question 33.
Write Kolbe’s electrolysis with equation.
Answer:
By the electrolysis of concentrated aqueous solution of sodium or potassium salt of monocarboxylic acid, alkanes are obtained at anode.

Electrolysis occurs as follows:
2CH₃COONa ⇌ 2CH₃COO^–+2Na⁺

At Anode: Acetate ion loose electron and become neutral and then dissociate to form alkane.
2CH₃COO^– 2e^– → 2CH₃COO → C₂H₆ + 2CO₂

At Cathode: Hydrogen ion gains electron to give hydrogen gas.
2Na⁺ + 2HOH + 2e^– → 2NaOH + H₂.

Question 34.
Give only equations for the following reactions :
(1) Calcium carbide with water.
(2) Reaction of bromine water on ethylene.
(3) On heating ethylene with alkaline KMnO4.
(4) On heating benzene with cone. HN03 and cone. H2SO4.
(5) On heating benzene with CH3Cl in presence of anhydrous AlCl3.
Answer:
1. Reaction of calcium carbide with water :
CaC₂ + 2H OH → CH ≡ CH + Ca(OH)₂

2. Reaction of ethylene with bromine water :

3. On heating ethylene with alk. KMnO4:

4. On heating benzene with conc. HNO₃ and conc. H2SO₄:

5. On heating benzene with CH₃Cl in presence of anhydrous AlCl₃:

Question 35.
How are the following obtained :
1. Acetaldehyde from Acetylene,
2. Mustard gas from Ethylene,
3. Ethane from Grignard reagent,
4. Cuprous acetylide from acetylene,
5. Methane from Aluminium carbide.
Answer:
1. Acetaldehyde from Acetylene :

2. Mustard gas from Ethylene:

3. Ethane from Grignard reagent:

4. Cuprous Acetylide from Acetylene :
CH≡ CH + Cu₂Cl₂ + 2NH₄OH → Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

5. Methane from Aluminium carbide :
Al₄C₃ + 12H₂O → 3CH₄+4Al(OH)₃.

Question 36.
A primary alkyl halide when subjected to Wurtz reaction forms CgHlg as the only product. By the monobromination of this alkane, an isomer of tertiary bromide is obtained. Identify the alkane and tertiary bromide.
Answer:
Since alkane CgHlg on monobromination forms an isomer of a tertiary bromide, thus tertiary hydrogen should be present in alkane. It is only possible when primary alkyl halide (Which participate in Wurtz reaction) contain a tertiary hydrogen.

Question 37.
Write the intermediate hydrocarbon radical formed by the monochlorination of 2-methyl propane. Which of these is more stable? Give reason also.
Answer:
2-methylpropane gives the following two types of radicals which are as follows :

Radical (I) is more stable, because it is 3° free radical and stabilizes nine hyper can- jugated structures (because it contains 9- α hydrogen).

Radical (II) is less stable because it is 1° free radical and it stabilizes only one hyperconjugated structure (because it contains only 1α -hydrogen).

Question 38.
Explain Dehydrohalogenation and Dehalogenation with example.
Answer:
Dehydrohalogenation: Elimination of a hydrogen halide molecule from an organic compound is known as dehydrohalogenation. During this reaction, hydrogen is eliminated from β-position. Therefore, it is known as β-elimination.
Example: On heating normal propyl chloride with alcoholic KOH, alkene is obtained.
CH₃– CH₂ – CH₂ – Cl + KOH → CH₃— CH = CH₂ + KCl + H₂O.

Dehalogenation: On heating vicinal dihalide with Zn powder in presence of methyl alcohol alkene is obtained. During this reaction, only halogen is eliminated, therefore it is known as dehalogenation reaction.

Question 39.
What happens when :
(1) On heating sodium acetate with soda lime.
(2) On heating 2-propanol with alumina at 300°C temperature.
(3) On passing acetylene through ammoniacal AgN03 solution.
(4) On heating ethyl iodide with sodium.
(5) By reacting hypochlorous acid on Ethylene.
Answer:
1. On heating sodium acetate with soda lime :

2. On heating 2-propanol with alumina at 300°C temperature :

3. On passing acetylene through ammoniacal AgNO₃ solution :
CH ≡ CH + 2 AgNO₃ + 2NH₄OH → Ag- C = C – Ag + 2NH₄NO₃ + 2H₂O

4. On heating ethyl iodide with sodium :

5. On reacting hypochlorous acid with ethylene :

Question 40.
Write notes on :
(1) Hydroboration
(2) Epoxy reaction.
Answer:
1. Hydroboration: When alkene is treated with diborane, double bond of alkene undergoes addition reaction and trialkyl borane is formed which on hydrolysis gives alcohol. This is known as Hydroboration reaction.

2. Epoxy reaction: Alkene oxide is known as Epoxide. Formation of epoxide by the oxidation of alkene is known as Epoxy reaction.
Example: Lower alkenes combine with oxygen at a temperature of200-400°C in the presence of silver catalyst to form epoxide.
Silver catalyst

Question 41.
In the alkane H3C — CH2 — C(CH3)2 — CH2— CH(CH3)2. Identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
Answer:

Question 42.
In the reaction of HI, HBr and HCl with propene same intermediate carbocation is formed. Bond energies of HCl, HBr and HI are mainly 430-5 kJmol^-1 363^-1 kJmol-1 and 296-8 kJmor¹. What will be the order of reactivity of halogen acids?
Answer:
Addition of alkene with halogen acids is an Electrophilic addition reaction.

Since first step is a slow step, thus it is a rate-determining step. Rate of this step, depends upon the availability of proton, which again depend on the dissociation energy of H-X molecules. Lower the bond dissociation energy of H – X molecule, higher is the reactivity of halogen acid. Thus, decreasing order of reactivity of halogen acid is as follows : HI(296-8kJ) > HBr(363-7kJ) > HCl(430-5kJ).

Question 43.
What is Cracking? Give an example.
Answer:
Cracking: When fuel oil or lubricating oil is heated at a high temperature of 770 K in absence of air C—C and C-H bonds break to form lower paraffins or olefins. The method is called cracking or pyrolysis and has great importance in petroleum refining. In short conversion of less volatile higher hydrocarbon into more volatile lower hydrocarbons is called Cracking.

Cracking is of two types :
(a) Thermal cracking: When diesel, oil etc. are heated at high temperature, petrol is obtained. By cracking of kerosene oil, gas is prepared.

(b) Catalytic cracking: In presence of catalyst (Al₂O₃, SiO₂ etc.) cracking is achieved at low temperature. By catalytic cracking of lubricating oil, petrol is obtained.
Uses:
1. Formation of gas from kerosene oil.
2. Formation of petroleum gas from petrol.

Question 44.
A hydrocarbon ‘A’ (Vapour density = 14) decolourises Bayer’s reagent and reacts follows :

Write the name and chemical formula of A, B, C and D.
Answer:
Hydrocarbon ‘A’ with vapour density 14 has molecular mass 28, therefore it is an alkene containing 2 ‘C’ and 4 ‘H’ i.e., CH₂ = CH₂.

Hydrocarbons Class 11 Important Questions Long Answer Type

Question 1.
Explain the laboratory method of preparation of ethylene under the follow¬ing heads:
(i) Chemical equation,
(ii) Labelled diagram,
(iii) Procedure,
(iv) Two main pre¬cautions.
Answer:
Laboratory preparation of alkene (ethylene): Ethylene is obtained in laboratory by heating ethyl alcohol with excess of cone. H₂SO₄ at 170°C.
(i) Chemical equation:

(ii) Labelled diagram :

(iii) Method: 50 cc of ethyl alcohol and 100 cc of conc.H₂SO₄ is taken in a flask, then 8 gm anhydrous Al₂ (SO₄)₃ and 50 gm of sand is added. The mixture of Al₂ (SO₄)₃ and sand checks up formation of foam and facilitates the reaction to take place at 140° C. Now the flask is fitted with a thermometer, an exit tube and a dropping funnel. Flask is placed on a sand bath and fixed with a stand.

The other end of exit tube dips in wash bottle containing NaOH. Another tube from wash bottle leads to a behive shelf placed in a trough of water. A water jar is inverted over behive shelf. Flask is heated at a temperature of 150° C and simultaneously mixture of alcohol and cone. H₂SO₄ is added dropwise into the flask. Along with ethylene, CO₂ (by oxidation of alcohol) and SO₂(by reduction of H₂SO₄) are also present as impurities in flask. These impurities get adsorbed in NaOH solution and pure ethylene is collected in gas jars by downward displacement of water. Ethylene prepared by this method, is pure.

(iv) Precautions :
1. H2S04 should be more in quantity otherwise ethylene gets converted to ether.
2. On mixing small amount of Al₂(SO₄)₃ in the mixture of ethyl alcohol and cone. H₂SO₄the reaction occurs at 140°C.

Question 2.
Explain the laboratory preparation of acetylene under the following heads :
(i) Chemical equation,
(ii) Labelled diagram,
(iii) Procedure,
(iv) Two main pre¬cautions.
Answer:
Laboratory method of preparation of acetylene: The first member of alkyne homologous series, i.e., ethyne or acetylene is prepared by dropping water on calcium carbide. Acetylene obtained by this method contains impurities of PH₃ and NH₃. These are removed by passing the gas through CuSO₄ solution.
(i) Chemical equation :
CaC₂ +2H₂O → Ca(OH)₂ +CH≡ CH
Calcium carbide

(ii) Labelled diagram :

(iii) Method: Calcium carbide is placed on sand taken in a flask. The flask is fitted with a dropping funnel and a delivery tube. The air in the flask is displaced by oil gas. Water is added dropwise from the dropping funnel. The reaction is vigorous and produces much heat. Acetylene is formed. It is passed through acidified CuSO₄ solution to remove impuri¬ties and collected by downward displacement of water.

(iv) Precautions : (i) Before starting the experiment, air in the flask is replaced by oil gas because acetylene forms explosive mixture with air.
(ii) Water is added dropwise.
(iii) Sand is kept at bottom in the flask and then CaC₂ kept over it.

Question 3.
Describe laboratory method of preparation of methane with labelled diagram.
Answer:
Alkanes are obtained by heating sodium salts of fatty acids with soda lime at 630K. Mixture of caustic soda and quick lime is known as soda lime. Only NaOH takes part in the reaction and CaO keeps NaOH dry and decreases its intensity by which reaction does not occur on glass.

Mixture of sodium acetate and soda-lime is taken in a hard glass tube fitted with a delivery tube with the help of a cork. Other end of the delivery tube is immersed in a gas jar filled with water. Glass tube containing the mixture is heated when methane gas is collected over water in the gas jar.

Question 4.
An unsaturated hydrocarbon ‘A’ joins two molecules of H2 and gets reduced. It then undergo ozonolysis and gives Butane-1,4-dial, Ethanal and Propanone. Write the structure and IUPAC name of compound ‘A’. Explain the reactions.
Answer:
Hydrocarbon ‘A’ can add two molecules of H₂. Thus, ‘A’ should be alkane, diene or Alkyne. After ozonolysis reduced ‘A’ gives three ‘species’ of which one is dialdehyde, this means the molecule breaks at both the ends. Thus, A should contain two double bonds. The three species are as follows :

Question 5.
Explain Acidic nature of Acetylene with Example.
Answer:
Acidic nature of acetylene can be explained on the basis of orbital structure. In C₂H₂ carbon atom is sp hybrid state, sp hybrid carbon is smaller and hence its electronegativity is greater than hydrogen. So it pulls electron pair closer to it.

Due to this polarisation of C-H bonds occurs and partial positive charge develops on hydrogen. Therefore, hydrogen is replaced in the form of proton, easily and thus acetylene exhibits acidic nature.

Example: 1. Reaction with AgNO₃: On passing acetylene gas through ammoniacal silver nitrate solution, white precipitate of silver acetylide is formed.

2. Reaction with ammoniacal cuprous chloride: On passing acetylene gas through ammoniacal cuprous chloride solution, red precipitate of cuprous acetylide is formed.

Question 6.
What is confirmation? Describe conformation found in ethane and cyclohexane.
Answer:
Conformation: C- C bonds present in alkanes are sigma (σ) bonds which are formed by axial overlap of two hybridised carbon atoms.
Distribution of electrons in the bond due to planar nature of molecular axis can present the movement of carbon atoms on all the four sides of the bond. Keeping one carbon atom stationary, the second carbon can be rotated on all the four sides of bond axis in space.

Due to this form both the carbon atoms have various states of regular tetrahedrally linked atoms or group, various such forms in space are called conformation.
On account of rotation between the carbon-carbon single bond various spatial ar¬rangements are possible which are called as conformation.

Confirmation of C₂H₆: Two main types of conformation of ethane are :
(i) Eclipsed conformation
(ii) Staggered conformation.
The two conformation of any molecule cannot be superimposed but can be converted into one another.
If the position of one carbon atom of ethane is fixed in space and the other carbon atom is rotated in all the four sides of single bond then various conformations of ethane are possible. Out of these, the conformation which has the lowest energy is called staggered and the one having greatest energy as the eclipsed conformation.

(i) Eclipsed conformation : In eclipsed conformation all the C-H bonds of one methyl group are covered up by C-H bonds of second methyl group.

(ii) Staggered conformation: In staggered conformations the H atoms of both the methyl groups occupy position at maximum distance out of eclipsed and staggered form. It is the staggered conformation which is more stable by 12.6 kJ mol^-1. Due to less difference in energies of staggered and eclipsed forms they change one into other readily and cannot be separated.

Question 7.
In the presence of peroxide, addition of HBr in propene follows Anti- Markownikoff’s rule, but HC1 and HI does not exhibit peroxide effect.
Answer:
Addition of HCl and HI in propene does not exhibit peroxide effect. This is because H-Cl bond (430.5 kJ mol^-1) being stronger than H-Br bond (363.7 kJ mol-1) does not cleave by free radical. On the other side H-I bond (296.8 kJ mol^-1) is weak and iodine instead of joining to free radical double bond combines to form iodine molecule.

This can also be discussed with the help of thermodynamic calculations. Free radicals of hydrogen halide in addition reactions undergo enthalpy change in the followings steps :

It is clear from the above data, that both the steps of addition of HBr are exothermic which is favourable condition for the reaction. But in HCl and H one step is endothermic which is not a favourable condition for any chemical reaction from thermodynamic point of view.

Question 8.
Explain conformational isomerism in cyclohexane.
Answer:
Conformation of cyclohexane: In addition to open chain alkanes the closed chain alkanes i.e., cycloparaffins also exhibit conformational isomerism. The bond angles in cyclopropane (C₃H₆) and cyclobutane (C₄H₈) respectively are 60° and 90° therefore both the rings are under strain and are more reactive. Further they are planar molecules.

The bond angle in cyclopentane is 108° which is quite close to tetrahedral bond angle but cyclohexane ring is free from strain and its each C-C-C bond angle is 109° 28′ hence it is stable. It can be shown by two conformation in space, which are called as chair conformation and boat conformation.

Chair conformation: If is the most stable conformation of cyclohexane in which all bond angles are equal to tetrahedron bond angles and all the adjacent carbon atoms of C-H bonds are in staggered position. This conformation is free from all sorts of strain and has less energy.

Boat conformation: This is the second conformation of cyclohexane and is less stable than the chair conformation. All the carbon (C-C) bonds are present tetrahedrally in space but the H atoms present on adjacent carbon atoms are in eclipsed state. In boat confor¬mation both the H atoms shown as H„ repel each other. There is no repulsion between the H atoms of chair conformation. There is difference of 44 kJ mol^-1 energy between the two conformations.

Question 9.
Write staggered and eclipsed forms of Newman and Sawhorse. Which form will be more stable and why?
Answer:
Staggered form of Ethane is about 12-5 kJ mol^-1 of energy more stable than the eclipsed form. This is because in staggered form the hydrogen atoms of the two carbon atoms are maximum distance apart whereas in eclipsed form they are very close (sometimes overlapped atoms). Thus, in staggered form has minimum repulsive force, minimum energy and maximum stability.

Question 10.
Explain Resonance structure of Benzene.
Answer:
Resonance theory: Bond length of each C – C bond in benzene is 1.39 Å which is intermediate between single bond (1.54Å) and double bond (1.34Å).
Thus each carbon atom in benzene is supposed to be linked by partial double bond. Thus it is supposed that benzene is a resonance hybrid of following structures :

Formula (I) and (II) are Kekule structure compared to other structure their contribution towards resonance is more. These structures participate 80% to resonance. Thus benzene is supposed to be the resonance hybrid of mainly these two Kekule structures.

Energy of Dewar’s structure is high due to presence of para bond. Therefore, it contributes only 20% to resonance whereas Kekule structure is more stable and contributes 80% to resonance.

Question 11.
Explain molecular orbital structure (Model) of Benzene.
Answer:
Orbital Model of Benzene: In benzene, each carbon is in state of sp² hybridization which means there are three sp2 orbitals on each carbon atom which form an angle of 120° and the fourth unhybrid pz orbital remains perpendicular to the plane of ring.

Two sp² orbitals of each carbon undergo axial overlap with sp² orbital of neighbouring carbon atoms to form sigma bonds. The third sp² orbital overlaps with s-orbital of hydrogen to form sigma bond. As a result of this overlapping, a regular hexagon is formed which has an internal angle of 120°. It has six C – C sigma bonds and six C – H sigma bonds.

Now, there remains one p_z-orbital in each carbon atom. These p_z-orbitals are parallel to each other and perpendicular to the plane of ring.

Question 12.
Explain the free radical mechanism of halogenation.
Answer:
Halogenation reaction in alkanes occur by free radical mechanism.
1. Chain initiation: The reaction is initiated by homolysis of chlorine molecule in presence of heat and chlorine free radical is obtained.

2. Chain propagation step: Chlorine free radical displaces hydrogen and alkyl free radical is obtained.


This methyl free radical reacts with chlorine molecule and forms alkyl halide and a chlorine free radical.
This free chlorine radical again reacts with alkane to give methyl free radical. In this way, the reaction is repeated.

3. Chain termination step: At the end both the free radicals react mutually and end the chain

Free radical of halogen attacks alkyl halide and form a new free radical.

This free radical reacts with halogen to form dihaloalkane. This reaction takes place slowly till all the hydrogen atoms get displaced.

Question 13.
Explain three examples of electrophilic substitution reactions in aromatic hydrocarbon with their mechanism.
Answer:
All electrophilic reactions complete in the following steps:
(i) Electrophilic reagent is obtained by the fission of reagent.

(ii) This electrophilic reagent attacks benzene nucleus and positive charge is produced on the ring which achieve stability by resonance.
Example: 1. Halogenation: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

(iii) A proton is removed from positively charged carbonium ion and product is formed.

Example: 1. Halogenatíon: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

Mechanisam:
(i) Cl₂+FeCl₃ → Cl⁺+FeCl₄


2. Nitration: Nitrobenzene is produced by the reaction of benzene with cone. HNO₃ and cone H₂SO₄.
Mechanism :
HNO₃ +2H₂SO₄ →NO₂⁺+ H₃O⁺ + 2HSO₄


3. Sulphonation: Benzene sulphonic acid is produced by heating benzene with cone H₂SO₄.
Mechanism:
(i) 2H₂SO₄ → SO₃ + HSO₄^– +H₃O⁺

Question 14.
Write IUPAC name of the following compounds :
(a) CH3CH = C(CH3)2
(b) CH2== CH — C = C —CH3



Answer:

Question 15.
Give Electronic specification of directive effect of groups.
Answer:
If a substituent group is already present in benzene ring then as a result of substitution reaction, the position of the incoming group is decided on the basis of the substituent present. This is known as directive effect of that group.
Directive effect of group on the basis of electronic concept:
1. ortho and para directing group : These are such groups which being already present on the benzene ring direct the incoming group to ortho and para place.
Example : – OH, – OCH₃, – CH₃, – NH₂.
These groups due to their electron releasing tendency increase the electron density at the ortho and para place.

Due to the increase in electron density, reactivity of benzene nucleus also increases and electrophilic reagents due to high electron density enter into these places.

2. meta directing effect: These are such groups which being already present on the benzene ring direct the incoming group to meta place.
Example : – N0₂, – COOH, – SO₃H, – CN.
These groups when present in benzene ring due to mesomeric effect pull the electrons as a result of which electron density decreases mainly at ortho and para place whereas it remains unaffected at meta position i.e. at met a place electron density is high. Therefore electrophilic reagent enters at meta place.

Question 16.
Explain the halogenation in benzene.
Answer:
Example: 1. Halogenatíon: At normal temperature and in presence of a halogen carrier benzene reacts with halogen to form halobenzene.

Mechanisam:
(i) Cl₂+FeCl₃ → Cl⁺+FeCl₄

2. Nitration: Nitrobenzene is produced by the reaction of benzene with cone. HNO₃ and cone H₂SO₄.
Mechanism :
HNO₃ +2H₂SO₄ →NO₂⁺+ H₃O⁺ + 2HSO₄

3. Sulphonation: Benzene sulphonic acid is produced by heating benzene with cone H₂SO₄.
Mechanism:
(i) 2H₂SO₄ → SO₃ + HSO₄^– +H₃O⁺

Question 17.
How will you convert benzene into the following :
(i) p-nitro bromobenzene
(ii) wi-nitrochlorobenzene
(iii) p-nitrotoluene
(iv) Acetophenone.
Answer:
(i) Bromo group is o/p directing whereas nitro group is m-directing. Thus, to obtain p-nitrobromobenzene bromination and nitration are performed.

(ii) To obtain m-nitro chlorobenzene, chlorination is performed after nitration. Chloro group is o/p directing and nitro group is m-directing.

(iii) To obtain nitro toluene, nitration is performed after Friedel-Crafts reaction.

(iv) To obtain acetophenone, Friedel-Crafts Acylation (-COR) is performed.

Question 18.
Write structural formula and IUPAC name of all possible position isomers of the following compounds in which number of double and triple bonds are repre¬sented :
(a) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond).
Answer:
(a) Isomers of C₄H₈ which possess double bond are as follows :

(b) Isomers of C₄H₈ which possess triple bond are as follows :

Question 19.
How will you convert the following compounds to benzene :
(i) Ethyne,
(ii) Ethene,
(iii) Hexane.
Answer:
(i) Benzene from Ethyne :

(ii) Benzene from Ethene:

(iii) Benzene from Hexane :

Question 20.
Which of the system is not aromatic? Give reason.
(i)

Answer:

Reason: Due to presence of sp3 hybridized carbon atom, molecule is not planar. It contains 6π-electrons but its π-electron of meta, does not surround all the carbon atoms. Thus, it is not aromatic.

Reason: Due to the presence of sp³-hybridised carbon atom, the molecule is not planar. It has 4π-electrons. Thus, molecule is not aromatic because its (4n + 2)π-electrons do not possess planar cyclic electron cloud. ”

Reason : Cyclo-octatetraene is non-planar with 87r-electrons. Thus, it is not aromatic.

Question 21.
Relative reactivity of 1°, 2°, 3° halogens towards chlorination is 1: 3.8: 5. Determine the percent of all monochlorinated products of 2-methyibutane.
Answer:

Relative amount of monochlorinated products=Amount of hydrogen x Relative reactivity 1° (For monochlorinated product) = 9 ×1=9 2° (For monochlorinated product) = 2×3.8 = 7.6 3° (For monochlorinated product) =1×5 = 5
Total number of monochlonnated compounds = 9+ 7.6 +5=21.6
% amount of 1° monochlonnated product = \(\frac{9 \times 100}{21 \cdot 6} \) = 41.6%
% amount of 20 monochlonnated product =\(\frac{7.6 \times 100}{21 \cdot 6} \) = 35.18%
% amount of 3° monochLorinated product =\(\frac{5 \times 100}{21 \cdot 6}\) = 23.15%.

Question 22.
Which product will be obtained as a result of the following reaction and why :

Answer:

When Friedel-Crafts reaction takes place with higher alkyl halide like n-propyl carbocation (1 ° carbocation) rearranges to more stable isopropyl carbocation (2° carbocation) as a result of which main product obtained is isopropyl benzene.

Question 23.
Arrange the following compounds in the decreasing order of their relative reactivity towards electrophilic reagent:

Answer:
-OCH₃ (Methoxy group) is electron releasing group. It increases the electron density over benzene nucleus due to Resonance effect (+R effect). Due to this anisole is relatively more reactive towards electrophilic reagents than benzene.

In case of aryl halides, halogens .due to their -I effect are extremely unreactive by which total electron density in benzene nucleus decreases. Due to this substitution decreases.
-NO₂ group is electron attractive group. This is due to strong -I effect decreases the electron density in benzene nucleus. Due to this nitrobenzene becomes less reactive. Thus, order of reactivity of these compounds towards electrophilic reagents is as follows :

Hydrocarbons Class 11 Important Questions Objective Type

1. Choose the correct answer:

1. Bond angle H-C-H in methane is:
(a) 100.5°
(b) 109.0
(c) 109°28’
(d) 180°.
Answer:
(c) 109°28’

Question 2.
C≡C contains:
(a) 3 σ bonds
(b) One σ and three π-bonds.
(c) 3 n-bonds
(d) One σ and 2π-bonds.
Answer:
(d) One σ and 2π-bonds.

Question 3.
Ethene has:
(a) Sσ and one z-bond
(b) 6σ-bonds
(c) 4σ and 2π-bonds
(d) Sσ-bonds.
Answer:
(a) Sσ and one z-bond

Question 4.
Which is most reactive:
(a) C₂H₂
(b) CH₄
(c) C₂H₄
(d) C₂H₆.
Answer:
(a) C₂H₂

Question 5.
Represents geometrical isomerism:
(a) But-2-ene
(b) But-2-yne
(c) Butan-2-ol
(d) Butanol.
Answer:
(a) But-2-ene

Question 6.
Molecular formula of all the following compounds is C4H8. One which represents geometrical isomerism:
(a)

(b)

(c)

(d)
Answer:

(a)

Question 7.
The hybridized orbital of carbon in benzene is :
(a) sp³
(b) sp²
(c) sp
(d) dsp³.
Answer:
(b) sp¹

Question 8.
C – C single bond carbon atoms is HC ≡ C – CH = CH2 have the hybridization : ’
(a) sp³– sp³
(b) sp²-sp³
(c) sp³ – sp
(d) sp- sp².
Answer:
(d) sp- sp².

Question 9.
The number of sigma bonds in 1-butene is :
(a) 8
(b) 10
(c) 11
(d) 12.
Answer:
(c) 11

Question 10.
The double bond present between two carbon atoms in ethylene is :
(a) Two perpendicular sigma bonds
(b) One sigma and one pi bond
(c) Two perpendicular pi bonds
(d) Two pi bonds at an angle of 60°.
Answer:
(b) One sigma and one pi bond

Question 11.
The reagent that differentiates ethylene from acetylene is :
(a) Aq. alkaline KMnO₄
(b) Cl₂ dissolved in CCl₄
(c) Ammoniacal Cu2Cl₂
(d) Cone. H2SO₄.
Answer:
(c) Ammoniacal Cu2Cl₂

Question 12.
In an engine knocking is produced when the fuel:
(a) Bums slowly
(b) Bums fast
(c) Contains water
(d) Is mixed with machine oil.
Answer:
(b) Bums fast

Question 13.
The compound that is mixed in fuel to reduce knocking is :
(a) PbBr₂
(b) ZnBr₂
(c) PbO
(d) TEL (Tetraethyl lead).
Answer:
(d) TEL (Tetraethyl lead).

Question 14.
The process which can be used to prepare methane is :
(a) Wurtz reaction
(b) Kolbe reaction
(c) Reduction of alkyl halide
(d) Hydrogenation of alkene.
Answer:
(c) Reduction of alkyl halide

Question 15.
Acetylene reacts with HCl in presence of HgCl₂ and forms :
(a) Methyl chloride
(b) Acetaldehyde
(c) Vinyl chloride
(d) Formaldehyde.
Answer:
(b) Acetaldehyde

Question 16.
Propyne reacts with aqueous H₂SO₄ in presence of HgSO₄. The main product is:
(a) Propanal
(b) Propyl hydrogen sulphate
(c) Acetone
(d) Propanol.
Answer:
(c) Acetone

Question 17.
The reagent to differentiate propene from propyne is :
(a) Bromine
(b) Alkaline KMnO₄
(c) Ammoniacal AgNO₃
(d) Ozone.
Answer:
(b) Alkaline KMnO₄

Question 18.
The reaction is called:
(a) Wurtz reaction
(b) Kolbe’s reaction
(c) Sabatier-Senderens reaction
(d) Carbylamine reaction.
Answer:
(c) Sabatier-Senderens reaction

Question 19.
Ammoniacal silver nitrate solution reacts with C₂H₂ to form :
(a) Silver mirror
(b) Silver oxide
(c) Silver formate
(d) Silver acetylide.
Answer:
(d) Silver acetylide.

Question 20.
An unknown compound A has a molecular formula C4H6. When A is treated with an excess of Br₂ a new substance B with formula C₄H₆Br₂ is formed. A forms white precipitate ammoniacal AgNO₃ solution. A may be :
(a) But-1 -yne
(b) But-2-yne
(c) But-1 -ene
(d) But-2-ene.
Answer:
(b) But-2-yne

Question 21.
Group which increase activity :
(a) o,p-directing group
(b) m-directing group
(c) NO₂
(d) None of these.
Answer:
(a) o,p -directing group

Question 22.
Which of the following alkane is not obtained by Wurtz reaction :
(a) CH₄
(b) C₂H₆
(C) C₃H₈
(d) C₄H₁₀.
Answer:
(a) CH₄

Question 23.
Which of the following is non-aromatic :
(a) Benzene
(b) Cyclo-octadiene
(c) Tropolium cation
(d) Cyclopentadienal cation.
Answer:
(d) Cyclopentadienal cation.

Question 24.
Which of the following is non-aromatic :
(a)
(b)
(c)
(d)
Answer:
(c)

Question 25.
Of the following whose dipole moment is zero:
(a) cis-2-butene
(b) trans- 1 -butene
(c) 1 -butene
(d) 2-methyl-1 -butene.
Answer:
(b) trans-1 -butene

Question 26.
Strongest acid is:
(a) HC ≡ CH
(b) C₆H₆
(c) C₂H₆
(d) CH₃OH.
Answer:
(d) CH₃OH.

Question 27.
Butene -1 is converted to butane:
(a) Pd/ H₂
(b) Zn / HCl
(c) Sn / HCl
(d) Zn – Hg.
Answer:
(a) Pd/ H₂

Question 28.
Number of carbon-carbon triple bond in CaC₂ is:
(a) One σ, one π
(b) One σ, two π
(c) One σ,\(1 \frac{1}{2} \) π
(d) One σ bond.
Answer:
(b) One σ, two π

Question 29.
Reagent which distinguishes 1-butyne and 2-butyne is:
(a) Br₂ in CCl₄
(b) Dil.H₂SO₄ / HgSO₄
(c) H₂ Lindlar’s reagent
(d) Ammoniated CaCl₂.
Answer:
(d) Ammoniated CaCl₂.

Question 30.
Is produced by the reaction of n-propyl bromide and alcoholic KOH:
(a) 1 -Butene
(b) 1 -Butanol
(c) 2-Butene
(d) 2-Butanol.
Answer:
(c) 2-Butene

2. Fill in the blanks:

1. Kerosene oil is a mixture of ……………….. .
Answer:
Alkane

2. Carbon-carbon bond length is minimum in ……………….. .
Answer:
Ethyne

3. ……………….. reagent is used to distinguish between propene and propyne.
Answer:
Baeyer’s (Alkaline KMnO₄)

4. Teflon is a polymer of ……………….. .
Answer:
Tetrafluoroethylene

5. Dehydration of ethanol forms ……………….. .
Answer:
Ethene

6. ……………….. is formed by addition of H₂ on Benzene.
Answer:
Cyclohexane

7. Geometrical isomerism is found in ……………….. .
Answer:
Alkene

8. Most stable configuration is in ……………….. form.
Answer:
Staggered

9. trans isomer is ……………….. stable than cis isomer.
Answer:
More

10. ……………….. is obtained by the reaction of aluminium carbide with water.
Answer:
CH₄

11.
Answer:
C₂H₄.

3. Match the following:

I.

‘A’	‘B’
1.	(a) Anti-Markownikoff’s rule
2.	(b) Baeyer’s reagent
3.Mixture of KMnO4 + KOH	(C) Kolbe reaction
4.	(d) Dehydration
5.	(e) Decarboxylation.

Answer:
1. (e) Decarboxylation
2. (d) Dehydration
3. (b) Baeyer’s reagent
4. (C) Kolbe reaction
5. (a) Anti-Markownikoff’s rule.

II.

Answer:
1. (b) Kolbe reaction
2. (d) By Wurtz reaction
3. (e) Methane
4. (a) Electrophilic substitution reaction
5. (c) sp² hybridization.

III.

Answer:
1. (e) Sidewise overlapping
2.(a) Electrophilic substitution
3. (f) O, p-group
4. (b) Unsaturated hydrocarbon
5. (c) Liquefied petroleum gas
6. (g) Compressed natural gas,
7.  (d) cis-trans isomers.

4. Answer in one word/sentence:

1. Full name of TEL is.
Answer:
Tetraethyl lead

2. Which product is formed when ethylene dibromide is heated with Zn powder?
Answer:
Ethylene

3. Name the smelling agent in LPG.
Answer:
Ethyl mercaptan

4. Ethane is formed by the electrolysis of aqueous solution of potassium acetate.
Answer:
Kolbe reaction

5. Name of the reaction CH₂ =   is
Answer:
Sabatier and Senderens reduction reaction

6. write the name of the products formed in the reaction.
Answer:
(A) CH ≡ CH,
(B) CH₃ – CHO,
(C) CH₃CH₂OH,

7. Name the reaction by which Alkane is prepared by the reaction of alkyl halide and sodium.
Answer:
Wurtz-Fittig reaction.

5. State true or False: 

1. Pi ( π) bond is stronger than sigma (σ) bond.
Answer:
False

2. Alkyl group in benzene is O-,ρ- directing.
Answer:
True

3. Benzene is very reactive due to the presence of double bond.
Answer:
False

4. Due to presence of directing group benzene nucleus becomes more reactive.
Answer:
False.