AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs

Students can go through AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs

→ The Central Tendencies are 3 types. They are

  1. Arithmetic Mean
  2. Median
  3. Mode

→ Information, available in the numerical form or verbal form or graphical form that helps in taking decisions or drawing conclusions is called Data.

→ Arithmetic mean of the ungrouped data = AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs 1 (short representation) where ∑xi represents the sum of all xis, where ‘i’ takes the values from 1 to n.

AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs

→ Arithmetic mean = Estimated mean + Average of deviations
AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs 2

→ Mean is used in the analysis of numerical data represented by unique value.

→ Median represents the middle value of the distribution arranged in order.

→ The median is used to analyse the numerical data, particularly useful when there are a few observations that are unlike mean, it is not affected by extreme values.

→ Mode is used to analyse both numerical and verbal data.

→ Mode is the most frequent observation of the given data. There may be more than one mode for the given data.

→ Representation of classified distinct observations of the data with frequencies is called ‘Frequency Distribution’ or ‘Distribution Table’.

→ Difference between upper and lower boundaries of a class is called length of the class denoted by ‘C’.

→ In a class the initial value and end value of each class is called the lower limit and upper limit respectively of that class.

AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs

→ The average of upper limit of a class and lower limit of successive class is called upper boundary of that class.

→ The average of the lower limit of a class and upper limit of preceding class is called the lower boundary of the class.
The progressive total of frequencies from the last class of the table to the lower boundary of particular class is called Greater than Cumulative Frequency (G.C.F).

→ The progressive total of frequencies from first class to the upper boundary of particular class is called Less than Cumulative Frequency (L.C.F.).

→ Histogram is a graphical representation of frequency distribution of exclusive class intervals. When the class intervals in a grouped frequency distribution are varying we need to construct rectangles in histogram on the basis of frequency density.
Frequency density = \(\frac{\text { Frequency of class }}{\text { Length of that class }}\) × Least class length in the data

→ Frequency polygon is a graphical representation of a frequency distribution (discrete/ continuous).

AP Board 8th Class Maths Notes Chapter 7 Frequency Distribution Tables and Graphs

→ Infrequency polygon or frequency curve, class marks or mid values of the classes are taken on X-axis and the corresponding frequencies on the Y-axis.

→ Area of frequency polygon and histogram drawn for the same data are equal.

→ A graph representing the cumulative frequencies of a grouped frequency distribution against the corresponding lower/upper boundaries of respective class intervals is called Cumulative Frequency Curve or “Ogive Curve”.

AP SSC 10th Class Maths Notes Chapter 13 Probability

Students can go through AP SSC 10th Class Maths Notes Chapter 13 Probability to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 13 Probability

→ Theory of probability has its origin date back to 16th century.

→ J. Cardan, an Italian physician and mathematician wrote the first book on probability named “The Book of Games of Chance”.

AP SSC 10th Class Maths Notes Chapter 13 Probability

→ James Bernoulli (1654 – 1705), A.De Moivre (1667 – 1754) and Pierre Simon Laplace (1749 – 1827) made significant contribution to the theory of probability.

→ Experimental or empirical probability : The probability estimated on the basis of results of an actual experiment is called experimental probability of empirical probability.
Eg : An unbiased coin is tossed for 1000 times, head turned up for 455 times and tail turned up 545 times, then the probability or likelyhood of getting a head is = \(\frac{455}{1000}\) = 0.455.

Thus experimental probability = \(\frac{\text { No. of trials in which the event happened }}{\text { Total no. of trials }}\)

→ Classical or Theoretical probability: Classical probability of an event (E) is defined Number as

P(E) = \(\frac{\text { Number of outcomes favourable to E }}{\text { No. of all possible outcomes of the experiment }}\)

This definition was given by ‘Pierre Simon Laplace’.
Eg: The probability of getting a head when a coin is tossed is given by Number of outcomes favourable to this event getting a head = 1 Number of all possible outcomes of this experiment = 2 (Head, Tail)

∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total events }}\) = \(\frac{1}{2}\)

Note: If an experiment is conducted for many number of times, then the experimental probability may become closer and closer to theoretical probability.
AP SSC 10th Class Maths Notes Chapter 13 Probability 1

AP SSC 10th Class Maths Notes Chapter 13 Probability

→ The probability of a sure event is 1.

→ The probability of an impossible event is zero.

→ The probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1.

→ An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

→ For any event E, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where E and \(\overline{\mathrm{E}}\) are complementary events.

→ Playing cards and their probability : A deck of playing cards consists of 52 cards which are divided into four suits of 13 cards each.
They are:
AP SSC 10th Class Maths Notes Chapter 13 Probability 2

→ The cards in each suit are:
AP SSC 10th Class Maths Notes Chapter 13 Probability 3
Eg : When a card is drawn at random from a deck of cards then

  • Getting a black or red card – equally likely exhaustive events.
  • Getting an ace or king – mutually exclusive.
    AP SSC 10th Class Maths Notes Chapter 13 Probability 4
  • Getting an ace or a hearts – not mutually exclusive since the hearts contain an ace.
    AP SSC 10th Class Maths Notes Chapter 13 Probability 5

AP SSC 10th Class Maths Notes Chapter 13 Probability

→ When a coin is tossed, the outcomes are H, T (Head, Tail).

→ When a dice is thrown the outcomes are 1, 2, 3, 4, 5 and 6.

→ When two dice are thrown, the outcomes are
AP SSC 10th Class Maths Notes Chapter 13 Probability 6

→ If a coin is tossed n-times or n – coins are tossed simultaneously, then the number of total outcomes = 2n.

→ If a dice is thrown for n – times or n – dice are thrown simultaneously then the number of total outcomes = 6n.

AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots

Students can go through AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots

→ The product of two same numbers is called its square.
Ex: 1) x × x = x2
2) 6 × 6 = 62 = 36

→ The digits in the units place of a square number are 0, 1, 4, 5, 6, 9.

→ If the digits 2,3, 7 or 8 are in the units place of an umber then it is not a perfect square.

AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots

→ If there are ‘n’ digits in a number then the no.of digits in its square = 2n or (2n -1).

→ Sum of ‘n ‘ consecutive odd numbers = n2

→ The square of any odd number say ‘n’ can be expressed as the sum of two consecutive numbers as
AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots 1

→ If a, b, c are any three positive integers and a2 + b2 = c2 then a, b, c are called Pythagorean triplets. Ex: (3, 4, 5) (5, 12, 13).

→ If a square number is expressed, as the product of two equal factors, then one of the factors is called the square root of that square number. Thus, the square root of 169 is 13. It can be expressed as √169 = 13 (symbol used for square root is √). Thus it is the inverse operation of squaring.

→ If the same number is multiplied itself by 3 times then it is called a cube of a number. Ex: cube of x = x × x × x = x3

AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots

→ If a cube number is expressed, as the product of 3 equal factors, then one of he factors is called the cube root of that number.
Symbol for cube root is \(\sqrt[3]{ }\)
Ex: \(\sqrt[3]{64}=\left(4^{3}\right)^{1 / 3}\) = 4


AP Board 8th Class Maths Notes Chapter 6 Square Roots and Cube Roots 2

AP Board 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion

Students can go through AP Board 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion

→ Two simple ratios are expressed like a single ratio as the ratio of product of antecedents to product of consequents and we call it compound ratio of the given two simple ratios.
a : b and c : d are any two ratios, then their compound ratio is \(\frac{a}{b}\) × \(\frac{c}{d}\) = \(\frac{ac}{bd}\) i.e. ac : bd.

→ A percentage(%) compares a number to 100. The word percent means “per every hundred” or “out of every hundred”. 100% = \(\frac{100}{100}\) it is also a fraction with denominator 100.

→ Discount is a decrease percent of marked price. Price reduction is called rebate or discount. It is calculated on marked price or list price.

AP Board 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion

→ Profit or loss is always calculated on cost price. Profit is an example of increase percent of cost price and loss is an example of decrease percent of cost price.

→ VAT will be charged on the selling price of an item and will be included in the bill.
VAT is an increase percent on selling price.

→ Simple interest is an increase percent on the principal.

→ Simple interest (I) = \(\frac{P \times T \times R}{100}\)
where P = Principal, T = Time inyears, R = Rate of interest.

→ Amount = Principal + Interest = P + \(\frac{P \times T \times R}{100}\) = P\(\left(1+\frac{T \times R}{100}\right)\)

→ Compound interest allows you to earn interest on interest.

→ Amount at the end of ‘n’ years using compound interest is A = P \(\left(1+\frac{R}{100}\right)^{n}\)

AP Board 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion

→ The time period after which interest is added to principal is called conversion period.
When interest is compounded halfyearly, there are two conversion periods in a year, each after 6 months. In such a case, ha If year rate will be half of the annual rate.

→ Note: 1.615 : 1 is called as golden ratio.
In ancient Greece, artists and architects believed there was a particular rectangular shape that looked very pleasing to the eye. For rectangles of this shape, the ratio of long side to the short side is roughly 1.615 : 1. This ratio is very close to what is known as golden ratio.

AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry

Students can go through AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry

→ If a person is looking at an object then the imaginary line joining the object and the eye of the observer is called the line of sight or ray of view.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 1

→ An imaginary line parallel to earth surface and passing through the point of observation is called the horizontal.

→ If the line of sight is above the horizontal then the angle between them is called “angle of elevation”.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 2

→ If the line of sight is below the horizontal then the angle between them is called the angle of depression.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 3

→ Useful hints to solve the problems:

  1. Draw a neat diagram of a right triangle or a combination of right triangles if necessary.
  2. Represent the data given on the triangle.
  3. Find the relation between known values and unknown values.
  4. Choose appropriate trigonometric ratio and solve for the unknown.

→ The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

→ To use this application of trigonometry, we should know the following terms.

→ The terms are Horizontal line, Line of Sight, Angle of Elevation and Angle of Depression.

→ Horizontal line: A line which is parallel to earth from observation point to object is called “horizontal line”.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 4

→ Line of Sight (or) Ray of Vision: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

→ Angle of Elevation: The line of sight is above the horizontal line then angle between the line of sight and the horizontal line is called “angle of elevation”.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 5
Note:

  1. If the angle of observer moves towards the perpendicular line (pole/tree/ building), then angle of elevation increases and if the observer moves away from the perpendicular line (pole/tree/building), then angle of elevation decreases.
  2. If height of tower is doubled and the distance between the observer and foot of the tower is also doubled, then the angle of elevation remains same.
  3. If the angle of elevation of sun above a tower decreases, then the length of shadow of a tower increases.

→ Angle of Depression: The line of sight is below the horizontal line then angle between the line of sight and the horizontal line is called angle of depression.
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 6
Note:

  1. The angle of elevation and depression are always acute angles.
  2. The angle of elevation of a point P as seen from a point ‘O’ is always equal to the angle of depression of ‘O’ as seen from P.

→ Points to be kept in mind:
I. Trigonometric ratios in a right triangle:
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 7
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 8

II. Trigonometric ratios of some specific angles:
AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry 9

→ Solving Procedure:
When we want to solve the problems of height and distances, we should consider the following :

  1. All the objects such as tower, trees, buildings, ships, mountains, etc. shall be considered as linear for mathematical convenience.
  2. The angle of elevation or angle of depression is considered with reference to the horizontal line.
  3. The height of the observer is neglected, if it is not given in the problem.
  4. To find heights and distances, we need to draw figures and with the help of these figures we can solve the problems.

AP Board 9th Class Maths Notes Chapter 4 Lines and Angles

Students can go through AP Board 9th Class Maths Notes Chapter 4 Lines and Angles to understand and remember the concepts easily.

AP State Board Syllabus 9th Class Maths Notes Chapter 4 Lines and Angles

→ A ray is a part of line. It begins at a point and goes on endlessly in a specified direction.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 1

→ A part of a line with two end points is called a line segment.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 2

AP Board 9th Class Maths Notes Chapter 4 Lines and Angles

→ Points on the same line are called collinear points.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 3

→ The angle is formed by rotating a ray from an initial position to a terminal position.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 4

→ One complete rotation makes an angle 360°.

→ Angles are named according to their measure.
Obtuse angle 90° < x < 180°.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 5

→ Straight angle y = 180°
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 6

→ Reflex angle 180° < z < 360°
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 7

→ If two lines have no common points, they are called parallel lines. In the figure l // m.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 8

→ If two lines have a common point then they are called intersecting lines.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 9

→ Three or more lines meet at a point are called concurrent lines.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 10

AP Board 9th Class Maths Notes Chapter 4 Lines and Angles

→ Two angles are said to be supplementary angles if their sum is 180°.
E.g.: (100°, 80°), (110°, 70°), (120°, 60°), (179°, 1°), (90°, 90°) etc.

→ The supplementary angle to x° is given by (180° – x°).

→ Two angles are said to be complementary if their sum is 90°.

→ The complementary angle to x° is (90° – x°).
E.g.: (89°, 1°), (70°, 20°), (60°, 30°) etc.

→ Two angles are said to be form a pair of adjacent angles if they have a common arm and lie on the either sides of the common arm.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 11
∠1 and ∠2 are a pair of adjacent angles with OB as their common arm.

→ A pair of adjacent angles are said to be a linear pair of angles if their sum is 180°. ∠1 + ∠2 = 180°
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 12

→ When a pair of lines meet at a point, they form four angles. The two pairs of angles which have no common arm are called vertically opposite angles.
In the figure (∠1, ∠3) and (∠2, ∠4) are the pairs of vertically opposite angles.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 13

→ When two lines intersect, the pairs of vertically opposite angles thus formed are equal a = c and b = d. (from the figure)
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 14

AP Board 9th Class Maths Notes Chapter 4 Lines and Angles

→ When a pair of lines intersected by a transversal, there forms eight angles.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 15

→ When a pair of parallel lines intersected by a transversal the pairs of a) alternate interior angles b) corresponding angles c) alternate exterior angles are equal and the interior / exterior angles on the same side of the transversal are supplementary.
AP Board 9th Class Maths Notes Chapter 4 Lines and Angles 16

→ Lines which are parallel to same line are parallel to each other.

AP Board 9th Class Maths Notes Chapter 4 Lines and Angles

→ The sum of the interior angles of a triangle is 180°.

→ If one side of a triangle is produced, then the exterior angle thus formed is equal to the sum of the two interior opposite angles.

AP SSC 10th Class Maths Notes Chapter 11 Trigonometry

Students can go through AP SSC 10th Class Maths Notes Chapter 11 Trigonometry to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 11 Trigonometry

→ In our daily life, we can measure the heights, distances and slopes by using some mathematical techniques.

→ The mathematical techniques which come under a branch of mathematics is called ‘trigonometry’.

→ “Trigonometry” is the study of relationships between the sides and angles of a triangle.

AP SSC 10th Class Maths Notes Chapter 11 Trigonometry

→ Early astronomers used to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in engineering and physical sciences are based on trigonometrical concepts.

→ Naming the sides in a right triangle:
Let’s take a right triangle ABC as shown in the figure.
AP SSC 10th Class Maths Notes Chapter 11 Trigonometry 1
Consider ∠CAB as A where angle ‘A’ is acute.
Since AC is the longest side, it is called “Hypotenuse”.

→ Now observe the position of side BC with respect to angle A. It is opposite to angle ‘A’ and we can call it as “Opposite side of angle A”.

→ And the remaining side AB can be called as “Adjacent side of angle A”.

→ Trigonometric Ratios:
The ratios of the sides of a right angled triangle with respect to its acute angles, are called Trigonometric ratios.

→ Consider a right angled triangle ABC having right angle at B as shown in the given figure.
AP SSC 10th Class Maths Notes Chapter 11 Trigonometry 2
Then, trigonometric ratios of the angle A in right angled triangle ABC are defined as follows:
AP SSC 10th Class Maths Notes Chapter 11 Trigonometry 3

→ There are three more ratios defined in trigonometry which are considered as multiplicative inverse of the above three ratios.

AP SSC 10th Class Maths Notes Chapter 11 Trigonometry

→ Multiplicative inverse of “sine A” is “cosecant A”.
i.e., cosec A = \(\frac{1}{\sin A}\) = \(\frac{\text { Hypotenuse }}{\text { Opposite side of the angle } A}\)

→ Multiplicative inverse of “cosine A” is “secant A”.
i.e., sec A = \(\frac{1}{\cos A}\) = \(\frac{\text { Hypotenuse }}{\text { Adjacent side of the angle } A}\)

→ Multiplicative inverse of “tangent A” is “cot A”.
i.e., cot A = \(\frac{1}{\tan A}\) = \(\frac{\text { Adjacent side of the angle } A}{\text { Opposite side of the angle } A}\)

Note:
i) The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangles, if the angle remains the same.
ii) Each trigonometric ratio is a real number and has no unit.
iii) “sin θ” is one symbol and sin, cos, tan etc., cannot be separated from θ.
iv) If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of angle can be easily determined.
v)
AP SSC 10th Class Maths Notes Chapter 11 Trigonometry 4

→ Trigonometric ratios of some specific angles:
The values of various trigonometric ratios of 0°, 30°, 45°, 60° and 90°.
AP SSC 10th Class Maths Notes Chapter 11 Trigonometry 5
Note:
i) The values of “sin θ” and “cos θ” always lie between ‘0’ and ‘1’.
ii) In the case of tan θ, the values increase from 0 to ∞ (not determinate).
iii) In the case of cot θ, the values decrease from ∞ to 0.
iv) In the case of cosec θ, the values decrease from ∞ to 1.
v) In the case of sec θ, the values increase from 1 to ∞.

AP SSC 10th Class Maths Notes Chapter 11 Trigonometry

→ Trigonometric ratios of complementary angles:
Two angles are said to be complementary, if their sum is equal to 90°.
In a right angled triangle, if ∠B = 90°, then ∠A + ∠C = 90° i.e., ∠A and ∠C form a pair of complementary angles.
If ‘θ’ is an acute angle, then we can prove that
sin (90 – θ) = cos θ
cos (90 – θ) = sin θ
tan (90 – θ) = cot θ
cot (90 – θ) = tan θ
sec (90 – θ) = cosec θ
cosec (90 – θ) = sec θ

→ Trigonometric Identity: An identity equation having trigonometric ratios of an angle is called trigonometric identity. It is true for all the values of the angles involved in it.
We have three major trigonometric identities. They are
i) sin2 A + cos2 A = 1
ii) sec2 A – tan2 A = 1
iii) cosec2 A – cot2 A = 1
Note: sin2 θ = (sin θ)2 but sin θ2 ≠ (sin θ)2

AP Board 9th Class Maths Notes Chapter 3 The Elements of Geometry

Students can go through AP Board 9th Class Maths Notes Chapter 3 The Elements of Geometry to understand and remember the concepts easily.

AP State Board Syllabus 9th Class Maths Notes Chapter 3 The Elements of Geometry

→ Geometry is structured on its building blocks namely point, line and plane.

→ In geometry there are undefined terms like point, plane and line.

→ Angles, circles and triangles are the examples for defined terms.

→ No better entrance exists than Euclid’s time honoured ‘Elements’.

AP Board 9th Class Maths Notes Chapter 3 The Elements of Geometry

→ In ‘The Elements’, Euclid developed a new system of thought which laid the foundation for the advancement of the geometry.

→ Some of the Euclid’s axioms are:

  • Things which are equal to same things are equal to one another.
  • If equals are added to equals, the wholes are equal.
  • If equals are subtracted from equals, the remainders are also equal.
  • Things which coincide with one another are equal to one another.
  • The whole is greater than part.
  • Things which are double of the same things are equal to one another.
  • Things which are halves of the same things are equal to one another.

→ Euclid’s postulates are
Postulate – 1 : To draw a straight line from any point to any point.
Postulate – 2 : A terminated line can be produced indefinitely.
Postulate – 3 : To describe a circle with any centre and radius.
Postulate – 4 : That all right angles are equal to one another.
Postulate – 5 : If a straight line falling on two straight lines makes the interior angles on the same side of it taken together is less than two right angles, then the two straight lines, if produced infinitely, meet on that side on which the sum of the angles is less than two right angles.

AP Board 9th Class Maths Notes Chapter 3 The Elements of Geometry

→ Equivalent versions of Euclid’s fifth postulate:

  • Through a point not on a given line, exactly one parallel line may be drawn to the given line – John Play Fair (1748 – 1819).
  • The sum of angles of any triangle is a constant and is equal to two right angles (Legendre).
  • There exists a pair of lines everywhere equidistant from one another (Posidominus).
  • If a straight line intersects any one of two parallel lines, then it will intersect the other also (Proclus).
  • The statements that were proved to be true are called propositions or theorems.
  • The statements neither proved nor disproved are called conjectures.
  • There are non-Euclidian geometries.

AP SSC 10th Class Maths Notes Chapter 10 Mensuration

Students can go through AP SSC 10th Class Maths Notes Chapter 10 Mensuration to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 10 Mensuration

→ A solid is a geometrical shape with three dimensions namely length, breadth and height.
Eg:
AP SSC 10th Class Maths Notes Chapter 10 Mensuration 1

AP SSC 10th Class Maths Notes Chapter 10 Mensuration

→ A solid has two types of area namely,
a) Lateral Surface Area (L.S.A.)
b) Total Surface Area (T.S.A,)

→ In general, L.S.A. of a solid is the product of its base perimeter and height.
Eg : L.S.A. of a cuboid = 2h(l + b)
L.S.A. of a cylinder = 2πrh

→ The T.S.A. of a solid is the sum of L.S.A. and the areas of its top and base.
Eg : T.S.A. of a cylinder = 2πrh + 2πr2
= 2πr(r + h)

→ In general, the volume of a solid is the product of its base area and height.
V = A. h
Eg: Volume of a cube = a2 . a = a3
Volume of a cylinder = πr2 . h = πr2h

AP SSC 10th Class Maths Notes Chapter 10 Mensuration

→ The volume of solid formed by joining two basic solids is the sum of volumes of the constituents.

→ Surface area of the combination of solids: In calculating the surface area of the solid which is a combination of two or more solids, we can’t add the surface areas of all its constituents, because some part of the surface area disappears in the process of joining them.

→ Surface areas and volume of different solid shapes:
AP SSC 10th Class Maths Notes Chapter 10 Mensuration 2
AP SSC 10th Class Maths Notes Chapter 10 Mensuration 3

AP SSC 10th Class Maths Notes Chapter 10 Mensuration

→ Some solid figures and their combination shapes:
AP SSC 10th Class Maths Notes Chapter 10 Mensuration 4
AP SSC 10th Class Maths Notes Chapter 10 Mensuration 5

AP Board 9th Class Maths Notes Chapter 2 Polynomials and Factorisation

Students can go through AP Board 9th Class Maths Notes Chapter 2 Polynomials and Factorisation to understand and remember the concepts easily.

AP State Board Syllabus 9th Class Maths Notes Chapter 2 Polynomials and Factorisation

→ An algebraic expression ¡n which the variables involved have only non-negative integral powers is called a polynomial.
E.g.: 5x3 – 2x + 8

→ Polynomials can be classified as monomials. binomials, trinomials and polynomials based on the number of terms it contains.

→ A polynomial with a single term is a monomial.
E.g.: 2x, -5x2, \(\frac{6}{7}\)x3 etc.

→ A polynomial with two terms is a binomial.
E.g.: 2x + 5y; -3x2 + 5x etc.

→ A polynomial with three terms is a trinomial.
E.g.: 3x2 + 5x – 8; 3x + 2y – 5z etc.

→ In general a polynomial may contain two or more than two terms.
E.g.: 5x + 8x2 – 3x3 + 7

AP Board 9th Class Maths Notes Chapter 2 Polynomials and Factorisation

→ Degree of a polynomial ¡s the heighest degree of its variable terms.
E.g.: Degree of 3x2 + 4xy3 + y2 is 4.

→ Degree of a constant term is considered as zero.
E.g.: Degree of 4 is zero.

→ The general form of a polynomial is a0xn + a1xn-1 + a2xn-2 …….. + an-1x + an where a0, a1, a2,…… an-1, an are constants and a0 ≠ 0. Its degree is ‘n’.

→ Polynomials are again classified based on their degrees.
AP Board 9th Class Maths Notes Chapter 2 Polynomials and Factorisation 1

→ The zero of a polynomial p(x) is the value of x at which p(x) becomes zero (i.e.) p(x) = 0.
E.g.: Zero of 3x – 5 is x = \(\frac{5}{3}\)

→ To find the zero of a polynomial we equate the polynomial to zero and solve for the value of the variable.
E.g.: To find zero of 7x + 8.
7x + 8 = 0
7x = – 8
x = \(\frac{-8}{7}\)

→ Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number. If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a). This is called the Remainder theorem.
E.g.: If p(x) = 4x3 + 3x + 8 then the remainder when it is divided by x – 1 is p(1).
i.e., p(1) = 4 + 3 + 8 = 15.

→ If p(x) is a polynomial of degree n ≥ 1 and ‘a’ is any real number, then
(i) (x – a) is a factor of p(x), if p(a) = 0
(ii) and its converse “if (x – a) is a factor of a polynomial p(x) then p(a) = 0. This is called Factor theorem”.
E.g.: Let p(x) = x2 – 5x + 6 and p(2) = 22 – 5(2)+ 6 = 0 and hence (x – 2) is a factor of p(x) conversely; p(x) = x2 + 7x + 10 and (x + 2) is a factor, then p(-2) = 0.

AP Board 9th Class Maths Notes Chapter 2 Polynomials and Factorisation

→ Algebraic identities

  • (x + y)2 = x2 + 2xy + y2
  • (x – y)2 = x2 – 2xy + y2
  • (x + y) (x-y) = x2 – y2
  • (x + a) (x + b) = x2 + (a + b) x + ab
  • (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
  • (x + y)3 = x3 + y3 + 3xy (x + y)
  • (x – y)3 = x3 – y3 – 3xy (x – y)
  • (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = x3 + y3 + z3 – 3xyz

AP Board 9th Class Maths Notes Chapter 1 Real Numbers

Students can go through AP Board 9th Class Maths Notes Chapter 1 Real Numbers to understand and remember the concepts easily.

AP State Board Syllabus 9th Class Maths Notes Chapter 1 Real Numbers

→ Numbers of the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0 are called rational numbers, represented by ‘Q’.

→ There are infinitely many rational numbers between any two integers.
E.g.: 3 < \(\frac{19}{6}\), \(\frac{20}{6}\), \(\frac{21}{6}\), \(\frac{22}{6}\), \(\frac{23}{6}\), ……. < 4

→ There are infinitely many rational numbers between any two rational numbers.
E.g.: \(\frac{3}{4}\) < \(\frac{29}{8}\) < \(\frac{71}{16}\) < \(\frac{81}{14}\) ……. < \(\frac{13}{2}\)

→ To find the decimal representation of a rational number we divide the numerator of a rational number by its denominator.
E.g.: The decimal representation of \(\frac{5}{6}\) is
AP Board 9th Class Maths Notes Chapter 1 Real Numbers 1
∴ \(\frac{5}{6}\) = 0.833 …. = 0.8 \(\overline{3}\)

AP Board 9th Class Maths Notes Chapter 1 Real Numbers

→ Every rational number can be expressed as a terminating decimal or as a non-terminating repeating decimal. Conversely every terminating decimal or non¬terminating recurring decimal can be expressed as a rational number.
E.g.: 1.6 \(\overline{2}\) = \(\frac{161}{99}\)

→ A rational number whose denominator consists of only 2’s or 5’s or a combination of 2’s and 5’s can be expressed as a terminating decimal.
E.g. : \(\frac{13}{32}\) can be expressed as a terminating decimal (∵ 32 = 2 × 2 × 2 × 2 × 2)
\(\frac{7}{125}\) can be expressed as a terminating decimal (∵ 125 = 5 × 5 × 5)
\(\frac{24}{40}\) can be expressed as a terminating decimal (∵ 40 = 2 × 2 × 2 × 5)

→ Numbers which can’t written in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0, are called irrational numbers.
E.g.: √2, √3, √5,….. etc.
The decimal form of an irrational number is neither terminating nor recurring decimal.

→ Irrational numbers can be represented on a number line using Pythagoras theorem.
E.g.: Represent √2 on a number line.
AP Board 9th Class Maths Notes Chapter 1 Real Numbers 2

→ If ‘n’ is a natural number which is not a perfect square, then √n is always an irrational number.
E.g.: 2, 3, 5, 7, 8, …… etc., are not perfect squares.
∴ √2, √3, √5, √7 and √8 are irrational numbers.

AP Board 9th Class Maths Notes Chapter 1 Real Numbers

→ We often write π as \(\frac{22}{7}\) there by π seems to be a rational number; but π is not a rational number.

→ The collection of rational numbers together with irrational numbers is called set of Real numbers.
R = Q ∪ S

→ If a and b are two positive rational numbers such that ab is not a perfect square, then , √ab is an irrational number between ‘a’ and b’.
E.g.: Consider any two rational numbers 7 and 4.
7 × 4 = 28 is not a perfect square; then √28 lies between 4 and 7.
i.e., 4 < √28 < 1

→ If ‘a’ is a rational number and ‘b’ is arty irrational number then a + b, a – b, a.b or \(\frac{a}{b}\) is an irrational number.
E.g.: Consider 7 and √5 then 7 + √5, 7 – √5, 7√5 and \(\frac{7}{\sqrt{5}}\)= are all irrational numbers.

→ If the product of any two irrational numbers is a rational number, then they are said to be the rationalising factor of each other.
E.g.: Consider any two irrational number 7√3 and 5√3.
7√3 × 5√3 = 7 × 5 × 3 = 105 a rational number.
Also 7√3 × √3 = 21 – a rational number.
5√3 × √3 = 15 – a rational number.
So the rationalising factor of an irrational number is not unique.

→ The general form of rationalising factor (R.F.) of (a ± √b} is (a ∓ √b). They are called conjugates to each other.

→ Laws of exponents:

i) am × an = am+n
e.g.: 54 . 5-3 = 54+(-3) = 51 = 5

ii) (am)n = amn
e.g.: (43)2 = 43×2 = 46

iii)
AP Board 9th Class Maths Notes Chapter 1 Real Numbers 4 = am-n if (m > n)
= 1 if m = n
= \(\frac{1}{a^{n-m}}\) if (m < n)
AP Board 9th Class Maths Notes Chapter 1 Real Numbers 3

AP Board 9th Class Maths Notes Chapter 1 Real Numbers

iv) am . bm = (ab)m
e.g.:(-5)3 . (2)3 = (-5 × 2)3 =(-10)3

v) \(\frac{1}{a^{n}}\) = a-n
e.g.: (6)-3 = \(\frac{1}{6^{3}}\) = \(\frac{1}{216}\)

vi) a0 = 1
e.g.: \(\left(\frac{-3}{4}\right)^{0}\) = 1
Where a, b are rationals and m, n are integers.

→ Let a, b be any two rational numbers such that a = bn then b = \(\sqrt[n]{a}\) = \((\mathrm{a})^{1 / \mathrm{n}}\)
Here ‘b’ is called nth root of a.
e.g.: 42 = 16 then \(16^{1 / 2}\) or \(\sqrt[2]{16}\)
34 = 81 then 3 = \(\sqrt[4]{81}\) or \((81)^{1 / 4}\)

→ Let ‘a’ be a positive number and n > 1 then \(\sqrt[n]{a}\) i.e., nth root of a is called a surd.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has a higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has a higher boiling point (391 K) than that hydrocarbon butane (309K).

Reason: In propanol, strong intermolecular hydrogen bonding is present between the molecules. But in the case of butane weak van der Waals force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation: .

  • Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
  • Hydro carbons are non polar and these dorv.t form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer:
Given molecular formula of monnhydric phenols is C7H8O. The no. of possible isomers with molecular formula C7H8O are three.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 1

Question 4.
Give the reagents used for the preparation of phenol from chiorobenzene.
Answer:
Phenol is prepared from chiorobenzene as follows. Reagents required are

  1. NaOH, 623K, 300 atm
  2. HCl.
    Chemical reaction :
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 2

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Only 1° – alcohols form Ethers on acid dehydration. But not 2° or 3°-alcohols.

Reason : In case of 2° or 3° alcohols steric hindrance arises. Due to this steric hindrance alkenes are formed but not Ethers.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 6.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I: When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 3
Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 4

Question 7.
Name the reagents used in the following reactions.

  1. Oxidation of primary alcohol to carboxylic acid
  2. Oxidation of primary alcohol to aldehyde.

Answer:

  1. The reagents used for the oxidation of 1° – alcohols to carboxylicacid are acidified K2Cr2O77 (or) Acidic/alkaline KMnO4 (or) Neutral KMnO4
  2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH2Cl2.

Question 8.
Write the equations for the following reactions.
i) Bromination of phenol to 2,4, 6-tribromophenol
ii) Benzyl alcohol to benzoic acid.
Answer:
i) Bromination of phenól to 2, 4, 6 tribromophenol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 5

Question 9.
IdentIfy the reactant needed to form t-.butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 6

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 10.
Write the structures for the following compounds

  1. Ethoxyethane
  2. Ethoxybutane
  3. Phenoxyethane

Answer:

  1. Ethoxyethane → CH3 – CH2 – O – CH2 – CH3
  2. Ethoxybutane → CH3 – CH2 – O – CH2 – CH2 – CH2 – CH3
  3. Phenoxyethane → C6H5 – O – CH2 – CH3

Short Answer Questions

Question 1.
Draw the structures of all isomeric alcohols of molecular formula C5H12O2 and give their IUFAC names and classify them as primary, secondary and tertiary alcohols.
Answer:

  • Given molecular formula of compound is C5H12O.
  • It has eight isomeric alcohols.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 7
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 8

In the above isomeric alcohols (i), (ii), (iii), (iv) and 1°-alcohols; (v), (vi) and (viii) are 2°- alcohols, (vii) is 3°-alcohol.

Question 2.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give rèason.
Answer:
While separating a mixture of ortho and para nitrophenols by steam distillation, ortho nitrophenol is steam volatile.

Reason: In ortho nitrophenol intra molecular hydrogen bonding is present and in case of para nitrophenol inter molecular hydrogen bonding is present. So O-nitrophenol is steam volatile.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 9

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 3.
Give the equations for the preparation of phenol from Cumene. [Mar. 14]
Answer:
Phenol.is prepared from Cumene as follows.

  1. Oxidation of Cumene to Cumene hydroperoxide.
  2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 10

Question 4.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of Ethene to yield ethanol involves 3—step mechanism.
Step – 1: In step – 1 formation of carbocation takes place by the protonation of ethene.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 11
Step – 2 : In step – 2 carbocation formed in the above step attacked by water.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 12
Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 13

Question 5.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 14

  •  In phenol hydroxyl group is attached to the Sp2 hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

  • The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
  • The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
  • Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 15

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 6.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 16
b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 17

Question 7.
Ethanol with H2SO4, at 443K forms ethene while at 413 K it forms ethoxy ethane. Explain the mechanism.
Answer:
Case – 1: Ethanol reacts with Cone. H2SO4 at 443K forms ethene
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 18
Mechanism:
Step – 1: Formation of protonated alcohol
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 19
Step 2 : Formation of carbo cation
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 20
Step 3 : Formation of ethene by elimination of a proton
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 21
Case – II : Ethanol reacts with Cone. H2SO4 at 413 K to form ethoxy ethane.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 22
Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 23

Question 8.
Account for the statement: Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Explanation : Consider ethanol, propane and methoxy methane which are having comparable molecular masses. The boiling points, molecular masses and structures of the above compounds mentioned below.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 24
The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 9.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH3 influences + R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophihic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m—position. So 0-and P-products are mainly formed during electrophillic substitution reactions.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 25

Question 10.
Write the products of the following reactions :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 26
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 27
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 28

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 29
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 30

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
i) 2, Methyl butan—ol
ii) 1-Phenylprpan-2-ol
iii) 3, 5-Dhuethylhexane-1, 3, 5-triol
iv) 2, 3-Diethylphenol
v) 1-Ethoxypropane
vi) 2-Ethoxy-3-methylpentane
vii) Cyclohexylmethanol
viii) 3-Chloromethylpentan-1-ol
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 31
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 32

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 3.
Write the equations for the preparation of phenol using benzene, conc. H2SO4 and NaOH. [Mar. 14]
Answer:
The equations for the preparation of phenol using conc.H2SO4 and NaoH as follows
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 33

Question 4.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H2O2 to form alcohols. This reaction is called as hydroboration-oxidation reaction.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 34

Question 5.
Write the IUPAC name of the following compounds:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 35
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 36

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 6.
How will you synthesise:
i) 1 – Phenylethanol from a suitable alkene
ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction.
iii) Pentan-1-ol using a suitable alkyl halide.
Answer:
i) Synthesis of 1-phenylethanol from a suitable alkene : When styrene undergo hydrolysis in presence of dil.H2SO4 to form 1-phenyl ethanol. It is an example of Marknowni koff’s rule.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 37
ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN2 reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 38
iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 39

Question 7.
Explain why-
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
ii) OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
Explanation: .
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 40

  • -NO2 is an electron withdrawing group and -OCH3 is electron releasing group.
  • By the presence of electron withdrawing group the phenoxide ion is more stabilized. By the presence of electron releasing group the phenoxide ion is less stabilized.
  • Due to high stability of phenoxide ion, acidic nature increases.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 41

ii) The -OH group attached to benzene ring activates it towards electrophilic substitution.

Explanation : When an electrophile is attacked, the – OH group exerts +R effect on the benzene ring. So electrodensity in the ring increases at ortho and para positions. When an electrophile attacks, substitution takes place at O and p-positions. So benzene ring activates towards electrophilic substitution.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 42

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 8.
Wth a suitable example write equations for the following:. [T.S. MAr. 19, 18; A.P. Mar. 18, 16] [A.P. Mar. 16]
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
Answer:
i) Kolbes reaction: Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophilhic substitution with CO2to form salicylic acid.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 43

ii) Relmer-Tlemann reactIon : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction. .
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 44

iii) Wilhiamsons ether synthesis:

  • This method is used for the preparation of symmetrical and unsymmetrical ethers.
  • The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamsons Synthesis.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 45

Question 9.
How are the following conversions carried out?
i) Benzyl chloride to Benzyl alcohol
ii) Ethyl magnesium bromide to Propan-1-ol
iii) 2-Butanone to 2-Butanol
Answer:
i) Conversion of Benzyl chloride to Benzyl alcohol : Ben.zyl chloride reacts with aq. NaOH to form benzyl alcohol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 46

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 47

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H4 to form 2-Butanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 48

Question 10.
Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
i) 1-Propoxypropane
ii) Ethoxybenzene
iii) 2-Methoxy-2-methylpropane
iv) 1 -Methoxyethane
Answer:
i) Preparation of 1-propoxy propane :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 49
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 50

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 11.
How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
Answer:
1 – Proponal reacts with Conc. H2SO4 at 413 K to form 1-propoxy propane.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 51

Question 12.
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH3 influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophillic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m-position. So O-and P-products are mainly formed during electrophillic substitution reactions.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 52

Question 13.
Write equations of the below given reactions:
i) Alkylation of anisole
ii) Nitration of anisole
iii) Friedel-Crafts acetylation of anisole
Answer:
i) Friedel crafts Alkylation of anisole :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 53
ii) Nitration of anisole
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 54

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 55
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 56
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 57

Question 15.
Explain why phenol with bromine water forms 2,4,6-tribromophenol while on reaction with bromine in CS2 at low temperatures forms para-bromophenol as the major product.
Answer:
a) Phenol under goes Bromination in presence of CS2 to form p-bromophenol as major product.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 58
b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 59
Explanation: In bromination of phenol, the polarisation of Br2 molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Textual Examples

Question 1.
Give IUPAC names of the following compounds:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 60
Solution:
i) 4-Chloro-2, 3-dimethylpentan-1-ol
ii) 2-Ethoxypropane
iii) 2, 6-Dimethyiphenol
iv) 1-Ethoxy-2-nitrocyclohexane

Question 2.
Give the structures and IUPAC names of the products expected from the following reactions :
a) Catalytic reduction of butanal.
b) Hydration of propene in the presence of dilute sulphuric acid.
c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 61

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 3.
Arrange the following sets of compounds in order of their increasing boiling points :
a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Solution:
a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 62

Question 4.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3, 5-dinitrophenol, phenol, 4-methylphenol.
Solution:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 5.
Write the structures of the major products expected from the following reactions:
a) Mononitratlon of 3-methylphenol .
b) Dinitratlon of 3methylphenol
c) Mononitration of phenyl methanoate.
Solution:
The combined influence of -OH and -CH3 groups determine the position of the incoming group.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 63

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 64
i) What would be the major product of this reaction?
ii) Write a suitable reaction for the preparation of t-butylethyl ether.
Solution:
i) The major product of the given reaction is 2-methylprop-1-ene.
It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 65

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 66
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 67

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 68
Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (y)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi)

Question 3.
Name the following compounds according to IUPAC system.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 69
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 70
Answer:
i) 3-Chioremethyl 2-isopropylpentan-1-ol
ii) 2, 5-Dimethylhexane-1, 3-dial
iii) 3-Bromocyclohexanol
iv) Hex-1-en-3-ol
v) 2-Bromo-3-methylbut-2-en-1-ol.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 71
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 72

Question 5.
Write structures of the products of the following reactions :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 73
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 74
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 75

Question 6.
Predict the major product of acid catalysed dehydration of

  1. 1-methylcyclohexanol and
  2. butan-1-ol.

Answer:

  1. 1-Methylcyclohexene
  2. A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give,secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 76

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers

Question 8.
Predict the products of the following reactions:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 77
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 78