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Wednesday, May 31, 2017

Missing Letter puzzle 6 [With Answers]

1. Which letter replaces the question mark?
Answer: M

Explanation: In each diagram, start with the upper left box, and move anti-clockwise around the others. Letters advance through the alphabet in steps of 7 for the upper left diagram, 8 for the middle, then 9, then 10.

2. Which letter replaces the question mark?
Answer: B


Explanation: Starting at the top and working down, letters move through the alphabet in steps of 8 letters, then 9, then 10 etc.

3. Which letter replaces the question mark?
Answer:  K


Explanation: Split the diagram in half horizontally and vertically. Start in the top left square of each quarter, and move clockwise around the other squares in the quarter. In each quarter, letters advance through the alphabet in steps of 2 for the upper left quarter, 3 for the upper right, 4 for the lower right and 5 for the lower left.

4. Which letter replaces the question mark?
Answer: U

Explanation: Start with the D on the outer left of the diagram, and move clockwise around the outer segments, then start on the N anti-clockwise around the inner segments. Letters advance through the alphabet in steps of 3, then 4, 5, and 6, repeating this sequence.

5. Which letter replaces the question mark?
Answer:  J

Explanation: In each circle, the reverse alphabetical value of the lower letter equals the sum of the alphabetical values of the top 2 letters.

Preparation of Amines

1. Reduction of nitro compounds:
Reactants: Nitro compounds
Reagent: H2 in the presence of finely divided nickel, palladium or platinum
Product: Amines
Reactants: Nitro compounds
Reagent: Metals in acidic medium. Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction.
Product: Amines
This method cannot be used when the molecule also contains some other easily hydrogenated group, such as a Carbon carbon double bond.

2. Ammonolysis of alkyl halides:
Reactants: alkyl or benzyl halide
Reagent: ethanolic solution of ammonia
Mechanism: nucleophilic substitution reaction
Product: Amines
CH3CH2CH2Br + NH3CH3CH2CH2NH2 + HBr
This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis.
The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.
The free amine can be obtained from the ammonium salt by treatment with a strong base:
R - NH⁺₃X⁻ + NaOH → R - NH₂ + H₂O + NaX⁻
The order of reactivity of halides with amines is RI > RBr >RCl.

3. Reduction of nitriles: This reaction is used for ascent of amine series, i.e. for preparation of amines containing one carbon atom more than the starting nitrile.
Reactants: Nitriles
Reagent: lithium aluminium hydride (LiAlH4)
Reaction: Reduction
Products: Amines
Reactants: Nitriles
Reagent: H2 in presence of catalyst Ni
Reaction: Reduction
Product: Amines
4. Reduction of amides:
Reactants: Amides
Reagent: Lithium aluminium hydride (LiAlH4)
Reaction: Reduction
Product: Amines
Reactants: Amides
Reagent: mixture of base and bromine (KOH + Br2)
Product: Hofmann Bromamide reaction.
Product: Amines
The reaction is as follows, RCONH2 + Br2 + 4KOH → RNH2 + K2CO3 + 2KBr + 2H2O
Here, the amine formed has one carbon less than that of the corresponding amide. Due to the loss of carbon atom, this reaction is also called as Hofmann degradation of amides.
Base abstracts an acidic N-H proton, yielding an anion.
The anion reacts with bromine in an α-substitution reaction to give an N-bromoamide.
Base abstraction of the remaining amide proton gives a bromoamide anion.
The bromoamide anion rearranges as the R group attached to the carbonyl carbon migrates to nitrogen at the same time the bromide ion leaves, giving n isocyanate.
The isocyanate adds water in a nucleophilic addition step to yield a carbamic acid (aka urethane).
The carbamic acid spontaneously loses CO2, yielding the amine product.
Apart from this, amides can be dehydrated by P2O5 to their corresponding nitriles and nitriles can then be reduced.
By this method you are retaining the number of carbon atoms in both amide and the amine
5. Gabriel phthalimide synthesis:
Reactants: Pthalimide and alkyl halide
Reagent: Ethanolic potassium hydroxide  
Product: Primary amine
Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide
Phthalimide on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

6. Schmidt reaction (IIT):
Reactants: Hydrozoic acid and carboxylic acid
Reagent: Sulphuric acid
Product: Amines
7. Curtius Reaction (IIT):
Reactants: Acid chloride and sodium azides
Reagent: Heat and water
Product: Amines
The isocyanate formed by reaction of acid chloride with sodium azides is decomposed with treatment of water and amines are obtained.

Sunday, May 28, 2017

Missing letter puzzle 6

1. Which letter replaces the question mark?
2. Which letter replaces the question mark?
3. Which letter replaces the question mark?
4. Which letter replaces the question mark?
5. Which letter replaces the question mark?

Uniform circular motion

If the particle moves in the circle with a uniform speed, we call it a uniform circular motion. In this case, dv/dt = 0 and equation gives .

Thus, the acceleration of the particle is in the direction of that is, towards the centre. The magnitude of the acceleration is α = ω²r.



Thus, if a particle moves in a circle of radius r with a constant speed v, its acceleration is v²/r directed towards the centre. This acceleration is called centripetal acceleration. Note that the speed remains constant, the direction continuously changes and hence the “velocity” changes and there is an acceleration during the motion.


The linear acceleration is .

This acceleration is directed towards the centre of the circle.

Non uniform circular motion: If the speed of the particle moving in a circle is not constant, the acceleration has both the radial and the tangential components. According to equation, the radial and the tangential acceleration are 

Thus, the component of the acceleration towards the centre is ω²r = - v²/r and the component along the tangent (along the direction of motion) is dv/dt. The magnitude of the acceleration is .
The direction of this resultant acceleration makes an angle α with the radius where .

Saturday, May 27, 2017

Puzzle 6 [With Answers]

Puzzle 1: Pick the odd one out:

Crabcake, Laughing, Calmness, Canopy, Stupid, Hijack, First, Deft, Happy

Answer: Happy

Explanation: All the words except "happy" contain three consecutive letters of the alphabet.

Crabcake, Laughing, Calmness, Canopy, Stupid, Hijack, First, Deft

Puzzle 2: Fill in the missing pair:

BF   CH   ?   HO   LT

Answer: EK

Puzzle 3: If BDG is equal to CFJ, then find the value of EGJ

Answer: FIM

Puzzle 4: Fill in the sequence:

5   6   9   15   ?   40

Answer: 25

Puzzle 5: Fill in the blanks:

3   3   5   4   4   ?   5

Answer: 3

Explanation: Number of letters in one, two, three, four, five, six, seven and so on.

Ordinary D.F equations, their order and degree

Differential equation: An equation containing an independent variable, dependent variable and differential coefficient of dependent variable with respect to independent variable is called a differential equation.

Order of a differential equation: The order of a differential Equation is the order of the highest order derivative appearing in the equation.
For example, in the equation , the order of highest order derivative is 2. So. It is a differential equation of order 2.

The equation  is of the order 32, It is a differential equation of order 2.

Degree of a differential equation: The degree of a differential equation is the degree of the highest order derivative, when differential coefficient are made free from radicals and fractions.

Example I: Consider the differential equation
In this equation the number of highest order derivative is 2. So it is a differential equation of degree 1.

Example II: Consider the differential equation

In this equation, the order of the highest order derivative is 3. And its power its power is 2. So it is a differential equation of order 3 and degree 2.

Friday, May 26, 2017

Puzzles 6


Puzzle 1: Pick the odd one out:

Crabcake, Laughing, Calmness, Canopy, Stupid, Hijack, First, Deft, Happy

Puzzle 2: Fill in the missing pair:

BF   CH   ?   HO   LT

Puzzle 3: If BDG is equal to CFJ, then find the value of EGJ

Puzzle 4: Fill in the sequence:

5   6   9   15   ?   40

Puzzle 5: Fill in the blanks:

3   3   5   4   4   ?   5

Nucleic Acids

Definition:
The particles in nucleus of a cell, responsible for heredity, are called chromosomes which are made up of proteins and nucleic acids.
Nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.
Examples: Deoxyribonucleic acid (DNA) & Ribonucleic acid (RNA).

Chemical Composition of Nucleic Acids:
Each nucleotide has three components: a 5 - carbon sugar, a phosphate group, and a nitrogenous base.
In DNA molecules, the sugar moiety is β - D - 2 - deoxyribose where as in RNA molecule, it is β -  D - ribose.
DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T).
RNA also contains four bases, the first three bases are same as in DNA but the fourth one is uracil (U).

Structure of Nucleic Acids:
A unit formed by the attachment of base to 1’ position of sugar is known as nucleoside, the sugar carbons are numbered as 1’, 2’, 3’, etc. in order to distinguish these from bases. When nucleoside is linked to phosphoric acid at 5’ - position of sugar moiety we get a nucleotide.
Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of pentose sugar.
A simplified version of nucleic acid chain is as shown below.
Information regarding the sequence of nucleotides in the chain of nucleic acid is called primary structure. Nucleic acids have a secondary structure also.
James Watson and Francis Crick gave a double strand helix structure for DNA. Two nucleic acid chains are wound about each other and held together by hydrogen bond between pairs of bases.
These are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

DNA Fingerprinting:
Every individual has unique fingerprints.
A sequence of bases on DNA is also unique for a person and information regarding this is called DNA fingerprinting.
It is same for every cell and cannot be altered by any known treatment.

Uses of DNA finger printing:
i) In forensic laboratories for identification of criminals.
ii) To determine paternity of individual
iii) To identify racial groups to rewrite biological evolution.

Biological functions of Nucleic acids:
DNA is the chemical basis of heredity and may be regarded as the reserve of genetic information.
DNA is responsible for maintaining the identity of different species of organisms over millions of years.
A DNA is capable of self-duplications during cell division and identical DNA strands are transferred to daughter cells.
Nucleic acids are responsible for protein synthesis in the cell.