Engineering Chemistry Notes Dr. Bathini Srinivas Ph.D (JNTUH) Unit I: Molecular structure and Theories of Bonding Topic1: Molecular orbitals Q1) Define degenerate orbitals? Give the examples. Ans. Orbital’s which is having same energy is called degenerate orbital’s. Examples Px, Py, Pz orbital having same energy and dxy, dyz, dzx, dx2-y2, dz2 orbital’s having same energy ,these orbital’s are called degenerate orbital’s. Q2) Write the difference between the molecular orbitals and atomic orbitals? Differences between Molecular Orbital and Atomic Orbital Molecular Orbital Atomic Orbital 1. An electron Molecular orbital is under the influence of two or more nuclei depending upon the number of atoms present in the molecule. 2. Molecular orbitals are 1. An electron in atomic orbital is under the influence of only one positive nucleus of the atom. 2. Atomic orbitals are inherent property of an atom. 3. They have simple shapes.
Transcript
Engineering Chemistry Notes
Dr. Bathini Srinivas
Ph.D (JNTUH)
Unit I: Molecular structure and Theories of Bonding
Topic1: Molecular orbitals
Q1) Define degenerate orbitals? Give the examples.
Ans. Orbital’s which is having same energy is called degenerate orbital’s. Examples Px, Py, Pz orbital having same energy and dxy, dyz, dzx, dx2-y2, dz2 orbital’s having same energy ,these orbital’s are called degenerate orbital’s.
Q2) Write the difference between the molecular orbitals and atomic orbitals?
Differences between Molecular Orbital and Atomic Orbital
Molecular Orbital Atomic Orbital
1. An electron Molecular orbital is
under the influence of two or more
nuclei depending upon the number
of atoms present in the molecule.
2. Molecular orbitals are formed by
combination of atomic orbitals
3. They have complex shapes.
1. An electron in atomic orbital is under the
influence of only one positive nucleus of
the atom.
2. Atomic orbitals are inherent property of an
atom.
3. They have simple shapes.
Q 3) Draw the shapes and structures of S, P,d orbital’s ?
S Orbital shape spherical, P orbital shape dumbel
Figure: d orbital shape double dumbel
Explain about molecular orbital diagram of N2 Molecule?
Nitrogen:
This molecule has ten electrons. The atomic orbitals combine to produce the following molecular
is paramagnetic or diamagnetic will be determined by the spin state. If there are unpaired
electrons, the complex is paramagnetic; if all electrons are paired, the complex is diamagnetic.
Q 10. Why do transition metal compounds forms complexes?
Sol. The transition metals are almost unique in their tendency to form coordination complexes. The tendency of cations of transition elements to form complexes is due to two factors;
1. These ions are very small in size and, therefore, have high positive charge density. This facilitates acceptance of lone pair of electrons from other molecules.
2 .They have vacant orbitals and these orbitals have the appropriate type of energy to accept lone pair of electrons.
Topic1. Hardness of waterQ1. Define hardness of water? What are the causes of hardness?
Hardness of water:
When soap comes in contact with hard water, sodium stearate will react with dissolved Ca and Mg salts and produce Ca-stearate or Mg- stearate which is white precipitate.
increase in stream production rate and (vi) The high levels of water in boilers.
Prevention of priming: - The priming is avoided by
(i)Fitting mechanical steam purifiers
(ii)Avoiding rapid change in steaming rate
(iii) Maintaining low water levels in boilers and
(iv)Efficient softening and filtration of boiler feed water.
b). Foaming: - Formation of stable bubbles at the surface of water in the boiler is calling
foaming. More foaming will cause more priming. It results with the formation of wet steam that
harms the boiler cylinder and turbine blades. Foaming is due to the presence of oil drops, grease
and some suspended solids.
Prevention of Foaming: Foaming can be avoided by
(1)Adding antifoaming chemicals like castor oil. The excess of castor oil addition can cause
foaming.
(2) Oil can be removed by adding sodium aluminates or alum.
(3) Replacing the water concentrated with impurities with fresh water.
Q 9. Explain the reasons of caustic embrittlement?
Caustic embrittlement:
Caustic embrittlement is a term used for the appearance for cracks inside the boiler, particularly
at those places which are under stress such as riverted joints due to the high concentration of
alkali leading to the failure of the boiler. The cracks have appearance of brittle fracture. Hence,
the failure is called 'caustic embrittlement'.
Reasons for the formation of caustic embrittlement:
During the softening process by lime soda process, free Na2CO3 is usually present in small
portion in soft water which decomposes to give sodium hydroxide and CO2 at high pressure of
the boilers.
Na2CO3 + H2O → 2NaOH + CO2
The precipitation of NaOH makes the boiler water 'caustic'. The NaOH containing water flows
into small pits and minute hair cracks present on the boiler. As the vapour evaporates, the
concentration of caustic soda increases progressively creating a concentration cell as given
below, thus dissolves in the iron of boiler as sodium ferrate.
(-) Iron at
bends, rivets and
joints
/ Concentrated
NaOH solution // Dilute NaOH
solution / Iron at plane (+)
surfaces
The iron at plane surfaces surrounded by dilute NaOH becomes cathodic while the iron at bends,
rivets, joints are surrounded by highly concentrated NaOH becomes anionic which consequently
decayed or corroded. The cracking of the boiler occurs particularly at stress parts like bends,
joints, rivets etc., causing the failure at bolier.
Thus, the cracks present at such places are intercrystalline, irregular running from one rivet to
another without joining each other. These cracks have the appearance of brittle fracture, hence,
known as caustic embrittlement.
Preventions of caustic embrittlement:
a) By using sodium phosphate as softening reagent instead of sodium carbonate, disodium
hydrogen phosphate is the best softening reagent because it not only forms complex with
Ca+2 and Mg+2 resulting the softening of water but also maintains pH of water 9-10. The
Phosphates used are trisodium phosphate, sodium dihydrogen phosphate etc.
b) By adding tanning or legnin to boiler water, which blocks the hair cracks and pits that are
present on the surface of the boiler plate, preventing the infiltration of the caustic soda
solution
c) By adding sodium sulphate to boiler water, which also blocks the hair cracks and pits that are
present on the surface of the boiler plate, preventing the infiltration of the caustic soda
solution. The amount of sodium sulphate added to the boiler be in the ratio [Na2SO4 conc. /
NaOH conc.] kept as 1:2 , 2:1 and 3:1 in boilers working as pressures upto 10, 20 and above
30 atmospheres respectively.
Disadvantages of caustic embrittlement:
The cracking or weakening of the boiler metal causes the failure of boiler.
Q10. Discuss the sludge and scale formation in boilers?
In boilers more water is removed in form of steam during boiling, hence boiler water gets
concentrated with dissolved salts and reaches saturation point. At this point the dissolved salts
precipitated and slowly settle on the inner walls of the boiler plate. The precipitation takes place
in two ways
a). Sludge formation:
The precipitation in the form of soft loose and slimy deposits formed comparatively in the colder
portions of boiler is called sludge.
Reasons for the formation of sludge:
The dissolved salts whose solubility is more in hot water and less in cold water produce sludges.
Ex: MgCO3, MgCl2, CaCl2, and MgSO4. The sludges were formed at comparatively colder
portions of the boiler and get collected where rate of flow of water is low.
Disadvantages of sludges:
i) Sludges are bad conductors of heat and results in wastage of heat and fuel.
ii) Sometimes sludges were entrapped in the scales and are deposited as scale, which causes more loss of efficiency of the boiler.
iii) Excessive sludge formation leads to the settling of sludge in slow circulation areas such as pipe connections, plug openings leading to the chocking of pipes.
Prevention of sludge formation:
i) By using soft water which is free from dissolved salts like MgCO3, MgCl2, CaCl2, and MgSO4
can prevent sludge formation.
b) Scale formation:
The precipitation in the form of hard deposits, stick very firmly on the inner walls of the boiler is called scales.
Reasons for the formation of sludge:
i) Decomposition of Ca(HCO3)2:
Due to the high temperature and pressure present in the bolers, the Ca(HCO3)2 salt decomposes
to CaCO3 ↓, an insoluble salt, forms scale in low pressure boilers.
Ca (HCO3)2 → CaCO3 + H2O + CO2
ii) Decomposition of CaSO4
CaSO4 is more soluble in cold water hence its solubility decreases and precipitates out to
produce hard scale on the surface of the boiler. The solubility of CaSO4 is 3200ppm at 150C,
reduces to 27ppm at 3200C and completely insoluble in super heated water. CaSO4 scale is very
hard, highly adherent and difficult to remove.
iii) Hydrolysis of magnesium salts:
Dissolved magnesium salts undergo hydrolysis at high temperature prevailing inside the boiler forming magnesium hydroxide precipitate, which form salt type of scale.
MgCl2 + 2H2O → Mg (OH)2 ↓ + 2HCl
iv) Presence of silica:
Silica present in small quantities deposits as calcium silicate (CaSiO3) or Magnesium silicate (MgSiO3). The deposits form hard scale very difficult to remove.
Disadvantages of scale formation:
i) wastage of fuel: Scales are bad conductors of heat due to which the flow of heat to inside
water is decreased hence excessive heating is required which increases fuel consumption causing
wastage of fuel. The wasage of fuel increases with increases of the thichness of the scale.
ii) Lowering the boiler safety: Due to scale formation overheating of the boiler done to maintain
the constant supply of steam. due to over heating the boilermaterial become softer and weaker,
which causes distortion of boiler.
iii) Decrease in efficiency: Scales deposited in the valves and condensers of the boiler cause
chocking which results in decrease in efficiency of the boiler.
iv) Danger of explosion: Because of the formation of the scales, the boiler plate faces higher
temperature outside and lesser temperature inside due to uneven heat transfer resulting cracks in
the layer of scales. Water passes through the crack and comes in contact with boiler plate having
high temperature. This causes formation of large amount of steam suddenly developing sudden
high pressure. This causes the explosion of the boiler.
Removal scales:
i) If the scale formed is soft it can be removed by a scrapper, wire brush etc.
ii) By giving thermal shocks done by heating the boiler to high temperature and suddenly
cooling with cold water, if the scale is brittle in nature.
iii) If the scale is very adherent and hard, chemical treatment must be given. Ex: CaCO 3 scale is
removed by washing the boiler plate with EDTA solution.
iv) Frequent blowdown operation can remove the scales, which are loosely adhering.
Prevention of Scale formation:
Scale formation can be prevented by softening water by following methods.
Unit-III
ELECTRO CHEMISTRY AND CORROSION
Topic 1: Galvanic Cells
Q 1.What is galvanic cell? Explain the construction and reactions of Daniel cell?
Galvanic cell:
Galvanic cell is a device in which chemical energy is converted into electrical energy.
These cells are called Electrochemical cells or voltaic cells. Daniel cell is an example for
galvanic cell.
This cell is made up of two half cells. One is oxidation or anodic half cell. The other is
reduction or catholic half cell. The first half cell consists of ‘Zn’ electrode dipped in
ZnSO4solution and second half cell consists of ‘Cu” electrode dipped in Cuso4 solution. Both
the half cells are connected externally by metallic conductor. And internally by ‘salt bridge’ salt
bridge is a U- tube containing concentrated solution of Kcl or NH4 NO3 in agar-agar gel
contained porous pot. It provides electrical contact between two solutions.
The following reactions take place in the cell.
At cathode:
Zn → Zn+2 +2e- (oxidation or de-elecronation)
At cathode:
Cu+2 +2e- → Cu (Reduction or electronatioin)
The movement of electrons from Zn to cu produces a current in the circuit.
K
The overall cell reaction is: Zn +Cu+2 → Zn+2 +Cu
The galvanic cell can be represented by
Zn / ZnSO4 // CuSO4/ Cu
The passage of electrons from one electrode to other causes the potential difference between
them which is called E.M.F.
Q2. Derive the Nernst equation? Explain the application of this equation?
Ans: We have considered only standard reduction potentials, which refer to solution
concentrations of 1M. It is common in the laboratory to work with solutions of lower
concentrations, and reduction potentials depend on the concentration of the solutions in the
electrochemical cells. The dependence is given by the Nernst equation. Temperature is another
variable in the equation, although normally experiments will be carried out at a specified
Oxidation half reaction, which results in fall of concentration of Cl- ions
If it acts as cathode, it involves reduction
Hg2Cl2 → Hg22+ + 2Cl−
Hg22+ + 2e− → 2Hg
--------------------------------
Hg2Cl2 + 2e−→ 2Cl−+ 2Hg
--------------------------------
Reduction half reaction, which results increase in concentration of Cl- ions. Thus, Calomel electrode is
reversible to Cl- ions. The reduction potential of calomel electrode is given by
Since [Hg] = [Hg2Cl2] = 1 and [Cl-] ≈ 4M, then ESCE = 0.242 V
The electrode potential of any other electrode on hydrogen scale can be measured when it is
combined with calomel electrode. The emf of such a cell is measured. From the value of
electrode potential of calomel electrode, the electrode potential of the other electrode can be
evaluated.
Advantages:
1. Its construction is very easy
2. Results of cell potential measurements are reproducible.
Disadvantages:
Since Hg2Cl2 breaks at 500C, it can’t be used above this temperature.
Q4. Differentiate metallic and electrolytic conductors?
Ans:
Metallic conductors Electrolytic conductors
1. Conductance is due to the flow of
electrons.
2. It does not result any chemical change.
3. Metallic conduction decreases with
increase in temperature.
4. It does not involve any transfer of matter.
1. Conductance is due to the movement of
ions in a solution.
2. Chemical reactions take place at the
electrodes.
3. Electrolytic conduction increases with
increase in temperature.
4. It involves transfer of matter.
Q5. Give the differences between Primary and Secondary cells?
Differences between Primary Secondary cells
1. These are non-rechargeable and meant
for a single use and to be discarded after
1. These are rechargeable and meant for
multi cycle use.
use.
2. Cell reaction is not reversible.
3. Cannot be rechargeable.
4. Less expensive.
5. Can be used as long as the materials are
active in their composition.
Eg: Leclanche cell, ‘Li’ Cells.
2. Cell reaction can be reversed.
3. Can be rechargeable.
4. Expensive.
5. Can be used again and again by
recharging the cell.
Eg; Lead- acid cell, Ni-cd cells.
Q6. Define the terms Equivalent conductance? With units.
Ans: It is defined as the conductance of all ions produced by the dissociation of Igm equivalent
of an electrolyte dissolved in certain volume ‘V’ of the solvent at const temperature
Units = = Ohm-1 cm2 eq
Q7. What is EMF of cell? How the emf of cell is calculated? Remember b E.M.F:-
Ans: The difference of potential which causes flow of electrons from an electrode of higher
potential to an electrode of lower potential is called Electro motive force (EMF) of the cell.
The E.M.F of galvanic cell is calculated by the reduction half – cell potentials using to following
ex. Ecell = E (right) - E(left)
Ecell EMF of the cell.
Eright reduction potential of right hand side electrode.
Eleft reduction potential of left hand side electrode.
Applications of EMF measurement:-
1. Potentiometric titrations can be carried out.
2. Transport number of ions can be determined.
3. PH can be measured.
4. Hydrolysis const, can be determined.
5. Solubility of sparingly soluble salts can be found.
Q8. The resistance of 0.1 N solution of an electrolyte is 40 ohms. If the distance between the
electrodes is 1.2cm and the area of cross-section is 2.4cm. Calculate the equivalent
conductivity.
Distance between electrodes l = 1.2cm
Area of cross-section a = 2.4cm2
Cell const. = 0.5cm-1
Normality of given solution = 0.1 N.
Resistance R = 40 ohms.
Specific conductance K = 0.0125
Equivalent Conductivity = 125 ohm-1cm2 eq-1
Q 9: Calculate the emf for the cell,
Ans: Zn/Zn+ // Ag+ /Ag given E0Zn+/Zn+2 / Zn = 0.762v and E0Ag+/Ag = 0.8 v
Given cell is Zn /Zn+2 //Ag+/Ag.
E0 = Zn+2/Zn = 0.762 v
E0 = Ag+/Ag =0.8 v
E0cell = E0right – E0left
= 0.8 – (-0.762) = 1.562 v.
Q10 : Calculate the cell Constant of a cell having a solution of concentration N/30 gm eq /li
of an electrolyte which showed the equivalent conductance of 120 Mhos cm2 eq- 1,
resistance 40 ohms.
Ans: Resistance R = 40 ohms.
Equivalent conductance of solution (A) = 120 mho cm2eq-1
Concentration of sol. N= gm eq/li
=0.033N.
Cell const =?
Equivalent conductance =
Specific cond. (K) = 0.00396
Cell constant = s p. conductance x Resistance
= 0.00396 x 40
= 0.1584 cm-1
Q 11. Write brief about working and construction of Glass Electrode?
Ans: Glass Electrode:
Credit for the first glass sensing pH electrode is given to Cremer, who first described it in his
paper published in 1906. Later several groups contributed for development of different ion
selective electrodes An Ionselective electrode (ISE) (also known as a specific ion electrode, or
SIE) is a sensor which converts the activity of a specific ion dissolved in a solution into an
electrical potential which can be measured by some potentio-metric devise like a voltmeter or pH
meter. As we know, the emf is theoretically dependent on the logarithm of the ionic activity
(concentration), in accordance with the Nernst equation. Basically a concentration cell is
developed with respect to the ion under observation. The sensing part of the electrode is usually
Made as an ion-specific membrane which is coupled with a reference electrode. So we need to
have different ISE s for different ions.
Glass Electrode: Most often used pH electrodes are called glass electrodes and belong to the
family of ISEs. They are sensitive only to H+ ions. Typical glass electrode is made of glass tube
engaged with small glass bubble sensitive to protons. Inside of the electrode is usually filled with
buffered solution of chlorides in which silver wire covered with silver chloride is immersed. pH
of internal solution varies- for example it can be 1.0(0.1M HCl) or 7.0 Active part of the
electrode is the glass bubble. While tube has strong and thick walls, bubble is made to be as thin
as possible.
Surface of the glass is protonated by both internal and external solution till equilibrium is
achieved. Both sides of the glass are charged by the adsorbed protons, this charge is responsible
for potential difference. This potential in turn is described by the Nernst equation and is directly
proportional to the pH difference between solutions on both sides of the glass.
The majority of pH electrodes available now a day are combination electrodes that have both
glass H+ ion sensitive electrode and reference electrode compartments, conveniently placed in
one housing.
Range of a pH glass electrode
The pH range at a constant concentration can be divided into 3 parts
Useful Working Range: Dependence of potential on pH has linear behavior and within which
such electrode really works as ion-selective electrode for pH and obeys Nernst equation.
Alkali error range: At very low concentration of hydrogen-ions (high values of pH) metal ions
interfere. In this situation dependence of the potential on pH become non-linear.
Acidic error range: At very high concentration of hydrogen-ions (low values of pH) the anions
plays a big role as interfering ions, in addition to destruction of the glass matrix used for making
glass bulb. Almost all commonly used glass electrodes have a working.
pH range from pH = 1 till
pH = 12. So specially designed electrodes should be used only for working in aggressive
conditions.
Unit IV: Stereochemistry, Reaction Mechanism and synthesis of drug molecules
Topic 1 StereochemistryQ1. Define Enantiomers and Diastereomers?
A). Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer. Diastereomerism occurs when two or more stereoisomer’s of a compound have different configurations at one or more (but not all) of the equivalent (related) stereo centers and are not mirror images of each other.
Enantiomers are chiral molecules that are mirror images of one another. Furthermore, the molecules are non-super imposable on one another. ... It is sometimes difficult to determine whether or not two molecules are Enantiomers. The enantiomers differ only in their spatial arrangements at the stereo center.
Consider 2-bromo-3-chlorobutane, which has stereo centers at C2 and C3. In general, a molecule with n stereo centers has 2n stereo isomers, so there are a total of four possibilities for 2-bromo-3-chlorobutane:
Q2. Indicate, with a suitable diagram, the potential energy changes during rotation about C (2) -C (3) bond of n-butane.
Q3. Name the elements of symmetry. Discuss, with the help of an example, an optically active compound without chirality.
The compound that has a plane of symmetry will show optical activity. The compound has to be non-planar. There are some compounds, which do not have a chiral carbon, that show optical activity. The best example is biphenyls. Take the example of the one above (the picture ). It should have been a planar compound ( obviously, each carbon on the benzene ring is sp2 hybridized) but, because of the repulsion between the two NO2 groups attached ( it is a big group and their electron clouds repel), one of the NO2 moves out of the plane, thus making the compound optically active. This is how a compound without chiral carbon becomes optically active.
Each of the four stereo isomers of 2-bromo-3-chlorobutane is chiral. There are two pairs of enantiomers. Any given molecule has its enantiomer; the two other molecules are its diastereomers.
An absolute configuration refers to the spatial arrangement of the atoms of a chiral molecular entity (or group) and its stereo chemical description e.g. R or S, referring to Rectus, or Sinister, respectively.
The precise arrangement of substituent’s at a stereogenic center is known as the absolute configuration of the molecule.
The arrangement of atoms in an optically active molecule, based on chemical interconversion from or to a known compound, is a relative configuration. Relative configurations of optically active compounds by chemically interconnecting them.
Topic 2: Reaction Mechanism and synthesis of drug moleculesQ5. Explain Reduction of carbonyl compound using LiAlH4
LiAlH4 is one of the good reducing agent , it reduce the formaldehyde to methyl alcohol, aldehyde to primary alcohol , ketones to secondary alchol
Reaction type: Nucleophilic Addition
Q6. Write about Electrophilic Addition Reactions?
A. Electrophillic addition :In organic chemistry, an electrophilic addition reaction is an addition reaction where, in a chemical compound, a π bond is broken and two new σ bonds are formed. The substrate of an electrophilic addition reaction must have a double bond or triple bond.
The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.
In step 2 of an electrophilic addition, the positively charged intermediate combines with (Y) that is electron-rich and usually an anion to form the second covalent bond.
Step 2 is the same nucleophilic attack process found in an SN1 reaction. The exact nature of the electrophile and the nature of the positively charged intermediate are not always clear and depend on reactants and reaction conditions.
In all asymmetric addition reactions to carbon, regioselectivity is important and often determined by Markovnikov's rule. Organoborane compounds give anti-Markovnikov additions. Electrophilic attack to an aromatic system results in electrophilic aromatic substitution rather than an addition reaction.
The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.
Q7. What is Aspirine? Write any two pharmaceutical applications.
Ans. Aspirin is a non-steroidal anti-inflammatory drug .Aspirin, or acetylsalicylic acid (ASA), is commonly used as a pain reliever for minor aches and pains and to reduce fever. It is also an anti-inflammatory drug and can be used as a blood thinner.People with a high risk of blood clots, stroke, and heart attack can use aspirin long-term in low doses.
Q8. What is the Markownikoff rule?
Ans. Markovnikov’s Rule: This rule states that an addition reaction of an asymmetric alkene by an asymmetric reagent the negative part of the reagent attacks the carbon containing the less number of hydrogen across the double bond.
Markovnikov’s rule is mostly carried out using HBr as it is a good reagent for this process, neither highly exothermic nor highly endothermic.
We obtain two products 2-Bromopropane and 1-Bromopropane.
according to Markovnikov’s rule the major product the major product will be decide by carbo cation stability, we see that between 2-Bromopropane and 1-Bromopropane if we remove the bromine atom, 2-Bromopropane will have a 2°2°carbocation and 1-Bromopropane will have 1°1° carbo cation. Hence the major product will be 2-Bromopropane.
2. Show, how SN2 reaction give rise to inverted product
In this rate of reaction depends on the concentration of substrate as well as concentration reagent. This is single step mechanism , Nucleophilic addition and elimination takes place simultaneously ,and inversion of the molecule takes place , this inversion is called Walden inversion .
Q9. Write the structure and synthesis of Paracetamol?
The original method for production involves the nitration of phenol with sodium nitrate gives a mixture of two isomers, from which the wanted 4-nitrophenol (bp 279 °C) can easily be separated by steam distillation. In this electrophilic aromatic substitution reaction, phenol's oxygen is strongly activating, thus the reaction requires only mild conditions as compared to nitration of benzene itself. The nitro group is then reduced to an amine, giving 4-aminophenol. Finally, the amine is acetylated with acetic anhydride.
10. Explain Reduction of carbonyl compound using LiAlH4
LiAlH4 is one of the good reducing agent , it reduce the formaldehyde to methyl alcohol, aldehyde to primary alcohol , ketones to secondary alchol
Step 1: The nucleophilic H in the hydride reagent adds to the electrophilic C in the polar carbonyl group in the aldehyde, electrons from the C=O move to the O creating an intermediate metal alkoxide complex. (note that all 4 of the H atoms can react)
Step 2: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex.
Unit V: Spectroscopic Techniques and ApplicationsTopic 1: IR spectroscopyQ1. IR spectra is often characterized as molecular finger-prints. Comment on it.
IR spectrum showing fingerprint region
The region to the right-hand side of the diagram (from about 1500 to 500 cm-1) usually contains a very complicated series of absorptions. These are mainly due to all manner of bending vibrations within the molecule. This is called the fingerprint region.
Q2. What is λmax in UV-visible spectrum?
The λmax in UV-visible spectrum is the wave length where the molecule has maximum absorption of light (maximum absorption coefficient)
UV-visible spectrum
Q3: A solution of X of concentration 0.010 mol dm–3 gives an absorbance of 0.5. What concentration is a solution of X which gives an absorbance reading of 0.25? Assume that the same optical cell is used for both readings.
Answer: Solution X concentration is (C1) = 10-2 mol dm-3
Absorbance (A1) = 0.5
For the same molecule if Absorbance (A2) is = 0.25
Than the concentration C2 is =?
From Beer Lamberts A = εCl
We can write A1/A2 = C1/C2
By substituting above values
0.5/0.25 = 10-2/C2
i.e C2 = 0.5x10-2 mol. dm-3
Q4. The four central lines in the high resolution υ =1←υ = 0 infrared spectrum of HCl37 occur at 2837.6, 2858.8, 2899.2 and 2918.6 cm-1. Deduce as much as possible about the molecule. Would the corresponding lines in HCl35 lie at the same spectral positions?
Answer: The band centre is at the average of the two central lines, i.e. 2879.0 cm-1.
This is equal to the fundamental frequency, ω0.
The 4B separation at the centre is 40.4 cm-1, giving a value of 10.1 cm-1 for B.
The moment of inertia is (I) = h/8π2Bc = 2.771x10-47 kg m2
The reduced mass is given by
Therefore,
Since the positions of the lines depend on the reduced mass, the lines for HCl35 will be at a different position. The reduced mass of HCl35 is smaller and hence its moment of inertia is smaller and rotational constant larger, so the lines will have larger separation. The fundamental vibration frequency will also be higher since it is inversely proportional to the square root of the reduced mass.
Q5. Explain batho-, hipso-, hyper- and hypochromic shifts with neat diagramme
Changes in chemical structure or the environment lead to changes in the absorption spectrum of molecules and materials. There are several terms that are commonly used to describe these shifts that you will see in the literature, and with which you should be familiar.
Bathochromic: a shift of a band to lower energy or longer wavelength (often called a red shift).
Hypsochromic: a shift of a band to higher energy or shorter wavelength (often called a blue shift).
Hyperchromic: an increase in the molar absorptivity.
Hypochromic: an decrease in the molar absorptivity.
UV-visible spectrum showing red, blue, hyper and hypo chromic shifts
Q6. Deduce the Beer-Lambert law for absorptivity and concentration.
Answer: According to this equation the Absorbance (A) is directly proportional to concentration (C) of the solution
i.e A α C ------------------------(i)
and the Absorbance is also directly proportional to path length of the light travelled (l)
i.e A α l ------------------------(ii)
From equation (i) and (ii)
A α Cl ------------------------(iii)
The equation can be written as
A = εCl
(where ‘ε’ is Absorptivity constant )
Topic2: Rotational and NMR spectroscopy
Q7. How many different types of H-atom environments are present in methyl alcohol? Also mention the ratio of peak areas due to –CH3 group and –OH group in NMR spectrum
Methanol NMR shows two types of peaks one belongs to –CH3 and another belongs to –OH The peak area ratios (number of hydrogens present on functional group -CH3: Number of hydrogens present on functional group –OH) i.e 3: 1
Q 8: What is Magnetic resonance imaging (MRI)? Describe the applications of MRI.
Magnetic resonance imaging is a scan that produces detailed pictures of organs and other internal body structures while a CT scan forms images inside of the body. CT scans use radiation, which may be harmful to the body, while MRIs do not. MRI’s cost more than CT scans
Q 9: What type of information is obtained by studying the UV, IR, H1-NMR.
Nuclear Magnetic Resonance (NMR) Spectroscopy is one of the most useful analytical techniques for determining the structure of an organic compound
UV/Vis spectroscopy is an absorption spectroscopy technique that utilizes electromagnetic radiation in the 10 nm to 700 nm range. The energy associated with light between these wavelengths can be absorbed by both non-bonding n-electrons and π-electrons residing within a molecular orbital.
The wavenumbers of the absorbed IR radiation are characteristic of many bonds, so IR spectroscopy can determine which functional groups are contained in the sample.
Q 10.What is the principle of Nuclear Magnetic Resonance (NMR) spectroscopy?
Answer: Nuclear magnetic resonance is defined as a condition when the frequency of the rotating magnetic field becomes equal to the frequency of the processing nucleus.
Principle of NMR:
The principle of nuclear magnetic resonance is based on the spins of atomic nuclei. The magnetic measurements depend upon the spin of unpaired electron whereas nuclear magnetic resonance measures magnetic effect caused by the spin of protons and neutrons. Both these nucleons have intrinsic angular momenta or spins and hence act as elementary magnet.
The existence of nuclear magnetism was revealed in the hyper fine structure of spectral lines. If the nucleus with a certain magnetic moment is placed in the magnetic field, we can observe the phenomenon of space quantization and for each allowed direction there will be a slightly different energy level.
Tutorial topicsQ1) What is the difference between temporary and permanent hardness of water?
Types of Hardness: Hardness in water is of two types.
(1) Temporary hardness and (2) permanent hardness
2. Temporary hardness: Temporary hardness is due to presence of dissolved bicarbonates of
calcium and magnesium salts present in water.
By boiling Temporary hardness can be removed [bicarbonates converts into carbonates as
precipitate].
Ex: Ca(HCO3)2 CaCO3↓+ CO2 +H2O
Mg(HCO3)2 Mg(OH)2↓+ 2CO2
Permanent Hardness: Permanent hardness is due to presence of chlorides [Cl-], sulphates
[SO42-] and Nitrates [NO3
-] of calcium, magnesium and other heavy metals.
This hardness cannot be removed easily by boiling. Hence, it is called permanent hardness.
Total hardness of water = Temporary hardness + Permanent Hardness
Q 2. Write the relationship between units of hardness of water.
The following are common units for hardness of water.
1. Parts per million [ppm]
2. Milligram per litre [mg/lt]
3. Degree Clark [0cl]
4. DegreeFrench [0Fr]
5. Milli equivalents per litre [m eq/lt]
6. Parts Per Million [PPM]:-The number of parts by weight of CaCO3 equivalents hardness
causing salt present in one million parts of water.
[One million = 10 lakhs [106]].
1ppm = 1 part CaCO3 eq. hardness in 106 parts of water.
7. Milligrams per Litre [mg/Lt]:-The number of milligrams of CaCO3 equivalent hardness
causing salt present in one litre of water. i.e 1mg/lit
As density of water is unity i.e. 1lit of water = 1Kg = 1000X1000 mgs (at 4oC)
= 106 parts
Hence 1 mg/lt = 1 ppm
8. Drgree Clark [0Cl]:- The number of parts of CaCO3 equivalent hardness causing salt present
in 70,000 parts of water.
10 Cl = 1 parts of CaCO3 eq .hardness per 70,000 parts of water.
1 ppm = 0.070Cl
9. Degree French [0Fr]:-The number of parts by weight of CaCO3 equivalent hardness causing
salt per 105 parts by weight of water.
10Fr = 1 parts of CaCO3 eq. hardness per 105 arts of water 1 ppm = 0.1 0Fr
10. Milli equivalent per Litre [Mg/Lt]:-The number of milli equivalent of CaCO3 eq.
Q 3. Which one is most widely used chlorination process?
Ans:
Chlorination:
Chlorination is the process of purifying the drinking water by producing a powerful Germicide
like hypochlorous acid. When this chlorine is mixed with water it produces Hypochlorous acid
which kills the Germs present in water.
H2O+Cl2→ HOCl + HCl
Chlorine is basic (means PH value is more than 7) disinfectant and is much effective over the
germs. Hence chlorine is widely used all over the world as a powerful disinfectant. Chlorinator is
an apparatus, which is used to purity the water by chlorination process.
Q 4. Why is Ion exchange process preferred over Zeolite-process for the softening?
Ion exchange process (or) deionization or demineralization:
Ion exchanges are of two types they are anionic and cationic. These are co-polymers of styrene
& divinyl benzene i.e., long chain organic polymers with a micro porous structure.
Cation exchange resins: The resins containing acidic functional groups such as -COOH, -SO3H
etc. are capable of exchanging their H+ ions with other cations are cation exchange resins ,
represented as RH+.
Anion exchange resins: The resins containing amino or quaternary ammonium or quaternary
phosphonium (or) Tertiary sulphonium groups, treated with “NaOH solution becomes capable of
exchanging their OH- ions with other anions. These are called as Anion exchanging resins
represented as ROH-
Process: The hard water is passed first through cation exchange column. It removes all the
cation (ca2+ & Mg2+) and equivalent amount of H+ icons are released from this column.
2RH+ + Ca2+ → R2Ca2+ + 2H+
After this the hard water is passed through anion exchange column, which removes all the anions
like SO42-, Cl-, CO3
2- etc and release equal amount of OH- from this column.
R1OH + Cl- → R1Cl + OH-
2R1OH +SO42- → R2
1SO4 +2OH-
The output water is also called as de-ionized water after this the ion exchanges get exhausted.
The cation exchanges are activated by mineral acid (HCl) and anion exchanges are activated by
dil NaOH solution.
R2Ca + 2H+ → 2RH + Ca+2
R21SO4 + 2OH- → 2R1OH + SO4
2-
Advantages:
(1) The process can be used to soften highly acidic or alkaline water.
(2) It produces water of very low hardness. So it is very good for treating for use in high
pressure boilers.
Disadvantages:-
(1) The equipment is costly and common expensive chemicals required.
(2) It water contains turbidity, and then output of this process is reduced. The turbidity must
below 10 ppm.
Q5) Discuss the sludge and scale formation in boilers?
In boilers more water is removed in form of steam during boiling, hence boiler water gets
concentrated with dissolved salts and reaches saturation point. At this point the dissolved salts
precipitated and slowly settle on the inner walls of the boiler plate. The precipitation takes place
in two ways
a). Sludge formation:
The precipitation in the form of soft loose and slimy deposits formed comparatively in the colder
portions of boiler is called sludge.
Reasons for the formation of sludge:
The dissolved salts whose solubility is more in hot water and less in cold water produce sludges.
Ex: MgCO3, MgCl2, CaCl2, and MgSO4. The sludges were formed at comparatively colder
portions of the boiler and get collected where rate of flow of water is low.
Disadvantages of sludges:
i) Sludges are bad conductors of heat and results in wastage of heat and fuel.
ii) Sometimes sludges were entrapped in the scales and are deposited as scale, which causes
more loss of efficiency of the boiler.
iii) Excessive sludge formation leads to the settling of sludge in slow circulation areas such as
pipe connections, plug openings leading to the chocking of pipes.
Prevention of sludge formation:
i) By using soft water which is free from dissolved salts like MgCO3, MgCl2, CaCl2, and MgSO4
can prevent sludge formation.
b) Scale formation:
The precipitation in the form of hard deposits, stick very firmly on the inner walls of the boiler is
called scales.
Reasons for the formation of sludge:
i) Decomposition of Ca(HCO3)2:
Due to the high temperature and pressure present in the bolers, the Ca(HCO3)2 salt decomposes
to CaCO3 ↓, an insoluble salt, forms scale in low pressure boilers.
Ca (HCO3)2 → CaCO3 + H2O + CO2
ii) Decomposition of CaSO4
CaSO4 is more soluble in cold water hence its solubility decreases and precipitates out to
produce hard scale on the surface of the boiler. The solubility of CaSO4 is 3200ppm at 150C,
reduces to 27ppm at 3200C and completely insoluble in super heated water. CaSO4 scale is very
hard, highly adherent and difficult to remove.
iii) Hydrolysis of magnesium salts:
Dissolved magnesium salts undergo hydrolysis at high temperature prevailing inside the boiler
forming magnesium hydroxide precipitate, which form salt type of scale.
MgCl2 + 2H2O → Mg (OH)2 ↓ + 2HCl
iv) Presence of silica:
Silica present in small quantities deposits as calcium silicate (CaSiO3) or Magnesium silicate
(MgSiO3). The deposits form hard scale very difficult to remove.
Disadvantages of scale formation:
i) wastage of fuel: Scales are bad conductors of heat due to which the flow of heat to inside
water is decreased hence excessive heating is required which increases fuel consumption causing
wastage of fuel. The wasage of fuel increases with increases of the thichness of the scale.
ii) Lowering the boiler safety: Due to scale formation overheating of the boiler done to maintain
the constant supply of steam. due to over heating the boilermaterial become softer and weaker,
which causes distortion of boiler.
iii) Decrease in efficiency: Scales deposited in the valves and condensers of the boiler cause
chocking which results in decrease in efficiency of the boiler.
iv) Danger of explosion: Because of the formation of the scales, the boiler plate faces higher
temperature outside and lesser temperature inside due to uneven heat transfer resulting cracks in
the layer of scales. Water passes through the crack and comes in contact with boiler plate having
high temperature. This causes formation of large amount of steam suddenly developing sudden
high pressure. This causes the explosion of the boiler.
Removal scales:
i) If the scale formed is soft it can be removed by a scrapper, wire brush etc.
ii) By giving thermal shocks done by heating the boiler to high temperature and suddenly
cooling with cold water, if the scale is brittle in nature.
iii) If the scale is very adherent and hard, chemical treatment must be given. Ex: CaCO 3 scale is
removed by washing the boiler plate with EDTA solution.
iv) Frequent blowdown operation can remove the scales, which are loosely adhering.
Prevention of Scale formation:
Scale formation can be prevented by softening water by following methods.
Q6. Write the structure and synthesis of Paracetamol?The original method for production involves the nitration of phenol with sodium nitrate gives a mixture of two isomers, from which the wanted 4-nitrophenol (bp 279 °C) can easily be separated by steam distillation. In this electrophilic aromatic substitution reaction, phenol's oxygen is strongly activating, thus the reaction requires only mild conditions as compared to nitration of benzene itself. The nitro group is then reduced to an amine, giving 4-aminophenol. Finally, the amine is acetylated with acetic anhydride.
Q5. Explain batho-, hipso-, hyper- and hypochromic shifts with neat diagramme
Changes in chemical structure or the environment lead to changes in the absorption spectrum of molecules and materials. There are several terms that are commonly used to describe these shifts that you will see in the literature, and with which you should be familiar.
Bathochromic: a shift of a band to lower energy or longer wavelength (often called a red shift).
Hypsochromic: a shift of a band to higher energy or shorter wavelength (often called a blue shift).
Hyperchromic: an increase in the molar absorptivity.
Hypochromic: an decrease in the molar absorptivity.
UV-visible spectrum showing red, blue, hyper and hypo chromic shifts
Unit wise Question Bank
Unit I: Molecular structure and Theories of Bonding
Two marks questions
Q1. Define degenerate orbitals? Give the examples.
Ans. Orbital’s which is having same energy is called degenerate orbital’s. Examples Px, Py, Pz
orbital having same energy and dxy, dyz, dzx, dx2-y2, dz2 orbital’s having same energy ,these
orbital’s are called degenerate orbital’s.
Q2. What is nodal plane?
Ans. Plane which is having probability of finding the electron is equal to zero is called nodal
plane, number of orbitals is equal to n, and number of nodal planes are (n-1)
Q3. What are the shapes and structures of S, P,d orbital’s ?
S Orbital shape spherical, P orbital shape dumbel
Figure: d orbital shape double dumbel
Q4. What is a Coordinate Bond?
Sol. When both the electrons are being shared between the atoms are contributed by one atom only, then this type of bond is called Coordination Bond. It is also called Dative Bond.
A coordinate bond established between two atoms; one of which has a complete octet with at least one pair of unshared electrons while the other is short of two electrons. The Coordinate Bond is shown by (a) sign.
Example: Formation of hydronium ion from water molecules. In this, oxygen atom in water molecule is the donor and hydrogen ion is the acceptor.
Q5. Why do transition metal compounds forms complexes?
Sol. The transition metals are almost unique in their tendency to form coordination complexes. The tendency of cations of transition elements to form complexes is due to two factors;
1. These ions are very small in size and, therefore, have high positive charge density. This facilitates acceptance of lone pair of electrons from other molecules.
2 .They have vacant orbitals and these orbitals have the appropriate type of energy to accept lone pair of electrons.
Three marks questions
Q1. Explain about hydrogen molecules?
The simplest molecule is hydrogen, which can be considered to be made up of two separate
protons and electrons. There are two molecular orbital’s for hydrogen, the lower energy orbital
has its greater electron density between the two nuclei. This is the bonding molecular orbital -
and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital
more stable than two separated atomic hydrogen orbital’s. The upper molecular orbital has a
node in the electronic wave function and the electron density is low between the two positively
charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and
such an orbital is called an anti bonding molecular orbital. Normally, the two electrons in
hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen
is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one
electron into its anti bonding orbital (*), and the atoms will separate. The energy levels in a
hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals
combine to form two molecular orbitals, one bonding () and one anti bonding (*). This is
shown graphically below
Q2. Write the difference between the molecular orbitals and atomic orbitals?
Differences between Molecular Orbital and Atomic Orbital
Molecular Orbital Atomic Orbital
4. An electron Molecular orbital is
under the influence of two or more
nuclei depending upon the number
of atoms present in the molecule.
5. Molecular orbitals are formed by
combination of atomic orbitals
6. They have complex shapes.
4. An electron in atomic orbital is under the
influence of only one positive nucleus of
the atom.
5. Atomic orbitals are inherent property of an
atom.
6. They have simple shapes.
Q3. Write the silent features of CFT
Important Features of Crystal Field theory are
1. Transition metal ion is surrounded by ligands with lone pair of electrons and the complex is a
combination of central ion surrounded by other ions or molecules or diploes i.e ligand
2. All types of ligands are regarded as point charges.
3. The interaction between the metal ion and the negative ends of anion (or ion dipoles) are
purely electrostatic i.e bond between the metal and ligand is considered 100 percent ionic.
4.The ligands surrounding the metal ion produce electric field influences the energies of the
orbitals of central metal ion particularly d-orbitals.
5. In the case of free metal ion all the five d-orbitals have the same energy. Such orbital having
the same energies are called degenerate orbital’s.
Q4. Write the applications of co-ordination compounds?
Coordination compounds are widely use now days. Some of their applications are listed below:
Extraction process of gold and silver
Used as a catalyst in many industrial processes.
Example: Nickel, Copper can be extracted by using hydrometallurgical process involving
coordination compounds.
Used in hardness of water.
Example: EDTA (Ethylenediaminetetraacetate) is used in the estimation of Ca+2 and Mg+2 in
hard water.
Cyanide complexes are used in electroplating.
Q5. Write a note on spectro chemical series?
The tool used often in calculations or problems regarding spin is called the spectro chemical
series. The spectro chemical series is a list that orders ligands on the basis of their field strength.
Ligands that have a low field strength, and thus high spin, are listed first and are followed by
ligands of higher field strength, and thus low spin. This trend also corresponds to the ligands
abilities to split d orbital energy levels. The ones at the beginning, such as I−, produce weak
splitting (small Δ) and are thus weak field ligands. The ligands toward the end of the series, such
as CN−, will produce strong splitting (large Δ) and thus are strong field ligands. A picture of the
Calculate temporary, permanent and total hardness?
Ans:
Hardness causing
salt (H.C.S)
Quantity (H.C.S) Mol.Wt.of (H.C.S)
CaCl2 71 111
MgSO4 48 120
Ca(HCO3)2 146 162
Mg(HCO3)2 42 146
NaOH 40 -
Temporary Hardness = Mg(HCO3)2 + Ca(HCO3)2
= 28.7 + 90.1 = 118.8ppm
Permanent Hardness =CaCl2 + MgSO4
= 64 + 40 = 104ppm
Total Hardness = Temporary Hardness + Permanent Hardness
= 118.8 + 104
= 222.8ppm.
NaOH (compounds of alkali metals) does not give hardness to water.
Q5. Mention the steps involved in the treatment of potable water
Treatment of water for municipal supply:Ans: The treatment of water for drinking purpose mainly includes colloidal impurities and harmful pathogenic bacteria. The following is the flow diagram of water treatment for domestic purpose and various stages involved in the purification, given as:
(a) Screening: In this, the water is passed through screens having number of holes in it to remove floating impurities.
(b) Sedimentation with coagulation:
In this, the suspended and colloidal impurities are allowed to settle under gravitation. The basic principle of this treatment is to allow water to flow at a very slow velocity so that the heavier particles coagulate like alums, sodium alluminate and salts of iron are added which produces gelatinous precipitates called floc. Floc attracts and helps the accumulation of the colloidal particles, resulting in setting of the colloidal particles.
(c) Sterilization and disinfection:
Destruction of harmful pathogenic bacteria from the drinking water is carried out of sterilization and disinfection.
The following are the methods adopted for domestic purpose.
(i) Boiling:
By boiling water for 15-20 minutes, harmful bacteria are killed. This is not possible for municipal supply of water. This method of sterilization is adapted for domestic purpose.
(ii) Passing ozone:
Ozone is an unstable isotope of oxygen, produces Nascent oxygen which is powerful disinfectant.
O3 → O2 + O
(iii) By ultraviolet light:
UV light is used as disinfectant for swimming pool water as no chemicals are used. It is safe for skin. In this process water is exposed to UV rays which are generated foe an electric mercury vapor lamp.
source of waterscreeningsedimentationcoagulationfilterationsterilizationbreak-point chlorination
Five marks questions with answersQ1. Write the experimental procedure for the determination of total hardness by EDTA
method.
Ans: Estimation of Hardness of Water by EDTA Method [Complexometric Method]:-
In this complexometric method, disodium salt of Ethylene diamine tetra acetic acid is
used as complexometric agent.
NaOOC-H2C CH2COOH
HOOC-H2C N-CH2-CH2-N CH2COONa
Basic principle:
When hard water comes in contact with EDTA at pH 9-10 (pH maintained by buffer solution
prepared by using NH4Cl + NH4OH) the Ca2+ and Mg2+ forms colourless stable complex with
EDTA.
Erichrome Black – T [EBT] indicator forms an unstable, wine red coloured complex with Ca2+
and Mg2+.
Ca2+ + EBT (Blue color) → [Ca-EBT] (un stable wine red color) (at pH = 9-10)
Mg2+ + EBT (Blue color) → [Mg-EBT] (un stable wine red color) (at pH = 9-10)
The wine red coloured, complex is titrated with EDTA, where EDTA replaces EBT from metal
indicator complex by releasing the blue coloured EBT indicator.
An absolute configuration refers to the spatial arrangement of the atoms of a chiral molecular entity (or group) and its stereo chemical description e.g. R or S,[1] referring to Rectus, or Sinister, respectively.
The precise arrangement of substituent’s at a stereogenic center is known as the absolute configuration of the molecule. The arrangement of atoms in an optically active molecule, based on chemical interconversion from or to a known compound, is a relative configuration. Relative configurations of optically active compounds by chemically interconnecting them.
Q2. Show, how SN2 reaction give rise to inverted product?
In this rate of reaction depends on the concentration of substrate as well as concentration reagent. This is single step mechanism , Nucleophilic addition and elimination takes place simultaneously ,and inversion of the molecule takes place , this inversion is called Walden inversion .
Q3. Indicate, with a suitable diagram, the potential energy changes during rotation about C (2) - C (3) bond of n-butane.
Q4. Name the elements of symmetry. Discuss, with the help of an example, an optically active compound without chirality.
The compound that has a plane of symmetry will show optical activity. The compound has to be non-planar. There are some compounds, which do not have a chiral carbon, that show optical activity. The best example is biphenyls. Take the example of the one above (the picture ). It should have been a planar compound ( obviously, each carbon on the benzene ring is sp2 hybridized) but, because of the repulsion between the two NO2 groups attached ( it is a big group and their electron clouds repel), one of the NO2 moves out of the plane, thus making the compound optically active. This is how a compound without chiral carbon becomes optically active.
Q5. What is the Markownikoff rule?
Ans. Markovnikov’s Rule: This rule states that an addition reaction of an asymmetric alkene by an asymmetric reagent the negative part of the reagent attacks the carbon containing the less number of hydrogen across the double bond.
Markovnikov’s rule is mostly carried out using HBr as it is a good reagent for this process, neither highly exothermic nor highly endothermic.
Propene reacts with HBr:
CH3−CH=CH2+HBr−−>CH3−CH(Br)−CH3+CH3−CH2−CH2(Br)CH3−CH=CH2+HBr−−>CH3−CH(Br)−CH3+CH3−CH2−CH2(Br)We obtain two products 2-Bromopropane and 1-Bromopropane.
according to Markovnikov’s rule the major product the major product will be decide by carbo cation stability, we see that between 2-Bromopropane and 1-Bromopropane if we remove the bromine atom, 2-Bromopropane will have a 2°2°carbocation and 1-Bromopropane will have 1°1° carbo cation. Hence the major product will be 2-Bromopropane.
Three Marks Questions:
Q1. Define Enantiomers and Diastereomers?
A). Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer. Diastereomerism occurs when two or more stereoisomer’s of a compound have different configurations at one or more (but not all) of the equivalent (related) stereo centers and are not mirror images of each other.
Enantiomers are chiral molecules that are mirror images of one another. Furthermore, the molecules are non-super imposable on one another. ... It is sometimes difficult to determine
whether or not two molecules are Enantiomers. The enantiomers differ only in their spatial arrangements at the stereo center.
Consider 2-bromo-3-chlorobutane, which has stereo centers at C2 and C3. In general, a molecule with n stereo centers has 2n stereo isomers, so there are a total of four possibilities for 2-bromo-3-chlorobutane:
Each of the four stereo isomers of 2-bromo-3-chlorobutane is chiral. There are two pairs of enantiomers. Any given molecule has its enantiomer; the two other molecules are its diastereomers.Q2. Write a note on Grignard addition on carbonyl compounds?The reaction is very fast, even at room temperature, because the carbonyl group has a highly electrophilic carbonyl carbon, which combines with the nucleophilic carbon of the Grignard reagent. Alkyl magnesium halides is called Grignard reagent.( R-MgX) .This reagent reacts with carbonyl compound , mechanism as shown below, following by hydrolysis secondary alcohols are formed.
Q3. Write the oxidation reactions of alcohol using KMnO4
Primary alcohol oxidation in presence of KMnO4 gives aldehydes further oxidation gives carboxylic acids
Secondary alcohols can be oxidised to ketones but no further:
Tertiary alcohols cannot be oxidised (no carbinol C-H)
Q4. What is Saytzeff rule? Explain with one example?
According to Saytzeff's rule (also Zaitsev's rule), during dehydration, more substituted alkene (olefin) is formed as a major product, since greater the substitution of double bond greater is the stability of alkenes’
* Sulfuric acid is a strong acid and ionizes to give a proton.
* Thus formed H+ ion attacks the -OH group. The OH2+ group formed is good leaving group
due to accumulation of positive charge on oxygen atom. Now the loss of H2O creates positive charge on the carbon atom and thus by forming a tertiary carbocation.
* Finally, one of the hydrogen adjacent to the positively charged carbon is removed as proton, H+ to make a double bond. However there are three adjacent hydrogen atoms (indicated by descriptors I, II & III). Hence three different alkenes are possible.
However according to Saytzeff's rule, highly substituted alkene, as shown below by the loss of H+(I), is formed as a major product.
Q5. What is Aspirine? Write any two pharmaceutical applications.
Aspirin is a non-steroidal anti-inflammatory drug .Aspirin, or acetylsalicylic acid (ASA), is commonly used as a pain reliever for minor aches and pains and to reduce fever. It is also an anti-inflammatory drug and can be used as a blood thinner.People with a high risk of blood clots, stroke, and heart attack can use aspirin long-term in low doses.
5 Marks Questions with answers
Q1. Write short notes on the following:a) Markownikoff’s rule b) Anti-Markownikoff’s rule
Markovnikov's rule (Markovnikov addition): In an addition reaction of a protic acid HX (hydrogen chloride, hydrogen bromide, or hydrogen iodide) to an alkene or alkyne, the hydrogen atom of HX becomes bonded to the carbon atom that had the greatest number of hydrogen atoms in the starting alkene or alkyne.
Anti-Markovnikov addition: In presence of peroxides an addition reaction of a generic Electrophiles HX to an alkene or alkyne, the hydrogen atom of HX becomes bonded to the carbon atom that had the least number of hydrogen atoms in the starting alkene or alkyne.
Q2. Write SN1 reactions with Mechanism?
Nucleophilic substitution reactions are ionic reactions that break and make chemical bonds by transfers of pairs of electrons. We illustrate this using a general representation of a nucleophilic substitution reaction in which a halogen (X) is replaced by a new nucliophile (:N).
R3C-X + -: N → R3C-N + -:X
The electron pair in the original C:X bond remains with the halogen (X) as that bond breaks, while the electron pair on -:N becomes the new C:N chemical bond.
Nucleophilic Substitution Mechanisms
The two major mechanisms for nucleophilic substitution are called SN1 and SN2. We describe them here using haloalkanes (R3C-X) as the substrates.
The SN1 Mechanism. The SN1 mechanism has two steps and an intermediate carbocation R3C+.
In the first step, the C-X bond in R3C-X breaks to give a negatively charged halide ion (-:X) and positively charged carbocation (R3C+). Carbocations are also called carbonium ions. In this ionization reaction (a reaction that forms ions), the electron pair in the C-X bond remains with the halogen (X) as the C-X bond breaks.
The intermediate carbocation reacts in the second step with an unshared electron pair on the species -:N to form the new C:N bond. We use the letter N to signify that -: N is a nucleophile. A nucleophile is a chemical species with an unshared pair of electrons that reacts with electron deficient centers such as the C+ atom in R3C+.
Nucleophiles always have an unshared electron pair that forms the new chemical bond, but they are not always negatively charged. When the nucleophile (:N) in an SN1 reaction is electrically neutral (uncharged), it reacts with the intermediate carbocation to give a positively charged product.
Arrows Show How the Electrons Move.
The Meaning of SN1. SN1 stands for Substitution (S) Nucleophilic (N) Unimolecular (1). The rate of reaction depends on the concentration of substrate (R3C-X) only.
Q3. Write about Electrophilic Addition Reactions?
Electrophillic addition :
In organic chemistry, an electrophilic addition reaction is an addition reaction where, in a chemical compound, a π bond is broken and two new σ bonds are formed. The substrate of an electrophilic addition reaction must have a double bond or triple bond.
The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.
In step 2 of an electrophilic addition, the positively charged intermediate combines with (Y) that is electron-rich and usually an anion to form the second covalent bond.Step 2 is the same nucleophilic attack process found in an SN1 reaction. The exact nature of the electrophile and the nature of the positively charged intermediate are not always clear and depend on reactants and reaction conditions.In all asymmetric addition reactions to carbon, regioselectivity is important and often determined by Markovnikov's rule. Organoborane compounds give anti-Markovnikov additions. Electrophilic attack to an aromatic system results in electrophilic aromatic substitution rather than an addition reaction.The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.
Q4. Explain Reduction of carbonyl compound using LiAlH4
LiAlH4 is one of the good reducing agent , it reduce the formaldehyde to methyl alcohol, aldehyde to primary alcohol , ketones to secondary alchol
Reaction type: Nucleophilic Addition
NUCLEOPHILIC ADDITION OF LiAlH4 TO AN ALDEHYDE
Step 1: The nucleophilic H in the hydride reagent adds to the electrophilic C in the polar carbonyl group in the aldehyde, electrons from the C=O move to the O creating an intermediate metal alkoxide complex. (note that all 4 of the H atoms can react)
Step 2: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex.
Q5. Write the structure and synthesis of Paracetamol?The original method for production involves the nitration of phenol with sodium nitrate gives a mixture of two isomers, from which the wanted 4-nitrophenol (bp 279 °C) can easily be separated by steam distillation. In this electrophilic aromatic substitution reaction, phenol's oxygen is strongly activating, thus the reaction requires only mild conditions as compared to nitration of benzene itself. The nitro group is then reduced to an amine, giving 4-aminophenol. Finally, the amine is acetylated with acetic anhydride.
Q1: Define the terms Chromophore and Auxochrome in UV spectroscopy.
An auxochrome is a functional group of atoms with one or more lone pairs of electrons when attached to a chromophore, alters both the wavelength and intensity of absorption.
Q2: Methane does not absorb IR energy. Why?
The molecules having zero dipole moment can’t absorb the IR radiation because the vibrations in the bond of that molecule doesn’t change the net dipole moment of the molecule
Q3: What is Magnetic resonance imaging (MRI)? Describe the applications of MRI.
Magnetic resonance imaging is a scan that produces detailed pictures of organs and other internal body structures while a CT scan forms images inside of the body. CT scans use radiation, which may be harmful to the body, while MRIs do not. MRI’s cost more than CT scans
Q4: Define IR spectroscopy.
Infrared spectroscopy (IR spectroscopy or vibrational spectroscopy) involves the interaction of infrared radiation with matter.
Q5: C-C double and triple bonds of ethane (CH2=CH2) and acetylene CH≡CH do not absorb IR energy. Why?
The molecules having zero dipole moment can’t absorb the IR radiation because the vibrations in the bond of that molecule doesn’t change the net dipole moment of the molecule, for the molecules ethane and acetylene the net dipole moment is zero.
3 marks Question:
Q1: IR spectra is often characterized as molecular finger-prints. Comment on it.
IR spectrum showing fingerprint region
The region to the right-hand side of the diagram (from about 1500 to 500 cm-1) usually contains a very complicated series of absorptions. These are mainly due to all manner of bending vibrations within the molecule. This is called the fingerprint region.
Q2: How many different types of H-atom environments are present in methyl alcohol? Also mention the ratio of peak areas due to –CH3 group and –OH group in NMR spectrum
Methanol NMR shows two types of peaks one belongs to –CH3 and another belongs to –OH The peak area ratios (number of hydrogens present on functional group -CH3: Number of hydrogens present on functional group –OH) i.e 3: 1
Q3: What type of information is obtained by studying the UV, IR, H1-NMR.
Nuclear Magnetic Resonance (NMR) Spectroscopy is one of the most useful analytical techniques for determining the structure of an organic compound
UV/Vis spectroscopy is an absorption spectroscopy technique that utilizes electromagnetic radiation in the 10 nm to 700 nm range. The energy associated with light between these wavelengths can be absorbed by both non-bonding n-electrons and π-electrons residing within a molecular orbital.
The wavenumbers of the absorbed IR radiation are characteristic of many bonds, so IR spectroscopy can determine which functional groups are contained in the sample
Q4: What is λmax in UV-visible spectrum?
The λmax in UV-visible spectrum is the wave length where the molecule has maximum absorption of light (maximum absorption coefficient)
UV-visible spectrum
Q5: A solution of X of concentration 0.010 mol dm–3 gives an absorbance of 0.5. What concentration is a solution of X which gives an absorbance reading of 0.25? Assume that the same optical cell is used for both readings.
Answer: Solution X concentration is (C1) = 10-2 mol dm-3
Absorbance (A1) = 0.5
For the same molecule if Absorbance (A2) is = 0.25
Than the concentration C2 is =?
From Beer Lamberts A = εCl
We can write A1/A2 = C1/C2
By substituting above values
0.5/0.25 = 10-2/C2
i.e C2 = 0.5x10-2 mol. dm-3
5 mark Questions with answers:
Q1. The four central lines in the high resolution υ =1←υ = 0 infrared spectrum of HCl37 occur at 2837.6, 2858.8, 2899.2 and 2918.6 cm-1. Deduce as much as possible about the molecule. Would the corresponding lines in HCl35 lie at the same spectral positions?
Answer: The band centre is at the average of the two central lines, i.e. 2879.0 cm-1.
This is equal to the fundamental frequency, ω0.
The 4B separation at the centre is 40.4 cm-1, giving a value of 10.1 cm-1 for B.
The moment of inertia is (I) = h/8π2Bc = 2.771x10-47 kg m2
The reduced mass is given by
Therefore,
Since the positions of the lines depend on the reduced mass, the lines for HCl35 will be at a different position. The reduced mass of HCl35 is smaller and hence its moment of inertia is smaller and rotational constant larger, so the lines will have larger separation. The fundamental vibration frequency will also be higher since it is inversely proportional to the square root of the reduced mass.
Q2. Explain batho-, hipso-, hyper- and hypochromic shifts with neat diagramme
Changes in chemical structure or the environment lead to changes in the absorption spectrum of molecules and materials. There are several terms that are commonly used to describe these shifts that you will see in the literature, and with which you should be familiar.
Bathochromic: a shift of a band to lower energy or longer wavelength (often called a red shift).
Hypsochromic: a shift of a band to higher energy or shorter wavelength (often called a blue shift).
Hyperchromic: an increase in the molar absorptivity.
Hypochromic: an decrease in the molar absorptivity.
UV-visible spectrum showing red, blue, hyper and hypo chromic shifts
Q3. Deduce the Beer-Lambert law for absorptivity and concentration.
Answer: According to this equation the Absorbance (A) is directly proportional to concentration (C) of the solution
i.e A α C ------------------------(i)
and the Absorbance is also directly proportional to path length of the light travelled (l)
i.e A α l ------------------------(ii)
From equation (i) and (ii)
A α Cl ------------------------(iii)
The equation can be written as
A = εCl
(where ‘ε’ is Absorptivity constant )
Q4. What is the principle of Nuclear Magnetic Resonance (NMR) spectroscopy?
Answer: Nuclear magnetic resonance is defined as a condition when the frequency of the rotating magnetic field becomes equal to the frequency of the processing nucleus.
Principle of NMR:
The principle of nuclear magnetic resonance is based on the spins of atomic nuclei. The magnetic measurements depend upon the spin of unpaired electron whereas nuclear magnetic resonance measures magnetic effect caused by the spin of protons and neutrons. Both these nucleons have intrinsic angular momenta or spins and hence act as elementary magnet.
The existence of nuclear magnetism was revealed in the hyper fine structure of spectral lines. If the nucleus with a certain magnetic moment is placed in the magnetic field, we can observe the phenomenon of space quantization and for each allowed direction there will be a slightly different energy level.
Q5. What is force constant (k)? how it relates to reduced mass (µ) of a molecule?
Answer: To help understand IR, it is useful to compare a vibrating bond to the physical model of a vibrating spring system. The spring system can be described by Hooke's Law, as shown in the equation given on the left.
Consider a bond and the connected atoms to be a spring with two masses attached. Using the force constant k (which reflects the stiffness of the spring) and the two masses m1 and m2, then the equation indicates how the frequency, u, of the absorption should change as the properties of the system change
Beyond syllabus topics with material
Sewage Water Treatment
Sewage Water: Sewage is a water-carried waste, in solution or suspension that known as domestic or municipal wastewater, it is characterized by volume or rate of flow, physical condition, chemical and toxic constituents, and its bacteriologic status. It consists mostly washing water, food preparation wastes, laundry wastes, soaps, detergents, and other waste products of normal living. It also contains wastes that result from industrial processes, surface runoff.
Type of Pollutants in the Sewage Water:
Organic pollutants and nutrients (Chemical): Sewage is a complex mixture of chemicals. These include high concentrations of ammonium, nitrate, phosphorus, high conductivity (due to high dissolved solids), high alkalinity, with pH typically ranging between 7 and 8. The organic matter of sewage is measured by determining its biological oxygen demand (BOD) or the chemical oxygen demand (COD). Sewage also contains the nutrients like nitrogen and phosphorus.
Pathogens and micro-pollutants: Sewage water mostly contains pathogenic microorganisms.
Treatment of sewage water:
Sewage treatment generally involves three stages, called primary, secondary and tertiary treatment.
Steps Involved in Treatment of Sewage Water:
Primary treatment consists of temporarily holding the sewage in an inactive basin where heavy solids can settle to the bottom while oil, grease and lighter solids float to the surface. The settled and floating materials are removed and the remaining liquid may be discharged or subjected to secondary treatment. Some sewage treatment plants that are connected to a combined sewer system have a bypass arrangement after the primary treatment unit.
Secondary treatment removes dissolved and suspended biological matter. Secondary treatment is typically performed by local water-borne microorganisms in a managed conditions. This is done by a) Trickling filter process or b) Active sludge process
a) Trickling filter process: It is a circular tank and is filled with coarse or crushed rocks. sewage is sprayed over this bed by using slowly rotating arms. When sewage percolating downwards, microorganisms present in the sewage grow on the surface of filtering media using organic material of the sewage as food. After completion of aerobic oxidation the treated sewage is taken in to settling tank and the sludge is removed. This process removes 80-85% BOD.
Figure: Trickilng Filters
b.) Activated sludge process: Activated sludge is biologically active sewage and it has large number of bacteria, which can easily oxidize the organic impurities.
The sewage effluent from primary treatment is mixed with the required amount of activated sludge. Then the mixture is aerated in the aeration tank. Under the condition organic impurities of the sewage get oxidized by the microorganisms. This process removes 90-95% of BOD.
Tertiary treatment is sometimes defined as anything more than primary and secondary treatment in order to allow rejection into a highly sensitive or delicate ecosystem (low-flow Rivers). Treated water is sometimes disinfected chemically or physically prior to discharge into a stream, river. In the tertiary treatment, the effluent introduced in to a flocculation tank where lime is added to remove phosphates. From the flocculation tank the effluent is led to ammonia stripping tower, pH is maintained to 11 and NH4
+ is converted gaseous NH3. Finally, this effluent is treated with disinfectants (Chlorine). If it is sufficiently clean, it can also be used for groundwater recharge or agricultural purposes.
