chemistry

Atomic Strucure, core points !!

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In 1913, Niels Bohr proposed a model of the atom. He proposed that the electrons in an atom could only be in certain orbits, or energy levels, around the nucleus. Refinement of Bohr theory led to the modern theory of atomic structure based on quantum mechanics.

Bohr's model is based on particle theory.

As wave-particle duality which says that all micromatter particle exhibit dualitya new model is proposed.

Debroglie p = h/lambda

Uncertainty principle delta(p)* delta(x) >=h/4pi

The quantum mechanics model

Principal quantum number-Shell,Azimuthal quantum number-sublevel,Magnetic quantum number-orbital, spin quantum number

The orbits are called as shells. The energy level of orbits or shells increases as they increase in distance from the nucleus of the atom. The orbits or shells are represented by numbers as 1,2,3,4,5,6 or 7. They are represented by letters as K,L,M,N,O,P,Q.

Sublevel of an Orbit

The energy levels, or orbits or shells are further divided into sublevels, or subshells. These subshells are designated by letters: s for the first possible sublevel, p for the second possible sublevel, d for the third, f for the fourth, g for the fifth, and from here on they simply go in alphabets.

The number of sublevels of each energy level is equal to the number of the energy levels. This means energy level 1, the K shell will have only one sub levels ? s sublevel. The energy level 2, the L shell will have 2 sub levels ? s and p.

Orbitals

Sublevels have further divisions called orbitals. Electrons are found in these orbitals. Each orbital contains two electrons.

?s? sublevel has only one orbital. ?p? sublevel has 3 orbitals. ?d? sublevel has 5 orbitals. ?f? sublevel has 7 orbitals.

The two electons in each orbital spin in different directions.

Shape of Orbitals
1. Spherical shape for s.
2. Dumbbell shape for orbitals of p.
3. Four-lobed shape for orbitals of d.
4. Complex shape for all orbitals of higher sublevels.

Pauli's exclusion principle: No two electrons can have all four same quantum numbers

Electrons occupy the lowest energy sublevels that are available. This is known as ?aufbau? order or principles.

Hund?s rule says that, for any set of orbitals of equal energy say p orbitals of orbit 2, there is one electron is each orbital before the second electron enters or occupies an orbital.

The energy level of some sublevels at higher orbits is less than the some sublevels at lower orbitals. This fact is to be kept in mind when electron configuration is determined for any atom. The increasing order of energy levels of sublevels is:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f*, 5d, 6p, 7s, 5f*, 6d, 7p, 8s

Visible lines in H atom
spectrum are called the
BALMER series.

Atomic Spectra
When one heats up a gas, it emits light of various wavelengths. For monatomic gases formed of only one kind of atom the emission spectra contains only light of particular wavelengths, and for hydrogen the spectra obeys a very simple relation:

1/lambda = RH[(1/nf²) -(1/ni²) ]

where ni and nf are positive non-zero integers with ni > nf and RH is a constant called Rydberg's constant:
RH = 1.097 x 10^7 m - 1. (2)

One of the first such series of lines discovered which obey the rule was found by Balmer, which corresponded to nf = 2 and is in the visible light region.

certain phenomena of light can be explained when light is considered as composed of particle. Certain other phenomena can be explained by considering light as a wave.

Similarly micromatter like electrons have particle characteristics and wave characteristics.

Shape of Orbitals
1. Spherical shape for s.
2. Dumbbell shape for orbitals of p.
3. Four-lobed shape for orbitals of d.
4. Complex shape for all orbitals of higher sublevels.

all trends for inorganic chemistry

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Periodic Table in a flash

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Hi Guys

I jus wanted to share a poem wid u it can help u in remembering all the elements in a particular group.

Group1: Helina ki rab se fariyaad.

Group2: Be magach ca sar baara

Group13: Balga Intel

Group14: Kasai ge ka son lead

Group15: Naya purana aaj sab bikega

Group16: Os(Us) se tepo

Group18(Noble metals): Heni ar kar xe run

What is Hyperconjugation ?

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Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a s-bond (usually C-H or C-C) with an adjacent empty (or partially filled) p-orbital or a p-orbital to give an extended molecular orbital that increases the stability of the system. Based on the valence bond model of bonding, hyperconjugation can be described as "double bond - no bond resonance" but it is not what we would "normally" call resonance, though the similarity is shown below. What is the key difference between hyperconjugation and resonance ? Hyperconjugation is a factor in explaining why increasing the number of alkyl substituents on a carbocation or radical centre leads to an increase in stability.

Let's consider how a methyl group is involved in hyperconjugation with a carbocation centre.
First we need to draw it to show the C-H s-bonds. Note that the empty p orbital associated with the positive charge at the carbocation centre is in the same plane (i.e. coplanar) with one of the C-H s-bonds (shown in blue.)
This geometry means the electrons in the s-bond can be stabilised by an interaction with the empty p-orbital of the carbocation centre. (this diagram shows the similarity with resonance and the structure on the right has the "double bond - no bond" character)

The stabilisation arises because the orbital interaction leads to the electrons being in a lower energy orbital

Of course, the C-C s-bond is free to rotate, and as it does so, each of the C-H s-bonds in turn undergoes the stabilising interaction.
So the ethyl cation has 3 C-H s-bonds that can be involved in hyperconjugation.
The more hyperconjuagtion there is, the greater the stabilisation of the system.
So for example, the t-butyl cation has 9 C-H s-bonds that can be involved in hyperconjugation. Hence (CH₃)₃C+ is more stable than CH₃CH₂+
The effect is not limited to C-H s-bonds, appropriate C-C s-bonds can also be involved in hyperconjugation.

MOLECULAR ORBITAL THEORY SIMPLIFIED

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TOTAL NUMBER OF ELECTRONS:

10 11 12 13 14 15 16 17 18

1 1.5 2 2.5 3 2.5 2 1.5 1 <--- BOND ORDER

D P P P D P P P D

D ----> Dimagnetic

P------> Paramagnetic

the bond order calculated by the formula will always give you the correct answer.

but as for the magnetic nature...90% will be correct....

their are some exceptions like B2

both are paramagnetic

Inorganic 4 ALL ORES from METALLURGY(SIMPLIFIED)

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SOME IMPORTANT METALS AND THEIR ORES:

1. SODIUM (Na):

Sodium Chloride NaCl (rock salt, table salt, common salt)

Sodium Carbonate Na₂CO₃ (Soda Ash)

Na₂CO₃.10H₂O(Washing Soda)

Na₂CO₃.H₂O(Crystal Carbonate)

Sodium Nitrate NaNO₃(chlie salt petre or chile nitre)

Borax Na₂B₄O₇.10H₂O(Tincal)

Sodium Sulphate Na₂SO₄.10H₂O(Glauber's Salt)

Cryolite Na₃AlF₆

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2.Potassium:

Potassium Chloride KCl(Sylvine)

Potassium Carbonate K₂CO₃(Pearl Ash)

Potassium Nitrate KNO₃(Indian Salt Petre,nitre or salt petre)

Carnallite KCl.MgCl₂.6H₂O

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3. Copper:

Cuprite Cu₂O (Ruby Copper)

Copper glance Cu₂S (Chalcocite)

Chalcopyrite CuFeS₂ or Cu₂S.Fe₂S₃(Copper pyrites)

Malachite CuCO₃.Cu(OH)₂

Azurite 2CuCO₃.Cu(OH)₂

_{_____________________________________________}

4.SIVER:

Native Silver Ag

Argentite Ag₂S

Horn silver AgCl (Kerargyrite)

Pyragyrite Ag₂S.Sb₂S₃(Ruby Silver)

________________________________________

5. GOLD:

Native Gold Au

Sylvanite A telluride of Gold and Silver

Bismuth Auride AuBi

Calaverite AuTe₂

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6.MAGNESIUM:

Magnesite MgCO₃

Dolomite MgCO₃.CaCO₃

Carnallite KCl.MgCl₂.6H₂O

Epsom salt MgSO₄.7H₂O

Brucite Mg(OH)₂

Kieserite MgSO₄.H₂O

Schonite K₂SO₄.MgSO₄.6H₂O

Asbestos CaSiO₃.3MgSiO₃

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7.CALCIUM:

Limestone CaCO₃(iceland spar, marble, chalk, calcite)

Gypsom CaSO₄.2H₂O

Anhydrite CaSO₄

Flouride CaF₂(Flourspar)

Phosphite Ca₃(PO₄)₂

Chlorapatite 3Ca₃(PO₄)₂.CaCl₂

Fluor apatite 3Ca₃(PO₄)₂.CaF₂

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8. STRONTIUM:

Strontianite SrCO₃

Celestine SrSO₄

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9.BARIUM:

Witherite BaCO₃

Heavy spar BaSO₄

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10.ZINC:

Zincite ZnO

Calamine ZnCO₃

Zinc Blende ZnS (Black Jack)

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11. CADMIUM:

Greenokite CdS

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12. MERCURY:

Cinnabar HgS

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13. ALUMINIUM:

Corundum Al₂O₃

Diaspore Al₂O_3.H₂O

Bauxite Al₂O_3.2H₂O

Alunite K₂SO₄.Al₂(SO₄)₃.4Al(OH)₃ (Alum stone)

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14. TIN:

Cassiterite SnO₂

_{____________________________________________________}

15. LEAD:

Galena PbS

Cerrusite PbCO₃

Matlockite PbCl₂

Anglesite PbSO₄

Lanarkite PbO.PbSO₄

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16. IRON:

Haemitite Fe₂O₃

Load Stone Fe₃O₄

Limonite 2Fe₂O₃.3H₂O

Siderite FeCO₃(Spathic iron ore)

Iron Pyrites FeS₂

Chalcopyrite CuFeS₂

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17.MANGANESE:

Pyrolusite MnO₂

Braunite Mn₂O₃

Hausmannite Mn₃O₄

________________________________________________

18. TITANIUM:

Rutile TiO₂

Illmenite FeTiO₃

important things usually left in inorganic.

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SECONDARY FORCES:

These forces decide the physical properties of the molecule like boiling point, meltiing point, refractive index, viscoscity and solubility

1. HYDROGEN BOND:

it is a type of dipole dipole interaction...

it is only shown by N, O , F

to learn it...

hydrogen ne FON (phone) lagaya to H-bonding hui.

NOTE: boiling point of water is greater than HF and NH3 exceptionally.

Q. Enthalpy of vapourization of H2O is greater than HF. WHY?

A. water evaporates as a monomer so we have to break all the H bonds but HF evaporates in the form of a dimer so we have to break less bonds.

(draw the diagram...it'll be clear to you at once)

2.INTRA MOLECULAR HYDROGEN BOND:

It is the hydrogen bond formed within the molecule:

example o-nitrophenol

meta and para isomers don't form intermolecular H-bond.

APPLICATIONS:

1. Boiling Point inter molecular H-bond

1/ intramolecular H bond Solubility

3. VISCOSCITY No. of H bonds Molecular weight

4. ACIDIC NATURE less H-bonds

Q.Paranitrophenol is more acidic than orthonitrophenol?

A. in orthonitrophenol due to intramolecular H-bonding a strong cyclic ring caleed as chelate is formd which provides extra stability to the molecule.

To phir khud hi soch lo wo apna H kyun dega.

kyunki agar usne H de diya to chelate ring toot jayegi.

1st disassociation of maleic acid is grtr dn fumaric acid because it acquires stability due to the chelate formation but the 2nd disassociation of maleic acid is less dn the fumaric acid

ALCOHOLS ARE SOLUBLE IN WATER DUE TO H-BOND.

small important things in inorganic

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APPLICATION OF DIPOLE DIPOLE FORCE:

diple- dipole boiling point

Molecule with 0 dipole moment shows minimum boiling point and with maximum boiling point shows maximum boiling point.

ex: Cis isomers have high boiling point than trans.

MELTING POINT SYMMETRY PACKING 1/size

ex: s-block have low Melting point while d-block have high MP

Trans isomers have higher melting point than cis due 2 symmetry

Para isomers have high MP due 2 symmetry.

WHY ICE FLOATS ON WATER?

Ice has tetrahedral arrangement of water in which one H2O moleule is surrounded by 4 other H2O molecules and leave large vacant spaces due to which its volume increases and density decreases.

WHAT ARE ELATHRATES OF CAGE COMPOUNDS?

when noble gases like Ar, Kr, Xe are passed into H2O and suddenly freezed then dese gases get trapped in the vacant spaces of ice. such compounds are called cage-compounds.

ION DIPOLE FORCE:

When ionic compound or polar covalent compound is dissolved in water or polar solvents then

1. cation is surrounded by negative part of the solvent by ion dipole force for Group 1 and Group 2 cations. Coordination No is 6 but 4 Li and Be it is 4.

due to absence of d-orbital

NaCl when hydrated gives [Na(H2O)6]

Anion is surrounded by positive part of the solvent by ion dipole force

Which is:

a)H(delta +ve) if solvent is polar protic (which can donate H) and here the ion dipole force is called as H bonding

b) other dn H if the solvent is polar protic solvent and the ion dipole force is not the H bond

CONCLUSION: solvation energy is more in case of polar protic solvents hence the stability for the formation of cation or anion is more.

Q. Why does Li doesn't form alum?

A. for the formation of alum the required coordination number for the cation is six but Li has coordination number 4.

TO CHECK SOLUBILITY:

1.HYDRATION ENERGY

2. ABILITY TO GET POLARIZED

3.PHYSICAL STATE

find the solubility order of iodine, bromine and Chlorine:

answer: Cl2I2

EXPLANATION: I2 is polarized to a greater extent but due to its large size and its solid state it is less soluble.

Br2 secondly polarizes to a greater extent ( less dn I) and due to itz liquid nature it is more soluble in water.

INSTANTANEOUS DIPOLE INDUCED DIPOLE FORCES:

In non polar gaseous molecules due to collison or friction momentarily dipole is produced which in turn produced the dipole produce the dipole in the neighbouring moleule.

Ex: all noble gases

this force only is called as VANDER WAAL FORCE.

order of Vander waal force:

more the vander waal force more easy is the liquification.

Boiling Point: NH3 > PH3

H2O > H2S

HF > HCl <>

such order is due to H-bonding

and the boiling point increases downward due to increase in vanderwaal force of attraction.

Vander Waal force exists in all states of matter.

Liquids/solids in which molecules are held together by weak V.W. forces are volatile/sublime

VOLATILE LIQUIDS ACT AS GOOD FUELS.

INORGANIC 5 HYBRIDIZATION and GEOMETRY SIMPLIFIED

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PS:THIS FORMULA WILL ALWAYS GIVE YOU THE CORRECT ANSWER

Just get through wid all this u'll find inorganic petty easy....

SHORTEST FORMULAE FOR HYBRIDIZATION:

No of valence electrons + No. of atoms attached to central atom

CONDITIONS:

1. Don't count the multiple bonded attached atoms.

2. Don't coun't odd or unpaired electrons.

after putting the values in the formula if ya get:

2 then compound is sp hybridized

3 then compound is sp2 hybridized

4 then compound is sp3/dsp2 hybridized

5 then compound is sp3d/dsp3 hybridized

6 then compound is sp3d2/d2sp3 hybridized

TO DIFFERENTIATE b/w sp3 and dsp2 hybridization and so on.

1. Easiest way is to check whether d-orbital present or not.

2. If the compound undergoes reaction with strong ligand than dsp2, dsp3,d2sp3 hybridization occurs.

3.If the compound undergoes reaction with weak ligand than sp3, sp3d, sp3d2 hybridization occurs.

TO IDENTIFY STRONG LIGAND AND WEAK LIGAND:

ALL NITROGEN CONTAINING except nitrate ion are strong ligands.

CO is exceptionally strong ligand...

ALL STRONG LIGANDS CAUSE PAIRING OF UNPAIRED ELECTRON IN D-ORBITAL.

GEOMETRIES:

REQUIRED (R): the total no. of atoms which are necessary for perfect geometry of that hybridization or number of hybrid orbitals.

AVAILABLE(A): Atoms attached.

R-A=no. lone pair

1.sp hybridization:

R=2, A=2

R-A=0

linear geometry, 0 dipole moment.

R=2, A=1

R-A=1 lone pair

so dipole moment is not 0.

2.sp2 hybridization:

a. R=3

-A=3

0 = no. of lone pairs.

Geometry: trigonal planar and dipole moment =0

b. R=3

A=2

lone pair = 1

Geometry: V shape/angular

c. R=3

A=1

lone pair = 2

Geometry: Linear

3. sp3

a. R=4

A=4

lone pair=0

Geometry: Tetrahedral

b. R=4

A=3

lone pair = 1

Geometry: pyramidal

c. R=4

A=2

lone pair = 2

Geometry: V-shape

d. R=4

A=1

lone pair = 3

Geometry: Linear

4.sp3d

a. R=5

A=5

lone pair = 0

Geometry: Trigonal Bipyramidal

b. R=5

A=4

lone pair = 1

Geometry: see-saw

c. R=5

A=3

lone pair = 2

Geometry: T-shape

d. R=5

A=2

lone pair = 3

Geometry: Linear

LONE PAIR CAUSE MAXIMUM repulsion so they will be placed at equitorial postion (larger bond angle)

5. sp3d2:

a. R=6

A=6

lone pair = 0

Geometry: square bipyramidal

b. R=6

A=5

lone pair = 1

Geometry:square pyramidal

c. R=6

A=4

lone pair = 2

Geometry : square planar.

PRACTICE PROBLEMS:

1. BeCl2 = 2+2 = 2 =sp Linear

2. CO2 = 4/2 =2 =sp Linear

3.XeOF2 = 8/2 + 2/2 = sp3d T-shape

4. XeF6 = 8/2 + 6/2 = Sp3d3 Distorted octahedral or caped octahedrl

SN2 mechanism

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Overview:

The general form of the S_N2 mechanism is as follows:

nuc: = nucleophile
X = leaving group (usually halide or tosylate)

The S_N2 reaction involves displacement of a leaving group (usually a halide or a tosylate), by a nucleophile. This reaction works the best with methyl and primary halides because bulky alkyl groups block the backside attack of the nucleophile, but the reaction does work with secondary halides (although it is usually accompanied by elimination), and will not react at all with tertiary halides. In the following example, the hydroxide ion is acting as the nucleophile and bromine is the leaving group:

Because of the backside attack of the nucleophile, inversion of configuration occurs.

Solvents: Protic solvents such as water and alcohols stabilize the nucleophile so much that it won't react. Therefore, a good polar aprotic solvent is required such as ethers and ketones and halogenated hydrocarbons.

Nucleophiles: A good nucleophile is required since it is involved in the rate-determining step.

Leaving groups: A good leaving group is required, such as a halide or a tosylate, since it is involved in the rate-determining step

Reactions of Phenols

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Compounds in which a hydroxyl group is bonded to an aromatic ring are called phenols. The chemical behavior of phenols is different in some respects from that of the alcohols, so it is sensible to treat them as a similar but characteristically distinct group. A corresponding difference in reactivity was observed in comparing aryl halides, such as bromobenzene, with alkyl halides, such as butyl bromide and tert-butyl chloride. Thus, nucleophilic substitution and elimination reactions were common for alkyl halides, but rare with aryl halides. This distinction carries over when comparing alcohols and phenols, so for all practical purposes substitution and/or elimination of the phenolic hydroxyl group does not occur.

1. Acidity of Phenols

On the other hand, substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols. This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group, as displayed in the following diagram. The alcohol cyclohexanol is shown for reference at the top left. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. The trinitro compound shown at the lower right is a very strong acid called picric acid.

Why is phenol a much stronger acid than cyclohexanol? To answer this question we must evaluate the manner in which an oxygen substituent interacts with the benzene ring. As noted in our earlier treatment of electrophilic aromatic substitution reactions, an oxygen substituent enhances the reactivity of the ring and favors electrophile attack at ortho and para sites. It was proposed that resonance delocalization of an oxygen non-bonded electron pair into the pi-electron system of the aromatic ring was responsible for this substituent effect. Formulas illustrating this electron delocalization will be displayed when the "Resonance Structures" button beneath the previous diagram is clicked. A similar set of resonance structures for the phenolate anion conjugate base appears below the phenol structures.
The resonance stabilization in these two cases is very different. An important principle of resonance is that charge separation diminishes the importance of canonical contributors to the resonance hybrid and reduces the overall stabilization. The contributing structures to the phenol hybrid all suffer charge separation, resulting in very modest stabilization of this compound. On the other hand, the phenolate anion is already charged, and the canonical contributors act to disperse the charge, resulting in a substantial stabilization of this species. The conjugate bases of simple alcohols are not stabilized by charge delocalization, so the acidity of these compounds is similar to that of water. An energy diagram showing the effect of resonance on cyclohexanol and phenol acidities is shown on the right. Since the resonance stabilization of the phenolate conjugate base is much greater than the stabilization of phenol itself, the acidity of phenol relative to cyclohexanol is increased. Supporting evidence that the phenolate negative charge is delocalized on the ortho and para carbons of the benzene ring comes from the influence of electron-withdrawing substituents at those sites. The additional resonance stabilization provided by ortho and para nitro substituents will be displayed by clicking the "Resonance Structures" button a second time. You may cycle through these illustrations by repeated clicking of the button.

2. Substitution of the Hydroxyl Hydrogen

As with the alcohols, the phenolic hydroxyl hydrogen is rather easily replaced by other substituents. For example, phenol reacts easily with acetic anhydride to give phenyl acetate. Likewise, the phenolate anion is an effective nucleophile in S_N2 reactions, as in the second example below.

C₆H₅?OH + (CH₃CO)₂O C₆H₅?O?COCH₃ + CH₃CO₂H

C₆H₅?O^(?) Na⁽⁺⁾ + CH₃CH₂CH₃?Br C₆H₅?O?CH₂CH₂CH₃ + NaBr

3. Electrophilic Substitution of the Phenol Aromatic Ring

The facility with which the aromatic ring of phenols and phenol ethers undergoes electrophilic substitution has been noted. Two examples are shown in the following diagram. The first shows the Friedel-Crafts synthesis of the food preservative BHT from para-cresol. The second reaction is interesting in that it further demonstrates the delocalization of charge that occurs in the phenolate anion. Carbon dioxide is a weak electrophile and normally does not react with aromatic compounds; however, the negative charge concentration on the phenolate ring enables the carboxylation reaction shown in the second step. The sodium salt of salicylic acid is the major ptoduct, and the preference for ortho substitution may reflect the influence of the sodium cation. This is called the Kolbe-Schmidt reaction, and it has served in the preparation of aspirin, as the last step illustrates.

4. Oxidation of Phenols

Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone (also known as 1,4-benzoquinone or simply quinone); an ortho isomer is also known. These compounds are easily reduced to their dihydroxybenzene analogs, and it is from these compounds that quinones are best prepared. Note that meta-quinones having similar structures do not exist. The redox equilibria between the dihydroxybenzenes hydroquinone and catechol and their quinone oxidation states are so facile that milder oxidants than chromate (Jones reagent) are generally preferred. One such oxidant is Fremy's salt, shown on the right. Reducing agents other than stannous chloride (e.g. NaBH₄) may be used for the reverse reaction.
The position of the quinone-hydroquinone redox equilibrium is proportional to the square of the hydrogen ion concentration, as shown by the following half-reactions (electrons are colored blue). The electrode potential for this interconversion may therefore be used to measure the pH of solutions.

Quinone + 2H⁽⁺⁾	2e^(?) ?2e^(?)	Hydroquinone

Although chromic acid oxidation of phenols having an unsubstituted para-position gives some p-quinone product, the reaction is complex and is not synthetically useful. It has been found that salcomine, a cobalt complex, binds oxygen reversibly in solution, and catalyzes the oxidation of various substituted phenols to the corresponding p-quinones. The structure of salcomine and an example of this reaction are shown in the following equation. The solvent of choice for these oxidations is usually methanol or dimethylformamide (DMF).

Isomers

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Isomers

Isomers are molecules with the same molecular formula, but different arrangements of atoms. There are different types of isomers, shown by the diagram on the right.

Functional Isomerism

Functional isomerism, an example of structural isomerism, occurs substances have the same molecular formula but different functional groups. This means that functional isomers belong to different homologous series.

ethanol	methoxymethane	C₂H₆O
		Alcohols have the hydroxyl group, ?OH. Ethers have the functional group R?O?R'.

propanal	propanone (acetone)	C₃H₆O
		Aldehydes and ketones both have the carbonyl group C=O. In ketones this is attached to two carbon atoms; in aldehydes it is attached to 1 or 2 hydrogen atoms.

Positional Isomerism

Positional isomerism, an example of structural isomerism, occurs when functional groups are in different positions on the same carbon chain.

butan-1-ol	butan-2-ol


but-1-ene	but-2-ene	Note: this is cis-but-2-ene, which has a geometric isomer called trans-but-2-ene (select here to find out more)


2-methylphenol	3-methylphenol	4-methylphenol

Chain Isomerism

Chain isomerism occurs when the way carbon atoms are linked together is different from compound to compound. It is an example of structural isomerism, and is also called nuclear isomerism.

pentane

2-methylbutane

2,2-dimethylpropane

THE GENERAL FEATURES OF TRANSITION METAL CHEMISTRY

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This page explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number), complex ion formation, coloured ions, and catalytic activity.

You will find some of this covered quite briefly on this page with links to other parts of the site where the topics are covered in more detail.

The electronic structures of transition metals

What is a transition metal?

The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms.

We'll explore d block elements first:

d block elements

You will remember that when you are building the Periodic Table and working out where to put the electrons, something odd happens after argon.

At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium.

Only after that do the 3d levels fill.

The elements in the Periodic Table which correspond to the d levels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below.

The electronic structures of the d block elements shown are:

Sc		[Ar] 3d¹4s²
Ti		[Ar] 3d²4s²
V		[Ar] 3d³4s²
Cr		[Ar] 3d⁵4s¹
Mn		[Ar] 3d⁵4s²
Fe		[Ar] 3d⁶4s²
Co		[Ar] 3d⁷4s²
Ni		[Ar] 3d⁸4s²
Cu		[Ar] 3d¹⁰4s¹
Zn		[Ar] 3d¹⁰4s²

You will notice that the pattern of filling isn't entirely tidy! It is broken at both chromium and copper.

Note: This is something that you are just going to have to accept. There is no simple explanation for it which is usable at this level. Any simple explanation which is given is faulty!

People sometimes say that a half-filled d level as in chromium (with one electron in each orbital) is stable, and so it is - sometimes! But you then have to look at why it is stable. The obvious explanation is that chromium takes up this structure because separating the electrons minimises the repulsions between them - otherwise it would take up some quite different structure.

But you only have to look at the electronic configuration of tungsten (W) to see that this apparently simple explanation doesn't always work. Tungsten has the same number of outer electrons as chromium, but its outer structure is different - 5d⁴6s². Again the electron repulsions must be minimised - otherwise it wouldn't take up this configuration. But in this case, it isn't true that the half-filled state is the most stable - it doesn't seem very reasonable, but it's a fact! The real explanation is going to be much more difficult than it seems at first sight.

Neither can you use the statement that a full d level (for example, in the copper case) is stable, unless you can come up with a proper explanation of why that is. You can't assume that looking nice and tidy is a good enough reason!

If you can't explain something properly, it is much better just to accept it than to make up faulty explanations which sound OK on the surface but don't stand up to scrutiny!

Transition metals

Not all d block elements count as transition metals! There are discrepancies between the various UK-based syllabuses, but the majority use the definition:

A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.

On the basis of this definition, scandium and zinc don't count as transition metals - even though they are members of the d block.

Scandium has the electronic structure [Ar] 3d¹4s². When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The Sc³⁺ ion has no d electrons and so doesn't meet the definition.

Zinc has the electronic structure [Ar] 3d¹⁰4s². When it forms ions, it always loses the two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d¹⁰. The zinc ion has full d levels and doesn't meet the definition either.

By contrast, copper, [Ar] 3d¹⁰4s¹, forms two ions. In the Cu⁺ ion the electronic structure is [Ar] 3d¹⁰. However, the more common Cu²⁺ ion has the structure [Ar] 3d⁹.

Copper is definitely a transition metal because the Cu²⁺ ion has an incomplete d level.

Transition metal ions

You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because in the empty atom, 4s orbitals have a lower energy than 3d orbitals.

However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital.

The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons.

Note: This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.

Remember this:

When d-block elements form ions, the 4s electrons are lost first.

To write the electronic structure for Co²⁺:

Co		[Ar] 3d⁷4s²
Co²⁺		[Ar] 3d⁷

The 2+ ion is formed by the loss of the two 4s electrons.

To write the electronic structure for V³⁺:

V		[Ar] 3d³4s²
V³⁺		[Ar] 3d²

The 4s electrons are lost first followed by one of the 3d electrons.

Note: You will find more examples of writing the electronic structures for d block ions, by following this link.

Use the BACK button on your browser to return quickly to this page.

Variable oxidation state (number)

One of the key features of transition metal chemistry is the wide range of oxidation states (oxidation numbers) that the metals can show.

Note: If you aren't sure about oxidation states, you really need to follow this link before you go on.

Use the BACK button on your browser to return quickly to this page.

It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like sulphur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals.

However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent.

Examples of variable oxidation states in the transition metals

Iron

Iron has two common oxidation states (+2 and +3) in, for example, Fe²⁺ and Fe³⁺. It also has a less common +6 oxidation state in the ferrate(VI) ion, FeO₄^2-.

Manganese

Manganese has a very wide range of oxidation states in its compounds. For example:

+2	in Mn²⁺
+3	in Mn₂O₃
+4	in MnO₂
+6	in MnO₄^2-
+7	in MnO₄^-

Other examples

You will find the above examples and others looked at in detail if you explore the chemistry of individual metals from the transition metal menu. There is a link to this menu at the bottom of the page.

Explaining the variable oxidation states in the transition metals

We'll look at the formation of simple ions like Fe²⁺ and Fe³⁺.

When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound.

There are several energy terms to think about, but the key ones are:

The amount of energy needed to ionise the metal (the sum of the various ionisation energies)

The amount of energy released when the compound forms. This will either be lattice enthalpy if you are thinking about solids, or the hydration enthalpies of the ions if you are thinking about solutions.

The more highly charged the ion, the more electrons you have to remove and the more ionisation energy you will have to provide.

But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion.

Thinking about a typical non-transition metal (calcium)

Calcium chloride is CaCl₂. Why is that?

If you tried to make CaCl, (containing a Ca⁺ ion), the overall process is slightly exothermic.

By making a Ca²⁺ ion instead, you have to supply more ionisation energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca²⁺ ions than there is if you only have a 1+ ion. The overall process is very exothermic.

Because the formation of CaCl₂ releases much more energy than making CaCl, then CaCl₂ is more stable - and so forms instead.

What about CaCl₃? This time you have to remove yet another electron from calcium.

The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionisation energy between the second and third electron removed.

Although there will be a gain in lattice enthalpy, it isn't anything like enough to compensate for the extra ionisation energy, and the overall process is very endothermic.

It definitely isn't energetically sensible to make CaCl₃!

Thinking about a typical transition metal (iron)

Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion.

Fe		[Ar] 3d⁶4s²
Fe²⁺		[Ar] 3d⁶
Fe³⁺		[Ar] 3d⁵

The 4s orbital and the 3d orbitals have very similar energies. There isn't a huge jump in the amount of energy you need to remove the third electron compared with the first and second.

The figures for the first three ionisation energies (in kJ mol^-1) for iron compared with those of calcium are:

metal	1st IE	2nd IE	3rd IE
Ca	590	1150	4940
Fe	762	1560	2960

There is an increase in ionisation energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. However, there is much less increase when you take the third electron from iron than from calcium.

In the iron case, the extra ionisation energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made.

The net effect of all this is that the overall enthalpy change isn't vastly different whether you make, say, FeCl₂ or FeCl₃. That means that it isn't too difficult to convert between the two compounds.

The formation of complex ions

What is a complex ion?

A complex ion has a metal ion at its centre with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by co-ordinate (dative covalent) bonds. (In some cases, the bonding is actually more complicated than that.)

The molecules or ions surrounding the central metal ion are called ligands.

Simple ligands include water, ammonia and chloride ions.

What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion.

Some examples of complex ions formed by transition metals

: [Fe(H₂O)₆]²⁺

[Co(NH₃)₆]²⁺

[Cr(OH)₆]^3-

[CuCl₄]^2-

Other metals also form complex ions - it isn't something that only transition metals do. Transition metals do, however, form a very wide range of complex ions.

The formation of coloured compounds

Some common examples

The diagrams show aproximate colours for some common transition metal complex ions.

You will find these and others discussed if you follow links to individual metals from the transition metal menu (link at the bottom of the page).

Alternatively, you could explore the complex ions menu (follow the link in the help box which has just disappeared off the top of the screen).

The origin of colour in the transition metal ions

When white light passes through a solution of one of these ions, or is reflected off it, some colours in the light are absorbed. The colour you see is how your eye perceives what is left.

Attaching ligands to a metal ion has an effect on the energies of the d orbitals. Light is absorbed as electrons move between one d orbital and another. This is explained in detail on another page.

Catalytic activity

Transition metals and their compounds are often good catalysts. A few of the more obvious cases are mentioned below, but you will find catalysis explored in detail elsewhere on the site (follow the link after the examples).

Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is expored in the main catalysis section.

Transition metals as catalysts

Iron in the Haber Process

The Haber Process combines hydrogen and nitrogen to make ammonia using an iron catalyst.

Nickel in the hydrogenation of C=C bonds

This reaction is at the heart of the manufacture of margarine from vegetable oils.

However, the simplest example is the reaction between ethene and hydrogen in the presence of a nickel catalyst.

Transition metal compounds as catalysts

Vanadium(V) oxide in the Contact Process

At the heart of the Contact Process is a reaction which converts sulphur dioxide into sulphur trioxide. Sulphur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst.

Iron ions in the reaction between persulphate ions and iodide ions

Persulphate ions (peroxodisulphate ions), S₂O₈^2-, are very powerful oxidising agents. Iodide ions are very easily oxidised to iodine. And yet the reaction between them in solution in water is very slow.

The reaction is catalysed by the presence of either iron(II) or iron(III) ions.

Short note on halogens

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The halogens or halogen elements are a series of nonmetal elements from Group 17 (old-style: VII or VIIA; Group 7 IUPAC Style) of the periodic table, comprising fluorine, F, chlorine, Cl, bromine, Br, iodine, I, astatine, At.

The group of halogens is the only group which contains elements in all three familiar states of matter at standard temperature and pressure.

Abundance

Owing to their high reactivity, the halogens are found in the environment only in compounds or as ions. Halide ions and oxoanions such as IO₃^? can be found in many minerals and in seawater. Halogenated organic compounds can also be found as natural products in living organisms. In their elemental forms, the halogens exist as diatomic molecules, but these only have a fleeting existence in nature and are much more common in the laboratory and in industry. At room temperature and pressure, fluorine and chlorine are gases, bromine is a liquid and iodine and astatine are solids; Group 7 is therefore the only periodic table group exhibiting all three states of matter.

Etymology

The term halogen originates from 18th century scientific French nomenclature based on erring adaptations of Greek roots; the Greek word halos meaning "salt", and genes meaning "production" ? referring to elements which produce a salt in union with a metal.

Properties

The halogens show a number of trends when moving down the group - for instance, decreasing electronegativity and reactivity, increasing melting and boiling point.

Halogen	Standard Atomic Weight (u)	Melting Point (K)	Boiling Point (K)	Electronegativity (Pauling)
Fluorine	18.998	53.53	85.03	3.98
Chlorine	35.453	171.6	239.11	3.16
Bromine	79.904	265.8	332.0	2.96
Iodine	126.904	386.85	457.4	2.66
Astatine	(210)	575	610 ?	2.2
Ununseptium	(291)*	*	*	*

* Ununseptium has not yet been discovered; values are either unknown if no value appears, or are estimates based on other similar chemicals.

Diatomic halogen molecules

halogen	molecule	structure	model	d(X?X) / pm (gas phase)	d(X?X) / pm (solid phase)
fluorine	F₂			143	149
chlorine	Cl₂			199	198
bromine	Br₂			228	227
iodine	I₂			266	272

Chemistry

Reactivity

Halogens are highly reactive, and as such can be harmful or lethal to biological organisms in sufficient quantities. Fluorine is the most reactive element in existence, attacking such inert materials as glass, and forming compounds with the heavier noble gases. It is a corrosive and highly toxic gas. The reactivity of fluorine is such that, if used or stored in laboratory glassware, it can react with glass in the presence of small amounts of water to form SiF₄. Thus flourine must be handled with substances such as Teflon, extremely dry glass, or metals such as copper or steel which form a protective layer of fluoride on their surface.

Both Chlorine and bromine are used as disinfectants for drinking water, swimming pools, fresh wounds, dishes, and surfaces. They kill bacteria and other potentially harmful microorganisms through a process known as sterilization. Their reactivity is also put to use in bleaching. Chlorine is the active ingredient of most fabric bleaches and is used in the production of most paper products.

Hydrogen halides

The halogens all form binary compounds with hydrogen, the hydrogen halides, HX (HF, HCl, HBr, HI), a series of particularly strong acids. When in aqueous solution, the hydrogen halides are known as hydrohalic acids. HAt, or "hydrastatic acid", should also qualify, but it is not typically included in discussions of hydrohalic acid due to astatine's extreme instability toward alpha decay.

Interhalogen compounds

The halogens react with each other to form interhalogen compounds. Diatomic interhalogen compounds (e.g. BrF, ICl, ClF) bear resemblance to the pure halogens in some respects. The properties and behaviour of a diatomic interhalogen compound tend to be intermediate between those of its parent halogens. Some properties, however, are found in neither parent halogen ? Cl₂ and I₂ are soluble in CCl₄ but ICl is not, since it is a polar molecule due to the relatively large electronegativity difference between I and Cl.

Organohalogen compounds

Many synthetic organic compounds such as plastic polymers, and a few natural ones, contain halogen atoms; these are known as halogenated compounds or organic halides. Chlorine is by far the most abundant of the halogens, and the only one needed in relatively large amounts (as chloride ions) by humans. For example, chloride ions play a key role in brain function by mediating the action of the inhibitory transmitter GABA and are also used by the body to produce stomach acid. Iodine is needed in trace amounts for the production of thyroid hormones such as thyroxine. On the other hand, neither fluorine nor bromine are believed to be essential for humans, although small amounts of fluoride can make tooth

Alkenes & elimination it is important

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Chapter 5 Notes: Alkene Structure, Elimination Reactions

Alkenes

C=C double bond
Cn H2n - general formula
unsaturated
not "saturated" with maximum hydrogens
unsaturated fats have some double bonds (easier to digest)
note that cycloalkanes also are Cn H2n

Unsaturated Compounds

alkenes & polyenes (dienes, trienes, etc.)
arenes - aromatic rings
alkynes - triple bonds
combinations - e.g., enynes

Bonding Trends

Bond Lengths in Angstroms (Bond Strengths in kcal/mole)
Compound	Hybrid	C-C bond	C-H bond
ethane	sp3	1.54 (88)	1.11 (98)
ethylene	sp2	1.34 (172)	1.10 (104)
acetylene	sp	1.21 (230)	1.08 (125)

multiple bonds are stronger
shorter bonds are stronger

Alkene Nomenclature

parent alkene is the longest continuous carbon chain that includes the double bond
number from the end that gives the double bond the lower number
use -en- instead of -an-
use only the first number of the double bond
name substituents as usual

Cycloalkene Nomenclature

double bond assumed at C1-C2
number in the direction that gives substituents lower numbers

3-methylcyclohexene

5,5-dimethyl-1,3-cyclopentadiene

Alkene Structure

C=C double bond is one sigma bond and one pi bond
sp2 hybridization (trigonal planar)
pi bond doesn't rotate
unlike ethane, ethene has no other conformers

Cis-Trans Isomers

another example of stereoisomers
2-butene has two isomers:

trans-2-butene

cis-2-butene

Cis-Trans Isomers

an alkene can have cis-trans isomers only when each C in the double bond is attached to 2 different groups

Stereochemistry Designation

cis or trans ?

note that cis-trans works unambiguously only for disubstituted alkenes
cis and trans can be used to follow the structure of the parent chain in some cases
E,Z designation is more general and always works

E,Z Designation

assign priorities to the two groups on each C of the double bond
( one high priority, one low priority for each C )
if the the two high priority groups are on the same side, it is Z
if the the two high priority groups are on opposite sides, it is E

Priority Rules

consider just each atom bonded to the C
higher atomic number is higher priority
for a tie, consider each atom bonded next in turn
for a double bond, consider it like two identical atoms

Practice Nomenclature

(E)-3,4-dimethyl-2-octene

(Z)

Relative Stabilities

greater substitution leads to greater stability
tetrasubstituted > trisubstituted > disubstituted > monosubstituted
trans typically more stable than cis

Elimination Reactions - Preparation of Alkenes

dehydrogenation (loss of H2 from an alkane)
dehydration (loss of H2O from an alcohol)
dehydrohalogenation (loss of HX from an alkyl halide)

Dehydration

regioselectivity - Zaitsev Rule (prefer more stable alkene)
stereoselectivity
mechanism - carbocation intermediate
reactivity: 3° > 2° > 1°
carbocation rearrangements - hydride shifts, alkyl shifts

Dehydrohalogenation

base-promoted elimination
E2 mechanism - concerted, bimolecular
stereoselectivity - anti elimination
E1 elimination for 3° alkyl halides in absence of base

very important note on benzene

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Aromatic Compounds

Aromaticity

pleasant odor
unusually low reactivity
- substitution, not addition
unusually stable
characteristic ring structurewith delocalized pi bonding

Benzene - stability

C6H6 (1,3,5-cyclohexatriene ?)
no typical C=C reactions
unreactive with HX, X2, KMnO4
reaction requires extreme conditions
when reaction does occur, it is substitution not addition

Benzene - structure

all C are sp2 (trigonal, 120° angles)
ideal for a planar hexagon
all C-C bonds are the same (139 pm)
compare C-C (154 pm), C=C (134 pm)
cyclic conjugated pi bonds are unusually stable (resonance)

Nomenclature of Aromatics

monosubstituted benzenes:
common names - see Table 5.1
disubstituted benzenes:
ortho (1,2-), meta (1,3-), para (1,4-)

p-nitrobenzoic acid / / / 2-chloro-6-ethylaniline

Nomenclature of Aromatics

group names:
phenyl C6H5
benzyl C6H5CH2

(E)-1-phenyl-1-butene

Electrophilic Aromatic Substitution

benzene can be made to react with very strong electrophiles (E+)
intermediate is a carbocation
(like addition to one of the pi bonds)
nucleophiles don't add to the cation
(H+ leaves, regenerates benzene ring)
reaction is substitution (E+ for H+)

Mechanism of Aromatic Substitution

Mechanism - why slower than alkenes

Ea for electrophilic attack on benzene is greater than Ea for electrophilic attack on an alkene
although the cation intermediate is delocalized and more stable than an alkyl cation, benzene is much more stable than an alkene

Mechanism - why substitution

the substitution product regains the aromatic stability
an addition product would be a conjugated diene, not as stable

Bromination of Benzene

electrophile is Br+
generated from Br2 + FeBr3

Chlorination of Benzene

electrophile is Cl+
generated from Cl2 + FeCl3

Nitration of Benzene

electrophile is NO2+
generated from H2SO4 + HNO3

Sulfonation of Benzene

electrophile is HSO3+
generated from H2SO4 + SO3

Friedel-Crafts Alkylation

electrophile is an alkyl cation (R+)
generated from RCl + AlCl3

Friedel-Crafts Acylation

electrophile is an acyl cation (RCO+)
generated from RCOCl + AlCl3

Substituent Effects

substituents on the benzene ring can affect the reaction in two ways:
reactivity - substituted benzene may react faster or slower than benzene itself reacts
orientation - the new group may be oriented ortho, meta, or para with respect to the original substituent

Reactivity Effects

activating - reaction is faster
observed with electron-donating groups that make the ring more electron-rich
deactivating - reaction is slower
observed with electron-withdrawing groups that make the ring less electron-rich

Orientation Effects

substituent already present on the benzene ring determines the location of the new group
ortho,para-directors: electron-donating groups direct the new group mainly to ortho & para
meta-directors: electron-withdrawing groups direct new group mainly meta

Ortho, Para Directors

the best cation is formed when the electrophile adds either ortho or para
(better than unsubstituted)

Meta Directors

the best cation is formed when the electrophile adds meta
(but this is worse than unsubstituted)

Classifying Substituents

activating and o,p-directing:
alkyl, aryl, O and N groups
deactivating and m-directing:
N+ groups, polar multiple bonds
deactivating but o,p-directing:
the halogens (F, Cl, Br, I)
(electron-withdrawing atoms, but lone pairs can stabilize the cation when it is ortho or para)

Oxidation of Side Chains

alkyl groups attached to aromatic rings are easily oxidized to carboxylic acids

Reduction of Aromatic Rings

under extreme conditions, a benzene ring can be hydrogenated to a cyclohexane ring

Polycyclic Aromatics

larger aromatic compounds can be made from fused benzene rings

naphthalene / / / anthracene

Heterocyclic Aromatics

some aromatic rings have atoms other than carbon

pyridine / / / pyrrole / / / furan

Synthetic Strategy

synthesis of complex compounds requires attention to the order in which groups are attached
retrosynthetic analysis - think backwards one step at a time
(What reaction could have made this target compound?)

Synthesis Example

target compound: p-nitrobenzoic acid

Synthesis Example

Graphite

extended sheets of benzene rings
electrically conductive
good lubricant

Fullerenes

curved closed form of 60 carbon atoms in benzene rings with intervening 5-membered rings
(soccer ball pattern)
subject of the 1996 Nobel Prize in chemistry

Excellent notes on optical isomer please c it

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STEREOISOMERISM - OPTICAL ISOMERISM Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of optical isomers in a molecule. What is stereoisomerism? What are isomers? Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is *not* a form of stereoisomerism, and is dealt with on a separate page.

What are stereoisomers? In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Optical isomerism is one form of stereoisomerism. Optical isomerism *Why optical* isomers?** Optical isomers are named like this because of their effect on plane polarised light.

Simple substances which show optical isomerism exist as two isomers known as *enantiomers. A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as the (+)* form. For example, one of the optical isomers (enantiomers) of the amino acid alanine is known as (+)alanine. A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is known as the (-) form. So the other enantiomer of alanine is known as or (-)alanine. If the solutions are equally concentrated the amount of rotation caused by the two isomers is exactly the same - but in opposite directions. When optically active substances are made in the lab, they often occur as a 50/50 mixture of the two enantiomers. This is known as a *racemic mixture* or *racemate*. It has no effect on plane polarised light.

How optical isomers arise The examples of organic optical isomers required at A' level all contain a carbon atom joined to four different groups. These two models each have the same groups joined to the central carbon atom, but still manage to be different: Obviously as they are drawn, the orange and blue groups aren't aligned the same way. Could you get them to align by rotating one of the molecules? The next diagram shows what happens if you rotate molecule B. They still aren't the same - and there is no way that you can rotate them so that they look exactly the same. These are isomers of each other. They are described as being *non-superimposable* in the sense that (if you imagine molecule B being turned into a ghostly version of itself) you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direction.

What happens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibility. The two models are aligned exactly as before, but the orange group has been replaced by another pink one. Rotating molecule B this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups attached to the central carbon are different. Chiral and achiral molecules The essential difference between the two examples we've looked at lies in the symmetry of the molecules. If there are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine slicing through the molecule, the left-hand side is an exact reflection of the right-hand side. Where there are four groups attached, there is no symmetry anywhere in the molecule. A molecule which has no plane of symmetry is described as *chiral. The carbon atom with the four different groups attached which causes this lack of symmetry is described as a chiral centre* or as an *asymmetric carbon atom. The molecule on the left above (with a plane of symmetry) is described as achiral. Only chiral molecules have optical isomers. The relationship between the enantiomers* One of the enantiomers is simply a *non-superimposable mirror image* of the other one. In other words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its mirror image) have a different spatial arrangement, and so can't be superimposed on each other. If an *achiral* molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in space, you could make the two look identical. It would be possible to superimpose the original molecule and its mirror image. Some real examples of optical isomers *Butan-2-ol* The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star. It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image.

Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups - look, for example, at the ethyl group at the top of the diagram. It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers. So which of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level purposes, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly. *2-hydroxypropanoic acid (lactic acid)* Once again the chiral centre is shown by a star. The two enantiomers are: It is important this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you joining it on to the central carbon wrongly. If you draw it like this in an exam, you won't get the mark for that isomer even if you have drawn everything else perfectly. *2-aminopropanoic acid (alanine)* This is typical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is replaced by -NH₂ The two enantiomers are: Only one of these isomers occurs naturally: the (+) form. You can't tell just by looking at the structures which this is. It has, however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand structure, and the way the groups are arranged around the central carbon atom is known as an L- configuration. Notice the use of the capital L. The other configuration is known as D-. So you may well find alanine described as *L-(+)alanine*. That means that it has this particular structure and rotates the plane of polarisation clockwise. Even if you know that a different compound has an arrangement of groups similar to alanine, you still can't say which way it will rotate the plane of polarisation. The other amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH₃ group), but some are (+) forms and others are (-) forms. It's quite common for natural systems to only work with one of the enantiomers of an optically active substance. It isn't too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with. In the lab, it is quite common to produce equal amounts of both forms of a compound when it is synthesised. This happens just by chance, and you tend to get racemic mixtures.

Oxidizing Reagents

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Introduction

one of the unique characteristics of carbon is that it has nine stable oxidation states. It should not be surprising that organic chemists have developed reagents that allow them to alter these oxidation levels. This topic presents a survey of some of those reagents.

Oxidizing Reagents

We have already seen several examples of such reagents in our discussion of the oxidation of alcohols. They are repeated here for the sake of completeness.

Chromic Acid

This reagent is prepared by mixing sodium or potassium dichromate with sulfuric acid as shown in Equation 1.

It is used to oxidize secondary alcohols to ketones:

It may also be used to oxidize primary alcohols to carboxylic acids. As Equation 3 indicates, the alcohol is initially oxidized to an aldehyde. Under the reaction conditions, a molecule of water adds to the carbonyl group to form a hydrate which is subsequently oxidized to the carboxylic acid.

Pyridinium Chlorochromate (PCC)

In order to prevent aldehydes from further oxidation, it is necessary to avoid the addition of water to the carbonyl group. PCC was developed as a non-aqueous alternative to chromic acid. Using this reagent, 2-phenylethanol may be oxidized to phenylacetaldehyde without subsequent oxidation to phenylacetic acid:

Potassium permanganate (KMnO₄) and Osmium tetroxide (OsO₄)

These reagents are used to convert alkenes into the corresponding 1,2-diols (glycols) by a process called syn hydroxylation. Equation 5 illustrates the process for the reaction of 1,2-dimethylcyclohexene with a dilute solution of potassium permanganate.

The reaction is thought to involve the formation of an intermediate cyclic permanganate ester which is readily hydrolysed under the reaction conditions to yield the 1,2-diol. A cyclic osmate ester is generated with OsO₄.

Exercise 1 Draw a Lewis structure for the permanganate ion, MnO₄^-. How many resonance structures can you draw for this ion?

Since aqueous KMnO₄ is purple, this reaction is often used as a qualitative test for the presence of an alkene: a dilute solution of permanganate is added to a sample of the unknown compound; if the color is discharged, the test is taken as positive. The formation of a grey-black precipitate of manganese dioxide confirms the analysis.

Ozone

Ozone, O₃, is an allotrope of oxygen. It is a highly reactive molecule that is generated by passing a stream of dioxygen over a high voltage electric discharge. (It is possible to smell ozone in the atmosphere after a lightning storm if the lightning has struck nearby.) It is not possible to draw a single structure for O₃ in which each oxygen atom has a filled valence shell and is, at the same time, uncharged. Rather, resonance theory describes the structure of this compound as a hybrid of the three resonance contributors shown in Figure 1.

Figure 1

Ozone: An Allotrope of Oxygen

In a process called ozonolysis, an alkene is treated with ozone to produce intermediates called ozonides, which are reduced directly, generally with zinc metal in acetic acid, to yield aldehydes or ketones, depending on the substituents attached to the double bond of the initial alkene. Equations 6 -8 provide three specific examples.

Note that an aromatic ring is resistant to ozone.

The value of ozonolysis lies in the structural insight it affords a chemist who is trying to determine the identity of an unknown compound. Figure 2 illustrates this idea.

Figure 2

Structural Elucidation with Ozonolysis

The unknown is degraded into smaller, simpler molecules that are more readily identified. Once identified, these fragments are then mentally reconnected by joining the carbonyl carbons together to create an alkene.

Exercise 2 Draw the structures of the alkenes that yielded the following aldehydes and/or ketones upon ozonolysis.

Reducing Agents

In our discussion of the oxidation of alcohols, we classified this process as a 1,2-elimination of the "elements of" dihydrogen. The reverse process, the 1,2-addition of the "elements of" dihydrogen to a multiple bond, constitutes a reduction. The reagents used for the 1,2-addition of the "elements of" dihydrogen to a multiple bond depend upon the nature of the multiple bond. For homonuclear multiple bonds, i.e. alkenes and alkynes, the most common method is called catalytic hydrogenation: a solution of the alkene or alkyne is mixed with dihydrogen gas in the presence of a catalytic quantity of a transition metal. For heteronuclear multiple bonds; aldehydes, ketones, nitriles, esters, etc., addition of the "elements of" dihydrogen is genarally accomplished in two steps, addition of hydride ion, :H^-, followed by addition of H⁺. Since the addition of hydride ion is rate determining, these reductions are called hydride ion reductions. We'll take a look at catalytic hydrogenation first.

Catalytic Hydrogenation

The catalyst most commonly used to reduce carbon-carbon multiple bonds consists of platinum metal dispersed over the surface of finely divided charcoal (Pt/C). Palladium and rhodium are also used (Pd/C and Rh/C). These reagents are available commercially. The catalyst is mixed with a solution of the alkene or alkyne dissolved in an inert solvent such as ethanol or diethyl ether. Since the reaction mixture is heterogeneous, it is important that the catalyst be dispersed over a large surface area in order to insure adequate contact between the reactants and the catalyst. The charcoal provides the required surface area.

The mechanistic details of catalytic hydrogenations are uncertain because of the difficulties associated with studying heterogeneous reactions. Figure 3 animates the generally accepted view of the process using 2-butyne as an example.

Figure 3

Catalytic Hydrogenation

Adsorption onto the catalytic surface brings the reactants into proximity. It also weakens the H-H and the C-C bonds, increasing the reactivity of the reactants. The details of the transfer of the hydrogen atoms from the platinum to the alkyne are uncertain. However, as the animation indicates, both hydrogen atoms add to the same side of the pi bond, leading to the formation of cis-2-butene. In other words, catalytic hydrogenation of alkenes and alkynes involves the syn addition of the "elements of" dihydrogen to the multiple bond. Equation 9 illustrates the syn addition of dihydrogen to (E)-3-chloro-2-phenyl-2-butene. Since the addition occurs to the top and bottom faces of the pi bond with equal probability, the reaction produces a racemic mixture of the two stereoisomers shown.

As reaction 10 indicates, catalytic hydrogenations are not restricted to alkenes and alkynes. Compounds containing multiple bonds between carbon and a heteroatom such as oxygen or nitrogen may also be reduced catalytically.

While the outcome depicted in reaction 10 may be desireable, it would be nice to have a method to hydrogenate heteronuclear multiple bonds while leaving carbon-carbon multiple bonds untouched. Such a method exits. It takes advantage of the fact that, unlike C-C multiple bonds, which are relatively non-polar, heteronuclear multiple bonds have a permanent bond dipole; the carbon is electron deficient while the heteroatom is electron rich. This means that the carbon atom of a heteronuclear multiple bond is inherently reactive toward negatively charged reagents, while the heteroatom is inherently reactive towards positively charged reagents. Figure 4 demonstrates this reality for negatively and positively charged hydrogen atoms.

Figure 4

Opposites Attract

The net result of the addition of :H^- and H⁺ to the multiple bond is the 1,2-addition of the "elements of" dihydrogen. Since it is not possible to have significant concentrations of :H^- and H⁺ in the same flask at the same time (why?), hydrogenation of heteronuclear multiple bonds is normally a 2-step process; hydride ion adds to the carbon atom in the first step, while a proton adds to the heteroatom in the second. The remainder of this topic will consider different reagents that act as a source of hydride ion.

Hydride Ion Reductions

Reagents that act as hydride ion donors all share one structural feature: They all contain at least one hydrogen atom that is bonded to another atom which is less electronegative than hydrogen. The greater the difference in electronegativity, the more reactive the reagent will be as a hydride donor. The most reactive source of hydride ion is lithium aluminum hydride, LiAlH₄. This material is a grey solid that reacts violently with protic solvents. Most commonly it is used as a suspension in a dry, inert solvent such as diethyl ether or THF. A solution of the compound to be reduced is added to this suspension and stirred vigorously until analysis indicates that all of the starting material has reacted. At this point the mixture is acidified by the careful addition of aqueous acid. Figure 5 illustrates these two steps for the reduction of acetophenone.

Figure 5

LiAlH₄ Reduction of Acetophenone

Note that all four hydrogen atoms attached to the aluminum in LiAlH₄ are active; one mole of LiAlH₄ will reduce four moles of the ketone.

LiAlH₄ is so reactive that it will reduce almost any type of heteronuclear multiple bond. It will even reduce carboxylic acids and esters to the corresponding primary alcohols as indicated in reactions 11 and 12, and it reduces amides to amines as shown in Equation 13.

Clearly these reactions are more complicated than the mechanism shown in Figure 5 would suggest. Elimination of water as well as reduction must be involved.

Before we consider less reactive hydride donors, let's revisit the reaction of the unsaturated ketone we considered in Equation 10. Figure 6 compares the catalytic hydrogenation of pent-4-en-2-one with its reduction by LiAlH₄.

Figure 6

Reduction of Homonuclear and Heteronuclear Multiple Bonds

Because simple, i.e. non-conjugated, double bonds are non-polar, they are non-reactive towards nucleophilic reagents.

Exercise 3 Why are alkenes non-reactive towards nucleophiles, but react readily with electrophiles?

While lithium aluminum hydride's high reactivity may be useful, it can also be a disadvantage if you want to selectively reduce one heteronuclear multiple bond in a compound that contains several. In that case it is desireable to have a reagent that is less reactive and more selective. One such reagent is sodium borohydride, NaBH₄. This material is a white solid. It is much less reactive than LiAlH₄. In fact, it is possible to reduce aldehydes, ketones, and esters with NaBH₄ even in protic solvents such as ethanol. It will not reduce carboxylic acids or amides. The mechanism of the reaction is very similar to that shown in Figure 5. Figure 7 compares the results of the reduction of a compound that contains both an ester and an amide group with LiAlH₄ and NaBH₄.

Figure 7

LiAlH₄ vs NaBH₄

Now let's consider another aspect of reactivity. For compounds that belong to the carboxylic acid family, in particular carboxylic acids, esters, and amides, the oxidation level of the carboxyl carbon is +3. As you can see from the reactions in Figure 7, the oxidation level of the carboxyl carbon decreases to -1 when the carboxyl group is reduced to a primary alcohol. The question then becomes, "Can you stop the reduction at an intemediate oxidation level?" The answer is yes. Equation 14 shows how.

Here the ester is reduced to an aldehyde. The oxidation level of the carbonyl carbon decreases from +3 to +1. The trick is to use a sterically hindered reducing agent that has only one active hydrogen. In this case the reagent is called diisobutyl aluminum hydride, sometimes abbreviated DIBAL. By using 1 equivalent of DIBAL at low temperatures it is possible to reduce the ester to the corresponding aldehyde without further reduction of the aldehyde to the primary alcohol. Use of more than 1 equivalent will lead to reduction of the aldehyde.

Finally, Table 1 summarizes the reactivities of the various reducing reagents we have considered in this topic.

Table 1

Hydride Reductions

Reducing Agent	Alkenes	Aldehydes	Ketones	Esters	Amides	Carboxylic Acids
H₂, Pt/C	alkanes	1^o alcohols	2^o alcohols	NR	NR	NR
LiAlH₄	NR	1^o alcohols	2^o alcohols	1^o alcohols	amines	1^o alcohols
NaBH₄	NR	1^o alcohols	2^o alcohols	1^o alcohols	NR	NR
DIBAL	....	1^o alcohols	2^o alcohols	aldehydes	aldehydes	1^o alcohols

How to Study Organic Chemistry

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Organic Chemistry is a challenging subject. It uses its own language and employs many very precise concepts yet without referring mathematical tools or aspects. Within first few hours of study we will be able to use the basic concepts to understand a lot about the molecular world around us.

The first difficulty a student encounters is the amount of study material available about organic molecules and their reactions any standard text book is not less then 1400 pages long, we students are expected to learn all this material without investing a considerable time and effort in studying it and if we do so we disproportionate our time allocations with other subjects eventually bringing ourselves under pressure to leave organic chemistry or compromise at other subjects.

The solution of above problem is that someone should work on our part to extract all relevant important matter and concept for us, this someone is our teacher. Most of standard text book (like I L Finar Morison Boyd) available in market are not oriented for JEE, preparation rather these are among the best books available for college students all over the world. The sequence of chapters or 100% content may not return us for the time we have invested. We need not follow line to line of the text but choose the desired component from the word index given at the back of book.

In many ways, learning organic is like learning another language. Make sure you are familiar with basic terms like, electrophile, nucleophile, base, substrate, carbocations, free radical electron releasing groups etc.

You should understand the nature of organic chemistry when a reaction occurs one must first know what reagents are the starting materials and what the final products are? The conversion of starting materials to the products will involve either breaking bonds, making bonds or both. The detailed sequence of which bonds are broken and formed, in what order, and the stereo chemical relationships of these bonds is called a mechanism for the reaction.

Our text book is organized primarily by types of compounds which contain a specified functional group, e.g. Collection of all compounds where ? OH group is connected to a carbon chain or structure are called alcohols.

When we study each type of functional group we will find that each reacts by only a few mechanistic paths and hence has a chemical personality of its own. Do not treat mechanism as just another thing to memorize, remember working organic chemist do not just repeat what is known. They use that knowledge to solve problems and discover new chemistry.

Understanding mechanism is the key to modern organic chemistry although we will be studying hundreds or thousands of reactions, these reactions occur via only a few fundamental mechanistic path ways. It is the recognition of the mechanistic similarities between different reactions that allow organic chemistry to be readily understood. Understanding mechanism will help make sense of thousand of facts that comprise organic chemistry.

IIT-JEE question appears as if it is newly framed for you but you will find at least some what related what you have done in mastering the mechanism the only challenge these questions pose is that you have to identify an appropriate mechanism component which operates there.

Anticipate what will be on the exam. Notice what the teacher spends time on in class. If you teacher assigns a specific problem makes sure you know how to work every one of these problems. Your teacher knows important concepts, mechanism or part of subject from which a question of JEE level can be framed, so they will even say "This problem will be on exam". If your teacher says this, believe it!

When you analyze your minor tests you should understand what you did wrong. You will need to know had to do it right next time because chemistry builds up from the base of knowledge, everything you learn at the beginning will be needed later for something more complicated. If you miss a concept on the first test it will be trouble for all coming minor tests.

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Hybridization	Lone pairs	Bond pairs	Shape	Example
sp	0	2	Linear (planar)	CO2
sp 2	0	3	Trigonal planar	BF3
sp 2	1	2	Bent or V shape	SO2
sp 3	0	4	Tetrahedral	CH4
sp 3	1	3	Trigonal Pyramidal or Pyramidal	NH3
sp 3	2	2	V Shape or Bent or Angular	H2O
sp 3	3	1	Linear	HF , HCl
sp 3 d	0	5	Trigonal Bipyramidal	PCl5
sp 3 d	1	4	Irregular tetrahedral or See saw	SF4
sp 3 d	2	3	T Shape	ICl3
sp 3 d	3	2	Linear	XeF2
sp 3 d 2	0	6	Square Bipyramidal or Octahedral	SF6
sp 3 d 2	1	5	Square pyramidal	IF5
sp 3 d 2	2	4	Square planar	XeF4
sp 3 d 3	0	7	Pentagonal Bipyramidal	IF7

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formulae for electrostates

current electricity

S-BLOCK

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Basic definations:
1A Alkali metals: Their oxides/hydroxides are strongest bases irresepective of periods.

2A Alkaline earth metals: Because of their carbonates they are most abundant in earth's crust and basic in nature.

PHYSICAL PROPERTIES:

1. Atomic size:

1A > 2A > d-block

2. Sonority is the property to produce sound for longer time due to collision b/w the atoms if once hammered

3.High conductivity: due to presence of free electrons,ions and low Ionization energy

4. LUSTROUS BEHAVIOUR: Due to oscialltion of free electron by absorption of energy from photons

5. Metallic bond strength decreases downward

Metallic bond strength 1/metallic character number of unpaired electron

6. Softness: It decreases left to right. 1A are softest metals due to weakest metallic bond strength or due to crystalline properties. Softness increases downward.

7. Brittleness: It decreases from right to left. Hence these s-block metals are very very less brittle.

Brittleness is the separtion of layers of the particles from the crystal structure by applying the force at a point.

U cn also undrstnd like this: Metal Jitna soft hoga utna hi mushkil hoga uski layers ko separate karna.

8. Crystal structure:

1A similar crystal lattice.

2A dissimilar crystal lattice.

1A physical properties show regular trend

2A phys. prop. show irregular trend.

9. 1A: melting and boiling point decrease downard regularly due to decrease in metallic bond strength

2A: MP and BP decreease dwnwrd irregularly due to different crystal structures.

The BP and MP of s-block metals is less than d block metals due to stronger metallic bond strength in d block elements.

10. DENSITY: It increases downward due to poor shielding effect and increase in atomic mass.

1A Li<K

2A Ca

Li, K and Na are lighter than water

there is SUDDEN EXPANSION in volume of K and Ca

therefore their density is low.

Be has high density because of different LATTICE STRUCTURE

The density of d-block metals is greater than s-block metals due to low atomic mass and high volume of s-block metals...

10. PHOTOELECTRIC EFFECT:

In all metals of s block only K, Rb, Cs give photoelectric effect. They are used in photoelectric effect due to high Ionization energy.

11. FLAME TEST:

s-block salts like NO3-, NO2-, SO42-etc...

are heated on bunsen burner the salt absorbs heat radiations and then releaes energy which belongs to visible part of the spectrum which produces colours. On the basis of colour we can decide the elements..

NOTE: Be and Mg dunno give flame test due to high IE.

Platinum wire is taken due to its inertness and SEMI-SOLID form is taken of the salt due to low IE.

Halides are taken because of emission of high volatile radiation.

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Li Crimson Red

Na Golden Yellow

K Purple/lilac/plane violet

Rb violet

Cs Blue

NOTE: LILAC colour of K has been asked in JEE

Be X

Mg X

Ca Brick

Sr Crimson Red

Ba Apple Green

NOTE: wavlength increases downward and frequency upward.

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All S-block compounds are dimagnetic(exceptions alwys dere)

FLAME TEST is significant for s-block elements because all elements are white in colour and water soluble...

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CHEMICAL PROPERTIES:

1. Chemical reactivity:

Q. What happens when Na is left open in air ?

A. Na in air gives Na₂O

Na₂O comes in contact of water present in atmosphere and it fomrs NaOH

NaOH comes in contact with carbon dioxide of air and forms Na₂CO₃.

Na forms carbonates at end.

Metals like Fe and Al loose their lusturous behaviour due to formation of oxides.

All s-block elements follow the same reaction.

Therefore they can't be kept free in air as they are too reactive.

Hence they are stored in organic solvents like Carbon Tetrachloride, pyridene, toulene, benzene except Be, Mg and Li

Be and Mg form stable layer of oxides. Thus they can be kept freely in water or air.

IMP: Li is stored in wax. it can't be STORED IN KEROSENE because its density is less than that of kerosene. ( IT FLOATS ON THE SURFACE OF KEROSENE)

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SOLUBILITY:

The solubility of compounds of s-block elements with NO3-,HCO3-, OH-, H-, O2- and many other anions INCREASES DOWN THE GROUP.

FOR all 1A compounds the SOLUBILITY WILL INCREASE DOWN THE GROUP (except for Chlorides...itz an exception cn't do much about it).

The same fate will follow for 2A compounds....

BUT THERE ARE IMPORTANT EXCEPTIONS OF COMPOUNDS WHICH HAVE TO BE LEARNED:

Carbonates, MnO4-, Sulphates.

in these compounds the solubility decreases downward (ONLY 2A)

The reason is in these compounds Lattice energy > Hydration energy.

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chemistry

Atomic Strucure, core points !!

all trends for inorganic chemistry

Periodic Table in a flash

What is Hyperconjugation ?

MOLECULAR ORBITAL THEORY SIMPLIFIED

Inorganic 4 ALL ORES from METALLURGY(SIMPLIFIED)

important things usually left in inorganic.

small important things in inorganic

INORGANIC 5 HYBRIDIZATION and GEOMETRY SIMPLIFIED

SN2 mechanism

Reactions of Phenols

1. Acidity of Phenols

2. Substitution of the Hydroxyl Hydrogen

3. Electrophilic Substitution of the Phenol Aromatic Ring

4. Oxidation of Phenols

Isomers

Isomers

Functional Isomerism

Positional Isomerism

Chain Isomerism

THE GENERAL FEATURES OF TRANSITION METAL CHEMISTRY

Short note on halogens

Abundance

Etymology

Properties

Diatomic halogen molecules

Chemistry

Reactivity

Hydrogen halides

Interhalogen compounds

Organohalogen compounds

Alkenes & elimination it is important

Chapter 5 Notes: Alkene Structure, Elimination Reactions

Alkenes

very important note on benzene

Aromatic Compounds

Excellent notes on optical isomer please c it

Oxidizing Reagents

Introduction

Oxidizing Reagents

Chromic Acid

Pyridinium Chlorochromate (PCC)

Potassium permanganate (KMnO4) and Osmium tetroxide (OsO4)

Ozone

Figure 1

Ozone: An Allotrope of Oxygen

Figure 2

Structural Elucidation with Ozonolysis

Reducing Agents

Catalytic Hydrogenation

Figure 3

Catalytic Hydrogenation

Figure 4

Opposites Attract

Hydride Ion Reductions

Figure 5

LiAlH4 Reduction of Acetophenone

Figure 6

Reduction of Homonuclear and Heteronuclear Multiple Bonds

Figure 7

LiAlH4 vs NaBH4

Table 1

Hydride Reductions

How to Study Organic Chemistry

S-BLOCK

Potassium permanganate (KMnO₄) and Osmium tetroxide (OsO₄)

LiAlH₄ Reduction of Acetophenone

LiAlH₄ vs NaBH₄