1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
(C6H5)3P – CH3 + C6H5Li ⎯→ (C6H5)3P – CH2 : – + Br + + C6H6 + LiBr – The first reaction is a nucleophilic substitution reaction. Triphenylphosphine is an excellent nucleophile and a weak base. It reacts readily with 1º and 2º alkyl halide by an SN2 mechanism to displace a halide ion from the alkyl halide to give an alkyltriphenylphosphonium salt. The second reaction is an acid-base reaction. A strong base (usually an alkyllithium or phenyllithium) removes a proton from the carbon that is attached to phosphorus to give the ylide. Phosphorus ylides can be represented as a hybrid of the two resonance structures shown here. Quantum mechanical calculations indicate that the contribution made by the first structure is relatively unimportant. R´´ (C6H5)3P = C (C6H5)3P – C : R´´´ –R´´ + R´´´ The mechanism of the Wittig reaction has been the subject of considerable study. An early mechanistic proposal suggested that the ylide, acting as a carbanion, attacks the carbonyl carbon of the aldehyde or ketone to form an unstable intermediate with separated charges called a betaine. In the next step, the betaine is envisioned as becoming an unstable four-membered cyclic system called an oxaphosphetane, which then spontaneously loses triphenylphosphine oxide to become an alkene. However, studies by E. Vedejs and others suggest that the betaine is not an intermediate and that the oxaphosphetane is formed directly by a cycloaddition reaction. The driving force for the Wittig reaction is the formation of the very strong (∆Hº = 540 kJ mol –1 ) phosphorus –oxygen bond in triphenylphosphine oxide. R–C + – R ´ R ´´ R ´ R ´´ :C–R ´ R – C – C – R´´´ :O: Aldehyde or ketone P(C6H5)3 + .. – :O: P(C6H5)3 + Ylide Betaine (may not be formed) Specific Example : R´ R Alkene (+diastereomer) R ´ R ´´ R – C – C – R´´´ :O .. – P(C6H5)3 Oxaphosphetane R´´ C = C + O = P(C6H5)3 R´´´ Triphenylphosphine oxide Methylenecyclohexane (86%) – + O + :CH2 – P(C6H5)3 CH2 + O=P(C6H5)3 XtraEdge for IIT-JEE 30 FEBRUARY 2012 CH2 O P(C6H5)3 – + CH2 O –P(C6H5)3 Michael Additions : Conjugate additions of enolate anions to α-β-unsaturated carbonyl compound are known generally as Michael additions. An example is the addition of cyclohexanone to C6H5CH=CHCOC6H5 : O O OH – O – – O C6H5CH=CH–CC6H5 O O +H3O + C6H5 CH CH δ– C—O δ– C6H5 C6H5 CH H C H C = O C6H5 The sequence that follows illustrates how a conjugate aldol addition (Michael addition) followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the Robinson anulation (ring-forming) reaction (after the English chemist, Sir Robert Robinon, who won the Nobel Prize in chemistry in 1947 for his research on naturally occurring compounds) : O O CH3 O CH3 CH2 OH + CH2 = CHCCH3 CH2 O O C 2-Methylcyclo- H3C O hexane-1, 3-dione – CH3OH (conjugate addition) Methyl vinyl ketone aldol base condensation (–H2O) O CH3 (65%) O
KEY CONCEPT Inorganic Chemistry Fundamentals Tetragonal distortion of octahedral complexes (Jahn- Teller distortion) : The shape of transition metal complexes are affected by whether the d orbitals are symmetrically or asymmetrically filled. Repulsion by six ligands in an octahedral complex splits the d orbitals on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a completely regular octahedron. The symmetrical arrangements of d electrons are shown in Table. Symmetrical electronic arrangements : Electronic configurat ion d 5 d 6 d 8 d 10 All other arrangements have an asymmetrical arrangement of d electrons. If the d electrons are asymmetrically arranged, they will repel some ligands in the complex more than others. Thus the structure is distorted because some ligands are prevented from approaching the metal. as closely as others. The eg orbitals point directly at the ligands. Thus asymmetric filling of the eg orbitals in some ligands being repelled more than others. This causes a significant distortion of the octahedral shape. In contrast the t2g orbitals do not point directly at the ligands, but point in between the ligand directions. Thus asymmetric filling of the t2g orbitals has only a very small effect on the stereochemistry. Distortion caused by asymmetric filling of the t2g orbitals is usually too small to measure. The electronic arrangements which will produce a large distortion are shown in Table. The two eg orbitals d 2 2 and d 2 are normally x − y z degenerate. However, if they are asymmetrically filled then this degeneracy is destroyed, and the two t2g CO-ORDINATION COMPOUND & METALLURGY eg orbitals are no longer equal in energy. If the d 2 z orbital contains one. Asymmetrical electronic arrangements : Electronic configurati on XtraEdge for IIT-JEE 31 FEBRUARY 2012 d 4 d 7 d 9 more electron than the d 2 2 orbital then the ligands x −y approaching along +z and –z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octahedron along the z axis. This is called tetragonal distortion. Strictly it should be called tetragonal elongation. This form of distortion is commonly obsered. If the d 2 2 orbital contains the extra electron, then x −y elongation will occur along the x and y axes. This means that the ligands approach more closely along the z-axis. Thus there will be four long bonds and two short bonds. This is equivalent to compressing the octahedron along the z axis, and is called tetragonal compression, and it is not possible to predict which will occur. For example, the crystal structure of CrF2 is a distorted rutile (TiO2) structure. Cr 2+ is octahedrally surrounded by six F – , and there are four Cr–F bonds of length 1.98 – 2.01 Å, and two longer bonds of length 2.43 Å. The octahedron is said to be tetragonally distorted. The electronic arrangement in Cr 2+ is d 4 . F – is a weak field ligand, and so the t2g level contains three electrons and the eg level contains one electron. The d 2 2 orbital has four lobes whilst x −y the d 2 orbital has only two lobes pointing at the z ligands. To minimize repulsion with the ligands, the single eg electron will occupy the d 2 orbital. This is z equivalent to splitting the degeneracy of the eg level so that d 2 is of lower energy, i.e. more stable, and z d 2 2 is of higher energy, i.e. less stable. Thus the x −y t2g eg
- Page 3 and 4: Volume - 7 Issue - 8 February, 2012
- Page 5 and 6: 25% increase in number of girls fro
- Page 7 and 8: Dr Amitabha Ghosh was the only Asia
- Page 9 and 10: Sol. (i) The capacitor A with diele
- Page 11 and 12: Sol. (i) Let m be the mass of elect
- Page 13 and 14: Step 5. 1 gram equivalent acid neut
- Page 15 and 16: = 2π ∫ π 2 / 3 2 t sec dt 2 t 1
- Page 17 and 18: (A) (C) A 2B0πqR 3 B 0 PR 3 2 ×
- Page 19 and 20: PHYSICS 1. AB and CD are two ideal
- Page 21 and 22: 4. In a Young's experiment, the upp
- Page 23 and 24: Matter Waves : Planck's quantum the
- Page 25 and 26: The intercept of V0 versus ν graph
- Page 27 and 28: PHYSICS FUNDAMENTAL FOR IIT-JEE The
- Page 29 and 30: etween total mass and number of mol
- Page 31: KEY CONCEPT Organic Chemistry Funda
- Page 35 and 36: 1. It is possible to supercool wate
- Page 37 and 38: z 1 K2 = = y( 0. 48M − z) 1. 26×
- Page 39 and 40: MATHEMATICAL CHALLENGES 1. as φ (a
- Page 41 and 42: 9. Point P (x, 1/2) under the given
- Page 43 and 44: 10. 11 ∴ n(E) = 10 + 9 + 8 + ...
- Page 45 and 46: dx (ii) ∫ 2 ax + bx + c This can
- Page 47 and 48: MATHS Functions with their Periods
- Page 49 and 50: a PHYSICS Questions 1 to 9 are mult
- Page 51 and 52: R P ⊗ B O ω0 14. Magnitude of cu
- Page 53 and 54: XtraEdge for IIT-JEE 51 FEBRUARY 20
- Page 55 and 56: XtraEdge for IIT-JEE 53 FEBRUARY 20
- Page 57 and 58: 7. NaOH can be prepared by two meth
- Page 59 and 60: (C) X = 100 ml of SO2 at 1 bar, 25
- Page 61 and 62: 21. Column-I Column-II (A) If y = 2
- Page 63 and 64: (A) potential energy (PE) increases
- Page 65 and 66: 2. In the reaction 3 Cu + 8 HNO3
- Page 67 and 68: (D) P A T B (S) Pressure is increas
- Page 69 and 70: XtraEdge for IIT-JEE 67 FEBRUARY 20
- Page 71 and 72: 14. Derive the expression for the c
- Page 73 and 74: (a) temperature and (b) Pressure ?
- Page 75 and 76: should be bought to meet the requir
- Page 77 and 78: This work done is stored inside the
- Page 79 and 80: Use : It can be used to accelerate
- Page 81 and 82: 10. (i) (ii) Benzene H CH3 - C = O
KEY CONCEPT<br />
Inorganic<br />
Chemistry<br />
Fundamentals<br />
Tetragonal distortion of octahedral complexes (Jahn-<br />
Teller distortion) :<br />
The shape of transition metal complexes are affected<br />
by whether the d orbitals are symmetrically or<br />
asymmetrically filled.<br />
Repulsion by six ligands in an octahedral complex<br />
splits the d orbitals on the central metal into t2g and eg<br />
levels. It follows that there is a corresponding<br />
repulsion between the d electrons and the ligands. If<br />
the d electrons are symmetrically arranged, they will<br />
repel all six ligands equally. Thus the structure will<br />
be a completely regular octahedron. The symmetrical<br />
arrangements of d electrons are shown in Table.<br />
Symmetrical electronic arrangements :<br />
Electronic<br />
configurat<br />
ion<br />
d 5<br />
d 6<br />
d 8<br />
d 10<br />
All other arrangements have an asymmetrical<br />
arrangement of d electrons. If the d electrons are<br />
asymmetrically arranged, they will repel some<br />
ligands in the complex more than others. Thus the<br />
structure is distorted because some ligands are<br />
prevented from approaching the metal.<br />
as closely as others. The eg orbitals point directly at<br />
the ligands. Thus asymmetric filling of the eg orbitals<br />
in some ligands being repelled more than others. This<br />
causes a significant distortion of the octahedral<br />
shape. In contrast the t2g orbitals do not point directly<br />
at the ligands, but point in between the ligand<br />
directions. Thus asymmetric filling of the t2g orbitals<br />
has only a very small effect on the stereochemistry.<br />
Distortion caused by asymmetric filling of the t2g<br />
orbitals is usually too small to measure. The<br />
electronic arrangements which will produce a large<br />
distortion are shown in Table.<br />
The two eg orbitals d 2 2 and d 2 are normally<br />
x − y<br />
z<br />
degenerate. However, if they are asymmetrically<br />
filled then this degeneracy is destroyed, and the two<br />
t2g<br />
CO-ORDINATION COMPOUND<br />
& METALLURGY<br />
eg<br />
orbitals are no longer equal in energy. If the d 2<br />
z<br />
orbital contains one.<br />
Asymmetrical electronic arrangements :<br />
Electronic<br />
configurati<br />
on<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 31 FEBRUARY <strong>2012</strong><br />
d 4<br />
d 7<br />
d 9<br />
more electron than the d 2 2 orbital then the ligands<br />
x −y<br />
approaching along +z and –z will encounter greater<br />
repulsion than the other four ligands. The repulsion<br />
and distortion result in elongation of the octahedron<br />
along the z axis. This is called tetragonal distortion.<br />
Strictly it should be called tetragonal elongation. This<br />
form of distortion is commonly obsered.<br />
If the d 2 2 orbital contains the extra electron, then<br />
x −y<br />
elongation will occur along the x and y axes. This<br />
means that the ligands approach more closely along<br />
the z-axis. Thus there will be four long bonds and<br />
two short bonds. This is equivalent to compressing<br />
the octahedron along the z axis, and is called<br />
tetragonal compression, and it is not possible to<br />
predict which will occur.<br />
For example, the crystal structure of CrF2 is a<br />
distorted rutile (TiO2) structure. Cr 2+ is octahedrally<br />
surrounded by six F – , and there are four Cr–F bonds<br />
of length <strong>1.</strong>98 – 2.01 Å, and two longer bonds of<br />
length 2.43 Å. The octahedron is said to be<br />
tetragonally distorted. The electronic arrangement in<br />
Cr 2+ is d 4 . F – is a weak field ligand, and so the t2g<br />
level contains three electrons and the eg level contains<br />
one electron. The d 2 2 orbital has four lobes whilst<br />
x −y<br />
the d 2 orbital has only two lobes pointing at the<br />
z<br />
ligands. To minimize repulsion with the ligands, the<br />
single eg electron will occupy the d 2 orbital. This is<br />
z<br />
equivalent to splitting the degeneracy of the eg level<br />
so that d 2 is of lower energy, i.e. more stable, and<br />
z<br />
d 2 2 is of higher energy, i.e. less stable. Thus the<br />
x −y<br />
t2g<br />
eg