According to crystal field theory, the bonding between the metal and the ligands is purely electrostatic. In an isolated metal ion the five d orbitals have the same energy (degenerate). When six ligands approach along the axes in an octahedral field, they repel the d orbitals unequally: the two orbitals pointing along the axes (dx2-y2 and dz2, called eg) are repelled more and rise in energy, while the three orbitals pointing between the axes (dxy, dyz, dzx, called t2g) are repelled less and are lowered in energy. This energy gap between t2g and eg is the crystal field splitting energy (delta o).

Haloalkanes are commonly prepared from alcohols by replacing the -OH group with a halogen. (1) With hydrogen halides: R-OH + HX -> R-X + H2O (reactivity of alcohols is 3 degrees > 2 degrees > 1 degree). (2) With phosphorus halides: 3R-OH + PCl3 -> 3R-Cl + H3PO3, and R-OH + PCl5 -> R-Cl + POCl3 + HCl. (3) With thionyl chloride (the best method, as the by-products are gases): R-OH + SOCl2 -> R-Cl + SO2 + HCl.

R-OH + SOCl2 -> R-Cl + SO2 + HCl

In the SN2 (substitution nucleophilic bimolecular) mechanism, the reaction occurs in a single step; the nucleophile attacks the carbon from the side opposite the leaving group, and bond breaking and bond making occur simultaneously. Its rate depends on both the substrate and the nucleophile, and it gives inversion of configuration (Walden inversion); it is favoured by primary halides. In the SN1 (substitution nucleophilic unimolecular) mechanism, the reaction occurs in two steps through a carbocation intermediate; its rate depends only on the substrate, and it usually gives a racemic mixture. It is favoured by tertiary halides and polar protic solvents.

(1) Acid-catalysed hydration of alkenes: an alkene adds water in the presence of dilute sulphuric acid following Markovnikov's rule to give an alcohol, e.g. CH2=CH2 + H2O -> CH3CH2OH. (2) Reduction of carbonyl compounds: aldehydes on reduction (with H2/Ni or NaBH4 or LiAlH4) give primary alcohols (R-CHO -> R-CH2OH), and ketones on reduction give secondary alcohols (R-CO-R' -> R-CH(OH)-R'). Alcohols can also be prepared from Grignard reagents reacting with carbonyl compounds.

R-CHO + 2[H] -> R-CH2OH

The carbonyl group (C=O) is polar because oxygen is more electronegative than carbon, so the carbon carries a partial positive charge and the oxygen a partial negative charge. This makes the carbonyl carbon electrophilic, so it is readily attacked by nucleophiles, leading to nucleophilic addition. For example, with hydrogen cyanide (HCN), the addition gives a cyanohydrin: R-CHO + HCN -> R-CH(OH)-CN. Aldehydes are more reactive than ketones towards nucleophilic addition due to less steric hindrance and a greater positive charge on the carbonyl carbon.

R-CHO + HCN -> R-CH(OH)-CN

Carboxylic acids are more acidic than phenols because the carboxylate ion formed after the loss of the proton is stabilised by resonance in which the negative charge is spread equally over two electronegative oxygen atoms, making it very stable. In the phenoxide ion the negative charge is mainly on one oxygen and partly on less electronegative ring carbons, so it is less stabilised. Electron-withdrawing groups (like -Cl or -NO2) increase acidity because they further disperse and stabilise the negative charge of the carboxylate ion (for example, chloroacetic acid is stronger than acetic acid), while electron-releasing groups decrease acidity.

(1) Reduction of nitro compounds: a nitro compound is reduced (with H2/Ni, or Sn/HCl, or Fe/HCl) to a primary amine; for example, nitrobenzene is reduced to aniline, C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O. (2) Hofmann's bromamide degradation: an amide is treated with bromine and a strong alkali (Br2 + NaOH/KOH) to give a primary amine with one carbon atom less than the amide, R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O.

R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O

Amines are basic because the nitrogen atom has a lone pair of electrons which it can donate to a proton (or to a Lewis acid), forming a bond; thus amines can accept a proton. In the gaseous phase (or a non-aqueous medium), only the inductive effect of the alkyl groups operates: more alkyl groups increase the electron density on nitrogen, so the basic strength follows the order tertiary > secondary > primary > ammonia. In aqueous solution the order changes because of the combined effect of inductive effect, solvation (hydrogen bonding of the cation) and steric hindrance.

Benzenediazonium chloride is prepared by diazotisation: aniline is treated with nitrous acid (formed from sodium nitrite and hydrochloric acid) at a low temperature of 0 to 5 degrees C, C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O. In the Sandmeyer reaction, the diazonium group is replaced by a halogen or cyanide using the corresponding copper(I) salt; for example, C6H5N2Cl + CuCl -> C6H5Cl + N2. Diazonium salts are very useful for introducing many groups into the benzene ring.

C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O

Proteins are naturally occurring biomolecules that are polymers of alpha-amino acids joined by peptide bonds (-CO-NH-); they are essential for growth and maintenance of the body and act as enzymes, hormones and structural materials. Denaturation of a protein is the process in which a protein loses its natural three-dimensional shape (its secondary and tertiary structure) due to heat, change in pH, or addition of chemicals, while the primary structure (sequence of amino acids) remains intact. On denaturation the protein loses its biological activity; for example, the coagulation of egg white on boiling.

Reflexive: for any integer a, a - a = 0, which is divisible by 3, so (a, a) in R. Symmetric: if (a - b) is divisible by 3, then (b - a) = -(a - b) is also divisible by 3, so aRb implies bRa. Transitive: if (a - b) and (b - c) are both divisible by 3, then their sum (a - b) + (b - c) = (a - c) is also divisible by 3, so aRb and bRc imply aRc. Since R is reflexive, symmetric and transitive, it is an equivalence relation.

One-one: suppose f(x1) = f(x2). Then 2x1 + 3 = 2x2 + 3, so 2x1 = 2x2 and x1 = x2; hence f is one-one. Onto: let y be any real number in the codomain. Solving y = 2x + 3 gives x = (y - 3)/2, which is a real number, and f((y - 3)/2) = y. So every y has a pre-image, and f is onto. Being both one-one and onto, f is a bijection.

Since f is a bijection, its inverse exists. Let y = f(x) = 2x + 3. Solve for x in terms of y: y - 3 = 2x, so x = (y - 3)/2. Therefore the inverse function is f inverse (y) = (y - 3)/2, or writing in terms of x, f inverse (x) = (x - 3)/2. We can check that f(f inverse (x)) = x.

f inverse (x) = (x - 3)/2

First combine tan inverse (2) + tan inverse (3). Using tan inverse (x) + tan inverse (y) = pi + tan inverse ((x + y)/(1 - xy)) when xy > 1: here x = 2, y = 3, xy = 6 > 1, so tan inverse (2) + tan inverse (3) = pi + tan inverse ((5)/(1 - 6)) = pi + tan inverse (-1) = pi - pi/4 = 3pi/4. Adding tan inverse (1) = pi/4: total = pi/4 + 3pi/4 = pi. Hence proved.

tan inverse x + tan inverse y = tan inverse((x + y)/(1 - xy))

To multiply, each element of a row of A is multiplied by the corresponding element of a column of B and summed. AB = [[(1)(2) + (2)(1), (1)(0) + (2)(3)], [(3)(2) + (4)(1), (3)(0) + (4)(3)]] = [[2 + 2, 0 + 6], [6 + 4, 0 + 12]] = [[4, 6], [10, 12]].

(AB)ij = sum of (row i of A)(column j of B)

Write A = (1/2)(A + A') + (1/2)(A - A'). Let P = (1/2)(A + A') and Q = (1/2)(A - A'). Then P' = (1/2)(A + A')' = (1/2)(A' + A) = P, so P is symmetric. And Q' = (1/2)(A - A')' = (1/2)(A' - A) = -(1/2)(A - A') = -Q, so Q is skew-symmetric. Since A = P + Q, every square matrix is the sum of a symmetric matrix P and a skew-symmetric matrix Q.

A = (1/2)(A + A') + (1/2)(A - A')

The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting: area = (1/2)|1(0 - 3) + 6(3 - 0) + 4(0 - 0)| = (1/2)|-3 + 18 + 0| = (1/2)(15) = 7.5 square units.

Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

First find the determinant: |A| = (2)(4) - (3)(1) = 8 - 3 = 5, which is non-zero, so the inverse exists. For a 2 x 2 matrix [[a, b],[c, d]], the adjoint is [[d, -b],[-c, a]], so adj A = [[4, -3],[-1, 2]]. Then the inverse is A inverse = (1/|A|) adj A = (1/5)[[4, -3],[-1, 2]] = [[4/5, -3/5],[-1/5, 2/5]].

A inverse = (1/|A|) adj A

Write D = determinant of coefficients = |[[2, 3],[1, 2]]| = (2)(2) - (3)(1) = 4 - 3 = 1. Dx (replace first column by constants) = |[[8, 3],[5, 2]]| = (8)(2) - (3)(5) = 16 - 15 = 1. Dy (replace second column by constants) = |[[2, 8],[1, 5]]| = (2)(5) - (8)(1) = 10 - 8 = 2. By Cramer's rule, x = Dx/D = 1/1 = 1 and y = Dy/D = 2/1 = 2. So x = 1, y = 2.

x = Dx/D, y = Dy/D

For continuity at x = 5, the left-hand limit, the right-hand limit and f(5) must all be equal. Left-hand value (using kx + 1): f(5) = 5k + 1. Right-hand limit (using 3x - 5): as x approaches 5, 3(5) - 5 = 10. For continuity, 5k + 1 = 10, so 5k = 9, giving k = 9/5.

LHL = RHL = f(a)