Organic Chemistry - Some Basic Principles and…

Structure, Classification and IUPAC Nomenclature

Hybridisation and counting sigma and pi bonds

Sigma and pi bond count

n_\sigma = (\text{number of atoms}) - 1 + (\text{number of rings}) \qquad n_\pi = n_{\text{double}} + 2\,n_{\text{triple}}

Choosing and numbering the parent chain

Order of numbering decisions

\text{principal group} \;\to\; \text{multiple bonds} \;\to\; \text{all prefixes (lowest set)} \;\to\; \text{alphabetical order}

Seniority of functional groups in IUPAC names

Class	Group	Suffix when senior	Prefix when not senior
Carboxylic acid	$\mathrm{-COOH}$	-oic acid	carboxy
Sulphonic acid	$\mathrm{-SO_3H}$	-sulphonic acid	sulpho
Ester	$\mathrm{-COOR}$	alkyl …-oate	alkoxycarbonyl
Acid chloride	$\mathrm{-COCl}$	-oyl chloride	chlorocarbonyl
Amide	$\mathrm{-CONH_2}$	-amide	carbamoyl
Nitrile	$\mathrm{-C{\equiv}N}$	-nitrile	cyano
Aldehyde	$\mathrm{-CHO}$	-al	formyl, or oxo when its carbon is in the chain The aldehyde outranks the ketone: a compound with both is named as an -al with an oxo prefix.
Ketone	$\mathrm{{>}C{=}O}$	-one	oxo
Alcohol	$\mathrm{-OH}$	-ol	hydroxy
Amine	$\mathrm{-NH_2}$	-amine	amino
Alkene, alkyne	C=C, C≡C	-ene, -yne	Never a prefix; always part of the parent name
Halide, nitro, ether	–X, $\mathrm{-NO_2}$ , –OR	Never a suffix	halo, nitro, alkoxy

Seniority falls from the top row down. The last row never takes the suffix.

Common traps

Branch carbons count too

A name such as 2,2-dimethylbutane carries two methyl carbons that the parent name hides. Each is a separate sp³ carbon. Write the full structure before you count.

A carbonyl or nitrile carbon is not sp³

The carbon of C=O, CHO and COOH is sp², and the carbon of C≡N is sp. Only a carbon with four single bonds is sp³.

One ring double bond does not make a ring aromatic

Cyclohexene has one C=C in a six-membered ring. It is alicyclic, not aromatic and not benzenoid.

The ketone does not outrank the aldehyde

–CHO comes before >C=O in the seniority list. An option that places the ketone first is wrong.

The nitrile sits between the amide and the aldehyde

The order runs

\mathrm{-CONH_2 > -CN > -CHO}

. An order that puts –CHO above –CN, or –CN above

\mathrm{-CONH_2}

, is wrong.

The junior group never takes the suffix

\mathrm{HOCH_2CH_2COCH_3}

is 4-hydroxybutan-2-one. A name such as 3-oxobutan-1-ol gives the suffix to the alcohol, which ranks below the ketone.

Prefixes go in alphabetical order, not locant order

1,1-Dimethyl-3-ethylcyclohexane lists the prefixes by number. The correct name is 3-ethyl-1,1-dimethylcyclohexane: e comes before m, and di is ignored.

Compare the whole locant set

2,3,6 beats 2,4,5 because the sets first differ at the second term, and 3 < 4. Adding up the locants, or looking only at the first one, gives the wrong numbering.

The OH carbon is C-1 in a ring

A cyclic alcohol with a ring double bond is named cyclohex-2-en-1-ol, never cyclohex-1-en-3-ol. The principal group takes C-1 before the double bond is numbered.

Structural Isomerism

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Counting structural isomers of a formula

Degree of unsaturation (rings + π bonds)

\text{DoU} = \dfrac{2C + 2 + N - H - X}{2}

Kinds of structural isomerism

Type	What differs	Example pair (one formula)
Chain	The carbon skeleton	Pentane and 2-methylbutane (C₅H₁₂)
Position	Where the group or multiple bond sits	Propan-1-ol and propan-2-ol (C₃H₈O)
Functional group	The functional group itself	Ethanol and methoxymethane (C₂H₆O)
Metamerism	The alkyl groups on either side of one group	Methoxypropane and ethoxyethane (C₄H₁₀O) Two ethers can be metamers. An ether and an alcohol cannot: they are functional isomers.
Ring-chain	A ring in place of a C=C	Cyclopropane and propene (C₃H₆)
Tautomerism	The position of a mobile hydrogen; the forms interconvert	Propanone and prop-1-en-2-ol (keto and enol)

Chain, position, functional, metamer and ring-chain isomers are separate compounds; tautomers exist in equilibrium.

Common traps

A moved double bond is position isomerism

But-1-ene and but-2-ene keep one straight four-carbon skeleton and one kind of group; only the C=C moves. They are position isomers, not chain or functional isomers.

Metamers keep one functional group

Metamerism needs the same group with different alkyls around it. An alcohol and an ether of one formula are functional isomers, never metamers.

No α-hydrogen, no tautomer

Keto–enol tautomerism moves a hydrogen from the carbon next to C=O onto the oxygen. A ketone whose α-carbons carry no hydrogen, such as 2,2,4,4-tetramethylpentan-3-one, cannot form an enol.

Isopropyl and propyl are different side chains

A three-carbon side chain attached through its end carbon gives propylbenzene; attached through its middle carbon it gives isopropylbenzene (cumene). Count both.

A renumbered chain is the same compound

2-Methylpentane and 4-methylpentane are one compound numbered from opposite ends. Always name each isomer properly before you add it to the count.

Nitrogen adds and halogen subtracts in the DoU formula

Each N adds one to the top of the degree-of-unsaturation formula and each halogen counts like a hydrogen. Leaving N out makes a nitrile formula look saturated.

Stereoisomerism and Conformations

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Geometrical isomerism and conformations

Condition for cis–trans isomerism

\mathrm{abC{=}Ccd}:\quad a \neq b \ \text{and}\ c \neq d

Chiral centres, meso compounds and optical purity

Optical purity

\text{optical purity}\ (\%) = \dfrac{\text{observed rotation}}{\text{rotation of the pure enantiomer}} \times 100

Counting stereoisomers

Maximum number of stereoisomers

N \le 2^{n}, \qquad n = \text{chiral centres} + \text{stereogenic C=C bonds}

Common traps

Two identical groups on one carbon cancel it

In 2-methylbut-2-ene,

\mathrm{(CH_3)_2C{=}CHCH_3}

, C-2 carries two methyl groups. Swapping them changes nothing, so there are no cis and trans forms.

The trans isomer usually melts higher

Trans alkenes pack more neatly in a crystal, so they generally have the HIGHER melting point. The cis isomer has the dipole moment and usually the higher boiling point.

Conformations are not separable isomers

Staggered and eclipsed ethane interconvert by rotation about a single bond. They are called rotamers or conformers, never enantiomers or geometrical isomers.

Deuterium is different from hydrogen

A

\mathrm{CH_2D}

group is not a

\mathrm{CH_3}

group, and a carbon carrying H and D can be a chiral centre. Treat D as a fourth, distinct group.

Two chiral centres do not guarantee optical activity

If the two halves of the molecule are identical and a mirror plane cuts between them, the compound is meso and optically inactive, as in meso-2,3-dibromobutane.

Compare configurations, not drawings

Two drawings that look like mirror images may be the same molecule turned round. Assign R or S to every centre in both before calling them enantiomers.

2ⁿ is a ceiling, not the answer

Check for identical halves before writing 2ⁿ. Tartaric acid has two chiral centres but only 3 stereoisomers, because the meso form is its own mirror image.

A double bond is a stereogenic unit too

A C=C with two different groups on each carbon doubles the count, just like a chiral carbon. Counting only the chiral carbons halves the answer.

Include stereoisomers only when asked

When a question says 'including stereoisomers', a chiral monochloro product counts as two. When it asks only for structural isomers, it counts as one.

Electronic Effects, Resonance and Acidity

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Acid strength from conjugate-base stability

Acid strength and conjugate base

\mathrm{HA \rightleftharpoons H^+ + A^-}\qquad \text{more stable } \mathrm{A^-} \;\Rightarrow\; \text{larger } K_a,\ \text{smaller } \mathrm{p}K_a,\ \text{weaker base } \mathrm{A^-}

Inductive, resonance, electromeric and hyperconjugation effects

Effect	Electrons move through	Permanent or temporary	Typical example
Inductive (I)	σ bonds, weakening with distance	Permanent	Cl pulls electrons along the chain in $\mathrm{ClCH_2COOH}$
Resonance (R or M)	π bonds and lone pairs on adjacent atoms	Permanent	$\mathrm{-NH_2}$ pushes its lone pair into the ring of aniline
Electromeric (E)	One π bond, shifted completely to one atom	Temporary; only while the reagent is present	The C=O of propanone as $\mathrm{CN^-}$ attacks The only one of the four that disappears when the reagent is taken away.
Hyperconjugation	A C–H σ bond into an adjacent empty p or π orbital	Permanent	The three C–H bonds of the $\mathrm{CH_3}$ group stabilise the C=C of propene

Resonance and the electromeric effect need a π system; the inductive effect needs only σ bonds.

Rules for the stability of resonance structures

Rule	More stable contributor	Less stable contributor
Neutral beats charge-separated	$\mathrm{CH_2{=}CH{-}Cl}$	$\mathrm{^-CH_2{-}CH{=}\overset{+}{Cl}}$
Every atom with a complete octet	$\mathrm{CH_3{-}\overset{+}{O}{=}CH_2}$	$\mathrm{CH_3{-}O{-}\overset{+}{C}H_2}$ (carbon with a sextet)
Negative charge on the more electronegative atom	$\mathrm{CH_2{=}CH{-}O^-}$	$\mathrm{^-CH_2{-}CH{=}O}$
Opposite charges close, like charges apart	Unlike charges on neighbouring atoms	Like charges on neighbouring atoms (the worst case)
No atom beyond an octet	Nitrogen with four bonds and a + charge, as in $\mathrm{-\overset{+}{N}({=}O)O^-}$	Nitrogen with five bonds: not a valid structure A 'resonance structure' with five bonds to N or C is simply wrong, however it is charged.

Apply the rules from the top; the first rule that separates two structures decides.

Common traps

Hyperconjugation is a permanent effect

Hyperconjugation needs no reagent: it is present in the ground state of every molecule with an α-C–H next to an empty or π orbital. A statement calling it temporary is false.

H⁺ shows a +E effect, not −E

When

\mathrm{H^+}

attacks a C=C, the π electrons move towards the carbon it bonds to. That is the +E effect. The −E effect goes with a nucleophile such as

\mathrm{CN^-}

.

Iodine is the weakest −I group of the halogens

The −I effect of a halogen follows its electronegativity, so –I withdraws least. It also withdraws less than –COOH, –CN and

\mathrm{-NO_2}

.

Five bonds to nitrogen is never allowed

In a nitro group the nitrogen has four bonds and a positive charge. A drawing that gives it five bonds breaks the octet rule and is not a resonance structure at all.

Charge separation costs stability

Among valid structures, the one without separated charges is the most stable. A charge-separated structure contributes less, even though it explains the dipole moment.

Conjugation shortens the single bond

In a conjugated enal, resonance gives the C–C bond between C=C and C=O some double-bond character. That bond is shorter than the same bond in the saturated aldehyde, not longer.

Basicity runs opposite to acidity

The anion of the strongest acid is the weakest base. Chloride, from the strong acid HCl, is a far weaker base than ethanoate, which is weaker than hydroxide.

An alkoxide is a stronger base than hydroxide

An alcohol is a slightly weaker acid than water because the +I alkyl group destabilises the alkoxide. So

\mathrm{RO^-}

is a stronger base than

\mathrm{OH^-}

.

Rank C–H acidity by s-character first

A C–H on an sp carbon is more acidic than one on an sp² carbon, whatever the size of the molecule. Only among sp³ C–H bonds do resonance and the inductive effect decide.

Reaction Intermediates, Bond Fission and Reagents

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Carbocation stability, hydride affinity and rearrangement

Hyperconjugation count and alkyl order

\text{hyperconjugating H} = \text{number of } \alpha\text{-H} \qquad 3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^+}

Bond fission, free radicals, carbanions and reagent types

Species	Formed by	Carbon: hybridisation and shape	Electrons on the carbon	Behaves as
Carbocation	Heterolysis; carbon loses the pair	sp², trigonal planar	6 (a sextet)	Electrophile
Carbanion	Heterolysis; carbon keeps the pair	sp³, pyramidal	8, with one lone pair	Nucleophile and base
Free radical	Homolysis	sp², nearly planar	7, with one unpaired electron	Neutral, very reactive; starts chain reactions Radical and carbocation stability follow the same alkyl order; carbanion stability runs the other way.

All three are short-lived intermediates; none is isolated in an ordinary reaction.

Common traps

More stable means LOWER hydride affinity

Hydride affinity measures how eagerly a cation grabs

\mathrm{H^-}

. A stable cation is not eager, so the order of hydride affinity is the reverse of the order of stability.

A meta donor cannot reach the charge

On a benzyl cation, the positive charge spreads only to the ortho and para ring carbons. An

\mathrm{-OCH_3}

at the meta position cannot donate its lone pair to the charge, so it helps far less than a para one.

The vinyl cation is not allylic

In

\mathrm{CH_2{=}CH^+}

the charge sits ON the double-bond carbon, which cannot spread it. In the allyl cation,

\mathrm{CH_2{=}CH{-}CH_2^+}

, it sits next to the double bond and is shared by resonance.

The methyl cation has no hyperconjugation

Hyperconjugation needs a C–H bond on the carbon NEXT to the cationic carbon.

\mathrm{CH_3^+}

has no such carbon, so it has no hyperconjugating hydrogen at all.

Carbanion stability runs opposite to carbocation stability

Alkyl groups push electrons towards the carbon. That helps a positive carbon and hurts a negative one, so the methyl carbanion is the most stable simple carbanion and the tertiary one the least.

Ionic reactions come from heterolysis

Homolysis gives neutral radicals, which lead to free-radical reactions. Only heterolysis gives the ions that ionic reactions need.

A π bond can make a nucleophile

Ethene has no lone pair and no charge, yet its π electrons attack electrophiles such as

\mathrm{H^+}

and

\mathrm{Br^+}

. Count alkenes when a question asks for nucleophiles.

Methods of Purification

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Choosing a distillation method

Method	Use when	Classic example
Simple distillation	Boiling points far apart, or a liquid from a non-volatile solid	Chloroform from aniline
Fractional distillation	Liquids with close boiling points	Crude oil into petrol, kerosene and diesel fractions
Distillation under reduced pressure	The liquid decomposes at its normal boiling point	Glycerol from spent lye
Steam distillation	Steam-volatile substance, immiscible with water	Aniline from an aniline–water mixture; o-nitrophenol from p-nitrophenol The substance distils below its own boiling point because the two vapour pressures add up.
Azeotropic distillation	A constant-boiling mixture that fractional distillation cannot split	Water removed from ethanol by adding benzene

Name the property that differs, and the method follows.

Crystallisation, sublimation and extraction

Method	Property exploited	Example
Crystallisation	Solubility in one solvent, hot against cold	Benzoic acid recrystallised from hot water
Fractional crystallisation	A small difference in solubility, used repeatedly	Two solids of similar solubility separated in stages
Sublimation	Solid changes directly to vapour	Naphthalene or camphor separated from sodium chloride
Differential extraction	Different solubility in two immiscible solvents	An organic compound extracted from water into ether
Acid–base extraction	An acid or base turned into a water-soluble salt	Phenol pulled from a toluene solution into aqueous NaOH NaOH extracts acids and phenols; dilute HCl extracts amines. Neither extracts a neutral compound.

All five work on solids or solutions, where the distillations cannot help.

Common traps

Lower pressure means a lower boiling point

Distillation under reduced pressure works because the boiling point FALLS as the pressure falls. The liquid then boils below the temperature at which it would decompose.

The rising vapour gets richer in the lower-boiling liquid

In a fractionating column the higher-boiling component condenses first and runs back down. The vapour that reaches the top is richer in the more volatile component.

Steam distillation needs a water-immiscible substance

Steam distillation works only if the substance does not mix with water and has an appreciable vapour pressure near 373 K. A water-soluble compound, or a non-volatile one, cannot be steam distilled.

Sublimation is about solid to vapour, not a low melting point

Sublimation separates a solid that passes straight into vapour from one that does not. A statement that sublimation is used for compounds with a low melting point is false.

NaOH does not extract an amine

An amine is a base, so a base cannot turn it into a salt. With aqueous NaOH, aniline stays in the organic layer along with any neutral compound.

The impurity matters as much as the compound

Methods of purification depend on the nature of the compound AND of the impurity. The same compound may need a different method when the impurity changes.

Chromatography

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Retardation factor and order of elution

Retardation factor

R_f = \dfrac{\text{distance moved by the compound from the base line}}{\text{distance moved by the solvent front from the base line}}

Adsorption and partition chromatography

Technique	Principle	Stationary phase	Mobile phase
Column chromatography	Adsorption	Silica gel or alumina packed in a glass column	A solvent (the eluant) run down the column
Thin-layer chromatography (TLC)	Adsorption	A thin layer of silica gel or alumina on a glass plate	A solvent rising up the plate by capillary action
Paper chromatography	Partition	Water held in the pores of the paper	A solvent rising up the paper The paper itself is only the support, not the stationary phase.

Column and TLC share a principle; paper chromatography is the odd one out.

Common traps

The paper is not the stationary phase

In paper chromatography the stationary phase is the water held in the paper's pores. A statement that the paper material itself is the stationary phase is false.

TLC works by adsorption, not partition

Thin-layer chromatography uses a solid adsorbent on a plate, just like column chromatography. Only paper chromatography, among the three, works by partition.

No visualising agent in the mobile phase

Spots are revealed after the plate is run: by UV light, iodine vapour or a spray. Adding a visualising agent to the solvent is not a method of locating spots.

Measure from the base line, not the plate edge

If a distance is given from the bottom of the plate, subtract the height of the base line first, for both the spot and the solvent front.

More polar means lower Rf on silica

A polar compound, such as an alcohol or an acid, sticks to polar silica gel and travels less. Its Rf is SMALLER than that of a non-polar compound in the same solvent.

The first compound out has the higher Rf

On a column, the compound that is adsorbed least runs fastest and elutes first. The same compound travels furthest on a TLC plate, so it has the higher Rf.

Lassaigne's Test and Estimation of Nitrogen

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Dumas method for nitrogen

Percentage of nitrogen (Dumas)

V_{\mathrm{STP}} = \dfrac{(p - f)\,V}{T} \times \dfrac{273}{760} \qquad \%\,\mathrm{N} = \dfrac{28}{22400} \times \dfrac{V_{\mathrm{STP}}\,(\text{mL})}{m\,(\text{g})} \times 100

Kjeldahl method for nitrogen

Percentage of nitrogen (Kjeldahl)

\%\,\mathrm{N} = \dfrac{1.4 \times M \times V \times b}{m} \qquad (b = \text{basicity of the acid},\ V \text{ in mL},\ m \text{ in g})

Lassaigne's test for nitrogen, sulphur, halogens and phosphorus

Element	In the extract as	Reagent	Positive result
Nitrogen	NaCN	$\mathrm{FeSO_4}$ , boil, then conc. $\mathrm{H_2SO_4}$	Prussian blue, $\mathrm{Fe_4[Fe(CN)_6]_3}$
Sulphur	$\mathrm{Na_2S}$	Sodium nitroprusside	Violet, $\mathrm{Na_4[Fe(CN)_5NOS]}$
Sulphur	$\mathrm{Na_2S}$	Ethanoic acid and lead acetate	Black precipitate of PbS
Nitrogen and sulphur together	NaSCN	$\mathrm{Fe^{3+}}$ (iron(III) chloride)	Blood red, $\mathrm{[Fe(SCN)]^{2+}}$ With excess sodium the thiocyanate breaks down, and the separate cyanide and sulphide tests work instead.
Chlorine	NaCl	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	White AgCl, soluble in ammonia
Bromine	NaBr	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Pale yellow AgBr, sparingly soluble in ammonia
Iodine	NaI	Boil with $\mathrm{HNO_3}$ , then $\mathrm{AgNO_3}$	Yellow AgI, insoluble in ammonia
Phosphorus	$\mathrm{Na_3PO_4}$ (after oxidation with $\mathrm{Na_2O_2}$ )	$\mathrm{HNO_3}$ and ammonium molybdate	Yellow $\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$

The fusion extract detects nitrogen, sulphur, the halogens and phosphorus, and nothing else.

Common traps

No carbon, no cyanide

The nitrogen test depends on NaCN, which needs carbon from the compound. Hydrazine and hydroxylamine contain nitrogen but no carbon, so they give no Prussian blue.

Nitric acid, not hydrochloric acid

The extract is acidified with

\mathrm{HNO_3}

before adding

\mathrm{AgNO_3}

. HCl would add chloride and give white AgCl whatever the compound contained.

Lassaigne's test cannot detect oxygen

The fusion extract shows N, S, halogens and P. A set of elements that includes oxygen or carbon is not what the sodium fusion extract detects.

Subtract the aqueous tension first

The collected nitrogen is saturated with water vapour. Use the measured pressure minus the aqueous tension before reducing the volume to STP; forgetting it raises the answer by a few per cent.

Nitrogen gas is 28, not 14

22400 mL of

\mathrm{N_2}

at STP weighs 28 g. Using 14 g halves the percentage.

Copper gauze, not copper oxide, reduces the oxides

Copper(II) oxide oxidises the compound; the heated copper gauze further along reduces any oxides of nitrogen back to

\mathrm{N_2}

.

Sulphuric acid is dibasic

One mole of

\mathrm{H_2SO_4}

neutralises two moles of

\mathrm{NH_3}

. Forgetting the factor of 2 gives exactly half the right percentage, and that value is always among the options.

Not for nitro, azo or ring nitrogen

Kjeldahl's method does not work for nitro and azo compounds or for pyridine, because their nitrogen is not fully converted into ammonium sulphate. Use Dumas' method for them.

Use the acid actually neutralised

When a question gives the total acid and the excess titrated back with alkali, the ammonia used only the difference. Subtract before you multiply by the basicity.

Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

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Combustion analysis for carbon, hydrogen and oxygen

Percentages of carbon and hydrogen

\%\,\mathrm{C} = \dfrac{12}{44} \times \dfrac{m_{\mathrm{CO_2}}}{m} \times 100 \qquad \%\,\mathrm{H} = \dfrac{2}{18} \times \dfrac{m_{\mathrm{H_2O}}}{m} \times 100

Carius method for halogens

Percentage of halogen (Carius)

\%\,\mathrm{X} = \dfrac{\text{atomic mass of X}}{\text{molar mass of AgX}} \times \dfrac{m_{\mathrm{AgX}}}{m} \times 100

Estimation of sulphur and phosphorus

Percentages of sulphur and phosphorus

\%\,\mathrm{S} = \dfrac{32}{233} \times \dfrac{m_{\mathrm{BaSO_4}}}{m} \times 100 \qquad \%\,\mathrm{P} = \dfrac{62}{222} \times \dfrac{m_{\mathrm{Mg_2P_2O_7}}}{m} \times 100

Common traps

Hydrogen is 2/18 of water, not 1/18

Each water molecule carries two hydrogen atoms. Using 1/18 halves the percentage of hydrogen and then throws off the oxygen found by difference.

Oxygen is never weighed directly

Combustion analysis gives only C and H. Oxygen comes from 100 minus everything else, so an error in carbon or hydrogen carries straight into it.

Match the tube to the gas

The

\mathrm{CaCl_2}

tube gains the mass of water and the potash (KOH) tube the mass of

\mathrm{CO_2}

. Swapping them swaps the two fractions, 12/44 and 2/18.

Divide by the silver halide's molar mass

The fraction is the halogen's atomic mass over the molar mass of the WHOLE precipitate: 35.5/143.5 for AgCl, not 35.5/108.

Keep the three silver halides apart

AgCl is 143.5, AgBr 188 and AgI 235 g mol⁻¹. Pairing bromine with 143.5 is the commonest slip, and it gives an answer that is often among the options.

Carius does not estimate nitrogen

The Carius tube is used for halogens, sulphur and phosphorus. Nitrogen is estimated by Dumas' or Kjeldahl's method.

Two phosphorus atoms per pyrophosphate

\mathrm{Mg_2P_2O_7}

holds two P atoms, so phosphorus is 62/222 of its mass. Using 31/222 halves the answer.

Use the molar mass the question gives

Some papers print an unusual molar mass for

\mathrm{BaSO_4}

or for the pyrophosphate. The key is worked with the printed value, so use it even if it differs from 233 or 222.

Sulphur is weighed as a barium salt

Silver nitrate traps halogens; barium chloride traps sulphate. A question that adds

\mathrm{BaCl_2}

is estimating sulphur, and the precipitate is

\mathrm{BaSO_4}

.

Organic Chemistry - Some Basic Principles and Techniques formulas

Structure, Classification and IUPAC Nomenclature

Hybridisation and counting sigma and pi bonds

Choosing and numbering the parent chain

Seniority of functional groups in IUPAC names

Common traps

Structural Isomerism

Counting structural isomers of a formula

Kinds of structural isomerism

Common traps

Stereoisomerism and Conformations

Geometrical isomerism and conformations

Chiral centres, meso compounds and optical purity

Counting stereoisomers

Common traps

Electronic Effects, Resonance and Acidity

Acid strength from conjugate-base stability

Inductive, resonance, electromeric and hyperconjugation effects

Rules for the stability of resonance structures

Common traps

Reaction Intermediates, Bond Fission and Reagents

Carbocation stability, hydride affinity and rearrangement

Bond fission, free radicals, carbanions and reagent types

Common traps

Methods of Purification

Choosing a distillation method

Crystallisation, sublimation and extraction

Common traps

Chromatography

Retardation factor and order of elution

Adsorption and partition chromatography

Common traps

Lassaigne's Test and Estimation of Nitrogen

Dumas method for nitrogen

Kjeldahl method for nitrogen

Lassaigne's test for nitrogen, sulphur, halogens and phosphorus

Common traps

Estimation of Carbon, Hydrogen, Halogens, Sulphur and Phosphorus

Combustion analysis for carbon, hydrogen and oxygen

Carius method for halogens

Estimation of sulphur and phosphorus

Common traps

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