JEE Chemistry Organic Chemistry Some Basic Principles and Techniques Notes Set 12

Classification and Basic Terms

Organic compounds are first split into open-chain (acyclic or aliphatic) and ring (cyclic) compounds. Cyclic compounds are alicyclic or aromatic, and each of these may be homocyclic (ring of carbon atoms only) or heterocyclic (ring containing another atom such as O, N or S).
Carbon forms an enormous number of compounds because of catenation: it bonds strongly to itself, forms stable multiple bonds through side-by-side p-orbital overlap (possible because the atom is small), and has a very strong C–C bond. Silicon, germanium, tin and lead show the property to a much smaller extent because their bonds are weaker.
A saturated compound has only sigma bonds; an unsaturated compound has at least one pi bond. A molecular formula gives the number of atoms; structural formulae can be written as expanded, condensed or bond-line forms. A homologous series is a family with the same functional group in which successive members differ by CH2.

Degree of Unsaturation (DU)

Each degree of unsaturation corresponds to the loss of two hydrogen atoms from the saturated formula: one double bond, one ring, or half a triple bond. A triple bond or two rings counts as 2.
For a compound CxHyNwXs with any number of oxygen atoms: \( DU=(x+1)-\frac{y+s-w}{2} \). Oxygen and other divalent atoms are ignored, halogens are counted like hydrogen and nitrogen is counted with a negative sign.
Example: C6H5Cl has DU = 7 − 3 = 4 (one benzene ring). A benzene ring always accounts for DU = 4, so a compound with DU below 4 cannot contain one.

Types of Carbon and Hydrogen

A carbon atom is primary, secondary, tertiary or quaternary according to whether it is bonded to one, two, three or four other carbon atoms. A hydrogen atom takes the degree of the carbon to which it is attached. In 2,2,3-trimethylpentane there are five primary carbons, one secondary, one tertiary and one quaternary carbon.
The number of different kinds of hydrogen decides the number of monosubstituted products, not the number of types of carbon.

Scheme of IUPAC Naming

A name is built as: secondary prefix (substituents, alphabetical) + primary prefix (cyclo) + word root (carbon count of the main chain) + primary suffix (ane, ene, yne) + secondary suffix (principal functional group).
Word roots: meth (1), eth (2), prop (3), but (4), pent (5), hex (6), hept (7), oct (8), non (9), dec (10).
Always-prefix groups: alkyl, alkoxy, halo (fluoro, chloro, bromo, iodo), nitro, nitroso. Locants are separated from words by hyphens and from each other by commas.

Naming Alkanes: Choosing and Numbering the Parent Chain

1. Choose the longest continuous chain. 2. If two chains are equally long, choose the one with more substituents. 3. If that also ties, choose the one whose substituents get the lowest locants.
Number from the end that gives the lowest locant set (compare term by term, the first difference decides). If the locant sets are identical, give the lower number to the substituent that comes first alphabetically. If even that ties, give the lower number to the substituent whose own branch has the lower locant.
Identical substituents are grouped with di-, tri-, tetra-. Prefixes are arranged alphabetically; di, tri, tetra (for simple groups), sec and tert are ignored while alphabetising, but iso, neo and the di/tri of a complex substituent are counted.

Substituent Groups (Radicals)

Removing one hydrogen gives an alkyl group CnH2n+1, an alkenyl group CnH2n−1 or an alkynyl group CnH2n−3. For complex substituents the carbon with the free valency is numbered 1 and the name is written in brackets, for example 5-(1,2-dimethylpropyl)nonane.
Retained names: n-propyl and isopropyl (1-methylethyl); n-butyl, sec-butyl, isobutyl (2-methylpropyl) and tert-butyl (1,1-dimethylethyl). Unsaturated groups: ethenyl (vinyl), prop-1-enyl, prop-2-enyl (allyl), 1-methylethenyl, prop-1-ynyl, prop-2-ynyl.
Aromatic groups: phenyl C6H5–, benzyl C6H5CH2–, benzal C6H5CH=, benzo C6H5C≡, tolyl CH3C6H4– (o, m, p), alpha-naphthyl (position 1) and beta-naphthyl (position 2).
Retained alkane names: iso applies to 4–6 carbon alkanes of the type (CH3)2CH–R with no other substituent (isobutane, isopentane, isohexane); neo applies to 5–6 carbon alkanes of the type (CH3)3C–R (neopentane, neohexane). Iso-octane is 2,2,4-trimethylpentane.

Unsaturated and Cyclic Hydrocarbons

For alkenes and alkynes the parent chain must contain the multiple bond even if a longer chain exists. Number to give the multiple bond the lowest locant, then the substituents. Endings: ene, diene, yne, diyne; when the first letter of a suffix is a vowel (a, e, i, o, u, y) the final e of the primary suffix is dropped (propane + ol gives propanol, but propane + diol keeps the e).
Cyclic hydrocarbons take the prefix cyclo. In a ring with only substituents the numbering gives the lowest locant set; when a multiple bond is present it receives the lowest number.

Compounds with Functional Groups

Selecting the main chain: it must contain the senior functional group, then the maximum number of such groups, then the multiple bonds, then the maximum number of carbons. Numbering gives the lowest locant to the functional group first, then multiple bonds, then substituents, then alphabetical order.
Approximate seniority: carboxylic acid > sulphonic acid > anhydride > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine. The most senior group is expressed as the suffix; all others are prefixes.
Suffix and prefix for common groups: COOH oic acid / carboxylic acid (prefix carboxy); SO3H sulphonic acid (sulpho); COOR alkyl oate (alkoxycarbonyl); COX oyl halide (halocarbonyl); CONH2 amide (carbamoyl); CN nitrile (cyano); CHO al / carbaldehyde (formyl or oxo); C=O one (oxo); OH ol (hydroxy); NH2 amine (amino).
When the carbon of the group is counted in the chain use oic acid, oate, oyl halide, amide, nitrile, al; when it is not (usually on a ring) use carboxylic acid, carboxylate, carbonyl halide, carboxamide, carbonitrile, carbaldehyde.
Examples: HCOOH methanoic acid; HOOC–(CH2)4–COOH hexanedioic acid; CH3COCl ethanoyl chloride; CH3CONHCH3 N-methylethanamide; CH3CH2CH2CN butanenitrile; OHC–CH2–CO–CH3 3-oxobutanal; CH3NH2 methanamine; CH3CH2CH2NHCH2CH3 N-ethylpropan-1-amine. A mixed anhydride lists the two acids alphabetically (ethanoic propanoic anhydride).

Ethers, Heterocycles and Replacement Names

Ethers are alkoxyalkanes: the smaller group becomes the alkoxy prefix and the larger group the parent (CH3OCH(CH3)2 is 2-methoxypropane). Polyethers may be named by replacement: a hetero atom inside the chain is shown by the prefixes oxa, thia, aza and selena with its locant, for example 3,6,9-trioxaundecane.
Three-membered cyclic ethers are named as oxiranes or as epoxy compounds (1,2-epoxypropane). In ring systems numbering starts from the senior functional group.

Aromatic Compounds

Benzenoid compounds contain at least one benzene ring. Disubstituted benzenes use o- (1,2), m- (1,3) and p- (1,4) or numerals; only three trisubstituted benzenes exist when all three groups are identical (1,2,3-, 1,2,4- and 1,3,5-).
Common names accepted in practice: toluene, xylene, cumene (isopropylbenzene), styrene (phenylethene), phenol, cresol, catechol (1,2-dihydroxybenzene), resorcinol (1,3-), hydroquinone (1,4-), aniline (benzenamine), benzaldehyde, benzoic acid, benzoyl chloride, benzonitrile, acetophenone (methyl phenyl ketone), benzophenone (diphenyl ketone) and aspirin (2-ethanoyloxybenzene-1-carboxylic acid).
When there are several substituents, rule out the group that gives the suffix, then give the lowest locant set; alphabetical order decides only a tie. Example: 1-chloro-4-methyl-2-nitrobenzene.

Worked Examples

Worked Example 1. Name \( \ce{CH3-CH2-CH(CH2CH2CH3)-CH(CH3)-CH3} \).
Answer:
Longest chain: from the isopropyl methyl through C2, C3 and the propyl group gives a 6-carbon chain: hexane. Substituents: methyl at C2 and ethyl at C3 (when numbered from the isopropyl end).
The other numbering gives locants {4,5}, which is higher.
3-Ethyl-2-methylhexane.

Worked Example 2. Find the degree of unsaturation of a compound of formula C7H6O that contains a benzene ring and suggest the compound.
Answer:
DU = (7 + 1) − 6/2 = 5.
Benzene ring = 4, so one more double bond is present: a C=O.
Benzaldehyde (benzenecarbaldehyde), C6H5CHO.

Worked Example 3. Give the IUPAC name of \( \ce{HOOC-CH2-CH(OH)-CH2-COOH} \).
Answer:
Both COOH groups are in the chain, so the parent is pentanedioic acid.
The hydroxy group (a prefix) is at C3.
3-Hydroxypentanedioic acid.