Organic Chemistry – Some Basic Principles and Techniques

From the chapter examples:

- **CH$_2$=C=O: left carbon is sp$^2$, middle carbon is sp.
- CH$_3$CH=CH$_2$: terminal CH$_3$ carbon is sp$^3$, the two double-bond carbons are sp$^2$.
- (CH$_3$)$_2$CO: both methyl carbons are sp$^3$ and carbonyl carbon is sp$^2$.
- CH$_2$=CHCN: the two alkene carbons are sp$^2$ and the nitrile carbon is sp.
- C$_6$H$_6$: all carbon atoms are sp$^2$**.

So the hybridisation states are:
- CH$_2$=C=O: **sp$^2$, sp**
- CH$_3$CH=CH$_2$: **sp$^3$, sp$^2$, sp$^2$**
- (CH$_3$)$_2$CO: **sp$^3$, sp$^3$, sp$^2$**
- CH$_2$=CHCN: **sp$^2$, sp$^2$, sp**
- C$_6$H$_6$: **sp$^2$ for all carbon atoms**

Count $c3$ bonds as all single bonds plus one $c3$ part of each multiple bond, and count $c0$ bonds from multiple bonds.

- **C$_6$H$_6$**: 6 C–C $c3$ bonds + 6 C–H $c3$ bonds = **12 $c3$, and 3 double bonds = 3 $c0$.
- C$_6$H$_{12}$ (cyclohexane): only single bonds, so 18 $c3$, 0 $c0$.
- CH$_2$Cl$_2$: 4 C–H/C–Cl single bonds = 4 $c3$, 0 $c0$.
- CH$_2$=C=CH$_2$**: two double bonds give 2 $c3$ and 2 $c0$, plus 4 C–H $c3$ bonds = **6 $c3$, 2 $c0$.
- CH$_3$NO$_2$**: one C–N $c3$, three C–H $c3$, and one N=O double bond contributes 1 $c3$ and 1 $c0$; total **5 $c3$, 1 $c0$.
- HCONHCH$_3$: H–C, C=O, C–N, N–H, N–C, and three C–H bonds give 7 $c3$; the C=O has 1 $c0$.

So the counts are:
C$_6$H$_6$ = 12 $c3$, 3 $c0$; C$_6$H$_{12}$ = 18 $c3$, 0 $c0$; CH$_2$Cl$_2$ = 4 $c3$, 0 $c0$; CH$_2$=C=CH$_2$ = 6 $c3$, 2 $c0$; CH$_3$NO$_2$ = 5 $c3$, 1 $c0$; HCONHCH$_3$ = 7 $c3$, 1 $c0$**.

From the given structures in the chapter, the bond-line formulas correspond to:

- Isopropyl alcohol: a three-carbon chain with OH on the middle carbon, i.e. **CH$_3$–CH(OH)–CH$_3$.
- 2,3-Dimethylbutanal: a four-carbon aldehyde chain with methyl groups on C-2 and C-3, i.e. CH$_3$–CH(CH$_3$)–CH(CH$_3$)–CHO.
- Heptan-4-one: a seven-carbon chain with a ketone at C-4, i.e. CH$_3$CH$_2$CH$_2$COCH$_2$CH$_2$CH$_3**.

In bond-line form, these are the corresponding zig-zag structures with the functional group placed at the correct carbon.

The question asks for IUPAC names of the compounds shown in the textbook figures. The source page gives these as examples in Problem 8.8:

- (i) 6-Methyloctan-3-ol
- (ii) Hexane-2,4-dione
- (iii) 5-Oxohexanoic acid
- (iv) Hexa-1,3-dien-5-yne

These are the systematic names derived in the worked solutions.

- (b) The correct name is 2,4,7-Trimethyloctane because numbering must give the lowest set of locants; 2,4,7 is lower than 2,5,7. - (c) The correct name is 2-Chloro-4-methylpentane because substituents are listed in alphabetical order and the lower locant is given to the substituent that comes first alphabetically when positions are equivalent. - (d) The correct name is But-3-yn-1-ol because the OH group gets priority and must get the lowest locant. So the correct choices are the first, first, first, and first option in each pair.

Form each homologous series by adding successive **$-CH_2-$ units.

### (a) Starting with HCOOH
This is the alkanoic acid series:
1. HCOOH
2. CH$_3$COOH
3. CH$_3$CH$_2$COOH
4. CH$_3$CH$_2$CH$_2$COOH
5. CH$_3$CH$_2$CH$_2$CH$_2$COOH**

### (b) Starting with CH$_3$COCH$_3$
This is the alkanone series:
1. **CH$_3$COCH$_3$
2. CH$_3$COCH$_2$CH$_3$
3. CH$_3$COCH$_2$CH$_2$CH$_3$
4. CH$_3$CO(CH$_2$)$_3$CH$_3$
5. CH$_3$CO(CH$_2$)$_4$CH$_3$

### (c) Starting with H-CH=CH
This represents the alkene series:
1. CH$_2$=CH$_2$
2. CH$_2$=CHCH$_3$
3. CH$_2$=CHCH$_2$CH$_3$
4. CH$_2$=CH(CH$_2$)$_2$CH$_3$
5. CH$_2$=CH(CH$_2$)$_3$CH$_3$**

From the chapter’s solved example, the compounds are:

- (a) 2,2,4-Trimethylpentane: condensed formula **CH$_3$C(CH$_3$)$_2$CH$_2$CH(CH$_3$)CH$_3$. Bond-line formula is the zig-zag pentane chain with methyl groups at C-2, C-2 and C-4. Functional group: none (alkane).
- (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid: condensed formula HOOCCH$_2$C(OH)(COOH)CH$_2$COOH or equivalently the textbook form for citric acid. Functional groups: hydroxyl and carboxylic acid groups.
- (c) Hexanedial: condensed formula OHC(CH$_2$)$_4$CHO. Bond-line formula is a six-carbon chain with aldehyde groups at both ends. Functional group: dialdehyde**.

So the key identifications are alkane, hydroxy-tricarboxylic acid, and dialdehyde.

Use the visible groups in each structure to name the functional group(s), such as -OH, -CHO, -COOH, -NH2, >C=O, -NO2, -X etc.

**O$_2$NCH$_2$CH$_2$O$^-$ is more stable.

Reason: The nitro group (-NO$_2$) is a strong electron-withdrawing group by the inductive effect. It pulls electron density away from the negative oxygen, helping to stabilise the anion. In CH$_3$CH$_2$O$^-$**, the ethyl group shows electron-donating tendency, which increases electron density on oxygen and makes the anion less stable.

Alkyl groups act as electron donors when attached to a **$c0$ system because of hyperconjugation. The C–H $c3$ electrons** of the alkyl group overlap with the adjacent $c0$ system and become partially delocalised. This shifts electron density toward the unsaturated system, so the alkyl group behaves as an electron-donating group.

The chapter asks for resonance structures of the following compounds. The important resonance contributors are:

- **(a) C$_6$H$_5$OH: phenol shows resonance between the oxygen lone pair and the benzene ring, giving charge-separated forms with negative charge at ortho and para positions and positive charge on oxygen.
- (b) C$_6$H$_5$NO$_2$: nitrobenzene shows resonance where the nitro group withdraws electron density from the ring; the ring bears positive charge at ortho/para positions and the nitro group is represented by equivalent canonical forms.
- (c) CH$_3$CH=CHCHO**: an $c0$-conjugated aldehyde; resonance shifts electrons from C=C toward C=O, giving charge-separated forms with negative charge on oxygen.
- **(d) C$_6$H$_5$CHO: benzaldehyde shows resonance between the carbonyl group and the ring.
- (e) C$_6$H$_5$CH$_2\dot{}$: benzyl radical is resonance-stabilised; the unpaired electron is delocalised to ortho and para positions.
- (f) CH$_3$CH=CHCH$_2\dot{}$**: allylic radical; the unpaired electron is delocalised over the allyl system.

Since the full curved-arrow diagrams are graphical, the answer here states the required resonance systems and the positions of delocalisation.

A nucleophile is a reagent that donates an electron pair to form a bond. It is nucleus-seeking. Examples: **OH$^-$, CN$^-$, HS$^-$, R$_3$C$^-$.

An electrophile is a reagent that accepts an electron pair to form a bond. It is electron-seeking. Examples: H$^+$, BF$_3$, CH$_3^+$, NO$_2^+$**.

So, nucleophiles are electron-rich species, and electrophiles are electron-deficient species.

**HO$^-$ is a nucleophile** because it has a lone pair of electrons and donates an electron pair to the electrophilic proton of CH$_3$COOH in the acid-base reaction.

Classify each reaction by the chapter’s reaction types:

- (a) CH$_3$CH$_2$Br + HS$^-$ → CH$_3$CH$_2$SH + Br$^-$: substitution reaction.
- (b) (CH$_3$)$_2$C=CH$_2$ + HCl → addition product: addition reaction.
- (c) CH$_3$CH$_2$Br + HO$^-$ → CH$_2$=CH$_2$ + H$_2$O + Br$^-$: elimination reaction.
- (d) rearranged product formation from alcohol + HBr: rearrangement reaction.

To answer, compare the structures as follows:

- If they differ in connectivity, they are structural isomers. - If they have the same connectivity but differ in spatial arrangement around a double bond or ring, they are geometrical isomers. - If only the placement of electrons changes without changing atom positions, they are resonance contributors. Use the actual structures in the textbook figure to decide which category applies.

From the chapter:

- **(a) CH$_3$O–OCH$_3$ → CH$_3\dot{O}$ + \dot{O}CH$_3$: homolytic cleavage, producing free radicals.
- (b) bond cleavage in the shown molecule with hydroxide leading to water: heterolytic cleavage, giving ionic intermediates.
- (c) Br → Br$^-$: heterolytic cleavage leading to an ion.
- (d) E$^+$ → E$^+$E: this represents formation/attack by an electrophile; as written in the exercise image it is meant to show electron flow with curved arrows, but the direct classification depends on the exact bond depicted. In the chapter context, the key idea is to identify whether a bond breaks by homolysis or heterolysis and whether the intermediate is a free radical, carbocation, or carbanion**.

The clearest textbook-style classification is: (a) homolysis/free radicals; the others involve heterolytic electron movement and ionic intermediates.

Inductive effect is the permanent polarisation of a $c3$ bond caused by the electronegativity of a substituent; the effect is transmitted through the carbon chain and decreases rapidly with distance.

Electromeric effect is a temporary effect shown by compounds having multiple bonds, in which the $c0$ electrons are completely transferred to one atom in the presence of an attacking reagent.

The acidity orders are explained by the inductive effect:

- **(a) Cl$_3$CCOOH > Cl$_2$CHCOOH > ClCH$_2$COOH: more Cl atoms exert a stronger -I effect, stabilising the carboxylate ion more and increasing acidity.
- (b) CH$_3$CH$_2$COOH > (CH$_3$)$_2$CHCOOH > (CH$_3$)$_3$C.COOH: alkyl groups have a +I effect**, which destabilises the carboxylate ion and decreases acidity as alkyl substitution increases.

The principles are:

- Crystallisation: based on difference in solubilities of a compound and impurities in a suitable solvent. Example: purification of benzoic acid.
- Distillation: based on difference in boiling points of liquids. Example: separation of chloroform and aniline.
- Chromatography: based on differential adsorption or partition of components between stationary and mobile phases. Example: separation of plant pigments by paper chromatography.

To separate two compounds with different solubilities in a solvent S, use crystallisation. Dissolve the impure mixture in a suitable solvent in which one compound is more soluble at high temperature and less soluble at room temperature. On cooling, the less soluble/pure compound crystallises out, while the more soluble impurity remains in the mother liquor.

- Distillation is used to separate volatile liquids from non-volatile impurities or liquids with sufficiently different boiling points.
- Distillation under reduced pressure is used for liquids having very high boiling points or those that decompose at or below their boiling points; reducing pressure lowers the boiling temperature.
- Steam distillation is used for steam-volatile substances immiscible with water; the mixture boils below the normal boiling point because vapour pressures add up to atmospheric pressure.

So the difference lies in the conditions used and the type of substances separated.