OC Chapter 03 — Question Bank

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Degree of unsaturation of C₈H₈ is:

A. 2

B. 3

C. 4

D. 5

✓ D

Saturated reference C₈H₁₈. Short by 10 H → 10/2 = 5 DU — consistent with a benzene ring (4) + one extra C=C (e.g. styrene).

Slide p.9

Each carbon of a C=C bond is hybridised:

A. sp

B. sp²

C. sp³

D. sp³d

✓ B

sp² → trigonal planar, 120°. The unhybridised p forms the π bond.

Slide p.4

A C=C double bond cannot rotate at room temperature because:

A. The σ bond is too short

B. There is no thermal energy in alkenes

C. Rotation would break the π bond (~64 kcal/mol)

D. sp² carbons are locked by VSEPR

✓ C

Rotation forces the two p orbitals perpendicular → π overlap destroyed → need ~64 kcal/mol. Thermal energy at 298 K is far too little.

Slide p.13

In cis-2-butene vs trans-2-butene, which has the higher boiling point?

A. trans (0.9 °C)

B. Same b.p.

C. trans (3.7 °C)

D. cis (3.7 °C)

✓ D

cis has small net dipole → weak dipole–dipole forces add to London → slightly higher b.p. trans dipoles cancel → lower b.p.

Slide p.12

In Z/E nomenclature, "Z" means the two highest-priority groups are:

A. On opposite sides of C=C

B. On the same side of C=C

D. Both axial

C. Both terminal

✓ B

Z = "zusammen" (German together) = same side. E = "entgegen" (opposite) = opposite sides.

Slide p.18

Propene + HCl gives mainly:

A. 1-chloropropane

B. 2-chloropropane

C. 1,2-dichloropropane

D. Propene (no reaction)

✓ B

Markovnikov: H goes to C with more H's (C1), Cl to more substituted C (C2). The intermediate carbocation (CH₃CH⁺CH₃, 2°) is more stable than CH₃CH₂CH⁺₂ (1°).

Slide pp.36–41

Propene + HBr + peroxide gives mainly:

A. 2-bromopropane

B. 1-bromopropane (anti-Markovnikov)

C. 1,2-dibromopropane

D. Propanal

✓ B

Peroxide effect switches HBr to radical chain mechanism. Br· adds first to the LESS substituted C (the more stable secondary radical forms after Br attacks C1) → anti-Markovnikov product.

Slide p.45

2-Methylpropene + H₂O / 10% H₂SO₄ gives:

A. Isobutane

B. n-Butanol

C. tert-Butyl alcohol (2-methyl-2-propanol)

D. Isobutyraldehyde

✓ C

Markovnikov hydration: OH ends up on the more substituted C → the 3° t-Bu cation is stabilised most → 3° alcohol.

Slide p.43

Cyclohexene + cold dilute KMnO₄ gives:

A. cis-1,2-cyclohexanediol

B. trans-1,2-cyclohexanediol

C. Adipic acid

D. Cyclohexanone

✓ A

Cold dilute KMnO₄ is the syn dihydroxylation reagent: both OH's are added to the same face → cis diol. Purple KMnO₄ decolourises and brown MnO₂ precipitates (Baeyer's test).

Slide p.56

Q10

2-Methyl-2-butene + hot conc. KMnO₄ / H⁺ gives:

A. 2-methyl-2,3-butanediol

B. Acetone + acetic acid

C. Acetaldehyde + acetone

D. 2-butanone

✓ B

Hot KMnO₄ cleaves C=C. The disubstituted C (CMe₂) → acetone (CH₃COCH₃). The mono-H-bearing C (CHMe) → acetic acid (CH₃COOH).

Slide p.57

Q11

1-Butene + (1) O₃ (2) Zn / H₃O⁺ gives:

A. Butanoic acid + CO₂

B. Propanal + formaldehyde

C. Propanal + propanal

D. Butan-2-one

✓ B

CH₃CH₂CH=CH₂. Each sp² C becomes a carbonyl. CH₃CH₂CH= → CH₃CH₂CHO (propanal). =CH₂ → HCHO (formaldehyde). Reductive Zn work-up gives aldehydes (not acids).

Slide pp.59–61

Q12

Benzene + Br₂ / FeBr₃ produces:

A. 1,2-Dibromocyclohexane (addition)

B. Bromocyclohexane

C. No reaction

D. Bromobenzene + HBr (substitution)

✓ D

Aromatic ring resists addition (would destroy aromaticity). Instead, FeBr₃ polarises Br₂ to give Br⁺ which substitutes an H on the ring (EAS).

Slide pp.92–93

Q13

In Friedel-Crafts alkylation, the active electrophile generated from R-Cl + AlCl₃ is:

A. R-AlCl₃

B. The carbocation R⁺ (or a tight ion pair)

D. AlCl₄⁻

C. A free radical R·

✓ B

AlCl₃ (Lewis acid) abstracts Cl⁻ from R-Cl, leaving R⁺ which attacks the aromatic π cloud. Because R⁺ can rearrange to a more stable cation, F-C alkylation often gives the rearranged product (e.g. 1-Br-propane + AlCl₃ on benzene → isopropylbenzene).

Slide pp.96–97

Q14

Nitration of benzene uses:

A. NaNO₂ / HCl

B. Conc. HNO₃ + conc. H₂SO₄ (generating NO₂⁺)

C. NO + O₂

D. N₂O₄ alone

✓ B

H₂SO₄ protonates HNO₃ and dehydrates it to NO₂⁺ (nitronium ion). NO₂⁺ is the electrophile that adds to the ring.

Slide p.94

Q15

Which group is an o/p director AND a deactivator?

A. –OH

B. –CH₃

C. –Cl

D. –NO₂

✓ C — halogens

Halogens are the unique exception: their lone pair gives resonance donation (so they direct o/p), but they pull electrons inductively (so they deactivate).

Q16

Which group is a meta director?

A. –NH₂

B. –OCH₃

C. –CH₂CH₃

D. –NO₂

✓ D

EW groups with π-acceptor character (–NO₂, –CN, –CHO, –COOH, –SO₃H) destabilise the o/p arenium ions and direct the new electrophile to the meta position.

Q17

Which test distinguishes 1-hexyne from 2-hexyne?

A. [Ag(NH₃)₂]NO₃ (white ppt with terminal alkyne only)

B. Br₂ / CCl₄

C. Hot KMnO₄

D. H₂ / Pt

✓ A

Only the terminal alkyne 1-hexyne has an acidic ≡C-H. [Ag(NH₃)₂]⁺ replaces it → white silver acetylide precipitate. 2-hexyne (internal) has no acidic H, no precipitate.

Slide p.74

Q18

Benzene resists addition reactions because:
2019 PP III.6 (inverse)

A. It has no π electrons

B. Addition would destroy the aromatic π system (~36 kcal/mol stabilisation)

C. C–H bonds are too strong

D. sp² carbons cannot be attacked

✓ B

Addition breaks aromaticity. EAS keeps it intact (electrophile in, H out) and is therefore preferred.

Q19

"ortho-, meta-, para-" correspond to which substitution patterns on a disubstituted benzene?

A. 1,2 / 1,3 / 1,4

A. 1,2 / 1,4 / 1,3

C. 1,3 / 1,2 / 1,4

D. 1,4 / 1,3 / 1,2

✓ A

o = 1,2 (adjacent); m = 1,3 (skip one); p = 1,4 (opposite).

Slide p.81

Q20

Which of these molecules is aromatic (Hückel's rule, 4n+2 π e⁻)?

A. Cyclobutadiene C₄H₄ (4 π e⁻)

B. 1,3-Cyclohexadiene (4 π e⁻, not fully conjugated ring)

C. Cyclooctatetraene C₈H₈ (8 π e⁻; non-planar)

D. Benzene C₆H₆ (6 π e⁻)

✓ D

6 = 4(1)+2 → aromatic. Cyclobutadiene fails (4 π e⁻, antiaromatic). 1,3-cyclohexadiene has sp³ carbons in the ring (not fully conjugated). Cyclooctatetraene puckers to avoid antiaromaticity.

D1Degree of unsaturation (DU)+

Total number of multiple bonds + rings in a molecule. Every two H's missing from the saturated reference = 1 DU. Formula DU = (2C + 2 + N − H − X) / 2 (ignore O). Benzene C₆H₆ has 4 DU (1 ring + 3 C=C). Used to narrow down structures from molecular formulas.

Slide p.8

D2cis/trans & Z/E isomerism+

cis/trans applies when each alkene C has one H + one "other" substituent: same side = cis, opposite = trans. Z/E applies generally: rank the two substituents on each alkene C by atomic number (CIP rules); if the higher-priority groups are on the same side → Z (zusammen); opposite → E (entgegen). Z/E and cis/trans don't always match for the same molecule because priority can put a low-Z substituent on the "trans" side.

Slide pp.10–18

D3Markovnikov's rule+

Empirical: when HX adds to an unsymmetric alkene, H goes to the C with more H's; X goes to the more substituted C. Mechanistic: protonation gives the more stable carbocation (3° > 2° > 1° > CH₃⁺), which the halide then attacks. Peroxide effect (HBr only): radical chain mechanism reverses the selectivity to anti-Markovnikov.

Slide pp.36–45

D4Ozonolysis+

Cleavage of a C=C by ozone (O₃) followed by reductive work-up (Zn/H₃O⁺ or Me₂S). Each sp² carbon of the original alkene becomes a carbonyl: if it carried an H → aldehyde; if it carried two C substituents → ketone. Terminal CH₂= → HCHO. Classic structure-determination tool: the carbonyl products map back uniquely to the alkene.

Slide pp.59–61

D5Aromaticity & Hückel's rule+

A molecule is aromatic if it is (i) cyclic, (ii) planar, (iii) fully conjugated (uninterrupted p orbital ring), (iv) contains (4n+2) π electrons (n = 0, 1, 2, …). Benzene (6 π) is the prototype. Aromaticity confers ~36 kcal/mol of extra stabilisation; this is why benzene resists addition reactions and instead undergoes electrophilic substitution. The antiaromatic alternative (4n π e⁻, e.g. cyclobutadiene) is destabilised.

Slide pp.77–79

D6Electrophilic Aromatic Substitution (EAS)+

The defining reaction of aromatic rings. An electrophile E⁺ (Br⁺, NO₂⁺, SO₃H⁺, R⁺, RCO⁺) attacks the π cloud, generating a cationic sigma-complex (arenium ion) with the charge delocalised over the ring. Loss of H⁺ from the sp³ carbon restores aromaticity, giving Ar-E + H-X. Five canonical EAS reactions: halogenation (X₂/FeX₃), nitration (HNO₃/H₂SO₄), sulfonation (SO₃/H₂SO₄), Friedel-Crafts alkylation (RX/AlCl₃), Friedel-Crafts acylation (RCOCl/AlCl₃).

Slide pp.88–97

Define degree of unsaturation. Calculate DU for (a) C₇H₁₄, (b) C₈H₈, (c) C₆H₁₀, and propose a plausible structure for each.

7 marks

Definition

DU = total number of multiple bonds + rings = (2C + 2 + N − H − X)/2.

(a) C₇H₁₄

Saturated ref = C₇H₁₆; short 2 H → DU = 1. Could be: 1-heptene (one C=C) or methylcyclohexane (one ring).

(b) C₈H₈

Saturated ref = C₈H₁₈; short 10 H → DU = 5. Consistent with benzene ring (4) + one extra C=C — e.g. styrene (C₆H₅CH=CH₂).

(c) C₆H₁₀

Saturated ref = C₆H₁₄; short 4 H → DU = 2. Could be hex-1-yne (one triple), 1,3-hexadiene (two C=C), or cyclohexene (1 ring + 1 C=C).

Marking (7): definition + formula (1) · correct DU for a/b/c (3) · one valid structure each (3).

State Markovnikov's rule, explain it mechanistically using HCl addition to propene, and describe the peroxide effect on HBr addition.

9 marks

Empirical rule

When H-X adds to an unsymmetric alkene, the H attaches to the C of the C=C that already bears the most H's; the X ends up on the more substituted C.

Mechanism with propene + HCl

The alkene π electrons attack H⁺, removing one alkene C from sp² to sp³ and leaving a carbocation on the other alkene C.
Two cations are possible: 2° isopropyl (CH₃CH⁺CH₃) if H lands on C1, or 1° n-propyl (CH₃CH₂CH⁺₂) if H lands on C2.
The 2° cation is more stable (hyperconjugation, 3°>2°>1°>Me) → lower TS energy → that pathway wins.
Cl⁻ attacks the 2° cation → 2-chloropropane.

Modern statement

HX adds so as to generate the more stable carbocation intermediate. This subsumes the empirical rule and explains exceptions (e.g. CF₃CH=CH₂ gives anti-Markovnikov because the trifluoromethyl destabilises the would-be 2° cation).

Peroxide effect (HBr only)

HBr + ROOR follows a radical chain instead of cationic addition. The peroxide initiates Br·; Br· adds to the less substituted C of the alkene (giving the more stable 2° radical at the more substituted C), then this radical abstracts H from H-Br. Net: H ends up on the more substituted C, Br on the less substituted → anti-Markovnikov. Only HBr shows this; HCl and HI do not (their bond energies don't fit the chain energetics).

Marking (9): empirical statement (1) · cation stability argument (3) · correct propene product (1) · modern statement (1) · peroxide mechanism + product (3).

Predict the principal products of ozonolysis (O₃ then Zn/H₃O⁺) for: (a) 2-methyl-2-butene, (b) cyclohexene, (c) 1,3-butadiene.

8 marks

The rule

Each alkene sp² C becomes C=O. If that C had an H, it becomes an aldehyde; if it had two C-substituents, it becomes a ketone. Reductive work-up gives carbonyls (not acids).

(a) 2-methyl-2-butene CH₃C(CH₃)=CHCH₃

Left C has 2 methyls → acetone (CH₃)₂C=O. Right C has H + Me → acetaldehyde CH₃CHO.

(b) Cyclohexene

Ring opens at C=C. Both sp² C's bore one H + one CH₂ chain → one molecule of hexanedial (OHC–(CH₂)₄–CHO).

(c) 1,3-Butadiene CH₂=CH–CH=CH₂

Both double bonds cleave: terminal CH₂ sites → HCHO. Middle CH=CH sites → one molecule of glyoxal (OHC–CHO). Net: 2 HCHO + 1 OHC-CHO.

Marking (8): rule stated (2) · (a) correct products (2) · (b) correct product (2) · (c) correct products (2).

Describe the five major Electrophilic Aromatic Substitution reactions of benzene. For each, give the reagent, the electrophile and the product.

10 marks

Halogenation

Reagent: Br₂ / FeBr₃ (or Cl₂ / FeCl₃). Electrophile: Br⁺ (or Cl⁺). Product: bromobenzene (or chlorobenzene) + HX.

Nitration

Reagent: conc. HNO₃ + conc. H₂SO₄. Electrophile: NO₂⁺ (nitronium). Product: nitrobenzene + H₂O.

Sulfonation

Reagent: fuming H₂SO₄ (H₂SO₄ + SO₃). Electrophile: SO₃H⁺ (or SO₃). Product: benzenesulfonic acid.

Friedel-Crafts alkylation

Reagent: R-Cl / AlCl₃. Electrophile: R⁺ (alkyl cation, prone to rearrangement). Product: alkylbenzene + HCl.

Friedel-Crafts acylation

Reagent: R-COCl / AlCl₃. Electrophile: R-C≡O⁺ (acylium ion). Product: aryl ketone Ar-CO-R + HCl. Doesn't rearrange (acylium is resonance-stabilised) so it's the cleaner version when you want an alkyl chain on the ring.

Marking (10): 2 marks per reaction (reagent + electrophile + product), 5 reactions.

Devise simple chemical tests to distinguish each compound in the group: 2-methylbutane, 3-methyl-1-butene, 3-methyl-1-butyne.

6 marks

Step 1 — Br₂ / CCl₄

2-Methylbutane is a saturated alkane → no decolourisation. The other two contain C=C or C≡C → red-brown Br₂ decolourises rapidly. This separates the alkane from the alkene/alkyne.

Step 2 — [Ag(NH₃)₂]NO₃ (Tollens-like, on the remaining two)

3-Methyl-1-butyne is a terminal alkyne with acidic ≡C-H → white silver acetylide precipitate. 3-Methyl-1-butene has no acidic H → no precipitate.

Summary table

2-methylbutane → Br₂ (−), Ag⁺ (−)
3-methyl-1-butene → Br₂ (+), Ag⁺ (−)
3-methyl-1-butyne → Br₂ (+), Ag⁺ (+)

Marking (6): Br₂ test logic (2) · Ag(NH₃)₂⁺ test logic (2) · correct identification of all three (2).

OC Chapter 03 — Unsaturated Hydrocarbons · Q-Bank