Exercises — Alkenes — preparation (dehydration, dehydrohalogenation, Zaitsev's rule), addition reactions

Reaction type: dehydration (a β-elimination). The two departing pieces are –OH (from the carbon bearing it, the α-carbon) and –H (from an adjacent carbon, the β-carbon). Removing an H and an OH from neighbouring carbons lets a new C=C π bond snap into place. Net loss = ${\text{H}}_2\text{O}$, hence de-hydr-ation. Trace it with curly arrows (Picture A): the β C–H electron pair swings in to become the π bond, while the C–OH₂⁺ electrons leave with water.

(a) Addition — Br atoms add across the double bond, destroying it (decolourisation is the visible sign). (b) Elimination — H and X leave adjacent carbons, building the double bond. (c) Addition — two H atoms add across C=C (hydrogenation), destroying the double bond. Rule of thumb: making an alkene = elimination; using an alkene = addition.

Alcoholic KOH → base → elimination (dehydrohalogenation). The Br sits on C2, so a β-H can come from C1 or C3.

• β-H from C1 → but-1-ene, ${\text{CH}}_2={\text{CH-CH}}_2{\text{-CH}}_3$ → 1 alkyl group on the C=C (monosubstituted).
• β-H from C3 → but-2-ene, ${\text{CH}}_3\text{-CH}={\text{CH-CH}}_3$ → 2 alkyl groups on the C=C (disubstituted).

Zaitsev's rule: the more substituted (more stable) alkene is the major product. Major product = but-2-ene, a disubstituted alkene (2 alkyl groups feeding the double bond via hyperconjugation + I).

Geometry (E/Z): but-2-ene has two versions because the C=C cannot rotate (the π bond locks it). If the two ${\text{CH}}_3$ groups sit on opposite sides = trans (E)-but-2-ene; on the same side = cis (Z)-but-2-ene. The bulky methyls repel less when opposite, so trans (E)-but-2-ene is the major geometric isomer. See the E/Z figure below.

Ionic, electrophilic addition → Markovnikov. The ${\text{H}}^{+}$ adds where it makes the more stable cation. Follow the curly arrows in the figure below:

1. The π electron pair reaches out and grabs ${\text{H}}^{+}$ (arrow from the double bond to H).
2. This leaves a cation on the other carbon.

• ${\text{H}}^{+}$ on C1 (terminal ${\text{CH}}_2$) → 2° cation on C2 (stable). Then ${\text{Br}}^{-}$ lends its lone pair to that carbon (arrow from Br⁻ to C2).
• ${\text{H}}^{+}$ on C2 → 1° cation on C1 (unstable) → minor.

The 2° pathway dominates → $\text{Br}$ ends on the more substituted carbon → 2-bromopropane, ${\text{CH}}_3{\text{-CHBr-CH}}_3$.

Peroxide switches the mechanism to free-radical (Kharasch effect). Now $\text{Br}\bullet$ adds first, and it adds so as to make the more stable carbon radical.

• $\text{Br}\bullet$ on C1 → radical on C2 = 2° radical (stable). Then that carbon grabs H.
• Result: Br on C1 (the carbon with more H's) = anti-Markovnikov = 1-bromopropane, ${\text{CH}}_3{\text{-CH}}_2{\text{-CH}}_2\text{Br}$.

Note: this reversal happens only for HBr. (In radicals we track single electrons, so half-headed "fish-hook" arrows would replace the full curly arrows of Picture A.)

Count alkyl groups directly attached to the C=C carbons.

• 2-methylbut-2-ene, $({\text{CH}}_3{)}_2\text{C}={\text{CH-CH}}_3$: the two double-bond carbons carry 3 alkyl groups (two ${\text{CH}}_3$ on one C, one ${\text{CH}}_3$ on the other) → trisubstituted.
• but-1-ene, ${\text{CH}}_2={\text{CH-CH}}_2{\text{CH}}_3$: only 1 alkyl group → monosubstituted.

Why substitution helps (Picture B): each alkyl group carries C–H bonds whose electrons can lean into the π system — that overlap is hyperconjugation. More alkyl groups = more of those overlapping bonds + more +I donation (electrons pushed toward the C=C) → lower energy. 2-methylbut-2-ene is more stable (3 vs 1 alkyl substituents).

Structure: ${\text{CH}}_3{\text{-CBr(CH}}_3){\text{-CH}}_2{\text{-CH}}_3$. Br is on C2, which is bonded to C1 (${\text{CH}}_3$), C3 (${\text{CH}}_2$) and a methyl branch. β-H's are available from C1 and from C3.

• β-H from C1 → 2-methylbut-1-ene, ${\text{CH}}_2={\text{C(CH}}_3){\text{-CH}}_2{\text{CH}}_3$ → C=C carries 2 alkyl groups (disubstituted).
• β-H from C3 → 2-methylbut-2-ene, $({\text{CH}}_3{)}_2\text{C}={\text{CH-CH}}_3$ → C=C carries 3 alkyl groups (trisubstituted).

Zaitsev picks the more substituted → 2-methylbut-2-ene (trisubstituted) is the major product.

Step 1 — dehydration. Propan-2-ol, ${\text{CH}}_3{\text{-CH(OH)-CH}}_3$, with conc. ${\text{H}}_2{\text{SO}}_4$, heat: ${\text{CH}}_3{\text{-CH(OH)-CH}}_3\stackrel{\text{conc. }{H}_{2}S{O}_4,\Delta }{\to }{\text{CH}}_3\text{-CH}={\text{CH}}_2+{\text{H}}_2\text{O}$ (Acid protonates OH → water leaves → 2° cation → loses β-H → propene. Trace this three-step E1 flow with curly arrows as in Picture A.) Step 2 — Markovnikov HBr addition (no peroxide): ${\text{CH}}_3\text{-CH}={\text{CH}}_2+\text{HBr}\to {\text{CH}}_3{\text{-CHBr-CH}}_3$ Br lands on the more substituted C (via the 2° cation). Product: 2-bromopropane.

Add HBr in the presence of organic peroxide: ${\text{CH}}_3\text{-CH}={\text{CH}}_2+\text{HBr}\stackrel{\text{peroxide}}{\to }{\text{CH}}_3{\text{-CH}}_2{\text{-CH}}_2\text{Br}$ The peroxide switches ionic → free-radical, and $\text{Br}\bullet$ adds first to give the stable 2° radical, dropping Br on the terminal carbon = anti-Markovnikov = 1-bromopropane. Single change: add peroxide (and it must be HBr, not HCl/HI).

${\text{H}}^{+}$ adds to C1 (terminal ${\text{CH}}_2$) giving a 2° cation on C2: ${\text{CH}}_3\text{-}\stackrel{+}{C}{\text{H-C(CH}}_3{)}_3$. But right next door sits a fully substituted quaternary carbon carrying methyls. A methyl group migrates (1,2-shift) from C3 to C2, converting the 2° cation into a 3° cation on C3 — far more stable (see Carbocations — stability and rearrangement). ${\text{Br}}^{-}$ then attacks the new 3° cation site. Result: rearranged skeleton → 2-bromo-2,3-dimethylbutane. The before/after atom map (which carbon holds the +, where the migrating ${\text{CH}}_3$ lands, where Br finally attaches) is drawn in the figure below.

This is carbocation rearrangement: whenever an ionic mechanism generates a cation that can become more stable by a 1,2-shift, it will.

Test for unsaturation: add reddish-brown bromine water. The alkene's π electrons attack ${\text{Br}}_2$ (via a cyclic bromonium ion), so ${\text{Br}}_2$ is consumed and the colour decolourises — positive test for a C=C. ${\text{CH}}_3\text{-CH}={\text{CH-CH}}_3+{\text{Br}}_2\to {\text{CH}}_3{\text{-CHBr-CHBr-CH}}_3$ Conversion to butane: catalytic hydrogenation (H₂, Ni/Pt/Pd), which adds H across the π bond (syn, saturating it): ${\text{CH}}_3\text{-CH}={\text{CH-CH}}_3+{\text{H}}_2\stackrel{\text{Ni}}{\to }{\text{CH}}_3{\text{-CH}}_2{\text{-CH}}_2{\text{-CH}}_3$ Product: butane (a saturated alkane).

Alcoholic KOH on an alkyl halide does elimination or substitution? ::: Elimination (base) — aqueous would substitute. Markovnikov: H adds to the carbon with more or fewer H's? ::: More H's (leaving the stable cation on the more substituted carbon). Peroxide effect reverses HX for which halide only? ::: HBr only. Zaitsev counts alkyl groups on what? ::: Only the two double-bond carbons. A 2° cation next to a bulky quaternary carbon may do what? ::: A 1,2-methyl shift to become a 3° cation. Why does but-2-ene have cis and trans forms? ::: The C=C cannot rotate, so groups are locked on the same (cis/Z) or opposite (trans/E) sides — trans is more stable.