4.(a)(i) Complete the above reaction and write the steps involved in the reaction. [norbornene + D–Br/H2O → (A) + (B)] 5…

(a)(i) DBr/H₂O addition to norbornene

Norbornene adds D⁺ from the exo (less hindered) face to one alkene carbon (C-3), giving a bridged, delocalised 2-norbornyl cation with D exo at C-3. The nucleophile then attacks the cation from the exo face, at C-2, as the endo face is shielded. Bromide gives exo-2-bromo-3-exo-deuterionorbornane (A) and water gives exo-norbornan-2-ol-3-exo-d (B), each as a racemate because the symmetrical bridged ion can be opened at either end. D and X are both exo, so the net addition is syn, unlike the anti addition to an ordinary cyclic alkene. Some D scrambling by Wagner–Meerwein or hydride shifts can also occur.

(a)(ii) Order of bromination rates in methanol

The rate-determining step forms a bromonium ion that has considerable carbocation character, so more alkyl substitution and better stabilisation give higher rates. The alkenes are (A) 2-methylpropene, (B) trans-but-2-ene, (C) cis-but-2-ene and (D) 3,3-dimethylbut-1-ene.

The ascending order is D < B < C < A. D is monosubstituted, and its bulky tert-butyl group adds steric hindrance and a −I effect, so it is slowest. The two but-2-enes are 1,2-disubstituted; the cis isomer is more strained and higher in ground-state energy than trans, so C reacts faster than B. A is 1,1-disubstituted and its bromonium ion opens towards a tertiary centre, so it is the fastest. The relative placement of B, C and D rests on reported rate trends.

(a)(iii) Hydroboration–oxidation of 1,1-diphenyl-2,2-diethylethene

The alkene is Ph₂C=C(Et)₂. Hydroboration is a concerted, syn, four-centre addition of B–H across C=C. Boron goes to the carbon that carries less partial positive charge, C(Et)₂, and hydrogen goes to the diphenyl carbon, which better stabilises the positive charge. (A) is the alkylborane Ph₂CH–C(Et)₂–BH₂. Because the alkene is tetrasubstituted, reaction stops at the mono- or dialkylborane stage.

In the second step HOO⁻ adds to boron, then the alkyl group migrates from B to O with retention of configuration, and hydrolysis releases the alcohol. (B) is 3-(diphenylmethyl)pentan-3-ol, Ph₂CH–C(OH)(Et)₂, also named 2-ethyl-1,1-diphenylbutan-2-ol. It has no stereocentre.

(b) Quaternary ammonium salt with NaNH₂/NH₃

The salt is N-benzyl-N-methyl with a benzo-fused eight-membered ring, as read from the drawing. Amide ion deprotonates to give an ammonium ylide. These are the classical Sommelet–Hauser conditions. The ylide forms at the N–CH₃ group, less stabilised but more accessible, in equilibrium with the more stabilised benzylic ylides.

The N–CH₂⁻ ylide undergoes a [2,3]-sigmatropic rearrangement with the exocyclic benzyl group. Its carbon bonds to the ortho carbon of the benzyl ring while the N–CH₂(benzyl) bond breaks, giving a non-aromatic isotoluene-type intermediate. A proton shift then restores the aromatic ring and the exocyclic CH₂ becomes CH₃. The product is the tertiary amine in which nitrogen, still in the eight-membered ring, carries an ortho-methylbenzyl (2-methylbenzyl) group in place of the original benzyl and methyl groups. The competing [1,2] Stevens shift is less favoured here.

(c)(i) Crossed aldol of butan-2-one and acetaldehyde

Six β-hydroxy carbonyl compounds are possible (ignoring stereoisomers), from three enolates and two acceptors.

Acetaldehyde enolate adds to acetaldehyde to give 3-hydroxybutanal, or to butanone to give 3-hydroxy-3-methylpentanal. The enolate of butanone at the CH₃ side adds to acetaldehyde to give 5-hydroxyhexan-3-one, or to butanone to give 5-hydroxy-5-methylheptan-3-one. The enolate at the CH₂ side adds to acetaldehyde to give 4-hydroxy-3-methylpentan-2-one, or to butanone to give 4-hydroxy-3,4-dimethylhexan-2-one. Aldehydes are better acceptors than ketones, so products from acetaldehyde as acceptor dominate in practice, and on heating these β-hydroxy compounds dehydrate to enones.

(c)(ii) Benzonitrile from benzene

Step 1: nitration, conc. HNO₃/conc. H₂SO₄ at about 50 °C, gives nitrobenzene. Step 2: reduction, Sn/HCl or H₂/Pd, gives aniline. Step 3: diazotisation, NaNO₂/HCl at 0–5 °C, gives benzenediazonium chloride. Step 4: the Sandmeyer reaction with CuCN (KCN/CuCN) replaces N₂⁺ by CN and releases N₂, giving benzonitrile. The diazonium salt must be kept cold and used at once, since it decomposes on warming.