(a) In the reaction R—COO⁻ + Br—CN → R—CN + Br⁻ + CO₂↑ What is the origin of —CN group in the product ? Explain by using isotopic…

(a) Origin of the Cyanide Group The reaction R—COO⁻ + Br—CN → R—CN + Br⁻ + CO₂ is a decarboxylative cyanation. The —CN group in the product R—CN originates exclusively from the cyanogen bromide (Br—CN), not from the carboxylate. This is proven via isotopic labelling. If Br—¹³C—N is used, the product R—¹³CN is formed, confirming the carbon comes from the reagent. Conversely, if R—¹³C—OO⁻ is used, the CO₂ evolved is ¹³CO₂, while the nitrile carbon remains ¹²C. The mechanism involves nucleophilic attack of the carboxylate oxygen on the electrophilic carbon of Br—CN, forming an acyl cyanide intermediate (R—CO—CN). This intermediate undergoes decarboxylation, where the C—CO bond breaks, releasing CO₂ and generating a carbanion or radical species that is trapped by the cyanide moiety or, more accurately in this specific context, the reaction proceeds via an S_N2-like displacement where the carboxylate acts as a nucleophile to displace bromide, but the key evidence is that the carbon skeleton of the nitrile is distinct from the carboxyl carbon.

(b) Reactivity of Quaternary Ammonium Salt The substrate is benzyltrimethylammonium salt (Ph—CH₂—N⁺(CH₃)₃ X⁻). At low temperature, treatment with sodium amide (NaNH₂) generates a benzylic ylide via deprotonation of the α-carbon. This ylide undergoes the Sommelet-Hauser rearrangement. The methyl group migrates from nitrogen to the benzylic carbon, forming a new C—C bond, yielding a tertiary amine with a methylated benzyl group (e.g., Ph—CH(CH₃)—N(CH₃)₂). At high temperature, the ylide is unstable and undergoes Hofmann elimination. The amide ion abstracts a proton from the β-carbon (the methyl groups on nitrogen), leading to the expulsion of the trimethylamine and formation of an alkene. Since the β-hydrogens are on the methyl groups, the product is methylene cyclohexane derivative or, in this specific benzyl case, the elimination occurs to form a styrene derivative if the β-carbon is the ring carbon, but typically for benzyltrimethylammonium, the elimination yields styrene (Ph—CH=CH₂) if the structure allows, or more commonly, the Hofmann product is the least substituted alkene. In this specific case, the β-carbons are the methyls on N, so elimination yields a methylene group on the nitrogen? No, Hofmann elimination requires a β-hydrogen on a carbon adjacent to the carbon bearing the leaving group. Here, the leaving group is on N. The α-carbon is CH₂. The β-carbons are the ring carbons (if we consider the ring) or the methyls on N? Actually, for R—N⁺(Me)₃, the β-hydrogens are on the R group. If R is benzyl, the β-hydrogens are on the phenyl ring (no elimination) or if we consider the methyls on N, they are not β to the C-N bond in the standard sense for alkene formation. Wait, the standard Hofmann elimination of benzyltrimethylammonium yields styrene. The base abstracts a proton from the benzylic position? No, that forms the ylide. For elimination, it must abstract a proton from a carbon adjacent to the benzylic carbon. In benzyltrimethylammonium, the adjacent carbons are the ipso-carbon of the ring (no H) or... actually, benzyltrimethylammonium salts typically undergo the Sommelet-Hauser rearrangement at low T. At high T, if elimination occurs, it would require a β-H. If the structure is strictly benzyl, there are no β-hydrogens on an sp³ carbon adjacent to the benzylic carbon except the ring carbons which are sp². Therefore, high-temperature treatment often leads to decomposition or further rearrangement, but in the context of standard exam questions, if a β-H is available (e.g., if it were a longer chain), Hofmann elimination gives the least substituted alkene. For benzyltrimethylammonium, the primary reaction is Sommelet-Hauser. If forced to discuss "high temperature" elimination, it implies the formation of the least substituted alkene from the ylide precursor if a β-H exists.

(c) Dehydration and Rearrangement The reactant is 2-(2-hydroxypropan-2-yl)cyclohexane. Treatment with BF₃ · OEt₂ (a Lewis acid) promotes dehydration.

1. Activation: BF₃ complexes with the -OH group, making it a good leaving group (-OH · BF₃⁻).
2. Carbocation Formation: Loss of water generates a tertiary carbocation at the side-chain carbon: Cyclohexyl—C⁺(CH₃)₂.
3. Rearrangement: This carbocation is adjacent to the cyclohexane ring. A 1,2-alkyl shift (ring expansion) or hydride shift can occur. However, the most stable cation is formed by a 1,2-hydride shift from the ring carbon (C1) to the side-chain cation. This places the positive charge on the ring carbon (C1), which is a tertiary carbocation within the ring.
4. Products:

• Product A (Minor): Direct elimination from the initial side-chain cation yields an exocyclic double bond: 1-(prop-1-en-2-yl)cyclohexane (or similar isomer).
• Product B (Major): Elimination from the ring carbocation (C1) yields an endocyclic double bond. The most stable alkene is the trisubstituted endocyclic alkene, 1-methylcyclohexene (if the shift was hydride) or a ring-expanded product. Given the structure, a 1,2-methyl shift from the side chain to the ring is less likely than hydride. The major product B is the more substituted, stable endocyclic alkene (e.g., 1-methylcyclohexene derivative if the methyls rearrange, or simply the most stable alkene formed by loss of a proton from C2 or C6 of the ring). The major product is the one with the more substituted double bond (Saytzeff product).

(d) Reactivity towards Nucleophiles with BF₃ BF₃ acts as a Lewis acid, coordinating to the carbonyl oxygen, increasing the electrophilicity of the carbonyl carbon. (i) p-trifluoromethyl benzaldehyde: The -CF₃ group is a strong electron-withdrawing group via the -I effect. It stabilizes the developing negative charge in the transition state and increases the partial positive charge on the carbonyl carbon. Thus, it is more reactive towards nucleophilic attack. (ii) p-tolualdehyde: The -CH₃ group is electron-donating via the +I effect and hyperconjugation. It destabilizes the positive charge on the carbonyl carbon and reduces its electrophilicity. Thus, it is less reactive than the trifluoromethyl derivative. Verdict: The rate of nucleophilic addition follows: p-CF₃-C_6H₄-CHO > p-CH₃-C_6H₄-CHO.

(e) Transformation Completion (i) Part (i) is missing from the provided text/figure description. (ii) Isomerization of cis-Decalin to trans-Decalin: The transformation involves the conversion of cis-1,2,3,4,4a,5,8,8a-octahydronaphthalene (cis-decalin) to trans-decalin.

• Mechanism: This is a conformational isomerization. Cis-decalin exists as a rapidly interconverting pair of chair-chair conformers where one ring junction hydrogen is axial and the other equatorial. Trans-decalin is a rigid, locked chair-chair conformer where both ring junction hydrogens are equatorial (or one up, one down in the plane).
• Process: Under thermal or catalytic conditions, the cis-decalin undergoes ring flipping. However, direct interconversion of cis to trans is forbidden by simple ring flipping because it would require passing through a high-energy twist-boat or boat conformation. Typically, this isomerization requires a catalyst (like acid or metal) that allows for bond rotation or a reversible addition-elimination mechanism, or high-temperature gas-phase isomerization. The driving force is the greater stability of the trans-isomer, where all ring junction hydrogens are equatorial, minimizing 1,3-diaxial interactions. The product is the thermodynamically more stable trans-decalin.