1. Provide the structures or reagents in each of the following questions. Explain your reasoning. a) Na/NH₃ reduces internal alkynes to (E)-double bonds. A is (E)-3-hexene. Syn 1,2-dihydroxylation of an (E)-double bond affords (±)-B, 3,4-hexanediol. Lindlar-catalyzed hydrogenation of internal alkynes gives (Z)-double bonds. C is (Z)-3-hexene. Epoxidation of C occurs in a syn fashion providing the cis-oxirane D. Acid-catalyzed opening of the epoxide in an anti attack of water on the protonated epoxide affords B. Both routes end at B because one pathway is trans reduction and syn addition of two hydroxyl groups. The other pathway is cis reduction and anti addition.

b) The hindered borane adds to the terminal alkyne to afford a vinylborane that has the boron attached to the terminal end of the double bond. Oxidation of the borane yields "1-hydroxy-1-pentene", which tautomerizes to the aldehyde A, pentanal. There are two Sia-B bonds. "Sia" means sec-isoamyl. [Amyl is a generic term for a five carbon chain.] B is 3-methyl-2-butanol. Mercury-catalyzed hydration of the terminal alkyne leads to 2-pentanone C via its enol, 2-hydroxy-1-pentene.

c) Lindlar reduction gives 3-methyl-1-buttene A. [No stereochemistry here.] Hydroboration in turn adds the elements of water in an anti-Markovnikov mode to give alcohol B. HBr effects S_N2 displacement of B to afford C, 1-bromobutane. Sodamide forms the anion of the terminal alkyne, which undergoes S_N2 displacement of 1-bromobutane to form 2,7-dimethyl-3-octyne, D.

d) Reduction to (E)-5-decene. Anti addition of Br₂.

e) No commentary

f) See c).

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2. Compounds A and B are constitutional isomers. Determine the structures and reagents A-O. Explain your reasoning. The transformation A ---> H is clearly a Lindlar catalyzed hydrogenation of an alkyne to a (Z)-double bond. Ozonolysis of H gives K, which is either propionaldehyde or acetone. Acetone cannot be K because H would have to be 2,3-dimethyl-2-butene (tetramethylethylene), an alkene that is not derivable from an alkyne. Thus, K = propionaldehyde, H = (Z)-3-hexene and A = 3-hexyne. J (cis-2,3-(1-propyl)oxirane) is formed in one step from H with MCPBA. Ozonolysis of A (there is no reduction step in the ozonolysis of an alkyne) gives a C₃ carboxylic acid M, namely propanoic acid (propionic acid). Oxidation of propionaldehyde K gives propionic acid M. Saturation of 3-hexyne with hydrogen gives n-hexane C, which is formed from B by the same procedure. Therefore, B must be 1-hexyne or 2-hexyne. Since permanganate oxidation affords CO₂, the remainder of the C₆ alkyne must be C₅ in the form of pentanoic acid (valeric acid) L. B is 1-hexyne. [Note: CO₂ is formed by over oxidation of formic acid (HCO₂H). 2-Hexyne would have given butanoic acid (butyric acid) and ethanoic acid (acetic acid)]. meso-E is 3,4-dibromohexane. Given the anti mechanism of bromine addition to a double bond, D is (E)-3-hexene, which is prepared by Na/NH₃ (reagent N) reduction of A. F is the trans stereoisomer of J, formed in a double inversion, two-step reaction. Reagent O is H₃O⁺, which opens the racemic epoxide to give meso-G, 3,4-hexandiol.

3. This problem bears many similarities to #2. It is presented as a word problem rather than as a chart. Two bottles are found on a laboratory shelf labeled "alkyne A" and "alkyne B". Hydrogenation of A or B over a platinum catalyst gives the same alkane C. A and B have the same connectivity of carbon atoms. Compound A reacts with H₂ in the presence of Lindlar's catalyst to form D. A forms a (Z)-double bond. Compound D reacts with O₃ to form a single compound E, C₃H₆O. D is a symmetrically substituted double bond (single monofunctional product) as is A a symmetrically substituted triple bond. E cannot be acetone because ozonolysis would have produced it from 2,3-dimethyl-2-butene (tetramethylethylene), which cannot be derived from an alkyne. E must be propionaldehyde, CH₃CH₂CHO. Therefore, D is (Z)-3-hexene and A is 3-hexyne. On the other hand, compound B reacts with aq. H₂SO₄ in the presence of HgSO₄ to give two ketones J and K. B is either 1-hexyne or 2-hexyne. 1-Hexyne would give only 2-hexanone upon Hg⁺⁺-catalyzed hydration. 2-Hexyne would give 2- and 3-hexanone. These two ketones are J and K; B is 2-hexyne. Under the same conditions, A gives only J. Since A, 3-hexyne, affords only J under these conditions, J must be 3-hexanone and K is 2-hexanone. Compound B also reacts with Na/NH₃ to give F, (F is (E)-2-hexene) which itself reacts with Br₂/H₂O to give a pair of constitutional isomers, G and H. Anti addition of the elements of HOBr gives G and H. See below. Treatment of either G or H with aqueous NaOH gives the same compound I, C₆H₁₂O, that is also formed by the reaction of F with peracid. Halohydrin formation followed by base treatment gives an even number of S_N2 inversions of stereochemistry: two. Peracid gives zero inversions. Therefore, the two reaction sequences give the same trans epoxide I. What are the structures of A-K? Explain and illustrate. [Note: G and H are not distinguished from one another. Pay attention to stereochemistry.]