Suggest a strategy to obtain the anti-Markovnikov alcohol selectively from an alkene, name reagents conditions and outline the key mechanistic idea. Recommend one synthetic route to convert an alkene

Here are the solutions to Question Two. QUESTION TWO a) Suggest a strategy to obtain the anti-Markovnikov alcohol selectively from an alkene. Name reagents conditions and outline the key mechanistic idea (4 marks) Strategy: Hydroboration-oxidation. This method adds a hydroxyl group to the less substituted carbon of the double bond (anti-Markovnikov addition). Reagents and Conditions: 1. Borane (BH_3) in tetrahydrofuran (THF) or diborane (B_2H_6). 2. Hydrogen peroxide (H_2O_2) in the presence of a base (e.g., NaOH). Key Mechanistic Idea: • In the first step, borane adds to the alkene in a syn fashion, with the boron atom attaching to the less sterically hindered (less substituted) carbon of the double bond and a hydrogen atom attaching to the more substituted carbon. This is a concerted, four-membered transition state. • In the second step, the alkylborane intermediate undergoes oxidation and hydrolysis, replacing the boron atom with a hydroxyl group, also with retention of stereochemistry. This results in the overall anti-Markovnikov addition of water across the double bond. b) The R&D team wants a diol instead of an alcohol. Recommend one synthetic route to convert an alkene into a diol and outline the mechanism (4 marks) Synthetic Route: Syn-dihydroxylation using osmium tetroxide (OsO_4) followed by a reducing agent. Reagents and Conditions: 1. OsO_4 (catalytic amount) and NMO (N-methylmorpholine N-oxide) as a co-oxidant, or OsO_4 (stoichiometric amount). 2. NaHSO_3/H_2O or H_2S (for workup if stoichiometric OsO_4 is used). Mechanism: Step 1: The alkene reacts with OsO_4 to form a cyclic osmate ester intermediate. This is a concerted syn addition, meaning both oxygen atoms from OsO_4 add to the same face of the double bond. Alkene + OsO_4 Cyclic Osmate Ester Step 2: The osmate ester is then hydrolyzed (e.g., by NaHSO_3/H_2O) to yield the syn-diol, where both hydroxyl groups are added to the same face of the original double bond. The osmium catalyst is regenerated. Cyclic Osmate Ester + H_2O NaHSO_3 Syn-diol + Osmium species c) Design a quick lab test to confirm whether the feedstock contains an alkene. State the expected observations and the conclusions that can be drawn (4 marks) Lab Test: Bromine water test. Procedure: Add a few drops of reddish-brown bromine water (Br_2 in water or CCl_4) to a small sample of the feedstock in a test tube. Shake gently. Expected Observations: • If an alkene is present, the reddish-brown color of the bromine water will rapidly disappear (decolorize). • If no alkene is present (e.g., an alkane), the reddish-brown color will persist. Conclusions: • Decolorization of bromine water: This indicates the presence of a carbon-carbon double bond (alkene) in the feedstock. The bromine undergoes an addition reaction across the double bond, consuming the colored Br_2 and forming a colorless dibromoalkane. • No change in color: This indicates the absence of a carbon-carbon double bond, suggesting the feedstock is saturated (e.g., an alkane) or contains functional groups that do not react with bromine under these conditions. Last free one today — make it count tomorrow, or type /upgrade for unlimited.