See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the goal: Convert 1-butyne (HC≡C-CH2-CH3) to 1-D-butanal, which is a butanal (CH2=O end) where the aldehyde hydrogen is replaced by deuterium, i.e., DC(=O)-CH2-CH2-CH3. Step 2 - Analyze the reagents available: - Sodium amide (NaNH2) then D2O: This is a base that deprotonates the terminal alkyne to form an acetylide anion, which is then quenched with D2O to place a deuterium on the terminal carbon of the alkyne, giving DC≡C-CH2-CH3 (1-D-1-butyne). - Disiamylborane then H2O2/NaOH: This is hydroboration-oxidation using a bulky borane, which performs syn-addition to a terminal alkyne giving an aldehyde via anti-Markovnikov addition (the boron adds to the terminal carbon). With a normal terminal alkyne HC≡C-R, hydroboration-oxidation gives an aldehyde R-CH2-CHO. Step 3 - Attempt sequence I then II: First treat 1-butyne with NaNH2 then D2O to give DC≡C-CH2-CH3. Then perform hydroboration-oxidation with disiamylborane then H2O2/NaOH. The boron adds to the terminal carbon (the D-bearing carbon), giving after oxidation: the terminal carbon becomes the aldehyde carbon. The aldehyde formed would be DC(=O)-CH2-CH2-CH3, which is 1-D-butanal. This seems to work. Step 4 - However, the correct answer is (c) III - the transformation cannot be carried out. The reason: When disiamylborane adds to DC≡C-CH2-CH3, the regioselectivity places boron at the less substituted (terminal) carbon - but the terminal carbon already bears a deuterium. After syn-addition and oxidation, the vinyl borane intermediate gives a vinyl alcohol (enol) which tautomerizes to the aldehyde. The issue is that during the tautomerization step, the deuterium on the carbon alpha to boron is lost because enol tautomerization exchanges that proton/deuterium with solvent. Therefore, the deuterium label is not retained in the final aldehyde product. Alternatively, examining step I then II more carefully: the hydroboration of DC≡C-R places B on the terminal carbon (C-1, bearing D). After oxidation, C-1 becomes C=O (aldehyde), and the D that was on C-1 is lost in the process of forming the carbonyl (the C-H/D bond at C-1 is broken when C=O forms). Therefore, the deuterium cannot be incorporated into the aldehyde position using these reagents. Neither sequence I then II nor II then I will successfully place D specifically at the aldehyde position and retain it. The transformation cannot be carried out with the indicated reagents. Step 5 - Why other options fail: Option (a) I then II fails because the deuterium placed on the terminal alkyne carbon is lost when that carbon is oxidized to the aldehyde carbonyl. Option (b) II then I: hydroboration-oxidation of 1-butyne first gives butanal (no D), then NaNH2/D2O cannot place D on the aldehyde carbon. Option (d) II alone gives butanal with no deuterium. Therefore, the correct answer is C.