Analyzing Scheme 1
The Battle of Nucleophiles
Welcome to a beautiful sequence of organic transformations! Let's break down Scheme 1 by first identifying our starting material, M. Looking closely at the skeletal structure, M is 3-butyn-1-ol (HO−CH2−CH2−C≡CH).
Notice that this molecule possesses two distinct acidic protons: the proton attached to the oxygen (alcohol) and the proton attached to the terminal alkyne. When we introduce an excess of sodium amide (NaNH2), a remarkably strong base, it deprotonates both sites! This yields a dianion containing both an alkoxide and an acetylide: NaO−CH2−CH2−C≡C−Na+.
Now comes the critical catch—regioselectivity. We add exactly one equivalent of ethyl iodide (CH3CH2I). Which nucleophile will attack first? To answer this, we must compare their basicities. The pKa of an alcohol is around 16, while the pKa of a terminal alkyne is around 25. This means the acetylide ion is a much stronger base and, consequently, a far superior nucleophile. The acetylide swiftly attacks the ethyl iodide, forming an internal alkyne: NaO−CH2−CH2−C≡C−CH2CH3.
Following this, we add one equivalent of methyl iodide (CH3I). Now it is the remaining alkoxide's turn to shine! It attacks the methyl iodide via an SN2 mechanism, forming a methyl ether. Our intermediate is now CH3O−CH2−CH2−C≡C−CH2CH3.
The Stereoselective Reduction
The final step of Scheme 1 employs hydrogen gas with Lindlar's catalyst. This is a classic, poisoned palladium catalyst specifically designed to reduce internal alkynes to cis-alkenes, preventing over-reduction to alkanes.
Thus, our product X is cis-1-methoxy-3-hexene. When we examine the given options, the visual representation in Option (A) perfectly matches this stereochemistry and connectivity!
Analyzing Scheme 2
The Hidden Epoxide
Moving on to Scheme 2, our starting material N is 1-butyne (CH3−CH2−C≡CH). The first step involves adding two equivalents of NaNH2. One equivalent deprotonates the terminal alkyne to form the acetylide (CH3−CH2−C≡C−). But why do we need a second equivalent?
The answer lies in the next reagent: 1-bromo-2-propanol. This molecule contains an acidic hydroxyl group. The leftover equivalent of NaNH2 deprotonates this OH group, forming an alkoxide that immediately undergoes an intramolecular SN2 reaction, kicking out the bromide ion to form an epoxide (propylene oxide) in situ!
Once the epoxide is formed, our waiting acetylide nucleophile attacks it. Under basic conditions, the nucleophile attacks the less sterically hindered carbon of the epoxide, yielding an elongated alkoxide intermediate.
The Final Oxidation and Isomer Check
We then add mild acid (H3O+) to protonate the alkoxide, giving us an alcohol. Next, we subject the molecule to hydrogen gas with Palladium on Carbon (H2,Pd/C). Unlike Lindlar's catalyst, this is a robust reducing agent that completely saturates the triple bond into a single bond, resulting in 2-heptanol.
Finally, we use Chromium trioxide (CrO3), a strong oxidizing agent, to oxidize the secondary alcohol into a ketone. Our final product Y is 2-heptanone (CH3−CH2−CH2−CH2−CH2−C(=O)−CH3).
Because Y contains a methyl group directly attached to a carbonyl carbon (CH3−C=O), it will yield a positive Iodoform test. Furthermore, if we count the atoms, both X and Y share the exact same molecular formula (C7H14O). Since they belong to different functional groups (an ether/alkene vs. a ketone), they are functional isomers. Therefore, statement (C) is the correct conclusion for product Y!
