Question

predict the product for the following reactions.
rxn i
rxn ii
rxn i rxn ii
a.

b.

c.

d.

Explanation:

Step1: Analyze Rxn I (Epoxide Hydration)

The starting material is a substituted epoxide. In acidic water ($\ce{H+/H2O}$), epoxides undergo ring - opening. The mechanism involves protonation of the epoxide oxygen, followed by nucleophilic attack by water. For a substituted epoxide, the regiochemistry (which carbon is attacked) and stereochemistry (retention or inversion) matter. The epoxide here has a methyl substituent on one of the epoxide carbons. In acidic conditions, the more substituted carbon (the one with the methyl group) is attacked (due to carbocation - like character development), and the stereochemistry: since the attack is on the carbon with the methyl group (which has a wedge in the starting epoxide), the product should have the - OH groups with the correct stereochemistry. The first product option for Rxn I (the one with $\ce{HO - C(CH3)-CH2 - OH}$ with the correct stereochemistry of the methyl and - OH) is the one where the methyl is on the same carbon as one - OH (the left - most structure in option C for Rxn I? Wait, no, let's re - examine. Wait, the starting epoxide for Rxn I has a methyl group on the epoxide carbon (wedge). In acidic hydrolysis, the nucleophilic attack (by water) on the epoxide carbon: the stereochemistry of the methyl group: since the epoxide ring is three - membered, the protonation and then attack. The product for Rxn I should be the diol where the two - OH groups are on adjacent carbons, and the methyl group is on the carbon with one - OH, with the correct stereochemistry. Looking at the options, for Rxn I, the correct product is the one where the methyl is on the carbon with the - OH (the structure with $\ce{HO - C(CH3)-CH2 - OH}$ with the methyl and one - OH on the same carbon, and the other - OH on the adjacent carbon, and the stereochemistry: the starting epoxide has a wedge on the methyl - bearing epoxide carbon, so the product should have the - OH on that carbon with the same stereochemistry? Wait, no, in epoxide ring - opening, when the epoxide is protonated, the ring opens, and the nucleophile (water) attacks. For a chiral epoxide, the stereochemistry: if the epoxide has a substituent (methyl) on one carbon, in acidic conditions, the attack is on the more substituted carbon (due to the development of partial positive charge, which is stabilized by the alkyl group). So the product for Rxn I should be the diol where the two - OH groups are on adjacent carbons, and the methyl group is on the carbon with one - OH, with the correct stereochemistry. Looking at the options, option C for Rxn I has the structure $\ce{HO - C(CH3)-CH2 - OH}$ with the methyl and - OH on the same carbon (the carbon with the methyl has a - OH, and the other carbon has a - OH), and the stereochemistry of the methyl: the starting epoxide had a wedge on the methyl - bearing epoxide carbon, so the product should have the - OH on that carbon with the same stereochemistry (the methyl is on a wedge - like position? Wait, no, the starting epoxide for Rxn I: the epoxide has a methyl group (wedge) on one carbon. After ring - opening, the carbon with the methyl group now has a - OH, and the stereochemistry: since the attack is from the opposite side of the epoxide ring? Wait, maybe I'm overcomplicating. Let's look at Rxn II.

Step2: Analyze Rxn II (Epoxide Ring - Opening with $\ce{NaOCH3}$ then $\ce{H2O}$)

$\ce{NaOCH3}$ is a strong base, so the $\ce{OCH3-}$ is a nucleophile. Epoxides react with nucleophiles in basic conditions (or with strong nucleophiles) via $\ce{S_{N}2}$ - like mechanism, where the nucleophile attacks the less substituted carbon (because the epoxide is a strained ring, and the less substituted carbon is more accessible for $\ce{S_{N}2}$ attack). The starting epoxide for Rxn II has a methyl group (dashed, so opposite stereochemistry to Rxn I's wedge) on one epoxide carbon. The nucleophile $\ce{OCH3-}$ will attack the less substituted carbon (the one without the methyl group? Wait, no, the epoxide carbons: one has a methyl (dashed) and the other is unsubstituted? Wait, the starting epoxide for Rxn II: the epoxide has a methyl group (dashed) on one carbon. So the two epoxide carbons: one with methyl (dashed) and one without. In basic conditions (with $\ce{OCH3-}$ as nucleophile), the nucleophile attacks the less substituted carbon (the one without the methyl group) via $\ce{S_{N}2}$, which involves inversion of stereochemistry at the attacked carbon. Then, after the first step (attack by $\ce{OCH3-}$), we have a product with $\ce{OCH3-}$ on one carbon and - OH (after $\ce{H2O}$ workup) on the other? Wait, no, the first step is $\ce{NaOCH3}$ (so $\ce{OCH3-}$ is the nucleophile), attacks the epoxide, opening the ring, forming an alkoxide, which is then protonated by $\ce{H2O}$ to form an alcohol. So the product for Rxn II should have $\ce{OCH3-}$ on one carbon and - OH on the adjacent carbon, with the correct stereochemistry. The starting epoxide for Rxn II has a dashed methyl group (so the epoxide carbon with methyl is on a dashed bond). When $\ce{OCH3-}$ attacks the other epoxide carbon (less substituted), the stereochemistry at that carbon is inverted. Then, the product for Rxn II should be the one with $\ce{HO - C(CH3)-CH2 - OCH3}$? No, wait, the options for Rxn II: the correct product should have $\ce{OCH3-}$ on the carbon that was attacked (less substituted), and - OH on the other, with the correct stereochemistry. Looking at the options, option C for Rxn II has the structure with $\ce{HO - C(CH3)-CH2 - OCH3}$ with the correct stereochemistry (the dashed lines for the methyl group, and the - OH and $\ce{OCH3-}$ on adjacent carbons). Wait, no, let's match Rxn I and Rxn II products.

Wait, Rxn I: acidic hydrolysis of epoxide, product is diol. The correct diol for Rxn I is the one where the two - OH groups are on adjacent carbons, and the methyl group is on the carbon with one - OH, with the stereochemistry matching the starting epoxide (wedge). So in option C, Rxn I product is $\ce{HO - C(CH3)-CH2 - OH}$ with the methyl and - OH on the same carbon (the carbon with methyl has a - OH, and the other - OH on the adjacent carbon), and the stereochemistry (the methyl is on a wedge - like position? Wait, the starting epoxide for Rxn I has a wedge methyl, so the product should have the - OH on that carbon with the same stereochemistry (wedge), and the other - OH on the adjacent carbon. Then Rxn II: basic conditions ( $\ce{NaOCH3}$), nucleophilic attack by $\ce{OCH3-}$ on the epoxide carbon (less substituted, the one without methyl), inversion of stereochemistry. The starting epoxide for Rxn II has a dashed methyl (so the epoxide carbon with methyl is on a dashed bond). So when $\ce{OCH3-}$ attacks the other epoxide carbon, the stereochemistry at that carbon is inverted, and the product has $\ce{OCH3-}$ on that carbon and - OH on the other, with the correct stereochemistry. So the correct pair is Rxn I: the diol with methyl and - OH on the same carbon (structure in option C for Rxn I), and Rxn II: the structure with $\ce{OCH3-}$ and - OH on adjacent carbons (structure in option C for Rxn II). So the correct option is C.

Answer:

C. $\boldsymbol{\ce{HO - C(CH3)-CH2 - OH}}$ (for Rxn I) and $\boldsymbol{\ce{HO - C(CH3)-CH2 - OCH3}}$ (for Rxn II) (the structures as shown in option C)