Let's break down the first reaction in **2 steps** as requested. --- ## **Step 1: Reaction Analysis** ### **Reaction Details** - Starting Material: Steroid-like moleculestructure given) - Reagents: 1. **(1) LiAlH4, Et2O, °C** 2. **(2) CH₃CH₂CHO** 3. **(3) H₃O⁺** ### **Explanation:** - **LiAlH4 (Lithium Aluminum Hydride)** is a strong reducing agent, usually used to reduce carbonyl compounds (like esters, carboxylic acids, ketones, aldehydes) to alcohols. - **Et2O (Diethyl Ether)** is a solvent. - **CH₃CH₂CHO** is propanal, an aldehyde. - **H₃O⁺** is used for aqueous workup, often to neutralize or protonate the product. #### **What Happens?** 1. **LiAlH4 Reduction:** If there is a carbonyl group (like a ketone or ester) in the molecule, it will be reduced to an alcohol. 2. **Reaction with Propanal:** The newly generated alkoxide (from the reduction) can attack propanal in a nucleophilic addition reaction (like in Grignard or organolithium reactions). 3. **Acid Workup:** Protonates the alkoxide to form a secondary or tertiary alcohol. --- #### **End of Step 1 Explanation** - **LiAlH4** reduces a carbonyl to an alcohol. - Alcohol (or alkoxide) reacts with **propanal** (nucleophilic addition). - **H₃O⁺** protonates the final product. --- ## **Step 2: Predicting the Product** 1. **Reduction:** - If the starting material contains a ketone (most likely at the C3 position in steroids), it is reduced to a secondary alcohol. 2. **Nucleophilic Addition:** - The alcohol is deprotonated by LiAlH4, then attacks propanal, forming a new carbon-carbon bond and a new alcohol group. 3. **Protonation:** - Acid workup gives a **tertiary alcohol** at that position. --- ## **Final Answer:** ### **Product Structure** - The original ketone (probably at C3) is converted to a **tertiary alcohol**. - The new substituent is a **CH(OH)CH₂CH₃** group attached to the former carbonyl carbon. **Summary:** **The product is the steroid backbone with a tertiary alcohol at the original ketone position, where the substituent is CH(OH)CH₂CH₃ (from addition of propanal).** --- ### **Stepwise Recap:** 1. **LiAlH4 reduction:** Ketone → Alcohol. 2. **Addition of propanal and acid workup:** Alcohol attacks propanal → Tertiary alcohol (with CH(OH)CH₂CH₃ group). --- Let me know if you want to break down any of the other reactions in a similar two-step manner!