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Give step-by-step solution with explanation and final answer:| Be Vs hori Spoidaitd | > 0 EO mwa:

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Give step-by-step solution with explanation and final answer:Uploaded Image| Be Vs hori Spoidaitd | > 0 EO mwa:

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Asked by: Rahul
Created at: 2026-03-01 06:22:14

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Let's break down the reaction step by step. ### **Step 1: Analyze the Starting Material** - The starting material contains a **cyclic ketone** (indicated by the double-bonded O). - There are two stereocenters already shown, and the question asks for careful attention to stereochemistry. ### **Step 2: Identify the Reagent** - The reagents are: 1. **PhMgBr** (Phenylmagnesium bromide, a Grignard reagent) 2. **H₂O** (workup step) A Grignard reagent adds to the carbonyl carbon of a ketone, forming a **tertiary alcohol**. ### **Step 3: Mechanism** - The **PhMgBr** acts as a nucleophile and attacks the carbonyl carbon. - The double bond to oxygen breaks, and the oxygen gets a negative charge. - After acid workup (H₂O), the oxygen is protonated, forming an alcohol. ### **Step 4: Stereochemistry** - Addition to the carbonyl carbon creates a new stereocenter. - Since the attack can happen from either face, both wedge and dash are possible, but usually, we show one product. ### **Step 5: Draw the Product** - The original carbonyl carbon now has: - An OH group (from the former C=O) - A phenyl group (from PhMgBr) - The rest of the ring (and original substituents) - **Retain the original stereochemistry of other centers.** - The newly formed alcohol and phenyl group at the former carbonyl carbon should be shown **one as wedge, one as dash** to reflect the new stereocenter. --- ## **Final Structure** I'll describe the product clearly: - The 6-membered ring stays intact. - The carbon that was double-bonded to O now has: - An OH group (up or down, let's pick up/wedge) - A phenyl group (down/dash) - The rest of the ring and substituents retain their positions. ### **Product Structure** ``` OH (wedge/up) | / \ --C-----C-- / \ | | C C \ / C---------C | Ph (dash/down) ``` - The two original substituents on the ring (**ethyl and methyl**) remain in their places as shown in your starting material. - The **OH** is on a wedge and **Ph** on a dash at the former carbonyl carbon, or vice versa (depending on the approach of the Grignard reagent). --- ## **Summary Table** | Step | Reactant/Intermediate | Product/Result | |-------------------|-------------------------------|-------------------------| | Starting material | Cyclohexanone derivative | | | Add PhMgBr | Grignard addition | Tertiary alkoxide | | Add H₂O | Protonation | **Tertiary alcohol** | | Stereochemistry | At former C=O carbon: | New stereocenter (OH/Ph)| --- ### **Key Points** - The product is a **tertiary alcohol**. - The **phenyl** group and **OH** are added to the carbonyl carbon. - Use wedge/dash notation to indicate the new stereochemistry. If you want a detailed drawn structure, let me know!

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