The correct sequence of reagents for the following conversion will be
(a) [Ag(NH3)2]+ OH-, H+ / CH3O

Q: Why can't we just use Grignard reagent directly on the reactant?

If we use Grignard directly, it will react with both the ketone and the aldehyde. The ketone will form a tertiary alcohol (adding one methyl), and the aldehyde will form a secondary alcohol (adding one methyl). But our target product has *two* methyl groups at the aldehyde position!

Q: Why do we use Tollen's reagent instead of $KMnO_4$?

$KMnO_4$ is a strong oxidizing agent. While it would oxidize the aldehyde to a carboxylic acid, it could also cause oxidative cleavage of the ring near the ketone under vigorous conditions. Tollen's reagent is mild and highly selective for aldehydes.

Q: What is the purpose of $CH_3OH / H^+$?

This is Fischer esterification. It converts the carboxylic acid into a methyl ester. We need an ester because esters react with *two* equivalents of Grignard reagent to form tertiary alcohols.

Q: How does the ester react with two equivalents of Grignard reagent?

The first equivalent of Grignard attacks the ester carbonyl, and the methoxy group ($-OCH_3$) leaves, forming a ketone intermediate. The second equivalent of Grignard then attacks this new ketone, forming the final tertiary alcohol.

Q: Does the Grignard reagent react with the ketone and the ester at the same time?

Yes, when we add excess Grignard reagent, it will attack all available electrophilic carbonyl carbons in the molecule simultaneously.

Question

The correct sequence of reagents for the following conversion will be

Options:

(A) $[Ag(NH_3)_2]^+ OH^-, H^+ / CH_3OH, CH_3MgBr$

(B) $CH_3MgBr, H^+ / CH_3OH, [Ag(NH_3)_2]^+ OH^-$

(C) $CH_3MgBr, [Ag(NH_3)_2]^+ OH^-, H^+ / CH_3OH$

(D) $[Ag(NH_3)_2]^+ OH^-, CH_3MgBr, H^+ / CH_3OH$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We are presented with a fascinating organic conversion. Our starting material is a cyclohexane ring adorned with two distinct functional groups: a **ketone** at the top and an **aldehyde** at the bottom.

Our target product is a molecule where both of these carbonyl groups have been transformed into **tertiary alcohols**.

Specifically, the ketone has gained one methyl group, while the aldehyde has gained **two methyl groups**. This observation is the key to unlocking the entire sequence!

### The Master Plan
To add methyl groups and form tertiary alcohols, the **Grignard reagent** ($CH_3MgBr$) is our weapon of choice.

However, we face a strategic dilemma. A Grignard reagent adds exactly *one* alkyl group to a ketone or an aldehyde. If we were to react our starting material directly with excess $CH_3MgBr$, the ketone would correctly form the desired tertiary alcohol. But the aldehyde would only form a secondary alcohol!

To add *two* methyl groups to the bottom carbon, we must first convert the aldehyde into an **ester**. Esters are unique because they react with two equivalents of a Grignard reagent to yield tertiary alcohols.

### Step 1: Selective Oxidation
Before we can form an ester, we need a carboxylic acid. We must oxidize the aldehyde, but we have to be incredibly careful.

If we use a strong oxidizing agent like $KMnO_4$, we risk oxidizing the ketone or even cleaving the ring. We need a gentle touch.

Enter **Tollen's reagent** ($[Ag(NH_3)_2]^+ OH^-$). This mild oxidizing agent is perfectly selective. It smoothly oxidizes the aldehyde group ($-CHO$) into a carboxylic acid group ($-COOH$) while leaving the ketone completely untouched.

### Step 2: Esterification
Now that we have our carboxylic acid, we can proceed to form the ester.

We treat the intermediate with methanol ($CH_3OH$) in the presence of an acid catalyst ($H^+$). This is a classic **Fischer esterification**.

The carboxylic acid group is converted into a methyl ester ($-COOCH_3$). Our molecule is now primed and ready for the final assault.

### Step 3: The Grignard Addition
With our ketone and ester in place, we unleash an excess of **methylmagnesium bromide** ($CH_3MgBr$).

The ketone reacts with one equivalent of the Grignard reagent. The nucleophilic methyl group attacks the carbonyl carbon, pushing the electrons up to the oxygen, which is subsequently protonated to form the tertiary alcohol.

Simultaneously, the ester reacts with *two* equivalents of the Grignard reagent. The first equivalent attacks the ester carbonyl, kicking out the methoxy leaving group to form a ketone intermediate. The second equivalent immediately attacks this new ketone, ultimately forming the second tertiary alcohol.

### Final Conclusion
By carefully orchestrating this sequence—selective oxidation, esterification, and finally Grignard addition—we successfully achieve the desired transformation.

The correct sequence of reagents is indeed $[Ag(NH_3)_2]^+ OH^-$, followed by $CH_3OH/H^+$, and finally $CH_3MgBr$. This perfectly matches option (a).

Animated Solution for Chemistry - Organic Chemistry: The correct sequence of reagents for the following conversion will be

Select Answer:

Visualized Solution

$A na l y z in g t h e C o n v er s i o n$

$I d e n t i f y in g t h e C han g es$

$S t e p 1 : S e l ec t i v e O x i d a t i o n$

$S t e p 2 : E s t er i f i c a t i o n$

$S t e p 3 : G r i g na r d A dd i t i o n$

$C o n c l u s i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Carbonyl Compounds

Analyzing the Setup We are presented with a fascinating organic conversion

The Master Plan

Step 1

Step 2

Step 3

Final Conclusion

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