In the following sequence of reactions, CH3CH2OH -(P+I2) A -(Mg, ether) B -(HCHO) C -(H2O) D the

Q: Why must we use dry ether when preparing the Grignard reagent?

Grignard reagents are extremely strong bases and will react violently with any source of acidic protons, including water. Dry ether provides a stable, aprotic solvent environment that solvates and stabilizes the magnesium complex without destroying it.

Q: What would happen if we used acetaldehyde instead of formaldehyde?

If you use acetaldehyde ($CH_3CHO$), the Grignard reagent would attack the carbonyl carbon, but the resulting product after hydrolysis would be a secondary alcohol (2-butanol) instead of a primary alcohol.

Q: Can we use $HCl$ instead of $P+I_2$ in the first step?

While $HCl$ can convert alcohols to alkyl chlorides, the reaction is very slow for primary alcohols like ethanol and usually requires a catalyst like $ZnCl_2$ (Lucas reagent). $P+I_2$ is much more efficient for making primary alkyl halides.

Q: Why does the carbon chain increase by exactly one carbon?

The Grignard reagent acts as a nucleophile and attacks the carbonyl carbon of formaldehyde. Since formaldehyde ($HCHO$) contains exactly one carbon atom, this single carbon is permanently added to the carbon skeleton of the original alkyl group.

Q: What is the byproduct formed in the final hydrolysis step?

When the alkoxide intermediate is hydrolyzed with water, the magnesium and iodine are released as basic magnesium iodide, $Mg(OH)I$, which is a stable inorganic byproduct.

Question

In the following sequence of reactions,
$CH_3CH_2OH \xrightarrow{P+I_2} A \xrightarrow[	ext{ether}]{Mg} B \xrightarrow{HCHO} C \xrightarrow{H_2O} D$
the compound 'D' is

Options:

(A) butanal

(B) n-butyl alcohol

(C) n-propyl alcohol

(D) propanal

Sigma Percentile · Accepted Answer

The journey from a simple alcohol to a more complex one is a classic tale in organic chemistry, often starring the versatile Grignard reagent. Let's break down this fascinating reaction sequence step-by-step.

### The First Transformation: Creating a Good Leaving Group

We start with ethyl alcohol ($CH_3CH_2OH$). The hydroxyl group ($-OH$) is a poor leaving group, so we need to convert it into something more reactive. Enter phosphorus and iodine ($P + I_2$).

When red phosphorus reacts with iodine, it forms phosphorus triiodide ($PI_3$) *in situ*. This reagent is excellent at substituting the hydroxyl group with an iodine atom.

$$CH_3CH_2OH \xrightarrow{P+I_2} CH_3CH_2I$$

Our product **A** is ethyl iodide. The iodine atom is a fantastic leaving group, setting the stage for the next act.

### Forging the Grignard Reagent

Now, we treat ethyl iodide with magnesium metal in the presence of dry ether. This is the textbook method for preparing a Grignard reagent. The magnesium inserts itself between the carbon and the iodine, reversing the polarity of the carbon atom (a phenomenon known as *umpolung*).

$$CH_3CH_2I \xrightarrow{Mg, 	ext{ether}} CH_3CH_2MgI$$

Product **B** is ethyl magnesium iodide. The carbon attached to magnesium is now highly nucleophilic, eager to attack an electrophile.

### The Nucleophilic Attack

Our electrophile is formaldehyde ($HCHO$). The carbonyl carbon in formaldehyde is electron-deficient due to the electronegative oxygen pulling electron density away.

The nucleophilic ethyl group from our Grignard reagent attacks this carbonyl carbon, pushing the pi electrons up onto the oxygen.

$$CH_3CH_2MgI + HCHO ightarrow CH_3CH_2CH_2OMgI$$

This forms an alkoxide intermediate, product **C**. Notice how the carbon chain has grown! We started with a two-carbon chain (ethyl) and added one more carbon from formaldehyde, creating a three-carbon framework.

### The Final Quench

The last step is simple hydrolysis. Adding water ($H_2O$) provides a proton to the negatively charged oxygen of the alkoxide, yielding the final alcohol.

$$CH_3CH_2CH_2OMgI \xrightarrow{H_2O} CH_3CH_2CH_2OH + Mg(OH)I$$

Our final product **D** is **n-propyl alcohol** (or 1-propanol).

**Key Takeaway:** Reacting a Grignard reagent with formaldehyde is a reliable method to synthesize a primary alcohol with a carbon chain extended by exactly one carbon atom.

Animated Solution for Chemistry - Organic Chemistry: In the following sequence of reactions, $C H_{3} C H_{2} O H P + I_{2} A M g ether B H C H O C H_{2} O D$ the compound 'D' is

Select Answer:

Visualized Solution

Reaction Sequence

Formation of Product A

Formation of Product B

Formation of Product C

Formation of Product D

Final Answer

The Sigma Insight: Alcohols, Phenols, Ethers

The First Transformation

Forging the Grignard Reagent

The Nucleophilic Attack

The Final Quench

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the following sequence of reactions, CH3​CH2​OHP+I2​​AMgether​BHCHO​CH2​O​D the compound 'D' is