Two compounds A and B with same molecular formula (C3H6O) undergo Grignard's reaction with methylmag

Q: Why does a tertiary alcohol react fastest in the Lucas test?

The Lucas test proceeds via an $S_N1$ mechanism. Tertiary alcohols form highly stable tertiary carbocations, leading to an immediate reaction and turbidity.

Q: Can a primary alcohol give a positive Iodoform test?

Generally no, with one exception: ethanol. Ethanol has the required $CH_3-CH(OH)-$ structure, but other primary alcohols do not.

Q: What is the reagent used in the Iodoform test?

The reagent is a mixture of iodine ($I_2$) and sodium hydroxide (NaOH), which forms sodium hypoiodite (NaOI) in situ.

Q: Why do we use Grignard reagents?

Grignard reagents are excellent nucleophiles used to form new carbon-carbon bonds. When they react with carbonyl compounds, they produce alcohols.

Q: What were the original compounds A and B?

Based on the products, compound A was likely propanal (which gives 2-butanol) and compound B was acetone (which gives 2-methyl-2-propanol).

Question

Two compounds $A$ and $B$ with same molecular formula ($C_3H_6O$) undergo Grignard's reaction with methylmagnesium bromide to give products $C$ and $D$. Products $C$ and $D$ show following chemical tests.

\begin{array}{lcc}
\hline 	ext{Test} & C & D \
\hline 	ext{Ceric ammonium nitrate test} & 	ext{Positive} & 	ext{Positive} \
	ext{Lucas test} & 	ext{Turbidity obtained after five minutes} & 	ext{Turbidity obtained immediately} \
	ext{Iodoform test} & 	ext{Positive} & 	ext{Negative} \
\hline
\end{array}

$C$ and $D$ respectively are

Options:

(A) $C = 	ext{H}_3	ext{C}-	ext{CH}_2-	ext{CH}(	ext{OH})-	ext{CH}_3 	ext{ ; } D = 	ext{H}_3	ext{C}-	ext{C}(	ext{OH})(	ext{CH}_3)-	ext{CH}_3$

(B) $C = 	ext{H}_3	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{CH}_2-	ext{OH} 	ext{ ; } D = 	ext{H}_3	ext{C}-	ext{CH}_2-	ext{CH}(	ext{OH})-	ext{CH}_3$

(C) $C = 	ext{H}_3	ext{C}-	ext{CH}_2-	ext{CH}_2-	ext{CH}_2-	ext{OH} 	ext{ ; } D = 	ext{H}_3	ext{C}-	ext{C}(	ext{OH})(	ext{CH}_3)-	ext{CH}_3$

(D) $C = 	ext{H}_3	ext{C}-	ext{C}(	ext{OH})(	ext{CH}_3)-	ext{CH}_3 	ext{ ; } D = 	ext{H}_3	ext{C}-	ext{CH}_2-	ext{CH}(	ext{OH})-	ext{CH}_3$

Sigma Percentile · Accepted Answer

### Decoding the Chemical Tests

Imagine you are a chemical detective, and you've just been handed two mysterious compounds, **C** and **D**. Both were born from a Grignard reaction, and now it's your job to figure out their exact identities using a series of classic qualitative tests. Let's break down the clues one by one.

The first piece of evidence is the **Ceric Ammonium Nitrate test**. Both compounds give a positive result. What does this tell us? This test is the gold standard for identifying alcohols. When an alcohol reacts with ceric ammonium nitrate, it forms a red complex. Since both **C** and **D** pass this test, we can confidently say that they both contain a hydroxyl ($-OH$) group.

### The Lucas Test: A Race Against Time

Next, we subject our compounds to the **Lucas test**. This test uses a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$ to differentiate between primary, secondary, and tertiary alcohols based on how quickly they form an insoluble alkyl chloride (which appears as turbidity).

Compound **C** gives turbidity after about five minutes. This moderate reaction rate is the signature behavior of a **secondary ($2^\circ$) alcohol**.

On the other hand, compound **D** turns turbid immediately! This lightning-fast reaction tells us that **D** is a highly reactive **tertiary ($3^\circ$) alcohol**, which forms a very stable tertiary carbocation intermediate.

### The Iodoform Test: Finding the Methyl Group

The final and most specific clue comes from the **Iodoform test**. This test uses iodine and sodium hydroxide to check for the presence of a specific structural feature: a methyl group directly attached to the carbon bearing the hydroxyl group (a $CH_3-CH(OH)-$ moiety).

Compound **C** gives a positive iodoform test, yielding a yellow precipitate. This confirms that **C** has that specific methyl group. Compound **D**, however, gives a negative test, meaning it lacks this structural feature.

### Putting It All Together

Now, let's look at our options and see which one fits our detective work perfectly.

We need **C** to be a secondary alcohol with a $CH_3-CH(OH)-$ group, and **D** to be a tertiary alcohol.

Looking at Option (a):
- **C** is **2-butanol** ($H_3C-CH_2-CH(OH)-CH_3$). It is a secondary alcohol and it clearly has the $CH_3-CH(OH)-$ group. It passes all criteria!
- **D** is **2-methyl-2-propanol** ($H_3C-C(OH)(CH_3)-CH_3$). It is a tertiary alcohol and does not have the $CH_3-CH(OH)-$ group. It also passes all criteria!

Therefore, we can confidently conclude that the correct structures are given in Option (a). The beauty of organic chemistry lies in how these simple, elegant qualitative tests can pinpoint the exact molecular structure of a compound!

Select Answer:

Visualized Solution

Ceric Ammonium Nitrate Test

Lucas Test

Iodoform Test

Evaluating Options

Conclusion

The Sigma Insight: Alcohols, Phenols, Ethers

Decoding the Chemical Tests

The Lucas Test

The Iodoform Test

Putting It All Together

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