Consider the following reactions, A --(Ag2O, Delta)-- ppt. A --(Hg2+/H+)-- B --(NaBH4)-- C --(ZnC

Q: Why doesn't option (b) react with $Ag_2O$?

Option (b) is But-2-yne, which is an internal alkyne. It lacks the acidic terminal hydrogen required to react with heavy metal ions like $Ag^+$ to form a precipitate.

Q: What is the role of $Hg^{2+}/H^+$ in the reaction?

This reagent mixture is used for the Kucherov reaction. It catalyzes the hydration of alkynes, adding water across the triple bond according to Markovnikov's rule to form an enol, which then tautomerizes into a more stable ketone or aldehyde.

Q: Why does ethyne give an aldehyde while other alkynes give ketones?

Ethyne is a symmetric, two-carbon alkyne. When water adds to it, it forms vinyl alcohol, which tautomerizes exclusively to acetaldehyde. All other alkynes have at least three carbons, and Markovnikov addition ensures the oxygen attaches to an internal carbon, forming a ketone.

Q: How does the Lucas test distinguish between alcohols?

The Lucas test relies on the $S_N1$ mechanism. Tertiary alcohols form highly stable $3^\circ$ carbocations and react immediately to form insoluble alkyl chlorides (turbidity). Secondary alcohols form less stable $2^\circ$ carbocations, taking about 5 minutes. Primary alcohols form highly unstable $1^\circ$ carbocations and do not react at room temperature.

Q: Can $NaBH_4$ reduce the triple bond of the alkyne directly?

No, $NaBH_4$ is a mild and selective reducing agent. It effectively reduces polar double bonds like the carbonyl group ($C=O$) in aldehydes and ketones but is not strong enough to reduce non-polar carbon-carbon double ($C=C$) or triple ($C \equiv C$) bonds.

Question

Consider the following reactions,
$A \xrightarrow[\Delta]{Ag_2O} 	ext{ppt}$
$A \xrightarrow{Hg^{2+}/H^+} B \xrightarrow{NaBH_4} C \xrightarrow[	ext{Conc. HCl}]{ZnCl_2} 	ext{Turbidity within 5 minutes}$
$A$ is

Options:

(A) $CH \equiv CH$

(B) $CH_3-C \equiv C-CH_3$

(C) $CH_3-C \equiv CH$

(D) $CH_2=CH_2$

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating chemical puzzle. We are given an unknown compound $A$ that undergoes two distinct reaction pathways, and our goal is to deduce its structure based on the qualitative tests it passes. This is a classic example of a sequence problem in organic chemistry, where we must act as molecular detectives, piecing together clues from different chapters.

### Analyzing the Setup

Let's break down the information given to us:
1. **Pathway 1:** Compound $A$ reacts with $Ag_2O$ upon heating to form a precipitate.
2. **Pathway 2:** Compound $A$ undergoes hydration with $Hg^{2+}/H^+$ to form $B$, which is then reduced by $NaBH_4$ to form $C$. Finally, $C$ reacts with $ZnCl_2$ and concentrated $HCl$ to produce turbidity within 5 minutes.

Our strategy will be to analyze the endpoints of both pathways and work our way backward to find the identity of $A$.

### The Silver Mirror Clue

The first major clue lies in Pathway 1. Compound $A$ reacts with Tollens' reagent ($Ag_2O$) to yield a precipitate.

This is a highly specific qualitative test. Among hydrocarbons, only **terminal alkynes** possess an acidic hydrogen atom attached to the $sp$-hybridized carbon. This acidic hydrogen can be replaced by heavy metal ions like $Ag^+$ or $Cu^+$ to form insoluble acetylides.

$$R-C \equiv C-H + Ag^+ \xrightarrow{\Delta} R-C \equiv C-Ag \downarrow + H^+$$

Looking at our options:
- (a) $CH \equiv CH$ (Ethyne) is a terminal alkyne.
- (b) $CH_3-C \equiv C-CH_3$ (But-2-yne) is an internal alkyne and will not react.
- (c) $CH_3-C \equiv CH$ (Propyne) is a terminal alkyne.
- (d) $CH_2=CH_2$ (Ethene) is an alkene and will not react.

We have successfully narrowed down our suspects to options (a) and (c).

### The Time Trial: Lucas Test

Now, let's shift our focus to the end of Pathway 2. Compound $C$ reacts with the Lucas reagent (anhydrous $ZnCl_2$ and concentrated $HCl$) to produce turbidity within exactly 5 minutes.

The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols based on their reactivity via the $S_N1$ mechanism. The appearance of turbidity (due to the formation of an insoluble alkyl chloride) follows a strict timeline:
- **Immediate turbidity:** Indicates a $3^\circ$ alcohol (highly stable carbocation).
- **Turbidity in 5 minutes:** Indicates a **$2^\circ$ alcohol**.
- **Turbidity only upon heating:** Indicates a $1^\circ$ alcohol.

Since the turbidity appears in 5 minutes, compound $C$ must be a secondary ($2^\circ$) alcohol.

### The Missing Link: Retrosynthesis

We know $C$ is a $2^\circ$ alcohol. How was it formed? It was produced by the reduction of compound $B$ using sodium borohydride ($NaBH_4$).

$NaBH_4$ is a mild reducing agent that selectively reduces aldehydes to $1^\circ$ alcohols and ketones to $2^\circ$ alcohols. Since $C$ is a $2^\circ$ alcohol, compound $B$ must be a **ketone**.

Now, let's connect $B$ back to $A$. Compound $B$ (a ketone) is formed by the hydration of alkyne $A$ using $Hg^{2+}/H^+$. This is known as the Kucherov reaction, which follows Markovnikov's rule.

Let's test our two remaining suspects:
**Suspect 1: Ethyne ($CH \equiv CH$)**
Hydration of ethyne yields acetaldehyde ($CH_3CHO$), an aldehyde. Reduction of acetaldehyde gives ethanol, a $1^\circ$ alcohol. A $1^\circ$ alcohol would not give turbidity in 5 minutes in the Lucas test. Thus, ethyne is incorrect.

**Suspect 2: Propyne ($CH_3-C \equiv CH$)**
Hydration of propyne follows Markovnikov's rule, placing the $OH$ group on the more substituted carbon, yielding acetone ($CH_3-CO-CH_3$), a ketone.

$$CH_3-C \equiv CH \xrightarrow{Hg^{2+}/H^+} CH_3-CO-CH_3$$

Reduction of acetone with $NaBH_4$ yields propan-2-ol ($CH_3-CH(OH)-CH_3$), which is indeed a $2^\circ$ alcohol!

$$CH_3-CO-CH_3 \xrightarrow{NaBH_4} CH_3-CH(OH)-CH_3$$

Propan-2-ol will react with the Lucas reagent to give turbidity in 5 minutes, perfectly matching all the given conditions.

### Final Calculation

All the pieces of the puzzle fit perfectly together. Compound $A$ must be propyne because it is a terminal alkyne that ultimately yields a secondary alcohol upon hydration and reduction.

**Final Answer:** The correct option is **(c) $CH_3-C \equiv CH$**.

Animated Solution for Chemistry - Organic Chemistry: Consider the following reactions, $A A g_{2} O Δ ppt$ $A H g^{2 +} / H^{+} B N a B H_{4} C Z n C l_{2} Conc. HCl Turbidity within 5 minutes$ $A$ is

Select Answer:

Visualized Solution

$A Pathways Products$

$A A g_{2} O Precipitate$

$C Z n C l_{2} / H C l Turbidity$

$B N a B H_{4} C$

$A H g^{2 +} / H^{+} B$

\text{Final Answer}

\text{Key Takeaways}

The Sigma Insight: Hydrocarbons

Analyzing the Setup

The Silver Mirror Clue

The Time Trial

The Missing Link

Final Calculation

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