Consider the following reactions:
(A) -[Ozonolysis] (B) + (C)
(C7H14)
'B' -[I2 + NaOH, Heat] yello

Q: Why can't compound B be a methyl ketone like acetone?

While acetone would give a positive iodoform test (yellow precipitate), it is a ketone and therefore would NOT give a positive Tollens' test (silver mirror). Only aldehydes give the silver mirror test. Acetaldehyde is the only aldehyde that also gives the iodoform test.

Q: Why can't compound C be 2-pentanone?

2-pentanone ($CH_3COCH_2CH_2CH_3$) is a methyl ketone. If C were 2-pentanone, it would have given a positive iodoform test (yellow precipitate). The problem explicitly states that C gives no yellow precipitate, ruling out 2-pentanone.

Q: What if compound D gave immediate turbidity in the Lucas test?

Immediate turbidity would mean D is a tertiary ($3^\circ$) alcohol. However, D is formed by the reduction of C (a carbonyl compound) using $LiAlH_4$. The reduction of aldehydes gives $1^\circ$ alcohols, and the reduction of ketones gives $2^\circ$ alcohols. You cannot get a $3^\circ$ alcohol from the simple reduction of a carbonyl group. Thus, immediate turbidity is chemically impossible in this specific sequence.

Q: How do we know the exact position of the double bond in A?

Ozonolysis cleaves the double bond and places an oxygen atom on each of the resulting fragments. By identifying the two fragments (Acetaldehyde and 3-Pentanone), we simply remove the oxygen atoms and reconnect those specific carbons with a double bond. This uniquely determines the position of the double bond.

Q: Could compound A have been a cyclic alkene?

No. The molecular formula of A is $C_7H_{14}$, which follows the general formula $C_nH_{2n}$ for an acyclic alkene with one double bond. Furthermore, ozonolysis of a cyclic alkene would yield a single molecule with two carbonyl groups (a dicarbonyl compound), rather than two separate fragments B and C.

Question

Consider the following reactions:
(A) $\xrightarrow{	ext{Ozonolysis}}$ (B) + (C)
$(C_7H_{14})$
'B' $\xrightarrow[\Delta]{(I_2 + NaOH)}$ yellow ppt.
$\xrightarrow[\Delta]{Ag_2O}$ Silver mirror
(C) $\xrightarrow[\Delta]{(I_2 + NaOH)}$ no yellow ppt.
$\xrightarrow[\Delta]{LiAlH_4}$ (D) $\xrightarrow{	ext{Anhydrous } ZnCl_2 	ext{ and conc. HCl}}$ Gives white turbidity within 5 minutes
(A) is

Sigma Percentile · Accepted Answer

This problem is a brilliant multi-concept puzzle that tests your mastery over several fundamental organic chemistry reactions: Ozonolysis, the Iodoform test, Tollens' test, and the Lucas test. Let's break it down step-by-step like a detective solving a mystery.

### Decoding Compound B

We are given an unknown alkene **A** with the molecular formula $C_7H_{14}$. Upon ozonolysis, it cleaves into two fragments, **B** and **C**.

The problem states that compound **B** gives a yellow precipitate when heated with $I_2$ and $NaOH$. This is the classic **Iodoform test**, which indicates the presence of a methyl ketone group ($-COCH_3$) or acetaldehyde.

However, the plot thickens! Compound **B** also gives a positive **Silver mirror test** with Tollens' reagent ($Ag_2O$). The Tollens' test is exclusively given by aldehydes.

So, we need a molecule that is an aldehyde AND gives a positive iodoform test. There is only one molecule in the entire universe of organic chemistry that satisfies both conditions: **Acetaldehyde** ($CH_3CHO$). Thus, we have successfully identified **B**.

### Decoding Compound C and D

Next, let's look at compound **C**. It gives a negative iodoform test, meaning it is definitely not a methyl ketone.

When **C** is reduced using Lithium Aluminum Hydride ($LiAlH_4$), it forms compound **D**. Compound **D** is then subjected to the **Lucas test** (Anhydrous $ZnCl_2$ and conc. $HCl$). It produces white turbidity within exactly 5 minutes.

In the Lucas test, immediate turbidity indicates a $3^\circ$ alcohol, turbidity in 5 minutes indicates a $2^\circ$ alcohol, and no turbidity at room temperature indicates a $1^\circ$ alcohol. Therefore, **D** is a **secondary ($2^\circ$) alcohol**.

Since the reduction of **C** yields a secondary alcohol, **C** must be a **ketone**.

### The Carbon Math

Now, let's do some simple arithmetic. The original alkene **A** had 7 carbon atoms ($C_7H_{14}$). We know that **B** (Acetaldehyde) has 2 carbon atoms.

This means compound **C** must contain the remaining carbons: $7 - 2 = 5$ carbons.

So, **C** is a 5-carbon ketone. But remember, it gave a negative iodoform test, so it cannot be 2-pentanone (which is a methyl ketone). The only other 5-carbon ketone is **3-pentanone** ($CH_3CH_2COCH_2CH_3$).

Thus, **C** is 3-pentanone, and its reduction product **D** is 3-pentanol.

### Reconstructing Alkene A

Now for the grand finale. We know the two products of ozonolysis: Acetaldehyde ($CH_3CHO$) and 3-Pentanone ($O=C(CH_2CH_3)_2$).

To find the structure of the original alkene **A**, we perform a "reverse ozonolysis". We align the oxygen atoms of the two carbonyl groups, remove them, and stitch the remaining carbon atoms together with a double bond.

$$CH_3CH=O + O=C(CH_2CH_3)_2 \longrightarrow CH_3CH=C(CH_2CH_3)_2$$

The resulting molecule is **3-ethylpent-2-ene**.

Looking at the given options, the skeletal structure in option (d) perfectly matches 3-ethylpent-2-ene, featuring a double bond with a methyl group on one side and two ethyl groups on the other. This confirms our logical deduction!

Select Answer:

Visualized Solution

$C_{7} H_{14} O_{3} B + C$

$B ⟹ C H_{3} C H O$

$C ⟹ Ketone$

$C ⟹ 3 -Pentanone$

$A ⟹ 3 -ethylpent-2-ene$

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Carbonyl Compounds

Decoding Compound B

Decoding Compound C and D

The Carbon Math

Reconstructing Alkene A

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