Animated Solution for Chemistry - Organic Chemistry: Consider the following reactions : $A\underset{\text{(ii) }{H}_3{O}^{+}}{\overset{\text{(i) }C{H}_{3}MgBr}{\to }}B\underset{573\text{ K}}{\overset{Cu}{\to }}\text{2-methyl- 2- butene}$ The mass percentage of carbon in A is ......

Have you ever tried solving a maze by starting from the finish line? In organic chemistry, this powerful technique is called retrosynthetic analysis. Instead of moving forward, we take the final product and work our way backward to discover the starting materials. Let's apply this strategy to our problem!

We are given a final product: 2-methyl-2-butene. This alkene was formed by heating compound B with copper ($Cu$) at $573\text{ K}$.

Now, what does copper do at such a high temperature? It's a classic reagent! When primary or secondary alcohols are passed over hot copper, they undergo dehydrogenation to form aldehydes and ketones. However, tertiary alcohols don't have an alpha-hydrogen to lose. Instead, they undergo dehydration (loss of a water molecule) to form alkenes.

Since our product is an alkene, compound B must be a tertiary alcohol. By adding a water molecule back across the double bond (following Markovnikov's rule to get the most stable tertiary carbocation intermediate), we can deduce the structure of B.

Compound B is 2-methylbutan-2-ol ($C{H}_3-C(OH)(C{H}_3)-C{H}_2-C{H}_3$).

Now that we know B, let's take another step back. Compound A reacts with methyl magnesium bromide ($C{H}_{3}MgBr$), a Grignard reagent, followed by acidic hydrolysis ($H_3{O}^{+}$) to form our tertiary alcohol B.

Grignard reagents are fantastic nucleophiles. They attack the electrophilic carbonyl carbon of aldehydes and ketones. A crucial rule to remember is: - Formaldehyde + Grignard $\to$ Primary Alcohol - Other Aldehydes + Grignard $\to$ Secondary Alcohol - Ketones + Grignard $\to$ Tertiary Alcohol

Since B is a tertiary alcohol, compound A must be a ketone. The Grignard reagent ($C{H}_{3}MgBr$) adds a methyl group ($C{H}_3$) to the carbonyl carbon. To find A, we simply remove one of the methyl groups attached to the hydroxyl-bearing carbon in B and restore the carbon-oxygen double bond.

This reveals that compound A is butan-2-one ($C{H}_3-CO-C{H}_2-C{H}_3$).

We've successfully unmasked compound A! The final step is a simple mole concept calculation. We need to find the mass percentage of carbon in butan-2-one.

First, let's write down the molecular formula of butan-2-one: $C_4{H}_{8}O$.

Next, we calculate the total molar mass: - Carbon: $4\times 12=48\text{ g/mol}$ - Hydrogen: $8\times 1=8\text{ g/mol}$ - Oxygen: $1\times 16=16\text{ g/mol}$ - Total Molar Mass $=48+8+16=72\text{ g/mol}$

The mass contributed by carbon alone is $48\text{ g/mol}$.

Finally, we calculate the mass percentage:

And there we have it! By thinking backward and applying fundamental organic chemistry principles, we seamlessly arrived at the final answer. Keep practicing retrosynthesis; it's like having a superpower in organic chemistry!