Question

What are X and Y in the following reaction sequence? $Ethanal \rightarrow C_4H_8 \xrightarrow{Br_2/CCl_4} X \xrightarrow{alc. KOH, NaNH_2} Y$

• A. $X=$ alcoholic KOH, $NaNH_2$; $Y=CH_3CH_2C \equiv CH$
• B. $X=$ alcoholic KOH, $NaNH_2$; $Y=CH_3C \equiv CCH_3$
• C. $X=$ alcoholic KOH; $Y=CH_3CH_2C \equiv CH$
• D. $X=$ aq KOH; $Y=CH_3C \equiv CCH_3$

Hint:

$NaNH_2$ is a strong base used to achieve double dehydrohalogenation for terminal alkynes.

Answer 1

The Correct Option is A

Step 1: Build the alkene from ethanal.
Ethanal is a two-carbon aldehyde. To reach $C_4H_8$, two such units are joined and reduced, giving a four-carbon alkene such as but-1-ene. This is the starting alkene for the next step.

Step 2: Add bromine across the double bond.
Treating the alkene with $Br_2$ in $CCl_4$ adds one bromine to each of the two double-bond carbons. This gives a vicinal dibromide, a compound with two bromines on neighbouring carbons.

Step 3: Identify the role of X.
The next arrow uses alcoholic KOH together with $NaNH_2$. These are strong bases that remove two molecules of $HBr$, one after the other, from the dibromide.

Step 4: First elimination.
Alcoholic KOH pulls off the first $HBr$, turning the dibromide into a bromoalkene. This sets up the second elimination.

Step 5: Second elimination to the alkyne.
The stronger base $NaNH_2$ then removes the second $HBr$, creating a carbon-carbon triple bond. Because the starting alkene was terminal, the product Y is the terminal alkyne $CH_3CH_2C\equiv CH$.

Step 6: State X and Y.
So X is the alcoholic KOH with $NaNH_2$ step and Y is the terminal alkyne.
\[ \boxed{X=\text{alcoholic KOH, } NaNH_2; \; Y=CH_3CH_2C\equiv CH} \]