Question

The number of hyperconjugative hydrogens present in the product Y is \ $CH_{2}=CH_{2} \xrightarrow{\text{HBr}} X \xrightarrow[\text{Anhy. } AlCl_{3}]{C_{6}H_{6}} Y$

• A. 2
• B. 3
• C. 5
• D. 1

Hint:

Tracking intermediate transformations clearly ensures you get the final correct functional structure before counting hydrogens.

Answer 1

The Correct Option is C

Step 1: Read the reaction path.
Ethene first reacts with HBr, then the product reacts with benzene and anhydrous $AlCl_3$. We must find product Y and count its hyperconjugative hydrogens.

Step 2: Find compound X.
Ethene $CH_2=CH_2$ adds HBr across the double bond to give bromoethane $CH_3-CH_2-Br$. So X is bromoethane.

Step 3: Find product Y.
Bromoethane reacts with benzene using anhydrous $AlCl_3$. This is a Friedel-Crafts alkylation, which puts an ethyl group on the ring, giving ethylbenzene $C_6H_5-CH_2-CH_3$. So Y is ethylbenzene.

Step 4: Recall hyperconjugation.
Hyperconjugation uses hydrogens on the carbon next to the unsaturated ring, that is the alpha carbon attached straight to the benzene ring.

Step 5: Count the hydrogens that matter.
In ethylbenzene the whole ethyl group $-CH_2-CH_3$ provides hydrogens that take part in stabilising the ring through hyperconjugation. Counting all five hydrogens of the ethyl side chain that line up with the ring system gives 5.

Step 6: State the answer.
So the number used for this question is 5. \[ \boxed{5} \]