Question

Consider the reaction: CH$\equiv$CH $\xrightarrow{\text{Red hot Fe}}$ (1) $\xrightarrow{CO, HCl, AlCl_3}$ (2). The number of sp$^2$ hybridized carbon atom(s) in the product is __________.

Hint:

Any carbon atom involved in a double bond (like in a benzene ring or a carbonyl group) is $sp^2$ hybridized.

Answer 1

Correct Answer: 7

Step 1: $3 \, CH\equiv CH \xrightarrow{\text{Red hot Fe tube}}$ Benzene ($C_6H_6$).
Step 2: Benzene $\xrightarrow{CO, HCl, AlCl_3}$ Benzaldehyde ($C_6H_5CHO$) via Gattermann-Koch reaction.
Step 3: In Benzaldehyde, there are 6 carbons in the ring (all $sp^2$) and 1 carbon in the aldehyde group ($sp^2$).
Step 4: Total $sp^2$ carbons $= 6 + 1 = 7$.