JEE PYQ: Hydrocarbons Question 10
Question 10 - 2021 (25 Feb Shift 1)
Consider the following chemical reaction.
$\text{HC} \equiv \text{CH} \xrightarrow{(1)\ \text{Red hot Fe tube},\ 873\text{ K}}$ $\xrightarrow{(2)\ \text{CO}+\text{HCl}/\text{AlCl}_3}$ Product
The number of $sp^2$ hybridized carbon atom(s) present in the product is:
Show Answer
Answer: (7)
Solution
$\text{HC}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube}}$ Benzene $\xrightarrow{\text{CO}+\text{HCl}/\text{AlCl}_3}$ Benzaldehyde (Gattermann-Koch reaction). All carbon atoms in benzaldehyde are $sp^2$ hybridised = 7.
BETA
