Consider the following chemical reaction.CH e | vedclass

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(D) The reaction proceeds in two steps:$1$. $3 CH \equiv CH \xrightarrow{	ext{Red hot Fe tube, } 873 	ext{ K}} C_6H_6$ (Benzene).$2$. $C_6H_6 + CO +

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$7$

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(D) The reaction proceeds in two steps:$1$. $3 CH \equiv CH \xrightarrow{	ext{Red hot Fe tube, } 873 	ext{ K}} C_6H_6$ (Benzene).$2$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$ (Benzaldehyde) via Gattermann-Koch reaction.In benzaldehyde $(C_6H_5CHO)$,there are $6$ carbon atoms in the benzene ring,all of which are $sp^{2}$ hybridized.The carbonyl carbon atom $(C=O)$ is also $sp^{2}$ hybridized.Therefore,the total number of $sp^{2}$ hybridized carbon atoms is $6 + 1 = 7$.

Consider the following chemical reaction.
$CH \equiv CH \xrightarrow[{(2) CO, HCl, AlCl_3}]{{(1) Red hot Fe tube, 873 K}} \text{Product}$
The number of $sp^{2}$ hybridized carbon atom$(s)$ present in the product is ..........

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The reaction sequence is as follows:
Benzene + $CH_3COCl$ $\xrightarrow{\text{Anhydrous } AlCl_3} (A)$ $\xrightarrow{(x)} \text{Ethylbenzene}$.
Reagent $(x)$ can be:

The product $(A)$ of the following reaction is:

Comment on the optical activity of the products. They are:

Assertion : The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses.
Reason : There is a weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions.

The products of acid hydrolysis of $P$ and $Q$ can be distinguished by:
$P = H_2C=C(CH_3)(OCOCH_3)$
$Q = H_3C-CH=CH(OCOCH_3)$

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Consider the following chemical reaction.$CH \equiv CH \xrightarrow[{(2) CO, HCl, AlCl_3}]{{(1) Red hot Fe tube, 873 K}} \text{Product}$The number of $sp^{2}$ hybridized carbon atom$(s)$ present in the product is ..........