Consider the following chemical reaction.CH e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(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 ..........

Similar Questions

Which of the following compounds will not react with $NaHSO_3$?

Consider the following reactions:
$A$ $\xrightarrow[(i) CH_3MgBr]{(ii) H_3O^{+}} B$ $\xrightarrow[573 \text{ K}]{Cu} 2\text{-methyl-2-butene}$
The mass percentage of carbon in $A$ is:

In the given reaction $CH_3-CH_2-C(=O)-CH_2-COOC_2H_5$ $\xrightarrow{[X]} (A)$ $\xrightarrow{(i) LiAlH_4, (ii) H_2O/H^{+}} CH_3-CH_2-C(=O)-CH_2-CH_2OH + C_2H_5OH$,$[X]$ will be:

Benzaldehyde and acetaldehyde can be differentiated by

The typical reaction of an aldehyde is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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 ..........