Consider the following sequence of reactions. | vedclass

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(B) $1$. The starting material is $p$-phenetidine ($4$-ethoxyaniline). Treatment with $NaNO_2/HCl$ at $0-5^{\circ}C$ yields the diazonium salt $(A)$,w

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$4$

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(B) $1$. The starting material is $p$-phenetidine ($4$-ethoxyaniline). Treatment with $NaNO_2/HCl$ at $0-5^{\circ}C$ yields the diazonium salt $(A)$,which is $4$-ethoxybenzenediazonium chloride.$2$. Coupling of $(A)$ with phenol in the presence of $NaOH$ followed by acidification gives the azo dye $(B)$,$4$-ethoxy-$4'$-hydroxyazobenzene.$3$. Treatment of $(B)$ with $NaOH$ followed by ethyl bromide $(CH_3CH_2Br)$ performs an $O$-alkylation of the phenolic $-OH$ group,yielding the final product $(C)$,$4,4'$-diethoxyazobenzene.$4$. The structure of $(C)$ is $CH_3CH_2-O-C_6H_4-N=N-C_6H_4-O-CH_2CH_3$.$5$. In this molecule,each ethoxy group $(-OCH_2CH_3)$ contains two $sp^{3}$ hybridized carbon atoms. Since there are two such groups,the total number of $sp^{3}$ hybridized carbon atoms is $2 	imes 2 = 4$.

Consider the following sequence of reactions. Total number of $sp^{3}$ hybridised carbon atoms in the major product $C$ formed is . . . . . . .

Similar Questions

Identify $M$ in the following sequence of reactions:

In the given reaction sequence,what is $Z$?
Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$

The major product of the reaction is:
$R-CONH_2 \xrightarrow{NaOBr} (A)$
(where $R$ is the chiral group shown in the image)

The product $(C)$ in the above reaction is:

Isoniazid is used in the treatment of tuberculosis. Identify the least basic $N$ atom present in it.

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