In the following reaction sequence,identify t | vedclass

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

Question

(B) The reaction sequence is as follows:$1$. Phenol $(C_6H_5OH)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to form phenyl

Vedclass · Accepted Answer

$4-$nitrophenyl benzoate

Vedclass · Answer

(B) The reaction sequence is as follows:$1$. Phenol $(C_6H_5OH)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to form phenyl benzoate $(X)$ via the Schotten-Baumann reaction.$2$. Phenyl benzoate $(C_6H_5COOC_6H_5)$ undergoes electrophilic aromatic substitution (nitration). The ester group $(-OCOC_6H_5)$ is ortho/para directing. Due to steric hindrance,the para-substituted product is the major product.$3$. The nitration occurs on the ring attached to the oxygen atom (the phenoxy ring) because it is more activated than the benzoyl ring. Thus,the major product $Y$ is $4-$nitrophenyl benzoate.

In the following reaction sequence,identify the major product $Y$:
$C_6H_5OH$ $\xrightarrow{C_6H_5COCl, \text{Base}} X$ $\xrightarrow{\text{Nitration}} Y$

Similar Questions

How can phenol and cyclohexanol be distinguished?

The major product of the following reaction sequence is:
Aniline $\xrightarrow[(i) NaNO_2, HCl, 273-278K]{(ii) H_2O, warm} \text{Product}$ $\xrightarrow[(iv) NaOH]{(iii) Br_2, (excess)} \text{Product}$ $\xrightarrow{(v) CH_3I} \text{Final Product}$

Assertion : Phenol on oxidation with $KMnO_4$ gives meso-tartaric acid.
Reason : Pure phenol is colourless but turns pink due to oxidation to phenoquinone.

Select the correct decreasing order of acid strength for the following compounds: $(a)$ $Ethanol$ $(b)$ $2-Methylpropan-2-ol$ $(c)$ $Phenol$ $(d)$ $p-Nitrophenol$.

The reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$ gives:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In the following reaction sequence,identify the major product $Y$:$C_6H_5OH$ $\xrightarrow{C_6H_5COCl, \text{Base}} X$ $\xrightarrow{\text{Nitration}} Y$