Coloration of Br_2/CCl_4 will be discharged b | vedclass

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

Question

(A) The decolorization of $Br_2/CCl_4$ is a characteristic test for the presence of an unsaturated carbon-carbon double bond $(C=C)$.Cinnamic acid $(C

Vedclass · Accepted Answer

Cinnamic acid

Vedclass · Answer

(A) The decolorization of $Br_2/CCl_4$ is a characteristic test for the presence of an unsaturated carbon-carbon double bond $(C=C)$.Cinnamic acid $(C_6H_5CH=CHCOOH)$ contains a carbon-carbon double bond in its side chain.It reacts with bromine in carbon tetrachloride via electrophilic addition to form dibromocinnamic acid,which is colorless.$C_6H_5CH=CHCOOH + Br_2 \xrightarrow{CCl_4} C_6H_5CH(Br)-CH(Br)COOH$Benzoic acid,$o-$phthalic acid,and acetophenone do not contain $C=C$ bonds that undergo this addition reaction under these conditions.

Coloration of $Br_2/CCl_4$ will be discharged by

Similar Questions

Find the stability order of the following compounds:

What are the products obtained from the hydroboration-oxidation of propene and its hydration in the presence of dilute sulfuric acid,respectively?

Oils are converted into fats by

$A$ single compound of the structure $CH_3-CO-CH_2-CH_2-CH_2-CHO$ is obtainable from the ozonolysis of which of the following cyclic compounds?

The reaction,$H_2C=CH_2$ (Ethylene) $+ H_2O$ (water) $\xrightarrow{H_3PO_4, 300 \ ^\circ C/60 \ atm}$ $C_2H_5OH$ (Ethyl alcohol) is called:

Vedclass Test Series

Exam Paper Generator

Online Exam Module