X is heated with soda lime and gives ethane. | vedclass

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

Question

(C) The reaction of a carboxylic acid with soda lime $(NaOH + CaO)$ is known as decarboxylation.In this process,the carboxylic acid loses a molecule o

Vedclass · Accepted Answer

Propanoic acid

Vedclass · Answer

(C) The reaction of a carboxylic acid with soda lime $(NaOH + CaO)$ is known as decarboxylation.In this process,the carboxylic acid loses a molecule of $CO_2$ to form an alkane with one carbon atom less than the parent acid.To obtain ethane ($C_2H_6$,which has $2$ carbon atoms),the starting carboxylic acid must have $3$ carbon atoms.Therefore,$X$ is propanoic acid $(C_2H_5COOH)$.Reaction: $C_2H_5COOH + NaOH \xrightarrow{CaO} C_2H_6 + Na_2CO_3$.

$X$ is heated with soda lime and gives ethane. $X$ is

Similar Questions

Derive an equation for $pK_{a}$ for carboxylic acid compounds. Also,give the relation between $pK_{a}$ and acidic strength.

Which of the following compounds would you expect to be the strongest carbon acid?

The correct order of increasing acid strength of the compounds is:
$(A) CH_3CO_2H$
$(B) MeOCH_2CO_2H$
$(C) CF_3CO_2H$
$(D) (CH_3)_2CHCO_2H$

Identify the reagents $A$ and $B$ respectively in the following reactions: $CH_3COOH$ $\xrightarrow{A} CH_3COCl$ $\xrightarrow{B} CH_3CHO$

Which of the following is the strongest acid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module