Identify the product $C$ in the following reaction:
$CH_{3}CH_{2}COOH$ $\xrightarrow{NaOH} A$ $\xrightarrow{NaOH, CaO, \Delta} B$ $\xrightarrow{HNO_{3}, \Delta} C$
$CH_{3}CH_{2}COOH$ $\xrightarrow{NaOH} A$ $\xrightarrow{NaOH, CaO, \Delta} B$ $\xrightarrow{HNO_{3}, \Delta} C$
- A$CH_{3}CH_{2}CH_{3}$
- B$CH_{2}=CH_{2}$
- C$CH_{3}CH_{2}NO_{2}$
- D$CH_{3}CH_{2}CH_{2}OH$
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Among the following reactions,the reaction that "does not" give an alkane product is:
DifficultTS EAMCET 2021
View SolutionAssertion $(A)$: Cyclohexane is the most stable cycloalkane.
Reason $(R)$: Cyclopropane and cyclobutane are less stable due to angle strain and torsional strain.
The correct answer is
Reason $(R)$: Cyclopropane and cyclobutane are less stable due to angle strain and torsional strain.
The correct answer is
Match the column and find the correct answer:
Column-$I$ Column-$II$ $i$. $n$-Butane $\to 2$-methylpropane $A$. Free radical substitution $ii$. $CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$ $B$. Wurtz reaction $iii$. $R-COONa \xrightarrow{\text{soda-lime}} R-H$ $C$. Isomerism $iv$. $R-X + Na \xrightarrow{\text{Ether}} R-R$ $D$. Decarboxylation
| Column-$I$ | Column-$II$ |
| $i$. $n$-Butane $\to 2$-methylpropane | $A$. Free radical substitution |
| $ii$. $CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$ | $B$. Wurtz reaction |
| $iii$. $R-COONa \xrightarrow{\text{soda-lime}} R-H$ | $C$. Isomerism |
| $iv$. $R-X + Na \xrightarrow{\text{Ether}} R-R$ | $D$. Decarboxylation |
Ethyl iodide and $n$-propyl iodide are allowed to undergo Wurtz reaction. The alkane which will not be obtained in this reaction is
Medium
View Solution$A$ mixture of two organic chlorine compounds was treated with sodium metal in ether solution. Isobutane was obtained as a product. The two chlorine compounds are
Difficult
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