Question

Which of the following reactions is used to prepare isobutane?
(A) Wurtz reaction of C₂H₅Br.
(B) Hydrolysis of n-butylmagnesium iodide.
(C) Reduction of propanol with red phosphorus and HI.
(D) Decarboxylation of 3-methylbutanoic acid.

Answer 1

Hint: ‘But’ means four carbon structures. ‘ane’ means all the carbon is connected through a single bond. Isobutane or i-butane is an isomer of butane having chemical formula \[CH{\left( {C{H_3}} \right)_3}\]. It is also called \[2\]-methyl butane.

Complete Step by Step Answer:
Option A: wurtz reaction is an organic reaction used for synthesis of alkanes in presence of sodium metal. It can be represented by general reaction as \[R - X + 2Na + R - X \to R - R + 2NaX\]
Thus, on undergoing a wurtz reaction of ethyl bromide, n-butane is obtained as a product along with sodium bromide.
\[{C_2}{H_5}Br{\text{ }} + {\text{ }}2Na{\text{ }} + {\text{ }}{C_2}{H_5}Br{\text{ }} \to {\text{ }}{C_4}{H_{10}}{\text{ }} + {\text{ }}2NaBr\]

Option B: Hydrolysis refers to reaction with water. Thus, when n-butyl magnesium iodide is treated with water, it results in formation of n butane.
\[C{H_3}C{H_2}C{H_2}C{H_2}MgI + {H_2}0 \to C{H_3}C{H_2}C{H_2}C{H_3} + Mg(OH)I\]

Option C: Reduction means addition of hydrogen or removal of oxygen. Thus, reduction of alcohol with help of reducing agents leads to formation of alkane. Thus, on reaction of propanol with \[HI\] and red phosphorus, normal butane or n-butane is formed and not isobutane.
\[C{H_3}C{H_2}C{H_2}C{H_2}OH\xrightarrow{{HI}}C{H_3}C{H_2}C{H_2}C{H_3}\]
Thus, in all the above reactions isobutene is not formed. Hence option A, B, C are incorrect.

Option D: decarboxylation means removal of carbon dioxide as a byproduct. It is usually from a carboxylic functional group.
\[C{H_3} - CH(C{H_3}) - C{H_2}COOH\xrightarrow{{NaOH,CaO}}C{H_3} - CH(C{H_3}) - C{H_3} + C{O_2}\]
Thus, isobutane is formed as a product during decarboxylation of \[3\]-methylbutanoic acid.
Hence, the correct answer is option (D).

Note: If we had used butanoic acid instead of using \[3\]-methyl butanoic acid then the product would be normal butane or simply butane instead of iso-butane. And if we had used \[2\]-methyl propyl bromide in option A then the product obtained would be iso butane. Thus, on changing the reactant we can get the desired product.