NEET Practice Questions (MCQs) with Answers & Solutions

The given reaction involves the conversion of butane (CH3-CH2-CH2-CH3) to 2-methylpropane (CH3-CH-CH3) using AlCl3 as a catalyst. This rearrangement of the carbon skeleton is known as isomerization, which is an important process in the petroleum industry.

CH3-Br (methyl bromide) can be used to prepare methane (CH4) and ethane (CH3-CH3) in a single step reaction by treating it with a reducing agent like LiAlH4 or Zn/HCl. The alkyl halide (CH3-Br) is reduced to the corresponding alkane by replacing the halogen with hydrogen.

On catalytic reduction with H2/Pt, there are three alkene isomers that can give n-butane as the product: 1-butene, cis-2-butene, and trans-2-butene. These three alkenes will undergo hydrogenation in the presence of the Pt catalyst and hydrogen gas to form n-butane.

(d) Di-imide is used to reduce pi-bond formed between like atom.

 For example $\to$-CH=CH-, -N=N-

It cannot be reduced following gps.

$$\begin{gathered} \mathrm{O} \mathrm{N} \\ \to -\stackrel{\left|\right|}{\mathrm{C}}-, -\stackrel{\left|\right|}{\mathrm{C}}-,-\mathrm{N}=\mathrm{O},-\mathrm{C}\equiv \mathrm{N}-\mathrm{NC} \end{gathered}$$

(c) Anti-Markovnikoff's addition take place.

(a) Reductive ozonolysis

In the given reaction, propene (CH3-CH=CH2) undergoes a deuterium (D) exchange reaction in the presence of deuterated tetrahydrofuran (THF:BD3) and a base. The deuterium atoms replace the hydrogen atoms at the α and β positions, leading to the formation of CH3-CHD-CH2D.