NEET – 2023

NaBH₄ is a mild reducing agent and reduces. Aldehyde into the primary alcohol and ketone into the secondary alcohol. In the given compound ketone group is present, thus, NaBH₄ reduces the ketone group into an alcohol group. 

In the second step, the molecule undergoes a dehydration reaction, and an alkene is obtained as a product. 

The reactions are as follows:

Thus, option first is the correct answer. 
 

When primary alcohol reacts with a strong oxidizing agent then an acid is formed as a product. When primary alcohol reacts with the mild oxidizing agent then aldehyde is formed as a product. 

Here, CrO₃ – H₂SO₄, K₂Cr₂O₇ + H₂SO₄, and KMnO₄ + KOH/H₃O⁺ are strong oxidizing agents thus, converts alcohol into acid. 

But Cu, 573 K, and CrO₃₊ (CH₃CO)₂O are the weak oxidizing agent and reduce primary alcohol to aldehyde. 

Thus, option third is the correct answer. 
 

Explanation:
The first reaction is a reduction reaction. When acid reacts with a reducing agent then alcohol is obtained as a product. 
Here, B₂H₆ and LiAlH₄ can be used as reducing agents. 

The second reaction is also known as the hydroboration-oxidation reaction. In this reaction, alkene reacts with diborane gives trialkyl boranes as addition product. This is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide. 

Thus, in both reactions X is B₂H₆. 

Thus, an option first is the correct answer. 
 

When alcohol reacts with HBr then substitution reaction takes place and the OH group is substituted by the Br atom. In the first step, the OH group is converted into an H₂O group, and a secondary carbocation is formed as a product. Then rearrangement of carbocation takes place and more stable tertiary carbocation is obtained. 
In the first step, tertiary carbocation reacts with bromide ion, and an alkyl halide is obtained as a product. 

The reaction with mechanism is as follows:

Thus, option second is the correct answer.