AP EAMCET · Chemistry · p Block Elements (Group 13 & 14)
The Product(s) of the reaction
\(\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow\) 'Products' is/are
- A \(\mathrm{B}_2 \mathrm{H}_4 \mathrm{I}_2+2 \mathrm{Nal}\)
- B \(\mathrm{B}_2 \mathrm{H}_6+\mathrm{NaH}+\mathrm{HI}\)
- C \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2\)
- D \(2 \mathrm{NaBH}_4 \mathrm{I}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{NaBH}_4\) is the mild reducing agent. When sodium borohydride react with iodine, it will produced diboron, sodium iodide and hydrogen gas. It involves oxidation of sodium borohydride with iodine in diglyme gives diborane. This approach is common in industrial…
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