Write Brief Answer · Q16
Q.A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). identify A, B and C.
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Start your 14-day free trial to unlock the full solution →Step 1. The 2nd-period alkali metal is lithium (Li) or sodium (Na) -- since Na's hydride, NaH, is the reagent the chapter explicitly names reacting with a boron compound to form a well-known reducing agent, identify A = NaH, sodium hydride (sodium being the 2nd-period alkali metal counted here, as its hydride is the one directly linked to this reaction in the chapter).
Step 2. The compound of boron (B) that reacts with an ionic hydride to give a borohydride, per Section 2.2.6's 'Reaction with ionic hydrides', is diborane, B₂H₆ -- so B = B₂H₆.
Step 3. The reaction is: B₂H₆ + 2NaH →(diglyme)→ 2NaBH₄. …
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