How many moles of $B H_{3} / T H F$ react with $1$ mole of $1$ -pentene?

0.25 m o l B H_{3} / T H F

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0.33 m o l B H_{3} / T H F

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0.75 m o l B H_{3} I T H F

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1 m o l B H_{3} / T H F

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Solution

The correct option is D $0.33 m o l B H_{3} / T H F$
The reaction of alkene with $B H_{3} / T H F$ is

$B H_{3} + 3 R - C H = C H_{2} ⟶ (R - C H_{2} - C H_{2} -)_{3} B ⟶ 3 R - C H_{2} - C H_{2} - O H$

One mole of $B H_{3} / T H F$ reacts with $3$ moles of alkene.

Therefore $1$ mole of alkene would react with $= \frac{1}{3} = 0.33$ moles of $B H_{3} / T H F$

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Similar questions

A

B H_{3} / T H F

B H_{3}

Which of the following compound would yield trialkyl borane as shown below when treated with $B H_{3}$ and THF ?

(A) (i) B D_{3} / T H F - -------- \to (i i) C H_{3} C O O H P e n t e n e (i) B D_{3} / T H F - -------- \to (i i) C H_{3} C O O D (B)

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