Which of these reagents would accomplish the following reduction?

N a B H_{4}

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L i A l H_{4}

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1

mole

H_{2}

, poisoned catalyst, low pressure

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H_{3} O^{+}

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Solution

The correct option is A $N a B H_{4}$
$L i A l H_{4}$ affects cyanide and double bond while $H_{3} O^{+}$ converts cyanide to carboxylic acid $H_{2}$ converts cyanide to amine but $N a B H_{4}$ only reduces ketone to alcohol.
$O$ $O H$
So, $N \equiv C - C H_{2} - | | C - C H_{2} - C H = C H_{2} N a B H_{4} - --- \to N \equiv C - C H_{2} - | C H - C H_{2} - C H = C H_{2}$

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Similar questions

Q. Which of the reagents shown below would accomplish the following transformations?

(A)	(B)
1) $H_{3} O^{+}$	$B H_{3} T H F; H_{2} O_{2} / N a O H$
2) NaOH	$B H_{3} T H F; H_{2} O_{2} / N a O H$
3) HBr in ether	$H g (O A c)_{2} / H_{2} O; N a B H_{4}$
4) $N a N H_{2}$	$H g (O A c)_{2} / H_{2} O_{2}; N a B H_{4}$

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Q. When D-fructose is reduced with

N a B H_{4} / H_{3} O^{+}

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N a B H_{4} / H_{3} O^{+}

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Which of these reagents would accomplish the following reduction?

The correct option is A NaBH4LiAlH4 affects cyanide and double bond while H3O+ converts cyanide to carboxylic acid H2 converts cyanide to amine but NaBH4 only reduces ketone to alcohol. O OHSo, N≡C−CH2−||C−CH2−CH=CH2NaBH4−−−−→N≡C−CH2−|CH−CH2−CH=CH2