RMgX $R C N - -- \to$ $(A) (i) R M g X - ----- \to (i i) H_{3} O^{+}$ (B)
B will be :

1^{\circ}

ROH

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2^{\circ}

ROH

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3^{\circ}

ROH

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alkene

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Solution

The correct option is D $3^{\circ}$ ROH
In this reaction, first Grignard reagent attacks on triple bond of cyanide; then again it will attack on double bond and forms $3^{0}$ alcohol. $R_{3} C - O H$
RMgX $R C N - -- \to$ $R_{2} C = N M g X (i) R M g X - ----- \to (i i) H_{3} O^{+}$ $R_{3} C - O H$

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