In the reaction, Ph₃ P MeI[X] n-BuLi[Y], the compounds X and Y, respectively are · ABC26IN0506

Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Cluster Compounds Wade Rules · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

In the reaction, $\mathrm{Ph}_{3} \mathrm{P} \xrightarrow{\mathrm{MeI}}[\mathrm{X}] \xrightarrow{n-\mathrm{BuLi}}[\mathrm{Y}]$, the compounds X and Y, respectively are
(a)$\left[\mathrm{Ph}_{3} \mathrm{P}(\mathrm{Me}) \mathrm{I}\right] ; \mathrm{Ph}_{3} \mathrm{P}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$
(b)$\left[\mathrm{Ph}_{3} \mathrm{P}(\mathrm{Me})\right][\mathrm{I}] ; \mathrm{Ph}_{3} \mathrm{P}=\mathrm{CH}_{2}$
(c)$\left[\mathrm{Ph}_{3} \mathrm{P}(\mathrm{Me})_{2}\right] ; \mathrm{Ph}_{3} \mathrm{P}=\mathrm{CH}_{2}$
(d)$\left[\mathrm{Ph}_{3} \mathrm{P}(\mathrm{Me})\right][\mathrm{I}] ; \mathrm{Ph}_{3} \mathrm{P}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Ph₃P + MeI → [Ph₃PMe]⁺I⁻ (X, phosphonium salt); n-BuLi deprotonates it to the Wittig ylide Ph₃P=CH₂ (Y).

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