The reaction of $PCl_3$ with $PhLi$ in 1:3 molar ratio yields X as one of the products, which on further treatment with $CH_3I$ gives Y. The reaction of Y with $n-BuLi$ gives product Z. The products X, Y and Z, respectively, are
To determine the products X, Y, and Z in this chemical reaction question, we need to analyze each step of the given reaction sequence:
- The reaction between \(PCl_3\) and \(PhLi\) (phenyl lithium) in a 1:3 molar ratio produces triphenylphosphine \(PPh_3\) (X). This occurs because each phenyl group from PhLi can replace each Cl in \(PCl_3\), resulting in the formation of \(PPh_3\).
- When the product X is treated with methyl iodide \(CH_3I\), it undergoes a reaction where the phosphorus atom in \(PPh_3\) forms a bond with a methyl group, creating a phosphonium salt \([Ph_3P(CH_3)]I\) (Y).
- Finally, upon reacting the product Y with \(n-BuLi\) (n-butyl lithium), a strong base, the phosphonium salt can undergo a deprotonation forming a phosphine ylide. Yields \(Ph_3P=CH_2\) (Z), following Wittig-type reaction mechanisms where a methylene group is generated at the phosphorus center.
Hence, the correct sequence of products X, Y, and Z are:
- X: \(PPh_3\)
- Y: \([Ph_3P(CH_3)]I\)
- Z: \(Ph_3P=CH_2\)
The correct answer is Option 3: \(PPh_3\), \([Ph_3P(CH_3)]I\) and \(Ph_3P=CH_2\).
Each step in this reaction illustrates the formation and subsequent chemical transformation of phosphorus-containing compounds, culminating in a phosphonium salt generating an ylide.
Important Questions from Main Group Elements and Their Compounds
- __________ nitrides are termed by transition + inner transition metals.
- A non-metal ‘X’ occurs in nature in free state. It combines with hydrogen to form a gas, which is colourless and highly toxic. Apart from this, ‘X’ is oxidized to ‘$\text{XO}_2$’ which is a colourless, pungent and irritating gas. ‘X’ represents :
For the following reaction sequence,
X and Y, respectively, are- Conversion of boron trifluoride to tetrafluoroborate accompanies
The correct statements regarding B among the following
(i) Nuclear spin of 11B is greater than that of 10B
(ii) The polarities of B - H bond and C-H bonds are opposite
(iii) Cross - section of neutron absorption for 10B is much more than that of 11B
(iv) B reacts with boiling aq. NaOH solution to form NaB(OH)4
