The stability order of the following singlet halocarbene is:
(A)$C{{F}_{2}}$>$CC{{l}_{2}}$>$CB{{r}_{2}}$>$C{{I}_{2}}$
(B) $C{{I}_{2}}$>$CB{{r}_{2}}$>$CC{{l}_{2}}$>$C{{F}_{2}}$
(C) $CC{{l}_{2}}$>$C{{F}_{2}}$>$CB{{r}_{2}}$>$C{{I}_{2}}$
(D) $C{{F}_{2}}$>$C{{I}_{2}}$>$CC{{l}_{2}}$> $CB{{r}_{2}}$
Answer
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Hint One type of carbon cations without a formal charge is carbine and the other valence electron in nonbonding pair orthogonal to the $\pi $system. Carbenes can act as both electrophile and nucleophile. Both triplet methylene and singlet methylene are $s{{p}^{2}}$ hybridizations with bent shape and bond angles are ${{125}^{0}}-{{140}^{0}}$ and ${{102}^{0}}$ respectively
Complete step by step solution:
The singlet carbene is the lowest energy than triplet carbene, which means singlet carbene is the excited state spice and triplet carbene is a more stable state. This stability of carbenes decreases in the inductive withdrawal while stabilizes on$\pi $- orbital donation capability.
The single carbene will stabilize with substituents that can donate electron pairs like halogens by delocalizing the pair into an empty p-orbital. Due to the unavailability of vacant orbital at carbon, no corresponding stabilization possible for triplet carbene. So, the singlet carbene is more stable than the triplet carbene when there is a substituent with lone pairs in the singlet carbene.
Therefore, the order of stability of singlet carbene as follows,
\[C{{l}_{2}}\]<$CB{{r}_{2}}$<$CC{{l}_{2}}$<$C{{F}_{2}}$<$C{{H}_{2}}$
The order of stability of triplet carbene as follows,
\[C{{l}_{2}}\]>$CB{{r}_{2}}$>$CC{{l}_{2}}$>$C{{F}_{2}}$>$C{{H}_{2}}$
Hence, the stability order of the following singlet halocarbene is,
\[C{{F}_{2}}\]>$CC{{l}_{2}}$>$CB{{r}_{2}}$>$C{{I}_{2}}$
Thus, the correct answer is option A.
Note: Singlet carbene cannot do things like triplet carbenes do, because triplet carbene is highly reactive diradical. But singlet carbenes are not radicals. There is one electron in the usually full orbital and one in the empty orbital called triplets which determine triplets reactivity with olefins in cyclopropanation reactions
Complete step by step solution:
The singlet carbene is the lowest energy than triplet carbene, which means singlet carbene is the excited state spice and triplet carbene is a more stable state. This stability of carbenes decreases in the inductive withdrawal while stabilizes on$\pi $- orbital donation capability.
The single carbene will stabilize with substituents that can donate electron pairs like halogens by delocalizing the pair into an empty p-orbital. Due to the unavailability of vacant orbital at carbon, no corresponding stabilization possible for triplet carbene. So, the singlet carbene is more stable than the triplet carbene when there is a substituent with lone pairs in the singlet carbene.
Therefore, the order of stability of singlet carbene as follows,
\[C{{l}_{2}}\]<$CB{{r}_{2}}$<$CC{{l}_{2}}$<$C{{F}_{2}}$<$C{{H}_{2}}$
The order of stability of triplet carbene as follows,
\[C{{l}_{2}}\]>$CB{{r}_{2}}$>$CC{{l}_{2}}$>$C{{F}_{2}}$>$C{{H}_{2}}$
Hence, the stability order of the following singlet halocarbene is,
\[C{{F}_{2}}\]>$CC{{l}_{2}}$>$CB{{r}_{2}}$>$C{{I}_{2}}$
Thus, the correct answer is option A.
Note: Singlet carbene cannot do things like triplet carbenes do, because triplet carbene is highly reactive diradical. But singlet carbenes are not radicals. There is one electron in the usually full orbital and one in the empty orbital called triplets which determine triplets reactivity with olefins in cyclopropanation reactions
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