Arrange the following intermediates in decreasing order of stability:
i. $ C{F_3} - \mathop {C{H_2}}\limits^ \oplus $
ii. $ CC{l_3} - \mathop {C{H_2}}\limits^ \oplus $
iii. $ CB{r_3} - \mathop {C{H_2}}\limits^ \oplus $
(A) $ i > iii > ii $
(B) $ ii > i > iii $
(C) $ ii > iii > i $
(D) $ iii > ii > i $
Answer
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Hint: A reaction intermediate or an intermediate is any chemical substance produced during the conversion of some reactant to a product. All the substances generated by one step and used for the succeeding step (reacts further) are considered intermediates. An intermediate, that has a carbon atom bearing a positive charge and three bonds instead of four.
Complete answer:
In the given intermediates $ C{F_3} - \mathop {C{H_2}}\limits^ \oplus $ , $ CC{l_3} - \mathop {C{H_2}}\limits^ \oplus $ $ CB{r_3} - \mathop {C{H_2}}\limits^ \oplus $ , $ \mathop {C{H_2}}\limits^ \oplus $ is common for all intermediates. So, when comparing all intermediates $ \mathop {C{H_2}}\limits^ \oplus $ is neglected.
In $ CC{l_3} $ (-), there are lone pairs on carbon and vacant d-orbital on chlorine which results in back bonding and the positive charge gets delocalised to the vacant d-orbital of chlorine (known as d-orbital resonance), whereas no such phenomenon can be exhibited by fluorine in case of $ C{F_3} $ (-) due to absence of any vacant orbital resonance. Hence, $ CC{l_3} $ (-) is more stable than $ C{F_3} $ (-). However, fluorine is more electronegative than chlorine but it lacks d-orbital which is present in Chlorine.
$ CB{r_3} $ (-) is more stable than $ CC{l_3} $ (-) due to the increased strength of the d-orbital contribution that can occur.
Thus $ CB{r_3} - \mathop {C{H_2}}\limits^ \oplus > CC{l_3} - \mathop {C{H_2}}\limits^ \oplus > C{F_3} - \mathop {C{H_2}}\limits^ \oplus $
Hence the correct option is D. $ iii > ii > i $
Note:
Note that the carbocation is an organic molecule that has a carbon atom bearing four bonds. The primary carbanion and methyl carbanion are the most stable carbanions. In general, the three types of stability are stable equilibrium, unstable equilibrium and neutral equilibrium.
Complete answer:
In the given intermediates $ C{F_3} - \mathop {C{H_2}}\limits^ \oplus $ , $ CC{l_3} - \mathop {C{H_2}}\limits^ \oplus $ $ CB{r_3} - \mathop {C{H_2}}\limits^ \oplus $ , $ \mathop {C{H_2}}\limits^ \oplus $ is common for all intermediates. So, when comparing all intermediates $ \mathop {C{H_2}}\limits^ \oplus $ is neglected.
In $ CC{l_3} $ (-), there are lone pairs on carbon and vacant d-orbital on chlorine which results in back bonding and the positive charge gets delocalised to the vacant d-orbital of chlorine (known as d-orbital resonance), whereas no such phenomenon can be exhibited by fluorine in case of $ C{F_3} $ (-) due to absence of any vacant orbital resonance. Hence, $ CC{l_3} $ (-) is more stable than $ C{F_3} $ (-). However, fluorine is more electronegative than chlorine but it lacks d-orbital which is present in Chlorine.
$ CB{r_3} $ (-) is more stable than $ CC{l_3} $ (-) due to the increased strength of the d-orbital contribution that can occur.
Thus $ CB{r_3} - \mathop {C{H_2}}\limits^ \oplus > CC{l_3} - \mathop {C{H_2}}\limits^ \oplus > C{F_3} - \mathop {C{H_2}}\limits^ \oplus $
Hence the correct option is D. $ iii > ii > i $
Note:
Note that the carbocation is an organic molecule that has a carbon atom bearing four bonds. The primary carbanion and methyl carbanion are the most stable carbanions. In general, the three types of stability are stable equilibrium, unstable equilibrium and neutral equilibrium.
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