"Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged atom involvement of neighbouring groups in hyperconjugation and resonance."
Which of the following ions is more stable? Use resonance to explain your answer.
Carbocation $(A)$ is more stable than carbocation (B). Carbocation. $(A)$ is more planar and hence is stabilised by resonance while carbocation (B) is non-planar and hence it does not undergo resonance. Further, double bond is more stable within the ring in comparison to outside the ring.
"Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged atom involvement of neighbouring groups in hyperconjugation and resonance."
The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.
In triphenylmethyl cation, due to resonance, the positive charge can move at both the $o-$ and $p$-position of each benzene ring. This is illustrated below
Since, there are three benzene rings, hence, there are, in all, nine resonance structures. Thus, triphenylmethyl cation is highly stable due to these nine resonance structures.
"Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged atom involvement of neighbouring groups in hyperconjugation and resonance."
Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.
2-methyl butane has four different sets of equivalent H -atoms.
Stability of carbocation decreases in the order $$3\Upsilon > 2\Upsilon > 1\Upsilon $$. So, III (3)carbocation) is most stable followed by II (2Ycarbocation). Out of I and IV (both are 1Ycarbocation) I has a $\mathrm{CH}_3$ group at $\beta$-carbon while II has a $\mathrm{CH}_3$ group at $\alpha$-carbon. As + -effect decreases with distance, hence IV is more stable than I. Therefore, the overall stability of these four carbocations increases in the order.
$$\text { I }<\text { IV }<\text { II }<\text { III }$$
Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne's extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid $\mathrm{FeSO}_4$ and dilute sulphuric acid to a part of Lassaigne's extract. Manish and Rajni obtained prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same Lassaigne's extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation. Also, write the chemical equations to explain the formation of compounds of different colours.
If the organic compound contains both N and S , then while fusion it may for form either a mixture of sodium cyanide $(\mathrm{NaCN})$ and sodium sulphide $\left(\mathrm{Na}_2 \mathrm{~S}\right)$ or sodium thiocyanate ( NaSCN ) depending on the amount of Na metal used. If Less sodium metal is used, only NaSCN is obtained.
This then gives red colour on reacting with $\mathrm{Fe}^{3+}$ ions (produced by oxidation of $\mathrm{Fe}^{2+}$ ions while preparing Lassaigne's extract) due to the formation of ferric thiocyanate.
$\mathrm{Fe}^{2+} \xrightarrow{\text { Aerial oxidation }} \mathrm{Fe}^{3+}$
$\mathrm{Fe}^{3+}+3 \mathrm{NaSCN} \longrightarrow \underset{\substack{\text { Ferric thiocyanate } \\ \text { (red) }}}{\mathrm{Fe}(\mathrm{SCN})_3}+3 \mathrm{Na}^{+}$
In case, excess of sodium metal is used, the initally formed sodium thiocyanate decomposes as follows:
$\underset{\begin{array}{c}\text { Sodium } \\ \text { thiocyanate }\end{array}}{\mathrm{NaSCN}}+2 \mathrm{Na} \xrightarrow{\Delta} \underset{\begin{array}{c}\text { Sodium } \\ \text { cyanide }\end{array}}{\mathrm{NaCN}}+\underset{\begin{array}{c}\text { Sodium } \\ \text { sulphide }\end{array}}{\mathrm{Na}_2 \mathrm{~S}}$
This NaCN then reacts with $\mathrm{FeSO}_4, \mathrm{Fe}^{3+}$ ions and NaCN , it gives prussian blue colour due to the formation of ferric ferrocyanide or iron (III) hexacyanoferrate (II).
$2 \mathrm{NaCN}+\mathrm{FeSO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{Fe}(\mathrm{CN})_2$
$\mathrm{Fe}(\mathrm{CN})_2+4 \mathrm{NaCN} \longrightarrow \underset{\substack{\text { Sodium hexacyano - } \\ \text { ferrate (II). }}}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]}$
$3 \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+4 \mathrm{Fe}^{3+} \longrightarrow \underset{\substack{\text { Iron (III) hexacyanoferrate } \\ \text { (II) }(\text { prussian blue) }}}{\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_2\right]_3}+12 \mathrm{Na}^{+}$
On the basis of above results, it is clear that Ramesh used less sodium and hence NaSCN formed in the Lassaigne's extract which gave red colouration due to $\mathrm{Fe}(\mathrm{SCN})_3$ formation while Manish and Rajni used excess sodium and hence NaCN formed in the Lassaigne's extract which gave prussian blue colour of $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$.
Name the compounds whose line formulae are given below.
