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.
Write structural formulae for compounds named as
(a) 1-bromoheptane
(b) 5-bromoheptanoic acid
Draw the resonance structures of the following compounds.
Identify the most stable species in the following set of ions giving reasons
(a) $\stackrel{+}{\mathrm{C}} \mathrm{H}_3, \stackrel{+}{\mathrm{C}} \mathrm{H}_2 \mathrm{Br}, \stackrel{+}{\mathrm{C}} \mathrm{HBr}_2, \stackrel{+}{\mathrm{C}} \mathrm{Br}_3$
(b) $\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3, \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2 \mathrm{Cl}^{\mathrm{C}} \stackrel{\ominus}{\mathrm{C}} \mathrm{HCl}_2, \stackrel{\ominus}{\mathrm{C}} \mathrm{Cl}_3$
(a) $\stackrel{+}{\mathrm{C}} \mathrm{H}_3$ is the most stable species because the replacement of H by Br increases positive charge ( $-I$-effect) on carbon atom and destabilises the species and, more the number of Br -atoms, less stable is the species.
(b) $\stackrel{\ominus}{\mathrm{C}} \mathrm{Cl}_3$ is the most stable species because on replacing H by Cl , negative charge on carbon is dispersed due to $-I$-effect of Cl and thus get reduced and species is stabilised. Further, more the number of Cl atoms, more is the dispersal of the negative charge and hence more stable is the species.