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.
Give three points of differences between inductive effect and resonance effect.
Difference between inductive effect and resonance effect is as follows
| Inductive effect | Resonance effect |
|---|---|
| Inductive effect involves $\sigma$ - electrons displacement and occurs only in saturated compounds. | It involves $\pi$ - electrons or lone pair of electrons and occurs only in unsaturated and conjugated system. |
| Inductive effect can move upto 3 - carbon atoms. | It is applicable all along the length of conjugated system. |
| In inductive effect, there is a slight displacement of electrons and thus only partial positive and negative charges appear. | In resonance effect, there is complete transfer of electrons and thus full positive and negative charges appear. |
Which of the following compounds will not exist as resonance hybrid. Give reason for your answer.
(a) $\mathrm{CH}_3 \mathrm{OH}$
(b) $\mathrm{R}-\mathrm{CONH}_2$
(c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2$
(a) $\mathrm{CH}_3 \mathrm{OH}$ As it lacks $\pi$-electrons hence it will not exist as resonance hybrid.
(b) $R-\mathrm{CONH}_2$ Due to the presence of $n$-electrons on N and $\pi$-electrons on $\mathrm{C}=\mathrm{O}$ bond, hence amide can be represented as a resonance hybrid of the following three resonating structures.
(c) $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2$ As the lone pair of electrons on the N -atom is not conjugated with the $\pi$-electrons of the double bond, thus, resonance is not possible and hence no resonance hybrid will exist.
Why does $\mathrm{SO}_3$ act as an electrophile?
Three highly electronegative oxygen atoms are attached to sulphur atom in $\mathrm{SO}_3$ which makes sulphur atom electron-deficient. Further, due to resonance, sulphur acquires positive charge. Both these factors, make $\mathrm{SO}_3$ an electrophile.
Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.
The structure having more covalent bonds in a resonating structure, has more stability. Further, there is charge separation in structure (II) and the terminal carbon has only a sextet of electrons in (II). These two factors makes structure (II) less stable.
Hence, I > II in terms of stability.