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.
By mistake, an alcohol (boiling point $97^{\circ} \mathrm{C}$ ) was mixed with a hydrocarbon (boiling point $68^{\circ} \mathrm{C}$ ). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
The difference in boiling point of two liquids is more than $20^{\circ} \mathrm{C}$. Hence, simple distillation can be used and since at the boiling point of low boiling liquid, the vapours would consist entirely of only low boiling liquid without any contamination of vapours of high boiling liquid and vice-versa. Thus, both the liquids can be distilled without any decomposition.
Which of the two structures $(A)$ and $(B)$ given below is more stabilised by resonance.
Explain $$\mathop {C{H_3}COOH}\limits_{(A)} $$ and $$\mathop {C{H_3}CO{O^\Theta }}\limits_{(B)} $$
Resonating structures of (A) and (B) are as follows
Structure (II) is less stable than structure (I) because later carries separation of positive and negative charges. Therefore, contribution of structure (II) is less than that of (I) towards the resonance hybrid of compound (A), i.e., $\mathrm{CH}_3 \mathrm{COOH}$. On contrary, structure (III) and (IV) are of equal energy and hence contribute equally towards the resonance hybrid of compound ( $B$ ). Therefore, structure $(B)$ is more stable than structure $(A)$ i.e., $\mathrm{CH}_3 \mathrm{COO}^{\ominus}$.