For testing halogens in an organic compound with $\mathrm{AgNO}_3$ solution, sodium extract (Lassaigne's test) is acidified with dilute $\mathrm{HNO}_3$. What will happen if a student acidifies the extract with dilute $\mathrm{H}_2 \mathrm{SO}_4$ in place of dilute $\mathrm{HNO}_3$ ?
On adding dilute $\mathrm{H}_2 \mathrm{SO}_4$ for testing halogens in an organic compound with $\mathrm{AgNO}_3$, white precipitate of $\mathrm{Ag}_2 \mathrm{SO}_4$ is formed. This will interfere with the test of chlorine and this $\mathrm{Ag}_2 \mathrm{SO}_4$ may be mistaken for white precipitate of chlorine as AgCl . Hence, dilute $\mathrm{HNO}_3$ is used instead of dilute $\mathrm{H}_2 \mathrm{SO}_4$.
What is the hybridisation of each carbon in $\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2$ ?
The given structure is of allene $\left(\mathrm{C}_3 \mathrm{H}_4\right)$
$$\mathrm{H}_2 \stackrel{1}{\mathrm{C}}=\stackrel{2}{\mathrm{C}}=\stackrel{3}{\mathrm{C}} \mathrm{H}_2$$
In allene, carbon atoms 1 and 3 are $s p^2$-hybridised as each one of them is joined by a double bond. And, carbon atom 2 is $s p$-hybridised as it has two double bonds at each of its side. Therefore, the two $\pi$-bonds are perpendicular to each other, in allene, as shown below.
$\mathrm{H}_a$ and $\mathrm{H}_b$ lie in the plane of paper while $\mathrm{H}_c$ and $\mathrm{H}_d$ lie in a plane perpendicular to the plane of the paper. Hence, the allene molecule as a whole is non-planar.
Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Electronegativity of carbon atom, also depends on the hybridisation of the carbon atom. Since, $s$-electrons are more strongly attracted by the nucleus than $p$-electrons, thus, electronegativity increases with increase in s-character of the hybridised orbital i.e.,
Thus, $s p$-hybridised carbon is the most electronegative carbon.
Show the polarisation of carbon-magnesium bond in the following structure.
$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Mg}-\mathrm{X}$$
Carbon (2.5) is more electronegative than magnesium (1.2) therefore, Mg acquires a partial positive charge while carbon attached to it acquires a partial negative charge.
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\stackrel{-\delta}{\mathrm{C}} \mathrm{H}_2 \leftarrow \stackrel{+\delta}{\mathrm{Mg}}-\mathrm{X}$
Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by
The two isomers which differ in the position of the functional group on the carbon skeleton are called position isomers and this phenomenon as position isomerism. Thus, (A) and (B) may be regarded as position isomers and further they cannot be regarded as metamers since metamers are those isomers which have different number of carbon atoms on either side of the functional group. But here, the number of carbon atoms on either side of sulphur atom (functional group) is the same, i.e., 1 and 3.
