A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is______.

We have phenol, $$\mathrm{C_6H_5OH}$$, dissolved in chloroform, $$\mathrm{CHCl_3}$$, and the mixture is treated with aqueous $$\mathrm{NaOH}$$. This set of reagents is characteristic of the Reimer-Tiemann reaction. In this reaction, phenol first forms the phenoxide ion under the basic conditions, and the chloroform generates the dichlorocarbene species $$\mathrm{:CCl_2}$$, which subsequently undergoes electrophilic substitution at the ortho position of the phenoxide ring. On hydrolysis of the intermediate, we obtain o-hydroxybenzaldehyde as the major product.

So, compound $$P$$ is o-hydroxybenzaldehyde (also called salicylaldehyde). Its molecular formula is

$$\mathrm{C_6H_4(OH)(CHO)} \;=\; \mathrm{C_7H_6O_2}.$$

Now we calculate the mass percentage of carbon in this compound. First, we write down the molar masses of all the constituent atoms:

$$\text{Atomic mass of C} = 12 \text{ u}, \qquad \text{Atomic mass of H} = 1 \text{ u}, \qquad \text{Atomic mass of O} = 16 \text{ u}.$$

The total molar mass of $$\mathrm{C_7H_6O_2}$$ is obtained by adding the contributions from every atom:

$$ \begin{aligned} M &= \bigl(7 \times 12\bigr) + \bigl(6 \times 1\bigr) + \bigl(2 \times 16\bigr) \\ &= 84 + 6 + 32 \\ &= 122 \text{ u}. \end{aligned} $$

The mass contributed by carbon alone is

$$m_C = 7 \times 12 = 84 \text{ u}.$$

Using the formula for percentage composition,

$$ \text{Percentage of C} = \frac{m_C}{M} \times 100 = \frac{84}{122} \times 100 = 68.85\% \approx 69\%. $$

So, the answer is $$69\%$$.