Question

Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false.

X (molecular formula, $C_7H_6O_2$) is an aromatic white solid which liberates colourless, odourless gas on reacting with $NaH^{14}CO_3$.

$S_1$: Only three of the five functional isomers of X (including 'X' itself) will give positive 2, 4-DNP test.

$S_2$: The liberated colourless, odourless gas will contaning radioactive $^{14}C$.

$S_3$: Except 'X', no other functional isomer will liberate colourless odourless gas with $NaH^{14}CO_3$.

$S_4$: The DU of higher homolog of 'X' will be four.

• A. TTTF
• B. FTTF
• C. FTTT
• D. TTFF

Answer 1

Answer: (A)

TTTF

Answer 2

The given molecule X has the molecular formula $C_7H_6O_2$. It is an aromatic white solid that reacts with $NaH^{14}CO_3$ to liberate a colourless, odourless gas. The reaction with bicarbonate is characteristic of carboxylic acids, which are strong enough acids to react with weak bases like bicarbonate, producing $\text{CO}_2$. The DU of $C_7H_6O_2$ is $(2 \times 7 + 2 - 6)/2 = (14 + 2 - 6)/2 = 10/2 = 5$. An aromatic structure with DU=5 and formula $C_7H_6O_2$ is likely benzoic acid ($C_6H_5COOH$). Benzoic acid is an aromatic white solid and its reaction with bicarbonate is $C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2$. The gas is $\text{CO}_2$, which is colourless and odourless. Thus, X is benzoic acid.

Now let's evaluate the statements:

$S_1$: Only three of the five functional isomers of X (including 'X' itself) will give positive 2, 4-DNP test. The plausible aromatic functional isomers of $C_7H_6O_2$ with DU=5 are:

1. Benzoic acid ($C_6H_5COOH$): Carboxylic acid. Does not give a positive 2,4-DNP test. (This is X)
2. Phenyl formate ($HCOOC_6H_5$): Ester. Does not give a positive 2,4-DNP test.
3. o-Hydroxybenzaldehyde ($o-HOC_6H_4CHO$): Aldehyde + Phenol. Gives a positive 2,4-DNP test due to the aldehyde group.
4. m-Hydroxybenzaldehyde ($m-HOC_6H_4CHO$): Aldehyde + Phenol. Gives a positive 2,4-DNP test due to the aldehyde group.
5. p-Hydroxybenzaldehyde ($p-HOC_6H_4CHO$): Aldehyde + Phenol. Gives a positive 2,4-DNP test due to the aldehyde group. These are 5 functional isomers. Out of these, the three hydroxybenzaldehyde isomers give a positive 2,4-DNP test. So, statement $S_1$ is true.

$S_2$: The liberated colourless, odourless gas will contaning radioactive $^{14}C$. The reaction is $C_6H_5COOH + NaH^{14}CO_3 \rightarrow C_6H_5COONa + H_2O + ^{14}CO_2$. The carbon dioxide gas is formed from the bicarbonate ion, which is labeled with $^{14}C$. Therefore, the liberated $\text{CO}_2$ will contain $^{14}C$. Statement $S_2$ is true.

$S_3$: Except 'X', no other functional isomer will liberate colourless odourless gas with $NaH^{14}CO_3$. Reaction with bicarbonate requires a sufficiently acidic compound, typically a carboxylic acid (pKa ~4-5) or a very strong phenol (like picric acid, pKa < 1). Phenols generally have pKa values around 10 and do not react with bicarbonate. Among the 5 isomers identified:

1. Benzoic acid (X): Carboxylic acid, reacts with $NaH^{14}CO_3$.
2. Phenyl formate: Ester, not acidic.
3. o-, m-, p- Hydroxybenzaldehyde: Contain a phenolic -OH group. Phenols are not acidic enough to react with bicarbonate. Therefore, only X (benzoic acid) among these functional isomers will react with $NaH^{14}CO_3$ to liberate $\text{CO}_2$. Statement $S_3$ is true.

$S_4$: The DU of higher homolog of 'X' will be four. X is benzoic acid ($C_6H_5COOH$). A higher homolog of an aromatic carboxylic acid would typically be a methylbenzoic acid ($CH_3C_6H_4COOH$, e.g., toluic acid) or phenylacetic acid ($C_6H_5CH_2COOH$). For $CH_3C_6H_4COOH$ ($C_8H_8O_2$): DU = $(2 \times 8 + 2 - 8)/2 = (16+2-8)/2 = 10/2 = 5$. For $C_6H_5CH_2COOH$ ($C_8H_8O_2$): DU = $(2 \times 8 + 2 - 8)/2 = (16+2-8)/2 = 10/2 = 5$. The DU of these higher homologs is 5 (4 for the benzene ring + 1 for the carboxyl C=O). Statement $S_4$ is false.

The correct order of initials T or F for statements $S_1, S_2, S_3, S_4$ is TTTF.

Comparing this with the given options: (A) TTTF (B) FTTF (C) FTTT (D) TTFF

The correct option is (A).