What is the correct IUPAC name of
To determine the correct IUPAC name, we need to follow the IUPAC nomenclature rules for substituted benzoic acids.
1. Identify the parent compound: The parent compound is benzoic acid, as it contains a benzene ring with a -COOH group. The -COOH group is assigned position 1.
2. Number the benzene ring: The numbering of the ring is done such that the substituent with the next priority gets the lowest possible number. The priority order for the substituents present is: -COOH (highest priority, assigned position 1)>-OH>-Br>-NO\( _2 \). Therefore, the -OH group should get the lowest possible number. Numbering clockwise gives -OH at position 2, -Br at position 3, and -NO\( _2 \) at position 5. Numbering counterclockwise would give -OH at position 6, which is higher. Thus, the clockwise numbering is correct.
3. Identify the substituents and their positions: - -Br (bromo) is at position 3. - -OH (hydroxy) is at position 2. - -NO\( _2 \) (nitro) is at position 5.
4. Arrange the substituents alphabetically: The alphabetical order of the prefixes is bromo, hydroxy, nitro.
5. Write the IUPAC name: Combining the position numbers and the alphabetically ordered prefixes with the parent name benzoic acid, we get: 3-bromo-2-hydroxy-5-nitrobenzoic acid. This matches option (1).
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: