The Friedel-Crafts reaction is a key method used in organic chemistry to attach substituents to an aromatic ring. For this reaction to occur, the aromatic compound must not contain strongly deactivating groups such as -NO2 or -NH2 which reduce the electron density on the ring required for electrophilic aromatic substitution.
Let's examine each compound one by one:
Counting the compounds that cannot undergo Friedel-Crafts reactions, we have: nitrobenzene, aniline, m-nitroaniline, and m-dinitrobenzene, totaling 4 compounds.
Thus, the number of compounds that cannot undergo Friedel-Crafts reactions, which is 4, falls within the given range (4,4).
Friedel-Crafts reactions require an aromatic compound and an electrophile, facilitated by a Lewis acid catalyst (e.g., AlCl$_3$). However,
certain compounds cannot undergo Friedel-Crafts reactions due to deactivating groups or coordination issues with the catalyst.
Toluene, xylene, and cumene: These are activated aromatic compounds and can undergo Friedel-Crafts reactions.
Chlorobenzene: Chlorine is an electron-withdrawing group but is ortho/para-directing; hence it can still undergo Friedel-Crafts reactions.
Nitrobenzene, m-nitroaniline, m-dinitrobenzene: Nitro groups are strongly deactivating, making the aromatic ring unreactive for Friedel-Crafts reactions.
Aniline: The amino group ($-\text{NH}_2$) coordinates with the Lewis acid catalyst (AlCl$_3$), deactivating the ring.
Compounds that cannot undergo Friedel-Crafts reactions:
Nitrobenzene, aniline, m-nitroaniline, m-dinitrobenzene (4 compounds).
Final Answer: (4)
Consider the following reaction of benzene. the percentage of oxygen is _______ %. (Nearest integer) 





The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}