When \(\mathrm{C_6H_5CHO}\) reacts with the mixture of \(\mathrm{HNO_3}\) and \(\mathrm{H_2SO_4}\) at 273-283K gives
\(m\)-Nitrobenzaldehyde
The \(\mathrm{-CHO}\) group withdraws electrons from the ring, so it deactivates the ring and is a meta-directing group.
Nitration with concentrated \(\mathrm{HNO_3 + H_2SO_4}\) at 273-283 K therefore puts the nitro group at the meta position.
\(\mathrm{C_6H_5CHO + HNO_3 \xrightarrow{H_2SO_4} \mathit{m}\text{-}O_2NC_6H_4CHO + H_2O}\)
Hence, \(m\)-Nitrobenzaldehyde.
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(Given: molar mass in $\text{g mol}^{-1}$ $\text{H} : 1, \text{C} : 12, \text{O} : 16$) - Grignard reagent $\text{RMgBr (P)}$ reacts with water and forms a gas $\text{(Q)}$. One gram of Q occupies $1.4 \text{ dm}^3$ at STP. (P) on reaction with dry ice in dry ether followed by $\text{H}_3\text{O}^+$ forms a compound $\text{(Z)}$. 0.1 mole of $\text{(Z)}$ will weigh _________ g. (Nearest integer)
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The major products U and V in the following reaction are
