$CH_3-CH(CH_3)-MgBr $
$ CH_3-CH_2-CH_2MgBr$
To prepare 2-methylpropan-1-ol using methanal and a Grignard reagent, we follow this reaction:
\( \text{HCHO} + \text{R–MgBr} \xrightarrow{\text{ether}} \text{R–CH}_2\text{OH} \)
The required product is 2-methylpropan-1-ol, which has the structure:
\( \text{CH}_3–\text{CH(CH}_3)–\text{CH}_2\text{OH} \)
This shows that the R group in the Grignard reagent should be isopropyl \( (\text{CH}_3–\text{CH(CH}_3)–) \).
So, the correct Grignard reagent is:
\( \text{CH}_3–\text{CH(CH}_3)–\text{MgBr} \)
Option (A): \( \text{CH}_3–\text{CH(CH}_3)–\text{MgBr} \)
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to: