One of the products formed in the following reaction is
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Grignard reagents are aggressive proton-seekers! If a compound contains an active hydrogen like in $\text{H}_2\text{O}$, R-OH, $\text{R-NH}_2$, or R-COOH, the Grignard reagent will immediately abstract it to form an alkane (R-H), bypassing any potential carbon-carbon coupling pathways.
Step 1: Identify the two reactants. The figure shows cyclohexylmagnesium bromide \(C_6H_{11}MgBr\) reacting with cyclohexylamine \(C_6H_{11}NH_2\). One is a Grignard reagent and the other carries an N-H group. Step 2: Recall what Grignard reagents do with acidic H. The carbon bonded to magnesium behaves as a strong base, a carbanion \(R^-\). Any active hydrogen on O, N or S is grabbed at once. This is the Zerewitinoff type acid-base reaction. Step 3: Spot the active hydrogen. Cyclohexylamine has N-H bonds. These protons are acidic enough to be pulled off by the cyclohexyl carbanion. Step 4: Write the proton transfer. \[ C_6H_{11}MgBr + C_6H_{11}NH_2 \rightarrow C_6H_{11}H + C_6H_{11}NHMgBr \] The reaction is simply an acid-base step, not a coupling. Step 5: Name the products. The carbanion picks up a proton to become cyclohexane \(C_6H_{12}\); the rest forms a magnesium amide salt. Step 6: Match the option. One of the products is plain cyclohexane, which is the structure in option 1 (FigA). \[ \boxed{\text{Cyclohexane } (C_6H_{12}),\ \text{FigA}} \]
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