Step 1: Identify the starting material and reagent in step 1. The reaction begins with 1-methylcyclohexene (a cyclic alkene with an endocyclic double bond and a methyl substituent at C1). The first reagent is HCl. Step 2: Apply Markovnikov's rule for HCl addition. HCl adds across the double bond following Markovnikov's rule: the proton ($H^+$) adds to the carbon bearing more hydrogens (C2), and the chloride ($Cl^-$) adds to the carbon bearing fewer hydrogens (C1, which already has the methyl group). This gives 1-chloro-1-methylcyclohexane (a tertiary alkyl chloride). Step 3: Write the product of step 1. Product of step 1: 1-chloro-1-methylcyclohexane. The chlorine is at the tertiary carbon (C1 bearing the methyl group). This is favoured because the tertiary carbocation intermediate at C1 is more stable than a secondary carbocation at C2. Step 4: Identify the reagent in step 2 - AgNO$_2$. The second reagent is silver nitrite ($AgNO_2$). This is an ambident nucleophile that can react through either the nitrogen atom or the oxygen atom. With alkyl halides, $AgNO_2$ reacts preferentially through the NITROGEN to give an alkyl nitrite (R-O-N=O) as the major product. Step 5: Predict the product of step 2. Wait, let us be precise: $AgNO_2$ (silver nitrite) predominantly gives nitroalkane ($R-NO_2$) when the nitrogen end attacks ($SN2$ pathway), but since this is a tertiary alkyl halide, the $SN1$ pathway favours attack by the oxygen end to give an alkyl nitrite ($R-O-N=O$). The image-based option 4 shows the alkyl nitrite product formed from the tertiary chloride. Step 6: Confirm the overall reaction sequence and answer. 1-Methylcyclohexene $\xrightarrow{HCl}$ 1-Chloro-1-methylcyclohexane $\xrightarrow{AgNO_2}$ Alkyl nitrite (option 4). The two-step sequence involves electrophilic addition followed by nucleophilic substitution at the tertiary carbon. \[ \boxed{\text{Option 4: Alkyl nitrite product}} \]
