In the following reaction sequence :
`R-OH overset(P+I_(2))rarr R-I overset(AgNO_(2))rarr RNO_(2) overset(HNO_(2))rarr ` no reaction.
The alcohol is a :

Options

A. primary alcohol
B. secondary alcohol
C. tertiary alcohol
D. phenol

Explanation

To solve the problem, we need to analyze the given reaction sequence step by step and determine the type of alcohol involved based on the reactions that occur.

### Step 1: Identify the Reaction Sequence
The reaction sequence provided is:
1. R-OH + P + I2 → R-I
2. R-I + AgNO2 → RNO2
3. RNO2 + HNO2 → No reaction

### Step 2: Analyze Each Step
- **Step 1:** The first reaction involves treating an alcohol (R-OH) with phosphorus (P) and iodine (I2). This is a substitution reaction where the hydroxyl group (OH) is replaced by iodine (I), resulting in an alkyl iodide (R-I). This reaction is characteristic of alcohols and can occur with primary, secondary, or tertiary alcohols.

- **Step 2:** The alkyl iodide (R-I) is then treated with silver nitrite (AgNO2). This reaction involves a nucleophilic substitution where the nitrite ion (NO2-) replaces the iodine atom, resulting in the formation of a nitro compound (RNO2). This step can also occur with primary and secondary alkyl halides.

- **Step 3:** The nitro compound (RNO2) is treated with nitrous acid (HNO2). In this step, we need to consider the structure of RNO2. If R is a primary or secondary alcohol, the reaction with HNO2 would typically lead to the formation of an oxime. However, the problem states that there is "no reaction" after this step.

### Step 3: Determine the Type of Alcohol
- **Primary Alcohol:** If R-OH were a primary alcohol, the sequence would yield an oxime after treatment with HNO2. This would typically lead to a subsequent reaction with NaOH, producing a colored compound (indicating a reaction). Since the problem states "no reaction," R cannot be a primary alcohol.

- **Secondary Alcohol:** If R-OH were a secondary alcohol, the same reasoning applies. The reaction with HNO2 would also yield an oxime, which would react further, leading to a colored product. Thus, R cannot be a secondary alcohol either.

- **Tertiary Alcohol:** If R-OH were a tertiary alcohol, the first two steps would proceed normally (substitution with iodine and then with nitrite). However, when treated with HNO2, the tertiary carbon would not have any hydrogen atoms to facilitate the formation of an oxime, resulting in "no reaction." This fits the condition given in the problem.

- **Phenol:** If R-OH were phenol, it would not react with phosphorus and iodine to form an iodide due to the instability of the positive charge on the sp2 hybridized carbon in the aromatic ring. Therefore, phenol cannot be the alcohol in this reaction sequence.

### Conclusion
Based on the analysis, the alcohol in the reaction sequence must be a tertiary alcohol, as it is the only type that results in "no reaction" when treated with HNO2.

### Final Answer
The alcohol is a **tertiary alcohol**.