Identify product C in the following reaction sequence :
`CH_(3)overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"C "-CH_(2)CH_(2)OH overset(K_(2)Cr_(2)O_(2):H(2)SO_(4))underset(H_(2)O," heat ")rarr A overset(SOCl_(2))rarr B overset((CH_(3))_(2)NH)underset("(2 mol)")rarr C overset(overset("1. "LiAlH_(4))("diethyl ether"))underset("2. "H_(2)O)rarrD`

Options

A. `CH_(3)overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"CC"H_(2)C=N`
B. `CH_(3)overset(CH_(3))overset("| ")underset(CH_(3))underset("| ")"CH"H_(2)CH_(2)N(CH_(3))_(2)`
C. `{:(" "CH_(3)" N"(CH_(3))_(2)),(" | |"),(CH_(3)C CH_(2)CHN(CH_(3))_(2)),(" |"),(" "CH_(3)):}`
D. `{:(" "CH_(3)),(" |"),(CH_(3)C CH_(2)CHN(CH_(3))_(2)),(" | |"),(" "CH_(3)" OH"):}`

Answer

To identify product C in the given reaction sequence, we will follow the steps outlined in the question and analyze the transformations that occur at each stage.

### Step-by-Step Solution:

1. **Starting Material**: The starting material is 2-methyl-1-butanol, represented as:
\[ \text{CH}_3\text{C}(\text{CH}_3)_2\text{CH}_2\text{CH}_2\text{OH} \]
This is a primary alcohol.

2. **Oxidation Reaction**: The primary alcohol undergoes oxidation using potassium dichromate (\(K_2Cr_2O_7\)) in sulfuric acid (\(H_2SO_4\)) and heat. This reaction converts the alcohol into a carbonyl compound (aldehyde). The product A is:
\[ \text{CH}_3\text{C}(\text{CH}_3)_2\text{CHO} \]
(2-methylbutanal).

**Hint**: Remember that primary alcohols are oxidized to aldehydes or carboxylic acids depending on the conditions.

3. **Conversion to Acid Chloride**: The aldehyde (product A) is then treated with thionyl chloride (\(SOCl_2\)). This converts the aldehyde into an acid chloride (product B):
\[ \text{CH}_3\text{C}(\text{CH}_3)_2\text{C(=O)Cl} \]
(2-methylbutanoyl chloride).

**Hint**: Acid chlorides are formed from carboxylic acids or aldehydes when treated with thionyl chloride.

4. **Nucleophilic Attack by Dimethylamine**: Product B is then reacted with 2 moles of dimethylamine (\((CH_3)_2NH\)). The dimethylamine attacks the carbonyl carbon of the acid chloride, leading to the formation of an amide (product C):
\[ \text{CH}_3\text{C}(\text{CH}_3)_2\text{C(=O)N(CH_3)_2} \]
(N,N-dimethyl-2-methylbutanamide).

**Hint**: Acid chlorides react with amines to form amides through nucleophilic acyl substitution.

5. **Reduction with Lithium Aluminum Hydride**: Although not asked for product D, it is important to note that if product C (the amide) is treated with lithium aluminum hydride (\(LiAlH_4\)) followed by hydrolysis, it would reduce the amide to an amine (product D):
\[ \text{CH}_3\text{C}(\text{CH}_3)_2\text{CH}_2\text{NH(CH}_3)_2 \]
(N,N-dimethyl-2-methylbutanamine).

**Hint**: Lithium aluminum hydride is a strong reducing agent that can reduce amides to amines.

### Final Answer:
Thus, the product C in the reaction sequence is:
\[ \text{N,N-dimethyl-2-methylbutanamide} \]