Predict the end product of the following sequence of reactions
`(CH_(3))_(2)COunderset(acid)overset((CH_(3))_(2)NH)rarrA overset(H_(2)//Ni)rarr` .

Options

A. `(CH_(3))_(2)`
B. `(CH_(3))_(2)CHN(CH_(3))_(2)`
C. `H_(2)C=overset(CH_(3))overset(|)C-N(CH_(3))_(2)`
D. `(CH_(3))COHN(CH_(3))_(2)`

Explanation

To predict the end product of the given sequence of reactions, we can break down the process step-by-step.

### Step 1: Identify the Reactants
The first reactant is propanone, which has the structure:
\[ \text{CH}_3\text{C(=O)CH}_3 \]
This is a ketone.

The second reactant is dimethylamine, which has the structure:
\[ \text{(CH}_3)_2\text{NH} \]
This is a secondary amine.

### Step 2: Reaction of Propanone with Dimethylamine
When propanone reacts with dimethylamine in the presence of an acid, it undergoes a condensation reaction. The ketone reacts with the amine to form an imine (or an enamine, depending on the conditions).

In this case, since propanone has alpha hydrogens, the reaction will lead to the formation of an enamine. The reaction can be summarized as follows:
1. The oxygen from the ketone (propanone) and one hydrogen from the amine will be removed, resulting in the formation of a double bond between the carbon and nitrogen.

The structure of the intermediate product (let's call it A) will be:
\[ \text{CH}_3\text{C=C(N(CH}_3)_2)} \]
This is an enamine.

### Step 3: Hydrogenation of the Enamine
The next step involves the hydrogenation of the enamine (A) using hydrogen gas (H₂) and a nickel catalyst (Ni). During hydrogenation, the double bond in the enamine will be reduced to a single bond, and additional hydrogen atoms will be added to the nitrogen and carbon atoms.

The reaction can be summarized as follows:
- The double bond between the carbon and nitrogen in the enamine will break.
- Each carbon involved in the double bond will gain one hydrogen atom.

The final product (let's call it B) will have the structure:
\[ \text{CH}_3\text{CH}_2\text{N(CH}_3)_2 \]
This compound is a tertiary amine.

### Final Product
Thus, the end product of the sequence of reactions is:
\[ \text{CH}_3\text{CH}_2\text{N(CH}_3)_2 \]