give answer in 2 step with explanation at the end of each step and final answer at the end:November 15, 2025 CHEMISTRY 212 | please print this PDF and answer the
Homework #6 questions in the spaces p

Question

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VIPSolutions · Accepted Answer

## Step 1: Understanding the Question

You are asked to show how **N-ethyl-N-methylpropan-2-amine** can be synthesized a **single step** from:
- A ketone
- An aldehyde
- Two different amides

You must choose the correct starting materials (one ketone, one aldehyde, and two amides) that, upon a single-step reaction, would yield **N-ethyl-N-methylpropan-2-amine**.

**Explanation:**  
N-ethyl-N-methylpropan-2-amine is a tertiary amine (all three groups on nitrogen are carbon-based: ethyl, methyl, and isopropyl). The synthesis will likely involve forming the C–N bonds via nucleophilic substitution or reductive amination.

---

## Step 2: Matching Starting Materials to Product

Let's break down the molecule:

- **N-ethyl:** –CH₂CH₃ attached to N
- **N-methyl:** –CH₃ attached to N
- **propan-2-amine backbone:** the N is attached to the 2-position of propane (isopropyl group: –CH(CH₃)₂)

So, the full structure is:  
**CH₃–CH(N(CH₂CH₃)(CH₃))–CH₃**

Now, consider how each class of starting material could contribute:

- **Ketone:** Reductive amination of a ketone with a secondary amine gives a tertiary amine.
- **Aldehyde:** Reductive amination with a secondary amine can also give a tertiary amine.
- **Amide:** Can be reduced to an amine (though usually gives a primary or secondary amine).

**The correct combinations are:**

**A. From a Ketone:**  
Start with **propan-2-one (acetone)** (CH₃COCH₃) and react with **N-ethylmethylamine** (ethylmethylamine) under reductive amination conditions.

**B. From an Aldehyde:**  
Start with **ethanal (acetaldehyde)** (CH₃CHO) and react with **N-methylisopropylamine** under reductive amination.

**C. From Amides:**  
- **Amide 1:** N-ethyl-N-methylpropanamide (reduce to the amine via LiAlH₄)
- **Amide 2:** N-methyl-N-isopropylacetamide (reduce to the amine via LiAlH₄)

---

## **Final Answer (Summary Table):**

| Starting Material      | Structure/Name                               | Product (single-step reduction or amination)   |
|-----------------------|----------------------------------------------|-----------------------------------------------|
| **A Ketone**          | Acetone + N-ethylmethylamine                 | N-ethyl-N-methylpropan-2-amine                |
| **An Aldehyde**       | Acetaldehyde + N-methylisopropylamine        | N-ethyl-N-methylpropan-2-amine                |
| **An Amide**          | N-ethyl-N-methylpropanamide                  | N-ethyl-N-methylpropan-2-amine                |
| **A different amide** | N-methyl-N-isopropylacetamide                | N-ethyl-N-methylpropan-2-amine                |

**Explanation:**  
Each starting material, when subjected to either reductive amination (for ketone and aldehyde) or reduction (for amides), will give the same tertiary amine product, **N-ethyl-N-methylpropan-2-amine**. The key is matching the carbon fragments that will end up on the nitrogen.

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