VIPSolutionsAct like a helpful tutor and exlain me :What alcohol is formed when the alkene below is treated with H,0 in the presence of H,S0,? H,C—CH, CH; | | H,80, FE a + H,0 —_ CH. — 5 CH)—CH, Change the structure in the drawing area below to show the product formed. If there is no reaction, check the No reaction box under the drawing area. = ® © o 0 BET is ED EE CH. -— 3 CH, —CHy
Question:
Act like a helpful tutor and exlain me :
What alcohol is formed when the alkene below is treated with H,0 in the presence of H,S0,?
H,C—CH, CH;
| | H,80,
FE a + H,0 —_
CH. —
5 CH)—CH,
Change the structure in the drawing area below to show the product formed.
If there is no reaction, check the No reaction box under the drawing area.
=
® ©
o 0
BET is
ED EE
CH. -—
3 CH, —CHy
This is AI generated Response.
Asked by: Alok Kumar
Created at: 2026-02-03 04:39:40
Asked by: Alok Kumar
Created at: 2026-02-03 04:39:40
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## Step-by-Step Explanation
Let's break down what's happening in this reaction:
### 1. **Type of Reaction**
The reaction shown is a **hydration of an alkene** using water (\( H_2O \)) in the presence of sulfuric acid (\( H_2SO_4 \)). This is an example of an **acid-catalyzed hydration**.
### 2. **Mechanism Overview**
- **Markovnikov's Rule** applies: the \( -OH \) group will add to the more substituted carbon of the double bond (the carbon with more alkyl groups attached).
- The double bond will break, and an \( -OH \) group will attach to one carbon, while an \( -H \) attaches to the other.
### 3. **Identifying the Alkene**
The double bond in your molecule is:
```
CH3
|
H3C—CH—C=C—CH—CH3
| |
CH3 CH2—CH3
```
Focus on the carbon atoms of the double bond.
### 4. **Application of Markovnikov’s Rule**
- The carbon on the left of the double bond is **more substituted** (it has more alkyl groups attached), so the **-OH group will add here**.
- The carbon on the right gets the hydrogen.
### 5. **Drawing the Product**
- Remove the double bond.
- Add **-OH** to the more substituted carbon.
- Add **-H** to the less substituted carbon.
So, your new structure should look like this:
```
CH3
|
H3C—CH—C—C—CH—CH3
| | |
CH3 OH CH2—CH3
```
### 6. **Summary of Steps**
1. Find the double bond.
2. Break the double bond.
3. Add **-OH** to the more substituted carbon.
4. Add **-H** to the less substituted carbon.
5. Redraw the structure accordingly.
---
## **In Short**
- **Reagent:** \( H_2O/H_2SO_4 \)
- **Type of reaction:** Acid-catalyzed hydration (Markovnikov addition)
- **Product:** Alcohol with -OH on the more substituted carbon of the former double bond.
If you need help drawing the structure or have more questions, let me know!
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