Solved NAMING AND DRAWING FUNCTIONAL GROUPS PRACTICE | Chegg.com - Free Printable

Problem Analysis:

The worksheet provided focuses on naming and drawing functional groups in organic chemistry, specifically alcohols and ethers. Below, I will solve each problem step by step.

---

Problem 1: Draw the following alcohols

#### a) Heptan-2-ol
- Heptan-2-ol: This is a heptane chain with an alcohol group (-OH) at carbon 2.
- Structure:
```
CH3-CH2-CH(OH)-CH2-CH2-CH2-CH3
```

#### b) 3-Methyloctan-1-ol
- 3-Methyloctan-1-ol: An octane chain with a methyl group (-CH3) at carbon 3 and an alcohol group (-OH) at carbon 1.
- Structure:
```
HO-CH2-CH2-C(CH3)-CH2-CH2-CH2-CH3
```

#### c) Cyclopropanol
- Cyclopropanol: A cyclopropane ring with an alcohol group (-OH) attached to one of the carbons.
- Structure:
```
OH
|
CH2-CH2-CH2
```

#### d) 2,4,6-Trichlorononan-2-ol
- 2,4,6-Trichlorononan-2-ol: A nonane chain with chlorine atoms at carbons 2, 4, and 6, and an alcohol group at carbon 2.
- Structure:
```
Cl-CH(OH)-CH2-Cl-CH2-CH2-Cl-CH2-CH2-CH3
```

#### e) Pentan-1,4-diol
- Pentan-1,4-diol: A pentane chain with alcohol groups (-OH) at carbons 1 and 4.
- Structure:
```
HO-CH2-CH2-CH(OH)-CH2-CH3
```

#### f) Benzene-1,3-diol
- Benzene-1,3-diol: A benzene ring with alcohol groups (-OH) at positions 1 and 3.
- Structure:
```
OH
|
H C - C - H
| |
| |
H C - C - H
|
OH
```

#### g) But-2-ene-1-ol
- But-2-ene-1-ol: A butene molecule (double bond between carbons 2 and 3) with an alcohol group at carbon 1.
- Structure:
```
HO-CH=CH-CH3
```

#### h) 4-Methylpent-2-yne-1-ol
- 4-Methylpent-2-yne-1-ol: A pentyne chain with a triple bond between carbons 2 and 3, a methyl group at carbon 4, and an alcohol group at carbon 1.
- Structure:
```
HO-CH=C=CH-CH(CH3)-CH3
```

#### i) 3,4-Dimethylcycloheptan-1-ol
- 3,4-Dimethylcycloheptan-1-ol: A cycloheptane ring with methyl groups at carbons 3 and 4 and an alcohol group at carbon 1.
- Structure:
```
CH3
|
CH2-CH-CH2-CH2-CH2-CH2-OH
| |
CH3 CH3
```

---

Problem 2: Name the following alcohols

#### a)
- Structure:
```
CH3-CH2-CH2-CH2-CH2-OH
```
- Name: Pentan-1-ol

#### b)
- Structure:
```
CH3-CH2-CH2-CH2-CH(OH)-CH3
```
- Name: Hexan-5-ol

#### c)
- Structure:
```
HO-CH2-CH2-Cl
```
- Name: 2-Chloroethanol

#### d)
- Structure:
```
CH3-CH(CH3)-CH2-CH2-CH2-OH
```
- Name: 3-Methylhexan-1-ol

#### e)
- Structure:
```
CH3-CH2-CH2-CH2-CH2-CH2-OH
```
- Name: Heptan-1-ol

#### f)
- Structure:
```
HO-CH2-CH=CH2
```
- Name: Prop-2-en-1-ol (or Allyl alcohol)

#### g)
- Structure:
```
OH
|
H C - C - H
| |
| |
H C = C - H
|
H
```
- Name: 2-Hexen-1-ol

#### h)
- Structure:
```
CH3-CH2-C≡C-CH2-CH2-OH
```
- Name: 5-Hexyn-1-ol

#### i)
- Structure:
```
CH3
/ \
CH3-C-C-CH2-CH2-OH
\ /
CH3
```
- Name: 2,2-Dimethylbutan-1-ol

---

Problem 3: Explain why propane used as a fuel in a BBQ is gas at room temperature, but 2-propanol used as rubbing alcohol is a liquid at room temperature.

- Propane (C₃H₈):
- Propane is a small, nonpolar molecule with weak van der Waals forces.
- Its molecular weight is low (44 g/mol), which results in very weak intermolecular attractions.
- At room temperature (~25°C), the kinetic energy of propane molecules is high enough to overcome these weak forces, causing it to exist as a gas.

- 2-Propanol (C₃H₈O):
- 2-Propanol has a hydroxyl (-OH) group, which allows for hydrogen bonding.
- Hydrogen bonding is a stronger intermolecular force compared to van der Waals forces.
- The presence of hydrogen bonding increases the intermolecular attraction, requiring more energy to vaporize the molecule.
- As a result, 2-propanol remains a liquid at room temperature.

---

Problem 4: Draw the following ethers

#### a) 1-Propoxypentane
- Structure:
```
CH3-CH2-CH2-O-CH2-CH2-CH3
```

#### b) 2-Ethoxybutane
- Structure:
```
CH3-CH(OCH2-CH3)-CH2-CH3
```

#### c) 1-Methoxy-4-chlorohexane
- Structure:
```
CH3-O-CH2-CH2-CH2-CH(Cl)-CH3
```

#### d) 3-Butoxy-2-dimethyloctane
- Structure:
```
CH3-CH(CH3)-CH(O-CH2-CH2-CH2-CH3)-CH2-CH2-CH2-CH3
```

#### e) 2-Propoxy-4-phenylheptane
- Structure:
```
CH3-CH(O-CH2-CH2-CH3)-CH2-CH2-C6H5-CH2-CH2-CH3
```

#### f) 1-Ethoxycyclopentane
- Structure:
```
O-CH2-CH3
|
CH2-CH2-CH2-CH2-CH2
```

---

Problem 5: Name the following ethers

#### a)
- Structure:
```
CH3-CH2-CH2-O-CH3
```
- Name: 1-Methoxypropane

#### b)
- Structure:
```
CH3-CH2-CH2-O-CH2-CH3
```
- Name: 1-Ethoxypropane

#### c)
- Structure:
```
CH3
|
CH3-CH2-O-CH2-CH2-CH3
```
- Name: 2-Methoxyhexane

---

Final Answers:

1. Drawings:
- a) `CH3-CH2-CH(OH)-CH2-CH2-CH2-CH3`
- b) `HO-CH2-CH2-C(CH3)-CH2-CH2-CH2-CH3`
- c) `OH-CH2-CH2-CH2`
- d) `Cl-CH(OH)-CH2-Cl-CH2-CH2-Cl-CH2-CH2-CH3`
- e) `HO-CH2-CH2-CH(OH)-CH2-CH3`
- f) `OH-CH-CH-CH2-CH2-CH2-CH2-OH`
- g) `HO-CH=CH-CH3`
- h) `HO-CH=C=CH-CH(CH3)-CH3`
- i) `CH3-CH2-CH(OH)-CH2-CH2-CH3`

2. Names:
- a) Pentan-1-ol
- b) Hexan-5-ol
- c) 2-Chloroethanol
- d) 3-Methylhexan-1-ol
- e) Heptan-1-ol
- f) Prop-2-en-1-ol
- g) 2-Hexen-1-ol
- h) 5-Hexyn-1-ol
- i) 2,2-Dimethylbutan-1-ol

3. Explanation:
- Propane is a gas due to weak van der Waals forces.
- 2-Propanol is a liquid due to strong hydrogen bonding.

4. Drawings:
- a) `CH3-CH2-CH2-O-CH2-CH2-CH3`
- b) `CH3-CH(OCH2-CH3)-CH2-CH3`
- c) `CH3-O-CH2-CH2-CH2-CH(Cl)-CH3`
- d) `CH3-CH(CH3)-CH(O-CH2-CH2-CH2-CH3)-CH2-CH2-CH2-CH3`
- e) `CH3-CH(O-CH2-CH2-CH3)-CH2-CH2-C6H5-CH2-CH2-CH3`
- f) `CH3-CH2-CH2-CH2-CH2-O-CH2-CH3`

5. Names:
- a) 1-Methoxypropane
- b) 1-Ethoxypropane
- c) 2-Methoxyhexane

---

Boxed Final Answer:
\[
\boxed{
\text{See detailed solutions above.}
}
\]