Question

draw (on paper) a lewis diagram for ch₃och₂ch₃.
use your diagram to answer the following questions.
the number of c-h bonds =
the number of o-h bonds is
the number of c-c single bonds =
the number of c-o single bonds =
the total number of lone pairs =

Explanation:

Step1: Analyze the structure of \( \text{CH}_3\text{OCH}_2\text{CH}_3 \)

The molecule is methoxyethane (dimethyl ether's isomer? No, it's \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \)). Let's break down each part:
- \( \text{CH}_3 \): 3 C - H bonds.
- \( \text{O} \): connected to \( \text{CH}_3 \) and \( \text{CH}_2 \).
- \( \text{CH}_2 \): 2 C - H bonds, and 1 C - C bond (with \( \text{CH}_3 \)).
- \( \text{CH}_3 \) (the last one): 3 C - H bonds.

Step2: Calculate C - H bonds

First \( \text{CH}_3 \): 3, \( \text{CH}_2 \): 2, last \( \text{CH}_3 \): 3. Total \( 3 + 2 + 3 = 8 \).

Step3: Calculate O - H bonds

In \( \text{CH}_3\text{OCH}_2\text{CH}_3 \), there are no O - H bonds (it's an ether, not an alcohol), so 0.

Step4: Calculate C - C bonds

Only one C - C bond (between \( \text{CH}_2 \) and the last \( \text{CH}_3 \)).

Step5: Calculate C - O bonds

One C - O bond (between \( \text{CH}_3 \) and O, and wait, no: \( \text{CH}_3 - \text{O} - \text{CH}_2 \), so two C - O bonds? Wait, no: \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) is \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \), so O is bonded to \( \text{CH}_3 \) (C - O) and \( \text{CH}_2 \) (C - O)? Wait, no, the formula is \( \text{CH}_3\text{OCH}_2\text{CH}_3 \), so the structure is \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \), so O has two single bonds: one to \( \text{CH}_3 \) (C - O) and one to \( \text{CH}_2 \) (C - O)? Wait, no, that would be \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \), so O is between \( \text{CH}_3 \) and \( \text{CH}_2 \), so two C - O bonds? Wait, no, let's count the atoms: C (3), O (1), C (2), C (3)? Wait, no, the formula is \( \text{C}_3\text{H}_8\text{O} \)? Wait, no: \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) is \( \text{C}_3\text{H}_8\text{O} \)? Wait, no, \( \text{CH}_3 \) (1 C), O (1), \( \text{CH}_2 \) (1 C), \( \text{CH}_3 \) (1 C): total 3 C. So \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \) has 3 C? Wait, no, \( \text{CH}_3 \) (C1), O, \( \text{CH}_2 \) (C2), \( \text{CH}_3 \) (C3). So C1 - O, O - C2, C2 - C3. So C - O bonds: 2 (C1 - O and C2 - O). Wait, I made a mistake earlier. Let's re - analyze:

Correct structure: \( \text{C}_1\text{H}_3 - \text{O} - \text{C}_2\text{H}_2 - \text{C}_3\text{H}_3 \)? No, \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) is \( \text{C}_1\text{H}_3 - \text{O} - \text{C}_2\text{H}_2 - \text{C}_3\text{H}_3 \)? Wait, no, \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) has 3 C atoms: \( \text{CH}_3 \) (C1), \( \text{O} \), \( \text{CH}_2 \) (C2), \( \text{CH}_3 \) (C3). So bonds:

- C1 - O (1)
- O - C2 (1)
- C2 - C3 (1)

C - H bonds:

- C1: 3
- C2: 2
- C3: 3
Total: \( 3 + 2 + 3 = 8 \)

O - H bonds: 0 (no - OH group)

C - C bonds: 1 (C2 - C3)

C - O bonds: 2 (C1 - O and O - C2)

Lone pairs on O: O has 6 valence electrons. It's bonded to two C atoms (2 bonds, 4 electrons), so 2 lone pairs (4 electrons? Wait, 6 - 2*2 = 2 lone pairs? Wait, no: O has 6 valence electrons. In the Lewis structure, O is bonded to two C atoms (single bonds), so it has two bonding pairs (4 electrons), so remaining electrons: 6 - 4 = 2 lone pairs? Wait, no, 6 valence electrons: each bond is 2 electrons, so two bonds (C - O) use 4 electrons, so 6 - 4 = 2 lone pairs? Wait, no, 6 - 2*2 = 2? Wait, no, 6 - (2*2) = 2? Wait, no, each bond is a pair, so two bonds (two pairs) use 4 electrons, so 6 - 4 = 2 lone pairs (one pair? Wait, no, 6 electrons: two bonding pairs (4 electrons) and two lone pairs (4 electrons)? Wait, I'm confused. Wait, O in \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) (ether) has two lone pairs. Let's confirm: the Lewis structure of ether (R - O - R'): O has two lone pairs. So total lone pairs: 2 (on O). Wait, but let's recalculate:

Valence electrons: C: 4, H: 1, O: 6.

Total valence electrons: \( 3\times4 + 8\times1 + 6 = 12 + 8 + 6 = 26 \).

Bonds: C - H: 8 (8 bonds, 16 electrons), C - C: 1 (2 electrons), C - O: 2 (4 electrons). Total bonding electrons: \( 16 + 2 + 4 = 22 \).

Lone pairs: \( 26 - 22 = 4 \) electrons, which is 2 lone pairs (since each lone pair is 2 electrons). Wait, no: 4 electrons is 2 lone pairs? Wait, 4 electrons is two lone pairs (each pair is 2 electrons). Wait, no, 4 electrons is two pairs? Wait, no, one lone pair is two electrons. So 4 electrons is two lone pairs. Wait, O has two lone pairs (4 electrons). So total lone pairs: 2 (on O).

Wait, let's start over with the Lewis structure:

1. Count valence electrons:
- \( \text{CH}_3\text{OCH}_2\text{CH}_3 \) has 3 C, 8 H, 1 O.
- Valence electrons: \( 3\times4 + 8\times1 + 1\times6 = 12 + 8 + 6 = 26 \).

2. Connect the atoms: \( \text{C} - \text{C} - \text{O} - \text{C} \)? No, the correct structure is \( \text{C} - \text{O} - \text{C} - \text{C} \) (with H atoms attached). Wait, the formula is \( \text{CH}_3\text{OCH}_2\text{CH}_3 \), so the structure is \( \text{CH}_3 - \text{O} - \text{CH}_2 - \text{CH}_3 \), which is \( \text{C}_1\text{H}_3 - \text{O} - \text{C}_2\text{H}_2 - \text{C}_3\text{H}_3 \).

3. Draw bonds:
- \( \text{C}_1 - \text{H} \): 3 bonds (6 electrons)
- \( \text{C}_1 - \text{O} \): 1 bond (2 electrons)
- \( \text{O} - \text{C}_2 \): 1 bond (2 electrons)
- \( \text{C}_2 - \text{H} \): 2 bonds (4 electrons)
- \( \text{C}_2 - \text{C}_3 \): 1 bond (2 electrons)
- \( \text{C}_3 - \text{H} \): 3 bonds (6 electrons)

Total bonding electrons: \( 6 + 2 + 2 + 4 + 2 + 6 = 22 \).

Lone pairs on O: \( 6 - 2 - 2 = 2 \) lone pairs (4 electrons).

So:

- Number of C - H bonds: 8
- Number of O - H bonds: 0
- Number of C - C bonds: 1
- Number of C - O bonds: 2
- Total number of lone pairs: 2 (on O)

Answer:

The number of C - H bonds = \( \boldsymbol{8} \)
The number of O - H bonds = \( \boldsymbol{0} \)
The number of C - C bonds = \( \boldsymbol{1} \)
The number of C - O bonds = \( \boldsymbol{2} \)
The total number of lone pairs = \( \boldsymbol{2} \)