QQuestionChemistry
QuestionChemistry
Draw the Lewis structure for ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\right)$. Ethanol molecules have one hydroxy group. Be certain you include any lone pairs.
12 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:: Identify the total number of valence electrons in the ethanol molecule.
In total, there are $2 \times 4 + 6 + 6 = 22$ valence electrons in an ethanol molecule.
Step 2:: Draw a skeletal structure for the ethanol molecule.
\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}
Carbon atoms tend to form 4 bonds, and hydrogen atoms form 1 bond. Oxygen atoms usually form 2 bonds and have 2 lone pairs of electrons. Start by placing the carbon, hydrogen, and oxygen atoms in a way that satisfies these bonding requirements.
Step 3:: Add remaining valence electrons as single bonds and lone pairs.
\begin{array}{cccc}
Add the remaining valence electrons to complete single bonds between the atoms and add lone pairs to the oxygen atom. & & \mathrm{H} & \ & & | & \ \mathrm{H}-\mathrm{C}-\mathrm{C}-\mathrm{C} & - & \mathrm{O} & -\mathrm{H} \ & & | & \ & & \mathrm{H} & \end{array} & & \mathrm{H} & \ & & | & \ \mathrm{H}-\mathrm{C}-\mathrm{C}-\mathrm{C} & - & \mathrm{O}: & :\mathrm{H} \ & & | & \ & & \mathrm{H} & \end{array}
Step 4:: Distribute the remaining valence electrons as lone pairs on the carbon and hydrogen atoms.
\begin{array}{cccc}
& & \mathrm{H} & \ & & | & \ \mathrm{H}-\mathrm{C}: & - & \mathrm{O}: & :\mathrm{H} \ & & | & \ \mathrm{H}-\mathrm{C}: & & & \end{array}
Step 5:: Ensure that each atom has an octet (8 electrons) or a duet (2 electrons) for hydrogen.
All atoms in the Lewis structure have an octet (carbon, oxygen, and hydrogen attached to oxygen) or a duet (hydrogen atoms not attached to oxygen) of electrons, so the structure is valid.
Final Answer
The Lewis structure for ethanol is: \begin{array}{cccc} & & \mathrm{H} & \ & & | & \ \mathrm{H}-\mathrm{C}: & - & \mathrm{O}: & :\mathrm{H} \ & & | & \ \mathrm{H}-\mathrm{C}: & & & \end{array}
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students