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Chapter 1: Problem 8

Draw Lewis structures and condensed structural formulas for the four alcohols with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\). Classify each alcohol as primary, secondary, or tertiary.

Short Answer

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Question: Draw the Lewis structures and condensed structural formulas for the four alcohols with the molecular formula C₄H₁₀O. Classify each alcohol as primary, secondary, or tertiary. Answer: 1. n-Butanol: Condensed structural formula is CH₃CH₂CH₂CH₂OH and it is a primary alcohol. 2. sec-Butanol: Condensed structural formula is CH₃CH₂CH(OH)CH₃ and it is a secondary alcohol. 3. tert-Butanol: Condensed structural formula is C(CH₃)₃OH and it is a tertiary alcohol. 4. Isobutanol: Condensed structural formula is CH₃CH(CH₃)CH₂OH and it is a primary alcohol.

Step by step solution

01

Listing the isomers of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\).

The molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) can have different arrangements of carbon and hydrogen atoms bonded to a hydroxyl group (OH). We can list the 4 possible isomers as follows: 1. n-Butanol 2. sec-Butanol 3. tert-Butanol 4. Isobutanol Now, let's draw the Lewis structures and condensed structural formulas for each of these isomers and classify them as primary, secondary, or tertiary alcohols.
02

Drawing Lewis structures and condensed structural formulas for n-Butanol and classify its type.

n-Butanol has a hydroxyl group attached to one end of the straight chain of carbon atoms. Its Lewis structure is as follows: H H H H | | | | H-C-C-C-C-O-H | | | | H H H H And its condensed structural formula is: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\). n-Butanol is classified as a primary alcohol because the carbon atom bonded to the hydroxyl group is only bonded to one other carbon atom.
03

Drawing Lewis structures and condensed structural formulas for sec-Butanol and classify its type.

sec-Butanol has a hydroxyl group attached to the second carbon atom in a straight carbon chain. Its Lewis structure is as follows: H H H H | | | | H-C-C-C-C-H | | | | H H O H | H Its condensed structural formula is: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}\). sec-Butanol is classified as a secondary alcohol because the carbon atom bonded to the hydroxyl group is bonded to two other carbon atoms.
04

Drawing Lewis structures and condensed structural formulas for tert-Butanol and classify its type.

tert-Butanol has a hydroxyl group attached to a carbon atom that is also bonded to three other carbon atoms. Its Lewis structure is as follows: H H H | | | H H C-C-C-H | | | C-O-H H H | H Its condensed structural formula is: \(\mathrm{C}(\mathrm{CH}_{3})_{3} \mathrm{OH}\). tert-Butanol is classified as a tertiary alcohol because the carbon atom bonded to the hydroxyl group is bonded to three other carbon atoms.
05

Drawing Lewis structures and condensed structural formulas for Isobutanol and classify its type.

Isobutanol has a hydroxyl group attached to a secondary carbon atom in a branched carbon chain. Its Lewis structure is as follows: H H H | | | H-C-C-C-H | | | H C O H | | H H Its condensed structural formula is: \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{CH}_{3})\mathrm{CH}_{2} \mathrm{OH}\). Isobutanol is classified as a primary alcohol because the carbon atom bonded to the hydroxyl group is only bonded to one other carbon atom. In summary, we have drawn the Lewis structures and condensed structural formulas for the four alcohols with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) and classified them as primary, secondary, or tertiary alcohols.

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Most popular questions from this chapter

State the orbital hybridization of each highlighted atom. (a) CC (b) C=C (c) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) (d) C=O (e) O=CO (f) CCO (g) CN (h) \(\mathrm{H}-\stackrel{*}{\mathrm{O}}-\mathrm{N}=\stackrel{*}{\mathrm{O}}\) (i) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}\)

Draw condensed structural formulas for the three ketones with the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\).

How many electrons are in the valence shell of each atom? (a) Carbon (b) Nitrogen (c) Chlorine (d) Aluminum

Describe the bonding in these molecules in terms of hybridization of \(\mathrm{C}\) and \(\mathrm{N}\), and the types of bonds between carbon and nitrogen, and if there are any lone pairs, describe what type of orbital contains these electrons. (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Draw structural formulas for the three secondary amines with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\).
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