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Chapter 22: Problem 52

Draw structural formulas for each of the following alcohols. Indicate whether the alcohol is primary, secondary, or tertiary. a. 1 -butanol c. 2 -methyl-1-butanol b. 2 -butanol d. 2-methyl-2-butanol

Short Answer

Expert verified
a. 1-butanol: CH3-CH2-CH2-CH2-OH (Primary) b. 2-butanol: CH3-CH(OH)-CH2-CH3 (Secondary) c. 2 -methyl-1-butanol: CH3-CH2-CH(CH3)-CH2-OH (Primary) d. 2-methyl-2-butanol: CH3-CH(OH)-CH(CH3)-CH3 (Tertiary)

Step by step solution

01

a. 1 -butanol

1-butanol is an alcohol with four carbon atoms. It has the molecular formula C4H9OH. The structure can be drawn as follows: CH3-CH2-CH2-CH2-OH 1-butanol is a primary alcohol because the hydroxyl group (OH) is attached to a carbon that is bonded to only one other carbon.
02

b. 2 -butanol

2-butanol is an alcohol with four carbon atoms. It has a molecular formula C4H9OH. The structure can be drawn as follows: CH3-CH(OH)-CH2-CH3 2-butanol is a secondary alcohol because the hydroxyl group (OH) is attached to a carbon that is bonded to two other carbon atoms.
03

c. 2 -methyl-1-butanol

2-methyl-1-butanol is an alcohol with five carbon atoms. It has the molecular formula C5H12O. The structure can be drawn as follows: CH3-CH2-CH(CH3)-CH2-OH 2-methyl-1-butanol is a primary alcohol because the hydroxyl group (OH) is attached to a carbon that is bonded to only one other carbon atom.
04

d. 2-methyl-2-butanol

2-methyl-2-butanol is an alcohol with five carbon atoms. It has the molecular formula C5H12O. The structure can be drawn as follows: CH3-CH(OH)-CH(CH3)-CH3 2-methyl-2-butanol is a tertiary alcohol because the hydroxyl group (OH) is attached to a carbon that is bonded to three other carbon atoms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Primary Alcohol
Primary alcohols are a type of alcohol where the hydroxyl group (OH) is attached to a carbon atom that is bonded to only one other carbon atom. This is a very defining characteristic and is what distinguishes primary alcohols from other alcohols.
  • The simplest example is methanol, where the carbon is bonded only to hydrogen atoms.
  • Primary alcohols typically have the general formula R-CH2OH, where R is a substituent group.
In the examples provided in the exercise, two primary alcohols were discussed:
  • 1-butanol: Here, the OH group is connected to an end carbon of the chain, which bonds with only one other carbon (R-CH2OH).
  • 2-methyl-1-butanol: Similar to 1-butanol, the OH group is on a terminal carbon, thus classifying it as primary.
Thus, the presence of a primary alcohol heavily depends on the position of the OH group in relation to the entire carbon chain.
Secondary Alcohol
Secondary alcohols feature the hydroxyl group (OH) attached to a carbon atom, which is, in turn, bonded to two other carbon atoms. This setup clearly differentiates secondary alcohols from their primary counterparts.
  • A classic example of a secondary alcohol is isopropanol, widely used as a disinfectant.
  • Secondary alcohols have the general formula R2CHOH, meaning two substituent groups are connected to the carbon with the OH group.
In the given exercise:
  • 2-butanol: This alcohol displays typical secondary structure, with the OH group attached to a central carbon connected to two other carbons forming a part of the backbone chain.
The nature of secondary alcohols lies in their unique structural position, making understanding them crucial for organic chemistry.
Tertiary Alcohol
Tertiary alcohols represent a further complexity where the hydroxyl-bearing carbon atom is bonded to three other carbon atoms. This distinct arrangement separates tertiary alcohols from both primary and secondary types.
  • They do not easily oxidize, a notable characteristic due to the lack of hydrogen atoms bonded to the OH-bearing carbon.
  • The general formula for tertiary alcohols is R3COH, with three substituent groups.
For instance, in the exercise, we have:
  • 2-methyl-2-butanol: The OH group in this structure is bonded to a carbon centrally located among three other carbons, clearly categorizing it as a tertiary alcohol.
Understanding the sophisticated bonding and the placement of functional groups in tertiary alcohols helps students highlight the diversity and implications of carbon connectivity in organic structures.

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Most popular questions from this chapter

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\), butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\), and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\) a. Draw two repeating units of ABS plastic assuming that the three monomer units react in a \(1: 1: 1\) mole ratio and react in the same order as the monomers listed above. b. A sample of ABS plastic contains \(8.80 \% \mathrm{~N}\) by mass. It took \(0.605 \mathrm{~g} \mathrm{Br}_{2}\) to react completely with a \(1.20\) -g sample of ABS plastic. What is the percent by mass of acrylonitrile, butadiene, and styrene in this polymer sample? c. ABS plastic does not react in a \(1: 1: 1\) mole ratio among the three monomer units. Using the results from part b, determine the relative numbers of the monomer units in this sample of ABS plastic.

Draw the five structural isomers of hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\).

Explain why methyl alcohol is soluble in water in all proportions, while stearyl alcohol \(\left[\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{OH}\right]\) is a waxy solid that is not soluble in water.

The following organic compounds cannot exist. Why? a. 2 -chloro-2-butyne b. 2 -methyl-2-propanone c. 1,1 -dimethylbenzene d. 2 -pentanal e. 3 -hexanoic acid f. 5,5 -dibromo-1-cyclobutanol

Draw cyclic structures for D-ribose and D-mannose.
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