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Chapter 16: Problem 350

The Infrared spectrum of cis-1,2-cyclopentanediol has an \(\mathrm{O}-\mathrm{H}\) stretching band at a lower frequency than for a free - OH group, and this band does not disappear even at high dilution. trans \(-1,2\) -Cyclopentanediol shows no such band. Can you suggest a possible explanation?

Short Answer

Expert verified
In cis-1,2-cyclopentanediol, the O-H stretching band appears at a lower frequency due to the presence of intramolecular hydrogen bonding between the two hydroxyl groups, which weakens the O-H bond. This band does not disappear at high dilution because intramolecular hydrogen bonding occurs within the same molecule, independent of concentration. In contrast, trans-1,2-cyclopentanediol does not exhibit such a band as the hydroxyl groups are too far apart in the trans-configuration, preventing the formation of intramolecular hydrogen bonding.

Step by step solution

01

Understand the cis- and trans- Isomers

cis-1,2-cyclopentanediol and trans-1,2-cyclopentanediol are isomers of a cyclopentane ring with two hydroxyl groups (-OH) at the 1,2-positions. In the cis-isomer, the two hydroxyl groups are on the same side of the cyclopentane ring, while in the trans-isomer, the two hydroxyl groups are on the opposite sides of the cyclopentane ring.
02

Analyze the O-H Stretching Frequency

In an IR spectrum, the stretching frequency of a bond is influenced by the bond strength and the masses of the atoms involved. In general, stronger bonds have higher stretching frequencies. For O-H groups, hydrogen bonding can significantly affect the stretching frequency. A hydrogen bond can form when a hydrogen atom is covalently bonded to an electronegative atom (such as oxygen) and interacts with another electronegative atom on a nearby molecule or different part of the same molecule.
03

Identify the Presence of Intramolecular Hydrogen Bonding

In the cis-1,2-cyclopentanediol molecule, the two hydroxyl groups are close to each other due to their cis-configuration. This proximity allows the formation of an intramolecular hydrogen bond between the oxygen atom of one hydroxyl group and the hydrogen atom of the other hydroxyl group. On the other hand, in trans-1,2-cyclopentanediol, the two hydroxyl groups are too far away from each other due to their trans-configuration and cannot form an intramolecular hydrogen bond.
04

Explain the Lower Stretching Frequency in cis Isomer

The presence of intramolecular hydrogen bonding in cis-1,2-cyclopentanediol results in a weaker O-H bond compared to a free -OH group. This is because the hydrogen atom involved in hydrogen bonding has its electron density partially shifted towards the oxygen atom it is interacting with (forming the hydrogen bond). Consequently, the stretching frequency of the hydrogen-bonded O-H group becomes lower than that of a free -OH group.
05

Explain the Persistence of the Band at High Dilution

In high dilution, there will be fewer cis-1,2-cyclopentanediol molecules in a given volume, which could potentially affect the hydrogen bonding between molecules (intermolecular hydrogen bonding). However, the key factor here is that the hydrogen bonding in cis-1,2-cyclopentanediol is intramolecular, which means it occurs within the same molecule. Therefore, it is independent of the concentration of cis-1,2-cyclopentanediol, and the band will not disappear even at high dilution.
06

Explain the Absence of the Band in trans Isomer

In the trans-1,2-cyclopentanediol molecule, there is no possibility for intramolecular hydrogen bonding between the two hydroxyl groups as they are too far away from each other due to their trans-configuration. As a result, there will be no O-H band in its IR spectrum resulting from intramolecular hydrogen bonding. In conclusion, the presence of intramolecular hydrogen bonding in cis-1,2-cyclopentanediol is responsible for the O-H stretching band being at a lower frequency than a free -OH group and not disappearing at high dilution. On the other hand, the absence of intramolecular hydrogen bonding in trans-1,2-cyclopentanediol explains the absence of such a band in its IR spectrum.

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Most popular questions from this chapter

Outline all steps in a possible laboratory synthesis of each of the following compounds from cyclohexanol and any necessary aliphatic, aromatic, or inorganic reagents. (a) cyclohexanone \(\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\right)\) (b) bromocyclohexane (c) 1 -methylcyclohexanol (d) 1 -methylcyclohexene (e) trans-2-methylcyclohexanol (f) cyclohexylmethylcarbinol (g) trans \(-1,2\) -dibromocyclohexane (h) cyclohexanecarboxylic acid (i) adipic acid \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{COOH}\)

(a) From the reaction of either \((2 \mathrm{R}, 3 \mathrm{~S})\) or \(\left(2 \mathrm{~S}, 3 \mathrm{R}_{7}\right)-\) 3 -bromo-2-butanol with HBr, the product is meso-2, 3 -dibromobutane. Show how this product is formed. Is the intermediate bromonium ion chiral? (b) Contrast the reaction of \((2 \mathrm{R}, 3 \mathrm{~S})-3\) -bromo \(-2\) -butanol with HBr and the reaction of the same substrate with sodium ethoxide.

Outline the synthesis of n-butyldimethylcarbinol (2-methy1-2-hexanol) from acetone and n-butylbromide.

A naive graduate student attempted the preparation of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDBrCH}_{3}\) from \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDOHCH}_{3}\) by heating the deuterio-alcohol with \(\mathrm{HBr}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\). He obtained a product having the correct boiling point, but a careful examination of the spectral properties by his research director showed that the product was a mixture of \(\mathrm{CH}_{3} \mathrm{CHDCHBrCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CDBrCH}_{3}\). What happened?

Sodium metal was added to tert-butyl alcohol and allowed to react. When the metal was consumed, ethyl bromide was added to the resulting mixture. Work-up of the reaction mixture yielded a compound of formula \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}\). In a similar experiment, sodium metal was allowed to react with ethanol. When tert-butyl bromide was added, a gas was evolved, and work-up of the remaining mixture gave ethanol as the only organic material. (a) Write equations for all reactions, (b) What familiar reaction type is involved in each case? (c) Why did the reactions take different courses?
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