The question asks one to write the equations corresponding to the oxidation of a primary alcohol and a secondary alcohol. Also one needs to specify the structure of each alcohol and the principal products of oxidation. Concept Introduction: Monohydric alcohols can be categorized into three types as primary, secondary and tertiary alcohols. AS shown above in primary alcohols carbon with the −OH group is bonded to only one carbon whereas in secondary alcohols, carbon with the −OH group is bonded to two other carbon atoms. In tertiary alcohols(not shown here), carbon with the −OH group is bonded to three other carbon atoms. In the alcohols the −OH bond is polarized placing partially negative charge on O atom while placing a partially positive charge on the alkyl carbon atom of the R-OH bond. Hence due to the inter molecular hydrogen bonds formed between alcohol molecules, their boiling points have higher values compared to the alkanes and ethers with comparable relative molecular masses.

Question

Want to see the full answer?

Answer 1

Question

Chapter 20, Problem 70QAP

Interpretation Introduction

Interpretation:

The question asks one to write the equations corresponding to the oxidation of a primary alcohol and a secondary alcohol. Also one needs to specify the structure of each alcohol and the principal products of oxidation.

Concept Introduction:

Monohydric alcohols can be categorized into three types as primary, secondary and tertiary alcohols.

AS shown above in primary alcohols carbon with the −OH group is bonded to only one carbon whereas in secondary alcohols, carbon with the −OH group is bonded to two other carbon atoms. In tertiary alcohols(not shown here), carbon with the −OH group is bonded to three other carbon atoms.

In the alcohols the −OH bond is polarized placing partially negative charge on O atom while placing a partially positive charge on the alkyl carbon atom of the R-OH bond. Hence due to the inter molecular hydrogen bonds formed between alcohol molecules, their boiling points have higher values compared to the alkanes and ethers with comparable relative molecular masses.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

b) Certain cyclic compounds are known to be conformationally similar to carbohydrates, although they are not themselves carbohydrates. One example is Compound C shown below, which could be imagined as adopting four possible conformations. In reality, however, only one of these is particularly stable. Circle the conformation you expect to be the most stable, and provide an explanation to justify your choice. For your explanation to be both convincing and correct, it must contain not only words, but also "cartoon" orbital drawings contrasting the four structures. Compound C Possible conformations (circle one): Дет

Lab Data The distance entered is out of the expected range. Check your calculations and conversion factors. Verify your distance. Will the gas cloud be closer to the cotton ball with HCI or NH3? Did you report your data to the correct number of significant figures? - X Experimental Set-up HCI-NH3 NH3-HCI Longer Tube Time elapsed (min) 5 (exact) 5 (exact) Distance between cotton balls (cm) 24.30 24.40 Distance to cloud (cm) 9.70 14.16 Distance traveled by HCI (cm) 9.70 9.80 Distance traveled by NH3 (cm) 14.60 14.50 Diffusion rate of HCI (cm/hr) 116 118 Diffusion rate of NH3 (cm/hr) 175.2 175.2 How to measure distance and calculate rate

For the titration of a divalent metal ion (M2+) with EDTA, the stoichiometry of the reaction is typically: 1:1 (one mole of EDTA per mole of metal ion) 2:1 (two moles of EDTA per mole of metal ion) 1:2 (one mole of EDTA per two moles of metal ion) None of the above