• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Physical Properties of Organic Compounds

Extracts from this document...

Introduction

Physical Properties of Organic Compounds 1) Explain the differences in boiling points between the straight chain alkanes. Since straight change alkanes experience van der Waal (London ) forces, the higher the number of carbons in the straight chain alkane, the higher the boiling point, as more carbons contributes to stronger and more forces. 2) Explain the differences in boiling points between the straight and branched chain alkane. The difference in boiling points between straight and branched alkanes can be explained by the decrease in dipole moments due to a shorter chain length in branched alkanes. 0 3) What is the functional group in alcohol? The functional group in alcohols is the hydroxyl group -OH. 4) Explain the differences in the boiling points between similar massed short chain(<5 C) alcohols and alkanes. Alkanes, have much lower boiling points then similar massed short chained alcohols due to its ability to form hydrogen bonds. Since hydrogen bonds are significantly stronger than London forces, it would take more heat to boil them. Hence the alcohol would have a much higher boiling point than the alkane. ...read more.

Middle

An OH bond is more polar than a NH bond because oxygen has a higher electronegativity than nitrogen, resulting in a higher electronegativity difference between the OH bond, which means oxygen has a greater pull on the electrons than nitrogen does in their respective bonds. Hence the OH bond is polar. 12) Draw the H-bonding possible for Ammonia (NH3) 13) Are mono, di and tri substituted amines capable of H-bonding? Mono substituted amines are possible of hydrogen bonding (between the lone pair of the very electronegative of the nitrogen and the slightly positive hydrogen of another molecule) , as our di substituted amines. However, tri substituted amines are not capable of hydrogen bonding. 14) Compare the boiling points between ethane, methylamine and methanol. Explain. Between ethane and methylamine, methylamine has the higher boiling point despite both having the same number of electrons. This is due the fact that methylamine is able to form a hydrogen bond with another methylamine molecule which ethane is unable to do so. ...read more.

Conclusion

Explain. The difference in boiling points of methoxymethane and ethanol is that ethanol has a hydrogen atom attached directly to an oxygen and so hydrogen bonding can occur between ethanol molecules. However, in methoxymethane, there aren't enough hydrogens for hydrogen bonds to form. Hence the boiling point of ethanol is significantly higher. The same concept can be applied to comparing methoxyethane with propanol. 22) Compare the cis to trans- 2-butene isomers and pentane to neopentane. Explain the differences in boiling and melting points. Between the cis and trans isomers 23) Compare the structures of cis-1,2-dichloroethene and trans - 1,2 -dichloroethene. Explain the differences in boiling and melting points The difference between the two can be explained by the fact that in cis form, there are 2 polar C-Cl dipole moments, giving an overall molecular dipole and so when it bonds with other molecules, there are dipole-dipole forces as well as London dispersion which increases the boiling point. However, in the trans form, the two C-Cl bond moments cancel and the molecule is rendered non-polar and so there are no dipole-dipole forces involved hence a lower boiling point. ?? ?? ?? ?? November 14, 2009 Sanjana Negi ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Analysis of the Nitrogen Content of Lawn Fertiliser

    Subsequently, the amount of ammonium ions initially present in the 250 cm3 volumetric flask can be found and then the mass of nitrogen ions can be calculated. From this, the percentage by mass of nitrogen in the sample of fertiliser can be found.

  2. Design: Investigating the boiling point different alcohols

    * The thermometer will be clamped throughout the experiment and I will make sure the bulb submerged in alcohol, but not touching the bottom of the test tube to read the temperature carefully.

  1. Melting and Freezing point of naphthalene

    In this case energy is lost. Based on our results, the experiment has contradicted our hypothesis that the melting point of naphthalene is 80� and the freezing point is 60�.

  2. Determining the effect of carbon chain length of an alcohol on its fuel efficiency

    0.53% Uncertainty of the mass of the ethanol = (0.01/0.40)*100 = 2.5% Total percentage uncertainties = (0.05+0.53+2.5)= 3.08% * 3.08% of -361kJ = 11.1 kJ mol-1 Thus C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ?H = -361.0 � 11.1 kJ mol-1 Calculation for Propanol: Initial Mass of the Spirit Burner =

  1. Aim To investigate the chemical properties of alcohol

    10 ml of water + sodium metal The metal melts to silvery balls and moves quickly over to the water surface. Rapid evolution of a gas that extinguishes a burning splint with a 'pop' sound. 3. 10 drops of ethanol + 10 drops of glacial ethanoic acid + 6

  2. Kinetics of the Acid-Catalyzed Iodination of Propanone

    Concentration of Iodine in solution (m mol dm-3 ) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 1.0 + 1.1% 46.14 40.48 49.68 56.74 57.59 1.0 + 1.1% 26.72 24.80 24.52 30.27 21.50 1.0 + 1.1% 14.17 10.02 12.69 20.35 22.66 0.50 + 1.1% 08.23 22.80 73.73 24.22 21.28 0.25 + 1.1% 09.04 10.98 12.73 14.39 12.41

  1. A Comparison of Strong and Weak Acids and Bases

    because weak acid only partially dissociates, thus less H+(aq) ions will be readily available. And finally, comes the weak base at 1138�s for 1.00M of NH3(aq). Table 2.4: The order in which the speed of reaction occurs between the acids and the CaCO3(s). Speed of Reaction Name of respective acid Concentration of the acid / mol dm-3 1 HNO3(aq)

  2. The aim of this experiment is to examine the enthalpy of combustion of the ...

    are the following : Methanol: - 726 kJ/mol Ethanol: - 1367 kJ/mol Propan-1-ol: - 2021 kJ/mol Butan-1-ol: - 2676 kJ/mol Pentan-1-ol: -3330.63 kJ/mol According to the results it is clear that there were some mistakes / errors in the experiment.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work