Skip to main content
Chemistry LibreTexts

1.E: Introduction to Organic Chemistry (Exercises)

  • Page ID
  • Exercise 1-1 How many different isomers are there of \(\ce{C_H_2Br_4}\)? (Assume free-rotating tetrahedral carbon and univalent hydrogen and bromine.) How could one determine which of these isomers is which by the substitution method?

    Exercise 1-2 A compound of formula \(\ce{C_3H_6Br_2}\) is found to give only a single substance, \(\ce{C_3H_5Br_3}\), on further substitution. What is the structure of the \(\ce{C_3H_6Br_2}\) isomer and of its substitution product?

    Exercise 1-3 A compound of formula \(\ce{C_5H_{12}}\) gives only a single monobromo substitution product of formula \(\ce{C_5H_{11}Br}\). What is the structure of this \(\ce{C_5H_{12}}\) isomer? (Notice that carbon can form both continuous chains and branched chains. Also notice that structures such as the following represent the same isomer because the bonds to carbon are tetrahedral and are free to rotate.)

    Exercise 1-4 A gaseous compound of formula \(\ce{C_2H_4}\) reacts with liquid bromine \(\left( \ce{Br_2} \right)\) to give a single \(\ce{C_2H_4Br_2}\) compound. The \(\ce{C_2H_4Br_2}\) so formed gives only one \(\ce{C_2H_3Br_3}\) substitution product. Deduce the structure of \(\ce{C_2H_4}\) and the bromo compounds derived from it. (This was a key problem for the early organic chemists.)

    Exercise 1-5 Three different dibromobenzenes are known, here represented by just one of the Kekulé structures, \(14\), \(15\), and \(16\):

    Show how the substitution method described in Section 1-1F could be used to determine which isomer is which and, in addition, establish the structures of the various possible tribromobenzenes of formula \(\ce{C_6H_3Br_3}\).

    Exercise 1-6 The German chemist Ladenburg, in 1868, suggested the prismatic formula \(17\) for benzene:

    Assuming the \(\ce{C-C}\) bonds of the prism all are the same length, determine how many mono-, di-, and tribromine-substituted isomers are possible for \(17\). Compare the results with those expected for benzene with structure \(13\). If you have molecular models of the ball-and-stick type, these will be very helpful. A simple alternative model for \(17\) would be a piece of stiff paper folded and fastened as in \(18\) to give a prism with three equal square faces.

    Exercise 1-7 The compound \(\ce{C_2H_5Br}\) reacts slowly with the compound \(\ce{CH_4O}\) to yield a single substance of formula \(\ce{C_3H_8O}\). Assuming normal valences throughout, write structural formulas for \(\ce{CH_4O}\) and the three different possible structural (not rotational) isomers of \(\ce{C_3H_8O}\) and show how the principle of least structural change favors one of them as the reaction product. What would you expect to be formed from each of these three \(\ce{C_3H_8O}\) isomers with strong hydrobromic acid?

    Exercise 1-8 There are a large number of known isomers of \(\ce{C_5H_{10}}\), and some of these are typically unsaturated, like ethene, while others are saturated, like ethane. One of the saturated isomers on bromine substitution gives only one compound of formula \(\ce{C_5H_9Br}\). Work out a structure for this isomer of \(\ce{C_5H_{10}}\) and its monobromo substitution product.

    Exercise 1-9 Lithium hydride could be written as either \(\ce{Li}^\oplus \colon \ce{H}^\ominus\) or \(\ce{H}^\oplus \colon \ce{Li}^\ominus\) depending on whether lithium or hydrogen is more electron-attracting. Explain why hydrogen is actually more electron-attracting, making the correct structure \(\ce{Li}^\oplus \colon \ce{H}^\ominus\).

    Exercise 1-10 An acid \(\left( \ce{HA} \right)\) can be defined as a substance that donates a proton to a base, for example water. The proton-donation reaction usually is an equilibrium reaction and is written as

    Predict which member of each of the following pairs of compounds would be the stronger acid. Give your reasons.

    a. \(\ce{LiH}\), \(\ce{HF}\)
    b. \(\ce{NH_3}\), \(\ce{H_2O}\)
    c. \(\ce{H_2O_2}\), \(\ce{H_2O}\)
    d. \(\ce{CH_4}\), \(\ce{CF_3H}\)

    Exercise 1-11 (This problem is in the nature of review of elementary inorganic chemistry and may require reference to a general chemistry book.) Write Lewis structures for each of the following compounds. Use distinct, correctly placed dots for the electrons. Mark all atoms that are not neutral with charges of the proper sign.

    a. ammonia, \(\ce{NH_3}\)
    b. ammonium bromide, \(\ce{NH_4Br}\)
    c. hydrogen cyanide, \(\ce{HCN}\)
    d. ozone (\(\angle \ce{O-O-O} = 120^\text{o}\))
    e. carbon dioxide, \(\ce{CO_2}\)
    f. hydrogen peroxide, \(\ce{HOOH}\)
    g. hydroxylamine, \(\ce{HONH_2}\)
    h. nitric acid, \(\ce{HNO_3}\)
    i. hydrogen sulfide, \(\ce{H_2S}\)
    j. boron trifluoride, \(\ce{BF_3}\)

    Exercise 1-12 Use ball-and-stick models or suitable three-dimensional drawings to determine which members of the following sets of formulas represent identical compounds, provided "free rotation" is considered to be possible around all single bonds (except when these bonds are present in a cyclic structure):




    Exercise 1-13 Write structures for all of the different monobromo substitution products (of \(\ce{Br}\) for \(\ce{H}\)) you would expect for each of the following compounds.







    Exercise 1-14 There are two isomers of \(\ce{C_3H_6}\) with normal carbon and hydrogen valences. Each adds bromine - one rapidly and the other very sluggishly - to give different isomers of \(\ce{C_3H_6Br_2}\). The \(\ce{C_3H_6Br_2}\) derived from the \(\ce{C_3H_6}\) isomer that reacts sluggishly with bromine can give just two different \(\ce{C_3H_5Br_3}\) isomers on further bromine substitution, whereas the other \(\ce{C_3H_6Br_2}\) compound can give three different \(\ce{C_3H_5Br_3}\) isomers on further substitution. What are the structures of the \(\ce{C_3H_6}\) isomers and their \(\ce{C_3H_6Br_2}\) addition products?

    Exercise 1-15* (Remember that here and elsewhere, * denotes a more difficult exercise.) The vast majority of organic substances are compounds of carbon with hydrogen, oxygen, nitrogen, or the halogens. Carbon and hydrogen can be determined in combustible compounds by burning a weighed sample in a stream of oxygen (Figure 1-1) and absorbing the resulting water and carbon dioxide in tubes containing anhydrous magnesium perchlorate and soda lime, respectively. The gain in weight of these tubes corresponds to the weights of the water and the carbon dioxide formed.

    Figure 1-1: Schematic representation of a combustion train for determination of carbon and hydrogen in combustible substances.

    The molecular weight of a moderately volatile substance can be determined by the historically important Victor Meyer procedure, by which the volume of gas produced by vaporization of a weighed sample of an unknown is measured at a given temperature (Figure 1-2). The relationship \(PV = nRT\) is used here, in which \(P\) is the pressure in \(\text{mm}\) of mercury, \(V\) is the volume in \(\text{mL}\), \(T\) is the absolute temperature in \(^\text{o} \text{K}\) \(\left[ = 273.15 + T \left( ^\text{o} \text{C} \right) \right]\), \(n\) is the number of moles, and \(R\) is the gas constant \(= \: 62,400\) in the units of \(\left( \text{mm} \: \ce{Hg} \times \text{mL} \right) / \left( \text{mol} \times ^\text{o} \text{K} \right)\). The number of moles, \(n\), equals \(m/M\) in which \(m\) is the weight of the sample and \(M\) is the gross molecular weight. An example of the use of the Victor Meyer method follows.

    Figure 1-2: Schematic diagram of a Victor Meyer apparatus for determination of the vapor density of a substance that is volatile at the oven temperature \(T_1\). The air displaced from the heated chamber by the volatilization of the sample in the bulb is measured in the gas burette at temperature \(T_2\) as the difference in the burette readings \(V_2\) and \(V_1\).

    A \(0.005372\)-\(\text{g}\) sample of a liquid carbon-hydrogen-oxygen compound on combustion gave \(0.01222 \: \text{g}\) of \(\ce{CO_2}\) and \(0.00499 \: \text{g}\) of \(\ce{H_2O}\). In the Victor Meyer method, \(0.0343 \: \text{g}\) of the compound expelled a quantity of air at \(100^\text{o}\) \(\left( 373 ^\text{o} \text{K} \right)\) which, when collected at \(27^\text{o}\) \(\left( 300^\text{o} \text{K} \right)\) and \(728 \: \text{mm} \: \ce{Hg}\), amounted to \(15.2 \: \text{mL}\).

    Show how these results lead to the empirical and molecular formula of \(\ce{C_3H_6O}\). Write at least five isomers that correspond to this formula with univalent \(\ce{H}\), divalent \(\ce{O}\), and tetravalent \(\ce{C}\).

    Exercise 1-16 Determine the molecular formula of a compound of molecular weight 80 and elemental percentage composition by weight of \(\ce{C} = 45.00\), \(\ce{H} = 7.50\), and \(\ce{F} = 47.45\). Write structures for all the possible isomers having this formula. (See Exercise 1-15 for a description of how percentage composition is determined by combustion experiments.)

    Exercise 1-17 Why is the boiling point of water \(\left( 100^\text{o} \right)\) substantially higher than the boiling point of methane \(\left( -161^\text{o} \right)\)?

    Exercise 1-18 Dimethylmercury, \(\ce{CH_3-Hg-CH_3}\), is a volatile compound of bp \(96^\text{o}\), whereas mercuric fluoride \(\ce{F-Hg-F}\) is a high-melting solid having mp \(570^\text{o}\). Explain what differences in bonding in the two substances are expected that can account for the great differences in physical properties.

    Exercise 1-19* There are four possible isomers of \(\ce{C_4H_9Br}\). Let us call two of these \(A\) and \(B\). Both \(A\) and \(B\) react with water to give the same isomer of \(\ce{C_4H_{10}O}\) and this isomer of \(\ce{C_4H_{10}O}\) reacts with strong \(\ce{HBr}\) to give back only \(A\). Substitution of \(A\) with bromine gives only one of the possible \(\ce{C_4H_8Br_2}\) isomers. Substitution of \(B\) with bromine gives three different \(\ce{C_4H_8Br_2}\) isomers, and one of these is identical with the \(\ce{C_4H_8Br_2}\) from the substitution of \(A\). Write structural formulas for \(A\) and \(B\), and the isomers of \(\ce{C_4H_8Br_2}\) formed from them with bromine, and for the isomers of \(\ce{C_4H_{10}O}\) expected to be formed from them with water. Indicate in which reaction the principle of least structural change breaks down.


    • John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."