Both 1 and 3 are reasonable contributing structures for an amide bond. D The following structure is called imidazolium. Which of the following statements about imidazolium are true? Imidazolium a. Both nitrogens are sp2 hybridized, and the lone pair of electrons is in 2p orbitals.
The nitrogen on the right is sp3 hybridized while the nitrogen on the left is sp2 hybridized, and the lone pair of electrons shown is in an sp3 hybrid orbital. The molecule has an identical contributing structure not shown.
The molecule has no reasonable contributing structures. Statements a. Both N atoms in imidazolium are sp2 hybridized and there is a symmetric contributing structure that moves the upper double bond and interconverts the locations of the plus charge and lone pair as shown above. Statements b. Electronic Structures of Atoms Problem 1.
After each atom is its atomic number in parentheses. The valence shell is the outermost occupied shell of an atom. A valence electron is an electron in the valence shell. Lewis Structures and Formal Charge Problem 1. Nitrogen is farther to the right than carbon in Period 2 of the Periodic Table. Thus, nitrogen is more electronegative than carbon. Thus, chlorine is more electronegative than bromine. Thus, oxygen is more electronegative than sulfur. Show all valence electrons. None of them contains a ring of atoms.
Then write the molecular formula of each compound. Lone pairs were added to the following structural formulas for clarity. Which structural formulas are incorrect, and which atoms in them have an incorrect number of bonds?
The molecules in a , b , d , and f are incorrect, because there are five bonds to the circled carbon atom, not four. Then assign formal charges as appropriate. The following structural formulas show all valence electrons and all formal charges for clarity. Assign formal charges in each structure as appropriate.
There is a positive formal charge in parts a , e , and f. There is a negative formal charge in parts b , c , and d. Polarity of Covalent Bonds Problem 1. Electronegativity increases from left to right across a period and from bottom to top of a column in the Periodic Table.
Thus, statement a is true, but b , c , and d are false. Electronegativity increases with increasing positive charge on the nucleus and with decreasing distance of the valence electrons from the nucleus. Fluorine is that element for which these two parameters lead to maximum electronegativity. The difference in electronegativities is given in parentheses underneath each answer. Bond Angles and Shapes of Molecules Problem 1. Approximate bond angles as predicted by valence-shell electron-pair repulsion are as shown.
Be certain to show all valence electrons on each. Ketone O C H3. Here, as in other problems of this type, it is important to have a system and to follow it.
As one way to proceed, first decide the number of different carbon skeletons that are possible. A little doodling with paper and pencil should convince you that there are only two.
There are three possible locations for it. For the first carbon skeleton, there are four possible locations of the -OH group; for the second carbon skeleton there are two possible locations; and for the third, there are also two possible locations.
Four of these compounds marked by a symbol are not stable and are in equilibrium with a more stable aldehyde or ketone. You need not be concerned, however, with this now. Just concentrate on drawing the required eight condensed structural formulas. OH OH. A tertiary alcohol is one in which the -OH group is on a tertiary carbon atom.
A tertiary carbon atom is one that is bonded to three other carbon atoms. To make it easier for you to see the patterns of carbon skeletons and functional groups, only carbon atoms and hydroxyl groups are shown in the following solutions. To complete these structural formulas, you need to supply enough hydrogen atoms to complete the tetravalence of each carbon.
There are three different carbon skeletons on which the -OH group can be placed: C C. C C Three alcohols are possible from the first carbon skeleton, four from the second carbon skeleton, and one from the third carbon skeleton. Following are structural formulas for the eight aldehydes with the molecular formula C6 H1 2O.
They are drawn starting with the aldehyde group and then attaching the remaining five carbons in a chain structure 1 , then four carbons in a chain and one carbon as a branch on the chain structures 2, 3, and 4 and finally three carbons in a chain and two carbons as branches structures 5, 6, 7, and 8. Following are structural formulas for the six ketones with the molecular formula C6 H1 2O.
They are drawn first with all combinations of one carbon to the left of the carbonyl group and four carbons to the right structures 1, 2, 3, and 4 and then with two carbons to the left and three carbons to the right structures 5 and 6. There are eight carboxylic acids of molecular formula C6 H1 2O2. They have the same carbon skeletons as the eight aldehydes of molecular formula C6 H1 2O shown in part b of this problem.
In place of the aldehyde group, substitute a carboxyl group. Start with unbranched carbon chains of all possible lengths, then add branching to complete the set.
Polar and Nonpolar Molecules Problem 1. Indicate which ones have a dipole moment and in what direction it is pointing. In the following diagrams, the C-H bond dipole moment has been left out because it is a nonpolar covalent bond. Which of the following is the most stable conformation of cisisopropylmethylcyclohexane? Instructions: - -Menthol is responsible for the characteristic flavor and taste of peppermint. The structure of - -menthol is shown below.
Use this information to answer the following question. Refer to instructions. On the chair template provided below, draw the two chair conformations that are in equilibrium for - -menthol.
On the templates provided, draw the two chair conformations that are in equilibrium for D-pinitol. Circle the most stable conformation. Instructions: MATCH a structure below to each of the following descriptions and place the letter corresponding to the structure in the blank. Instructions: Consider the conformations of 2-methylbutane shown below to answer the following questions. ANS: B Which of the following structures represents trans- 1,2-dimethylcyclohexane? ANS: B Which of the following is the most stable conformation of cis- 1-isopropylmethylcyclohexane?
ANS: C Which of the following is the most stable conformation of trans- 1-ethylmethylcyclohexane? ANS: D Instructions: - -Menthol is responsible for the characteristic flavor and taste of peppermint. On the chair template provided below, draw the two chair conformations that are in equilibrium for - -menthol. ANS: D-Pinitol is an interesting hexahydroxycyclohexane, whose structure is shown below.
ANS: This skeletal structure corresponds to the molecular formula: a. Refer to Instructions. ANS: C Draw a structure corresponding to the following name: cis sec -butylethylcyclopentane ANS: Instructions: Consider the conformations of 2-methylbutane shown below to answer the following questions.
Which of the structures represents the most stable conformation of 2-methylbutane? Which of the structures represents the least stable conformation of 2-methylbutane? There are two staggered conformations of 1,2-dichloroethane. Draw them. ANS: Draw the Newman projection for the specified conformations for rotation about the C-C bond of 1,2-dichloroethane. There are two eclipsed conformations of 1,2-dichloroethane. ANS: Alkene chemistry is dominated by what type of reaction? If more than one major organic product is expected, draw each one.
ANS: Predict the major organic product s in the reaction below. The following reaction is carried out in cyclohexane with the application of heat. Write the complete equation for the reaction below. ANS: What alkene would you use to prepare the following alkyl halide?
ANS: Only one alkene would be possible: Instructions: Predict the structure of the alkene you would use to prepare the folloiwng compounds. Predict: ANS: Predict. ANS: Predict.
ANS: The product s of the reaction when carried out in an organic solvent would be: a. The reaction mixture would contain a majority of which isomeric product? Which product is the Markovnikov product?
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