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All Textbook Solutions for Pushing Electrons

1. Hydrogen is a Group I element and each hydrogen will contribute I valence electron. Carbon is a Group (Roman numeral) element and each carbon will contribute ___ (number) electrons. Every oxygen atom in a compound will contribute valence electrons. Chloromethane has the molecular formula . Its skeleton is and the number of valence electrons may be determined as follows. There are three hydrogen atoms, each of which contributes 1 valence electron; the single carbon contributes 4 electrons; and the single chlorine atom contributes 7 electrons, making a total of 14 . A convenient tabular form of this calculation is Methanol has the molecular formula CH4O. Its skeleton is HHCOHH Each of four hydrogen atoms contributes ___ valence electron, the carbon atom contributes ___, and the oxygen atom contributes ___, making a total of ___. H=C=O=3. The skeleton of chloromethane is __________________ The central carbon atom is bonded to each of the other atoms by a shared electron pair (represented by a straight line, ___) giving Now, each hydrogen has two electrons and the carbon atom has eight. However, chlorine must be provided with unshared electrons (represented by pairs of dots, ) to complete its octet, thus 4. Methanol’s skeleton is Connecting all bonded atoms by means of an electron pair (single bond) gives ___________________ and completing the octet of oxygen with two pairs of dots gives ___________________ 5. The structure for chloromethane is It contains 14 valence electrons. The number of available valence electrons in chloromethane is ___. The Lewis structure is correct. 6EQ 7. Dimethyl ether No. of electrons in structure _____ No. of valence electrons _____ Structure is ___________ (correct, incorrect) Methylamine (CH5N) No. of electrons in structure _____ No. of valence electrons _____ Structure is ___________Methanethiol (CH4S) No. of electrons in structure _____ No. of valence electrons _____ Structure is ___________Methylal (C3H8O2) No. of electrons in structure _____ No. of valence electrons _____ Structure is ___________11EQ Adding electrons to the skeleton by making single bonds between all bonded atoms gives Each hydrogen atom now has a pair of electrons, but each carbon has only 6 electrons. Adding a pair of electrons to each carbon gives the trial structure The number of electrons in the trial structure is ____. Since this number exceeds the number of available valence electrons, the structure is incorrect. This is done by removing an unshared pair from each of two adjacent atoms and adding one electron pair as a second bond between the atoms. Each such operation reduces the number of electrons in the trial structure by two. Removing the unshared pairs of electrons on the carbon atoms and adding a second carbon–carbon bond gives a structure in which there are four electrons involved in a double bond between the carbon atoms. The trial structure now contains electrons ____ and is correct. 14EQ 15EQ 16EQ The skeleton of acetyl chloride is . Write the best Lewis structure for acetyl chloride by following the procedure in the previous problem. Three constitutional isomers exist for the formula . One is pentane. Draw the other two. A number of constitutional isomers exist for the formula . One is acetone, seen in problem 33. Propose structures for five other isomers. Using the method outlined above, derive the structures for the following compounds. The unbounded skeletons are provided. Propyne (C3H4) No. of electrons in structure _____ No. of valence electrons _____ Structure is ___________ (correct, incorrect) (Second trial) (third trial) ______ ______ ___________________ ______ _____________________ ______ Structure is _____________ Structure is __________________21EQ 22EQ 23EQ 24EQ The skeleton of benzyldimethylamine is The number of available valence electrons is as follows: from the benzyl group, ___; from each of two methyl groups, ___; and, from the nitrogen atom, ___, for a total of ___. Filling in the skeleton with single bonds gives ____________________________ and adding the appropriate unshared pair gives ____________________________ The number of electrons in the functional group of this structure is ___, and the structure is (correct, incorrect). The skeleton is benzaldoxime is The number of valence electrons is as follows: from the phenyl group. ___; from each of two hydrogens, ___; from the carbon atom, ___; from the nitrogen atom, ___; and from the oxygen atom, ___, for a total of ___. Filling in the skeleton with single bonds and adding the appropriate unshared pairs gives _____________________ The number of electrons in the functional group of this trial structure is ___, which is ___ too many. Removing an unshared pair from carbon and nitrogen and adding a second bond between them gives _____________________ in which carbon, nitrogen, and oxygen have their customary valences of ___, ___, and respectively. The alternative structure with a double bond between nitrogen and oxygen is _____________________ This structure is not acceptable because it requires carbon and oxygen to exhibit the unfamiliar valences of ___ and ___. Derive Lewis structures for the compounds below.27EQ Derive Lewis structures for the compounds below. Furan 29EQ Derive Lewis structures for the compounds below. Azobenzene 31EQ Derive Lewis structures for the compounds below. Ethyl crotonate (For the purpose of drawing Lewis structures of compounds with more than one functional group, each group can be treated independently.)The Lewis structure of acetone is Circling the carbonyl carbon, i.e., the carbon atom attached to oxygen, and its octet gives Circling the oxygen atom and its octet gives Thus, atoms share electrons in making bonds, and a pair of electrons may be included in the octet of two different atoms. When computing the formal charge on an atom, the number of electrons that belong to that atom is compared with the number of electrons the atom would have in the unbonded and neutral state. If the two numbers are the same, the formal charge on the atom is zero. In a Lewis structure both electrons in an unshared pair belong to the atom, and one of every pair of shared (bonding) electrons belongs to the atom.Chloromethane has the Lewis structure _______________________________ The carbon atom is sharing 4 electron pairs. In each shared pair the carbon atom “owns” 1 electron. The number of electrons that “belong” to carbon is ___. Carbon, being a Group ___ element would have 4 , outer shell electrons in the unbonded, neutral state. Therefore, the carbon atom in chloromethane has a formal charge of zero. In the Lewis structure for chloromethane, the chlorine atom is sharing _____ electron pair and “owns” _____ of those electrons. Also, the chlorine atom possesses two electrons from each of _____ unshared pairs. The total number of electrons that belong to chlorine is 7 . Chlorine is a Group ____ element. The formal charge on chlorine in chloromethane is ____. 36EQ The oxygen atom in acetone possesses ____ unshared pairs and ____ shared pairs of electrons. The number of electrons that belong to oxygen is ____. Oxygen is a Group ____ element. The formal charge on oxygen in acetone is ____.Nitrobenzene has the skeleton The number of available valence electrons is: from the phenyl group, ____; from the nitrogen atom, ____; and from each of two oxygen atoms, ____, for a total of ____. Filling in all single bonds and adding the appropriate unshared pairs gives ______________________________ The functional group of this structure contains ____ electrons. Therefore, unshared pairs are removed from the nitrogen atom and one of the oxygen atoms; a double bond is added giving ______________________________ which has the correct number of electrons. In this structure the nitrogen atom is sharing ____ pairs of electrons. From each shared pair the nitrogen owns ____ electron for a total of ____. Nitrogen is a Group ____ element and would have outer shell electrons in the unbonded, neutral state. Since the nitrogen atom in nitrobenzene has one fewer electron than it would in the neutral state, it has a formal charge of +1 . This is added to the Lewis structure as + giving Since an electron is negatively charged, a shortage of one electron results in a single positive charge on an atom. Conversely, an excess of one electron results in a single negative charge on an atom. When an atom in a Lewis structure “owns” two fewer electrons than it would have in the neutral, unbonded state, it is denoted by ++ or +2 and, conversely, or – 2. 39EQ Compute and add on the formal charges I these Lewis structures. Pyridine N-oxide 41EQ 42EQ 43EQ 44EQ 45EQ 46EQ 47EQ Compute and add on the formal charges in these Lewis structures. Trimethylamine oxide 49EQ 50EQ The n-propyl cation can be formed from a molecule such as When the C–Cl bond is broken so that both electrons leave with Cl, the fragments formed are The carbon atom that had been attached to Cl is now sharing ____ electron pairs. In each shared pair the carbon atom owns ____ electron. The number of electrons that belong to carbon is ____. The formal charge on the carbon atom is ____. The correct Lewis structure for the n-propyl cation is 52EQ 53EQ Methanol, CH3OH, is a compound in which the formal charge on all the atoms is zero. Consider what results when a proton, H+, becomes bonded to methanol by way of one of the unshared electron pairs on the oxygen atom, i.e., This time we use the curved arrow to signify bond making. Now a pair of unshared electrons on oxygen is pushed toward the region between the oxygen atom and the hydrogen ion. It becomes an OH covalent bond. In the resulting structure the oxygen atom owns one electron from each of ____ shared pairs and two electrons from ____ unshared pair. The total number of electrons that belong to the oxygen atom is ____. Oxygen is a Group ____ element. Since the number of electrons that the oxygen atom owns in this structure is one fewer that it would have in the neutral, unbonded state, the charge on oxygen is ____. The correct Lewis structure for the conjugate acid of methanol is _________________________ Once again charge is conserved. Joining a neutral molecule and a cation must yield a cation. Again the formal charge distribution on the resulting ion is predictable from the arrow. Electrons are pushed away from the oxygen atom, leaving it with a positive charge. Electrons are pushed toward the hydrogen ion, neutralizing its erstwhile positive charg.When a proton becomes bonded to diethyl ether, by way of one of the unshared electron pairs on the oxygen atom, the result is In this structure the oxygen atom owns one electron from each of ____ shared pairs and two electrons from ____ unshared pair. The total number of electrons that belong to oxygen is ____. The formal charge on oxygen is ____. The correct Lewis structure for the conjugate acid of diethyl ether is ___________________________ Tetrahydrofuran has the structure When a proton becomes bonded to tetrahydrofuran the result is ___________________________ In this structure the oxygen atom owns one electron from each of ____ shared pairs and two electrons from ____ unshared pair. The formal charge on the oxygen atom is ____. The Lewis structure for the conjugate acid of tetrahydrofuran is ___________________________ 57EQ 58EQ The structure of pyridine is When a proton becomes bonded to the nitrogen atom by way of its unshared electron pair, the result is _____________________________ The carbon atom owns one electron from each of shared pairs and two electrons from unshared pair. The number of electrons that belong to carbon is . Carbon is a Group element. Since the carbon atom has one more electron than it would in the neutral, unbonded state, it has a formal charge of –1 . The Lewis structure for the methyl anion is The lithium fragment must have a formal charge of The n-butyl anion can be formed from When the CLi bond is broken so that both electrons remain with the carbon atom, the result is +Li+The isobutyl anion can be formed from When the CNa bond is broken so that both electrons remain with the carbon atom, the result is +Na+63EQ Ethanol, , is a compound in which the formal charge on all the atoms is zero. Under certain conditions the bond can be broken so that both electrons remain with the oxygen atom. The products are In this structure the oxygen owns one electron from shared pair and two electrons from each of unshared pairs. The total number of electrons belonging to oxygen is Oxygen is a Group element. The formal charge on the oxygen atom is . The correct Lewis structure for the ethoxide ion is Note that the other fragment, the proton, leaves with a formal charge of +1. The loss of a proton attached to the oxygen atom of t-butyl alcohol results in A very strong base can remove a proton from methylamine: 67EQ 68EQ 69EQ The homolysis of the OO bond in diacetyl peroxide gives two acetyl radicals with the Lewis structure The formal charge on this radical is .71EQ 72EQ 73EQ 74EQ 75EQ Heterolytic cleavage of the CO bond to yield a carbocation.77EQ 78EQ 79EQ 80EQ One Lewis structure for the 2-butenyl cation is CH3CH=CHC+H2. A new resonance structure can be generated by pushing the pi electrons to the receptor.2EQ One structure for the conjugate acid of acetone is The electrons in the carbon–oxygen double bond are pushable electrons, and the atom is a receptor. Similarly, a resonance structure for the conjugate acid of 2-butanone can be written. Thus, It is crucial that these arrows indicate precisely what electrons are being pushed where. Therefore, after doing these exercises check the answers carefully to see that you have inserted the arrows properly.5EQ Pairs of unshared electrons can be pushed. One Lewis structure for the methoxy-methyl cation is . The structure contains a pair of pushable electrons, namely, the unshared electrons on the atom. The structure also contains a positively charged atom that can act as a . A second resonance structure can be generated by pushing the unshared electrons to the receptor. Thus, It is not possible to push electrons toward the other carbon, because it is not a receptor. If you tried to push electrons to this carbon, you would generate a pentavalent carbon, which is not possible. One structure for the acetoxonium ion is Clearly, the receptor is the positively charged atom. However, there are available two different kinds of pushable electrons, namely, pi electrons in the carbon-oxygen double bond or unshared electrons on the oxygen atom. The proper choice in this case is dictated by the following considerations. In structure (3) the carbon atom possesses a formal positive charge and also has only six electrons in its outer shell (it lacks a stable octet of electrons). If the pi electrons are pushed, namely, structure (4) is obtained in which the oxygen possesses a formal positive charge but, in addition, both the oxygen and the carbon atom lack a stable octet. This structure will be very unstable. Structure (4), despite the fact that it can be generated by properly pushing electrons, is not included in the resonance hybrid for the acetoxonium ion. If, on the other hand, the unshared electrons on the oxygen are pushed, a more acceptable structure is obtained, namely, In structure (5) the oxygen atom possesses a formal positive charge, and both the carbon and the oxygen atom have a stable octet. Structure (5) is included in the resonance hybrid.8EQ There are no important resonance structures for the isopropyl cation CH3C+HCH3 because there are no in the structure. There are no important resonance structures for dimethyl ether CH3OCH3 because, although there are pushable electrons on the oxygen atom, there is no . There are no important resonance structures for the 5-pentenyl cation H2C=CHCH2CH2C+H2 because the pushable electrons and the receptor are separated by two methylene groups. Thus, the electrons have no way to get to the receptor.10EQ 11EQ 12EQ 13EQ 14EQ 15EQ 16EQ 17EQ The cyclohexane carboxylate anion has a Lewis structure Pushing a pair of unshared electrons away from the negatively charged oxygen atom and, at the same time, pushing a pair of pi electrons toward the other oxygen will generate a second resonance structure. Thus,One Lewis structure for the enolate anion of acetaldehyde is Pushing the pair of unshared electrons on the carbon atom away from the center of negative charge and pushing the pi electrons of the carbon-oxygen double bond to the oxygen atom generates a second resonance structure. Thus, 20EQ 21EQ 22EQ 23EQ 24EQ 25EQ 26EQ 27EQ 28EQ 29EQ 30EQ 31EQ 32EQ 33EQ 34EQ 35EQ 36EQ 37EQ 38EQ 39EQ 40EQ 41EQ 42EQ 43EQ 44EQ 45EQ 46EQ 47EQ 48EQ 49EQ 50EQ 51EQ 52EQ 53EQ 54EQ 55EQ 56EQ 57EQ 58EQ 59EQ 60EQ 61EQ 62EQ 63EQ 64EQ 1EQ 2EQ 3EQ 4EQ 5EQ 6EQ Here are some exercises in sigma bond breaking. Supply the arrows and the products. Remember that charge must be conserved. That is, if a neutral molecule is dissociated, the algebraic sum of the charges on the products must equal zero. If a positively charged ion is dissociated, the algebraic sum of the charges on the products must equal +1.8EQ 9EQ 10EQ 11EQ 12EQ 13EQ 14EQ 15EQ 16EQ 17EQ 18EQ 19EQ 20EQ 21EQ 22EQ 23EQ 24EQ 25EQ 26EQ 27EQ 28EQ 29EQ 30EQ The reaction just described is reversible. Deprotonation of the conjugate acid of an organic base by water provides another example of simultaneous making and breaking of sigma bonds. Thus, in the deprotonation of anilinium ion by water, the base is water, which has unshared electrons on the ________ atom. The acid is ________ ion. A pair of ________ electrons on the oxygen atom of water is pushed toward the ________ atom. Simultaneously, the pair of ________ electrons between the hydrogen and ________ atom of the anilinium ion is pushed toward the ________ atom. Thus, the oxygen- ________ sigma bond is made and a hydrogen- ________ sigma bond is broken. The nitrogen atom, which possessed a positive charge, is now ________, and the oxygen atom, which was neutral, now possesses a formal ________ charge.32EQ 33EQ 34EQ 35EQ 36EQ 37EQ 38EQ 39EQ 40EQ 41EQ 42EQ 43EQ 44EQ 45EQ 46EQ 47EQ 48EQ 49EQ 50EQ 51EQ 52EQ 53EQ 54EQ 55EQ 56EQ 57EQ 58EQ 59EQ 60EQ 61EQ 62EQ 63EQ 64EQ 65EQ 66EQ 67EQ 68EQ 69EQ 70EQ 71EQ 72EQ 73EQ 74EQ 75EQ 76EQ 77EQ 78EQ 79EQ 80EQ 81EQ 82EQ 83EQ 84EQ 85EQ 86EQ 87EQ 88EQ 89EQ 90EQ 91EQ 92EQ 93EQ 94EQ 95EQ 96EQ 97EQ 98EQ 99EQ 100EQ 101EQ 102EQ 103EQ 104EQ 105EQ 106EQ 107EQ 108EQ 109EQ 110EQ 111EQ 112EQ 113EQ 114EQ 115EQ 116EQ 117EQ 118EQ 119EQ 120EQ 121EQ 122EQ 123EQ 124EQ 125EQ 126EQ 127EQ 128EQ 129EQ 130EQ 131EQ 132EQ 133EQ 134EQ 135EQ 136EQ 137EQ 138EQ 139EQ 140EQ 141EQ 142EQ 143EQ 144EQ 145EQ 146EQ 147EQ 148EQ 149EQ 150EQ 151EQ 152EQ 153EQ 1EQ 2EQ 3EQ

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