Practice Drawing Lewis Structures Organic Chemistry

1.E: Structure and Bonding in Organic Molecules (Exercises)

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These are homework exercises to accompany Chapter 1 of Vollhardt and Schore's "Organic Chemistry" Textmap.

1.8: Hybrid Orbitals: Bonding in Complex Molecules and Practice Problems

Q25

Draw the Lewis Structures of the following molecules, and if necessary assign charges.

BrF
ClCN
SOCl₂
CH₃CH₂NH₂
CH₃CH₂OCH₂CH₃
(CH₃)₂NNH₂
CH₃NCNCH₃
CH₃COOH
N₂O₄

Q26

In problem 25 identify the polar covalent bonds with δ+ and δ- on the atoms according to their electronegativity.

Q27

Draw the Lewis Structures of the following species, and if necessary assign charges.

H₃O⁺
NH₄ ⁺
CH₃CH₂ ^-
CH₃CH₂ ⁺
CH₃CH₂
CH₃CH
H^-
CH₃COO^-

Q28

Add charges to the following Lewis Structures, if neutral do not add any charges.

(a) (b) (c)

(d) (e) (f)

Q29

Sulfuric Acid is a known strong acid, it's deprotonated form SO₄ ^2-, has several resonance structures. Draw the remaining two using electron pushing arrows.

Acetic acid is a common household acid once deprotonated it forms an anion with resonance structures possible. Draw these with electron-pushing arrows.

Q30

In problem 25 several of the species have possible resonance structures. Draw the electron-pushing arrows and resulting resonance structures for 25.c, 25.h, 25.i. Identify the major contributor, and give reasons.

Q31

Draw all possible resonance structures of the following species.

CH₂COCH₃ ^-
CH₂N₂
CO
CH₂ON(CH₃)₂
SCN^-
NO₃ ^2-
O₃

Q32

Draw all possible resonance structures of the following. Based on the resonance structures, which carbonyl carbon, marked in red, would be the most electron deficient?

Q33

Draw the Lewis structure for each compound.

Fluorine and the Fluoride anion
GeH₄ and GeO
H₂O₂ (Hydrogen Peroxide)
Phosphine (PH₃) and Ammonia (NH₃)

Q34

Draw the molecular orbital diagrams of H₂ ⁺, He₂ ²⁺, and He₂, and indicate the bond order. Compare and contrast the molecular orbital diagrams of He₂ ⁺ and He_2,which is more favorable?

Q35

Predict the geometry for each red-highlighted atom and give the hybridization of that center.

(a) (b) (c) (d) (e) (f)

Q36

In problem 35 indicate the hybrid orbitals' interactions (sp³, sp², sp) and the atomic orbitals used (s, p).

Q37

Draw the hybridized orbitals for each of the highlighted carbons in problem 35.

Q38

Determine the hybridization of each carbon based on the shape of each molecule.

(a) CH₃Br (b) CH₃CH₂OH (c) H₃CHCCH (d) CH₃ ^- (e) NH₄ ⁺

Q39

Draw the Kekule structures for the following molecules, an example is given below.

Example. CH₃CH₂CH₂CH₃

(a) CH₃CH₂OH (b) (CH₃)₂CH₂OH (c) (CH₃)₂NCHO (d) CH₂ClCH₂Cl (e) HOOCCH₂COOH

Q40

Draw the Kekule structures for the following molecules.

(a) (b) (c) (d) (e) (f)

Q41

Convert the following structures into condensed formulas

(a) (b) (c)

Q42

Give the condensed structures for the following Kekulé structures.

(a) (b) (c) (d) (e) (f)

Q43

In the previous problem, 42, draw the Kekulé structures as bond line structures.

Q44

Draw the following condensed formulas as hashed-wedged line structures.

CH₃CH₂OCH₂CN
CH₂Cl₂
(CH3CH₂)₃P
CH₂CHCH₂SiH₃

Q45

Draw all possible constitutional isomers, in bond-line form, ofIthe following molecules.

(a) C₃H₈ (b) C₆H₁₄ (c) C₄H₁₀

Q46

Draw line-bond structures of the following pairs of species and include all charges. Are any of the pairs resonance forms of the other species in the pair? If not explain the difference

(a) H₃CCHCCH₂and H₂CCCHCH₃ (b) HCN and H₃CNH₂ (c) CH₃CH₂COH and CH₃CH₂CHOH

Q47

The following boron-halogen compounds have resonance structures possible. Draw them and in each compound indicate which structure is favored (hint: the octet rule takes precedent). Also, compare the relative energy levels of the three favored resonance structures, which is favored/lowest in energy?

(a) (CH₃)₂BF (b) (CH₃)₂BCl (c) (CH₃)₂BBr

Q48

Using the formula for the degrees of each angle, [(n-2)*180]/n, where n equals the number of sides, determine the angles in each of the following shapes. If the given shapes were considered bond-line structures, order them from highest to lowest favorability based on the angles (The best overlap in sp³ hybridized carbon is at 104.5^o). Compare this answer to the following ring strain energies of each: cyclopropane (triangle)- 27.5 kcal/mol cyclobutane (square)- 26.3 kcal/mol cyclopentane (pentagon)- 6.2 kcal/mol cyclohexane (hexagon)-0.1 kcal/mol.

Q49

Give the hybridization orbital that the lone pair of each of the following species belongs in. Also, given that the 2p orbitals are higher in energy than the 2s orbitals, order the species from lowest to highest energy according to the s/p character. (sp³= 1/4s and 3/4p) (sp²= 1/3s and 2/3p) (sp= 1/2s and 1/2p).

(a) CH₃CH₂CH₂ ^- (b) CH₃CHCH^-(c) CH₃CC^-

Based on their relative energy, which species would it be most favorable to undergo the following dissociation (HA represents the neutral hydrocarbon)?

Q50

In each set which of the following positively polarized carbon bond atoms is the most electropositive?

(a) (i)CHCl₃ (ii) CHCl₂ (iii) CCl₄

(b) (i) ClCH₂OCH₂Cl (ii) CH₃OCH₂Cl

Q51

The following molecule is a common insecticide, cyfluthrin. It is known to be very effective against invertibrates, but it is not nearly as toxic to mammals. In the following highlight one (a) highly polarized covalent single bond (b) one highly polarized double bond (c) one sp²hybridized carbon atom (d) one sp hybridized carbon atom and (e) one conjugated system.

Q52

In the following reaction does the hybridization of the highlighted carbon atom change? Give the hybridization before and after.

Q53

A combustion elemental analysis finds that a compound is made up of 84% carbon and 16% hydrogen (C=12.0, H=1.0, O=16.0). Which of the following could be the compound tested?

(a) CH₄O (b) C₆H₁₄O₂ (c) C₁₄H₂₂ (d) C₇H₁₆

Q54

As drawn, which of the following is a correct formal charge assignment?

(a) +1 on F (b) -1 on N (c) +2 on N (d) -1 on B (e) +1 on B

Q55

The highlighted bond is formed by the interaction between what two orbitals?

Q56

Which highlighted carbon center has the smallest bond angles?

(a) CO₂ (b) CH₂Cl₂ (c) HCCH (d) CO

Q57

Which pair of structures are resonance hybrids?

(a) (b) (c) (d)

Solutions

S25.

(a) (b) (c) (d)

(e) (f) The lone pair is shared between the two N atoms.

(g) (h) (i)

S26

(a) (b) (c) (d)

(e) (f) (g)

(h) (i)

S27

(a) (b) (c) (d)

(e) (f) (g) (h)

S28

(a) (b) (c)

(d) (e) neutral (f)

S29

Sulfuric acid has two more possible resonance structures. Later, resonance/charge distribution will be mentioned; the resonance structures possible for the deprotonated sulfuric acid are one reason for why it is so strongly acidic.

Acetic acid has one additional resonance structure possible, again a reason for its moderate acidity.

S30

(h) Again, the resonance structure with no charges is the major contributor.

(i) Each resonance structure is equivalent, so there is no "major contributor".

S31

(a) (b)

(e) (f)

(g)

S32

The first structure has two possible resonance structures, further distributing electrons more among the carbonyl carbon. However, the second structure only has one possible resonance structure. This distributes electrons to a lesser extent among the carbonyl carbon. Thus, in the second structure, the aldehyde, the carbon marked in red is more electron deficient than in the carboxylic acid.

S34

H₂ ⁺ has a bond order of 1/2, He₂ ⁺ has a bond order of 1/2, and He₂ has a bond order of 0. The zero bond order indicates that He₂ is not favorable.

S35

dichloroethane: The highlighted carbon center is tetrahedral with a hybridization of sp³; it is bonded to four other groups.
acetic acid: The carbon center is trigonal planar with a hybridization of sp²; it is bonded to three groups and has no lone pairs.
propene: The carbon center is trigonal planar with a hybridization sp²; it is bonded to three groups and has no lone pairs.
acetylene: The carbon center is linear with a hybridization of sp; it is bonded to two groups with no lone pairs.
triethylamine: The nitrogen center is tetrahedral with a hybridization of sp³; it is bonded to three groups with one lone pair.
triethylammonium ion: The nitrogen center is tetrahedral with a hybridization of sp³; it is bonded to four groups with no lone pairs. All are sigma bonds.

S36

The highlighted carbon forms bonds with one chlorine, two hydrogen atoms, and one other carbon. Each of these atoms connect to the central carbon forming sp³-sp³ hybridized orbitals. The carbon and hydrogen both use electrons from the s orbital, while the bonding electrons from the chlorine are from the p orbitals. All are sigma bonds.
The central carbon is sp² hybridized, and it is bonded to one carbon and two oxygen atoms. The bond to the carbon involves an sp³-sp² overlap with the s orbital on the adjacent carbon involved in bonding, sigma bonding. The carbon oxygen double bond is slightly more complicated. The oxygen is less likely to undergo orbital hybridization. So, the C(sp²)-O(p) sigma bond is formed along with the interaction of the second oxygen p orbital with the carbon p orbital to yield a pi bond. The carbon oxygen bond is similar forming a sigma bond through the C(sp²)-O(p) interaction.
The central carbon is sp₂ hybridized and bonded to two other carbons and one hydrogen atom. The bond between the hydrogen atom involves the C(sp²)-H(s) interaction yielding a sigma bond. The C=C double bond involves the C(sp²)-C(sp²) interaction creating a pi bond. The C-C single bond is a sigma bond involving the C(sp²)-C(sp³) hybrid orbitals.
The highlighted carbon is bonded to one hydrogen and one carbon atom. The C-H single bond is made up of the C(sp)-H(s) interaction yielding a sigma bond. The C-C triple bond is made up of the C(sp)-C(sp) hybrid orbitals yielding a sigma bond, s orbitals, and two pi bonds, using the two p orbitals.
The highlighted nitrogen is bonded to three identical carbon groups and has one lone pair. Each of the carbon atoms is sp³ hybridized. So, a N(sp³)-C(sp³) yields a sigma bond.
The highlighted nitrogen is bonded to three identical carbon atoms and one hydrogen. There are no lone pairs on the nitrogen, and it is sp³ hybridized. The N-C bonds are the same as in part (e), N(sp³)-C(sp³) yielding a sigma bond. The N-H bond is formed from the N(sp³)-H(s) interaction, also a sigma bond.

S37

(a) (b) (c)

(d) (e) (f)

S38

Each carbon's hybridization is listed from left to right in the chemical formula.

tetrahedral, sp³
tetrahedral, sp³ ; tetrahedral, sp³
tetrahedral, sp³ ; linear, sp ; linear, sp
tetrahedral, sp³
tetrahedral, sp³

S40

(a) (b) (c) (d)

(e) (f)

S41

(a) (CH₃)₂CHNH₂ (b) CClH₂CH₂Cl (c) CH₃I

S42

(a) CH₃CH₂NH₂ (b) CH₃ C=O CH₃ (c) CH₃SCH₂C₂OH (d) CF₃CHOHCF₃ (e) CH₂CCH₂ (f) CH₂CH C=O OCH₃

S46

(a) the two structures are identical (b) These are not resonance structures; there additional H atoms. (c) These are not resonance structures; there are three additional H atoms.

S47

In each case the double bonded resonance structure is preferred, due to the octet rule being fulfilled.

The lowest energy structure can be determined by thinking about the electronegativity of each halogen atom. The higher the electronegativity, the less favorable a positive charge is. The order of the electronegativity is as follows, F>Cl>Br, and so the reverse is the order of favorability. (CH₃)₂BF (b) (CH₃)₂BCl (c) (CH₃)₂BBr

S48

Triangle- 60^o Square- 90^o Pentagon- 108^o Hexagon- 120^o

This would lead to the expectation that the order of favorability is Pentagon>Square>Hexagon>Triangle. However, the order differs and is in fact Hexagon>Pentagon>Square>Triangle.

In reality this is due to different conformations of the shapes, so that the hexagon and pentagon are no longer flat/planar. This will be covered in more detail later on.

S49

CH₃CH₂CH₂ ^- - sp³ CH₃CHCH^-- sp² CH₃CC^- - sp

The order of energy is as follows, or increasing p character. CH₃CC^-<CH₃CHCH^-<CH₃CH₂CH₂ ^-

The reaction would be most favorable if the A^- is lowest in energy, so, the order of favorability is the reverse of above. CH₃CH₂CH₂ ^-<CH₃CHCH^-<CH₃CC^-

S50

(a) CCl₄

(b) ClCH₂OCH₂Cl

(c) (CH₃)₃C⁺ The additional hydrocarbons on the ethyl group do marginally hyperconjugate and donate electrons to the center carbon, making it less electropositive.

S51

(a) In this structure any sp³ carbon connected to a heteroatom would be a polarized bond.

(b) In this structure this is the only highly polarized double bond.

(d) This is the only sp hybridized carbon.

(e) These two benzene rings are conjugated system.

S52

The hybridization changes from sp² to sp³.

S53

(d) 84/12=7 Carbon and 14/1=14 Hydrogen

S54

(d) is correct, boron typically has 3 valence electrons. It currently is bonded to four different atoms; it has an octet. 8/2=4 electrons, and 3-4=-1 formal charge.

S55

The bond is formed by the interaction between the s orbital of the hydrogen and the sp² orbital of the carbon.

S56

(b) This is the only sp³ hybridized highlighted carbon, characterized by bond angles of 104.5^o. Which is less than both sp² and sp, 120^o and 180^o respectively.

S57

(c) The number of electrons is the same and all atoms are in the same locations. Placement of charges is important to look at; in the pairs with no charges present then some atoms were moved around.