Concept explainers
a) Acrylonitrile, C3H3N, which contains a carbon-carbon double bond and a carbon- nitrogen triple bond.
Interpretation:
Structure of acrylonitrile, C3H3N, with a carbon-carbon double bond and a carbon- nitrogen triple bond along with lone pair of electrons is to be drawn.
Concept introduction:
A covalent bond is formed by mutual sharing of two electrons between the atoms, each atom giving one electron for sharing. Such a covalent bond, that is, a pair of shared electrons is represented as a line between the atoms, for example as A-B. The number of covalent bonds formed by an atom depends on the number of electrons that the atom requires for making the octet in its valence shell. Valence electrons that are not used for bonding are called lone-pair of electrons or nonbonding electrons.
Answer to Problem 33AP
Structure of acrylonitrile, C3H3N.
Explanation of Solution
Structure of acrylonitrile, C3H3N, with a carbon-carbon double bond and a carbon- nitrogen triple bond along with lone pair of electrons is required. Carbon with four valence electrons can form four covalent bonds. Nitrogen with five valence electrons can form three covalent bonds while hydrogen with one valence electron an form one covalent bond. There are three carbons in acrylonitrile molecule. Two carbons are joined by a double bond and the third carbon is involved in forming triple bond with nitrogen and a single bond with second carbon. Out of the five valence electrons available, nitrogen has utilized only three electrons in forming the triple bond. Therefore a lone pair of electron remains on nitrogen. The hydrogen atoms are distributed on different carbon atoms depending upon their valence requirements giving the structure as
Structure of acrylonitrile, C3H3N, with a carbon-carbon double bond and a carbon- nitrogen triple bond along with lone pair of electrons.
b) Ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbon atoms
Interpretation:
Structure of ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbon atoms along with lone pair of electrons is to be drawn.
Concept introduction:
A covalent bond is formed by mutual sharing of two electrons between the atoms, each atom giving one electron for sharing. Such a covalent bond, that is, a pair of shared electrons is represented as a line between the atoms, for example as A-B. The number of covalent bonds formed by an atom depends on the number of electrons that the atom requires for making the octet in its valence shell. Valence electrons that are not used for bonding are called lone-pair of electrons or nonbonding electrons.
Answer to Problem 33AP
Structure of ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbon atoms along with lone pair of electrons is
Explanation of Solution
Structure of ethyl methyl ether, C3H8O, with an oxygen atom bonded to two carbon atoms along with lone pair of electrons is required. Carbon with four valence electrons can form four covalent bonds. Oxygen with six valence electrons can form two covalent bonds while hydrogen with one valence electron an form one covalent bond. There are three carbons in ethyl methyl ether molecule. Since the oxygen atom is bonded to two carbon atoms, the third carbon must be attached to any one of the carbons bonded to oxygen atom. Hence the skeleton structure of ethyl methyl ether will be C-C-O-C. Out of the six valence electrons available, oxygen has utilized only two electrons in forming the bonds. Therefore two lone pairs of electrons remain on oxygen atom. The hydrogen atoms are distributed on different carbon atoms depending upon their valence requirements giving the structure as H3C-CH2-O-CH3.
Structure of ethyl methyl ether, C3H8O, which contains an oxygen atom bonded to two carbon atoms along with lone pair of electrons is
c) Butane, C4H10, which contains a chain of four carbon atoms.
Interpretation:
Structure of butane, C4H10, which contains a chain of four carbon atoms along with lone pair of electrons is to be drawn.
Concept introduction:
A covalent bond is formed by mutual sharing of two electrons between the atoms, each atom giving one electron for sharing. Such a covalent bond, that is, a pair of shared electrons is represented as a line between the atoms, for example as A-B. The number of covalent bonds formed by an atom depends on the number of electrons that the atom requires for making the octet in its valence shell. Valence electrons that are not used for bonding are called lone-pair of electrons or nonbonding electrons.
Answer to Problem 33AP
Structure of butane, C4H10.
Explanation of Solution
Structure of butane, C4H10, which contains a chain of four carbon atoms along with lone pair of electrons is
d) Cyclohexene, C6H10, which contains a ring of six carbon atoms and one carbon-carbon double bond.
Interpretation:
Structure of cyclohexene, C6H10, which contains a ring of six carbon atoms and one carbon-carbon double bond along with lone pair of electrons is to be drawn.
Concept introduction:
A covalent bond is formed by mutual sharing of two electrons between the atoms, each atom giving one electron for sharing. Such a covalent bond, that is, a pair of shared electrons is represented as a line between the atoms, for example as A-B. The number of covalent bonds formed by an atom depends on the number of electrons that the atom requires for making the octet in its valence shell. Valence electrons that are not used for bonding are called lone-pair of electrons or nonbonding electrons.
Answer to Problem 33AP
Structure of cyclohexene, C6H10, which contains a ring of six carbon atoms and one carbon-carbon double bond.
Explanation of Solution
Structure of cyclohexene, C6H10, with a ring of six carbon atoms and one carbon-carbon double bond along with lone pair of electrons is required. Carbon with four valence electrons can form four covalent bonds while hydrogen with one valence electron an form only one covalent bond. The carbons are to be arranged in the form a ring with two carbons attached through a double bond and two single bonds and others through four by single bonds. The ten hydrogen atoms are distributed on these six carbons satisfying their valence requirements. Thus no lone pair of electrons remains on either carbon or hydrogen. The structure of cyclohexene is
Want to see more full solutions like this?
Chapter 1 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- What is the stepwise mechanism for this reaction?arrow_forward32. Consider a two-state system in which the low energy level is 300 J mol 1 and the higher energy level is 800 J mol 1, and the temperature is 300 K. Find the population of each level. Hint: Pay attention to your units. A. What is the partition function for this system? B. What are the populations of each level? Now instead, consider a system with energy levels of 0 J mol C. Now what is the partition function? D. And what are the populations of the two levels? E. Finally, repeat the second calculation at 500 K. and 500 J mol 1 at 300 K. F. What do you notice about the populations as you increase the temperature? At what temperature would you expect the states to have equal populations?arrow_forward30. We will derive the forms of the molecular partition functions for atoms and molecules shortly in class, but the partition function that describes the translational and rotational motion of a homonuclear diatomic molecule is given by Itrans (V,T) = = 2πmkBT h² V grot (T) 4π²IKBT h² Where h is Planck's constant and I is molecular moment of inertia. The overall partition function is qmolec Qtrans qrot. Find the energy, enthalpy, entropy, and Helmholtz free energy for the translational and rotational modes of 1 mole of oxygen molecules and 1 mole of iodine molecules at 50 K and at 300 K and with a volume of 1 m³. Here is some useful data: Moment of inertia: I2 I 7.46 x 10- 45 kg m² 2 O2 I 1.91 x 101 -46 kg m²arrow_forward
- K for each reaction step. Be sure to account for all bond-breaking and bond-making steps. HI HaC Drawing Arrows! H3C OCH3 H 4 59°F Mostly sunny H CH3 HO O CH3 'C' CH3 Select to Add Arrows CH3 1 L H&C. OCH3 H H H H Select to Add Arrows Q Search Problem 30 of 20 H. H3C + :0: H CH3 CH3 20 H2C Undo Reset Done DELLarrow_forwardDraw the principal organic product of the following reaction.arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided structures, draw the curved arrows that epict the mechanistic steps for the proton transfer between a hydronium ion and a pi bond. Draw any missing organic structures in the empty boxes. Be sure to account for all lone-pairs and charges as well as bond-breaking and bond-making steps. 2 56°F Mostly cloudy F1 Drawing Arrows > Q Search F2 F3 F4 ▷11 H. H : CI: H + Undo Reset Done DELLarrow_forward
- Calculate the chemical shifts in 13C and 1H NMR for 4-chloropropiophenone ? Write structure and label hydrogens and carbons. Draw out the benzene ring structure when doing itarrow_forward1) Calculate the longest and shortest wavelengths in the Lyman and Paschen series. 2) Calculate the ionization energy of He* and L2+ ions in their ground states. 3) Calculate the kinetic energy of the electron emitted upon irradiation of a H-atom in ground state by a 50-nm radiation.arrow_forwardCalculate the ionization energy of He+ and Li²+ ions in their ground states. Thannnxxxxx sirrr Ahehehehehejh27278283-4;*; shebehebbw $+$;$-;$-28283773838 hahhehdvaarrow_forward
- Plleeaasseee solllveeee question 3 andd thankss sirr, don't solve it by AI plleeaasseee don't use AIarrow_forwardCalculate the chemical shifts in 13C and 1H NMR for 4-chloropropiophenone ? Write structure and label hydrogens and carbonsarrow_forwardPlease sirrr soollveee these parts pleaseeee and thank youuuuuarrow_forward
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning