
Concept explainers
Interpretation:
The hybrid orbitals of carbon atoms in the given molecules are to be determined.
Concept introduction:
Hybridization is the combining of atomic orbitals to form hybrid orbitals.
To determine hybridization of an atom, first draw the Lewis structure of the molecule.
Find the number of electron domains around an atom so as to get the number of hybrid orbitals used by the atom for bonding.
When atomic orbitals combine, they form equal number of hybrid orbitals.
The s orbital combines with one, two, or three p orbitals to form

Answer to Problem 36QP
Solution:
a)
b)
c)
d)
e)
a)
Explanation of Solution
The Lewis structure of
In this, the two carbon atoms are bonded to each other and the three hydrogen atoms by single bonds. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in both carbon atoms
b)
The Lewis structure of
The carbon on the left is bonded to three hydrogen atoms by single bonds and to a carbon atom by a single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in this carbon
The central carbon is bonded to one carbon atom by a single bond, to another by double bond and to a hydrogen by a single bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict thepresence of three hybrid orbitals. Thus, hybridization of the central carbon is
The carbon on the right is bonded to two hydrogen atoms by a single bond and to a carbon by a double bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict thepresence of three hybrid orbitals. Thus, hybridization of the carbon on the right is
Thus, the hybridizations of the carbon atoms, from left to right in the molecule, are
c)
The Lewis structure of
The carbon on left is bonded to three hydrogen atoms by single bonds and to a carbon atom by single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in this carbon,
The two carbon atoms in the center are bonded to each other by double bonds and to a carbon by a single bond. Thus, there are two electron domains around each carbon atom. Two electron domains depict the presence of two hybrid orbitals. Thus, in these two carbons, sp hybrid orbitals are present.
The carbon on the right is bonded to two hydrogen atoms by single bonds, to a carbon atom by a single bond and to an oxygen by a single bond. Thus, there are four electron domains around this carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon,
Thus, the hybridizations of the carbon atoms from left to right in the molecule are
d)
The Lewis structure of
The carbon on the left is bonded to three hydrogen atoms by single bonds and to a carbon atom by a single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon
The carbon on the right is bonded to a hydrogen atom by a single bond and to an oxygen atom by a double bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict the presence of three hybrid orbitals. Thus, the hybridization of the carbon atom on the right is
Thus, the hybridizations of carbon atoms from left to right in the molecule are
e)
The Lewis structure of
The carbon on left is bonded to three hydrogen atoms by single bonds and to carbon atom by single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon
The carbon on the right is bonded to an oxygen atom by a double bond and to another oxygen atom by a single bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict the presence of three hybrid orbitals. Thus, the hybridization of the carbonatomon the right is
Thus, the hybridizations of carbon atoms from left to right in the molecule are
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Chapter 9 Solutions
Chemistry
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. . If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T X O O лет-ле HO OH HO OH This transformation can't be done in one step.arrow_forwardDetermine the structures of the missing organic molecules in the following reaction: X+H₂O H* H+ Y OH OH Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. X Sarrow_forwardPredict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. No reaction. HO. O :☐ + G Na O.H Click and drag to start drawing a structure. XS xs H₂Oarrow_forward
- What are the angles a and b in the actual molecule of which this is a Lewis structure? H H C H- a -H b H Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal groups may have slightly different sizes. a = b = 0 °arrow_forwardWhat are the angles a and b in the actual molecule of which this is a Lewis structure? :0: HCOH a Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes. a = 0 b=0° Sarrow_forwardDetermine the structures of the missing organic molecules in the following reaction: + H₂O +H OH O OH +H OH X Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structure of the missing organic molecule X. Click and drag to start drawing a structure.arrow_forward
- Identify the missing organic reactant in the following reaction: x + x O OH H* + ☑- X H+ O O Х Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H₂O) are not shown. In the drawing area below, draw the skeletal ("line") structure of the missing organic reactant X. Click and drag to start drawing a structure. Carrow_forwardCH3O OH OH O hemiacetal O acetal O neither O 0 O hemiacetal acetal neither OH hemiacetal O acetal O neither CH2 O-CH2-CH3 CH3-C-OH O hemiacetal O acetal CH3-CH2-CH2-0-c-O-CH2-CH2-CH3 O neither HO-CH2 ? 000 Ar Barrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 2 2. n-BuLi 3 Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure.arrow_forward
- Predict the products of this organic reaction: NaBH3CN + NH2 ? H+ Click and drag to start drawing a structure. ×arrow_forwardPredict the organic products that form in the reaction below: + OH +H H+ ➤ ☑ X - Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. Garrow_forwardPredict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ marrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning

