What are the other two reasons for choosing (b) in Example 9.7? Example 9.7 Formaldehyde (CH 2 O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound. Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines. Solution The two possible skeletal structures are First we draw the Lewis structures for each of these possibilities: To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
What are the other two reasons for choosing (b) in Example 9.7? Example 9.7 Formaldehyde (CH 2 O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound. Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines. Solution The two possible skeletal structures are First we draw the Lewis structures for each of these possibilities: To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
Solution Summary: The author explains the two Lewis structures for formaldehyde labelled as ‘a’ and ‘b’ in example 9.7 of the book.
What are the other two reasons for choosing (b) in Example 9.7?
Example 9.7
Formaldehyde (CH2O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory specimens. Draw the most likely Lewis structure for the compound.
Strategy A plausible Lewis structure should satisfy the octet rule for all the elements, except H, and have the formal charges (if any) distributed according to electronegativity guidelines.
Solution The two possible skeletal structures are
First we draw the Lewis structures for each of these possibilities:
To show the formal charges, we follow the procedure given in Example 9.6. In (a) the C atom has a total of five electrons (one lone pair plus three electrons from the breaking of a single and a double bond). Because C has four valence electrons, the formal charge on the atom is 4 − 5 = −1. The O atom has a total of five electrons (one lone pair and three electrons from the breaking of a single and a double bond). Because O has six valence electrons, the formal charge on the atom is 6 − 5 = +1. In (b) the C atom has a total of four electrons from the breaking of two single bonds and a double bond, so its formal charge is 4 − 4 = 0. The O atom has a total of six electrons (two lone pairs and two electrons from the breaking of the double bond). Therefore, the formal charge on the atom is 6 − 6 = 0. Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges.
Check In each case make sure that the total number of valence electrons is 12. Can you suggest two other reasons why (a) is less plausible?
Q7.
a. Draw the line-bond structure of the major product for the following reaction, if a reaction
occurs, assume monohalogenation.
b. Calculate the product ratios using the following information (hint: use the number of
hydrogens in each category present to calculate the ratios).
Chlorination: 1° Reactivity=1
2° Reactivity=4
Heat
+ Cl2
3° Reactivity=5
Please correct answer and don't use hand rating and don't use Ai solution
Q10: Alkane halogenation
a. Give the name and structures of the five isomeric hexanes.
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Chem 0310 Organic Chemistry 1 Recitations
b. For each isomer, give all the free radical monochlorination and monobromination products
that are structurally isomeric.
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