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© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia.

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Presentation on theme: "© 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia."— Presentation transcript:

1 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.1 Naming a Ketone Given Its Structure The compound is a ketone, as shown by the single carbonyl group bonded to two alkyl groups: an ethyl group on the left (CH 3 CH 2 —) and a propyl group on the right (—CH 2 CH 2 CH 3 ). The IUPAC system identifies and numbers carbon chains to indicate where the carbonyl group is located, counting in the direction that gives the carbonyl carbon the lowest number possible. Analysis Give both the systematic (IUPAC) name and the common name for the following compound: The IUPAC name is 3-hexanone. The common name is ethyl propyl ketone. Solution The common name uses the names of the two alkyl groups.

2 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.2 Writing the Products of a Carbonyl Reduction First, draw the structure of the starting material, showing the double bond in the carbonyl group. Then rewrite the structure showing only a single bond between C and O, along with partial bonds to both C and O: Analysis What product would you obtain by reduction of benzaldehyde? The product obtained is benzyl alcohol. Solution Finally, attach hydrogen atoms to the two partial bonds and rewrite the product.

3 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.3 Predicting the Products of Hemiacetal and Acetal Formation First, rewrite the structure showing only a single bond between C and O, along with partial bonds to both C and O: Analysis Write the structure of the intermediate hemiacetal and the acetal final product formed in the following reaction: Next, add 1 molecule of the alcohol (CH 3 OH in this case) by attaching —H to the oxygen partial bond and —OCH 3 to the carbon partial bond. This yields the hemiacetal intermediate: Finally, replace the —OH group of the hemiacetal with an —OCH 3 from a second molecule of alcohol.

4 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson The reaction produces the acetal and water: Solution Worked Example 16.3 Predicting the Products of Hemiacetal and Acetal Formation Continued

5 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.4 Identification of Hemiacetals To identify a hemiacetal, look for a carbon atom with single bonds to 2 oxygen atoms, one an —OH group and one an —OR group. Note that the O of the —OR group can be part of a ring. Analysis Which of the following compounds are hemiacetals? Compound (a) contains 2 O atoms, but they are bonded to different C atoms; it is not a hemiacetal. Compound (b) has 1 ring C atom bonded to 2 oxygen atoms, one in the substituent —OH group and one bonded to the rest of the ring, which is the R group; it is a cyclic hemiacetal. Compound (c) also contains a C atom bonded to one —OH group and one —OR group, so it too is a hemiacetal. Solution

6 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.5 Identification of Acetals As in identifying hemiacetals, look for a carbon atom that has single bonds to 2 oxygen atoms, but in this case both of them will be —OR groups. Note that the O of the —OR group can be part of a ring. Analysis Which of the following compounds are acetals? In (a), the central carbon atom is bonded to one —CH 3, one —H, and two —OCH 2 CH 3 groups, so the compound is an acetal. Compound (b) does have a carbon atom bonded to 2 oxygen atoms, but one of the bonds is a double bond rather than a single bond, so this is not an acetal. Compound (c) has an oxygen atom in a ring, making it also part of an —OR group, where R is the ring. Since one of the carbons connected to the O in the ring is also connected to an —OCH 2 CH 3 group, compound (c) is an acetal. Solution

7 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 16.6 Writing the Products Obtained from Acetal Hydrolysis The products are the aldehyde or ketone plus 2 molecules of the alcohol from which the acetal could have been formed. First, identify the two C—O acetal bonds, redrawing the structure if necessary: Analysis Write the structure of the aldehyde or ketone that forms by hydrolysis of the following acetal: Next, break the H—OH bond and one of the acetal C — OR bonds (in this case, it does not matter which one); move the water OH to the acetal carbon to form the hemiacetal and the water H to the OR to form one molecule of HOR:

8 © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson In this example, the product is an aldehyde. Solution Remove the H and OR groups from the hemiacetal, and change the C—O single bond to a C=O double bond to give carbon the four bonds it must have. Combine the H and OR you removed from the second alcohol molecule. Worked Example 16.6 Writing the Products Obtained from Acetal Hydrolysis Continued


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