15 silberberg 8e ISMChapter 15 PDF

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Summary

CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON PROBLEMS 15 a) Plan: For compounds with seven carbons, first start with 7 C atoms in a straight chain. Then make all arrangements with 6 carbons in a straight chain and 1 carbon branched off the chain. Examine the structures to make su...

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CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON FOLLOW–UP PROBLEMS 15.1A

a) Plan: For compounds with seven carbons, first start with 7 C atoms in a straight chain. Then make all arrangements with 6 carbons in a straight chain and 1 carbon branched off the chain. Examine the structures to make sure they are all different. Continue in the same manner with 5 C atoms in chain and 2 branched off the chain, then 4 C atoms in chain and 3 branched off, and 3 C atoms in chain and 4 branched off. Examine all structures to guarantee there are no duplicates. Solution: Seven carbons in chain: H H H H H H H

H

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

(1) Six carbons in chain: H

H

C

H H

H H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

H

H

H

H H

H

H

C

C

H

H

C

H H

H

H

C

C

C

C

H

H

H

H

H

(2) (3) If the methyl group (—CH3) is moved to the fourth carbon from the left, the structure is the same as (3). If the methyl group is moved to the fifth carbon from the left, the structure is the same as (2). In addition, if the methyl group is moved to the sixth carbon from the left, the resulting structure has 7 carbons in a chain and is the same as (1).

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15-1

Five carbons in chain: H

H H

C

H H H

H

H

C

H H

H

C

C

C

C

H

H C

H

H

H

H

C

C

C

H

H

H

C

H H

H

C

C

H

H

H

H

H H

C

H

H H (5)

(4) H H

C

H H

H H

H

H

H

H

C

C

C

C

C

H

H

H

H

H

H

C

(6)

H

H

H

H

C

C

H

H H

H

C

C

C H

H H

H

C

C

H

H

H

H (7)

H

H

H

C

H

C

H

H

C

C

H

H

H H H

H

C

C

C

H

H

H

H

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15-2

Starting with a methyl group on the second carbon from the left, add another methyl group in a systematic pattern. Structure (4) has the second methyl on the second carbon, (5) on the third carbon, (6) on the 4th carbon. These are all the unique possibilities with the first methyl group on the second carbon. Any other variation will produce a structure identical to a structure already drawn. Next, move the first methyl group to the third carbon where the only unique placement for the second methyl group is on the third carbon as shown in structure (7). Which structure above would be the same as placing the first methyl group on the third carbon and the second methyl group on the fourth carbon? Are there any more arrangements with a 5-carbon chain? Think of attaching the two extra carbons as an ethyl group (—CH2CH3). If the ethyl group is attached to the second carbon from the left, the longest chain becomes 6 carbons instead of 5 and the structure is the same as (3). If the ethyl group is instead attached to the third carbon, the longest chain is still 5 carbons and the ethyl group is a side chain to give structure (8). 4 carbons in chain: H H H H H

C

H

H H

C

C

H H

H

C

C

C

C

H

H

H

H

(9) H Only this arrangement of 3 methyl groups branching off the two inner carbons in a 4 C chain produces a unique structure. Attaching 4 methyl groups off a 3 C chain is impossible without making the chain longer. So, the above nine structures are all the arrangements. b) Plan: For a five-carbon compound start with 5 C atoms in a chain and place the triple bond in as many unique places as possible. Solution: H H H H H H

H

C

C

C

C

C

H

H

H

H H

C

C

C

C

C

H

H H (2) 4 C atoms in chain: There are two unique placements for the triple bond: one between the first and second carbons and one between the second and third carbons in the chain. (1)

C

C

C

C

H

C

C

C

C

The fifth carbon is added as a methyl group branched off the chain. With the triple bond between the first and second carbons, the methyl group is attached to the third carbon to give structure (3).

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15-3

H H

C

H H

H

C

C

C

C

(3)

H

H

H

With the triple bond between the second and third carbons, the methyl group cannot be attached to either the second or the third carbons because that would create 5 bonds to a carbon and carbon has only 4 bonds. Thus, a unique structure cannot be formed with a 4 C chain where the triple bond is between the second and third carbons. There are only 3 structures with 5 carbon atoms, one triple bond, and no rings. 15.1B

a) Plan: Start with a ring consisting of 4 atoms. Make sure to include the double bond in the ring. Then work with a ring consisting of 3 atoms and 1 carbon branched off the ring. Move the position of the double bond relative to the branch to draw the 3 remaining structures. The double bond can occur within the ring or between a ring carbon and the branch carbon. Solution: Four carbons in the ring: H H H C C H

C

C H

H

Three carbons in the ring: H

H

H

C

H H

C C

C

H

C

C

H

C

H C

H

H

C C

H

C

C

H

H

H H H H H b) Plan: For a four-carbon compound start with 4 C atoms in a chain and place the two double bonds in as many unique combinations as possible (there are 2). It will not be possible to draw a molecule with 3 C atoms in a chain, 1 branch, and 2 double bonds without a C atom having more than 4 bonds.

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15-4

Solution: H

H

H

C

C

C

H 15.2A

H

H

H

H

H

C

C

C

C

C

H

H

H

Plan: Examine the structure for chain length and side groups; then use the structure to write the name. In part d), examine the structure for carbons that are bonded to four different groups. Those are the chiral carbons. Solution: a) 3,3-diethylpentane. There are 5 single-bonded carbons in the main chain and two ethyl groups attached to carbon #3 of that chain. The end of the name, pentane, indicates a 5 C chain (pent- represents 5 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds). 3,3-diethyl- means there are two ethyl groups and that each ethyl group is attached to carbon #3. CH 3

CH HC 3

CH

2

2

C

CH

2

CH

3

CH2 CH3 b) 1-ethyl-2-methylcyclobutane. The main “chain” is a four-membered ring with an ethyl group and a methyl group attached as branches. The name cyclobutane describes the 4 C ring (cyclobut-) with only single bonds (ane). The 1-ethyl- indicates that an ethyl group (–CH2CH3) is attached to carbon #1. The 2-methyl-indicates that a methyl group (–CH3) is attached to carbon #2. The lower number carbon is assigned to the ethyl group because it precedes methyl alphabetically. CH3

H2 C H3C

CH HC

CH2 C H2

c) trans-3-methyl-3-hexene. There are 6 carbons in the main chain, with a double bond between carbons #3 and #4 (counting from the right hand side). There is also a methyl group attached to carbon #3. The name 3-hexene describes the main chain (hex- indicates that there are 6 carbons in the chain, -ene indicates that there is a double bond in the chain, 3- indicates that the double bond starts at carbon #3). The name 3-methyl- indicates that there is a methyl group on carbon #3. Finally, because there is a double bond, we have to determine whether the groups bonded to the double bond are in a cis- (same side of the double bond) or trans- (opposite sides of the double bond) configuration. In this structure, the longer carbon chains attached to the double bond are on opposite sides of the double bond, so we add the prefix trans- to the name of the compound.

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15-5

CH

3

H2 C HC

CH 3

C C H

3

C H

2

d) 1-methyl-2-propylcyclopentane. The main “chain” in the molecule is a 5 C ring with only single bonds. A methyl group and a propyl group are attached to the ring. The name cyclopentane describes a 5-membered ring (cyclopent-) with all single bonds (-ane). The name 1-methyl- indicates that a methyl group is attached to the ring at the #1 position.The name 2-propyl- indicates that a propyl group is attached to the ring at the #2 position. The lower number carbon is assigned to the methyl group because it precedes propyl alphabetically. The chiral carbons (those bonded to four different groups) are marked with asterisks. H2 H2 * H3C C C

* H3C 15.2B

Plan: Analyze the name for chain length and side groups; then draw the structure. Solution: a) 3-ethyl-3-methyloctane. The end of the name, octane, indicates an 8 C chain (oct- represents 8 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds). 3-ethyl means an ethyl group (–CH2CH3) attached to carbon #3 and 3-methyl means a methyl group (–CH3) attached to carbon #3. CH 3

H3 C

CH2

C

CH2

CH2

CH2

CH2

CH3

CH2 CH

3

b) 1-ethyl-3-propylcyclohexane. The hexane indicates 6 C chain (hex-) and only single bonds (-ane). The cycloindicates that the 6 carbons are in a ring. The 1-ethyl indicates that an ethyl group (–CH2CH3) is attached to carbon #1. Select any carbon atom in the ring as carbon #1 since all the carbon atoms in the ring are equivalent. The 3propyl indicates that a propyl group (–CH2CH2CH3) is attached to carbon #3. CH2 CH3

CH2 CH2 CH3 c) 3,3-diethyl-1-hexyne. The 1-hexyne indicates a 6 C chain with a triple bond (-yne) between C #1 and C #2. The 3,3-diethyl means two (di-) ethyl groups (–CH2CH3) both attached to C #3.

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15-6

CH3 CH2 H

C

C

C

CH2

CH2

CH3

CH2 CH3 d) trans-3-methyl-3-heptene. The 3-heptene indicates a 7 C chain (hept-) with one double bond (-ene) between the 3rd and 4th carbons. The 3-methyl indicates a methyl group (–CH3) attached to carbon #3. The trans indicates that the arrangement around the two carbons in the double bond gives the two smaller groups on opposite sides of the double bond. Therefore, the smaller group on the third carbon (which would be the methyl group) is above the double bond while the smaller group, H, on the fourth carbon is below the double bond. HC 3 CH CH CH 2 3 2 C HC 3

CH

2

C H

15.3A

Plan: In an addition reaction, atoms are added to the carbon(s) in a double bond. Atoms are removed in an elimination reaction, resulting in a product with a double bond. In a substitution reaction, an atom or group of atoms substitutes for another one in the reactant. Solution: a) In this reaction, the Br on the carbon chain is replaced with a hydroxyl group while the carbons maintain the same number of bonds. This is a substitution reaction. b) The methyl group on the left-hand side of the structure is replaced by a hydrogen atom while the carbons maintain the same number of bonds. This is a substitution reaction. c) Water(in the form of a –H group and an –OH group) is added to the double bond, resulting in the product having three more atoms. Additionally, there is a second O bonded to the C in the product compared to the reactant. This is an addition reaction.

15.3B

Plan: a) An addition reaction involves breaking a multiple bond, in this case the double bond in 2-butene, and adding the other reactant to the carbons in the double bond. The reactant Cl2 will add –Cl to one of the carbons and –Cl to the other carbon. b) A substitution reaction involves removing one atom or group from a carbon chain and replacing it with another atom or group. For 1-bromopropane the bromine will be replaced by hydroxide. c) An elimination reaction involves removing two atoms or groups, one from each of two adjacent carbon atoms, and forming a double bond between the two carbon atoms. For 2methyl2propanol, the –OH group from the center carbon and a hydrogen from one of the terminal carbons will be removed and a double bond formed between the two carbon atoms.

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15-7

Solution: (a)

CH3

CH

CH

CH 3

+

Cl

Cl

CH

CH

Cl

2

CH3

CH 3

(b)

(c)

CH 3 CH3

C

CH CH 3 CH

O

15.4A

3

C

3

CH 2

+

H O 2

H

Plan: Determine the functional group(s) of the organic reactant(s) and then examine any inorganic reactant(s) to decide on the reaction type. Solution: 2– a) This reaction is an oxidation (elimination) reaction because Cr2O7 and H2SO4 are oxidizing agents. An alcohol group (–OH) is oxidized to a ketone group and a single bond between C and O is converted to a double bond. O

Cl b) This is a substitution reaction because the bromine atoms on the organic reactant (an alkyl halide) can be replaced with the cyanide groups from the inorganic reactant.

N

15.4B

C

H2 C

H2 C

H2 C

C

N

Plan: Examine any inorganic compounds and the organic product to determine the organic reactant. Look for differences between the reactants and products, if possible. Solution: a) The organic reactant must contain a single bond between the second and third carbons and the second and third carbons each have an additional group: an H atom on one carbon and a Cl atom on the other carbon.

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15-8

HC 3

H

C H3

C

C

CH3 CH

3

or

H3 C

CH2

C

CH3

H Cl Cl b) The third carbon from the left in the product contains the acid group. Therefore, in the reactant this carbon should have an alcohol group. H

H3C

CH2

C

OH

H 15.5A

Plan: In part a), LiAlH4 and H2O act as reducing agents to convert a ketone or aldehyde to an alcohol. To form the product, change the carbonyl group, =O, to an –OH group. In part b) ethyllithium, CH3CH2–Li, and H2O react to convert a ketone or aldehyde to form the alcohol and to add an ethyl group to the carbonyl carbon. To form the product, find the carbonyl carbon, add an ethyl group (from CH3CH2–Li) to the carbon and change the carbonyl group, =O, to a –OH bond. Solution: a) OH

b)

H2 C

CH3

OH 15.5B

Plan: The oxidizing agents in reaction a) indicate that the ketone group (=O) has been oxidized from an alcohol. To form the reactant, replace the C=O with an alcohol group, (C–OH). In reaction b) the reactants CH3CH2–Li and H2O indicate that a ketone or aldehyde reacts to form the alcohol. To form the reactant find the carbon with the alcohol group, remove the ethyl group that came from CH3CH2–Li and chan...