Chapter 10. BrCH 2 CH 2 CH 2 CCH 2 Br CH 3. CH 3 CCH 2 CH 2 Cl CH 3 CHCH 2 CH 2 CHCH Give IUPAC names for the following alkyl halides:

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1 hapter Give IUPA names for the following alkyl halides: (a), I 1-iodobutane 3 (b), l 1-chloro-3-methylbutane (c), ,5-Dibromo-2,2-dimethylpentane l (d), l 1,3-Dichloro-3-methylbutane I 2 2 I (e), Ethyl-1,4-diiodopentane (f), l 2-omo-5-chlorohexane 10.2 Draw structures corresponding to the following IUPA names: (a), 2-hloro-3,3-dimethylhexane l (b), 3,3-Dichloro-2-methylhexane l l (c), 3-omo-3-ethylpentane (d), 1,1-Dibromo-4-isopropylcyclohexane

2 (e), 4-sec-Butyl-2-chlorononane l (f), 1,1-Dibromo-4-tert-butylcyclohexane 10.3 Draw and name all monochloro products you would expect to obtain from radical chlorination of 2-methylpentane. Which, if any, are chiral? hν l2 l l l l l 1-chloro-2-methylpentane (chiral) 2-chloro-2-methylpetane 3-chloro-2-methylpentane (chiral) 2-chloro-4-methylpentane (chiral) 1-chloro-4-methylpentane 10.4 Taking the relative reactivities of 1,2,3 hydrogen atoms into account, what products would you expect to obtian from monochlorination of 2-methylbutane? What would the approximate percentage of each product be?

3 l2 hν l l 6/26 10/26 l l 7/26 3/ Draw three resonance forms for the cyclohexadienyl radical. yclohexadienyl radical 10.6 The major product of the reaction of methylenecyclohexane with N-bromosuccinimide is 1-(bromomethyl)cyclohexene. Explain. 2 2 light, l 4 Major product

4 Allylic position More hindered Less hindered light, l Major product ne more reason is the relative stability of alkene What products would you expect from the reaction of the following alkenes with? If more than one product is formed, show the structures of all. (a)

5 (b) A B D From the four radicals above, we can get four products drawn below: A main B D

6 lethethetherow do you prepare the following alkyl halides from the corresponding alcohols? (a) l l 3 3 r (b) P 2 3E r (c) P 3 E (d)

7 the 3 l l ler Just how strong a base would you expect a Grignard reagents to be? Look at Table 8.1, and the predict whether the following reactions will occur as written. (The pk a of N 3 is 35) (a) 3Mg 4 Mg (b) 3Mg N N Mg The Grignard reagent can be thought of as the magnesium salt of a hydrocarbon acid. Because hydrocarbons are very weak (pk a s in the range 44 to 60), it is very strong base. All the two reactions can occur ow might you replace a halogen substituent by a deuterium atom if you wanted to prepare a deuterated compound?? D Mg D Mg D ow would you carry out the following transformations using an organocopper coupling reaction? More than one step is required in each case.? (a) 3

8 l4 (3)2uLi 3 (b) ? Li Pentane Li Li Li 2 ui ( ) u Li ( ) u Li - 2 (c) ? Li ( ) u

9 B3,TF 2 22, PB Li Pentane Li Li Li ui ( ) 2 u Li - LiI ( ) 2 u Li ank each of the following series of compounds in order of increasing oxidation level: (a) l l xidation level order: < = < (b) 3 N 3 2 N 2 N N 2 xidation level order: 3 2 N 2 < N N 2 < 3 N Tell whether each of the following reaction is an oxidation, a reduction, or neither. Explain your answer.

10 (a) 3 2 NaB (b) 1. B 3 2. Na, 2 2 (a) Because of breaking one -, so it is reduction. (b) Because of forming one - and forming one -, so it is neither reduction nor oxidation Give an IUPA name for each of the following alkyl halides (yellow-green=l): (a) cis-1-hloro-3-methyl-cyclohexane (b) 4-chloro-2-methyl-2-heptene

11 10.17 Name the following alkyl halides according to IUPA rules: 3 3 (a) ,4-dibromo-2,6-dimethyl-heptane I (b) iodo-2-hexene l (c) 3 2-bromo-4-hloro-2,5-dimethyl-hexane 2 (d) bromomethylhexane (e) l bromo-6-chloro-2-hexyne Draw structures corresponding to the following IUPA names: (a) 2,3-Dichloro-4-methylhexane l 3 l (b) 4-omo-4-ethyl-2-methylhexane (c) 3-Iodo-2,2,4,4-tetramethylpentane

12 I (d) cis-1-bromo-2-ethylcyclopentane : Draw and name the monochlorination products you might obtain by radical chlorination of 2-methylbutane. Which of the products are chiral? Are any of the products optically active? As there are four kinds of hydrogen atoms, so there should be four kinds of products, witch are (a):1-chloro-2-methylbutane, (b):2-chloro-2-methylbutane, (c):3-chloro-2-methylbutane and (d):4-chloro-2-methylbutane. The products (a) and (c) are chiral. And the products (a) and (c) are optically active : A chemist requires a large amount of 1-bromo-2-pentene as starting material for a synthesis and decides to carry out an allylic bromination reaction: l 4 What is wrong with this synthesis plan? What side products would form in addition to the desired product? l 4 Mixture will be obtained What product(s) would you expect from the reaction of 1-methylcyclohexene with? Would you use this reaction as part of a synthesis? 3 l 4?

13 3 3 The products will be and As a part of synthesis I will not use this reaction. Because from this reaction I will get two mixed products. But usually in synthesis we only need one single product ow would you prepare the following compounds, starting with cyclopentene and any other reagents needed? (a) hlorocyclopentane (b) Methylcyclopentane (c) 3- omocyclopentene (d) yclopentanol (e) yclopentylcyclopentane (f) 1,3-yclopentadiene (a) l l I I ( 3 ) 2 uli 3 (b) (c) (d) (e) 1.B 3,TF , I I I 2Li Li ui u Li Pentane Base (f) Predict the product(s) of the following reactions:

14 (a) (c) 3? l 4? (b) (d) Sl 2? P3? (e) (f) (g) (a) (b) Mg 3 A? 2 Li Pentane A? ui ( 3 ) 2 uli Sl 2 l 4 3 B? B?? 2 l S 2 l N (c) (d) P3 3 3 (e) (f) 2Li Pentane (g) Mg Li Li Mg ui 2 3 P 3 Mg() ( ) 2 uli LiI ( 3 ) 2 uli Li 3 u (S)-3-Methylhexane undergoes radical bromination to yield optically inactive 3-bromo-3-methylhexane as the major product. Is the product chiral? What conclusions can you draw about the radical intermediate? 2 3 Two stereoisomers of the product are formed ()-3-bromo-3-methylhexane and

15 (S)-3-bromo-3-methylhexane. They are all chiral and form in a ratio of 1: onclusions: The radical intermediate is and the two sides have equal chance to be attacked It s planar Assume that you have carried out a radical chlorination reaction on ()-2-chloropentane and have isolated (in low yield) 2,4-dichloropentane. ow many stereoisomers of the product are formed and in what ratio? Are any of the isomers optically active? (See Problem 10.24) l l l l 2 l l ()-2-chloropentane l l l l (S) () () () 2,4-Dichloro-pentane 2,4-Dichloro-pentane Α Β Two stereoisomers of the product are formedin 1 : 1 ratio. The first one (A) is a meso compound, so it is optically inactive. The second one (B) is optically active alculate Δ for the reactions of l and with 4, and then draw a reaction energy diagram showing both processes. Which reaction is likely to be faster? Product bonds formed eactant bonds broken -l D=351kJ/mol - D=438kJ/mol -l D=432kJ/mol l-l D=243kJ/mol Total D=783kJ/mol Total D=681kJ/mol Δ= =-102kJ/mol Product bonds formed eactant bonds broken

16 - D=293kJ/mol - D=438kJ/mol - D=366kJ/mol - D=193kJ/mol Total D=659kJ/mol Total D=631kJ/mol Δ= =-28kJ/mol The reaction energy diagram: Energy eaction energy diagram of holomethane eaction process Energy eaction energy diagram of omomethane eaction process The reaction between holorine radical and methane is much faster Use the bond dissociation energies listed in Table 5.3 on page 154 to calculate 0 for the reactions of l. and. with a secondary hydrogen atom of propane. Which reaction would you expect to be more selective? For l. 0 = =62kj/mol For. 0 = =127kj/mol It is clear that l. has higher reactivity while. has lower reactivity; thus according to lower reactivity higher selectivity, the reaction of. with a secondary hydrogen atom of propane is more selective What product(s) would you expect from the reaction of 1, 4-hexdiene with? What is the structure of the most stable radical intermediate? The reaction will be following:

17 major That s because of relative stability of conjugated diene Alkyl benzenes such as toluence (methylbenzene) react with to give products in which bromine substitution has occurred at the position next to the aromatic ring (the benzylic position). Explain, based on the bond dissociation energies in Table 5.3 The reason is show bellow: l kJ/mol nearly 464kJ/mol The carbon hydrogen bond of the methyl has the lowest bond dissociation energy, so this bond is most likely to be broken and yields the most stable radical : Draw resonance structures for the benzyl radical, 6 5 2, the intermediate produced in the bromination reaction of toluene (Problem 10.29)

18 10.32 Draw resonance structure for the following species: (a) 3 2 (b) (c) 3 N (a). (b). (c). N N ank the compounds in each of the following series in order of increasing oxidation level: (a)

19 N 2 l l (b) a) = < < l b) N 2 = < l< Which of the following compounds have the same oxidation level and which have different levels? compound 1, 2, 4 have the same oxidation level and compound 3, 5 has different oxidation level Tell whether each of the following reactions is an oxidation or reduction: a. 3 2 r3 3 xidation b c. N 3 N Neither 1.Mg

20 eduction ow would you carry out the following syntheses? (a) Butylcyclohexane from cyclohexene (b) Butylcyclohexane from cyclohexanol (c) Butylcyclohexane from cyclohexane ( ) 2 uli In ether u Li P 3 ( ) 2 uli In ether u Li 2 eat/hv ( ) 2 uli In ether u Li The syntheses shown here are unlikely to occur as written. What is wrong with each? (a) F 1. Mg 2.3

21 (b) 2 L (c) F 3 (3)2uLi (a) rganofluorides rarely react with magnesium. So there is no alkyl magnesium fluorides. (b) The reaction reacts as follow: 2 L is more stable than 3 because of Zaitsev s rule. (c) Gilman reagents couldn t react with fluorides because of high density of electron on fluorine atom. So this reaction could not occur Why do you suppose it s not possible to prepare a Gringard reagent from a bromo alcohol such as 4-bromo-1-pentanol? Mg Mg

22 Give another example of a molecule that is unlikely to form a Grignard reagent. If 4-bromo-1-pentanol can form a Gringard reagent, it will nucleophilic attack the active atom in, so it s not possible to prepare a Gringard reagent. Such as form a Grignard reagent. which has active is unlikely to Addition of to a double bond with an ether substiutent occurs regiospecifically to give a product in which the and are bonded to the same carbon: 3 3 Draw two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed. Answer: more stable form Phenols, compounds that have an group bonded to a benzene ring, are relatively acidic because their anions are stabilized by resonance. Draw resonance structures for the phenoxide ion. phenoxide ion

23 10.41 Alkyl halides can be reduced to alkanes by a radical reaction with tributyltin hydride, ( 4 9 ) 3 Sn, in the presence of light (hv): X ( 4 9 ) 3 Sn ( 4 9 ) 3 SnX solution: Initiation step: ( 4 9 ) 3 Sn ( 4 9 ) 3 Sn Propagation steps: ( 4 9 ) 3 Sn X X Termination steps: ( 4 9 ) 3 Sn ( 4 9 ) 3 Sn ( 4 9 ) 3 Sn Sn( 4 9 ) 3 ( 4 9 ) 3 Sn X ( 4 9 ) 3 Sn X X X X X X X ( 4 9 ) 3 Sn ( 4 9 ) 3 Sn X ( 4 9 ) 3 Sn ( 4 9 ) 3 Sn X Identify the reagents a-c in the following scheme:

24 3 a b c a, 1) B 3, 2) -, 2 2 ; b, P 3 ; c, 1) Li, 2) ui, 3) 3 I Tertiary alkyl halides, 3 X, undergo spontaneous dissociation to yield a carboncation, 3.Which do you think reacts faster,( 3 ) 3,or 2 =( 3 ) 2? Explain Solutions: 2 =( 3 ) 2.Because of double bond s conjugate effect. So the carboncation is more stable.and the G is smaller.and as a result the reaction is faster arboxylic acids ( 2 ; pka 5) are approximately times more acidic than alcohols (; pka 16). In other words, a carboxylate ion ( - 2 ) is more stable than an alkoxide ion( - ). Explain, using resonance. The carboxylate ion has two resonance forms, which is rather than the alkoxide ion,only one Form. Thus the - 2 is much stable than -.