Organic Chemistry, 7e by L. G. Wade, Jr. Chapter 6 Alkyl Halides: Nucleophilic Substitution and Elimination Christine Hermann Radford University Radford,

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Organic Chemistry, 7e by L. G. Wade, Jr. Chapter 6 Alkyl Halides: Nucleophilic Substitution and Elimination Christine Hermann Radford University Radford, VA Copyright © 2010 Pearson Education, Inc.

6.1 Give the IUPAC name for (CH 3 ) 2 CBrCH 2 CH(CH 3 ) 2. a.2-Bromo-2,4-dimethylpentane b.4-Bromo-2,4-dimethylpentane c.2-Bromopentane d.2-Bromoheptane e.Isobutyl isopentyl bromide

6.1 Answer a.2-Bromo-2,4-dimethylpentane b.4-Bromo-2,4-dimethylpentane c.2-Bromopentane d.2-Bromoheptane e.Isobutyl isopentyl bromide Pentane is the longest continuous chain.

6.2 Identify the vinyl chloride. a.ClCH 2 CH=CH 2 b.ClCH=CH 2 c.C 6 H 5 Cl d.CH 3 CH 2 Cl e.CH 3 CH(Cl)CH 3

6.2 Answer a.ClCH 2 CH=CH 2 b.ClCH=CH 2 c.C 6 H 5 Cl d.CH 3 CH 2 Cl e.CH 3 CH(Cl)CH 3 A vinyl halide has a Cl attached to the doubly bonded carbon.

6.3 Identify the secondary alkyl halide. a.CH 3 Br b.CH 3 CH 2 Br c.CH 3 CH(Br)CH 3 d.(CH 3 ) 3 CBr

6.3 Answer a.CH 3 Br b.CH 3 CH 2 Br c.CH 3 CH(Br)CH 3 d.(CH 3 ) 3 CBr In a secondary alkyl halide, the halogen is attached to a carbon, which is attached to two other carbons.

6.4 Give the intermolecular attractions between alkyl halides. a.Hydrogen bonding b.London forces c.Dipole–dipole d.None

6.4 Answer a.Hydrogen bonding b.London forces c.Dipole–dipole d.None London forces are the strongest intermolecular attractions in alkyl halides.

6.5 a.CH 2 BrCH 2 CH=CH 2 b.CH 3 CH(Br)CH=CH 2 c.CH 3 CH 2 CBr=CH 2 d.CH 3 CH 2 CH=CHBr

6.5 Answer a.CH 2 BrCH 2 CH=CH 2 b.CH 3 CH(Br)CH=CH 2 c.CH 3 CH 2 CBr=CH 2 d.CH 3 CH 2 CH=CHBr Br substitutes for H on the allylic carbon.

6.6 a.CH 3 CH 2 CHCl 2 b.CH 3 CH(Cl)CHCl c.CH 3 C(Cl) 2 CH 3 d.CH 3 C(Cl)=CH 2 e.CH 3 CH=CHCl

6.6 Answer a.CH 3 CH 2 CHCl 2 b.CH 3 CH(Cl)CHCl c.CH 3 C(Cl) 2 CH 3 d.CH 3 C(Cl)=CH 2 e.CH 3 CH=CHCl The HCl adds via Markovnikov orientation.

6.7 Identify the strongest nucleophile. a.CH 3 OH b.(CH 3 ) 3 COH c.CH 3 O - d.(CH 3 ) 3 CO -

6.7 Answer a.CH 3 OH b.(CH 3 ) 3 COH c.CH 3 O - d.(CH 3 ) 3 CO - Negative nucleophiles are stronger than neutral nucleophiles. Bulky nucleophiles are weaker.

6.8 Where does the nucleophile attack in an S N 2 mechanism? a.Backside b.Frontside c.Both sides d.Sideways

6.8 Answer a.Backside b.Frontside c.Both sides d.Sideways The nucleophile attacks on the side opposite from the leaving group.

6.9 Give the rate equation for an S N 2 mechanism. a. k[RX][RX] b. k[RX][Nu] c. k[Nu][Nu] d. k[RX][B]

6.9 Answer a. k[RX][RX] b. k[RX][Nu] c. k[Nu][Nu] d. k[RX][B] The rate equation includes both the alkyl halide and the nucleophile.

6.10 Identify the alkyl halide that reacts the fastest in a S N 1 mechanism. a.CH 3 Cl b.CH 3 CH 2 Cl c.(CH 3 ) 2 CHCl d.(CH 3 ) 3 CCl

6.10 Answer a.CH 3 Cl b.CH 3 CH 2 Cl c.(CH 3 ) 2 CHCl d.(CH 3 ) 3 CCl Tertiary alkyl halides react the fastest.

6.11 Identify the best solvent for a S N 1 reaction. a.CH 3 OCH 3 b.(CH 3 ) 3 N c.CH 3 CH 2 OH d.CCl 4

6.11 Answer a.CH 3 OCH 3 b.(CH 3 ) 3 N c.CH 3 CH 2 OH d.CCl 4 A protic solvent is the best solvent for a SN1 reaction. A protic solvent has hydrogen bonding.

6.12 In an S N 1 reaction, a 2 o carbocation is formed. What happens next? a.Rearrangement may occur to form a 1 o carbocation. b.Rearrangement may occur to form a 3 o carbocation. c.The nucleophile attacks. d.The leaving group departs.

6.12 Answer a.Rearrangement may occur to form a 1 o carbocation. b.Rearrangement may occur to form a 3 o carbocation. c.The nucleophile attacks. d.The leaving group departs. A more stable carbocation will be formed.

6.13 What is the order for an E1 reaction? a.Zero order b.First order c.Second order d.Third order

6.13 Answer a.Zero order b.First order c.Second order d.Third order E1 reactions follow first order kinetics.

6.14 Identify the alkyl halide that reacts the fastest in an E1 mechanism. a.CH 3 Cl b.CH 3 CH 2 Cl c.(CH 3 ) 2 CHCl d.(CH 3 ) 3 CCl

6.14 Answer a.CH 3 Cl b.CH 3 CH 2 Cl c.(CH 3 ) 2 CHCl d.(CH 3 ) 3 CCl A tertiary alkyl halide reacts the fastest.

6.15 According to Zaitsev’s rule, what alkene is formed in the highest yield? a.Least substituted alkene b.Most substituted alkene c.Does not matter. d.A diene.

6.15 Answer a.Least substituted alkene b.Most substituted alkene c.Does not matter. d.A diene. In elimination reactions, the most substituted alkene usually predominates.

6.16 What stereochemistry is needed for an E2 reaction? a.Anti-coplanar arrangement with H and X on adjacent carbons. b.Syn-coplanar arrangement with H and X in adjacent carbons. c.H and X are on the same carbon. d.H and X are two carbons apart.

6.16 Answer a.Anti-coplanar arrangement with H and X on adjacent carbons. b.Syn-coplanar arrangement with H and X in adjacent carbons. c.H and X are on the same carbon. d.H and X are two carbons apart. H and X must be staggered for an E2 reaction.

6.17 Describe an E2 mechanism. a.Not stereospecific and not concerted. b.Stereospecific and not concerted. c.Stereospecific and concerted. d.Not stereospecific and concerted.

6.17 Answer a.Not stereospecific and not concerted. b.Stereospecific and not concerted. c.Stereospecific and concerted. d.Not stereospecific and concerted.

6.18 What conditions are best for an E2 mechanism? a.Weak acid b.Strong acid c.Weak base d.Strong base

6.18 Answer a.Weak acid b.Strong acid c.Weak base d.Strong base

6.19 Why can’t CH 3 Br undergo an E2 reaction? a.Bromide is a poor leaving group. b.The base cannot attack. c.CH 3 Br is solvated in the solvent. d.Two carbons are needed to make a double bond.

6.19 Answer a.Bromide is a poor leaving group. b.The base cannot attack. c.CH 3 Br is solvated in the solvent. d.Two carbons are needed to make a double bond.

6.20 Give the major mode of reaction for primary alkyl halides. a.S N 1 b.S N 2 c.E1 d.E2

6.20 Answer a.S N 1 b.S N 2 c.E1 d.E2