Problem Set Chapter 13 Solutions 13.27 Draw and name the eight isomeric alcohols with formula C 5 H 12 O. Which are chiral? February 28, 2013 To figure out which ones have a chiral center, look for a carbon with four different things bound to it. Structures B, F, and G are chiral. 13.28 Which of the eight alcohols you identified in Problem 13.27 react with CrO 3 in aqueous acid? Show the products you would expect from each reaction. CrO 3 reacts with primary alcohols to generate carboxylic acids, meaning that compound A, C, F, G, and H should all react with CrO 3 : 13.31 Predict the products of the following ether cleavage reactions: 13.32 How would you prepare the following ethers?
13.33 What products would you obtain from reaction of pentan-1-ol with the following reagents? (a) PBr 3 (b) SOCl 2 (c) CrO 3, H 2 O, H 2 SO 4 (d) Dess-Martin reagent 13.34 How would you prepare the following compounds from 2-phenyl-ethanol? More than one step may be required. (a) Styrene (PhCH=CH 2 ) (b) Phenylacetaldehyde (PhCH 2 CHO) (c) Phenylacetic acid (PhCH 2 CO 2 H) (d) Benzoic acid (e) Ethylbenzene (f) 1-phenylethanol
13.36 What Grignard reagent and what carbonyl compound might you start with to prepare the following alcohols?
13.39 How would you carry out the following transformations? More than one step may be required. 13.41 When 4-chlorobutan-1-ol is treated with a strong base such as sodium hydride, NaH, tetrahydrofuran is produced. Suggest a mechanism.
13.44 Methyl aryl ethers, such as anisole, are cleaved to iodomethane and a phenoxide ion by treatment with LiI in hot DMF. Propose a mechanism for this reaction. 13.48 Acid-catalyzed dehydration of 2,2-dimethylcyclohexanol yields a mixture of 1,2- dimethylcyclohexene and isopropylidene cyclopentane. Propose a mechanism to account for the formation of both products. OH H + + isopropylidene cyclopentane 13.49 Epoxides react with Grignard reagents to yield alcohols. Propose a mechanism. O CH 3 - OH CH 3 The Grignard reagent effectively acts as a nucleophilic carbanion. It attacks the epoxide to do an SN2 attack on the carbon it is attacking, leading to inversion of configuration at that carbon. The carbon it is not attacking retains the alcohol and its original configuration to yield the desired product.
13.50 How would you prepare the following substances from cyclopentanol? More than one step may be required. (a) Cyclopentanone (b) Cyclopentene (c) 1-Methylcyclopentanol (d) trans-2-methylcyclopentanol 13.57 2,3-Dimethylbutane-2,3-diol has the common name pinacol. On heating with aqueous acid, pinacol rearranges to pinacolone, 3,3-dimethyl-butan-2-one. Suggest a mechanism. 13.59 Propose a synthesis of bicyclohexylidene, starting from cyclohexanone as the only source of carbon. MgBr O OH
13.61 Identify the reagents a-f in the following scheme. 13.66 Anethole, C 10 H 12 O, a major constituent of the oil of anise, has the 1 H NMR spectrum shown. On oxidation with Na 2 Cr 2 O 7, anethole yields p-methoxybenzoic acid. What is the structure of anethole? Assign all peaks in the NMR spectrum, and account for the observed splitting patterns. The peaks at 7.23 ppm and 6.82 ppm correspond to a para-substituted aromatic ring. The singlet at 3.76 ppm corresponds to a methoxy group. Oxidation yields p-methoxybenzoic acid: Now you just have to figure out what the R group is. In order to have the appropriate molecular formula, it has to include 3 carbons and 5 hydrogens. The doublet in the 1 H NMR at 6.36 ppm is likely from an alkene (one proton) and the multiplet at 6.09 ppm is likely from the other side of the alkene. The doublet at 1.84 ppm is from a methyl group attached to an alkene. Putting all of the puzzle pieces together gives the following structure: 13.69 Compound A, C 5 H 10 O, is one of the basic building blocks of nature. All steroids and many other naturally occurring compounds are built from compound A. Spectroscopic analysis of A yields the following information: IR: 3400 cm -1 ; 1640 cm -1 1 H NMR: 1.63 δ (3H, singlet); 1.70 δ (3 H, singlet); 3.83 δ (1H, broad singlet); 4.15 δ (2H, doublet, J = 7 Hz); 5.70 δ (1H, triplet, J = 7 Hz) (a) From the IR spectrum, what is the nature of the oxygen-containing functional group? Alcohol! (b) What kinds of protons are responsible for the NMR absorptions listed? Singlet at 1.63 ppm = methyl group not attached to any neighbors Singlet at 1.70 ppm = another methyl group without neighbors
3.83 ppm = likely an OH group 4.15 ppm = A CH2 group that is attached to an alkene with one neighboring proton 5.70 ppm = An alkene proton attached to a CH2 group (c) Propose a structure for A.