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To Deprotonate Alcohols, You Should Use Sodium Hydride (Nah) As the Base

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To Deprotonate Alcohols, You Should Use Sodium Hydride (Nah) As the Base CHEM 341 Problem Set 20 Key 1.) Fill in the boxes: To deprotonate alcohols, you should use sodium hydride (NaH) as the base. Part a. has two possible starting alcohols. Part b. only has one possible starting alcohol. Why? In part a, the two carbons attached to the oxygen are primary allowing either to be used as a good alpha carbon for SN2 reactions. In part b, one carbon is primary and the other is secondary. The secondary can only serve as an alpha carbon in an SN2 reaction when moderately or weakly basic nucleophiles are used. Strongly basic nucleophiles will give E2 products as the major pathway. a.) NaH base HO O alcohol alkoxide Br halide O a.) NaH base OH O alcohol alkoxide Br halide O b.) NaH HO base O optically optically active active alcohol alkoxide Br halide O optically active 2.) To deprotonate phenols, thiols or carboxylic acids, you should use sodium hydroxide (NaOH) or potassium carbonate as the base. Sodium hydride is more expensive, much more dangerous, and is well overkill. Why can you use sodium hydroxide to deprotonate phenols, thiols, and carboxylic acids but not use it to deprotonate alcohols? Alcohols have about the same acidity as water (pKa ~ 15). As a result, you end of with a mixture of nucleophiles (an alkoxide and hydroxide). Phenols (pKa ~ 10), thiols (pKa ~ 10), and carboxylic acids (pKa ~ 5) are more acidic than water, and as result the phenol, thiol, and carboxylic acids are deprotonated so your largest concentration of nucleophiles is the one that you wish. The stronger the acid, the weaker the base that is needed to deprotonate it. 3.) Fill in the boxes: a.) NaOH HS base S thiol optically active optically active Br optically halide active S optically active b.) K2CO3 base OH O O O carboxylic acid Br optically halide active O optically active O 4.) To deprotonate terminal alkynes, sodium hydroxide will not work. Instead, you should use sodium amide (NaNH2). Why won’t sodium hydroxide work? The pKa of terminal alkynes is ~ 25. Sodium hydroxide (conjugate pKa ~ 15) is not strong enough to deprotonate it. Fill in the boxes: a.) NaNH2 base Na optically active optically active terminal alkyne Br halide optically active b.) H NaNH 2 Br Na H NaNH2 Br Na b.) H NaNH2 Na Br H NaNH2 Br Na 5.) For the following reactions, specify the major pathway of the reaction E1, E2, SN1, or SN2 above the arrow. Also, give the major carbon containing product(s). In the triangle, put an Y or N if the products are optically active. Optically active? A. SN2 N S Br S Br B. O SN2 Y O D D optically active C. SN2 O Br Y O D D optically active D. E2 Br + D N O D optically active O E. SN2 Br O Y O D D O optically active F. E2 Br Y O D D optically active G. SN1 S HS D Y Br D optically active 5.) continued… For the following reactions, specify the major pathway of the reaction E1, E2, SN1, or SN2 above the arrow. Also, give the major carbon containing product(s). In the triangle, put a Y or N if the products are optically active. Optically active? H. SN2 N Br O O O I. O SN2 Y Br D D optically active J. SN2 O O Y Br D D optically active K. Na SN2 N Br L. N Na SN2 N Br N M. Na SN2 N Br N C N N. Na E2 Br N O. E2 Na N Br N 6.) The following ethers can be made from an alkoxide (RO-) and alkyl bromide (RBr). Fill in the appropriate boxes. Alkoxide Alkyl Bromide a. O O Br b. O O Br c. O O Br Me d. O O Br optically active optically active e. Br optically O O active optically active f. O Br Br O O either combination 7a.) Draw the mechanism for the following reaction. NaBH4 EtOH O OH Either mechanism is acceptable. 7b.) If NaBD4 was used instead of NaBH4 in the reaction above, what would the product be? D OH racemic 8.) Predict the product(s) of the following reactions. Underneath each starting material, indicate what type of symmetry. Is there a correlation to the symmetry element and the number of products? a. NaBH4 O EtOH OH plane of symmetry one product axis of symmetry b. NaBH4 OH O EtOH two products that are racemic enantiomers plane of symmetry (plane of paper) c. NaBH4 O EtOH OH OH plane of symmetry (perpendicular to plane of paper) two products that are diastereomers d. NaBH4 O EtOH OH one product optically active optically active If there is an axis of symmetry through the carbonyl, then you get just one product. .
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