Alkyl Halides (Lab 13) – Flashcards
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1st wash: We want to keep it cold to avoid an E1 competition reaction because E1 and SN1 share the same step, and avoiding heat avoids elimination. 2nd wash: Neutralizes any HCl left. Salts (plus any t-butyl alcohol left) in aq layer. 3rd wash: Scavenges any remaining water-soluble compounds.
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in the separation of t-butyl chloride from the reaction mixture, what is the purpose of the following washes: cold water sodium bicarbonate water
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When an anhydrous salt such as CaCl2 or Na2SO4 is added it absorbes the water and the organic phase becomes clear.
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the addition of anhydrous calcium chloride to the crude product obtained from the separation procedure often results in the product becoming clear (transparent); what causes this clearing
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It is really not just a "primary" carbon it is a "benzylic" carbon. It is much more reactive because of the benzene ring.
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benzyl chloride, though a primary halide, readily forms a precipitate when treated with 2% ethanolic silver nitrate, explain
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the remaining 20% of carbocations in t-butyl undergo a competing side reaction because the water present in the HCl act as a base and removes a proton from the t-butyl cation, thus yielding an alkene. because no carbocation is involved in n-butyl, very little elimination competes with the substitution reaction and the yield is then higher.
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explain why more elimination occurs in competition with substitution in the preparation of t-butyl chloride than in the preparation of n-butyl bromide
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3 prime to 1 prime, confused about vinyl, allyl, and aryl chlorides.
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arrange halides tested in order of decreasing reactivity toward Ag+, (SN1)
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SN1 reactions favor 3 prime over 2 prime over 1 prime carbocations that form
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give an explanation for the order of reactivity observed for the three saturated alkyl halides with Ag+
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the carbocation formed is resonance stabilized
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explain the unusual position of allyl chloride in view of the fact that it is a primary halide
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carbon involved is a sp2 carbon, meaning more percent s character therefore a shorter, stronger bond therefore a harder bond to work with
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account for the low reactivity of chlorobenzene toward Ag+
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Sodium iodide is an ionic compound, any water in the experiment will cause the NaI to dissolve into Na+ and I- and plus give you a false result.
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why must acetone be anhydrous and the test tubes scrupulously dry for the test with sodium iodide
