Alkyl Halides (Lab 13)

in the separation of t-butyl chloride from the reaction mixture, what is the purpose of the following washes:
cold water
sodium bicarbonate
water
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.

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
When an anhydrous salt such as CaCl2 or Na2SO4 is added it absorbes the water and the organic phase becomes clear.
benzyl chloride, though a primary halide, readily forms a precipitate when treated with 2% ethanolic silver nitrate, explain
It is really not just a “primary” carbon it is a “benzylic” carbon. It is much more reactive because of the benzene ring.
explain why more elimination occurs in competition with substitution in the preparation of t-butyl chloride than in the preparation of n-butyl bromide
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.
arrange halides tested in order of decreasing reactivity toward Ag+, (SN1)
3 prime to 1 prime, confused about vinyl, allyl, and aryl chlorides.
give an explanation for the order of reactivity observed for the three saturated alkyl halides with Ag+
SN1 reactions favor 3 prime over 2 prime over 1 prime carbocations that form
explain the unusual position of allyl chloride in view of the fact that it is a primary halide
the carbocation formed is resonance stabilized
account for the low reactivity of chlorobenzene toward Ag+
carbon involved is a sp2 carbon, meaning more percent s character therefore a shorter, stronger bond therefore a harder bond to work with
why must acetone be anhydrous and the test tubes scrupulously dry for the test with sodium iodide
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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