Question

19. Which sequence of reagents could be used in the following Wittig synthesis starting with isopropyl bromide? (CH3)2CHBr -> [Cyclopentylideneisopropylidene image] C4H9Li (C6H5)3P [Cyclopentanone image]=O 1 2 3 A) 1 then 2, then 3 B) 1 then 3, then 2 C) 2 then 1, then 3 D) 2 then 3, then 1 20. The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways. As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides. What are the products from these reactions? CH3 H3C-C-Br CH3 II <- THF [Lithium cyclohexenolate image] -> H3C-I / THF -> I A) I is 2-methylcyclohexanone; II is 2-t-butylcyclohexanone B) I is 1-methoxycyclohexene; II is 1-t-butoxycyclohexene C) I is 2-methylcyclohexanone; II is 1-t-butoxycyclohexene D) I is 2-methylcyclohexanone; II is a mixture of cyclohexanone and 2-methylpropene

          19. Which sequence of reagents could be used in the following Wittig synthesis starting with isopropyl bromide?

(CH3)2CHBr -> [Cyclopentylideneisopropylidene image]

C4H9Li (C6H5)3P [Cyclopentanone image]=O
1 2 3

A) 1 then 2, then 3 B) 1 then 3, then 2
C) 2 then 1, then 3 D) 2 then 3, then 1

20. The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways. As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides. What are the products from these reactions?

CH3
H3C-C-Br
CH3
II <- THF [Lithium cyclohexenolate image] -> H3C-I / THF -> I

A) I is 2-methylcyclohexanone; II is 2-t-butylcyclohexanone
B) I is 1-methoxycyclohexene; II is 1-t-butoxycyclohexene
C) I is 2-methylcyclohexanone; II is 1-t-butoxycyclohexene
D) I is 2-methylcyclohexanone; II is a mixture of cyclohexanone and 2-methylpropene

19. Which sequence of reagents could be used in the following Wittig synthesis starting with isopropyl bromide?

(CH3)2CHBr -> [Cyclopentylideneisopropylidene image]

C4H9Li (C6H5)3P [Cyclopentanone image]=O
1 2 3

A) 1 then 2, then 3 B) 1 then 3, then 2
C) 2 then 1, then 3 D) 2 then 3, then 1

20. The lithium enolate base from cyclohexanone reacts with alkyl halides, often in different ways. As shown here, methyl iodide and tert-butyl bromide react to give different organic products, I and II, together with lithium halides. What are the products from these reactions?

CH3
H3C-C-Br
CH3
II <- THF [Lithium cyclohexenolate image] -> H3C-I / THF -> I

A) I is 2-methylcyclohexanone; II is 2-t-butylcyclohexanone
B) I is 1-methoxycyclohexene; II is 1-t-butoxycyclohexene
C) I is 2-methylcyclohexanone; II is 1-t-butoxycyclohexene
D) I is 2-methylcyclohexanone; II is a mixture of cyclohexanone and 2-methylpropene

Added by Pamela F.

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