Chapter 9: Q13. (page 347)
Question: Draw the structure of each transition state in the two-step mechanism for the reaction.
CH3CH2OH + H2So4 → CH3CH=CH2+ H2O
Short Answer
Answer
The mechanism is shown below.

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Chapter 9: Q13. (page 347)
Question: Draw the structure of each transition state in the two-step mechanism for the reaction.
CH3CH2OH + H2So4 → CH3CH=CH2+ H2O
Answer
The mechanism is shown below.

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Question: Prepare each compound from cyclopentanol. More than one step may be needed.
a.
b. 
c. 
d. 
Question: Explain the following observation. When 3-methylbutan-2-ol is treated with HBr, a single alkylbromide is isolated, resulting from a 1,2-shift. When 2-methylpropan-1-ol is treated with HBr, no rearrangement occurs to form an alkyl bromide.
Question: Answer each question using the ball-and-stick model of compound A.

a. Give the IUPAC name for A, including R,Sdesignations for stereogenic centers.
b. Classify A as a alcohol.
c. Draw a stereoisomer for A and give its IUPAC name.
d. Draw a constitutional isomer that contains an OH group and give its IUPAC name.
e. Draw a constitutional isomer that contains an ether and give its IUPAC name.
f. Draw the products formed (including stereochemistry) when A is
treated with each reagent: [1] NaH; [2] ; [3] , pyridine;
[4] HCl; [5] ,pyridine; [6] TsCl, pyridine.
Question: Show how a 1,2-shift forms a more stable carbocation from each intermediate.
a.
b. 
c. 
Question: Draw the products of each reaction, indicating the stereochemistry around any stereogenic centers.
a. 
b. 
c.
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