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Chapter 8: Q.50 (page 328)

Draw the products formed when each dihalide is treated with excess NaNH2.

a.

b.

c.

d.

Short Answer

Expert verified

The products formed after the treatment with NaNH2 is shown below:

Step by step solution

01

Type of the base used

NaNH2is a strong base.

When used in excess, it tends to abstract all the β protons since, in these substrates, two leaving groups are present. As a result, the elimination product formed gives an alkyne.

02

Removal of β hydrogens

Removal of excess βhydrogens gives the elimination product. Based on the steps, the products formed after the treatment with NaNH2 is shown below:

Representation of the possible alkynes from a

Representation of the possible alkynes from b

Representation of the possible alkynes from c

Representation of the possible alkynes from d

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Most popular questions from this chapter

For which double bonds are stereoisomers possible?

a.

b.

c.

How does each of the following changes affect the rate of an E1 reaction?

a. doubling [RX]

b. doubling [B:]

c. changing the halide from (CH3)3CBrto CH3CH2CH2Br

d. changing the leaving group from Cl–toBr–

e. changing the solvent from DMSO to CH3OH

Draw a stepwise mechanism for the following reaction. The four-membered ring in the starting material and product is called aβ-lactam. This functional group confers biological activity on penicillin and many related antibiotics, as is discussed in Chapter 22. (Hint: The mechanism begins withβelimination and involves only two steps.)

∆H°values obtained for a series of similar reactions are one set of experimental data used to determine the relative stability of alkenes. Explain how the following data suggest that cis-but-2-ene is more stable than but-1-ene (Section 12.3A).

What alkyl chloride affords the following alkene exclusively under E2 reaction conditions?
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