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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 尾 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 尾hydrogens 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

Which double bonds in the following natural products can exhibit stereoisomerism? Nerolidol is isolated from the angel鈥檚 trumpet plant, caryophyllene is present in hemp, and humulene comes from hops.

a.

b.

c.

Draw the alkynes formed when each dihalide is treated with excess base.

a.

b.

c.

d.

What is the major product formed when each alkyl halide is treated with each of the following reagents: [1] NaOCOCH3 ; [2] NaOCH3; [3]KOC(CH3)3? If it is not possible to predict the major product, identify the products in the mixture and the mechanism by which each is formed.

a.

b.

c.

(a) Draw all products formed by treatment of each dibromide (A and B) with one equivalent ofNaNH2 .(b) Label pairs of diastereomers and constitutional isomers.

Draw all products, including stereoisomers, in each reaction.

a.

b.

c.

d.
See all solutions

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