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Chapter 8: Q.21 (page 319)

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

a.

b.

c.

d.

Short Answer

Expert verified

The products formed after the treatment NaNH2 is shown below:

Step by step solution

01

Type of the base used

NaNH2 is a strong base. When used in excess, it tends to abstract all theβ 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 β hydrogens gives rise to the elimination product. Different types of β hydrogens give different types of products. But here, different types of β hydrogens give a single product due to the formation of alkyne.

Representation of beta hydrogens

03

Formation of alkynes

Based on the steps, the products formed after the treatment with NaNH2 s shown below:-

Representation of possible products in a

Representation of possible products in b

Representation of possible products in c

Representation of possible products in d

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Most popular questions from this chapter

What is the major E2 elimination product formed from each alkyl halide?

a.

b.

Draw the structure of a dihalide that could be used to prepare each alkyne. There may be more than one possible dihalide.

a.

b.

c.

Draw all products, including stereoisomers, in each reaction.

a.

b.

c.

d.

Draw the major product formed when (R)-1-chloro-3-methylpentane is treated with each reagent: (a)NaOCH2CH3 ; (b) KCN; (c) DBU.

Draw a stepwise mechanism for the following reaction.
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