Question

Draw the structures of organic compounds A and B. Indicate stereochemistry where applicable. 1) BuLi H-C C-H 2) CH3CH2Br (1 equiv.) HC CH2Cl2 Br2 (1 equiv.) Pd/C There is additional feedback available! View this feedback by clicking on the bottom divider bar. Click on the divider bar again to hide the additional feedback. Close A Previous Give up & View Solution Try Again Next Exit

Answer 1

Concepts and reason

The concept used to solve this question is to draw the structures according to the reactions of alkynes. Alkynes are organic compounds with at least one triple bond. They involve different kind of reactions like addition reactions (bromination, hydrobromination, hydroboration, and hydration), hydrogenation, ozonolysis, and so on. The terminal alkyne contains an acidic proton which can be deprotonated by a base to give an alkynide ion which involve in alkylation to give alkylated alkyne.

Fundamentals

The organic compounds with at least one triple bond are called as alkynes. There are two types of alkenes exist. 1) Terminal alkyne. 2) Internal alkyne. In terminal alkynes, the triple bond present at one of the end of the chain whereas, in internal alkynes, the triple bond present at somewhere except at the end.

Reactions of alkynes:

1|) Terminal alkynes when reacted with strong base, deprotonation takes place to give conjugate base of alkyne which is called as an alkynide ion, which is a stabilized ion because the lone pair of electrons occupies sp-hybridized orbital. Alkynide ions are considered as nucleophiles and they attack on the electrophilic site. When reacted with alkyl halides, alkynide ions undergo alkylation.

2) Alkynes undergoes bromination to give two different brominated products depends on the equivalents of bromine used in the reaction. In presence of one equivalent of bromine, 1,2-dibromo product is formed whereas in presence of excess bromine, tetrabromo product is formed.

3) Alkynes on catalytic hydrogenation with hydrogen gas in presence palladium carbon gives saturated compound.

A)

The given starting material is acetylene which is shown below.

$h—c=c—H $

Given that acetylene is reacting with two reagents to give compound A. The reaction is shown below.

In the first step, the acidic proton of acetylene is removed by BuLi to give lithium acetylide which on reacting with ethyl bromide yields compound A.

Therefore, the structure of compound A is as follows:

B)

From step 1, the structure of the compound A is as follows:

In the second step 2, compound A is reacting with 1 equivalent of ${\rm{B}}{{\rm{r}}_{\rm{2}}}$ in dichloromethane. According to reactions of alkynes, the alkyne on treatment with 1 equivalent of ${\rm{B}}{{\rm{r}}_{\rm{2}}}$ gives 1,2-dibromo product.

In the given reaction, compound B (but-1-yne) on treatment with one equivalent of ${\rm{B}}{{\rm{r}}_{\rm{2}}}$ yields 1,2-dibromobut-1-ene.

The reaction of bromination of compound A is shown below.

Stereochemistry of the product: The high priorities groups are on the opposite side of the double bond, so the stereochemistry of the compound B is ‘E’.

Therefore, the structure of compound B is as follows:

Ans: Part A

Part A

Answer

The structure of compound A is as follows:

Part B

The structure of compound B is as follows: