Predict the organic product of the following reaction and include hydrogen atoms in your structure

Predict the organic product of

Predict the organic product of

Predict the organic product of


Predict the organic product of the following reactionsand include hydrogen atoms in your structure.


---> The structure I have drawn in the first box is incorrect.

Answer

General guidance

Concepts and reason
This problem is based on the concept of acidic hydration and hydrohalogenation of alkenes.

Alkenes have high electron density due to the presence of pi-bond. As a result, alkenes are highly reactive towards the chemical species which are electron deficient in nature such as acids.

Fundamentals

In the reaction of alkenes with acids, alkenes attacks on the acids in order to get protonated which results in the formation of the most stable cation and then the chemical species having high electron density attacks on this cation.

Step-by-step

Step 1 of 6

Part a

The reaction is alkene is with the sulfuric acid (protonation) is given below:

H
CH3
H2SO4
→ H3C
CH3
CH₃

This reaction follows the Markovnikov’s rule which states that the hydrogen atom will be added to the olefinic carbon having more number of hydrogens in order to form most stable carbocation.

Step 2 of 6

The reaction of carbocation with water molecule is given below:

CH3
Нc—с
сня н
CHE
H20 HCC
—> НС—С
—o5
H3C
-c
— ОН
CH3
н-н- Hс
Сн,
CH₃

Part a

The structure of the product is as follows:

Нc— c
CH3
— ОН
Сн,


In the reaction water molecule attacks on the carbocation and forms an oxonium ion which on deprotonation result in the formation of a tertiary alcohol.

Step 3 of 6

Part b

The reaction is alkene is with the sulfuric acid (protonation) is given below:

H
CH3
H2SO4
→ H3C
CH3
CH₃

This reaction follows the Markovnikov’s rule which states that the hydrogen atom will be added to the olefinic carbon having more number of hydrogens in order to form most stable carbocation.

Step 4 of 6

The reaction of carbocation with methanol is given below:

CH,
CH3 CH3
CH3
H2C—
C
H3C-OH
→ H30—6-03
CH3
H nH,c
+H3C—C— 0-CH3
CHE
CH,

Part b

The structure of the product is given below:

H3C—6
CH3
0-CHg
Сн,


In the reaction methanol molecule attacks on the carbocation and forms an oxonium ion which on deprotonation result in the formation of an ether.

Step 5 of 6

Part c

The reaction is alkene is with the hydrogen bromide (protonation) is given below:

HzC
CH3 HzC CH3
H3C—C—CH=ch, HBP, H3c—<-ch+chz— CHC
| CHỊ Học CH3
-
HAC

This reaction follows the Markovnikov’s rule which states that the hydrogen atom will be added to the olefinic carbon having more number of hydrogens in order to form a secondary carbocation which rearranges to tertiary carbocation because tertiary carbocation is more stable.

Step 6 of 6

The reaction of carbocation with bromide ion is given below:

Hzc ✓
CHz
Rr
CHą
Br.
C-HC
Hzc BT
C-HC
Học
CH,
HzC
CH3

Part c

The structure of the product is as follows:

HÁC BỊ
CHỊ
нс
Сн,


In this reaction, bromide ion attacks on the tertiary carbocation and forms a carbon halogen bond in order to form a bromoalkene.

Answer

Part a

The structure of the product is as follows:

Нc— c
CH3
— ОН
Сн,

Part b

The structure of the product is given below:

H3C—6
CH3
0-CHg
Сн,

Part c

The structure of the product is as follows:

HÁC BỊ
CHỊ
нс
Сн,

Hottest videos

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts