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Question: Write structural formulae of organic compounds ‘A’, ‘B’, ‘C’, and ‘D’ in the following sequence of r...

Write structural formulae of organic compounds ‘A’, ‘B’, ‘C’, and ‘D’ in the following sequence of reactions.
A'A' +CH3MgBrH2OCH3CH2CH(OH)CH3conc.H2SO4BBr2Calc.KOHD+\,C{{H}_{3}}MgBr\,\xrightarrow{{{H}_{2}}O}\,C{{H}_{3}}C{{H}_{2}}CH(OH)C{{H}_{3}}\,\,\xrightarrow{conc.{{H}_{2}}S{{O}_{4}}}'B'\,\xrightarrow{B{{r}_{2}}}\,'C'\,\xrightarrow{alc.KOH}\,'D'

Explanation

Solution

Before solving the question, we should first know about Grignard reagents. The chemical compound in which there is a combination of halogen, organic group, and Mg is known as Grignard Reagent. They are formed when magnesium is reacted with alkyl halides. They are considered as good nucleophiles.

Complete step by step solution: The compounds with a structureCHO-CHO containing a functional group have a carbonyl center in which the carbon atom forms a bond with the hydrogen atom and with an alkyl group, This is known as Aldehyde.
The compound with a structure R2C=O{{R}_{2}}C=O where R can be any carbon-containing substituents is a functional group known as ketones.
The compound which has one (OH)(-OH) hydroxyl functional group bonded with one saturated carbon atom is known as alcohol.

The reaction of Aldehydes and Ketones form alcohol with the help of a Grignard reagent. Compound A is reacted with the Grignard reagent in the presence of water. The secondary alcohol which is formed is of 3 carbons. It shows that compound A is ‘Propanol’.
CH3CH2CHO+CH3MgBrH2OCH3CH2CH(OH)CH3C{{H}_{3}}C{{H}_{2}}CHO\,+\,C{{H}_{3}}MgBr\,\xrightarrow{{{H}_{2}}O}\,\,C{{H}_{3}}C{{H}_{2}}CH(OH)C{{H}_{3}}
A
2-butanol comes under the category of secondary alcohol is reacted with sulphuric acid which results in the formation of trans-2-butene as it is undergoing dehydration with H2SO4{{H}_{2}}S{{O}_{4}}. So, Compound B is ‘trans-2-butene’.
CH3CH2CH(OH)CH3concH2SO4CH3CH=CHCH3C{{H}_{3}}C{{H}_{2}}CH(OH)C{{H}_{3}}\xrightarrow{conc\,\,{{H}_{2}}S{{O}_{4}}}\,C{{H}_{3}}CH=CHC{{H}_{3}}
B
Then the Butene is reacted with Bromine forms 2,3 dibromo butane. So, Compound C is ‘2,3 dibromo butane’.
CH3CH=CHCH3Br2CH3BrCHCHBrCH3C{{H}_{3}}CH=CHC{{H}_{3}}\xrightarrow{B{{r}_{2}}}\,C{{H}_{3}}BrCH-CHBrC{{H}_{3}}
C
2,3 dibromo is reacted with alcoholic KOH which converts the alkane to alkene i.e 2,3 dibromo gets converted into but-2-ene. So Compound D is ‘But-2-ene’.
CH3BrCHCHBrCH3alcKOHCH3CH=CBrCH3C{{H}_{3}}BrCH-CHBrC{{H}_{3}}\,\xrightarrow{alc\,KOH}\,C{{H}_{3}}CH=CBrC{{H}_{3}}
D
Compound A = ‘Propanol’
Compound B = ‘trans-2-butene’
Compound C = ‘2,3 dibromo butane’
Compound D = ‘But-2-ene’$

Note: Organic compounds are the chemical compounds in which carbon is covalently bonded to other elements like Hydrogen etc. Cellulose, fructose are some basic examples of organic compounds. They are found all over the world and have numerous uses.