Question

A molecule of formula C7H12 reacts consecutively as shown below to produce propyl cyclohexane- carboxylate. Identify...

A molecule of formula C7H12 reacts consecutively a

A molecule of formula C7H12 reacts consecutively as shown below to produce propyl cyclohexane- carboxylate. Identify the structure of the C7H12 molecule and the materials that are produced after each step. Draw only the organic material.

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Answer #1
Concepts and reason

The concept used to solve this question is to utilize the functioning of given reagents in the reaction sequence to determine the structure of the unknown compound. The reagent BH3{\rm{B}}{{\rm{H}}_{\rm{3}}} followed by H2O2/OH{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}/{\rm{O}}{{\rm{H}}^ - } is used in the hydroboration reaction followed by oxidation reaction of an alkene. The reagent CrO3/H2SO4{\rm{Cr}}{{\rm{O}}_{\rm{3}}}/{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} is used in the oxidation of alcohol to carboxylic acid. The reagent LiOH is a base which is used in the deprotonation of carboxylic acid. The reagent bromopropane is used as an alkylating agent.

Fundamentals

Organic chemistry is a branch of chemistry which deals with the synthesis of scaffolds using different reagents by taking specific synthetic route. Every reagent has a specific application which can be used in the synthesis for the functional group transformation. The functional group in the compounds decides the reactivity of the compounds.

The reagent BH3{\rm{B}}{{\rm{H}}_{\rm{3}}} followed by H2O2/OH{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}/{\rm{O}}{{\rm{H}}^ - } is used in the hydroboration reaction followed by oxidation reaction of an alkene to give an alcohol. Therefore, when this reagent is used the starting material should have an alkene functional group that means the starting material should have at least one double bond.

The reagent CrO3/H2SO4{\rm{Cr}}{{\rm{O}}_{\rm{3}}}/{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} is used in the oxidation of alcohols. Therefore, when this reagent is used the starting material should have an alcohol functional group.

The reagent LiOH is a base which is used in the deprotonation of acidic proton. Therefore, when this reagent is used the starting material should have an acidic proton in it.

The reagent bromopropane is used as an alkylating agent. This reagent can be used in alkylation of an organic compound by following the nucleophilic substitution reaction mechanism.

From the above description of reagents, the following the reaction sequence can be obtained.

1. BHZ
Croz
H2SO4
Alkene
Alcohol
Carboxylic acid
2. H2O2
ОН
LiOH
Lithium salt of
Br carboxylic acid
propyl cyclohexanecarboxy

From the above scheme, the functional group transformations taken place in the side chain only not on the cyclohexane ring.

The given molecular formula is C7H12{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{{\rm{12}}}} .

The double bond equivalence can be calculated as follows:

DBE=C+1HN+X2=7+1122=2\begin{array}{l}\\{\rm{DBE}}\,{\rm{ = }}\,{\rm{C}}\,{\rm{ + }}\,{\rm{1}} - \frac{{{\rm{H}} - {\rm{N}}\,{\rm{ + }}\,{\rm{X}}}}{2}\\\\\,\,\,\,\,\,\,\,\,\,\,\, = \,7 + \,1\, - \frac{{12}}{2}\\\\\,\,\,\,\,\,\,\,\,\,\,\, = \,2\\\end{array}

The DBE = 2 indicates that the compound may have two double bonds or 1 triple bond or 1 one double bond and a ring.

As in the final compound, the cyclohexane ring is present so the starting material must contain a cyclohexane ring with one double bond.

Also, the reagent used in the first step is BH3{\rm{B}}{{\rm{H}}_{\rm{3}}} followed by H2O2/OH{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}/{\rm{O}}{{\rm{H}}^ - } which is used in the hydroboration reaction followed by oxidation reaction of an alkene to give an alcohol.

Therefore, it is an alkene.

As in the above mentioned scheme, the functional group transformations taken place in the side chain only so the starting material should contain a terminal double bond which satisfies all the reaction conditions.

From the above points, the expected structure of the starting materail with molecular formula C7H12{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{{\rm{12}}}} is as follows:

methylenecyclohexane
Chemical Formula: C,H12

The structure of the starting material from step 1 is as follows:

methylenecyclohexane

When methylenecyclohexane is reacted with BH3{\rm{B}}{{\rm{H}}_{\rm{3}}} followed by H2O2/OH{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}/{\rm{O}}{{\rm{H}}^ - } undergoes hydroboration followed by oxidation to give primary alcohol as product.

1. BHZ
OH
2. H2O2
OH
methylenecyclohexane
cyclohexylmethanol

Therefore, the structure of the second compound is as follows:

cyclohexylmethanol

The structure of the second compound from step 2 is as follows:

cyclohexylmethanol

When cyclohexylmethanol is reacted with CrO3/H2SO4{\rm{Cr}}{{\rm{O}}_{\rm{3}}}/{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} undergoes oxidation to give carboxylic acid as product.

ГОН
Cro
H2SO4
OH
cyclohexylmethanol
cyclohexanecarboxylic acid

Therefore, the structure of the third compound is as follows:

TOH
cyclohexanecarboxylic acid

The structure of the third compound from step 3 is as follows:

When cyclohexanecarboxylic acid is reacted with LiOH undergoes deprotonation to give lithium salt cyclohexanecarboxylic acid as product.

LiOH
OH
cyclohexanecarboxylic acid
lithium cyclohexanecarboxylate

Therefore, the structure of the fourth compound is as follows:

OLI
lithium cyclohexanecarboxylate

Ans: Part a

Part a

Answer

The structure of the compound with molecular formula C7H12{{\rm{C}}_{\rm{7}}}{{\rm{H}}_{{\rm{12}}}} is as follows:

methylenecyclohexane

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Answer #2

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answered by: Jazzy00088
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