
Alcohols react with Grignard reagent to form:
A. Alkanes
B. Alkenes
C. Alkynes
D. All of these
Answer
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Hint: Grignard reagent is an organometallic compound that consists of an organic molecule along with a metal generally magnesium and a halogen. It undergoes nucleophilic substitution reactions. It readily reacts with a carbonyl or an oxygen group. The general formula of a Grignard is R – Mg – X, where R is alkyl group, Mg is magnesium and X is any halogen.
Complete answer:
Grignard reagents are organometallic species with formula R-Mg-X. The reaction of Grignard reagents with alcohols is a type of acid – base reaction. The reaction involves the acidic nature of the O – H bond of the alcohols, as the O – H bond has removal of hydrogen.
So, the reaction happens as the hydrogen from alcohols get removed and the negative species formed get attached on the metal halide of the Grignard reagent, while the proton gets attached to the alkyl chain, this results in the formation of an alkane.
The reaction is:
$C{{H}_{3}}-O-H+{{C}_{2}}{{H}_{5}}-Mg-Br\to {{C}_{2}}{{H}_{6}}+MgBrOH$
Methyl alcohol reacts with ethyl magnesium bromide to form ethane and MgBrOH.
So, alcohols react with Grignard reagent to form alkanes.
Hence, option A is correct.
Note:
MgBrOH is called a basic bromide, which is an intermediate compound. Grignard reagents are formed by the reaction of haloalkane and Mg in the presence of ether. The reactions is, ${{C}_{2}}{{H}_{5}}Br+Mg\xrightarrow{ethoxyethane}{{C}_{2}}{{H}_{5}}MgBr$. Grignard reagents can be used in reactions that need to increase the hydrocarbon chains.
Complete answer:
Grignard reagents are organometallic species with formula R-Mg-X. The reaction of Grignard reagents with alcohols is a type of acid – base reaction. The reaction involves the acidic nature of the O – H bond of the alcohols, as the O – H bond has removal of hydrogen.
So, the reaction happens as the hydrogen from alcohols get removed and the negative species formed get attached on the metal halide of the Grignard reagent, while the proton gets attached to the alkyl chain, this results in the formation of an alkane.
The reaction is:
$C{{H}_{3}}-O-H+{{C}_{2}}{{H}_{5}}-Mg-Br\to {{C}_{2}}{{H}_{6}}+MgBrOH$
Methyl alcohol reacts with ethyl magnesium bromide to form ethane and MgBrOH.
So, alcohols react with Grignard reagent to form alkanes.
Hence, option A is correct.
Note:
MgBrOH is called a basic bromide, which is an intermediate compound. Grignard reagents are formed by the reaction of haloalkane and Mg in the presence of ether. The reactions is, ${{C}_{2}}{{H}_{5}}Br+Mg\xrightarrow{ethoxyethane}{{C}_{2}}{{H}_{5}}MgBr$. Grignard reagents can be used in reactions that need to increase the hydrocarbon chains.
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